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STU DENT SOLUTIONS MAN UAL J. Richard Christman Professor Emeritus U.,S. Coast Guard Academy FUNDAMENTALS OF PHYSICS Eighth Edition David Halliday (Jnivers iQ of P itts burgh Robert Resnick Rens s elaer Polytechnic Institute Jearl Walker Cleveland State Univers iQ John Wiley & Sons, Inc.
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Cover Image: @ Eric Heller/Photo ResearchersBicentennial Logo Design: Richard J. PacificoCopyright @ 2008 John Wiley & Sons, Inc. All rights reserved.No part of this publication may be reproduced, stored in a retrieval system or transmittedin any form or by any means, electronic, mechantcal, photocopying, recording, scanning,or otherwise, except as perrnitted under Sections 107 or 108 of the I 97 6 lJnited StatesCopyright Act, without either the prior written permission of the Publisher, orauthorizationthrough payment of the appropriate per-copy fee to the CopyrightClearance Center, Inc ., 222 Rosewood Drive, Danvers, MA 01923, or on the web atwww.copyright.com. Requests to the Publisher for permission should be addressed to thePermissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ07030-5714, (201) 148-6011, fax (201) 748-6008, or online athttp //www. wilelz. c om/ go/p ermi : ss ions.To order books or for customer service, please call 1-800-CALL-WILEY (225-5945).rsBN- 13 978- 0-47 r-779s8-2Printed in the United States of Amerca10 9 8 7 6 s 4 3 2 |Printed and bound by Bind-Rite Graphics.
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PREFACEThis solutions manual is designed for use with the textbook Fundamentals of Physics, eighthedition, by David Halliday, Robert Resnick, and Jearl Walker. Its primary puqpose is to showstudents by example how to solve various types of problems given at the ends of chapters in thetext.Most of the solutions start from definitions or fundamental relationships and the final equationis derived. This technique highlights the fundamentals and at the same time gives students theopportunity to review the mathematical steps required to obtain a solution. The mere pluggingof numbers into equations derived in the text is avoided for the most part. We hope students willlearn to examine any assumptions that are made in setting up and solving each problem.Problems in this manual were selected by Jearl Walker. Their solutions are the responsibility ofthe author alone.The author is extremely grateful to Geraldine Osnato, who oversaw this project, and to hercapable assistant Aly Rentrop. For their help and encouragement, special thanks go to the goodpeople of Wiley who saw this manual through production. The author is especially thankful forthe dedicated work of Karen Christman, who carefully read and coffected an earlier version ofthis manual. He is also grateful for the encouragement and strong support of his wife, MaryEllen Christman.J. Richard ChristmanProfessor EmeritusIJ.S. Coast Guard AcademyNew London, CT 06320
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Chapter L 1yd - (0.9144mX106 pmlm) - 9.144 x 105 pm.3use the given conversion factors.(a) The distance d, in rods is (4.0 furlongsX2Ol . 168 m/furlong) d - 4.0 furlongs - 5.0292mf rod(b) The distance in chains is (40 furlongsX20l 168 m/furlong) d - 4.Lfurrongs - : 4|chains . 20.L7 mlchain-I(a) The circumference of a sphere of radius R is given by 2r R. Substitute R - (6.37 x106mX10-tk^lm)should obtain 4.00 x 104km.(b) The surface area of a sphere is given by 4trR2, so the surface area of Earth is 4n(6.37 x103 k*)(c) The volume of a sphere is given by (4nlrR3, so the volume of Earth is G"13X6.37 x103 k*)3 _ 1.08 x 1012 km3.t7None of the clocks advance by exactly 24h in a 24-h period but this is not the most importantcriterion for judging their quality for measuring time intervals. What is important is that theclock advance by the same amount in each 24-h period. The clock reading can then easily beadjusted to give the correct interval. If the clock reading jumps around from one 24-h period toanother, it cannot be coffected since it would impossible to tell what the coffection should be.The followittg table gives the coffections (in seconds) that must be applied to the reading oneach clock for each 24-h period. The entries were determined by subtracting the clock readingat the end of the interval from the clock reading at the beginning. Chapter I
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CLOCK Sun. Mon. Tues. Wed. Thurs. Fri. -Mon. -Tues. -Wed. -Thurs. -Fri. -Sat A -16 -16 -15 -17 -15 -15 B -3 +5 -10 +5 +6 -7 C -58 -58 -s8 -s8 -58 -58 D +67 +67 +67 +67 +67 +67 E +70 +55 +2 +20 +10 +10Clocks C and D are the most consistent. For each clock the same coffection must be appliedfor each period. The coffection for clock C is less than the coffection for clock D, so we judgeclock C to be the best and clock D to be the next best. The coffection that must be applied toclock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to *10 s, for clockE it is in the range from -70 s to -2s. After C and D, A has the smallest range of correction,B has the next smallest range, and E has the greatest range. From best the worst, the ranking ofthe clocks is C, D, A, B, E.21(a) Convert grams to kilograms and cubic centimeters to cubic meters: 1 g _ I x 10- kg and1cm3 - (1 x To-2 m)3 rs-(1 e)(#) (,%) t :rx1o3kg(b) Divide the mass (in kilograms) of the water by the time (in seconds) taken to drain it. Themass is the product of the volume of water and its density: M - (5700m3X1 x 103 kg/nt)5.70 x 106kg. The time is t - (10.0hX3600s/h) - 3.60 x 104 s, so the mass flow rate n is M- 5.70 x 106kg- R - t 3.60 x 104 s 1 58 kg/s "3s(a) The amount of fuel she believes she needs is (750mi)l(0mif gal): 18.8gal. This is actuallythe number of IJ.K. gallons she needs although she believes it is the number of IJ.S. gallons.(b) The ratio of the U.K. gallon to the U.S. gallon is (4.545963 IL)l(3.785 3060L) : 1.201.The number of U.S. gallons she actually needs is (18.8 IJ.K. galx 1.201 U.S gallu.K. gal) : 22.5 IJ.S. gal .39The volume of a cord of wood is V - (8 ftx 4 tt)(4 ft) : I28ft3. IJse l ftAppendix D) to obtain V - I28ft3X0 .3048 ft)3 - 3.62m3. Thus 1.0 m3 of wood coffespondsto (l 13.62) cord - 0 .28 cord. ^l2 Chapter I
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4T(a) The difference in the total amount between 73 freight tons and 73 displacement tons is (8 barrel bulk/freight ton)(73 freight ton) - (7 barrel bulk/displacement ton)(73 displacementton) : 73 barrel bulk .Now l banel bulk - 0.141 5m3 - (0.!41 5m3)(28.378U.S. bushel): 4.01 5U.S. bushel ,SO 73barceI bulk - (73 barrel bulk)(4.01 5 U.S. bushellbarcel bulk) - 293 U.S. bushel .(b) The difference in the total amount between 73 register tons and 73 displacement tons is (z}barrel bulk/register ton)(7 3 register ton) - (7 barrel bulk/displacement ton)(73 displacementton) : 949 barrel bulk.Thus g4:gbarrel bulk - (949barrel bulk)(4.01 5U.S. bushellbarcel bulk):3810U.S. bushel .45 m-1x10-3km, 0.12 ATJ lmin.57(a) We want to convert parsecs to astronomical units. The distancebetween two points on a circle of radius r is d - 2r stn 0 12, where0 is the angle subtended by the radtal lines to the points. See thefigure to the right. Thus r - d,lz sin 0 12 and t /-" l pc - ^",., - 2.06 x 105 AU, ., " 012 2 sin( I" 12)where !" - - (2.78 x 10-4) was used. Finally (I13600)" 1 AU - (1 AU) 1Q.06 x 105 AU lpc) : 4.9 x 10-u p, .(b) A light year is (1.86 x 10s mrlsxl.0 yX3 65.3 daly)Q|hld)(3600 s/h) - 5 .87 x 1012 mtand l Au : 929 x J l.5g x 1o- ly. 5.87xlot2milly- },9^6 Tt,; LJ) Chapter I
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Chapter 21(a) The average velocity during any time interval is the displacement during that interval dividedby the duration of the interval: ?)av - Lr I Lt, where Lr is the displacement and Lt is the timeinterval. In this case the interval is divided into two parts. During the first part the displacementis Lr 1 : 40 km and the time interval is : (40km) Ltr (30 k*/h) -1.33h.During the second part the displacement is Lrz: 40km and the time intenral is (4gq) Ltz-(60 l*ttr - o 67h Both displacements arc in the same direction, so the total displacement is Lr40km +40km:80km. The total time interval is Lt: Lt1 * Ltz- 1.33h+0.67h:2.00h.The average velocity is (80km) : nl avg 40k*/h. (2.0 h)(b) The average speed is the total distance traveled divided by the time. In this case the totaldistance is the magnitude of the total displacement, so the average speed is 40 km fh.(c) Assume the automobile passes the origin at r g0time t : 0. Then its coordinate as a function of Gm)time is as shown as the solid lines on the graph 60 -- --Q- - --J -- -r -to the right. The average velocity is the slope of 40the dotted line. 20 0.5 1.0 1.5 2.0 t (h)-5Substitute,inturn, t- I,2,3, and4sintotheexpression r(t):3t-4*+#,where r isinmeters and t is in seconds: (a) r(Is): (3^lsxl s) - (mls2xl r)2 + (l^11311 s;3 - 0 (b) r(zs) : (3mlsx2 s) - (4mls(2s)2 + Qmls;12 s) - -2m (c) r(3s): (3^lsx3 s) - (4mls2x3 s)2 + Omlr311: r13 -0 (d) r(4s): (3mlsx4s) - (4mls2x4s)2 + Q^lr3;1+s;3(e) The displacement during an interval is the coordinate at the end of the interval minusthe coordinate at the beginning. For the interval from t -- 0 to t - 4 s, the displacement isLr - r(4s) - tr(0) - I2m- 0 - +12m. The displacement is in the positive r direction.4 Chapter 2
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(0 The average velocity during an interval is defined as the displacement over the intervaldivided by the duration of the interval: uavs- L, I Lt. For the interval from t- 2s to t-4sthe displacement is Lr - r(4 s) r(2 s) - I2m (-2m) - I4m and the time intenral isLt: 4s - 2s : 2s. Thus Lr uavg:E- l4m -7m/s.(d) The solid curye on the graph to the right ^ r r2.oshows the coordrnate r as a function of time. (m)The slope of the dotted line is the average 9.0velocitybetween t- 2.0s and t- 4.0s. 6.0 3.0 0.0 -3.019If ur is the velocity at the beginning of a time interval (at time t)and u2 is the velocity at theend (at tz), then the average acceleration in the interval is given by eavTake h : 0, u1 : 18 m/s, t2 - 2.4 s, and u2 auus -30 m/s - 18 m/s - t -zo ^lr 2.4s .The negative sign indicates that the acceleration is opposite to the original direction of travel.25(a) Solve u- us* at for t: t - (u - uo) f a. Substitute u-0.1(3.0 x 108 mls):3.0 x I07 ml s,u0 : 0, and e,:9.8(b) Evaluate r ^lr. The result is t - 3.06 x 106 s. This is !,zmonths.4.6 x 1013 m.27Solveu2 -u3+2a(r-r0 fora. TakeffO:0. Then a-(r-ril|2tr. Useu0:1.50x 105 mls,u - 5.70 x 106m/s, and r - 1.0cm - 0.010m. The result is (5.70 x 106m/s)2 - (1.50 x 105m/s)2 a_ - 1.62x l0r, ^/r, . 2(0.010m)33(a) Take frs- 0, and solve r- ust * *ot for a: a_ 2(r ust)f t2. Substitute r- 24.0m,Ug - 56.0k*/h - 15.55 m/s, and t - 2.00 s. The result is 2lzq.0m- (15.55m1sx2.00s)] 1 rr ^ r^Z &- Chapter 2 5
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The negative sign indicates that the acceleration is opposite the direction of motion of the car.The car is slowing down.(b) Evaluate u(30.3 km lh).45(a) Take the A axis to be positive in the upward direction and take t - 0 and A - 0 at the pointfrom which the wrench was dropped. If h is the height from which it was dropped, then theground is at A : -h. Solve u2 - uzo + 2gh for h: u2 ul h- 2sSubstitute uo : 0, 1) : -24mls, and g - g.8 m/s2: (24mls)2 :29.4m. h- 2(9.8 m/s2)(b) Solve 1) : uo - gt for t: (to -u) 24mls t- -2.45s. 9.8 ^l12(c) t (s) t (s) 2 2 0 0 a-10 u -10 (-)-zo (m/s) _20 -30 -30The acceleration is constant until the wrench t (s)hits the ground: a- -g.8 m/s2. Its graph is 0as shown on the right. a-5 (-is2) -10 _1547(a) At the highest point the velocity of the ball is instantaneously zero. Take the A axis to beupward, set u in u2 and solve for u$ us- EA. Substitute g6 Chapter 2
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A:50mtoget 2(9.8^ls2x50m) - 31 m/s.(b) It will be in the atr until A-0 again. Solve A - uot t7t for t. Since A the twosolutions aret:0 and t : 2uo I g . Rej ect the first and accept the second: t-_ .L 2uo 2(31 m/s) - -6.4s. I 9.8 ^lt(c) a60 u40 (m) (n/s) 40 20 20 0 _20 68 t (s) -40The acceleration is constant while the ball is in t (s)flight: a, 0on the right. a-5 (*/r) -10 _1549(a) Take the A axis to be upward and place the origin on the ground, under the balloon. Sincethe package is dropped, its initial velocity is the same as the velocity of the balloon, +lTm/s.The initial coordinate of the package is ao: 80 m; when it hits the ground its coordinate is zero.Solve a : Uo + uot - *gt for t: t-uo+ L r :J.-TD-c I 9.8 ^lt ;-i- ll V (9.8 mls2)2 g.g m/s2where the positive solution was used. A negative value for t coffesponds to a time before thepackage was dropped.(b) Use t):7)0 - gt: l}mls - (9.8 mls2x5.4s) - -.41 m/s. Its speed is 4Imf s.51The speed of the boat is given by u6 _ d lt, where d is the distance of the boat from the bridgewhen the key is dropped (IZm) andt is the time the key takes in falling. To calculate t, put the Chapter 2
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origin of the coordinate system at the point where the key is dropped and take the y axis to bepositive in the upward direction. Take the time to be zero at the instant the key is dropped. Youwant to compute the time t when A - -45 m. Since the initial velocity of the key is zera, thecoordinate of the key is given by A - -+g*. Thus 2(-45 m) -3.03s.This means 12m U6: - 4.0 m/s . 3.03 s--I5First find the velocity of the ball just before it hits the ground. During contact with the groundits average acceleration is given by Lu aavg , Nwhere Lu is the change in its velocity durittg contact and Lt is the time of contact.To find the velocity just before contact take the A axis to be positive in the upward directionand put the origin at the point where the ball is dropped. Take the time t to be zero when it isdropped. The ball hits the ground when A - -15.0m. Its velocity then is found fromu2- -ga,SO - -l -ze.g^ls2x-15.0m) : -17.rmls.The negative sign is used since the ball is traveling downward at the time of contact.The average acceleration during contact with the ground is rt: - (- 17 .I m/s) *avs0 zo.o x 1o-3 s 857 ^lt2The positive sign indicates it is upward.89The velocity at time t is given by u: f adt: f S.tdt:2.5t2+C, where e is aconstantofintegration. use the condition that u- +ITmlsatt- 2.0stoobtain C:u-2.5t2 - ITmls-2.5(2.0 s)2 - 7.0 m/s. The velocity att - 4.0 s * 7.0^lsr2.5t2 - 7.0^ls*2.5(4.0 r)2 : 47 m/s.9l(a) First convert the final velocity to meters per second: 1) : (60 k*/hx1000 m/km)1Q600 s/h) - 16.7mf s. The average acceleration is ult - (16.7^ls)/(5.4s):3.1 ^lr.(b) Since the initial velocity is zero,the distance traveled is r - Lot - Ltl .L ^ls2x5 .4 r)2 - 45 m.8 Chapter 2
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(c) Solve n - *ot for t. The result is 2(0.25 103 m) - 13 s.97The driving time before the change in speed limit was t6 - Lr luo, where Lr is the distance and u6 is the original speed limit. The driving time after the change is to - A^r lro, where I)q isthe new speed limit. The time saved is t6- te- Lr(*;) | I )-r2h. -(700km)(0.62ramilkm)( milh LL ) ss 65 mi lh)This is about t h and TZmrn.99Let t be the time to reach the highest point and us be the initial velocity. The velocity at thehighest point is zero, so 0 - us - gt and us- gt. Thus H- uot i7t: gtz ig*where the substitution was made for ns. Let H2 be the second height. It is given by H2 :LgQt)2 - 2gt2 4H. The balls must be thrown to four times the original height.I07(a) Suppose the iceboat has coordinate Ut at time h and coordinate Az at time t2. If a is theacceleration of the iceboat and tre is its velocity att - 0, thenAr: ?rltt+Lot? andy2: u0t2**t7.Solve these simultaneously for a and us. The results are 2(azh - aftz) t1t2(t2 - tt) ug:ffiandTake h - 2.0s and tz: 3.0s. The graph indicates that at: 16m and Uz - 27 m. These valuesyielda, - 2.0^lt2 and us - 6.0m/s.(b) The velocity of the iceboat at t - 3.0s is ,u: uo* at:6.0 mls * (2.0m1s2x3.0s) - L}mf s.(c) The coordinate at the end of 3.0 s is az - 27 m. The coordinate at the end of 6.0 s isUz: uotz+ |"*3- (6.0^lsx6.0s) + l(2.0m1s2x6.0r) - 72m. The distance traveled during thesecond 3.0-s interval is Uz - Uz:72m - 27 m- 45m. Chapter 2
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Chapter 31 a: : :(a)use lmt to obtain e, (-25.0 m)2 + (+40.0 m)2 47 .2m.(b) The tangent of the angle between the vector and the positive r axis is aa 4oo m tano - clr- -25.0 mThe inverse tangent is -58.0o or -58.0" + 180 - I22". The first angle has a positive cosineand a negative sine. It is not correct. The second angle has a negative cosine and a positive sine.It is correct for a vector with a negative r component and a positive A component.3The r component is given by ar - (7 .3 m) cos 250o -- -2.5 m and the A component is givenby a,athe components can also be computed using ar - -Q.3 m) cos 70" and e,a - -Q,3 m) sin 70o .It is also 20" from the negative A axis, so you might also use a,n - -Q.3 ) sin 20" and a,a :-(7.3 m) cos 20". These expressions give the same results.7(a) The magnitude of the displacement is the distancefrom one corner to the diametrically opposite corner:d - / (3.00 m)2 + (3 .70 m)2 + (4.30 m)2see this, look at the diagram of the room, with thedisplacement vector shown. The length of the diag-onal across the floor, under the displacement vector,is given by the Pythagorean theorem: L- tM,where (. is the length and u is the width of the room.Now this diagonal and the room height form a right tri-angle with the displacement vector as the hypotenuse,so the length of the displacement vector is given by d-{L2+h2-(b), (c), and (d) The displacement vector is along the straight line from the beginning to the endpoint of the trip. Since a straight line is the shortest distance between two points the length ofthe path cannot be less than the magnitude of the displacement. It can be greater, however. Thefly might, for example, crawl along the edges of the room. Its displacement would be the samebut the path length would be (. + ut * h" The path length is the same as the magnitude of thedisplacement if the fly flies along the displacement vector.10 Chapter 3
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(e) Take the r axis to be out of the page, the y axis to be to the right, and the z axis to be upward.Then the r component of the displacement is u) : 3.70 m, the A component of the displacementis 4.30m, and the z component is 3.00m. Thus i- (3.70m)i +(4.30m)j+(3.00m)t. You maywrite an equally correct answer by interchanging the length, width, and height.(f) Suppose the path of the fly is as shown by thedotted lines on the upper diagram. Pretend there isa hinge where the front wall of the room joins thefloor and lay the wall down as shown on the lowerdiagram. The shortest walking distance between the l- /-lower left back of the room and the upper right t. /front corner is the dotted straight line shown on the ./ / ..tdiagram. Its length is ( , T umtn w (3.70m * 3.00 m)2 + (4.30 m)2 - 7 .96m . h2(a) Let d,+6. Then r, : an*b* - i- 4.0m- 13m - -g.0m andrr: &y*b, - 3.0m +7.0m -10m. Thus r-: (-9.0m)i*(10m)j.(b) The magnitude of the resultant is r : ^ lr2 + ,2 - (-9.0m)2 + (10m)2 _ 13 m. Y*a(c) The angle 0 between the resultant and the positive r axis is given by tan? - ,a lr* -(10 m)l(-9.0m) - -1.1. 0 is either -48o or I32". The first angle has a positive cosine anda negative sine while the second angle has a negative cosine and positive sine. Since the ncomponent of the resultant is negative and the A component is positive, 0 - I32o .t7(a) and (b) The vector d has a magnitude 10.0m and makes the angle 30o with the positive naxis, so its components are a,n- (10.0m)cos30o - 8.67m and aa: (10.0m)sin30o - 5.00m.The vector 6hur a magnitude of 10.0m and makes an angle of 135o with the positive n axis,so its components are b*- (10.0m)cos 135o - -7.07 m and ba _ (10.0m) sin 135oThe components of the sum are rn: a* * b*- 8.67 m - 7.07 m- 1.60m and ra: ea * ba:5.0m*7.07 m- 12.1 m.(c) The magnitude of iis r - ym: (1.60m)2 + (12.I - I2.2m.(d) The tangent of the angle 0 between i and the positive r axis is given by tan? _ rr lr* -(lz.Lm)l(l.60m) : 7.56. e is either 82.5o or 262.5o. The first angle has a positive cosine anda positive sine and so is the correct answer. Chapter 3 11
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39Since ab cos Q : a*b* * oab, r a"br, a*b*+a,abr+&"b, cos d: abThe magnitudes of the vectors given in the problem are e, : (3.0)2+(3.0)2+(3.0)2b- (2.0)2 + (1.0)2 + (3.0)2 :3.7. The angle between them is found from (3.0X2.0) + (3.0X1.0) + (3.0X3.0) cos Q: _ 0 .926 (5.2)(3.7)and the angle is d:22o .43(a) and (b) The vector d, is along the r axis, so its r component is arcomponent is zero.(c) and (d) The r component of 6 is b* - b cos 0 -- (4.00 m) cos 30.0ocomponent is ba - b sin 0 - (4.00 m) sin 30.0o - 2.00 m.(e) and (f) The r component of c*is cr -- ccos(0 + 90")- (10.0m)cos I20" - -5.00m and theA component is ca: csin(9 + 90o): (10.0m)sin I20o - 8.66m.(g) and (h) In terms of components cr : pa* * qb* and ca : paa + eba. Solve these equationssimultaneously for p and q. The result is P:ffi- bac* - b*ca (3.46 mX8 .66m) - (2.00 mX-5.00 m) --667and q:ffi- arca -(3.00 - aac* mXS .66 m) -43447(a) The scalar product is d . b - abcos d : (10X6.0) cos 60o : 30(b) The magnitude of the vector product is ld*61 - ab sin d: (10X6.0) sin 60o : 52 .51Take the r axis to run west to east and the A axis to run south to north, with the origin at thestarting point. Let f,dest - (90.0 k-) j be the position of the destination and rrthe position of the sailor after the first leg of his journey and iz be the remaining displacement12 Chapter j
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required to complete the journey. The total journey is the vector sum of the two parts, so?dest : rI + iz and n:ri,,, iz - (90 km)i (50.0 km) i. - -The magnitude of the remaining trip is T2 + r?a (50.0 km)2 + (50.0 km)2 _ 103 km . ?.*The tangent of the angle with the positiye r direction is tan d : r2a lrr* - (90.0 km) 160.0 km) -1.80. The angle is either 60.9 or 180+ 60.9o :24Io. Since the sailor must sail northwest toreach his destination the correct angle is 24Io . This is equivalent to 60.9" north of west.7lAccording to the problem statement i+ E - 6.0i+ 1.0j and A- E - -4.0i +7.0j. Add theseto obtain 2A-2.0i+ 8.0j and then A- 1.0i++.0j. The magnitude of Ais (1.0)2 + (4.0)2 Chapter 3 13
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Chapter 47The average velocity is the total displacement divided by the time interval. The total displacementi is the sum of three displacements, each calcul ated as the product of a velocity and a timeinterval. The first has a magnitude of (60.0 km lh)(40.0 min) l$0.0 min lh)direction is east. If we take the r axis to be toward the east and the A axis to be toward thenorth, then this displacement is i1 - (40.0 km) i.The second displacement has a magnitude of (60.0 km lh)(20.0 min) 1rc0.0 min lh) : 20.0 km. Itsdirection is 50.0o east of north, so it may be written iz - (20.0km) sin50.0i + (20.0km)cos 50.0"i - (15.3 km)i* (12.9k*)i.The third displacement has a magnitude of (60.0 km lh)(50.0 min) l(60.0 min lh) _ 50.0 km. Itsdirection is west, so the displacement may be written fi - (-50 km) i. The total displacement is r-: r-r+ i2+ i, - (40.0km)i+ (15.3km)i+ (I2.9km)j - (50k*)i - (5.3 km) i + (12.9 km) j .The total time for the trip is 40 min * 20 mtn + 50 mininterval to obtain an average velocity of du,,e - (2.9k^lh) i + (7 .05k*/h)i. The magnitude ofthe average velocity is lr-uu*l : -7.6km/sand the angle O it makes with the positive r axis satisfies tanQ:!:.^l,n - 2.43 2.9 k-/h .The angle is d : 68o.11(a) The velocity is the derivative of the position vector with respect to time: 6 - *(i. 4*i +,t) : 8,i + rin meters per second for t in seconds.(b) The acceleration is the derivative of the velocity with respect to time: d-* (r,i.o) :8i14 Chapter 4
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in meters per second squared.17(a) The velocity of the particle at any time t is given by d - 6o + dt, where do is the initialvelocity and d is the acceleration. The r component is trr:uyrl a,nt - 3.00 mls - (1.00 mls2)tand the g component rs uy - tr1a* aat - -(0.500 mlsz)t When the particle reaches its maximumr coordinate trr:0. This means 3.00 mls - (1.00 mlst - 0 or t - 3.00s. The A component ofthe velocity atthis time is rra - (-0.500 mls2x3.00s): -1.50m/s. Thus 6 - (-1.50 mls)j.(b) The coordinates of the particle at any time t are r- nyrt + *.o*t and A - uyat + *,oot . Att - 3.00 s their values are r:(3.00 mlsx3.00s) - )ft00 mls2x3.00s)2 - 4.50mand : I _ a -;(o.5oo m/s2x3.oo s)2 -2.25m .Thus r: (4.50 m) i - (2.25m)i.29(a) Take the y axis to be upward and the r axis to be horizontal. Place the origin at thepoint where the diver leaves the platforrn. The cornponents of the divers initial velocity areuyn - 3.00m/s and uya - 0. At t - 0.800 s the horizontal distance of the diver from the platforrnis :x : uy*t - (2.00 mlsx0.800 s) : 1.60 m.(b) The drivers a coordinate is a- -+g*- -+(9.8 mls2x0.800s)2 - -3. 13m. The distanceabove the water surface is 10.0m - 3.13 m - 6.86m.(c) The driver strikes the water when A: -10.0m. The time he strikes is 2(- 10.0 m) - 1.43 sand the horizontal distance from the platform is r:1)0rt - (2.00 mlsxl .43 s): 2.86m.31(a) Since the projectile is released its initial velocity is the same as the velocity of the plane atthe time of release. Take the A axis to be upward and the r axis to be horizontal. Place theorigin at the point of release and take the time to be zero at release. Let r and y (: -730 m) bethe coordinates of the point on the ground where the projectile hits and let t be the time when ithits. Then : 1. A -uot cos 96 1gt ,,where 0o - 53.0o. This equation gives y+ -730 m + |(g.go mls2x5.oo s)2 ug: -T- *gt, / :202m/s. f cos 0g (5.00 s) cos(53.0o)(b) The horizontal distance traveled is r: uotsin0s - (202*1sX5.00 s) sin(53.0") : 806m. Chapter 4 15
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(c) and (d) The n component of the velocity is t)r : t)0 sin 06 - (202^ls) sin(53.0o) : 161 m/sand the A component is ua: -uscos 06 - gt: -(202mls) cos(53o) - (9.80 mls2x5.00 s) : -17Imf s .39Take the A axis to be upward and the r axis to the horizontal. Place the origin at the firingpoint, let the time be zero at firing, and let 0o be the firing angle. If the target is a distance d,away, then its coordinates are r- dandA:0. The kinematic equations are d:uotcos0s and0 - uotsin 0o - Lgt. Elimrnate t and solve for de. The first equation gives t - dlurcosgs. Thisexpression is substituted into the second equation to obtain 2rB sin 0s cos 9s - gd,- 0. [Jse thetrigonometric identity sin 9s cos ds _ | sin(zlil to obtain ufr sin (200 - gd,or QSmfl2)(45] m) sin(200): 4- ufi (460 mls)z - 2.12x 10-3 .The firing angle is 0o: 0.0606o. If the gun is aimed at a point a distance (. above the target,then tan2o: (, 1d, or (, - dtan?s - (45.7 m)tan0.0606 - 0.0484m - 4.84cm.47You want to know how high the ball is from the ground when its horizontal distance from homeplate is 97.5 m. To calculate this quantity you need to know the components of the initial velocityof the ball. [Jse the range information. Put the origin at the point where the ball is hit, take the yaxis to be upward and the r axis to be horizontal . If r (: 107 m) and y (- 0) are the coordinatesof the ball when it lands, then tr : uyrt and 0 - uyat - *St, where t is the time of flight of theball. The second equation gives t - 2uoa I g and this is substituted into the first equation. Use1)0n - uya, which is true since the initial angle is 0o : 45o . The result is r : 2ua I g. Thus gr (9.8 mls2xl07m) :22.9mf u,a - 2 sNow take r and A to be the coordinates when the ball is at the fence. Again ra: uyat-Lgt. The time to reach the fence is given by t - rluo* - (97.5m)lQ2,9mf s) : 4.26s.When this is substituted into the second equation the result is 1. a: utat - ,gt - (22.9^lsx4 .26s) - )e.8^ls2x4 .26s)2:8.63m.Since the ball started I.22m above the ground, it is 8.63 m + 1,.22m : 9.85 m above the groundwhen it gets to the fence and it is 9.85 m - 7 .32m - 2.53 m above the top of the fence. It goesover the fence.l6 Chapter 4
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51Take the A axis to be upward and the r axis to be horizontal. Place the origin at the point wherethe ball is kicked, or the ground, and take the time to be zero at the instant it is kicked. r andA are the coordinates of ball at the goal post. You want to find the kicking angle 0s so thatA- 3.44m when rand Ainto the second the result is gn2 y: rtan0s 2rocos2 gsYou may solve this by trial and effor: systematically try values of 0g until you find the two thatsatisfy the equation. A little manipulation, however, will give you an algebraic solutioll.Use the trigonometric identity 1/ cos 0o: 1 + tan2 0s to obtain 1 ar2 . I -^^2 tft tan2 os - ntanoo + + ;T- Y oThis is a quadratic equation for tanfls. To simplify writing the solutior, let c - ig*lr| -ifq.B0 mls2x50*)t lQ5mls)z - 19.6m. Then the quadratic equation becomes ctanz 0sr tan?o + y + c : 0. It has the solution tan 0o r* + 4(A + c)c 50m * (50 m)2 - 4(3.44 m + 19.6 mX 19.6 m) 2(19.6 m)The two solutions are tan?sand 0son the goal post.s3Let h be the height of a step and u be the width. To hit step n, the ball must fall a distancenh and travel horizontally a distance between (n - l)u and nu. Take the origin of a coordinatesystem to be at the point where the ball leaves the top of the stairway. Take the U axis to bepositive in the upward direction and the r axis to be horizontal. The coordinates of the ball attime t are given by nthe level of step n:The r coordinate then is 2n(0.203 m) n-uor (r.szmls) - (0.30e n)fr,. Chapter 4 t7
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Try values of n until you find one for which r lu is less than n but greater than n 1. Fornnlw - 2. 15. This is also greater than rL. For n- 3, r_0.535m and rlu - 2.64. This is lessthan n and greater than n - 1. The ball hits the third step.67To calcul ate the centripetal acceleration of the stone you need to know its speed while it is beingwhirled around. This the same as its initial speed when it flies off. [Jse the kinematic equationsof projectile motion to find that speed. Take the A axis to be upward and the r axis to behorrzontal. Place the origin at the point where the stone leaves its circular orbit and take the timeto be zeto when this occurs. Then the coordinates of the stone when it is a projectile are givenby r- uot and A_ -+g*. It hits the ground when r- 10m and A- -2.0m. Note that theinitial velocity is horizontal. Solve the second equation for the time: t : {:di. Substitutethis expression into the first equation and solve for uoi ug:r _9 - (10m) 15.7 m/s . 2y 2(-2.0m)The magnitude of the centripetal acceleration is a,: u2 lr - (15.7*1il2 l(1 .5m): 160 ,. ^/73(a) Take the positive r direction to be to the east and the positive A direction to be to the north.The velocity of ship A is given by d t : -t? knots) sin 4s")r i + r (zknots) cos 4s " r i : _rr")LT] I1and the velocity of ship B is given by dn ""|;;:li : -tQ8 knots) sin 40r i - tQ8 knots) cos 40"r i : _ll; ;LT:] I _rrllnffThe velocity of ship A relative to ship B is 6tn:6A-6s - (1 .0 knots) i + 1f 8. knots) j .The magnitude is uln: u2tnr*Io, (1 .0 knots)z + (38.4 knots)2 - 38.4 knots(b)The angle 0 that 6te makes with the positive r axis is UAB A , 38.4 knots -. 0 - tan-1 - tan-I 1.0 knots - 88.5o UAB rThis direction is 1.5o east of north.18 Chapter 4
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(c) The time t for the separation to become d is given by t d lr tu. Since a knot is a nauticalmile, t - (l60nautical miles) lQS. knots): 4.2h.(d) Ship B will be 1.5o west of south, relative to ship A.75Relative to the cur the velocity of the snowflakes has a vertical component of 8.0 m/s anda horizontal component of 50 km/htan 0 - unlu, - (13.9^ls)/(8.0 mls): I.74. The angle is 60o.77Since the raindrops fall vertic aIIy relative to the train, the honzontal component of the velocity ofa raindrop LS u1" 30^ls, the same as the speed of the train. If u, is the vertical component of the -velocity and 0 is the angle between the direction of motion and the vertical, then tan 0 uh/ur. -Thus t)u:uhf tan? - (30 mls)ltan70" - 10.9mf s. The speed of a raindrop is u - G*,?: (30 mls)2 + (10.9m1s)z - 32mf s.91(a) Take the positive A axis to be downward and place the origin at the firing point. Then the Acoordinate of the bullet is given by A - i7t. If t is the time of flight and A is the distance thebullet hits below the target, then 2(0.019 m) : 6.3 x 10-2 s(b) The mvzzle velocity is the initial velocity of the bullet. It is hortzontal. If r is the honzontaldistance to the target, then r : uOt and 30m Ug -y- t 6.3 x 10-2 s - 4.8 x 102 mlst07(a) Use A : uyat - *gt and na - t)ga - gt, where the origin is at the point where the ball is hit,the positive y direction is upward, and uya is the vertical component of the initial velocity. Atthe highest point ua:0, so uya - gt and a: *gt: ifg.8^ls2x3.0r)t - 44m.(b) Set the time to zero when the ball is at its highest point. The vertical component of the initialvelocity is then zero and the balls initral y coordinate ts 44m. The ball reaches the fence at timet- 2.5 s. Then its height above the ground is a: Uo_ Lst :44m- itg.8^ls(2.5 s)t(c) Since the ball takes 5.5 s to travel the horizontal distance of 97 .5 m to the fence, the horizontalcomponent of the initial velocity is 1)0r: Q7.5m)l(5.5 s) - 17.7 mf s. Since the ball took 3.0 sto rise from the ground to its highest point it must take the same time, 3.0s to fall from thehighest point to the ground. Thus it hits the ground 0.50 s after clearing the fence. The pointwhere it hits is (17 .7 s) : 8.9 m. ^1sX0.50 Chapter 4 T9
24.
111 :(a) The position vector of the particle is given by i 6ot + |dtz, where ds is the velocity at timet - 0 and d is the acceleration. The r component of this equation is r : ?Jont* +e*tT. Sinc a uo*0 this becomes r -*.o*t . The solution for t is t- - 2rfa* l zesm)l(4.0 : uyat + *ort - (8 .0 ^ls)The a coordinate then is A s) + *tz.o^1s2x3.81 s)2 : 45 m. ^1sX3.81(b) The r component of the velocity is ur: rr0* + e*t - (4.0 mls2x3.81s) - 15.2mls andthe a component is rra: uya + Lort: 8.0 mls * (2.0m1s2x3.81 s) - 15.6mf s. The speed is?r: l-j (15 .2^ls)2 + (15 .6mls)2 : 22m/s. lr"*r;--12t(a) and (b) Take the r axis to be from west to east and the A axis to be from south to north. Sum the two displacements from A to the resting place. The first is Lfij sin 37o) : (60 km) i+1+S km) j and the second is Liz- -(65 k*)i. The sum is Ar - (60 km) i-(20k*)i. The magnitude of the total displacement is Lrthe tangent of the angle it makes with the east is tan? - (-20 km)/(60 km) - -0 .33. The angleis 18" south of east.(c) and (d) The total time for the trip and rest is 50h+ 35h+ 5.0h:90h, so the magnitude ofthe ayeruge velocity is (63 km) lQ0 h) - 0.70 kmlh. The average velocity is in the same directionas the displacement, l8o south of east.(e) The average speed is the distance traveled divided by the elapsed time. The distance is75 km + 65 km - 140 km, so the average speed is (140 k*) lQ0 h) : 1.5 km/h.(0 and (g) The camel has I20h 90 h - 30 h to get from the resting place to B. If Lin isthe displacement of B from A and A4.r, is the displacement of the resting place from A, thedisplacement of the camel during this time is LrB- Lr-rest: (90km)i-(60km)i- (-20k*)j:(30k-)i+1zok-)i.Themagnitudeofthedisp1acementis-36kmandthe magnitude of the average velocity must be (36 km) lQO h) _ I.zk*/h. The angle 0 that theaverage velocity must make with the east is given by tan Qangle is 34" north of east.20 Chapter 4
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Chapter 5-5Label the two forces Ft and Fr. According to Newtons second law, F, + Fr: md,F, - rnd, - Fr. In unit vector notation Ft - (20.0 N) i and d - -(1 2^lrt;1ritr 30o) i - (I2mlr;1.os 30o)j - -6.0*lrt) i - (10. a^lr) jThus Fz - (z.0ks)(-6.o^lrt)i +(2.0ks)(-1l. mlrt)i - e0.0N)i - (-32N)i - (2rr.Di(b) and (c) The magnitude of F2 is F2: @- (-32N)2 + (-21 X;z - 3gN. Theangle that F2makes with the positive r axis is given by tan? - FzalFr* - (21 N) lQ2N) - 0.656.The angle is either 33" or 33o + 180 - 213o. Since both the r and A components are negativethe coffect result is 2I3". You could also take the angle to be 180o - 213o - -I47o.13In all three cases the scale is not accelerating, which means that the two cords exert forces ofequal magnitude on it. The scale reads the magnitude of either of these forces. In each casethe magnitude of the tension force of the cord attached to the salami must be the same as themagnitude of the weight of the salami. You know this because the salami is not accelerating.Thus the scale reading is ffig,where mtsthe mass of the salami. Its value is (11.0kgX9 .8*lr2) -108N.t9(a) The free-body diagram is shown in Fig. 5 - 16 of the text. Since the acceleration of theblock is zero, the components of the Newtons second law equation yield T - mg sin 0:0 andl7/r, mg cos 0(8.5 kexg .8^lrtl rin 30" : 42 N.(b) Solve the second equation for ,F^/: Fry - mgcos 0 - (8.5kg)(9.8^lrt;ros30o :72N.(c) When the string is cut it no longer exerts a force on the block and the block accelerates . The rcomponent of the second law becomes -mgsin e - me,, so a,: - gsin 0 - -(9.8^lr;ritt30o-4.9mf s2. The negative sign indicates the acceleration is down the plane.25According to Newtons second law F - ma, where F is the magnitude of the force, a" is themagnitude of the acceleration, and m is the mass. The acceleration can be found using theequations for constant-acceleration motion. Solve u : lJ1 * at for a: e, : u lt The final velocity Chapter 5 2l
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is u - (1600 korlhx1000m/km)1p600 s/h) : 444mls, so o (444mls)/(1.8 s) : : 247 and ^lr ^lst) -the magnitude of the force is Ir - (500 kgQaT I .2 x 105 N.29The acceleration of the electron is vertical and for all practical puqposes the only force actingon it is the electric force. The force of gravity is much smaller. Take the fr axis to be in thedirection of the initial velocity and the A axis to be in the direction of the electrical force. Placethe origin at the initial position of the electron. Since the force and acceleration are constanttheappropriateequations arer_ uot and a- Lot: ifgl*)*, where F- ma, wasusedtosubstitute for the acceleration a. The time taken by the electron to travel a distance n (: 30 mm)horizontally is t - r luo and its deflection in the direction of the force is 4.5 x 1 g-t6 y ,a2 1F 30x 10-3 m a: ;- /.m (#):;( 9.11 x 10-3t kg )( 1.2 x 107 m/s )35The free-body diagram is shown at the right. F* is the normal forceof the plane on the block and mj ts the force of gravity on the block.Take the positive r axis to be down the plane, in the direction ofthe acceleration, and the positive A axis to be in the direction of thenormal force. The r component of Newtons second law is thenmg sin 0 - ma) so the acceleration is a : g stn 0 .(a) Place the origin at the bottom of the plane. The equations formotion alongthe r axis are r: uot+)atz and u: n0*at. The blockstops when u : 0.According to the second equation, this is at the time t - -us f a. The coordinate when it stops is 2--1 3) *)"(+) fr : tJ.( -u0 6 A 2 a, 2 gstn1 2 t_ (-3.s0 mls)2 : -t.t8m | tq .B^lrt; Ltn ritt 32.0" ] (b) The time is t Ug Ug -3.50 m/s 0.674s. g srn? (9.g mlrt;,in 32.0"(c) Now set r: 0 and solve tr : uot + atz for t. The result is , 2uo 2uo 2(-3.50 m/s) t::: a g stn? (9.8 mlrt; ritt 32.0o - 1.35 s.The velocity is u - uol_ at: uo* gtsrnl - -3.50 mls + (9.8 mls2xl.35 s)sin 32o - 3.50 mls,22 Chapter 5
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as expected since there is no friction. The velocity is down the plane.45The free-body dtagrams for the links are drawn below. The force affrows are not to scale. 4onr{ F:onz{ ll-rrrJl--rr-rll fLon:{ fton+{ FA lltrl t o Ior I ltl ltl mgy mgililr on msll F3 on+ msll F+ on s Link I Link 2 Link 3 Link 4 Link 5(a) The links are numbered from bottom to top. The forces on the bottom link are the force ofgravity md, downward, and the force F, onr of hnk2, upward. Take the positive direction to beupward. Then Newtons second law for this link is Fz on 1 - mg : ma. Thus Fz onl : m(a+ g) - (0.100kgX2.50 mlr+ 9.8 mlr) - I.23N.(b) The forces on the second link arc the force of gravity md, downward, the force Fr on zof tink I, downward, and the force Fz onz of link 3, upward. According to Newtons thirdlaw Ft onz has the same magnitude as Fzonr. Newtons second law for the second link isF3 onz - fl on2 - mg : mQ,, SO Fz onz- m(&+ g)+ Fr onz- (0.100kg)(2.50 mlr+9.8 mls2)+ 1.23N - 2.46N,where Newtons third law was used to substitute the value of F2on I for Ft onz.(c) The forces on the third link are the force of gravity mj, downward, the force Fz on z of link2, downward, and the force F+ on z of link 4, upward. Newtons second law for this link isF+onZ- F2on3 -mg:TnA, SO F+on3 - m(a+ g)+ Fzon3 - (0.100NX2.50 mlr+9.8 mls2)+ 2.46N - 3.69N,where Newtons third law was used to substitute the value of F3 on2 for F2 on3.(d) The forces on the fourth link are the force of gravity mj, downward, the force Ft on + oflink 3, downward, and the force Fs on + of link 5, upward. Newtons second law for this link isFS on + - F3 on 4 - mg - mA, SO Fs on+ - m(a+ g)+ Ft on4 - (0.100kgX2.50 mlr+ 9.8 mls2) + 3.6gN - 4.92N,where Newtons third law was used to substitute the value of Faon 3 for F3 on4.(e) The forces on the top link arc the force of gravrty mrt, downward, the force Fq on s oflink 4, downward, and the applied force F, upward. Newtons second law for the top link isF - F+ on S - mg - nLa, SO F -m(a+ g)+ Fqon5 - (0.100k9(2.50 mlrt+9.8 mls2)+ 4.92N- 6.15N,where Newtons third law as used to substitute the value of Fs on 4 for F+ on s. Chapter 5 23
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(0 Each link has the same mass and the same acceleration, so the same net force acts on eachof them: Fnet - ma - (0.100 kg(z.50 mls2) - 0.25 N.s3(a) The free-body dragrams are shown to the right. F is the appliedforce and f is the force of block 1 on block 2. Note that F isapplie,C only to block 1 and that block 2 exerts the force - i onblock 1. Newtons third law has thereby been taken into account.Newtons second law for block 1 is F f - TrL1a, where o, isthe acceleration. The second law for block 2 is f -- TrL2e,. Sincethe blocks move together they have the same acceleration and thesame symbol is used in both equations. Use the second equationto obtain an expression for a: a - f l*r. Substitute into the firstequation to get F - f : TTLI f l*r. Solve for f : n Fmz Q.2NXl.2kg) f - m1 * TTL2 z.3kg + r.zk|- - 1 1 N llrr(b) If F is applied to blo ck2 instead of block 1, the force of contactis i- Fml Q.2NX2.3 kg) a_ TrLl I mz_ 2.3 kg + l.2kg t .t(c) The acceleration of the blocks is the same in the two cases. Since the contact force f isthe only horizontal force on one of the blocks it must be just right to give that block the same ---.1N.acceleration as the block to which F is applied. In the second case the contact force acceleratesa more massive block than in the first, so it must be larger.57(a) Take the positive direction to be upward for both the monkey and the package. Suppose themonkey pulls downward on the rope with a force of magnitude F. According to Newtons thirdlaw, the rope pulls upward on the monkey with a force of the same magnitude, so Newtonssecond law for the monkey is F - mrng - mrnarn, where mrn is the mass of the monkey ande,rn is its acceleration. Since the rope is massless F is the tension in the rope. The rope pullsupward on the package with a force of magnitude F, so Newtons second law for the packageis F + Flr - mpg : mpap, where mp is the mass of the package, a,p is its acceleration, and Fxis the normal force of the ground on it.Now suppose F is the minimum force required to lift the package. Then Flr - 0 and e,,p - 0.According to the second law equation for the package, this means F - mpg. Substitute wpg forF in the second law equation for the monkey, then solve for am. You should obtain : F - rrlrng : (mo - m,n)g (15 kg - 10kg)(9.8 An - _ r kg ^lr - +Y mf s 2 aTn ,rr-, *-, lo 24 Chapter 5
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(b) Newtons second law equations are F -mpg: ffipap for the package and F -rrLrng : TrL,n,arnfor the monkey. If the acceleration of the package is downward, then the acceleration of themonkey is upward, so o??" - -ap. Solve the first equation for -F: p - mp(g + op) : ffip(g - arn).Substitute the result into the second equation and solve for arni anL- (TY - *)g / mp*Tftm. 15kg+10kg(c) The result is positive, indicating that the acceleration of the monkey is upward.(d) Solve the second law equation for the package to obtain p- Trl,p(g - a,nr): (15kgX9.8mf s2 -2.0*ls2) - 120N61The forces on the balloon arc the force of gravity mj, down , and the force of the atr Fo, up.Take the positive direction to be up. When the mass is M (before the ballast is thrown out)the acceleration is downward and Newtons second law is Fo M g - - M a. After the ballastis thrown out the mass is lVI wL, where m is the mass of the ballast, and the accelerationis upward. Newtons second law is Fo (M m)g - (M m)a. The first equation givesFo: M(g a) and the second gives M(g a) (M m)g - (M m)a. Solve for m:m:ZMal(g + a).73 rTake the axis to be horizontal and positive in the direction that the crate slides. ThenFcosd facceleration (the only nonvanishing component). In part (a) the acceleration is e,r : F cos? - f - (450N)cos38" - 125N n n -^^ 1^2 -ut+m/s A m 310kg In part (b) m: Wlg _- (3 10N) lQ.8^ls) - 31.6kg and Fcos0-f (450N)cos38o 125N ,1 ..r r2 nr m 36.1 kg79Let F be the maglitude of the force, a4 (: L2.0mls2) be the acceleration of object I, and a,2(:3.30 mlt; be the acceleration of obj ect2. According to Newtons second law the masses arerrLl: Flol and TrL2- Ff e,2.(a) The acceleration of an object of mass rTL2 - mr is (L- F - - +O m/ S TrL2 - Tft1 or u, I2.0 Chapter 5 25
30.
(b) The acceleration of an object of mass TtLl 1 m2 is A- F _ F eta,z (12.0 ^ls2x3.3o *lt) - 2.6mf s2 Tft2 * mr (F lor) + (F lo) a1 a aZ 12.0 mlr + 3.3 0 mf s29l(a) Both pieces arc station dry, so you know that the net force on each of them is zero. The forceson the bottom piece are the downward force of gravity, with magnitude Tftzg, and the upwardtension force of the bottom cord, with magnitude T6. Since the net force is zero, T6: TTL27 : (4.5kgX9 .8^lr2) - 44N.(b) The forces on the top piece are the downward force of gravity, with magnitude r(Ltg, thedownward tension force of the bottom cord, with magnitude Ta, and the upward force tension ofthe top cord, with magnitude ?7. Since the net force is zero, T1 : Tb t mrg :44 N + (3.5 kg)(9.8 mlst) - 78 N .(c) The forces on the bottom piece are the downward force of gravity, with magnitude rrLsg, andthe upward tension force of the middle cord, with magnitude Trn Since the net force is zeto, : : ^ls) - T,n TTLsg (5.5 kgXg .8 54 N.(d) The forces on the top piece are the downward force of gravity, with magnitude w4g, theupward tension force of the top cord, with magnitude T7 (: I99 N), and the downward tensionforce of the middle cord, with magnitude Trn Since the net force is zero) T,n: Tt - Tft3g : IggN - (4.8 kg)(9 .8^ls2) - 152N.9s(a) According to Newtons second law the magnitude of the net force on the rider is It - ma, :(60.0kg)(3.0 mls) - 1.80 x 102N.(9) Take the t force to__b. thl vector sum of the force of the motorcycle and the force of Earth: t Fnet: Fr, + Fn. Thus Frn: 4r.t - Fs. Now the net force is parallel to the ramp and thereforemakes the angle 0 (: 10) with the horrzontal, so Fnt_ (F cos 0)i+(lrsin 0)j, where the r axisis taken to be horrzonta| and,the A axLS is taken to be vefitcal. The force of Earth is F" - -mgj,so F,- - (F cos il?+ (F sin o + milj.Thus F,n* - It cos 0 - (1.80 x I02 N) cos 10oand Frna - (1.80 x 102) sin 10o + (60.0kg)(g.g mls2) - 6.19 x 102 N.The magnitude of the force of the motorcycle is -6.44 x TA -r rn t02 N)2 r02 N)2 102N.26 Chapter 5
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99The free-body diagrams for the two boxes are shown below. Flrr FxzHere T is the tension in the cord, FxTr is the norrnal force of the left incline on box 1, and FNzis the norrnal force of the right incline on box 2. Different coordinate system are used for thetwo boxes but the positive r direction are chosen so that the accelerations of the boxes have thesame sign. The r component of Newtons second law for box 1 gives T - TTLI7 stn01 : TTL1& andthe r component of the law for box 2 gives mzg stn 02 - T : TTL27. These equations are solvedsimultaneously for T. The result is mtmzg (3.0 kgx2.0 kgX9.8 m/s2) : T: (sin 91 * sin oz) (sin 30o + sin 60.) 16 N ?TL1 * mZ 3.0 kg + 2.0kg ^101Free-body diagrams for the two tins are shown onthe right. T is the tension in the cord and Fy is Tthe normal force of the incline on tin 1. The posi-tive n direction for tin I is chosen to be down theincline and the positive r direction for tin 2 is cho-sen to be downward. The sign of the accelerations ofthe two tins arc both then positive. Newtons secondlaw for tin 1 gives T + Tft1g stn {3 - ma, and for tin 2 rgives mzg T F - TrL2cL. The second equation issolvedforT,withtheresultT- TrL2@_ a)-F-(2.0kgX9.8^lrt-5.5m1s2)-6.0N-2.6N.The first Newtons law equation is solved for sin p, with the result sin C:mra-T mts (1.0kgX9.8 m/s)The angle is I7o Chapter 5 27
32.
Chapter 51(a) The free-body diagram for the bureau is shown on the right.F is the applied force, "f- i, the force of friction, fi" is the normalforce of the flooq and mj is the force of gravity. Take the r axisto be honzontal and the A axis to be vertical. Assume the bureaudoes not move and write the Newtons second law equations. Ther componentis F- f :0 andthe A componentis Fl/ -rng:0.The force of friction is then equal in magnitude to the appliedforce: fforce of gravity: ,Fl/ - mg. As F increases, f increases untilf : ltrrFx. Then the bureau starts to move. The minimum forcethat must be applied to start the bureau moving is It - FrFx : ltr,mg - (0 .45)(45 kgX9.8 mls2) - 2.0 x I02 N .(b) The equation for F is the same but the mass is now 45 kg - 1,7 kg - 28 kg. Thus F - F,mg - (0 .45)(28 kgX9.8 mlst) - 1.2 x T02 N .3(a) The free-body dragram for the crate is shown on the right. F isthe force of the person on the crate,,f-ir the force of friction, Fr itthe noffnal force of the floor, and mj ts the force of gravity. Themagnitude of the force of friction is given by f - l.rrcFw, whereltr* is the coefficient of kinetic friction. The vertical componentof Newtons second law is used to find the normal force. Sincethe vertical component of the acceleration is zero, Fl/ - mg - 0and Fl/ - mg. Thus f: F;Fw : ltkmg - (0.35X55 kgX9.8 mls2) - l.g x I02 N .(b) IJse the horizontal component of Newtons second law to find the acceleration. SinceF-f:ffia, (F - f) (220N - 18eN) a: - - 0.5 6mf s2 m 55kg28 Chapter 6
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13(a) The free-body diagram for the crate is shown on the right. f isthe tension force of the rope on the crate, Fn is the norrnal force ofthe floor on the crate, rnfi rs the force of gravity, and ,i ir the forceof friction. Take the r axis to be horizontal on the right and the Aaxis to be vertically upward. Assume the crate is motionless. The rcomponent of Newtons second law is then 7 cos e - f : 0 and the Acomponent is Tsin 0+F^/ *mg:0, where 0 (: 15") is the angle be-tween the rope and the horizontal. The first equation gives / - T cos Iand the second gives Fx - mg - T sin 9. If the crate is to remain atrest, / must be less than lr"Fx, or T cos? <the tension force is sufficient to just start the crate moving T cos 0 - lrr(mg 7 sin d). Solvefor T: T : Fmg cos 0 + p,, stn0 cos 15o + 0.50 sin 15o(b) The second law equations for the moving crate are 7 cos 0 f : ma and -Fl/ + T sin 0mg:0. Now ff - p,n(mg 7 sin 0). This expression is substituted for f in the first equation to obtain? cos 0 - pn(mg - T sin 0) : ffia, so the acceleration is T(cos 0 + p,7" sin 0) a: -ltt 9 *Its numerical value ls e, = - (0.35X9 .8^ls2) - 1.3 ^lr23The free-body diagrams for block B and for the knotjust above block ,4 arc shown on the right. T1 is themagnitude of the tension force of the rope pulling on blo ck B , Tz is the magnitude of the tension force of the other rope, f is the magnitude of the force of friction exerted by the horizontal surface on block B, FAr is the magnitude of the norrnal force exertedby the surface on block B, We is the weight ofblock A, and W n is the weight of block B . 0 (- 30") is the angle between the second rope and the horizontal.For each object take the r axis to be horizontal and the A axis to be vertical. The fr componentof Newtons second law for block B is then T1 f :0 and the A component is Fy Wn:0.The r aomponent of Newtons second law for the knot is T2 eos 0 - Tr - 0 and the A componentis Tzsin? Wt:0. Eliminate the tension forces and find expressions for f and F11r in terms Chapter 6 29
34.
of Wt and Wn, then select Wt so f : F"Fx. The second Newtons law equation givesFyr - Ws immediately. The third gives Tz - Tr I cos 9. Substitute this expression into thefourth equation to obtain T1to obtain f - Waf tan?. For the blocks to remain stationary f must be less than F"Fx orWaltan? <Solve for We: We: FrWntanT - (0.25)(711N) tan3Oo - 1.0 x I02 N.27(a) The free-body dragrams for the two blocks areshown on the right. T is the magnitude of the ten-sion force of the string , Fx t is the magnitude of thenormal force on block A, Fw s is the magnitude ofthe nonnal force on blo ck B , f a is the magnitude ofthe friction force on block A, f e is the magnitudeof the friction force on block B, TTLy is the mass ofblock A, and TTLs is the mass of block B . 0 is theangle of the incline (30). We have assumed that theincline goes down from right to left and that blockA is leading. It is the 3.6-N block.For each block take the r axis to be down the plane and the A axis to be in the direction of thenormal force. For block A the r component of Newtons second law is mtgsin0- fa-T:TrLAa,gand the A component is Fnt-Tftagcos0:0.Here a,s is the acceleration of the block. The magnitude of the frictional force is ft : FkAFxt : FkAmtgcos 0,where Fxa- mtg cos9, from the second equation, is substituted. Fnt is the coefficient ofkinetic friction for block A. When the expression for f a is substituted into the first equation theresult is Tftsg sin 0 - FntThsg cos 0 - T - ?rLga,n .The same analysis applied to block B leads to Trlpg sin 0 - FnsTrlyg cos 0 + T - rTL Ba, p .We must first find out if the rope is taut or slack. Assume the blocks are not joined by a rope andcalcul ate the acceleration of each. If the acceleration of A is greater than the acceleration of B,then the rope is taut when it is attached. If the acceleration of B is greater than the accelerationof A, then even when the rope is attached B gains speed at a greater rate than A and the ropeis slack.30 Chapter 6
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Set T:0 in the equation you derived above and solve for as and a,p. The results arc a,A- g(sin0 - ltp7cos 0): (9.8 mlrt;1ritt30o - 0.10cos30") : 4.05^ltand a,s - g(sin0 - ltn6 coS 0): (9.8 mlr;1ritr30o - 0 .20cos30o): 3.20mlrt.We have learned that when the blocks are joined, the rope is taut, the tension force is not zero,and the two blocks have the same acceleration.Now go back to mAg sin 0 - Fntmeg cos 0 -T - TrL4a, and ThBgsrn) - FnsmBgcos 0 +T -TTL6a, where a has been substituted for both a,s and a,s. Solve the first expression for T,substitute the result into the second, and solve for a. The result is a: gsin g llnaml* H"nnm6 a - rYL4 g cos ^ + mB N)l (9.g N) + (020X72 - (9 .g^lrt; 30o ,i,r [(010X36 L 3.6N+7.2N l-/ mlrr; ,os 30o - 3.5 ^lrt .Strictly speakirg, values of the masses rather than weights should be substituted, but the factorg cancels from the numerator and denominator.(b) IJse ffLsg sin 0 - Fneme7 cos 0 - T - rTLAa to find the tension force of the rope: T - mAg sin 0 - FntTTLsgcos e - TtLAa, - (3.6 N) sin 30o - (0. 10X3.6 N) cos 30o - (3.6 N/9.8 mls2x3 .49 mlst) - 0.21 N .35Let the magnitude of the frictional force be au, where e, : 70 N . s/m. Take the direction of theboats motion to be positive. Newtons second law is then -aa : m du ldt Thus f du: Ju, u mJo /where us is the velocity at time zero and u is the velocity at time t. The integrals can beevaluated, with the result -uat tn*- -A.Take 1) : ,ol2 and solve for f : loookg t -m rn2 - 70N.s/m hz - g.9s. a49(a) At the highest point the seat pushes up on the student with a force of magnitude F^/ (: 556 N).Earth pulls down with a force of magnitude W (: 667 N). The seat is pushing up with a forcethat is smaller than the students weight in magnitude. The student feels light at the highestpoint. Chapter 6 31
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(b) When the student is at the highest point, the net force toward the center of the circular orbitis W - Fy and, according to Newtons second law, this must be mulR, where u is the speed ofthe student and R is theradius of the orbit. Thus mu2f R-W - f^i:667N- 556N- 111].{.The force of the seat when the student is at the lowest point is upward, so the net force towardthe center of the circle is F^i W and lr^/ W - mu2 I R. Solve for Fn: ,.,/,/- 667N- 77gN. rytW:ltlN+(c) At the highest point W F^/ : mu2 f R, so f^imu2 lR increases by a factor of 4, to 444N. Then F^/ :667 N - 444N - 223NI. :(d) At the lowest point W + Fl/ muI R, so F^/ -W-mu2fR,so-F1/ :mu2 I R is still 444 N, F^r 667 N + 444N 1.11 - x 103 N.53The free-body diagram for the plane is shown on the right" F is themagnitude of the lift on the wings and m LS the mass of the plane. Since thewings are tilted by 40" to the honzontal and the lift force is perpendicularto the wings, the angle 0 is 50o. The center of the circular orbit is to theright of the plane, the dashed line along r being a portion of the radius.Take the r axis to be to the right and the A axis to be upward. Thenthe r component of Newtons second law is F cos g - mu2 I R and the Acomponent is F sin I mg - 0, where R is the radius of the orbit. Thefirst equation gives It - mu2 I Rcos 0 and when this is substituted into thesecond, (*rlR)tan? - mg results. Solve for R: tr n12 R- tanl. IThe speed of the plane is tr :480 kmfh - 133 mls, so _ 033 mls)2 : 2.2 x 103 m R tansoo . 9.8 mls59(a) The free-body dragram for the ball is shown on the right. f,is the tension force of the upper string , ip is the tension forceof the lower string, and m is the mass of the ball. Note thatthe tension force of the upper string is greater than the tensionforce of the lower string. It must balance the downward pullof gravity and the force of the lower string.32 Chapter 6
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Take the r axis to be to the left, toward the center of the circular orbit, and the A axis to beupward. Since the magnitude of the acceleration is a -- ,2 I R, the r component of Newtonssecond law is Tucosg+ Tacos0-ry,where u is the speed of the ball and R is the radius of its orbit. The A component is 2,, sin 0 - Tzsrn? - mg - 0.The second equation gives the tension force of the lower string: Tt : Tu - mg f stn 0 . Since thetriangle is equilateral 0 - 30o. Thus (13a kgX9"9 m/s2) Tr: 35 N - - 8.74N . srn 30o(b) The net force is radially inward and has magnitude Fnet, str - (7, + TD cos 08.74N) cos 30" - 37.9 N.(c) [Jse Fn"t,str: nr,uz f R. The radius of the orbit is [(1.70m) l2)Jtan30" - 1.47 m. Thus RFn"t, str (1.a7 mX37.eN) 1.3 4kg - 6.45 mls .65The first sentence of the problem statement tells us that the maximum force of static frictionbetween the two block is f ,,^u*:12N.When the force F is applied the only horizontal force on the upper block is the frictional force ofthe lower block, which has magnitude f and is in the forward direction. According to Newtonsthird law the upper block exerts a force of magnitude f on the lower block and this force is inthe rearward direction. The net force on the lower block is F - f .Since the blocks move together their accelerations are the same. Newtons second law for theupper block gives f - Tftt& and the second law for the lower block gives F - f : TTL6a, where cris the common acceleration. The first equation gives a - f l*t Use this to substitute for a, inthe second equation and obtain F - f : (mal^t)f .Thus Ii-(r*e) f mt/ If f has its maximum value then F has its maximum value, So the maximum force that can beapplied with the block moving together is F- (t.ffi) lzN):27NThe acceleration is then LL: f l2N ._ ,) TrLl 4.0 N Chapter 6 33
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77(a) The force of friction is the only horizontal force on the bicycle and provides the centripetalforce need for the bicycle to round the circle. The magnitude of this force is f : mu2 f r, wherem is the mass of the bicycle and rider together, u is the speed of the bicycle, and r is the radiusof the circle. Thus : (85.0 kgXg.oo ml s)2 f - ,n 25.0m(b) In addition to the frictional force the road also pushes up with a nonnal force that is equalin magnitude to the weight of the bicycle and rider together. The magnitude of this force is.F1r/ - mg: (85.0kg)(9.8 The frictional and normal forces are pe{pendicular to ^lt)-833N.each other, so the magnitude of the net force of the road on the bicycle is 4r"t - , E vL * ** - Va (275 N)2 + (833 N)2 - 877 N.81The free-body diagrams are shown on theright. T is the tension in the cord, F* isthe norrnal force of the incline on block A,^F* u is the normal force of the platfonn onblock B, I is the angle that the incline makeswith the horizontal (which is also the anglebetween the normal force and the vertic aI),and ,f-ir the frictional force of the platforrnon block B . The r axis for each block is alsoshown.The r component of Newtons second law for block A gives mgsinO T- Trlye,, the frcomponent of the second law for block B is T fFxn - TTLsg - 0.Note that the blocks have the same acceleration.The magnitude of the frictional force LS ptrFxs: ltkTftsg,where ?rLpg was substituted for Fxn,and the r component for B becomes T - Fnmng - TTLsa,. The equations mgsin 0 - T - TLnaand T - Fnmng: TTLsa are solved simultaneously for T and a. The results are + (4.0 kgX2.0 kg)(sin 30o + 0.50) t _ TftlTrLB(sin 0 Fil rF r rJl 1 *^+r", 10roand (4.0kg) sin3Oo - (0.50x2.0kg) ., , r 2 e,- Masin 0 - Fnrns g: :1.6m/s ms+ mB r 4.0k9 + 2.0k98s(a) If u is the speed of the car, m ts its mass, and r is the radius of the curve, then the magnitudeof the frictional force on the tires of the car must be f : muz f r or else the cat does not negotiatethe curve. Since m - W I g, where W is the weight of the car, r. wu2 (10.7 x 103 N(13.4m1s)2 . ,_,1 1 a gr JL-J./rLzlWl. (9.g mls2x61.0 m)34 Chapter 6
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(b) The norrnal force of the road on the car is l7^/ - W and the maximum possible force of staticfriction is .f", -u* - FrFxforce that is required is less than the maximum possible, the car successfully rounds the curve.9tLet F be the magnitude of the applied force and f be the magnitude of the frictional force.Assume the cabinet does not move. Then its acceleration is zero and, according to Newtonssecond law, F - f . The norrnal force is F1,maximum force of static friction is Fr,maxif F is less than 378 N the cabinet does not move and the frictional force is fgreater than 378 N, then the cabinet does move and the frictional force is f(0.56X556 N) - 311 N.(a) The cabinet does not move and f : 222I.(b) The cabinet does not move and f : 334 N.(c) The cabinet does move and f : 311 N.(d) The cabinet does move and f : 311 N.(e) The cabinet moves in attempts (c) and (d).99(a) The free-body diagram for the block is shown on the right. Themagnitude of the frictional force is denoted by f , the magnitude -. Fl/of the normal force is denoted by lrl/ , and the angle between theincline and the horizontal is denoted by 0. Since the block is sliditgdown the incline the frictional force is up the incline. The positive /0r direction is taken to be down the incline. For the block when frit is sliding with constant velocity the n component of Newtonssecond law gives mg sin 0 f - 0 and the A component givesmg cos 0 -Fl/ - 0. The second equation gives F1/ - mg cos9, so the magnitude of the frictionalforce is fobtain mg sin 0 - Tftuamg cos 0 : 0. Thus the coefficient of kinetic friction is Fp - tan 0 .When the block is sliditrg up the incline the frictional force has the same magnitude but is directeddown the plane. The r component of the second law equation becomes mg sin 0 + Fnmg cos e -rrLa, where a, is the acceleration of the block. Thus a - (sin 0 + [L7"cos0)g - 29 sind, where tan?was substituted for Fn and tan 0 - sin 0 f cos 0 was used.If d is the displacement of an object with constant acceleration a, us is its initial speed and uis its final speed, then u2 ulr: Zad. Set u equal to zero and a, equal to 29 stn? and obtaind - -ufil2n -- -ufrlag sin9. The negative sign indicates that the displacement is up the plane.(b) Since the coefficient of static friction is greater than the coefficient of kinetic friction themaximum possible static frictional force is greater than the actual frictional force and the blockremains at rest once it stops. Chapter 6 3s
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105The box is subjected to two horizontal forces: the applied force of the worker, with magnitudeF, and the frictional force, with magnitude f .Newtons second law gives F - f : ma) where mis the mass of the box and a is the magnitude of its acceleration. The magnitude of the frictionalforce is fthe normal force of the floor. In this case F1y - mg and fbecomes F - Ltnmg - me so ltt - (lr - ma) l*g.Let u be the final speed of the box and d, be the distance it moves. Then u2 - Zad, anda,- u2 lza- (1.0 mlilz l2(1 .4m):0.3 57mf s2. The coefficient of kinetic friction is F - ma (85 N- (a0kgX0.357 m/s2) ltt - m9 - (40 kgX9.8 m/s2) 0.1836 Chapter 6
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Chapter 73(a) Use Eq. 2-16: u2 - u|+Zar, where us is the initial velocity, u is the final velocity, r is thedisplacement, and a is the acceleration. This equation yields(b) The initial kinetic energy is K,;,- i*rtThe final kinetic energy is Ky iO.67 x 10-27 kg)(2.9 x r07 ^ls)2:6.9 x 10-13J. - **r:The change in kinetic energy is LK :6.9 x 10-13 J - 4.8 x 10-13 J :2.I x 10-13 J.t7(a) Let F be the magnitude of the force exerted by the cable on the astronaut. The force ofthe cable is upward and the force of gravity is mg is downward. Furtherrnore, the accelerationof the astronaut is glI0, upward. According to Newtons second law, F mg: mgl10, soFdone by F is ILmgd,- 11(72 kgX9.8 m/s2X15 m) Wp: Fd - - l.16 x 104J. 10 10(b) The force of gravity has magnitude mg and is opposite in direction to the displacement.Since cos 180" - - 1, it does work ws(c) The net work done is W :1.16x 104J- 1.06x 10aJ - 1.1x 103 J. Since the astronaut startedfrom rest the work-kinetic energy theorem tells us that this must be her final kinetic energy.(d) Since K - i*rher final speed is 2K 2(lJ x 103 J) u- - 5.3 m/s . mt9(a) Let F be the magnitude of the force of the cord on the block. This force is upward,while the force of gravity, with magnitude M g, is downward. The acceleration is g 14, down.Take the downward direction to be positive. Then Newtons second law is Mg F- Mgl4,soFWp Chapter 7 37
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(b) The force of gravity is in the same direction as the displacement, so it does work Ws : M gd.(c) The net work done on the block is Wr: -3MSdl4+ Mgd - Il[gd,l4. Since the block staftsfrom rest this is its kinetic energy K after it is lowered a distance d.(d) Since K- LtW r, where u is the speed, u:Vt2K tsd tw:Vzafter the block is lowered a distance d. The result found in (c) w-as used.29(a) As the body moves along the r axis from n6 - 4.0 m the work done by theforce is w: F*d,r: -6rd,n - -3 *1"_,- -3(*r - *?) l_"n I_"nAccording to the work-kinetic energy theorem, this is the change in the kinetic energy: W:A,K:L*@?-r?),where u6 is the initial velocity (at t) and u.s is the final velocity (at r f). The theorem yields l2w= 2(-21 J) u1: I * +i: 2.}ks + (g.o m/s)2 - 6.6mf s. ,)2(b) The velocityof the paftrcle is uy - 5.O^ls when it is at r - n y. Solve the work-kinetic energytheorem for n y. The net work done on the particle is W -3(*tr *), so the work-kineticenergy theorem yields -3 (*r - rl) : **(r? - ,?). Thus - , t-+[(s.0m/s)2 - (8 .O^ls)r] + (3.0m), - 4.7 m. V6N35(a) The graph shows F as a function of r tf rs is positive. F(*)The work is negative as the object moves from r FsrSince the area of a triangle is ] (Uur.)(altitude), the work 0done from r:.0 to n: n0 is -+(roXFo) and the work donefrom rThe net work is the sum, which is zero. -Fs38 Chapter 7
43.
(b) The integral for the work is P*o w: Io"o ro(X t) dr:Fs(* ")1, -043The power associated with force f is given by P : F . 6, where u* is the velocity of the objecton which the force acts. Let d (: 3 7") be the angle between the force and the honzontal. ThenP - F.6- Fucos d- 022NX5.O^ls)cos 37":4.9 x I02W.45(a) The power is given by P - Fu and the work done by f from time t1 to time t2 is given by rtz rtz w - Jr, P dt: ,, dt. Jr,Since F is the net force, the magnitude of the acceleration is a - F l* and, since the initialvelocity is us: 0, the velocity as a function of time is given by , : uo * at - (F lm)t Thus w: nt^t " l (tr l *)t dt - )rr l *)e3 . J tt t?)For f1 - 0 and t2 : Yv i [tso xl2 I (t.o s)2 - 0 83 J 14/ z l5kg L l(b) For dr - 1.0 s and t2 - 2.0 s, w :1 [(s oryfl lrz.o , L lsLs If(zu r), - (1.0 s)r] - 2.s r.(c) For C1 - 2.0 s and tz - 3.0 s, w:1 fry] z lsk [(:0,), -(20,),] - 421 L(d) Substitute u- (Flm)t into P- Fu to obtain P- Ttztl* for the power at any time t. Atthe end of the third second P - (5.0 N)2(3.0 s) 115 kg - 5.0 W.47The net work Wnet is the sum of the work Wu done by gravrty on the elevator, the work W.done by gravity on the counterweight, and the work W, done by the motor on the system:Wn t - W" + W. + W". Since the elevator moves at constant velocity, its kinetic energy doesnot change and according to the work-kinetic energy theorem the net work done is zere. Thismeans Wu + W"* W" Chapter 7 39
44.
gravity on it is W"- -TrLegd- -(1200kgX9.8*lr;1S+m): -6.35 x 105J. The counterweightmoves downward the same distance, so the work done by gravity on it is W" : rn"gd -(950kgX9.8 mlr;1S+m) - 5.03 x 105J. Since Wr: 0, the work done by the motor on thesystem is W, _ -W" - W. - 6.35 x 105 J - 5.03 x 105 J - 1.32 x 105 J. This work is done in atime interval of At - 3.0 min - 180 s, so the power supplied by the motor to lift the elevator is p -%- 132 x 1o5J Lt 180 s -7.35x rozw.63(a) Take the positive r direction to be in the direction of travel of the cart. In time Lt thecart moves a distance Lr - u Lt, where u is its speed. The work done is W - F* Lr(Fu cos 0) LL where f is the force of the horse and 0 is the angle it makes with the horizontal.Now 6.0 milh600s, so It - t(40 lbXS.Sftls)cos30ol(600s): 1.8 x 105 ft.lb.(b) The averagepower is P - F*u - Fucos 0 -(401bX8.}ftfs)cos30o:3.0 x 102ft.lb/s.Since 1ft-Ibls - 1.818x 10-hp, thepoweris P - (3.0x 102ft.lblsxl.818x 10-thplft. lb/s) -0.55 hp.69(a) The applied force f is in the direction of the displacement i, so the work done by the forceis We : Fd -- (209 NX1.50m) - 314J.(b) The crate rises is distance Ly - d stn 0, where 0 is the angle that the incline makes withthe horizontal. The work done by the gravitational force of Earth is Ws-(25.0kg)(9.8 mls2xl.50m) sin 25.0"- -155 J.(c) The nonnal force of the incline on the crate is perpendicular to the displacement of the crate,so this force does no work.(d) The net work done on the qate is Wnet: We - 3I4J - 155 J - 159J. "Ws7lLet Wr (: 1 10 N) be the first weight hung on the scale and r be the elongation of the springwith this weight on it. Let Wz (- 240 N) be the second weight hung on the scale and n2 be theelongation of the spring when this weight is hung on it. In each case the spring pulls upwardwith a force that is equal to the weight hung on rt, so according to Hookes law Wr -- krt andWz: krz, where k is the spring constant. Now 11 and 12 are not the readings on the scales butnz- nr is the difference of the scale readings. Subtract the two Hookes law equations to obtainWz Wr: k(*z - rr). Thus _ ,u Wz /v,/ WtL f v 240N .<r-T-rJ - 110N IILTL . a W.J n rV t/Irr. n2-rr - Z 60 x 10-3m -40 x 10-3m -When W1 is hung on the spring the elongation is n1 - Wrlk: (110N)/(6.5 x 103].1/m)1.7 x l0-2m: 17mm. The reading on the scale is 40mm- ITmm:23mm.40 Chapter 7
45.
(b) When the third weight is hung from the spring the elongation of the spring is r - 30 mm -23mm:7.0mm. The weight is W : kr : (6.5 x 103 N/r")(7.0 x 10-3 m) : 45 N.73The elevator is moving upward with constant velocity, so the force F that is moving it must beequal in magnitude to the total weight of the elevator and load. That is, F _ Wrorur _ Mtotutg,where Mrorut is the total mass. The power required is P - Fu, where u is the speed of theelevator. Thus P - Mrorurgrr - (4500kg+ 1800kgX9.8^1s2X3.80 mls): 2.35 x 10sW.77(a)) Since the wind is steady the acceleration of the lunchbox is constant and x: - ust -r *.ot,where us is the velocity at time zero and a, is the acceleration. According to the graph thecoordinate is about 0.40m at time t-0.50s, so us- (0.40 m)l(0.50s) - 0.80m/s. The kinetic :energy at t -- 0 is Ko i*rfi(b) At t - 5.0 s the velocity is zero) so the kinetic energy is zero.(c) According to the work-kinetic energy theorem the work done by the wind force is the changein the kinetic energy, which is -0 .64 J. Chapter 7 4l
46.
Chapter 83(a) The folce of gravity is constant, so the work it does is given by W : F . i, where F is theforce and d is the displacement. The force is vertically downward and has magnitude mg, wherem is the mass of the flake, so this reduces to W _ mgh, where h is the height from which theflake falls. This is equal to the radius r of the bowl. Thus W : mgr - (2.00 x 10-t gX9.8 mlt1122.0 x r0-2m): 4.31 x 10-3 J.(b) The force of gravity is conservative, so the change in gravitational potential energy of theflake-Earth system is the negative of the work done: A(J: - -W : -4.31 x 10-3 J.(c) The potential energy when the flake is at the top is greater than when it is at the bottom bylAt/|. If U - 0 at the bottom, then [J: - +4.31 x 10-3 J at the top.(d) If Lr - 0 at the top, then U: - -4.31 x 10-3 J at the bottom.(e) All the answers are proportional to the mass of the flake. If the mass is doubled, all answersare doubled.-3The potential energy stored by the spring is given by (Ji - Ln*, where k is the spring constantand r is the displacement of the end of the spring from its position when the spring is inequilibrium. Thus 2[-I K- ., , 2(2s J) 8.9 x 103 N/*. ro (0.075 m)22(a) Neglect any work done by the force of friction and by air resistance. Then the only forcethat does work is the force of gravity, a consenative force. Let Ki be the kinetic energy ofthe truck at the bottom of the ramp and let K y be its kinetic energy at the top. Let [.Li be the gravrtational potential energy of the truck-Earth system when the truck is at the bottom and lett-I y be the gravitational potential energy when it is at the top. Then K 1 + U fpotential energy is taken to be zero when the truck is at the bottom, then LI I : ffigh, where h isthe final height of the truck above its initial position. Kithe truck, and Ky - 0 since the truck comes to rest. Thus mgh - *.*rt and h- u2 l2g.Substitute 1) - 130 korlh - 3 6.Im/s to obtain : (36 .I mls)2 : It, 66.5 m. 2(9.8 m/s2)If L is the length of the ramp, then Isinl5o :66.5m or L-(66.5m)lsin15" -257m.42 Chapter I
47.
The truck is not a particle-like object since its wheels turn and the cylinders of its motor move.However, if there is no frictional force between the tires and the roadway, these moving partshave no influence on the rate with which the truck slows. If there is friction, then when thedriver takes his foot off the gas pedal the tires exert a forward frictional force on the road andthe road exerts a backward frictional force of the same magnitude on the truck. This, along withair resistance, helps slow the truck. The frictional force is greater if the driver shifts to a lowergear.(b) The answers do not depend on the mass of the truck. They remain the same if the mass isreduced.(c) If the speed is decreased h and L both decrease. In fact, h is proportional to the square ofthe speed. If u is half its former value, then h is one-fourth its former value.11(a) The only force that does work as the flake falls is the force of gravity and it is a conservativeforce. If Ki is the kinetic energy of the flake at the edge of the bowl, K y is its kinetic energy atthe bottom, Ut is the gravitational potential energy of the flake-Earth system with the flake at thetop, and Uy is the gravitational potential energy with it at the bottom, then K1+Uf : Kt+U,i,.Take the potential energy to be zero at the bottom of the bowl. Then the potentral energy at thetop is Ut,: mgr, where r is the radius of the bowl and m is the mass of the flake. Ki - 0 sincethe flake starts from rest. Since the problem asks for the speed at the bottom, write *,*r forK y, The energy conservation equation becomes mgr - i*r, so ,tr: B - lrfr.s^ls2xo.zzom): 2.08m/s.(b) Note that the expression for the speed (u - M) does not contain the mass of the flake.The speed would be the same, 2.08 mls, regardless of the mass of the flake.(c) The final kinetic energy is given by Kf -- K +Ut-LIy. Since Ki is greater than before, K1is greater. This means the final speed of the flake is greater.15(a) Take the gravitational potenttal energy of the marble-Earth system to be zero at the positionof the marble when the spring is compressed. The gravitational potential energy when the marbleis at the top of its flight is then U s - mgh,, where h is the height of the highest point. This ish - 20 m. Thus (.Is - (5.0 x 10-t gX9.8 mlt11ZOm) : 0.98 J.(b) Before firing the marble is at rest and is again at rest at the top of its traJectory. Both the forceof the spring and the force of gravity, the only two forces acting, are conservative. Conservationof mechanical energy is expressed as LU n + At/" : 0, where [-I s is the gravitational potentialenergy and Lt" is the spring potential energy. This means L[]r- -LUn- -0.98J.(c) Take the spring potential energy to be zero when the spring has its equilibrium length. Thenits initial potential energy is (-1 - 0.98 J. This must be *.tt*, where k is the spring constant and " Chapter 8 43
48.
n is the initial compression. Solve for k: , K- 2(J" ., 2(0.e8 J) : 3.1 x L02 N/-. ro (0.080 m)231Information given in the second sentence allows us to compute the spring constant. Solve F - krfor k: F 27ON k -;- ffi- 1.35 x 101 N/-.(a) Now consider the block sliding down the incline. If it starts from rest at a height h abovethe point where it momentarily comes to rest, its initial kinetic energy is zero and the initialgravitational potential energy of the block-Earth system is rngh, where m is the mass of theblock. We have taken the zero of gravitational potenttal energy to be at the point where theblock comes to rest. We also take the initial potentral energy stored in the spring to be zero.Suppose the block compresses the spring a distance r before coming momentarily to rest. Thenthe final kinetic energy is zero, the final gravitational potential energy is zero, and the final springpotenttal energy is *tt*. The incline is frictionless and the normal force it exerts on the block Lrtt*, sodoes no work, so mechanical energy is conserved. This means mgh - , kr2 h- 2*g- (1 .35 x 104 N/-)(0.055 m)2 , I F -0.174m.If the block traveled down a length of incline equal to [., then (. srn3Oo(0.174m)f sin3Oo - 0.35m.(b) Just before it touches the spring it is 0.055 m away from the place where it comes to restand so is a vertical distance h - (0.055 m) sin3Oo - 0.0275m above its final position. Thegravitational potential energy is then mgh - ( r2kg)(9.8 mls2xo .0275 m) - 3.23 J .On the other hand, its initial potential energy is msh - ( l2kg)(9.8 mls2x0 .174m) : 20.5 J.The difference is its final kinetic energy: Ky :20.5 J - 3.23 J - 17.2 I. Its final speed is 2Kr 2(17 .2 J) m 12kg - 1.7 m/s .45(a) The force exerted by the rope is constant, so the work it does is Wforce and i ir the displacement. Thus W : Fdcos 0 - (7 .68 - 30. 1 J NX4.06 m) cos 15.0o .(b) The increase in thermal energy is LE6 - f d - (7 .42NX4.06 m) : 30. I J.44 Chapter B
49.
(c) We can use Newtons second law of motion to obtain the frictional and normal forces, thenuse Fttand the A axis nofinal to the floor. The r component of Newtons second law is Fcos 0 - f :0and the u component is F^r + F sin g mg _ 0, where m is the mass of the block, Fw is thenofinal force of the floor, F is the force exerted by the rope, and 0 is the angle between thatforce and the horizontal. The first equation gives f: F cos 0 - (7 .68 N) cos 15.0o - 7.42 Nand the second gives F,^/ - mg - F sin 0 - (3.57kgX9.8 mls2) - Q.68N)sin 15.0o -33.0N.Thus Fr,47(a) Take the initial glavitational potential energy to be LLi - 0. Then the final gravitationalpotential energy is U y - -mg L, where L is the length of the tree. The change is U y Ut, _-mgL - -(25kgX9 .8^ls2x I2m) - -2.9 x 103 J.(b) The kinetic energy is K - L*r: itz5kgX5.6 mls)2:3.9 x 102J.(c) The changes in the mechanical and thermal energies must sum to zero. Since the change inthermal energy is LErn: f L, where f is the magnitude of the average frictional force, L lZm69The change in the potential energy of the block-Earth system as the block goes from A to B isthe same for the two cases and, since mechanical energy is conserved, the change in the kineticenergy of the block is the same. The change in the kinetic energy is .1 LK:)*tL - ,2n) : ;*1Q.60 ^ls)2 - (2.00 mls)21 - (1.381lkg)m.For the second trial Ks : KA+ L,K - I*f+.00 mls)2 + (1.3s J/kg)m- (9.38 J lkg)m 2and the speed at B is 2Kslm 2(e.38 J lke) - 4.33 m/s .7SSince the blocks start from rest and the mechanical energy of the system consisting of the blocksand Earth is conserved, the final kinetic energy is the negative of the change in potential energy. Chapter 8 45
50.
If block B falls a distance d, block A moves a distance d, along the incline and rises a verticaldistance d srn 0 , where 0 is the angle of the incline. Thus K - -A[_I - - t?*egd,) + (mtgd sin?)] - gdlmn - mssin 0l - (9.8^1s2X0.25m)l2.0kg - (1.0kg)sin30"l - 3 .7 J.83(a) IJse conservation of mechanical energy. Let Ki be the initial kinetic energy, K y be the finalkinetic energy, and (. be the compression of the spring. Then the change in kinetic energy isLK : Kf - Ki. The block travels the distance d+1. along the incline and the vertical componentof its displacement has magnitude (d + [) srn?, where 0 is the angle of the incline. Thus thechange in the potential energy is LLI - mg(d,+ l)sin 0 + *ttl, where k is the spring constant.Since mechanical energy is conserved the final kinetic energy of the block is Kf : K,i, - mg(d,+ ,t) sinl - !,re 2 t6 J - (1 .0 kgX9.8 m/sX0.60 m + 0.20m) sin 40o ltroo N/*)(0.20 m)z - 7.0 J . 2(b) Now K y - 0 and K,; is the unknown. Conservation of mechanical energy gives Kr: mg(d,+/) sin 0 + )*n - (1.0 kgXg .8^1s2X0.60 m * 0.40 m) sin 40o .;Q00N/*)(0.40 m)2 : zzJ .87Neither the kinetic energy or the potentral energy changes, so conservation of energy tells us thatthe change in the total thermal energy is equal to the work done by the applied force: LE.;,,: W .If F is the magnitude of the applied force and d is the distance the cube travels, then W : F d.The thermal energies of the cube and the floor both change, so LEtn- L4rn,cube + A.Etrr, floorand L4tn,floor : Fd - L4*r,cube - (15NX3.0m) - 20J:25J .109(a) Take the potential energy of the ball-E arth system to be zero when the ball is at the bottomof its swing. Then the initial potenttal energy is 2mg L, where m is the mass of the ball andL is length of the rod. The initial kinetic energy is zero since the ball is at rest. Write *,*u,where u is the speed of the ball, for the final kinetic energy, at the bottom of the swing. Sincemechanical energy is conserved 2mg L : L*r and u:zEL - 2 (9.8 mls2xo .62m) : 4.9mls.46 Chapter 8
51.
(b) At the bottom of the swing the force of gravity is downward and the tension force of the rodis upward. If T is the magnitude of the tension force, Newtons second law is T -mg: m?rlL,SO T -mg*mulf -mg+4mg - 5mg - 5(0.0g}kgX9.8^lst) - 4.5N.(c) The diagram on the right is the free-body diagram for the ballwhen the tension force of the rod has the same magnitude as the forceof gravity. We wish to solve for 0. The component of the force of .- //gravity along the radial direction is mg cos 0 and is outward. Thenet inward force is T - mg cos0 and, according to Newtons secondlaw this must equal muT f L, where u is the speed of the ball. ThusT--muT lL*mgcosLWe now need to find the speed of the ball in terms of0. Take the potential energy to be zero when the rod ishorizontal. Since it starts from rest its kinetic energy is Lcos?also zeto. As can be seen on the diagram on the right,when the rod makes the angle 0 with the vertical, theball has dropped through a vertical distance L cos 0. Thepotential energy is then -mg L cos g. Write the kinetic energy as *.*r and the conservation ofenergy equation as 0 - -mgLcos 0+**u. Thus mu2 - 2mgLcos9. Substitute this expressioninto the equation developed above for T: T - Zmg cos 0 + mg cos0- 3mg cosd. According tothe condition of the problem, this must be equal to ffig, so 3 mg cos 0 - mg, or cos 0 - I 13. Thismeans 0 :7Io.(d) Notice that the mass of the ball cancels from the equation for cos 0, so 0 does not depend onthe mass. The answer to (c) remains the same.111(a) At the top of its flight the velocity of the ball has only a horizontal component and this isthe same as the horizontal component of the initial velocity. Let ug be the initial speed and 0obe the angle with which the ball is thrown. Then the kinetic energy at the top of the flight is K -i*t 6coS 0oi)2_ ]fto x 10-t g)t(s .0^ls)cos30"1: |.2J.(b) [Jse conservation of mechanical energy. When the ball goes from the window to the point3.0m below, the potential energy changes by LU- -mgd, where d (: 3.0m) is the distancefrom the window to the lower point. Since mechanical energy is conserved the change inthe kinetic energy is +mgd. If us is the initial speed and u is the speed at the lower point,I*u - /rt .0^ls)2 + 2(9.8^ls2x3.0m - t1m/s.(c) and (d) Notice that the mass cancels from the conservation of energy equation and none ofthe quantities in that equation depend on the initial angle, so the answer to part (b) does notdepend on the mass or the angle. Chapter I 47
52.
ttg(a) After the cue loses contact with the disk all the kinetic energy of the disk is converted tothermal energy. Thus the increase in thermal energy of the disk and court is LEn_ i*uito .42kgX4 .2mls)2 - 3.7 J.(b) The change in thermal energy is LEtn : f d, where f is the force of friction of the court onthe disk and d is the distance the disk travels. Thus f : L0tnf d - (3.7 J)lIZm) - 0.31N. Overthe entire l4m the increase in thermal energy is LErn - (0.31NX14N) - 4.3J.(c) A11 of the energy transferred from the cue to the disk ends up as thermal energy so the workdone by the cue is 4.3 J .t2t(a) and (b) The force is the negative of the slope of the curve. Take the potential energy to be-2.8J when the particle is at tr:1.0m and -17.5 J when the particle is at r:4.0m. Then ( -r7 - .3 J) (-2.9 J) F-- 40ffi --.4.8N.The magnitude is 4.8N and it is in the positive r direction.(c) and (d) When the particle is at r - 2.0m the potential energy is about tIthe kinetic energy is K - *.^r - *f2.0 kg)(- I .5 mf s)zE^"when the particle is at r- 1.5m and r- 13.5m. The particle moves between these twocoordinates.(e) When the particle is at renergy is K - E*"" (Ji - (-5.5 J) - (-17.5 J) : lzJ. The speed of the particle is 2(r2 I) 2.0 kg) - 3.5 m/s.I23(a) Let d, be the distance the ear travels. Its vertical position lowers by dstn?, where 0 is theangle of the incline, so the potentral energy changes by - Ltl - mgd sin 9, where m is themass of the car. The kinetic energy changes by LKspeed and u y is the final speed. The change in the mech aiical.tt.tgy is LE^"" : LK + A (J: --nlgd,sin 0*+*(r?-r?).Convert the given speeds to meters per second. They zte ui: 8.33 m/sand uy:11.1 mls, so AEn,, - -(1500 kgXg .8^ls2x50 m) sin 5.0o .:(1500 kg) [{t t. t ^ls)- (8 .33mlr)t] - -2.4 x 104J.The mechanical energy decreases by 2.4 x 104 J.48 Chapter 8
53.
(b) The change in mechanical energy is given by - f d, where f is the magnitude of the frictionalforce. Thus f : -LE^""f d- -(-2.4 x 104 Dl(50m): 4.7 x 102N.127(a) The potential energy does not change, so the change in the mechanical energy is equal to thechange in the kinetic energy. Let m be the mass of the block, u0 be its speed at the beginningof the acceleration period and u be its speed at the end. Then LE^r" - lm lr, - ,oliCtske) ltlo ^ls)2 - (t0mlr)] :6.0 x 103 J.(b) The ayeruge rate with which energy is transferred to the block is the total energy transferreddivided by the time for the transferraL since the acceleration a, is constant the time is givenby LtAE-. Lt - (6.0 x 103 Dl(10 s) : 6.0 x 102 w. "l(c) and (d) If the accelerating force has magnitude F, then the instantaneous rate of energytransfer is given by Pfor F. For u- 10m/s, P- (15kgX2.0^ls2x10m/s) - 3.0 x 102W and for u- 30m/s,P - (15 kgX2 .0^lrtx:0*ls) : 9.0 x 102 W131The kinetic energy gained per unit time is equal to the potentral energy lost per unit time. Ifmass L,m of water passes over the falls the gain in kinetic energy is LK - L,m gh, where h isthe height of the falls. The rate of production of electrical energy is D_3L^mgh 3, l4t:io200^lSX1000kl^X9.8^lS2X100m):8.8X108w.133(a) When the ball is at D the potenttal energy is mg L greater than when it is at A and the kineticenergy is **r3 less. Since mechanical energy is conserved, -+mu|+mgL - 0 andus- 1/ffi.(b) Let T be the tension in the rod when the ball is at B and Iet u be the speed of the ball then.Newtons second law gives T - mgu2 . When the ball is at B the potential energy is m,g L less than when it is at A and the kineticenergy is greater by **@ - r2i, so i*(r - ufi) - mgL - 0 and u2 - u20+2gL - 4gL, whereZgL was substituted for uzr. Thus T - mgtmulL - mg+4mg - 5mg.(c) When the ball is at C the potentral energy is the same as when it is at A and the kineticenergy is **r3 less. A11 of the kinetic energy is converted to thermal energy. The decrease inmechanical energy is i*r|: **zgL - mgL, where 2gL was substituted for u2o.(d) When the ball has settled at B the potential energy is mg L less than when it started at A andthe kinetic energy is i*r?, - mg L less. The mechanical energy has decreased by 2mg L. Chapter I 49
54.
Chapter 915You need to find the coordinates of the point where the shell explodes and the velocity of thefragment that does not fall straight down. These become the initial conditions for a projectilemotion problem to determine where it lands.Consider first the motion of the shell from firing to the time of the explosion. Place theorigin at the firing point, take the r axis to be horizontal, and take the A axis to be verticallyupward. The A component of the velocity is given by u - aya gt and this is zero at timet - uya I g - (ro I il sin gs, where us is the initial speed and 0s is the firing angle. The coordinatesof the highest point on the traJectory are (20 mls)2 tr : uort - uotcos gs -6 sin 9s cos 9s - n sin 60o cos 60o I 9.8 mlsand 1 A: ulat - 19, + 6 ,t2: ,i stn- 9s sin2 osSince no horrzontal forces act, the horizontal component of the velocity of the center of mass isconstant. At the highest point the velocity of the shell is ?re coS 06, in the positive n direction.This is the velocity of the center of mass. Let M be the mass of the shell and let Vo be thevelocity of the fragment that does not fall straight down. Then the velocity of the center of massis given by MVolzM :Vof 2, since the masses of the fragments are the same. Since the velocityof the center of mass is constant, ue cos 0s : Vof 2. This means Vo : 2u0 cos 0o : 2(20 mls) cos 60o : 20 m/s .Now consider a projectile launched horrzontally at time t - 0 with a speed of 20 m/s from thepoint with coordinates frs : I7.7 m, Ao : 15.3m. Its A coordinate is given by A- Uo trgt,andwhenit1andsthisisZero.Thetimeoflandingist:Mandthef,:coordinateofthelanding point is 2ao r-ro*Vot:fr1+Vo I - 17.7 m+ (20m/s) 53 m.23(a) Take the initial direction of motion to be positive and let J be the magnitude of the impulse,m be the mass of the ball, ui be the initial velocity of the ball, and u y be the final velocity ofthe ball. The impulse is in the negative r direction and the impulse-momentum theorem yields-J : TrLUf - TrLt)i. Solve for uy to obtain *rt* t (0.40kgX14m/s) - 32.4N s _67m/s u y- m - 0.40kg - .The final speed of the ball is 67 mf s.50 Chapter 9
55.
(b) The negative sign indicates that the direction of the velocity is opposite to the initial directionof travel. That is, it is in the negative r direction.(c) The magnitude of the average force is Fave- Jf Lt - (32.4N. s)l(27 x l0-3 s - I.2x 103N.(d) The impulse is in the negative r direction, the same as the force.35(a) Take the force to be in the positive direction, at least for earlier times. Then the impulse is Jr- ftoxlo-3 rf,-xL- ftoxto-3 t t -,,n .r061tz l cr.o x .t0e;t3^r ""^ - g.oN . s. 30x l0-3 - ro, , x ,.,,6. ,? f ,^, ^^ 1 Li,u.o loThe impulse is in the positive direction.(b) Since J - FavE Lt, where FavE is the average force and Lt is the duration of the kick, vg Fa,n: I ev Lt 3.oxlo-3s(c) To find time at which the maximum force occurs set the derivative of .F with respect to timeequal to zeto and solve for f. The result is t - 1.5 x 10-3 s. At that time the force is 4r,u,. - (6.0 x 106X1.5 x 10-) - (2.0 x 10eX1.5 x 10-3)2 : 4.5 x 103 N .(d) During the kick the ball gains momentum equal to the impulse. Since it starts from rest, itsmomentum just after the players foot loses contact is p - J . Let m be the mass of the ball andu be its speed as it leaves the foot. Then, since n : p lm, l): J _ 9.0N.s :20mlS. m 0.45 kg39No external forces with horizontal components act on the man-stone system and the verticalforces sum to zeto, so the total momentum of the system is conserved. Since the man andthe stone are initially at rest the total momentum is zero both before and after the stone iskicked. Let ms be the mass of the stone and l)s be its velocity after it is kicked; let mrn bethe mass of the man and I)Tn be his velocity after he kicks the stone. Then msl), * mrnlJnl: 0and u,rrlL)mman moves in the direction opposite to the direction of motion of the stone.47(a) Let m be the mass and uti be the velocity of the body before the explosion. Let TTL1, TrLz,and ff be the masses of the fragments. (The mass of the third fragment is 6.00 kg.) Write Chapter 9 51
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uti for the velocity of fragment l, -uzi for the velocity of fragment 2, and u3ri + q, j for thevelocity of fragment 3. Since the original body and two of the fragments all move in the rAplane the third fragment must also move in that plane. Conservation of linear momentum leadsto muni: nllrtj - TTL2uzi+ m3rhri+ m3qa j, or (mut* TTL2v2 - Tr?r3r) i - (mftr -t rfuu3a)i : 0.The r component of this equation gives ,u3r- Tftu,t* mzur: (20.0kgX200m/s) + (4.00kgX500m/s) _ 1.00 x 103 m/s. Tr 6.0 kgThe A component gives (100 kgX100 m/s) u3a : -TTLtur -- - -r67m/s Tr . 6.0 kgThus dz - (1.00 x 103 mls)i- (167mls)j. The velocity has amagnitude of 1.01 x 103 mls andis 9.48" below the r axis.(b) The initial kinetic energy is 111 KiThe final kinetic energy is 1.1.,1 Ky: ;*tr? +;rrL2u3+ r*tr3 :1 t1ro.0 kg)(100 mls)2 + (4.00 kgXsoo m/s)2 + (6.00 kextor a^^)t] 2L - 3.63 x 106J.The energy released in the explosion is 3.63 x 106 J - 4.00 x 10s J - 3.23 x 105 J.6l(a) Letmr be the mass of the cartthat is originallymoving, uybe its velocitybefore the collision,and urf be its velocity after the collision. Let Tft2 be the mass of the cart that is originally at restand uzf be its velocity after the collision. Then, according to Eq. 9-67, urf-- n" ,*utt. " mt t TTLzSolve for rft2 to obtain Tn2: - ttf TTLI - / l 2^ls - 016t"A) Lt; (0340kg) (0.340 kg) - 0099 Ikg 0.099 u, r ur f ( / .(b) The velocity of the second caft is given by Eq . 9 -68: uzr: Lrrn: t kg + 0.099 kgll (l .zmls):1.9 mls. arrL2 Tft1 10.34052 Chapter 9
57.
(c) The speed of the center of mass is : Trlluu * Tft2uzt (0.340 kgX I.zm/s) ucom ffi - 0 30kgi.0 Jrrftg- _ 0 .93 m/ s -n t .Values for the initial velocities were used but the same result is obtained if values for the finalvelocities are used. The acceleration of the center of mass is zero.63(a) Let rTrl be the mass of the body that is originally moving, uy be its velocity before thecollisior, and urf be its velocity after the collisioll. Let Tft2 be the mass of the body that isoriginally at rest and uzf be its velocity after the collisioll. Then, according to Eq. 9-67, utr - - TTLI TfL2 uyi, " mt*mZ -Solve for TtL2 to obtain uu - utf lll2: lll,1 ut f + LtlliSubstitute urf : ?,)rnl4 to obtain rrL2 - 3*tl5 - 3(2.0k9 15: I.2kg.(b) The speed of the center of mass is TTt tuYi * mzuz, (2.0 kgX4.0 m/s) ucom - TTII + m2 - 2.0 kg + I.2kg - 2.5 m/s.77(a) The thrust of the rocket is given by T - Rur"r, where R is the rate of fuel consumption andurcr is the speed of the exhaust gas relative to the rocket. For this problem Rurct- 3.27 x 103 mls, so T - (480kelsx3.27 x 103 mls): I.57 x 106N.(b) The mass of fuel ejected is given by Mna: RLt, where Lt is the time interval of thebum. Thus Mna: (480 kels)(250 s)My: Mt - M6et: 2.55 x 105kg - 1.20 x 105kg - 1.35 x l05kg.(c) Since the initial speed is zero, the final speed is given by Eq.9-88: u1:Urcl"ffi-(3.27X103mls)/0nffi-2.08X103m/s.79(a) Take the r axis to be positive to the right in Fig. 9-72 of the text and take the A axis tobe perpendicular to that direction. Consider first the slow barge and suppose the mass of coalshoveled in time Lt is LM. If ?r-" is the velocity of the barge and 0 is the velocity of the coalas it leaves the barge, then the change in the momentum of the coal-barge system during thisinterval is Af - A M(t - u-"). The momentum of the coal changed from 6rLM to i tlt andthe momentum of the barge did not change. The force that must be exerted on the barge to keep Chapter 9 53
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its velocity constant is F,- Lf ILt- (LM ILD(0 d"). Now i"shoveledpetpendicularlytothelengthoftheboatthen 0 - uri*Urj. ThusF, -(LMlLt)Uoj.Ua is the slight transverse speed the coal must be given to get it from one barge to the other.It is not given in the problem statement, so we assume it is so small it may be neglected. Theforce that must be applied to the slower barge is essentially zero.Now consider the faster barge, which receives coal with mass LM. Initially the coal has velocityt but after it comes to rest relative to the barge its velocity is d y , the same as the velocity ofthe barge. The momentum of the coal changes from LMi to LMdy and the momentum of thebarge does not change. The force that must be applied to the barge is Ff - (LMlLt)(df t).Now dy: ufi and 0 - uri*Urj, so the r component of the force is Ff*The rate with which coal is shoveled is converted from kg/min to kg/s in the first factor and thebarge speeds are converted from km/h to m/s by the last factor.(b) The A component of the force that is applied to the faster barge is Ftr: -(LMlLt)fLa. IfUa is small, Ff , is essentially zero.9t(a) If m is the mass of a pellet and u is its velocity as it hits the wall, then its momentum isp - n Lu - (2.0 x 10- t gX500 mls) : 1.0 kg . ml s, toward the wall.(b) The kinetic energy of a pellet is K - *.*r: *Q.O x 10-3 kgX500m/s)2 :2.5 x 102 J.(c) The force on the wall is given by the rate at which momentum is transferred from thepellets to the wall. Since the pellets do not reboutrd, each pellet that hits transfers momentumptransferred is p Altr tEavg (1.0kg . mlsxl0 s-t) - 10N. A -The force on the wall is in the direction of the initial velocity of the pellets.(d) If Af is the time interval for a pellet to be brought to rest by the wall, then the average forceexerted on the wall by a pellet is lokgm/s :r.7 Fn,*: P - vs a, 0.6 x x 103N. 1o-3 sThe force is in the direction of the initial velocity of the pellet.(e) In part (d) the force is averaged over the time a pellet is in contact with the wall, while inpart (c) it is averaged over the time for many pellets to hit the wall. Most of this time no pelletis in contact with the wall, so the average force in part (c) is much less than the average forcein (d).93(a) The initial momentum of the car is Fn: Tft6t - (1400kgX5 .3^ls)j - (7400 kg .mfs) j and thefinal momentum is Ft - (7400kg.mls)i. The impulse on it equals the change in its momentum:J- - Ff - Fr, - (7400 kg . mlsxi - i).54 Chapter 9
59.
(b) The initial momentum of the car it Fn - (7400kg .mfs)i and the final momentum rs Ft:0.The impulse acting on it is i - Ft - Ft,(c) The average force on the car is Fuue -AF_T A, A 4.6 s (1600NXi - j)and its magnitude is Favs- (1600N/t - 2300N.(d) The average force is Favg (-2.1 x lo4Di Lt 350 x 10-3 sand its magnitude is FavE : 2.1 x 104 N.(e) The average force is given above in unit vector notation. Its r and A components have equalmagnitudes. The r component is positive and the A component is negative, so the force is 45"below the positive r axis.97Let TTL p be the mass of the freight car and u p be its initial velocity. Let rrls be the mass of thecaboose and u be the common final velocity of the two when they are coupled. Conservation of thetotal momentum of the two-car system leads to mp.u7r: (mpimg)u, so u : ?rFTftpl(*r+mc).The initial kinetic energy of the system is Ki: !*rr, "2Jand the final kinetic energy is 1 ffi2pr2p Ky ^?.-rT (*e+mc)2 2 (me+mc)Since 27% of the original kinetic energy is lost K y - 0.73K,i,, or I (0.73) (:?rLp,") , (me + mc)Following some obvious cancellations this becomes rftpl(^e + mc)ms - (0.27 10.73)me - 0.37,mp - (0.37)(3. 18 x 104kg) - 1.18 x 104kg.101(a) Let ut,i be the speed of ball 1 before the collisions and ur f be its speed afterwards. Let uz.fbe the speed of ball 2 after the collision. Let m be the mass of each ball. Then conservationof momentum leads to the r component equation rrLU1,;- mutylcos?l * muzl cos02 and the Acomponent equation 0 _ -mury sin0l * muzy srn02. The masses cancel from these equations.The r component equation gives urf cosgl - uti-uzf cos02 and the y component equation gives Chapter 9 55
60.
ur sin dl fobtain u?r - (uzf sin 0)2 + (urr - uzf cos 0r)- [(1.1^ls)sin60o]+12.2mls - (1.1 mls)cos60"lt - 3 .62m2 lr .The speed is utf: @: r.gmls.(b) Divide ur1 sin01 - u2f stn02by ,ry cos91 :t)ti,- uzf cos02 and use tan01- (sin 0)l@os01)to obtain uzf srn?z tan01: - - 0 577 urt - uzf cos 02 Q.zmls) - (1.1 mls) cos 60The angle is 0r - 30o.(c) The initial kinetic energy is L*r?,i : )rn(z.2mf s)2 - 2.42m, itt joules tf m is in kilograms.The final kinetic energy is i*r? + Lrrr - L*(l .g mls)2 + i*(I .I mf s)2 - 2.4m, in joules tf m fis in kilograms. The collision is elastic (at least to the number of significant digits given in theproblem).107(a) The acceleration of the center of mass of the two-particle system is the net external force onparticles of the system divided by the total mass of the system: --+ CLcom _ 1", - [(-4.00N)i + (5.00Nj] + [(2.00N)i + (-4.00$j] Tft1 1 m2 2.00 x 10-kg + 4.00 x 10-3 kg - (-3 .33 x r02 mlrt) i + ( r.67 x r02 mlr) j .Since the acceleration is constant and the center of mass is initially at rest, the displacementduring the interval is A4o-: |a1tt1: )X-3.33 x t02mlrt)i+( r.67 x 102 mlrt)jl(2.oo x t0-r)t - ( -6.67 x 10-4 m) i + (3 .33 x 10-4 m)j .The magnitude of the displacement is lal (-6.67 x 10-4 m)2 + (3.33 x 10-4 m)2 - 7.45 x 10-4m.(b) If 0 is the angle made by the displacement and the positive r direction then tan? - (3.33 xl0-4 $l(-6.67 x l0-4m) - -4.99 and 0- -26.5o or 153o. Since the displacement has anegative r component and a positive A component the correct answer is 153o.(c) The velocity of the center of mass is u-.o*:d"o Lt- t(-3.33 x r02mlr)i+(r.67 x r02mlst)il(2.O0 x 10-s) - (-0 .667 mls)i + (0 .333^ls)j56 Chapter 9
61.
and the kinetic energy of the center of mass is l(,o*: lr*1 * rrl2)u?o :ir*1 * mz)@.o **r?o^r) - )rr.00 x 10-3 km + 4.00 x 10-t g)t(0 .667 mls)2 + (0.3 33mls)21 . .67 x 10-3 J .113Let M be the mass of the sled and us be its initial speed. Let M* be the mass of water scooped "up and u y be the final speed of the sled and the water it contains. Before the water is scoopedup the momentum of the sled-water system is Mruo and afterwards it is (Mr*M-)rf. The finalspeed is u7: M"uo Qgookgx25g*/9 lgom/s. A/Ir+M-: , -You should recognuze this as a completely inelastic collision between the sled and the water.115(a) Put the origin at the center of Earth. Then the distance rcom of the center of mass of ther,arth-Moon system is given by TTL11qr11t1 Tcom: mM+mE,where TTL 11,1 is the mass of the Moon , TTL E is the rnass of Earth, and r y is their separation. Thesevalues are given in Appendix C. The numerical result is (7 .36 x 1022 kgX3 .82 x 108 m) rcom: -4.64 x 106m.(b) The radius of Earth is Rn -- 6.37 x 106m, so r"o*l Rn: 0.73 and the distance from thecenter of earth to the center of mass of the earth-Moon system is 73% of Earths radius.tt7(a) The thrust is T _ Rurer, where R is the mass rate of fuel consumption and nrct is thespeed of the fuel relative to the rocket. This should be equal to the gravitational force M g,where M is the mass of the rocket (including fuel). Thus Rurer - M g and R(6100 kgx9.8 m/ s l(1200 mls) : 50 kgls.(b) Now Rur"t - M g - M a, where a, is the acceleration. This means M(g + a) (6100 kgX9.8 m/s2 + 21 m/s2) R_ - t.6x nrer - 1200 m/s ..2kg/s .129Write Eq. 9-68 in the form uzf - 2mrtnl(m+ M), where m is the mass of the incident objectand M is the mass of the target. Solve for M: uzf)- (3.0kgX2(8.0m/s) (6.0m/s)l _ M _m(Zuyi - - 5.0kg. uzf 6.0 mls Chapter 9 57
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Chapter L013Take the time t to be zero at the start of the interval. Then at the end of the interval t- 4.0 s,and the angle of rotation is 0 - clot + |at2. Solve for aoi et_ *t, 0 tzoraa - *e.0radls2)(4.0 s)2Now use u) : ws * at to find the time when the wheel is at rest (ar - 0): -L _ tis 24 radf s a 3.0 ,uW: -8os That is, the wheel started from rest 8.0 s before the start of the 4.0 s interval.2t(a) Use I rev - 2r rad and I min - 60 s to obtain u):ffi 200 rev (200 rev)(2n radfrev)(b) The speed of a point on the rim is given by u- u)r, where r is the radius of the flywheeland w must be in radians per second. Thus t): (20.9rad/sX0.60m: I2.5m/s.(c) If a is the angular velocity at time t, us is the angular velocity at t: 0, and o is the angularacceleration, then since the angular acceleration is constant a : Loy * at and (1000rev/min) -(200rev/min) e.:a - ao : - g00 revf minz t 1.0 min(d) The flywheel turns through the angle 0, which is 0 - aat.:a* - (200rev fmin)(1.0 min) .;(800 revf mi,rX1.0 min)2 : 600 rev.29(a) Earth makes one rotation per day and 1d is (24hx3600s/h) - 8.64 x 104 s, so the angularspeed of Earth is (2nrad)l(8.64 x 104 s) : 7.3 x 10-5 rad/s.(b) Use u - ar, where r is the radius of its orbit. A point on Earth at a latitude of 40o goesaround a circle of radius r: Rcos 40", where R is the radius of Earth (6.37 x 106m). Its speedis rtr - a(Rcos 40") : (7.27 x 10-s radfsx6.37 x 106 m)(cos40o) : 3.6 x 102 m/s.(c) At the equator (and all other points on Earth) the value of r,,, is the same (7 .3 x 10-5 rad/s).58 chapter I0
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(d) The latitude is 0o and the speed is u - wR - (7.3 x 10-s rudfsx6.37 x 106 *) - 4.6x102 m/s.33The kinetic energy is given by K : itr, where I is the rotational inertia and w is the angularvelocity. [Jse (602revf min)(Zrradf rcv) u): :63.0rad f s. 60 s/minThen r-#: ffi:r23ke m235Use the parallel axis theorem: J 1ron+Mh2, where /.o* is the rotational inertia about aparallel -axis through the center of mass, M is the mass, and h is the distance between the two axes. In thiscase the axis throughthe centerof mass is atthe 0.50mmark, so h- 0.50m-0.20m-0.30m.Now /.o,n: iuOSO r- 4.67 x LT-z kg.m2 +(0.56kgX0.30 m)2:9.7 x 10-kg.mZ37Since the rotational inerlia of a cylinder of mass M and radius R is Ienergy of a cylinder when it rotates with angular vel ocrty w is K -)tr:Iu^2o2.(a) For the first cylinder K - lf t .z5kgX0 .25 m)2e35 radf s)2- 1.1 x 103 J . 4,(b) For the second K - ltt .zskgX0 .7 5 m)2e35 rud,f s)2 : 4, 9.7 x 103 J .4l[Jse the parallel-axis theorem. According to Table l0-2, the rotational inertia of a uniform slababout an axis through the center and pe{pendicular to the large faces is given by lv[ I, : rom + b2) O(o . Chapter I0 59
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A parallel axis through a corner is a distance h - (o l2)2 + 2) from the center, so (b l2)2 J -/ro* + Mh2 + b2) + + b21 - +@+b2) +(o, J 0.r72kg - 3 [(0.035 m)2 + (0.084 *)] : 4.7 x l0-4 kg . mZ45Two forces act on the ball, the force of the rod and the forceof gravity. No torque about the pivot point is associated withthe force of the rod since that force is along the line fromthe pivot point to the ball. As can be seen from ttre diagram,the component of the force of gravity that is pe{pendicular tothe rod is rng sin 0, so if (. is the length of the rod then thetorque associated with this force has magnitude r : rngt sin 0 -(0.75kgX9 .8^lr2x 1.25m) sin 30o - 4.6N . ffi. For the positionof the ball shown the torque is counterclockwise.47Take a torque that tends to cause a counterclockwise rotation from rest to be positive and atorque that tends to cause a clockwise rotation from rest to be negative. Thus a positive torqueof magnitude rrFrsin 91 is associated with Ft and a negative torque of magnitude rzFzsin d2 isassociated with F2. Both of these are about O. The net torque about O is T : rtFt sin 91 - r2F2stn02 - (l .30 m)(a .20 N) sin 7 5.0" - (2. 15 mX4.90 N) sin 60.0o - -3.85 N ln .49(a) Use the kinematic equation a - clo t at, where es is the initial angular velocity, a is thefinal angular velocity, a is the angular acceleration, and t is the time. This gives :a-eo: 6.20radfs t 220 x 10-3 s - z8.2radlrt.(b) If I is the rotational inertia of the diver, then according to Newtons second law for rotation,themagnitudeofthetorqueactingonheris r: Ia- (L2.0kg.m2)(28.2radls2) - 3.38x102N.m.63Let (. be the length of the stick. Since its center of mass is (, 12 from either end, its initialpotential energy is L*gL where m is its mass, and its initial kinetic energy is zero. Its finalpotential energy is zero and its final kinetic energy is it r, where I is its rotational inertiafor rotation about an axis through one end and a is its angular velocity just before it hits thefloor.Conservationofenergyyie1ds}*gt-it,,)ofu-_/@.Thefreeendofthestick60 Chapter I0
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is a distance (. from the rotation axis, so its speed as it hits the floor is u msl3 I I.According to Table I0-2, f - **t, So u - Fnt. - 3(9 .8^1s2Xtr.00 m) : 5.42m/s .69(a) Choose clockwise rotation of the pulley to be positive and take its angular position 0 to bezero at time t - 0. Then the angular position at time t is 0 - **t, where a is its angularacceleration. Thus (x:20 l* - 2(1 .30rad) lQl .0 x 10-r) - 3 .14 x I02 radf s2.(b) The string does not slip on the pulley, so the acceleration of either block is a,(0.0240 m)(3 .14 x 102 rudls) - 7 .54^lt .(c) The forces on the hanging block are the tension force of the string and the gravitationalforce of Earth. Newtons second law for this block gives mg - T1(6.20kex9 .8mf s2 - 7.S4mlst) - r4.0N.(d) The net torque on the pulley is r(Tt Tz), so r(Tr T) - I a, where I is the rotationalinertia of the pulley. Thus Ia Tz: Tt r 0.024m79Use conservation of energy. Take the potential energy to be zero when the rod is horizontal.If L is the length of the rod, the center of mass of the rod is initially a distance (I sin q 12above the pin and the initial potential energy is LL : mgl(sin 0) 12, where m is the mass of therod. The initial kinetic energy is zero since the rod starts from rest. The final kinetic energy isrotational and is given by K I - Lt r?, where I is the rotational inertia of the rod and a y is itsangular speed as it passes the horizontal. The conservation law gives mg(L sin 0) 12 - Lt r?, sout.s ms(L sin 0) I I .We now need the rotational inertia for rotation about the pin. According to Table l0-2 therotational inertia of the rod about an axis through the center of mass is /ro-The parallel-axis theorem tells us that the rotational inertia for rotation about the pin is I -Oll})mLz +mQl2) - (Il3)*L2. The angular speed as the rod passes the horizontal is 3ms(I sin d) 39 srn? 3(9 .8^lr; ritr 40" - 3.1rad f s.87Take the positive direction to be toward the right for the block and take clockwise to be thepositive direction of rotation for the wheel. Let T be the tension force of the cord. The horizontalcomponent of Newtons second law for the block gives P - T Chapter I0 6l
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the block and a is its acceleration. The torque on the wheel is Tr, where r is the radius of thewheel, so Newtons second law for rotation gives Tr: Ia, where I is the rotational inertia ofthe wheel and a is its angular acceleration. Since the cord does not slip on the wheel , e,: re,.When this substitution is made for a, is the equation for the block, the result is P - T - Tnre,?so T - P mra. IJse this to substitute for T in the equation for the wheel. The result isPr - mr2a: Ia and the solution for a is Pr (L: *r- r r (3.0NX0.20m) J.vr/v.z.vLLL) A / 1t 2 mrT + I (2.0 kgXO .20 m)2 + 0.050 kg . m2 - 4.6radf S"89(a) For constant angular acceleration a : u)0+ at, so a - (a - ,il|t. Take w :0 and to obtainthe units requested use t - (30 s)/(60 s/min) : 0.50 min. Then e, : - 33 33 rev/min - _-66.7 rev mrn2 f . 0.50 minThe negative sign indicates that the direction of the angular acceleration is opposite that of theangular velocity.(b) The angle through which the turntable turns is 0 - aot.:a* - (3 3.33revfmin)(0.50min) + )e66.7 revfmin2x0.50min)29l(a) According to Table 1,0-2, the rotational inertia of a uniform solid cylinder about its centralaxis is given by Ic: *tW R, where M is its mass and R is its radius. For a hoop with mass Mand radius Rn Table I0-2 gives 111 - M RL for the rotational inertia. If the two bodies havethe same mass, then they will have the same rotational inertia if R2 f 2 - Rr, or Rn - Rf ,n.(b) You want the rotational inertia to be given by I- M k2 , where M ts the mass of the arbitrarybodyan.dkistheradiusoftheequivalenthoop.Thusk-ffi.115(a) The kinetic energy of the box is given by K6: *.*r, where nL is its mass andu is its speed.Thus the speed of the box is u- M and, since the cord does not slip on the wheel, theangular speed of the wheel is cr, :ulr: W. The rotational kinetic energy of the wheelis K-: )rr: :loJ. :t#: #:(b) IJse conservation of energy. As the box falls a distance h the potential energy of the box-wheel-Earth-mount system changes by LU - -mgh and the change in the kinetic energy isLK -6.0J+ 10J- 16J. Since LK + Lt-I-0, h- LKlmg0.27 m.62 Chapter I0
67.
Chapter LL-Tifr. work required to stop the hoop is the negative of the initial kinetic energy of the hoop.The initial kinetic energy is given by K - )trz + i*r, where I is its rotational inertia, mis its mass, a is its angular speed about its center of mass, and u is the speed of its center ofmass. The rotational inertia of the hoop is given by I - mRz, where R is its radius. Since thehoop rolls without sliding the angular speed and the speed of the center of mass are related bya-ulR,Thus 1 / u2 + I K -;*R (trr) ;*r2 - ml,2 - (r40keX0.150m/s)2 - 3. 15 J. LThe work required is W : -3. 15 J.t7(a) An expression for the acceleration is derived in the text and appears as Eq. I 1 - 13: 0com 1 * /,o,, lMRS)where M is the mass of the yo-yo, 1ro* is its rotational inertia about the center, and Ro is theradius of its axle. The upward direction is taken to be positive. Substitute Iro - 950 g . cmz,M - 120 g, fto : 0.3 2 cm, and g -- 980 cmf s2 to obtain acom: :13cm lr.(b) Solve the kinematic equation Ucom: i.o"o t2 for t and substitute !/com: I20 cm: 2(120 cm) t- -4.4s. 13 cm lr(c) As it reaches the end of the string its linear speed is ucom55 cm/s.(d) The translational kinetic energy is K - **r?r*: }fo.I20kgX0.55 mls)z- 1.8 x 10-2J.(e) The angular speed is given by u) : u"o l Ro and the rotational kinetic energy is K - *.Iro*r*I"o u?r*l Ra(0 The angular speed is u) - uro lRo - (0.55 mls)|Q.2 x 10-*) - 1.7 x 102rcdls - 27rcvf s.23(a) Let F - F*i* Frj and i - ri+ aj.Then i - ix It - (ri+Ail x (F"i+ Frj) - (*F, - -/A.rnA UF*)k. Chapter I I 63
68.
The last result can be obtained by multiplying out the quantities in parentheses and using i * i : t,Jxl--l (, i x i : 0, and j x j - 0. Numerically,/AnnAA F - t(3.0m)(6.0N) - (4.0mX-8.0N)l t - (s0N . m)k. li * f1(b) Use the definition of the vector product:T.andFwhentheyatedrawnwiththeirtailsattheSamepoint.NowT:ffi: (3.0 m)z + (4.0 m)2 - 5.0m and F - (-8.0N)2 + (6.0N)2rF 50 N . m, the same as the magnitude of the vector product. This meanssinQ: l and d:90o29(a) [Jse f - mrx 6, where r- is the position vector of the object, u- is its velocity vector, and mis its mass. The position and velocity vectors have nonvanishing r and z components, so theyarc written r_ ri+ rk and 6 - u*it urk. Evaluate the vector product term by term, makingsure to keep the order of the factors intact: - ("i + zk) x (u*i + u"t ; - *r*i x i + ru"i x tt + zu*kx i + zu"kx t r x,i .Now use i x i:0,i xt- -j,kxi:+j,and kxk- 0 to obtain Fxd-(-rr"+zu*)jThus | - m(-nuz + zu)j - (0.2s ke) l-fzo mXs.o mls) + (-2.0 mX-s.o m/s)] j-0.(b) [Jse F - i x F,with Ii - Fj. f - @?+ zk) x (Fi) :trFixj+zFkx i- rPk- zFi - (2.0mX4.0N)i - (-2.0mX4.0N)k - (8.0N .m)i + (8.0N .m)t.33(a) The angular momentum is given by the vector product ivector of the particle and t7 is its velocity. Since the position and velocity vectors are in the nAplane we may write v-_ ri+ yj and 6 - u*i* rri.Thus r- xd- (ri+ yj) x (r*i+ urj) - *r*i x i+ ru, i x j + au*j x ?+ yurj x jIJseixi- 0,?xj-k,i x i - -k, and j x j - 0 to obtain A/ rx6-(rua-UU*)k.64 Chapter I l
69.
Thus :;flr, ll?il, --60^rs)-(8mXsm/s)l t- etTx r02r,g m2rlr)t(b) The torque is given by i - ix f . Since the force has only an r component we may writeF - F*i and i - (*i+ail x (F*i; - -uF*t- -(8.0mX-7.0N)t- (56N.m)t.(c) According to Newtons second law for rotation, f - a[1at, so the time rate of change of theangular momentum is 56 kg . nf lt, in the positive z direction.37(a) Sineer: dLldt,the average torque acting during anyinterval is givenby rav- (Lf -Lt)lLt,where Li is the initial angular momentum, L y is the final angular momentum, and At is thetime interval. Thus : 0.800 kg . m2 ls - 3.00 kg . # It N .m. ,avg 1.50 s - -1 .47In this case the negative sign simply indicates that the direction of the torque is opposite thedirection of the initial angular momentum, which is taken to be positive.(b) The angle turned is 0 : ast * i*t. If the angular acceleration a is uniform, then so is thetorqueand e,:rlI. Furtherrnore, es:Ltf I, so 0_Lit+Lrtz I 0.140kg.m2(c) The work done on the wheel is W : r0 : (-l .47 N . m)(20.3 rad) : -29.8 J.(d) The average power is the work done by the flywheel (the negative of the work done on theflywheel) divided by the time interval: Pon:-Y avg Lt Lt 1.50s43(a) No external torques act on the system consisting of the man, bricks, and platforrll, so thetotal angular momentum of that system is conserved. Let Ii be the initial rotational inertia of thesystem and let I y be the final rotational inertia . If a,; is the initial angular velocity and w y is thefinal angular velocity, then INt: Ifaf and 60 kg m2 f ! ui: ( u) r- r;) u);- ,v/s) - 3.6 rev/s [z.Okg ,re )(rztevfs): 36rev/s . Chapter I I 65
70.
(b) The initial kinetic energy is Ki: LInr?, the final kinetic energy is Ky: ittrf , and theirratio is Kf t4: (2.okg-mz)(3.6rcvf s)2- K,;- Itw? (6.0 kg . m2X1.Zrev f s)2 3.0 .(c) The man did work in decreasing the rotational inertia by pulling the bricks closer to his body.This energy came from the mans store of internal energy.45(a) No external torques act on the system consisting of the two wheels, so its total angularmomentum is conserved. Let 11 be the rotational inertia of the wheel that is originally spinningand 12 be the rotational inertia of the wheel that is initially at rest. If w,i is the initial angularvelocity of the first wheel and w I is the common final angular velocity of each wheel, thenItwt: (1r + Iz)of and I1 uY - IJ IrQi Substitute 12:2Ir and a,; - 800rev f mrnto obtain w.S :267 rev/min.(b) The initial kinetic energy is Ki: LItr? and the final kinetic energy is Ky: i(tt + I2)wzy.The fraction lost is AK Kt-Kf fru?-(t,+tr)a?- wzn - 3? Ki - Ki - Ip? wl 0.667 (800 rev f mrn)z49No external torques act on the system consisting of the train and wheel. The total angularmomentum of the system is initially zero and remains zero. Let I (- M R) be the rotationalinertia of the wheel. Its final angular momentum is L*- Iw: MRzw, where M is the massof the wheel. The speed of the track is w R and the speed of the train is a R - u. The angularmomentum of the tratn is Lt - m(a R - u)R, where m is its mass. The direction of rotation ofthe track is taken to be positive. If the train is moving slowly relative to the track, its velocityand angular momentum are positive; if it is moving fast its velocity and angular momentum arenegative. Conservation of angular momentum yields 0 - M R2w + m(wR a)R. When thisequation is solved for u, the result is muR MU lttt + *1r*: $f + dRSubstitute M - I.Lm, R - 0.43 m, and n : 0. 15 m/s to obtain u): (1.1mt m)(0.43 m) -0.l|radfs.66 Chapter I I
71.
67(a) If we consider a short time interval from just before the wad hits to just after it hits and sticks,we may use the principle of conservation of angular momentum. The initial angular momentumis the angular momentum of the falling putty wad. The wad initially moves along a line that isd,lZ distant from the axis of rotation, where d is the length of the rod. The angular momentum ofthe wad is mudf 2. After the wad sticks, the rod has angular velocrty w and angular momentumIw, where I is the rotational inertia of the system consisting of the rod with the two balls andthe wad at its end. Conservation of angular momentum yields mudf 2- Iw. If M is the massof one of the balls, J - (2M +m)(d,12). Wher:nLud,lz - (2M +m)(dl2)zw is solved for w, theresult is Zmu u)-W:iU 2(0.0500 kg)(3.00 m/s) o,iimt -ol48tadfs(b) The initial kinetic energy is Kiratio is KylKu- IwZ l*r. When I -lZtW+m)d,l4 and u,):2mulQM+m)d, ut. substituted,this becomes K1 0.0s00 kga _ n n l^t - U.VLLJ K,i 2M * m 2(2.00 kg) + 0.0500 kg(c) As the rod rotates the sum of the kinetic and potential energies of the Earth-rod-wad systemis conserved. If one of the balls is lowered a distance h, the other is raised the same distanceand the sum of the potential energies of the balls does not change. We need consider only thepotential energy of the putty wad. It moves through a 90o arc to reach the lowest point on itspath, gaining kinetic energy and losing gravitational potential energy as it goes. It then swingsup through an angle 0, losing kinetic energy and gaining potentral energy, until it momentarilycomes to rest. Take the lowest point on the path to be the zero of potential energy. It starts adistance d,lz above this point, so its initial potential energy is U,i: mgd,lz. If it swings throughthe angle 0, measured from its lowest point, then its final position is (d l2)(I - cos 9) above thelowest point and its final potential energy is LI 1 - mg(dlz)(I - cos0). The initial kinetic energyis the sum of the kinetic energies of the balls and wad: KiAt its final position the rod is instantaneously stopped, so the final kinetic energy is K y : 0.Conservation of energy yields mgd,l2+ *fZtW + m)(d,12)2w2: rng(d,|z)(I - cos0). When thisequation is solved for cos 0, the result is cos o - -1 2 (ry)G)u2 (0 1*"radtil,:-o0226 2 [ ](ry)The result for 0 is 9l .3o. The total angle of the swing is 90" +91 .3o Chapter I I 67
72.
73(a) and (b) The dtagram on the right shows the parti-cles and their lines of motion. The origin is markedO and may be anywhere. The angular momentum of tparticle t has magnrtude (.r: mur1 sin 01 : mu(d+h) d,and it is into the page. The angular momentum of par-ticle 2 has magnitude (.2 +is out of the page. The net angular momentum has hmagnitude L t t0-4 kgX5 .46mlsx0 .0420m) : 6.65x t0-s kg .m2 lrand is into the page. This result is independent of thelocation of the origin.(c) and (d) Suppose particle 2 is traveling to the right. Then L - mu(d,+h)+muh - mu(d+2h).This result depends on h, the distance from the origin to one of the lines of motion. If the originis midway between the lines of motion, then h - -d,12 and L:0.77Use conservation of energy. If the wheel moves a distance d along the incline its center of massdrops a vertical distance h - d stn 0, where 0 is the angle of the incline. The potential energyof the Earth-wheel system changes by LLI : -mgh - -mgd sin 0, where m is the rnass of thewheel. The change in the kinetic energy is L^K : lm,u!"m + )to, where ucom is the final speedof the center of mass, u is the final angular speed of the wheel, and I is the rotational inertiaof the wheel. Since the wheel rolls without sliding ucom : u)r, where r is the radius of the axle.Thus A,K : Lr(mr2 + /). Since energy is conserved mgd,sin 0 - io(*r2 + /) and 2 2mgd,sin 0 2(10.0 kgXg .8^ls1(2.00 m) sin 30o u-: : : m,r2 + I (l0.0kg)(0.200 )2 +(0.600kg .m2)(a) The rotational kinetic energy is Krot: itr:ltO.600kg. m2X196 ndz lt):58.8J.(b) The square of the speed of the center of mass is ,?o - a2r2: (196 rud2 lrtx0.200m)2 -7.84m2 lr and the translational kinetic energy is Krrunr- **r.r^: lCf 0.0kgX7.84 m2 lr)39.2 J.85(a) In terms of the radius of gyration k the rotational inertia of the meffy-go-round is I- M k2and its value is (180kgX0.910*)t - l49kg . m2.(b) Recall that an object moving along a straight line has angular momentum about any pointthat is not on the line. Its magnitude is mud, where m rs the mass of the object, u is the speed ofthe obj ect, and d is the distance from the origin to the line of motion. In particular, the angularmomentum of the child about the center of the meffy-go-round is L"- mu&, where R is theradius of the meffy-go-round. Its value is (44.0kg)(3.00 mls)(l .20m): 158kg m2 ls.68 Chapter I I
73.
(c) No external torques act on the system consisting of the child and the merry-go-round, sothe total angular momentum of the system is conserved. The initial angular momentum is givenby muR; the final angular momentum is given by (/ + mRz)u, where a is the final commonangular velocity of the merry-go-round and child. Thus mu R - (/ + mR2)a and muR : w I+mRz a- 158 kg . m2 ls s r4gkg. m2+(44.0kg)(r.zlm)z -0744radf87The car is moving along the n axis, going in the negative r direction. Let r- be the vector fromthe reference point to the particle and u* be the velocity of the car. Then the angular momentumof the car is given by the vector product mi x 6, which is -mAu, t since d has only an rcomponent and, in all cases, r- has only r and A components. According to Newtons secondlaw in angular form the torque is the rate of change of the angular momentum.(a) and (b) The reference point is at the origin, so A - 0 and f - 0. The angular momentum isconstant so the torque is also zero.(c) The reference point is a distance lil - 5.0m from the r axis so the magnitude of the angularmomentum is ( - (3.0kgX2 .0^ls4)f3(5.0m: (30kg. m2 Ito)tt. Since y is positive the angularmomentum is in the negative z direction. Thus f - -(30 kg . m2 f sa1t3 k.(d) The torque is .m,lsa)t3 t<l - #L-(30kg -(90kg . nl f sa1t2k.(e) and (f) The reference point is the same distance from the r axis as in parts (c) and (d), so the the angular momentum and torque are the same. Now, howeveq A is negatiy., soTagnitudes ofI ad, i are in the positive z direction. Thus [- +(30kg .m2 |to)tt k and i - +(90kg. m2 f satzk.95Two horizontal forces act on the cylinder: the applied force Fupp in the positive r directionand the frictional force / along the r axis. Newtons second law for the center of mass isFapp + f * : mc;catrt, nt where nL is the mass of the cylinder and acom n is the r component of theacceleration of its center of mass. 4oo acts the top of the cylinder and i acts at the bottom,both at the rim, so the magnitude of the net torque on the cylinder is R(F - f and Newtons ")second law for rotation is R(F f ") - f a, where I is the rotational inertia of the cylinderand a is its angular acceleration. Since the cylinder rolls without sliding the acceleration of thecenter of mass and the angular acceleration ate related by ocomsimultaneously for ocom, e,) and f*. The solutions are 2R2 Fapp Ocom : mR2+I) 2RFupp Q.: mR2+I) Chapter II 69
74.
and r 2mR2 Fupp Jn mRz+I -r n According to Table l0-2 the rotational inertia of the cylinder for rotation about its central axisis given by I : i^R. Substitution of this expression leads to 4Fapp 4(12N) 1r- rZ acom: r.om/s G- 3(lokg) - ) 4(12N) e,:4Fapp: I o raol s 3mR 3(10 kg)(O. 1o m)-1r--^sr-Z )and Furo- f ,- 33 + - 4.0N.Sjnce f" is positive, the frictional force is in the positive r direction and can be writtenf: (4.0 N) i.70 Chapter I l
75.
Chapter L2-5Three forces act on the sphere: the tension force f of the rope(which is along the rope), the force of the wall Fr (which is hori-zontally away from the wall), and the force of gravity mj (which isdownward). Since the sphere is in equilibrium they sum to zero. Let0 be the angle between the rope and the vertical. Then the verticalcomponent of Newtons second is 7 cos 0 - mg - 0. The hoizontal component is trl/ - T sin 0 :0. (a) Solve the first equation for T: T - mg f cos 0. Substitute cos 0 -Lltffi to obtain +r2 2y l rng (0.85 kgX9.8 m/s T L 0.080 m - 9.4N.(b) Solve the second equation for Fli: F1r - T srn?. Use sin0,l/t2+12to obtain F1u. -TrL2+12 L NTT+F L 0.080 m 4.4N.7The board is in equilibrium, so the sum of the forces and the sum of the torques on it are eachzero. Place the r axis along the diving board. Take the upward direction to be positive. Takethe vertical component of the force of the left pedestal to be F1 and suppose this pedestal is atn - 0. Take the vertical component of the force of the right pedestal to be F2 and suppose thispedestal is at r: d. Let W be the weight of the diver, applied at tr: L. Set the expression forthe sum of the forces equal to zero: F1* Fz W :0.Set the expression for the torque about the right pedestal equal to zero: &d+w(L - d) :0.(a) and (b) The second equation gives F1: -+-: (H) (s8oN): -t2x ro3NThe result is negative, indicating that this force is downward. Chapter 12 7l
76.
(c) and (d) The first equation gives F2: W - il -580N+ 1.2 x 103N- 1.8 x 103N.The result is positive, indicating that this force is upward.(e) and (0 The force of the diving board on the left pedestal is upward (opposite to the forceof the pedestal on the diving board), so this pedestal is being stretched. The force of the divingboard on the right pedestal is downward, so this pedestal is being compressed.11Place the r axis along the meter stick, with the originat the zero position on the scale. The forces on it areshown on the diagram to the right. The coins are atn : nt (: 0.120 m) and m is their total mass. The knifeedge is at rmass of the meter stick is M and the force of gravityacts at the center of the stick, fr : n3 (: 0.500m).Since the meter stick is in equilibrium the sum of thetorques about rz must vanish: M g(U - r) - mg(*z - nr) - 0. Thus, Ms M-fr2-frt m:( - 0.455 m 0.120m (10.0 g) : 74.4 g. fr3-fr2 0.500m-0.455m2tConsider the wheel as it leaves the lower floor. There is nolonger a force of the floor on the wheel , and the only forceson it are the force F applied horizontally at the axle, the forceof gravity mg vertically downward at the center of the wheel,and the force of the step corner, shown as the two componentsfn and fr. If the minimum force is applied the wheel does notaccelerate, so both the total force and the total torque on it arezeto.Calculate the torque around the step corner. Look at the seconddiagram to see that the distance from the line of F to thecorner is r h, where r is the radius of the wheel and h isthe height of the step. The distance from the line of mg to rf othe corner is S r2+? -hY - ffi. Thus F(r h)-mgffi - o. The solution for F is 2rh - h2 F: mg r-h 2(0.0600 mX0.0300 m) - (0.0300 m)2.. 0.800kgX9 .8*ls) - 13.6N 0.0600m - 0.0300m72 Chapter I2
77.
33(a) Examine the box when it is about to tip. Since it willrotate about the lower right edge, that is where the normalforce of the floor is applied. This force is labeled Fl,. onthe diagram to the right. The force of friction is denotedby f , the applied force by F, and the force of gravity byW. Note that the force of gravity is applied at the centerof the box. When the minimum force is applied the boxdoes not accelerate, so the sum of the horizontal forcecomponents vanishes: F-f:0,the sum of the vertical force components vanishes: Fl,. -W:0,and the sum of the torques vanishes: FL-W-0. 2Here L is the length of a side of the box and the origin was chosen to be at the lower right edge.Solve the torque equation for F: p 89oN : -W - 22 445N.(b) The coefficient of static friction must be large enough that the box does not slip. The box is on :the verge of slipping tf lt, - f I F*. According to the equations of equilibrium Fn W - 890 NIand f : It - 445 N, so F, - (445 N) l$90 N) - 0.50.(c) The box can be rolled with a smaller applied force ifthe force points upward as well as to the right. Let 0 bethe angle the force makes with the horizontal. The torqueequation then becomes F Lcos 0 + F Lsin 0 -W L 12 : 0,with the solution W It_ 2(cos 0 + sin0)You want cos0+ sin0 to have the largest possible value. This occurs if 0_ 45o, a result youcan prove by setting the derivative of cos 0+ sin0 equal to zero and solvingfor 0. The minimumforce needed is F- W 89ON 4 cos 45o 4 cos 45" - 315 N.43(a) The shear stress is given by F lA, where F is the magnitude of the force applie d parullel toone face of the aluminum rod and A is the cross-sectional area of the rod. In this case F is the Chapter 12 73
78.
weight of the object hung on the end: F - ffig, where m is the mass of the object. If r is theradius of the rod then A - nr2. Thus the shear stress is F mg (1200 kgX9.8 m/s2) . :6.5 x ro6N/*t. i- #-(b) The shear modulus G is given by FIA G- LrlL)where L is the protrusion of the rod and A,r is its vertical deflection at its end. Thus (6.5 x 106N/m2X0.053m)-1 L^r:(Fl!)L_: 1 x 10-5m. G 3.0 x 1010 N/m2f-55(a) The forces acting on the bucket are the force of gravity, down, and the tension force of cableA, up. Since the bucket is in equilibrium and its weight is Wg8.01 x 103N, the tension force of cable A is Tt:8.01 x 103N.(b) [Jse the coordinate axes defined in the diagram. Cable A makes an angle of 66" with thenegative A axis, cable B makes an angle of 27" with the positive U axis, and cable C is alongthe r axis. The A components of the forces must sum to zero since the knot is in equilibrium.This means Ts cos 27" - Ttcos 66" - 0 and rs:ffiffi rt: (#) (8 or x lo3N)- 3 6sx ro3r{(c) The r components must also sum to zero. This means Tg *Tn sin27" -Tasin 66o - 0 andTs:TAsin 66" - T6sin 27" - (8.01 x 103N)sin 66" - (3.65 x 103N)sin 27o - 5.66 x 103N.61(a) The volume of the slab is (43mxl2m)(2.5m)- 1.29 x 103m3 and, since the density is3.2 x 103 kgl*t, its mass is m 4.13 x 106 kg. Thecomponent of the gravitational force parallel to the bedrock surface is mg sin d106 kgXg .8^lrt; ritr 26"(b) The maximum possible force of static friction is f ^u*- FFw, where F, is the coefficientof static friction and Fli is the norrnal force of the bedrock surface on the slab. Newtonssecond law (with the acceleration equal to zerc) gives the norrnal force as mgcosO, so f^* -F,mg cos 0 - (0.3 9)(4. 13 x 106 kgX9 .8 ro s26" ^lrt;(c) The bolts must support a total shearing force of 1 .77 x I07 N - I.42 x lo7 N - 3.5 x 106N.Each bolt ean support a shearing force of (3.6 x 108 N/*t )((6.4 x 10-4 m2 :2.3 x 105 N so the74 Chapter I2
79.
number of bolts required is (3.5 x 106N/(2.3 x 10sN): 15.2. Round up to the nearest integer:16 bolts are required.63Let Ts be the tension in the horizontal cord, f be the magnitude of the frictional force on blockA, .Fl/ be the magnitude of the normal force on that block, and ?rL4 be the mass of that block.Assume block A is stationary. Then Newtons second law for block A gives Ta f - 0 andFl/ - m.,q,g ltrFx, whete F, is the coefficient of static friction between the block and the table surface. Thismeans Tn must be less than Frmeg.Now consider block B. Let Ts be the tension in the cord attached to it and M6 be its mass.Newtons second gives T6 - wL6g:0, so Ts: TrLp!.Next consider the knot where the three cords join and let T be the tension in the third cord.Newtons second law gives 7 sin 0 - TaTequation gives Ta -T sin 0 - ffLpgsin 0f cos0 - TTLsgtan?.If block B does not slip Tn pgtan? must be less than FTTTLA7. Since TTL7 is the greatest it can bewithout block A slippitrg, TtLs tan 0 - Fsmag or v |f," ITLy 10 kg65The force diagram for the rod is shown on the right. T is the tensionforce of the rope and Fn is the force of the hinge. The angle u is1 80" 0r 0z: I 80" 60o - 0z: I20" 02. The net torque about thehinge is TL sina - mg(Llz)sin01 and this must be zero if the rod is tobe in equilibrium. Thus srnu-ms(L rL sin 91 ms(L 12) sin :. 12) _ 91 sin 01 )since T1 - mglz. This means a: 0t and 02: I20o - 60" : 60o81 DThe force diagram for the cube is shown on the right. Fr is the T -,noffnal force of the floor on the cube, ,f ir the force of friction I I rh,lof the flooq and m is the mass of the cube. Assume the cubeis stationary but it is about to tip. Only the lower right edgeof the cube exerts a force on the floor and the line of action I I I tl tt I I f, -t <- ___lof the normal force is through the right side of the cube. Thehorizontal somponent of Newtons second law for the center of !*g o Chapter 12 75
80.
mass is P fP(. - mgL12, where (. is the distance between O and the point of application of P.(a) The cube slides if f is greater than FFw, where F, is the coefficient of static frictionbetween the floor and the cube. According to the Newtons second law equations f : P and,Fl,. : nLg. Thus sliding occurs if P >for the cube to slide but not tip as P increases Fsrng must be less than mgLl2l. ot Lr, must beless than Ll2(, - (8.0 cm) 12(7.0 cm) - 0.57.(b) The cube tips before it slides if P(. >greater than mgLl2(. ot Lr, must be greater than Llz(. - 0.57.85The force diagram for the ladder is shown on the right. Fs is the forceof the ground on the ladder, F* is the force of the wall, and W is theweight of the ladder. The horizontal component of Newtons second lawgives F + Fn* - F*- 0 and the vertical component gives Fno W - 0. FnSet the net torque about the point where the ladder touches the wall equalto zero. The honzontal component of the force of the ground has a lever arrn of h, the vertical component of the force of the ground has a leveraffn of (., where (. is the distance from the foot of the ladder to the wall,the applied force has a lever arrn of (1 dlL)h, where L is the length of the ladder, and thegravitational force has a lever ann of (. 12. Thus Fn*h - Fnrl+ Fh(l - d,lL) + WtlT - 0. Ther axis was taken to be horizontal with the positive r direction to the right and the A axis wastaken to be vertical with the positive direction upward.The vertical component of the second law gives F ga - W : 200 N. The torque equation gives Fn*Now (. _ J(1om)2 - (8.0m)2 : 6.0m, so ((. lh)Fno - (6.0 m)l(8.0m)(200N) : 150N, Id,lL - 1 (2.0m)110m)-0.80, and (dlzh)W_ (6.0m)(200N/2(8"0m) - 75N. This means Dtgn 150N-(0.80)F-75N: 75N-(0.80),F.(a) If .F - 50N, Fn*:75N - (0.80)(50N) - 35N and Fn - (35N)i+1ZO0Ni.(b) If F:150N, Fn*:75N- (0.80)(150N;- -45N and Fn - (-45N)i+(200Dj.(c) When the ladder is on the verge of slipping the frictional force is to the left, in the negativer direction. Its magnitude is (0.80)F 75N. If the ladder does not slip this must be lessthan prFga: (0.38X200N - 76N. The applied force must be less than (75N+ 76N) 1Q.80)1.9 x 102 N. Thus the applied force that will just start the ladder moving is I.9 x 102 NI.76 Chapter 12
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Chapter L31The magnitude of the force of one particle on the other is given by It - Gmflt Lz I ,, where TrLland Tfi,2 are the masses, T is their separation, and G is the universal gravitational constant. Solvefor r: GmtffLz ^ 1 rc.67 x 0- 1r N . m2 lke(s .2ke)e.4 kg) 1 F7At the point where the forces balance GM"*lr?- GMr*|r3, where M" is the mass of Earth,M, is the mass of the Sun, m is the mass of the space probe, rr is the distance from the centerof Earth to the probe, and 12 is the distance from the center of the Sun to the probe. Substitute12 - d - rt, where d is the distance from the center of Earth to the center of the Sun, to find M"- M" ,2, (d - ,t)Take the positive square root of both sides, then solve for 11 . A little algebra yields _ dM (lso x toem) T1 M+tM - I .99 x 1030 kg + 5.98 x 1024 kg - 2.60 x 108 m.Values for A[., Mr, and d can be found in Appendix C.17The gravitational acceleration is given by as: GMf r2, where M is the mass of Earth and ris the distance from Earths center. Substitute r- R+ h, where R is the radius of Earth and his the altitude, to obtain as: GMI(ft+ h). Solve for h. You should get h-According to Appendix C of the text, R- 6.37 x 106m and M - 5.98 x 1024 kg, so @-R. (6.67 x 10-ll m3 lr . kgX5.98 x L024 kg) h: -6.37 x 106m- 2.6 x 106m.29(a) The density of a uniform sphere is given by p 3M l4nR3, where : M is its mass and R isits radius. The ratio of the density of Mars to the density of Earth is 3 j A 5 I i 0, k* Pnr 0 .6 5 I g+ km pn Mp Rtu X - 0.74. Chapter 13 77
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(b) The value of an at the surface of a planet is given by ag: GMIR, so the value for Mars is 2 Mxa lVL R1 r tF. 0 .6 5 r I g+ km A7I/I- ,r- t a (9.8 mlr2) - 3.8 ^lt WHagE:011 M 1 75 X 0, k*(c) If u is the escape speed, then, for a particle of mass m **r: GryandFor Mars u-VFu.67 x 10- , F ,1 :5.0x10m/s.37(a) Use the principle of conservation of energy. Initially the particle is at the surface of theasteroid and has potentral energy [-Li: -GMmlR, where M is the mass of the asteroid, R isits radius, and m is the mass of the particle being fired upward. The initial kinetic energy isi*r. The parttcle just escapes if its kinetic energy is zeto when it is infinitely far from theasteroid. The final potential and kinetic energies are both zero. Consenration of energy yields-GMmlR+ i*r:0. Replace GMIR with agR, where a,e is the gravitational accelerationat the surface. Then the energy equation becomes -a,sB+ *r- 0. Solve for u: 2anR 2(3.0 m/s ;1soo x 103m):1.7 x 103m1 s(b) Initially the particle is at the surface; the potential energy is [Ii: -GMrnlR and the kineticenergy is Kicomes to rest. The final potential energy is tJy: -GMml(R+ h) and the final kinetic energyis K,; - 0. Conservation of energy yields GMm I GMm -;+;*,2: R+hReplace GM with onR2 and cancel m in the energy equation to obtain onR2 -asR+)*: (R+ h)The solution for h is 2anR2 h- -R 2onR - u2 z(g.o m/s2)(soo x to3 m)2 - 2(3.0^lrt;1soo x (500 x 103 m) 103 m) - (1000 mls)2 :2.5 x 105 m.78 Chapter 13
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(c) Initially the particle is a distance h above the surface and is at rest. The potenttal energy isU, - -GMml@+ h) and the initial kinetic energy is K,; - 0. Just before it hits the asteroidthe potentral energy is U y -GMmlR.Write *.*rr for the final kinetic energy. Conservationof energy yields GMm GMm 1. + - mL)" R+h R 2Replace GM with onR2 and cancel m to obtain onR R+h esR*lrThe solution for u is 2anR2 u- 2(3.0^lr;1soo x 103 m)2 2(3.0^lrt;1soo x 103 m) - 500 x 103m+ 1000 x 103m -- 1..4 x 103 m/s .39(a) The momentum of the two-star system is conserved, and since the stars have the same mass,their speeds and kinetic energies are the same. use the principle of conservation of energy. Theinitial potential energy is Ut,- -GMlrn where M is the mass of either star and ri is theirinitial center-to-center separation. The initial kinetic energy is zero since the stars are at rest.The final potential energy is U1: -zGMlrn since the final separation is ,n12. Write Muz forthe final kinetic energy of the system. This is the sum of two terms, each of which is *M r .Consenration of energy yields _GMz __ _2GIVI2 + Muz Ti T6The solution for u is (6.67 x 10-rr m3 lr.kgX103o kg) - 8.2 x 104 m/s . 10lo m(b) Now the final separation of the centers is ry - 2R: 2 x 105 m, where R is the radiusof either of the stars. The final potentral energy is given by LI 1 - -GMlrf and the energyequation becomes -GMlrn: -GMlrt + Mu2. The solution for u is tlrt x 10-1rm3 lr keXl030ke) - (^iffi mt*) /,u.67 -1.8 x 107m/s. Chapter 13 79
84.
45Let l/ be the number of stars in the galaxy, M be the mass of the Sun, and r be the radius of thegalaxy. The total mass in the galaxy is IY M and the magnitude of the gravitational force actingon the Sun is F - G M lr. The force points toward the galactrc center. The magnitude ofthe Suns acceleration is a: 12 lR, where u is its speed. If T is the period of the Suns motionaround the galacttc center then u- 2nRlT and a, -- 4n2RlT. Newtons second law yields GIY MI R : 4r2 M RlT2. The solution for tf is 4r2 R3 iv- GTzMThe period is 2.5 x 10t y, which is 7.88 x 101t r, so 4n2(22 x 7020 m)3 : rv - (6.67 5.1 x 1010 )(7.88 x I01t r)t (2.0 x 1030 kg)4t(a) The greatest distance between the satellite and E,arths center (the apogee distance) is Ro:6.37 x 106m+360 x 103 m - 6.73 x 106m. The least distance (perigee distance) is Rp - 6.37 x106m+180x 103m - 6.55 x 106m. Here 6.37 x 106m is the radius of Earth. Look at Fig. 13-13to see thatthe semimajoraxis is o : (Ro+R|)f 2 - (6.73x 106m+6.55x 106 lrri;12 - 6.64x 106m.(b) The apogee and perigee distances ate related to the eccentricity e by Ro - o( I + e) andRpRo - Rp - 2ae. Thus e: Ro- Rp Ro- Rn 2a Ro * Rp 6,73x 106m- 6.55 x 106m 6.73 x 106 m * 6.55 x 106 m -0.0136.6t(a) IJse the law of periods: T2 : (4nIGM)r3, where M is the mass of the Sun (1.99 x 1030kg)and r is the radius of the orbit. The radius of the orbit is twice the radius of Earths orbit:r - 2r. - 2(1 50 x 10em):300 x 10em. Thus 4r213 T_ 4n2(300 x 10e m)3 (6.67 x 10- 11 m3 I , . kgX r .gg x 1030 kg)Divide by (365 dlfiQ{hldX60 minlhx60 s/min) to obtain T - 2.8y.(b) The kinetic energy of any asteroid or planet in a circular orbit of radius r is given byK - G M m l2r, where m is the mass of the asteroid or planet. Notice that it is proportional tom and inversely proportional to r. The ratio of the kinetic energy of the asteroid to the kinetic80 Chapter 13
85.
energy of Earth is K I K. - (m lnLe)(r.l i. Substitute m - 2.0 x lT-am" and r :2r" to obtainKIK.7s(a) Keplers law of periods gives 4r213 4n2(4.20 x 107 m)3 -2.15 x 104s. (6.67 x 10-11N .nP lkeX9.50 x l02s kg)(b) The craft goes a distance2nr tnaperiod, so its speed is us :2r(4.20x 107 m)l(2.L5x 104 s):1.23 x 104 mls.(c) The new speed is ,u:0.98u0 - (0.98XI.23 x 104 mls): l.2I x 104 mls.(d) The kinetic energy of the crafi is K - **r: *tz000kgXr.2 x 104 mls)z -2.20 x 1011 J.(e) The gravitational potential energy of the planet-craft system is u--GW r _-_ t^_o/ 4.20X107mwhere the potential energy was taken to be zero when the craft is far from the planet.(0 The mechanical energy of the planet-craft system is E - K+U - 2.20x 1011 J -4.53 x 1911 1--2.33 x 10ll J.(g) The mechanical energy of a satellite is given by E : -GmM l2a, where a is the semimajoraxis. Thus GmM (6.67 x l0-11N . m2 lke(3000kgX9.50 x t02s kg) o"- 2E 2(-2.33 x 10rr J) - 4.08 x I07 m.(h) and (i) The new period is 4r2 a3 4n2(4.08 x 107 m)3 x 104s. (6.67 x 10-11 N .m2 lkgX9.50 x l02s kg) -2.06Thechangeintheperiodis 2.06 x 104s- 2.I5 x 104s- -9 x 103s. Theperiodforthesecondorbit is smaller by 9 x 102(4.20 x 107 m)3 s.79Use 7;t - GmrTTlrnlr, where ms is the mass of the satellite, mrn is the mass of the meteor, andr is the distance between their centers. The distance between centers is r: R+d - 15m+3m -18 m. Here R is the radius of the satellite and d is the distance from its surface to the center ofthe meteor. Thus ,b- 6.67 x 10-rr N _ 7;1 .m2 lke)Q0kgX7.0 kg): x l0-ll N. Z.9 ^ Chapter 13 8l
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83(a) The centripetal acceleration of either star is given by o, - u2r, where w is the angular speedand r is the radius of the orbit. Since the distance between the stars is 2r the gravitational forceof one on the other is Gmz IQD, where m is the mass of either star. Newtons second lawgives Gmz IQD : murLr. Thus I Gm I x 1o-rrN .mlkgl3o. lorkgf a- 2 ,t Tt -;l ^Fu, -22 x 10-7rcdfs(b) As the meteoroid goes from the center of the two-star system to far away the kinetic energychanges by LK- -+mu2 and the potentral energy changes by Ltf - ZGmMlr, where M isthe mass of the meteoroid and u is its speed when it is at the center of the two-star system. Sinceenergy is conserved LK + A(J : 0 and 4(6.67 x 10-rr N .m2 lke(3.0 x 1030kg) : 8.9 x 104 m/s. 1.0 x 10ll m87(a) Since energy is conserved it is the same throughout the motion and there is no variation.(b) The potential energy at the closest distance (perihelion) is (Je: -GMnMs :-(6 (5.98 x lo24 kgX I .99 x 1030 kg) .67 x 1o-11 N .m2 lkg) Tp 1.47 x 101r m - -5 .40 x 1033 Jand at the furthest distance (aphelion) is (5.98 x T024 kgX | .99 x 1030 kg) IJo ra L.52 x 10rr mThe difference is 1 .8 x 1032 J.(c) Since energy is conserved the vanation in the kinetic energy must be the same as the variationin the potenttal energy, 1.8 x 1032 J.(d) The semimajor axis is a: (ro{ra)12 - (1.47 x 108km+1 .495 x 108 k^)12 - 1.50 x 108km.The kinetic energy at perihelion is 1l Ke: GMnMs _l [; 2")Now x 10ll m - 3 .46 x 10-12 m-1so Ke - (6.67 x 1o-rr N .m2 lkg)(s.99 x 1024 kgX r .gg x 1030 kgX3 .46 x l0- m- ) 12 I - 2.74 x 1033 Jand the speed is u,p 2KIME 2(2.7 4 x 1033 J)l (5.98 x 1024 kg) - 3 .02 x 104 m/s.82 Chapter 13
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Since angular momentum is conserved uprp: nara and the speed at aphelion is t)a: uprolro(3.A2 x 104 mlsxl .47 x 10rr rri;l!.sz x 10rt *) - 2.93 x 104m/s. The varration is 3.02 x104 mls - 2.93 x 104 mls - 9.0 x I02 m/s.93Each star is a distance r from the central star and a distance 2r from the other orbiting star, soit is attracted toward the center of its orbit with a force of magnitude rt-GM: 12 *ry (2r)2 4r2 r-rAccording to Newtons second law this must equal the product of the mass and centripetalacceleration ,2 I,. Each star travels a distance 2rr in a time equal to the period ?, so ?r :2rrf T,and the centripetal acceleration is 4n2 r lT2. ThusThe solution for ? is 4rr3 /2 Chapter 13 83
88.
Chapter 141The air inside pushes outward with a force given by piA, where Pi is the pressure inside theroom and A is the area of the window. Similarly, the air on the outside pushes inward witha force given by PoA, where Po is the pressure outside. The magnitude of the net force isF = (Pi - Po)A. Since 1 atm = 1.013 x 105 Pa, F = (1.0 atm - 0.96 atm)(1.013 x 105 Pa/atm)(3.4 m)(2.1 m) = 2.9 x 104 N .J.The change in the pressure is the force applied by the nurse divided by the cross-sectional areaof the syringe: F F 42N 5 D.p = A = 7fR2 = 7f(1.1 X 1O- 2 m)2 = 1.1 x 10 Pa.11The pressure P at the depth d of the hatch cover is Po + pgd, where p is the density of oceanwater and Po is atmospheric pressure. The downward force of the water on the hatch coveris (Po + pgd)A, where A is the area of the cover. If the air in the submarine is at atmo-spheric pressure then it exerts an upward force of PoA. The minimum force that must beapplied by the crew to open the cover has magnitude F = (Po + pgd)A - PoA = pgdA =(l024kg/m3 )(9.8 m/s 2)(100m)(1.2 m)(0.60m) = 7.2 x 105 N.19When the levels are the same the height of the liquid is h = (hI + h2)/2, where hl and h2 arethe original heights. Suppose hI is greater than h2. The final situation can then be achieved bytaking liquid with volume A(h l - h) and mass pA(h l - h), in the first vessel, and lowering it adistance h - h2. The work done by the force of gravity is W = pA(h l - h)g(h - h2) Substituteh = (hI + h2)/2 to obtain W = ~pgA(hl - h2)2 = ~(1.30 X 103 kg/m3 )(9.8 m/s2)(4.00 x 10- 4 m2)(1.56 m - 0.854 m)2 = 0.635 J.27(a) Use the expression for the variation of pressure with height in an incompressible fluid: P2 =PI - pg(Y2 - YI). Take YI to be at the surface of Earth, where the pressure is PI = 1.01 X 105 Pa,84 Chapter 14
89.
and Uz to be at the top of the atmosphere, where the pressure ts pz: 0. Take the density to be1.3 kg/*. Then, v Uz-At :79 x 103m- 7.gkm. p9 (1.3 kel^Xq.gr;7s)(b) Let h be the height of the atmosphere. Since the density varies with altitude, you must usethe integral Pz: Pt Ir^ ps da .Take p: po(l -Alh), where po is the density at Earths surface. This expression predicts thatp:poatA-0 and p: 0 atA-h. Assume g isuniformfrom A-0 to A-h. Nowtheintegralcan be evaluated: : pr - : pr lr^ Poe(r I) Pz da * pogh .Since p2 : 0, this means 2pr 2(1.01 x 105 Pa) ar.vr r-r r*/ h_Lyr: I = - 16 X 103m:16km. ^v Pog (1.3 kgl^Xq.8 m/s2)31(a) The anchor is completely submerged. It appears to be lighter than its actual weight becausethe water is pushing up on it with a buoyant force of p*gV , where p* is the density of waterand V is the volume of the anchor. Its effective weight (in water) is Weff: W - p*gV, whereW is its actual weight (the force of gravity). Thus v - w"rc- 2o=oN -w P*9 (gg8 kg/mXq.8 m/s2) - 2.045 x lo-2 m3 = .The density of water was obtained from Table I4-I of the text.(b) The mass of the anchor is m - pV, where p is the density of iron. Its weight in arr isW : mg: pgv - (7870kelmXg .8^ls(2.045 x r0-2m3) - 1.58 x 103 N.35(a) Let V be the volume of the block. Then, the submerged volume is Vr:2Vf3. Accordingto Archimedes principle the weight of the displaced water is equal to the weight of the block,so p-V, : pbV , where pu is the density of water, and pa is the density of the block. SubstituteV"- 2Vl3 to obtain pa- 2p*13- 2(gg8kg/mt)lZ: 6.7 x }zkgl^t. The density of waterwas obtained from Table l4-l of the text.(b) If po is the density of the oil, then Archimedes principle yields poV, - paV. SubstituteV, - 0.90V to obtain po: pbf 0.90: 7.4 x I02kgl^t.37(a) The force of gravity mg is balanced by the buoyant force of the liquid pgV": mgHere rra is the mass of the sphere, p is the density of the liquid, and V, is the submerged volume. Chapter 14 85
90.
Thus msphere, or V" 4nl lJ?, : 6 Ol- ro (?) (8oo kel^xo oeom)3 - r zkgAir in the hollow sphere, if ufry, has been neglected.(b) The density prn of the materral, assumed to be uniform, is given by p,* : m lV , where m isthe mass of the sphere and V is its volume. If ri is the inner radius, the volume is 4r V- -r?)- T [(0.090 m)3 - (0.080 m)3] - 9.09 x 10-a m3The density is m L.22kg x r v: n- 9.09 x lo-a m3 1.3 103 kgl^t49Use the equation of continuity. Let u1 be the speed of the water in the hose and u2 be its speedas it leaves one of the holes . Let A1 be the cross-sectional area of the hose. If there are l/ holesyou may think of the water in the hose as l/ tubes of flow, each of which goes through a singlehole. The cross-sectional area of each tube of flow is At lN . If A2 is the area of a hole theequation of continuity becomes urAtlN: uzAz. Thus u2: (AtllVAr)ur - (R2 llYr)ut, whereR is the radius of the hose and r is the radius of a hole. Thus u2: fi (0.9s cm)2 Tfur,s3Suppose that a mass A^m of water is pumped in time Lt The pump increases the potentialenergy of the water by A^mgh, where h is the vertical distance through which it is lifted,and increases its kinetic energy by i/r*u2, where u is its final speed. The work it does isLW : L*gh+ ).Lmuz and its power is r - Lw - Lnz ( on* 1.,t) P N Ar "," )Now the rate of mass flow is L* lAf - pAu, where p is the density of water and A is theatea of the hose. The area of the hose is ApAuobtained from Table I4-I of the text. Thus P pAu I gh+ / 1rt) - 2/ " l-. (50 m/s)2 - (1.s 7 kgls) I (9 .B^1s2X3.0 m) * /Jt LLL) I L 2 I :66W.86 Chapter I4
91.
--33(t) Use the equation of continuity: Apr : Azuz. Here A1 is th e cross-sectional arca ofthe pipe at the top and u1 is the speed of the water there; A2 is the cross-sectional areaof the pipe at the bottom and u2 is the speed of the water there. Thus u2 : lg.0 cm2)/(8.0 cm2)] (5.0 mls) z.Sm/s.(b) Use the Bernoulli equation: h + Lpu?+ pghr : pz+ Lprr* pghz, where p is the density ofwater, fu is its initial altitude, and hz is its final altitude. Thus pz: p1 * ,l) + )ofr? - - hz) ps(hr 2 [(s. o mls)2 - (2.5 m/s)2] + (998 ke l^tx9 .8^ls2xlo m) : 2.6 x 10s Pa.The density of water was obtained from Table I4-l of the text.59(a) Use the Bernoulli equation: h+ |pu?+ pgfu: pz+ lpu2z* pghz, where fu is the height ofthe water in the tank, h is the pressure there, and u1 is the speed of the water there; h2 is thealtitude of the hole, pz is the pressure there, and u2 is the speed of the water there. p is thedensity of water. The pressure at the top of the tank and at the hole is atmospheric, So pr : p2.Since the tank is large we may neglect the water speed at the top; it is much smaller than thespeed at the hole. The Bernoulli equation then becomes pghl : ipr? + pghz and U2: 2g(hr 2(g.8*ls2xo.3o m) : 2.42m/s .The flow rate is Azuz- (6.5 x 10-4 m2)(2.42m1s): 1.6 x 10-*/r.(b) Use the equation of continuity: A2u2: Azut, where A3: Azl2 and rh is the water speedwhere the cross-sectional area of the stream is half its cross-sectional area at the hole. Thusu3 : (Az I At)uz : Zuz - 4.84 ml s. The water is in free fall and we wish to know how far it hasfallen when its speed is doubled to 4.84 m/s. Since the pressure is the same throughout the fall,Lputr + pghz h2 h3: 1 : 4- (4.84mf s)2 - (2.42mf s)2 - 0.90m. 2s 2e.8 m/s2)67(a) The continuity equation yields Au - aV and Bernoullis equation yields ipr - Lp + )pvz,where Lp: pz - pr. The first equation gives V- (Ala)u. Use this to substitute for V in thesecond equation. You should obtain *pr : Lp + )p(Ala)2rr2 . Solve for u. The result is 2Lp #) Chapter 14 87
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(b) Substitute values to obtain t):V -3.06m/s.The density of water was obtained from Table l4-l of the text. The flow rate is Au(64 x 10-4 m2X3.06mls):2.0 x I0-2m3/s.75Let p (: 998 kg/-) be the density of water and pt (: kg/-) be the density of the other 800liquid. Let d*r, be the length of the water column on the left side, d*n (: 10.9cm) be thelength of the water column on the right side, and ds (: 8.0 cm) be the length of the columnof the other liquid. The pressure at the bottom of the tube is given by po + pd,t * p*d*L,where po is atmospheric pressure, and by po * p*d- n These expressions must be equal, sopo * pdt, * p-d*r : p0 * p*d*n The solution for d*r is d-r- P*d-n-Ptdt: -3.5gcm. P* 998 kgl^tBefore the other liquid is poured into the tube the length of the water column on the left side isthe same as the water column on the right side, namely 10.0cm. After the liquid is poured it is3.59cm. The length decreases by 10cm - 3.59cm - 6.41 cm. The volume of water that flowsout of the right arrn is T(L 50 cm)2 (6.41 cm) : 45.3 cm3.88 Chapter 14
93.
Chapter L53(a) The amplitude is half the range of the displacement, or trrn: 1 .0 rrlln.(b) The maximum speedu* is relatedto the amplitude trmby n*: u)frrn, where a is the angularfrequency. Since ur - 2nf ,where / is the frequency, trrn:2rf rrn:2r(l20Hz)(l .0x 1g-: m):0.7 5 m/s.(c) The maximum acceleration is arn5.7 x 102 . ^/r27(a) The motion repeats every 0.500 s so the period must be T * 0.500 s.(b) The frequency is the reciprocal of the period: f : IIT - IlQ.500s):2.00H2.(c) The angular frequency is a - 2nf :2r(2.00H2) - l2.6radf s.(d)Theangu1arfrequencyisre1atedtothespringconstantkandthemaSSmbya_ffi,Sok : mw2 - (0.500 kgX 12.57 radls) : 79.0 N/m.(e) If trtn is the amplitude, the maximum speed is t)rn _ urfrrn - 02.57 rud1sx0.350 m) _4.40 m/s.(0 The maximum force is exerted when the displacement is a maximum and its magnitude isgiven by Frn2The magnitude of the maximum acceleration is given by arn: tt2trrn, where a is the angularfrequency and nnl is the amplifude. The angular frequency for which the maximum accelerationis9isgivenby,-MandthecoffeSpondingfrequencyisgivenby Aa r2n t- : I l-s 9.8 ^lt :498H2. 2"y .-,For frequencies greater than 498zHz the acceleration exceeds g for some part of the motion.t7The maximum force that can be exerted by the surface must be less than FFN or else the blockwill not follow the surface in its motion. Here, F" is the coefficient of static friction and F71/is the normal force exerted by the surface on the block. Since the block does not acceleratevertic ally, you know that Flr : mg, where m is the mass of the block. If the block follows thetable and moves in simple harmonic motion, the magnitude of the maximum force exerted on itis given by tr - ma,m- mw2rTnacceleration ) u) is the angular frequency, and f is the frequency. The relationship a : 2n f wasused to obtain the last form. Chapter 15 89
94.
Substitute It - m(2nf)r,n and F^r - mg into F <largest amplitude for which the block does not slip is rnl-(2n f)2: (osoX?8 T/s1) - o.o3 F9 . (2n x 2.0Ht)2 1 m.A larger amplitude requires a trargw force at the end points of the motion. The surface cannotsupply the larger force and the block slips.t9(a) Let A / 2trt n1 -zcos ( , /be the coordinate as a function of time for particle 1 and 12: Iror(+.;)be the coordinate as a function of time for particle 2. Here T is the period. Note that since therange of the motion is A, the amplitudes are both A 12. The arguments of the cosine functionsare in radians.Particle 1 is at one end of its path (*t - Al2) when tZntlT+116 - 0 or t- -f lI2. That is, particle 1 lags particle 2 by one-twelfth a period. Wewant the coordinates of the particles at t - 0.50 s. They are x o50 s) nt :4 ro, (2zr I 2v) l.5s )and fr2:^costlznxo.5os l--0.4334. A -+=zr 2 1.5s 6/Their separation at that time is n1 - n2: -0.2504+ 0.433A - 0.1834.(b) The velocities of the particles are given by (Znt ?,1 -drt:4 T-sm t dt /and d" -z- dt T Ltt- 6/ TEvaluate these expressions for t - 0.50 s. You will find they are both negative, indicating thatthe particles are moving in the same direction.27When the block is at the end of its path and is momentarily stopped, its displacement is equal tothe amplitude and all the energy is potential in nature. If the spring potential energy is taken tobe zero when the block is at its equilibrium position, then E - lrrl: )r.t x t02N/,o) (0.024m)2 : 3.7 x t0- 2 I .90 Chapter 15
95.
29 -(a) and (b) The total energy is given by E itt*,-, where k is the spring constant and rrn isthe amplitude. When r : i*r- the potential energy is f-Jt - *tt* - ttt*?". The ratio is 1 v- itt*,"- E 1 4 itt*,"The fraction of the energy that is kinetic is K E-U E E -1 Y-1 1- 844 3-(c) Since E - *tr*h, and fJ: - *tt*, UIE: 12 l*,". Solve ,l*?*- Llz for tr. You should get:L - :rrnlt/r.39(a) Take the angular displacement of the wheel to be 0 - ?rn cos(2nt lT), where 0,n is theamplitude and T is the period. Differentiate with respect to time to find the angular velocity:O - -(2n lT)0,- sin(2zrt lD. The symbol O is used for the angular velocity of the wheel so itis not confused with the angular frequency. The maximum angular velocity is Qrn 39.5 radf s . T 0.500 s(b) When 0 - n12, then 010,"- l12, cos(ZntlT): I12, and sin(2zrtlT)- :@:tZlz,where the trigonometric identity cosz 4+ sin2 A - I was used. Thus o - a_ -+ o*sin T (r): (#) Qrrad) (f) : -3 4.2radf s .The negative sign is not significant. During another portion of the cycle its angular speed is+34.2radf s when its angular displacement is n l2rad.(c) The angular acceleration is d2 0 / 2n2 dtr: I t ,) 0,n cos(2rt lT) -When 0-n14, a- (J* 2 (;): - 1 24 radlr (osooAgain the negative sign is not significant. / Chapter 15 9l
96.
43(a) A uniform disk pivoted at its center has a rotational inertia of + M R2, where M is its massand R is its radius. See Table I0-2. The disk of this problem rotates about a point thatis displaced from its center by R + L, where L is the length of the rod, so, according to theparallel-axis theorem, its rotational inertia is Lm n2 + M (L + R)2. The rod is pivoted at one endand has a rotational inertia of mL13, where m is its mass. The total rotational inertia of thedisk and rod is I - *mn2 + M(L+ R)2 + |mL2: 1f0.500kgX0.100 m)2 +(0.500kgX0.500m+0.100m)z + 1to .2J0kgX0.500 m)2 - 0 .205kg . m2.(b) Put the origin at the pivot. The center of mass of the disk rs,(.a - L+R - 0.500m+0.100m -0.600m away and the center of mass of the rod is (.r- L12: (0.500m)12:0.250m away, onthe same line. The distance from the pivot point to the center of mass of the disk-rod system is t _ M la t d_-ffi:m(., (0.500 kgx0.600 m) + (0.270 kgX0 .250 m) _ n A _u.+Tilr.(c) The period of oscillation is 0.205 kg . m2 T :2r -1.50s. + m)sd (0.500 kg + 0 .270kgX9 .8^ls2x0 .447 m)51If the torque exerted by the spring on the rod is proportional to the angle of rotation of the rod andif the torque tends to pull the rod toward its equilibrium orientation, then the rod will oscillatein simple harmonic motion. If r -- -C0, where T is the torque, 0 is the angle of rotation, ande is a constant of proportionality, then the angular frequency of oscillation is atheperiodisT-2nfa-2rrre,whereIistherotationa1inertiaoftherod.Thep1anistofind the torque as a function of 0 and identify the constant e in terms of given quantities. Thisimmediately gives the period in terms of given quantities.Let (.s be the distance from the pivot point to the wall. This is also the equilibrium length of thespring. Suppose the rod turns through the angle 0, with the left end moving away from the wall.If L is the length of the rod, this end is now (Llz)sind fuither from the wall and has moved(Ll2)(1 -cos 0)to the right. The spring length is now {ff lDz(I _- cos q2 +Vo+ (LlT)sin 0f,If the angle e is small we may approximate cos I with 1 and sin 0 with 0 in radians. Then thelength of the spring is given by (,0+ L0 12 and its elongation is Lr: L0 12. The force it exertson the rod has magnitude It - k Lr- kL?f 2, where k is the spring constant. Since 0 is smallwe may approximate the torque exerted by the spring on the rod by r: -FLf 2, where the pivotpoint was taken as the origin. Thus r - -(kL lqg. The constant of proportionality C thatrelates the torque and angle of rotation is e - kL2 f 4.The rotational inertia for a rod pivoted at its center is I - mLz ll2, where m is its mass. SeeTable L0-2. Thus the period of oscillation is t--t"l tr -^ ,:2r tr1 mL2 l12 kL2 l4 0.600 kg 3(18s0 N/*) - 0.0653 s92 Chapter 15
97.
57(a) You want to solve e-bt/2* logarithm of both sides to obtain-btlZm (2* lb)1n3, where the sign wasreversed when the argument of the logarithm was replaced by its reciprocal. Thus t_ ffih3-14.3s(b) The angular frequency is w,- m *V 4rrr: 8.00 1.50 kg N/* (0.230 kg/s)2 4(1.50 kg) 2.31 radf s .The period is T(14.3 s)/(2.72 s) : 5.27 .75(a) The frequency for small amplitude oscillations is f : (l lzr)ffi,, where L is the length ofthe pendulum. This gives f - (Ilzr)tl (9.80^lsl(2.0m):0.3 5Hz.(b) The forces acing on the pendulum are the tension force 7 of the rod and the force of gravitymj. Newtons second law yields f + mfi : md, where m is the mass and d is the accelerationof the pendulum. Let d _ d,. + d, where d,. is the acceleration of the elevator and u" is theacceleration of the pendulum relative to the elevator. Newtons second law can then be writtenm(d - d") + f : md,. Relative to the elevator the motion is exactly the same as it would be inan inertial frame where the acceleration due to gravity is d - d,". Since j and d," are along thesame line and in opposite directions we can find the frequency for small amplitude oscillationsbyreplacing g with g*a" in the expression f -(Ilzr)FgIL Thus f^T r2n ml 9.8 mlrt + 2.0^lt - 0.39H2. VL(c) Now the acceleration due to gravity and the acceleration of the elevator are in the samedirection and have the same magnitude. That is, d - d," - 0. To find the frequency for smallamplitude oscillations, replace g with zero in f - (llzr)ffi,. The result is zero. Thependulum does not oscillate.83[Jse uTLstroke, or 0.38 m. Thus unl: 2n(3.0 Hz)(0.38 m) : 7 "zm/s.89(a) The spring stretches until the magnitude of its upward force on the block equals themagnitude of the downward force of gravity: ky - mg, where A is the elongation of thespring at equilibrium, k is the spring constant, and m is the mass of the block. Thusk - mg la - (1 .3 kgX9.8 */ sz)l (0.096 m) : 133 N/-. Chapter 15 93
98.
(b)TheperiodisgivenbyT:Ilf:2nlu):2r1ffik_2rr-0.62S.(c) The frequency is f : IIT - Il0.62s - L.6Hz.(d) The block oscillates in simple harmonic motion about the equilibrium point determined bythe forces of the spring and gravity. It is started from rest 5.0cm below the equilibrium point sothe amplitude is 5.0 cm.(e) The block has maximum speed as it passes the equilibrium point. At the initial position, theblock is not moving but it has potenttal energy U,i,When the block is atthe equilibrium point, the elong atioirof the spring is A- g.6cm and thepotential energy is 1, I I (,1 y- -msa. ;ka^Write the equation for conservation of energy as [-Li: ,rl **rand solve for u: 2(Lh - tI r) 2(-0.44 J + 0.61 - 0.51 m f s. m9t(a) The frequency of oscillation is - 3.2H2.(b) Because mechanical energy is conserved the maxlmum kinetic energy of the block has thesame value as the maximum potential energy stored in the spring, so i*u?^ l.2kg , : 0.26 [m. rrn -- l-m N Zurn: Affi(5.2mls)(c) The position of the block is given by *: nmcos(c,,,t+0 where n?n - 0.26m and u) - 2nf -2r(3.2H2):2rudf s. Since r-0 at time t-0, the phase constant O must be either +rl2 or -r, 12. The velocity at t-0 is given by -urfrrn sin@ and this is positive, so 6 must be -TT12,The function is tr : (0.26m) cos[(20 radls)f - Tr l2].94 Chapter 15
99.
Chapter L515The wave speed u is given by , - muwhere r is the tension in the rope and LL is the linearmass density of the rope. The linear mass density is the mass per unit length of rope: m p: i- 0.0600 kg - o.o3ookg f m. ,Jo*Thus 0.0300 kg lm - I29 m/s.t7(a) In the expression given for y, the quantrty A* is the amplitude and so is 0.1Zmm.(b) The wave speed is given by u - tFn, where r is the tension in the string and Lr is thelinear mass density of the string, so the wavelength is t- u tFrtp -c .tand the angular wave number is l4lm-r(c) The frequency is f :100H2, so the angular frequency is- 2nf :2r(100H2):628rad/s. a(d) The positive sign is used since the wave is traveling in the negative r direction.2t(a) Read the amplitude from the graph. It is the displacement at the peak and is about 5.0 cm.(b) Read the wavelength from the graph. The curye crosses A - 0 at about r : 15 cm and agatnwith the same slope at about tr- 55cm, so )- 55cm- 15cm:40cm-0.40m.(c) The wave speed iswhere T is the tension in the string and pr, is the linear mass density of the string. Thus u- 10-3kglm - L2mf s. Chapter 16 95
100.
(d) The frequency is v 12m/s f = - = = 30Hz A 0.40mand the period is 1 1 T = - = - - = 0.033 s . f 30Hz(e) The maximum string speed is Um = WYm = 27i}Ym = 2IT(30Hz)(5.0cm) = 940cm/s = 9.4m/s.(f) The angular wave number is 27i 27i k = - = = 16m- I . A 0.40m(g) The angular frequency is w = 2IT f = 2IT(30 Hz) = 1.9 X 102 rad/s.(h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 X 10- 2 m. The formulafor the displacement gives y(O,O) = Ym sin ¢. We wish to select ¢ so that 5.0 x 10- 2 sin ¢ =4.0 X 10- 2. The solution is either 0.93 rad or 2.21 rad. In the first case the function has apositive slope at x = 0 and matches the graph. In the second case it has negative slope and doesnot match the graph. We select ¢ = 0.93 rad.(i) A positive sign appears in front of w because the wave is moving in the negative ;]; direction.31The displacement of the string is given by Y = Ym sin(kx - wt) + Ym sin(kx - wt + ¢) = 2Ym cos(~¢) sin(kx - wt + ~d»,where ¢ = IT /2. The amplitude is A = 2Ym cos(~¢) = 2Ym cos (IT /4) = 1.41Ym·35The phasor diagram is shown to the right: Yl m and Y2mrepresent the original waves and Ym represents the resultantwave. The phasors corresponding to the two constituent wavesmake an angle of 90° with each other, so the triangle is a righttriangle. The Pythagorean theorem gives y~ = Y?m Yim + = (3.0 cm)2 + (4.0 cm)2 = 25 cm2 .Thus Ym = 5.0 cm.41Possible wavelengths are given by A = 2L / n, where L is the length of the wire and n is aninteger. The corresponding frequencies are given by f = v / A = nv /2L, where v is the wave96 Chapter 16
101.
speed. The wave speed is given by u - tm,-is llrLlM, of the T is the:tension in the wire, M the mass where M Ltt is the linear mass density of the wire and wire. p, I was used toobtain the last form. Thus ^TL T r2L + LM (10.0 mX0.100 kg) - n(7.glHr)(a) For : 1, f - 7 .9I Hz. rL(b) For ft:2, f : 15.8 Hz.(c) For fr:3, f :23.7H2.43(a) The wave speed is given by u - mt, where r is the tension in the string and p is thelinear mass density of the string. Since the mass density is the mass per unit length, F : M I L,where M is the mass of the string and L is its length. Thus (96.0 NXS.40 m) 0.l20kg - 82.0 m/s .(b) The longest possible wavelength . for a standing wave is related to the length of the stringby L - ,f2, so )- 2L-2(8.40m): 16.8m.(c) The frequency is f- ul^ - (82.0^ls)/(16.8m) - 4.88Hz,.47(a) Thc resonant wavelengths are given by - zLf n, where L is the length of the string and n ^is an integer, and the resonant frequencies are given by f - ul^ - nnlzL, where u is the wavespeed. Suppose the lower frequency is associated with the integer n,. Then since there are noresonant frequencies betweeo, the higher frequency is associated with n*1. That is, fr: TLt)l2Lis the lower frequency and fz: (n * l)u lzL is the higher. The ratio of the frequencies is fz: n+r finThe solution for n is rL : -f, r fz h 420H2-3I5Hz f - uf 2L - hl, - (3 l5Hz)13The lowest possible resonant frequency is(b) The longest possible wavelength is ) _ 2L. If f is the lowest possible frequency thenu - xf - 2Lf - 2(0.7s mX105 Hr) - 158 m/s.53The waves have the same amplitude, the same angular frequency, and the same angular wavenumber, but they travel in opposite directions. Chapter 16 97
102.
(a) The amplitude of each of the constituent waves is half the amplitude of the standing wave or0.50 cm.(b) Since the standing wave has three loops the string is three half-wavelengths long. If L is thelength of the string and is the wavelength, then L -3^12, or , - 2L13:2(3.0m)13 -2.0m. ^The angular wave number is k - 2nl) - 2nl(2.0m):3.1m-l.(c) If u is the wave speed, then the frequency is u 3u 3(100 m/s) I: )- n- ,(3.0"r)The angular frequency is u):2nf :2r(50H2) - 3.1 x I02rudf s.(d) Since the first wave travels in the negative r direction, the second wave must travel in thepositive r direction and the sign in front of u must be a negative sign.6t(a) The phasor dtagram is shown to the right: Ut, Uz, and Usrepresent the original waves and Urn represents the resultantwave. The honzontal component of the resultant is arnhUt Az - Ur arl3 - 2yrf 3. The vertical component isU*, : U2 : Ar f 2. The amplitude of the resultant is I Urn:m: 5 : ,al : o 0.83at(b) The phase constant for the result ant is 6 _tan-r u*, _ tan-r ( -1 3 arnh ^M^ -tan-r:-0.644rad:37" 2Yr 13 / 4(c) The resultant wave is : 5 stn(kr - wt + 0 .644 rad) A . 6AtThe graph below shows the wave at trme / - 0. As time goes on it moves to the right with speedt) : a lk. -Un -Ut98 Chapter 16
103.
69(a) Take the form of the displacement to be A@,t): arnstn(kr - at). The speed of a point onthe cord is u(r, t) - 0y l 0twave speed, on the other hand, is given by, - - alk. The ratio is ^lf urn : aArn 2narn n - Ngrn: 7 u alK(b) The ratio of the speeds depends only on the ratio of the amplitude to the wavelength.Different waves on different cords have the same ratio of speeds if they have the same amplitudeand wavelength, regardless of the wave speeds , linear densities of the cords, and the tensions inthe cords.77(a) If r is the tension in the wire and LL is its linear mass density, then the wave speed isa mlL, where m is the mass of the wire and L is itslength. Thus (r20NX1.s0m) : I44mls . 8.70 x 10-3 kg(b) A one-loop standing wave has two nodes, one a each end, and these are half a wavelengthapart, so the wavelength is - 2L - 2(1 .50m) - 3.00m. ^A two-loop standing wave has three nodes, one at each end and one at the midpoint. Since thenodes are half a wavelength apaft, the wavelength is , - L - 1.50 m.(c) and (d) The frequency is f - ulx. For the one-loop wave f : (I44mls)l(3.00m):48.0H2and for the two-loop wave f : (I44mls)/(1.50m) :96.6H2.87(a) The transverse rope velocity is given by , : aUrn, where a is the angular frequency and Arnis the amplitude. The angular frequency is a - 2n f, where f is the frequency. Thus uu 5.0 m/s oa ant : a : Znf 2r(5.0 Hz) - 0.I6m.(b) The wave speed is u - trn where r is the tension in the rope and LL is the linear massdensity of the rope. The linear mass density is l.L- mlL, where m is the mass of the rope andL is its length. The wave speed is , where is the wavelength. Since the rope is vibrating inits fundamental mode ) - 2L Thus ^f ^ r-puz-irxrf(c) Thc general forrn for the displacement at coordrnate r and time t for a standing wave that hasnodes atn -0and r- Iis A:Amstn(2rrlDsrn(2nft. Here Un isthemaximumdisplacementof any of the points along the rope. Since, in this case, the rope is vibrating in its fundamental Chapter 16 99
104.
mode Urn is the maximum displacement of the point at its center and thus has the value calculatedin part (a). The displacement at any coordinate r is a@,t):(0.16m)sin|#")sin|2rr(5.0Hz)t]-(0.I6m)sin[(1.6m-)"]sin[(31s-,)].89(a) The wave speed is given by u- mu where r is the tension in the rubber band and lr isthe linear mass density of the rubber band. According to Hookes law the tension is r - k L(..The length of the stretched rubber band is (.+ L(,, so the linear mass density is l.L: ml!+ Lt).The wave speed is k L(,(t + Lt)(b) The time for a pulse to travel the length of the rubber band is (.+ al m((+ Lt) t_T_((,+Lt) k L(,((,+ L!)[f L(. is much less than (, we may neglect the LI in the numerator. Then m[. KNwhich is proportional to L l161-.(c) It L(. is much greater than (, we may in the numerator. Then u:which is independent of L(..100 Chapter 16
105.
Chapter L7-5Let t y be the time for the stone to fall to the water and t, be the time for the sound of the splashto travel from the water to the top of the well. Then the total time elapsed from dropping thestone to hearing the splash is t_ ty *tr. It d is the depth of the well, then the kinematics offree fall gives d : *gt?, or t yor f" : d/ur.Thus the total time is 2d d t- g I usThis equation is to be solved for d. Rewrite it as d t- usand square both sides to obtain 2d, . I -lL L -z-t d*a as u"NIow multiply by gu? and realrange to get gd2 - Zu,(gt I rr s)d + gu?* : 0 .This is a quadratic equation for d. Its solutions are 2u r(gt | u s) * ar?@t + ur)2 4g2u?t2 d- 2sThe physical solution must yield d : 0 for t : 0, so we take the solution with the negative signin front of the square root. Once values are substituted the result d-40.7mis obtained.1_If d, is the distance from the location of the earthquake to the seismograph and u" is the speed ofthe S waves, then the time for these waves to reach the seismograph is ts: d/ur. Similarly, thetime for P waves to reach the seismograph is tp: d/up.The time delay is Af d d d(up - u - us up -- u sup ") ) Chapter 17 101
106.
SO Lt (4.5knlsx8.0 km/sX3.0 minx60 s/min) u-_ .l lrsup 1 ^ x 103 km.Notice that values for the speeds were substituted as given, in km/s, but that the value for thetime delay was converted from minutes to seconds.2(a) Use - ulf, where u is the speed of sound in air and f is the frequency. Thus ^ _ 343mls 7 62 x 10-5 m 4.5 x 106 H;- /t -(b) Now -- ulf , where u is the speed of sound in tissue. The frequency is the same for air ^and tissue. Thus _ 1500 m/s 3 33 x 1o-4 m lt - 4.5 x 106 Hr-t9Let Lr be the distance from the closer speaker to the listener. The distance from the otherspeaker to the listener is L2: d is the distance between the speakers. The lWphase difference at the listener ls , 2n(Lz - Lt)where is the wavelength. ^(a) For a minimum in intensity at the listener, 0 (2n + l)n, where n is an integer. - Thus" - 2(Lz L) lQn + 1). The frequency is t u (2n + l), (2n + 1X3 a3 mls) J_ r ) 2lrry-1,1 2l -37sm] - (2n+ 1X343 Ht). ITo obtain the lowest frequency for which a minimum occurs set n equal to 0. The frequency isf^in,l : 343H2.(b) To obtain the second lowest frequency set n equal to 1. This means multiply F*i* I by 3.(c) To obtain the third lowest frequency set n equal to 2. This means multiply lr",in, I by 5.For a maximum in intensrty at the listener, d : 2nn, where n is any positive integer. Thus )-:lEa-r]and u :::- nu n(343 m/s) n(686 Hz) . (3.7 5 m)2 + (2.00 m)2 - m lW-11 3.7 5(d) To obtain the lowest frequency for which a maximum occurs set n equal to 1. The frequencyis f^u*,I : 686 Hz.102 Chapter 17
107.
(e) To obtain the second lowest frequency set n equal to 2. This means multiply f-in, 1 by 2.(0 To obtain the third lowest frequency set n equal to 3. This means multiply .F*6, t by 3.25The intensity is the rate of energy flow per unit area pe{pendicular to the flow. The rate at whichenergy flows across every sphere centered at the source is the same, regardless of the sphereradius, and is the same as the power output of the source . If P is the power output and I is theintensity adistance r fromthesource,then P- IA:4rr2 I,where A(:4rr2) isthesurface area of a sphere of radius r. Thus P - 4r(2.50 m)2(1.91 x 10-4W/ttr) - 1.50 x I0-2W.29(a) Let 11 be the original intensity and 12 be the final intensity. The original sound level ishreference intensity. Since Cz : Cr + 30 dB, (10 dB) Iog(I2l Iil - (10 dB) log(/t I IO + 30 dB ,or (10 dB) tog(I2l Iil - (10 dB) log(/r I Iil - 30 dB .Divide by 10dB and use log(I2lIil - log(hlIo): Iog(I2lI) to obtain log(I2lIr):3. Now useeach side as an exponent of 10 and recogntze that 1glog(/2 /I) : Izl I,The result is IzlIt: 103. The intensity is multiplied by a factor of 1.0 x 103.(b) The pressure amplitude is proportional to the square root of the intensity so it is multipliedby a factor of .,4000 :32.43(a) The string is fixed at both ends and, when vibratittg at its lowest resonant frequency, exactlyhalf a wavelength fits between the ends . If L is the length of the string and l is the wavelength,then - 2L The frequency is f - ulx- ul2L, where u is the speed of waves on the string.Thus ^ : 2L f - 2(0.220 m)(920 Hz) : 405 m/s. u(b) The wave speed is given by u - mt where r is the tension in the string and pt is thelinear mass density of the string. If M is the mass of the string, then p : M I L since the stringis uniform. Thus r: truz: f*:W(4os ^ls)z-5e6N.(c) The wavelength is ) - 2L - 2(A.220m) : 0.440m. Chopter 17 103
108.
(d) The frequency of the sound wave in arc is the same as the frequency of oscillation of thestring. The wavelength is different because the wave speed is different. If u o is the speed ofsound in air the wavelength in atr is , ua 343m/s) ta l- ^ ,l 92AHz - 0 .373 m.45(a) Since the pipe is open at both ends there are displacement antinodes at both ends and aninteger number of half-wavelengths fit into the length of the pipe. If L is the pipe length and A isthe wavelength then - 2Lln, where n is an integer. If u is the speed of sound then the resonantfrequencies are givenbV f :nf ^ Now L-0.457m, so f :n(344mls)12Q.457m)- ^-nuf2L.376.4nH2. To find the resonant frequencies that lie between 1000H2 and 2000H2, first setf - 1000Hz and solve for rL, then set f :2000H2 and agaun solve for rL. You should get 2.66and 5.32. This means ft:3,4, and 5 are the appropriate values of n There are three resonancefrequencies in the given range.(b) For fr:3, f :3(376.4H27 - Il29Hz.(c) For fr: 4, f : 4(376.4H2) - 1506 Hz.47The string is fixedatboth ends so the resonantwavelengths are given by, - 2Lln,where L is thelength of the string and n is an integer. The resonant frequencies are given bV f - n l A: nu f 2L,where t is the wave speed on the string. Now tr : tmt where T is the tension in the string andp is the linear mass density of the string. Thus f : (nlzL),,mr. Suppose the lower frequencyis associated with TL : TL1 and the higher frequency is associated with TL : rL1 + I . There are noresonant frequencies between so you know that the integers associated with the given frequenciesdiffer by 1. Thus fi - (nrlzL)tmt and TLr*l F TLt tr 1 n 1 ! fz l-t,This means fz h - (llzL)mt and r : 4L2 ttffz f )t - 4(0.300 m)2(0.650 x 10-t kgl*Xl3 20Hz - 880 Ht) :45.3N.53Each wire is vibrating in its fundamental mode so the wavelength is twice the length of the wire :() 2L) and the frequency is ullT f: ) 2Ly p) : l-where u (: mD is the wave speed for the wire, T is the tension in the wire, and p is thelinear mass density of the wire.104 Chapter 17
109.
rSuppose the tension in one wire is and the oscillation frequency of that wire is f r. The tensionin the other wire is r *Lr and its frequency is fz. You want to calculate Lr for 600 Hz lr fi:and fz: 606 Hz.Now f.-1 rt n E LLandSO fz Lr ft TThis means Lr- T G), 1 - (g6Ht 600 Hrl I- oo2o --05(a) The expression for the Doppler shifted frequency is .f-fry * a uswhere f is the unshifted frequency, u is the speed of sound, u p is the speed of the detector (theuncle), and us is the speed of the source (the locomotive). All speeds are relative to the air. Theuncle is at rest with respect to the air, so ?/p : 0. The speed of the source is u gthe locomotive is moving away from the uncle the frequency decreases and we use the positivesign in the denominator. Thus f 343mls : 48s.8 Hz. U f Ug 343mls + 10.00m/s )(b) The girl is now the detector. Relative to the atr she is moving with speed u ptoward the source. This tends to increase the frequency and we use the positive sign in thenumerator. The source is moving at us: 10.00m/s away from the girl. This tends to decreasethe frequency and we use the positive sign in the denominator. Thus (u + u n) - (u + u il andf - f : 5oo.oHz.(c) Relative to the arr the locomotive is m m bat u,,,S =20).0(0 mtl: away from the uncle. I-Jse 10v o LO 10.0r sthe positive sign in the denominator. Relat irre:t,or thrv<?1 the un l1erirs moving at up: 10.00m/s tive da rr eunIC in tne 1m( ltor. htoward the locomotive. Use the positive si gnrir the nlu1 erat<or. TT S 1 hu * up // 3^l f1 00n t: ,3, f- f u ,(: 4 3mlls + 0.0( ml : -(5oo.oHz) I A5imlt, + n^0( ml ;:) S 486.2H2. u * ug ,t 3^l - 00" ls gat(d) Relative to the air the locomotive is m vir b,at U1s = 20. )0n ls away from the girl and the lov inio :0.0( m/ s Y- tgirl is moving at u p th e koco)mlol V .IJ e th positive signs in both the he locor rtirve. Jse he cnumerator and the denominator. Thus (u + Ur, ) - =(urf lJq)an f f - 500 .0H2. n) (o,( r) lnd Chapter 17 105
110.
69(a) The half angle 0 of the Mach cone is given by sin 0 - u lu s, where u is the speed of sound andus is the speed of the plane. Since us: I.5u, sin 0 -ull.Su: If L.5. Thismeans 0-42o.(b) Let h be the altitude of the plane and suppose the Mach coneintersects Earths surface a distance d behind the plane. The situ-ation is shown on the diagram to the right, with P indicating theplane and O indicating the observer. The cone angle is related toh and d,by tan?- hld so d - hf tan?. The shock wave reachesO in the time the plane takes to fly the distance d: t - dl, - hlutan? - (5000 n)11.5(331^ls)tan 42o :11s.77(a) Use the Doppler shift equation, which gives the detected frequency J . f- f U LUgwhere f is the emitted frequency, u is the speed of sound, Vn is the speed of the detector, andug is the speed of the source. You are the detector and are station?ry, so upthe source and is moving away from you, which lowers the frequency, so the positive sign isused in the denominator. Thus 330 m/s :9.7 x I02Hz. f- (1000H2) 330 mls + 10m/s(b) The detector is the cliff, which is stationary. The source is the siren, which is moving towardthe cliff. This increases the frequency and the minus sign is used in the denominator. Thus 330 mls f- (1000H2) - 10m/s 330 mls(c) The beat frequency is fb"ut:1.0 x 103 Hz - 9.7 x l02Hz:60Hz,.81(a) The rate with which sound energy is passing through the surface of a sphere with radius ris P - 4nrz I, where I is the intensity. Since energy is conserved this must be the power of thesource. Thus P - 4r(0.0080 W/*Xt0 -) _ 10 W.(b) The sound inrensity is f - Pf 4rr2 - (10w/4r(5.0*)2 - 0.032Wfm2.(c) The sound level is 0 : (10d8 log(Illo), where Is is the standard reference intensity, whichis given as 10-tzWl^in the text. At a point 10m from the source it is C : (10 dR)Ios 0.0080W lmz 12 :99 dB 10-106 Chapter 17
111.
85(a) The intensity is given by I : *prrsl, where p is the density of the medium, u is the speedof sound, a is the angular frequency, and sTn is the displacement amplifude. The displacementand pressure amplitudes are related by Lprn - pl)usrn, so srn: Lprnlpw and f - (Lprn)z lLpr.For waves of the same frequency the ratio of the intensity for propagation in water to the intensityfor propagation in air is I-- (Lp,,- pouo h lo-," I p*u- where the subscript a, denotes arc and the subscript ?t) denotes water.Since Io : I-, LPrn* /t0.998 x 103 kel^Xr 4v2mls) - se7 LPr,,o V e.2rr.g7ffir; ^tThe speeds of sound are given in Table 17 -I and the densities are given in Table l4-1.(b) Now Lprn* : Lpnlo, so I* parra : Q.2tkglm3)(343mls) _ z.g1 x 10-4 h - puuu (0.998 x 103 kglmXt Yzm/s) -v87(a) When the right side of the instrument is pulled out a distance d the path length for soundwaves increases by 2d. Since the interference pattern changes from a minimum to the nextmaximum, this distance must be half a wavelength of the sound. So 2d: where . is thewavelength. Thus , ^12,(343mls)/a(0.O165 m) : 5.2 x 103 Hz.(b) The displacement amplitude is proportional to the square root of the intensity (see Eq. 17 -27).Write t/Tconstant of proportionality. At the minimum, interference is destructive and the displacementamplitude is the difference in the amplitudes of the individual waves 1 srnwhere the subscripts indicate the paths of the waves. At the maximum, the waves interfereconstructively and the displacement amplitude is the sum of the amplitudes of the individualwaves i srnfor ssAn and ssBD. Add the equations to obtain ss7.n- (1f00 + m)lzcsubtract them to obtain ssso: (vm- vC00)l2C: I}f C. The ratio of the amplitudes issstnl ts"r:2.(c) Any energy losses, such as might be caused by frictional forces of the walls on the ak inthe tubes, result in a decrease in the displacement amplitude. Those losses are greater on path Bsince it is longer than path A.101(a) The frequency is increased by reflection from the flowing blood, so the blood must be flowingto the right, with a positive velocity component in the direction of the original source of theultrasound. Chapter t 7 107
112.
(b) Use the Doppler shift equation twice. It is f - f U f.Ugwhere fis the emitted frequency, u is the speed of soutrd, Vn is the speed of the detector, andu s is the speed of the source. First, take the source to be the ultrasound generator, which isstationdry, and the detector to be the blood. Thus us - 0 and up _ u6eos0, where u6 is thespeed of the blood. Since the detected frequency is greater than the generator frequency, use theplus sign in the numerator. Thus * uacos? f - fu uIn the next step the blood is the source and f is the emitted frequency. The generator is thedetector and is stationary. Take up:0 and us:lrbcos0. The detected frequency is higher, sowe use the minus sign in the denominator of the Doppler shift equation. The detected frequencyis r,,-r,#-r(ry) (#) :rffi,,where f was replaced by the expression developed previously. The solution for u6 is ub:(ffi) (#) :( )(#) -oeom/s(c) If 0 increases,g decreases. This means the numerator of the expression for f cos " decreasesand the denominator increases. Both changes result in a decrease in f " .108 Chapter 17
113.
Chapter L8ISince a volume is the product of three lengths, the change in volume due to a temperature changeLT is given by LVlinear expansion. See Eq. 18-11. Since V- (4nlrB3, where R is the original radius of thesphere, AV - 3u (+R) - (23 x to-6 lc")Gil(lO cm)3(100 Co) : 29 cm3 3 ) ^rThe value for the coefficient of linear expansion was obtained from Table l8-2.15If V. is the original volume of the cup, e.a is the coefficient of linear expansion of aluminum, andLT is the temperature increase, then the change in the volume of the cup is LV.See Eq. 18- 11. If P is the coefficient of volume expansion for glycerin then the change in thevolume of glycerin is LVnthe original volume of the cup. The volume of glycerin that spills is LVn LV" - (p - 3ao)V"LT to-u lc)) (1oo cm3x6 c") - o .2|cm32tConsider half the bar. Its original length rs (.s : L0 f 2 and its length after the temperature increaseis ( : to + qh LT . The old position of the half-bar, its new position, and the distan ee n that oneend is displaced form a right triangle, with a hypotenuse of length (., one side of length (0, andthe other side of length r. The Pythagorean theorem yields 12 : ,(.2 - 1z0 - tl * a LT)z - [3.Since the change in length is small we may approximate (1 * crLT)2 by I +2uLT, where thesmall term (a LT)z was neglected. Then 12 - tA + 2lla LT - 4 - 2[?ra LTand n:hrffi:ry :7.5x to-2m.25The melting point of silver is 1235 K, so the temperature of the silver must first be raised from15.0oC (- 288 K) to 1235 K. If m is the mass of the silver and c is its specific heat, this requiresenergy Q - cm(Ty -7,;): (236Jlkg.KX0.130kgXI235oC -288oC) - 2.gI x 104J Chapter I B 109
114.
Now the silver at its melting point must be melted. If L e is the heat of fusion for silver thisrequires Q:mLp-(0.130kgX105 x 103J/kg)- 1,36 x 104J.Thetotal energyrequiredas heat is 2.9I x 104J+ 1.36 x 104J- 4.27 x 104J. The specificheatof silver can be found in Table 18-3 and its heat of fusion can be found in Table 18-4.27Mass nL (: 0.100kg) of water, with specific heat c (: 4I90Jlkg.K), is raised from an initialtemperature T,i (: 23"C) to its boiling point Ty (: 100C). The heat input is given by A_cm(71 - Tt). This must be the power output of the heater P multiplied by the time t; A : Pt.Thus r__a cm(Ty -T) (4190 llkg.K)(0.100kgx100C - 23"C) t-;4t(a) There are three possibilities:1. None of the ice melts and the water-ice system reaches thermal equilibrium at a temperaturethat is at or below the melting point of ice.2. The system reaches thermal equilibrium at the melting point of ice, with some of the icemelted.3. All of the ice melts and the system reaches thermal equilibrium at a temperature at or abovethe melting point of ice.First suppose that no ice melts. The temperature of the water decreases from Twt (: 25"C) tosome final temperature Ty and the temperature of the ice increases from Tp (: -15"C) to TyIf m,yy is the mass of the water and cys is its specific heat then the water loses energy Q:cyrTftyy(Twt,-Tf)If m7 is the mass of the ice and c7 is its specific heat then the ice absorbs energy Q:crTrrtlf-Ttt),Since no energy is lost these two energies must be the same and cyrTTl,14r(Twt, - Ty) : cttrLtQf - Ttr) .The solution for the final temperature is rn CyrTTLryyTWt t ctTTLTTtf ,WW (4190 J/kg . KX0.200 kgX25"C) + (2220 J/kg . KX0.100 kgX- 15"C) - (4190 I lks . K)(0.200 ke) + (2220 I lkg. KXO.100 ke)This is above the melting point of ice, so at least some of the ice must have melted. Thecalculation just completed does not take into account the melting of the ice and is in elror.110 Chapter 18
115.
Now assume the water and ice reach thermal equilibrium at Ty : OoC, with mass m (< *r) ofthe ice melted. The magnitude of the energy lost by the water is Q : cyyITLltgTwt ,and the energy absorbed by the ice is Q : crTTLr(O - Ty) + mLp ,where L p is the heat of fusion for water. The first term is the energy required to warrn all theice from its initial temperature to OoC and the second term is the energy required to melt massm of the ice. The energy lost by the water equals the energy garned by the ice, so CyrrTLyrf Wt: -CTtTLlTtt + mL p .This equation can be solved for the mass m of ice melted: C14r Tft1,yTW t a C I Tft 1T t t m- Le (4t90 I lks . K)(0.200 kgx2s"C) + (2220 I lkg. KXO.100 kg)(- 15"C) 333 x 103Jlke : 5.3 x I0-2 kg - 53 g.Since the total mass of ice present initially was 100 g, there is enough ice to bring the watertemperature down to 0oC. This is the solution: the ice and water reach thermal equilibrium at atemperature of OoC with 53 g of ice melted.(b) Now there is less than 53 g of ice present initially. All the ice melts and the final temperatureis above the melting point of ice. The energy lost by the water is Q-cyrTTlyr(Twt,-Tf)and the energy absorbed by the ice and the water it becomes when it melts is Q : cr?rLr(0 - Ty) + cyrTTLlQf - 0) * mrL7, .The first term is the energy required to raise the temperature of the ice to 0o C, the second termis the energy required to raise the temperature of the melted ice from 0"C to Ty, and the thirdterm is the energy required to melt all the ice. Since the two energies are equal, cyrTTlryy(Twt, - Ty) : ctTTLr?Tp,) + cyrwLTTf + TrLTLp .The solution for T1 is cwmwTwt* ctmtTtt - mtLP T1: cyr(mw + *iSubstitute given values to obtain T1 - 2.5oC. Chapter t I 111
116.
43The internal energy is the same at the beginning and end of a cycle, so the energy Q absorbedas heat equals the work done: Q = W. Over the portion of the cycle from A to B the pressurep is a linear function of the volume V and we may write p = a + bV, where a = (10/3) Pa andb = (20/3 Pa/m3 . The coefficients a and b were chosen so that p = 10 Pa when V = 1.0 m3 andp = 30 Pa when V = 4.0 m 3 . The work done by the gas during this portion of the cycle isThe BC portion of the cycle is at constant pressure and the work done by the gas is W BC =P ~ V = (30 Pa)(1.0 m3 - 4.0 m3 ) = -90 J. The CA portion of the cycle is at constant volume, sono work is done. The total work done by the gas is W = W AB + W BC + W CA = 60 J - 90 J + 0 =-30J and the total energy absorbed as heat is Q = W = -30J. This means the gas loses 30J ofenergy in the form of heat.49(a) The change in internal energy ~Eint is the same for path iaf and path ibf. According to thefirst law of thermodynamics, ~Eint = Q - W, where Q is the energy absorbed as heat and W isthe work done by the system. Along iaf ~Eint = Q - W = 50cal- 20 cal = 30 cal. Along ibfW = Q - ~Eint = 36cal- 30 cal = 6 cal.(b) Since the curved path is traversed from f to i the change in internal energy is - 30 cal andQ = ~Eint + W = -30 cal- 13 cal = -43 cal.(c) Let ~Eint = Eint, J - E int, i. Then E int , J = ~Eint + E int, i = 30 cal + 10 cal = 40 cal.(d) and (e) The work WbJ for the path bf is zero, so QbJ = E int, J - E int, b = 40 cal - 22 cal =18 cal. For the path ibf Q = 36 cal so Qib = Q - QbJ = 36 cal- 18 cal = 18 cal.51The rate of heat flow is given by TH-Tc P cond = kA L where k is the thermal conductivity of copper (401 W /m· K), A is the cross-sectional area (in aplane perpendicular to the flow), L is the distance along the direction of flow between the pointswhere the temperature is T Hand Tc. Thus P cond = (40 I W /m . K)(90.0 x 10- 4 m 2 )(l25°C - 10.0°C) = 1.66 X 103 J / s . 0.250mThe thermal conductivity can be found in Table 18-6 of the text.112 Chapter 18
117.
--65Let h be the thickness of the slab and A be its area. Then the rate of heat flow through the slabis p i 4A(TH-Tc) rcono - ) hwhere k is the thermal conductivity of ice, Tu is the temperature of the water (0C), and Tc isthe temperature of the air above the ice (-10"C). The energy leaving the water freezes tt, theenergy required to freeze mass m of water beitrg A : L pm, where L p is the heat of fusion forwater. Differentiate with respect to time and recogntze that dQ ldt: Pcod to obtain dm Pco,ro - LFE.Now the mass of the ice is given by m - pAh, where p is the density of ice and h is thethickness of the ice slab, so dm I dt - pA(dh I dt) and Prono : LppA#Equate the two expressions for Pro,,o and solve for dh I dt: dh k(Tn- Tc) dt LFphSince l calk- (0.0040caIls.cm.KX4.186Jlcal)10 x l0-2mf cm)- I.674W/-.K. The SIvalue forthe tdensity of ice is p:O.gzglt - 0 .g2 x 103 kgl^t.Thus dh x 0.40 cmfh. ^ls - 1.1 10-u dt (333 x 103 Ilkg(o.92 x 103 kel^Xo.osom)73(a) The work done in process 1 is Wt - .ptVt, so according to the first law of thermod5mamicsthe change in the internal energy is AEnt : A-Wr: lprV,-4.ptV - 6.}p,iV. The work donein process 2 canbe computed as the sum of the areas of a triangle (with altitude 3pnl2-pn: p,;f 2andbase 4.0V) andrectangle (with sidespi and 4.0V,): Wz: *Onl2)(4.0V)+4.prV - 5.p,iV.The change in the internal energy is the same so the energy transferred to the gas a heat isQ: LBint+Wz - 6.oprV + 5.op,iV(b) The change in internal energy is the same for all processes that start at state a and end atstate b, namely 6.}ptV.75At the colder temperature the volume of the disk is V - n R2 (., where R is its radius and (. is itsthickness. After it is heated its volume is n(ft+Aft)r({+ Ll): rR(,+rR2 L(,+2rR(, LR+. . .,where terms that are proportional to the products of small quantities are neglected. Thus the Chapter 18 113
118.
change in the volume is LV - nR2 L(. + 2rR(. LR. Now AR - Ra LT and L!,where LT is the change in temperature and a (: 3.2 x 10-6 x 10-u lC" from Table 18-2) isthe coefficient of lin ear expansion for Pyrex. Thus Lv - 3r R2 ta LT - 37r(0.0800 m),(0.00500 m) (3 .2 x 10- 6 c"x60.0oc - 10.0"c) - 4.83 x 10-8 m3 .77Let pt be the in itial pressure, V be the initial volume, and Vy be the final volume. Then thework done by th e gas is :23J.81 # pdv: t:: avz dv -- rri vn)-The magnitude of the energy transferred as heat from the aluminum is given by Q : ?rL4ct(Ta -D and the magnitude of the energy transferred as heat into the water is given by Q - mu)c*(T -T-), wher:- cs is the specific heat of aluminum (900 Ilkg.K from table 18-3), cu) is the specificheat of water (4190 J lkg. K from the same table) , Tt is the initial temperature of the aluminum,T* is the initial temperature of the water, and T is the common final temperature. The solutionfor T is Trl,gcaTA I TTL,C-T- T- rTLgCl * ffiusCu (2.50 kgX900 J lke. KX92.0C) + (8.00 kgXa190 J lke. KX5.00.C) (2.s0 kgX900 I lke. K) + (8.00 kgX4190 J lke. K) - 10.5oC .Note that it is not necessary to convert the temperatures to the Kelvin scale.83Let m be the mass of the ice cube and crce be its specific heat. Then the energy required to bringthe ice cube temperature to 0o C is Q: LT - (0.700 kg)(2220 J lkg . K)(150 K) - 2.331 x 10s J. TTLCiceThis is less than the 6.993 x 105 J that are supplied, so the ice cube is brought to OoC and all orpaft of it melts. The specific heat for ice can be found in Table 18-3.If L e is the neat of fusion for water, then the energy required to melt all the ice is mLp -(0.700kgx333 x 103 Ilkg-2.331 x 10sJ.This less than the 6.993 x 105 J 2.33I x 10s J -- 4.662 x 10s J that arc avatlable, so allthe ice melts. The heat of fusion for water can be found in Table l8-4. Now energy E -4.662x10sJ-2.33Ix10sJ-2.33IxlOsJisusedtoraisethetemperatureofthewaterfromOoC. The final temperature is ry1 t _ E 2.33Ix105J : /9-5oC " . TTLC*a1sy (0.700 kgX4190 J lkg. K)ll4 Chapter I B
119.
Chapter 197(a) Solve the ideal gas law pV : nRT for n. First, convert the temperature to the Kelvin scale:T - 4A.0 + 273. 15 - 313.15K. Also convert the volume to m3: 1000cm3 - 1000 x 10-6m3.Then pv (1.01 x 10spaX1000 x 10-6m3) _ 3.gg x l-zmol . RT- (8.3lJlmol .KX3 13.15K) TL:(b) Solve the ideal gas law pV : nRT for T: rn _ pv (1.06 x 10s paX1500 x 10-u rn) :493 1 - ^(., K _ .ono L. - 220" rr r.r "R-13Suppose the gas expands from volume V to volume during the isothermal portion of theprocess. The work it does is w: [Jvn pdv -nR, l:: Y -nRrtn,where the ideal gas law pV - nRT was used to replace p with nRT f V. Now Vand Vy - nRT lpt, so Vr lV: pi,lpr. Also replace nRT with ptV to obtain W:ptVlnU. pfSince the initial gauge pressure is 1 .03 x 10s Pa, pt: 1 .03 x 105 Pa+ 1 .01 3 x 105 Pa - 2.04 x 10s Pa.The final pressure is atmospheric pressure: pf - 1.013 x 105 Pa. Thus 204 x 105-Pa w : (2.04x 105 pa)(O .r40m3)rr, - 2.00 L x 104J. 1.013 x lOsPaDuring the constant pressure portion of the process the work done by the gas is W : p ilV -Vf).Notice that the gas starts in a state with pressure p f, so this is the pressure throughout this portionof the process. Also note that the volume decreases from Vy to V. Now Vy : pV lp f, so 14/-Pf(u #)-(pf-Pt)v (1.013 x 10spa - 2.04 x 105pa)(O.r40m3) - -l .44 x 104J.The total work done by the gas over the entire process is W5.60 x 103 J. Chapter 19 115
120.
t9According to kinetic theory the rrns speed iswhere T is the temperature on the Kelvin scale and M is the molar mass. See Eq. 19-34.According to Table 19-I, the molar mass of molecular hydrogen is 2.02 gf mol _ 2.02 x10-3 kg/mol, so 3(9.31 J/mol KX2._15) n] UrmS : 2.02 x 10-3 kg/mol - : 1.8 ^ x r02m/s .29(a) According to Eq. 19-25, the mean free path for molecules in a gas is given by 1 ) tDrd,lYlVwhere d is the diameter of a molecule and l/ is the number of molecules in volume I/. Substituted - 2.0 x 10-10 m and IY lV - 1 x 106 molecules/m3 to obtain )- -6 x 1012m. fr"(2.0 x 10-to tn;z(l x 106 m-3)(b) At this altitude most of the gas particles are in orbit around Earth and do not suffer random-rzrng collisions. The mean free path has little physical significance.35(a) The average speed is v fu T:- lr)where the sum is over the speeds of the particles and tf is the number of particles. Thus _ (2.0 + 3.0 +4.0+ 5.0+ 6.0 + 7.0 + 8.0+ 9.0 + 10.0 + 11.0)km/s _ u 6.5km ls. 10(b) The rrns speed is given by ol vrmsNow Du + (7.0) + (g .0)+ (9.0)2 + (10.0), + (11.0)2] km2 lr - 505 km lr ,116 Chapter 19
121.
SO 505km2 f s2 10 - 7.1 km fs.4l(a) The distribution function gives the fraction of particles with speeds between u and u * du, soits integral over all speeds is unity: f p(r)du:1. Evaluate the integral by calculating the areaunder the curve in Fig . 19 -24. The area of the triangular portion is half the product of the baseand altitude, or *oro. The area of the rectangular portion is the product of the sides, or &u0.Thus f p(r)du: *ouo+auy: ].orr, So lroro- l and auo-2f3.(b) The average speed is given by ol uavg IP@)duFor the triangular portion of the distribution P(u) : au lro and the contribution of this portion is du: au3 :0,,, 2 * lr"o " 3where 2l3us was substituted for a. P(r): ^a: in the a, rectangular portion and the contribution ofthis portion is : a, Juo ["0 Lu dn |er, - rfi): lrA:,uoThus 1)av ir, 1 uo _ I.Zug and uuueluo : 1.2,(c) The mean-square speed is given by ,kr- l*P@)duThe contribution of the triangular section is A^1. ,du: 4uo uo ,,J * 1," bThe contribution of the rectangular portion is o : n I,n" ,, du: I rsrd - rl) !ra: l raThus urms ffi - 1.3luo and ?rrms lro - 1.3.(d) The number of particles with speeds between 1.5us and 2uo is given by N [1."19,0 f g) du.The integral is easy to evaluate since P(u): a, throughout the range of integrattoll. Thus thenumber of particles with speeds in the given range is IYa(2.0u0 1.510) - 0.5lYaus- IVl3,where 2l3us was substituted for a. The fraction of particles in the given range is 0.33. Chapter 19 Il7
122.
45When the temperature changes by LT the internal energy of the first gas changes by nrCr LT,the internal energy of the second gas changes by nzCz LT, and the internal energy of thethird gas changes by ntCz LT. The change in the internal energy of the composite gas isLEin,specific heat of the mixture. Thus e- ntet*nzez+nzCz tnZ4n3T1,1 (2.40mo1)(12.0 J/mol K) + (1.50mo1)(12.8 J/mol . K) + (3.20mo1)(20.0J lmol K) 2.40mol + 1.50 mol + 3 .20mol - 15.8 J/mol . K .53(a) Since the process is at constant pressure energy transferred as heat to the gas is given bya - nCp LT, where n is the number of moles in the g&s, Cp is the molar specific heat atconstant pressure, and LT is the increase in temperature. For a diatomic ideal gas with rotatingmolecules Cp Q : 77 LT : mol)(8.3 IaJ lmol . KX60.0 K) - 6.98 x 103 J . ,nR ,(4.00See Table I9-3 for the expression for Cp.(b) The change in the internal energy is given by LEin - nCv LT, where Cv is the specificheat at constant volume. For a diatomic ideal gas with rotating molecules Cv - lrn, so LE,nt:1nn, LT : vJ lm,ol . KX60.0K) - 4.gg x 103 J. 2 ]f^.00mo1)(8.3See Table l9-3 for the expression for Cy.(c) According to the first law of thermodynamics, L4int: Q W , so W : A - AEi,,, - 6.98 x 103 J - 4.gg x 103 J - l.gg x 103 J .(d) The change in the total translational kinetic energy is LK :3=rn LT :t=fo.00mol)(8.3 14J lmol . KX60.0K) - 2.gg x 103 J. 22,67Let po be the density of air suffounding the balloon and p n be the density of hot air in theballoon. The buoyant force on the balloon has magnitude pogV, where V is the volume of theenvelope, and it is upward. The magnitude of the force of gravity is W + pngV , where W is theweight of the basket and envelope combined, and it is downward. The second term is the weight118 Chapter 19
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of the hot air. If ft., (: 2.67 kN) is the net force on the balloon then Fnet: pagv - W - pngV ,SO Pn9 - pogV -W - Fnt- (tt.qX/m3XZ.tS x 103 m3) - 2.45 x 103 N - 2.67 x lo3 N V 2.18 x 103 m3 - 9.55 N/*tThe ideal gas law tell us that the number of moles of hot air per unit volume is nlv - plRT,where p is the pressurc and T is the temperature. Multiply by the molar mass M of air to obt arnpn - pM lRT. The pressure is atmospheric pressure (1.01 x 105 Pa) so the temperature shouldbe n1 I _ pM (1.01 x 10s Pa)(0 .028 kg/mol) . t - Rpn - (8.3 r J lmol . KX9.55 N/m) -2<AL69The molar specific heat of a monatomic ideal gas is e v - (3 l2)R, where R is the universal gasconstant. Let n be the number of moles of gas and LT be the change in temperature. Then thechange in the internal energy is AEnt - nCv LT - (3 lz)nRLT - (3 l2)(2.00mo1)(8.31Jlmol .KXl5.0K) - 374J .Since the process is adrabatic the energy transferred as heat is Q : 0.According to the first law of thermodynamics the work done by the gas is W - A LEn--374 LSince the gas is monatomic the internal energy is translational kinetic energy. The numberof atoms is the product of the number of moles and the Avogadro constant: iV - nlYe(2.00 moI)( 6.02 x 1023 atoms lmol- 1) kinetic energyperatomis (374J)l(I .20 x l}z4atoms):3.11 x 1022Jfatom.71The mean free path is given by I - Iltnnd,2(lVlv),where d, is the diameter of a molecule andl/ is the number of molecules in volume V . According to the ideal gas law IY lV - p lkf, wherep is the pressure, T is the temperature on the Kelvin scale, and k is the Boltzmann constant.Thus -y- - J1nd2p J2"(290 x 10- 12 m)2(2.00 atm)(1.01 x 10s Palatm) "JThe average time between collisions is T - lruue, where uavs is the average speed of themo1ecu1es.Thisisgivenby,Uavg-@,whereRistheuniversa1gaSconstantandMis the molar mass (see Eq. 19-30). The molar mass of oxygen is 32.0 x 10-3 kg/mol (see table19- 1), so ol 8(8.31I lmI. KX400 K) uavg : r(32.0 x 10-3 kg/mol) -514mls. Chapter 19 ll9
124.
The average time between collisions is (7.31 x 10-8-)/(5 l amls) - L.42 x 10-e s and thefrequency of collision is the reciprocal of this or 7 .04 x 10e collisions/s.77(a) Let p,i be the initial pressure, V be the initial volume, p f be the final pressure, and Vy be thefinal volume. According to the ideal gas law pV : p fVf since the initial and final temperaturesare the same. Thus pr: #f Qzatm) - 8oatm. fiu,-(b) The final temperature is the same as the initial temperature: 300 K.(c) Since p : nRT f V , the work done by the gas is r, : [" p d,v : ry dv - nRr tn (+) : pivtn (+) J vn, I:, (32atm)(1.01 x 105 pafatmxl.0LXl.00 x 10-3m3 lL)tn(#i) :4.5x 103J.(d) Since the process is now adtabatic, ptv: : pfvi, wherc l is the ratio of the heat capacity atconstant pressure to the heat capacrty at constant volume. Since the gas is monatomtc ^l - 11667 .The final pressure is Pr: (vt - rY = Pt -- (r,L)tuu,r2atm):3.zatm. r, ) oor,(e) Let T,; be the initial temperature and Ty be the final temperature. Then according to the idealgas law nR: pVlft,: pfVr lTt and rf^: rr 4r,- (32 atmxlg-? (32-t*)(l oL) JVr (3oo K) : r2oK . ffi"(0 The first law of thermodynamics tells us that the change in the internal energy Afint is equalto the negative of the work done by the gas during an adiabatic process. The change in theinternal energy is L4rnt_ nCvlf T) -- (CvlR)(pf - p), where Cv is the molar specificheat for constant volume processes. The ideal gas law was used to write the equation in thelast form. For a monatomic ideal gas Cv: Ql2)R, So W- -L4int- -(3 lz)(p{t -ptv).Now pfvf - (3.2atm)(4.0LXl.01 x 10sPafatmxl.O x 10-3m3 lL)- 1.29 x 103J and ptV(32arm)(1.0LXl.01 x 10s Pulatm)(l.0 x 10-3m3 lL) - 3.23 x 103J, so w:-:(r.2gx l03J- 3.23x 103D:z.gx 103J.(g) Now 1 : 1.4, so pf120 Chapter 19
125.
(h) The final temperature is now tf^: rr-1 fi7n: (a6atmxa.O-L(300K): r70K. ffi" @JwL(i) The molar specific heat is Cv - 5 f2, so the work done by the gas is W: -]@fVf - ptV).Now prVr: (4.6atm)(4.0LXl.01 x l05Pulatm)(l.0 x 10-3m3 lL)-1.86 x 103J and pV isstill 3.23 x 103 J, so w : -:(1.86 x l03J - 3.23 x 103D: 3.4x 103J.81(a) Since a molecule must have some speed, the integral of P(u) over all value of u must be LThus Cuz d,u : (I l3)Crl: 1 lo"oand C - 3ro.(b) The average speed of the particles ls P(u)u d,u : 3rit u3 d,u - (3 l4)ro . Io"o lo"o(c) The average of the square of the speed is (rr)uus : P(u)u2 d, : 3u; o - (3 l s)r" lo"o Iro oso the nns speed is urms- l@r: (3 ls)r" 0.77 5us.83(a) According to the ideal gas law p : nRT f V , so the work done by the gas as the volume goesfrom Vt to Vy is tv: pd,v: I:, I:: rydv-nRrtn(V) - (3.s0mol)(8.3 rllmol .KX283K)ln f #) # ) -) +.oo LLLThe temperature was converted from degrees Celsius to kelvins.(b) The internal energy of an ideal gas does not change unless the temperature changes, soaccording to the first law of thermodynamics the energy transferred as heat is a-2.37 x 103 J. Chapter 19 l2I
126.
Chapter 20-3(a) Since the gas is ideal, its pressurc p is given in terms of the number of moles TL, the volumeV, and the temperature T by p : nRT fV . The work done by the gas during the isothermalexpansion is rvz rvz w: Jv, pdv: nRr Jn +:nRrI"# ISubstitute Vz - 2.0014 to obtain W : nRT In2 - (4.00 mol)(8.3 14J lmol . KX400 K) ln} - 9.22 x 103 J .(b) Since the expansion is isothermal, the change in entropy is given by A,S - /tt lT) dQ -Q lT , where A is the energy absorbed as heat. According to the first law of thermodyramics,L4rnt- A W . Now the internal energy of an ideal gas depends only on the temperature andnot on the pressure and volume. Since the expansion is isothermal, LEin - 0 and Q : W . Thus 9.22 x 103J :23.IJfK. AS: Y- 400 K(c) A,S - 0 for all reversible adiabatic processes.1(a) The energy that leaves the aluminum as heat has magnitude Q : Thaco(Toi, - Tf), where mais the mass of the aluminum, ca is the specific heat of aluminum, Toi is the initial temperatureof the aluminum, and Ty is the final temperature of the aluminum-water system. The energy thatenters the water as heat has magnitude Q: mu)c*(Tf -T*a), where mu) is the mass of the wateqcu) is the specific heat of water, and T-t is the initial temperature of the water. The two energiesare the same in magnitude since no energy is lost. Thus maco(Toi - Tf) : ffiusc-(Tf - T*t) and macoToi, * TfLuc-T-,i, T1 maco I TTLucuThe specific heat of aluminum is 900 J lk1. K and the specific heat of water is 4190 J lkg. K.Thus (0.200 kg)(900 J/kg . KX100 C) + (0.0500 kgx4190 J lke. KX20" C) Ty (0.200 kg)(900 J/kg . K) + (0.0s00 kg)(4190 I lke. K) - 5J.00 C .This is equivalent to 330 K.122 Chapter 20
127.
(b) Now temperatures must be given in kelvinsi Tot, - 393K, T*r:293K, andTy -330K. Forthe aluminum, dQ : maco d,T and the change in entropy is ASo : IE:ffio"" l::,r:TTlacotn+ - (0 .200kgxe0 0Ilkg.K)h:# - -22.rJfK. 373K(c) The entropy change for the water is AS- : IE:m1D"- J[: #:TTL,uc-rn+, T*i 33K : - (0.0500 kg)(41 90 I lke. K) 1,r 2g3K *24.g J fK.(d) The change in the total entropy of the aluminum-water system is A,S - AS" + AS- --22.1 J fK+ 24.9 I fK - *2.8 J/K.25(a) The efficiency is t-Tn-Tc Ts (235 + 273) KNote that a temperature difference has the same value on the Kelvin and Celsius scales. Sincethe temperatures in the equation must be in kelvins, the temperature in the denominator wasconverted to the Kelvin scale.(b) Since the efficiency is given by t0.236(6.30 x 104D:1.49 x 104J.29(a) Energy is added as heat during the portion of the process from a to b. This portion occursat constant volume (V), so Qrn - nCv LT, where Cv is the molar specific heat for constantvolume processes. The gas is a monatomic ideal g?s, so Cv - )n and the ideal gas law givesATgiven. We need to find po. Now po is the same as pc and points c and b are connected by anadiabatic process. Thus p"Vl : pbvl and po: pc: (#) ,r- (*) (1 013 x 106pa) - 3 .167 x 104paThe energy added as heat is 3 Qrn: t.013 x 106Pa -- 3.167 x 104PaXl.00 x 10-3 m3): r.47 x 103 J. z =f Chapter 20 123
128.
(b) Energy leaves the gas as heat during the portion of the process from c to a. This is a constantpressure process, so 5 5 Qout = nOp I::lT = 2(Pa Va - Pc Vc) = "2Pa(Va - Vc) = ~(3.167 X 104 Pa)(-7.00)(1.00 X 1O- 3 m3 ) = -5.54 X 102 J,where Op is the molar specific heat for constant pressure processes. The substitutions Va - Vc =Va - 8.00Va = -7.00Va and Op = ~R were made.(c) For a complete cycle, the change in the internal energy is zero and W = Q = 1.47 X 103 J -5.54 X 102 J = 9.18 X 102 1.(d) The efficiency is E = W/Qin = (9.18 X 102 J)/(1.47 X 103 J) = 0.624.37An ideal refrigerator working between a hot reservoir at temperature TH and a cold reservoirat temperature Tc has a coefficient of performance K that is given by K = Tc/(TH - Tc).For the refrigerator of this problem, TH = 96° F = 309K and Tc = 70° F = 294K, so K =(294 K)/(309 K - 294 K) = 19.6. The coefficient of performance is the energy Qc drawn fromthe cold reservoir as heat divided by the work done: K = IQcl/IWI. Thus IQcl = KIWI =(19.6)(1.0 J) = 201.39The coefficient of performance for a refrigerator is given by K = IQcl/IWI, where Qc is theenergy absorbed from the cold reservoir as heat and W is the work done by the refrigerator, anegative value. The first law of thermodynamics yields Q + Qc - W = 0 for an integer number Hof cycles. Here Q H is the energy ejected to the hot reservoir as heat. Thus Qc = W - Q Q H· His negative and greater in magnitude than W, so IQcl = IQHI -IWI. Thus K = IQHI-IWI IWI .The solution for IWI is IWI = IQHI/(K + 1). In one hour, IWI = 7.54MJ = 1.57MJ. 3.8 + 1The rate at which work is done is (1.57 x 106 J)/(3600 s) = 440 W.(a) Suppose there are nL molecules in the left third of the box, nc molecules in the center third,and nR molecules in the right third. There are N! arrangements of the N molecules, but nL!are simply rearrangements of the n L molecules in the right third, nc! are rearrangements of thenc molecules in the center third, and nR! are rearrangements of the nR molecules in the rightthird. These rearrangements do not produce a new configuration. Thus the multiplicity is N! W=----- nL! nc! nR!124 Chapter 20
129.
(b) If half the molecules are in the right half of the box and the other half are in the left half ofthe box, then the multiplicity is N! W B = -(N-/-;-2-)!-( N-/-:-2-)! .If one-third of the molecules are in each third of the box, then the multiplicity is N! WA = (N/3)! (N/3)! (N/3)! .The ratio is WA (N/2)!(N/2)! WB (N/3)! (N/3)! (N/3)! .(c) For N = 100, SO! SO! = 4.16 x 1016. 33!33!34!49(a) and (b) The most probable speed is given by v p = J 2RT/ M and the rms speed is given byV rms = J3RT / 111, where T is the temperature on the Kelvin scale, M is the molar mass, and Ris the universal gas constant. See Eqs. 19-34 and 19-3S. Thus Llv = (J3 - J2) JRT/M.According to Table 19-1 the molar mass of nitrogen is 0.028 kg/mol. For T = 2S0 K, (8.31 J /mol . K)(2S0 K) 8 / ---------,-----,---- = 7m s 0.028 kg/moland for T = SOO K, (8.31 J/mol· K)(SOOK) = l.2 x 102m/s 0.028 kg/mol(c) Thc energy transferred as heat when the temperature changes by the infinitesimal dT atconstant volume is dQ = nCv dT, where n is the number of moles and C v is the molar specificheat for constant volume processes. Thus the entropy change is LlS= J dQ - T = iTf --dT=nCvln (Tf) . T.2 nCv T - . T~Here Ti is the initial temperature and T f is the final temperature. Since nitrogen is diatomicwith rotating molecules C v = SR/2 (see Table 19-3), LlS = (S/2)nRln ( Tf) Ti = (S/2)(l.Smol)(8.31 J/mol· K)ln (SOOK) 2S0K = 22J/K.55The temperature of the ice is raised to O°C, then the ice melts and the temperature of the resultingwater is raised to 40°C. As the temperature of the ice is raised the infinitesimal dT the energy Chapter 20 125
130.
transferred to it as heat is dQ = mCice dT, where Cice is the specific heat of ice and Tn is the massof the ice. The entropy change is tJ.S 1 = J iTf dQ -T = Ti mCice - - dT = me In -T T ·lce (Tf) ~Table 18 - 3 gives the specific heat of ice as 2220 J /kg . K. The initial temperature on the Kelvinscale is Ti = -20° + 273 ° = 253 K and the final temperature is T f = 273 K, so 273 K) tJ.S 1 = (0.600kg)(2220J/kg· K)ln ( 253K = 101 J/K.The heat of fusion of water is L f = 333 x 103 J /kg, so the entropy change on melting is tJ.S2 = mL f = (0.600kg)(333 x 103 J/kg) = 732J/K. T 273KThe initial temperature of the water is Ti = 273 K and its final temperature is T f = 40° + 273° =313 K. The specific heat of water is Cwater = 4190 J /kg . K, so the change in the entropy of thewater as its temperature is raised is tJ.S3 = mCwater ln (i) = (0.600kg)(4190J/kg· K)ln (~~~~) = 344J/K.The change in entropy for the complete process is tJ.S = tJ.S 1 + tJ.S2 + tJ.S3 = 101 J /K + 732 J /K +344J/K = 1.18 x 103 J/K.63(a) The coefficient of performance of a refrigerator is given bywhere IQL I is the energy extracted as heat from the low temperature reservoir and IQH I is theenergy transferred as heat to the high temperature reservoir. In this case the low temperaturereservoir is the interior of the refrigerator and the high temperature reservoir is the room. Thesolution for IQHI is K +1 4.60 + 1 IQHI = IQLI- K = (35.0kJ) 4.60 = 42.6kJ.(b) Over a cycle the change in the internal energy of the system is zero, so according to the firstlaw of thermodynamics the work done per cycle is IWI = IQHI - IQLI. Thus K = IQLI/IWIand IWI = IQLI = 35.0kJ = 7.61 kJ. K 4.60126 Chapter 20
131.
67(a) and (b) If there are l/ particles in all, with n in one box and lf n in the other, themultiplicity is W - tf ! lnl(.Af n)t. The least multiplicity occurs for n -0 or n - .Af and isW:.^f!/ff!O! - 1. The greatest multiplicity occurs for rL: (ff - Dlz or n- (l/ + l) 12 and is W:." tt [(r lz)(n - 1)] !10 l2)(lr + 1)l ! For l[: 3 this is W :3!lIl2! : 3 and for l/: 5 this is W :5!12!3!: 10.(c) and (d) The entropy is given by ,S - klnW , where k is the Boltzmann constant. The greatestentropy occurs if the multiplicity is the greatest. For lf1.5 x T0-23 JIK and for lr:5 it is Sr - (1.38 x 10-23 JllfJln 10 - 3.2 x 10-23 JlK. Chapter 20 127
132.
Chapter 2L1The magnitude of the force that either charge exerts on the other is given by n- 1 r +r..,lqtllqrl T2where r is the distance between them. Thus lqtllqrl 4nesF t- . /(s.gg x 10eN.m2 lc)(26.0 x 10-6c)(47.0 x 10-6c) 1. - t/ v 5.70 N5The magnitude of the force of either of the charges on the other is given by r - 1 n q(Q-q) 4"r, rzwhere r is the distance between the charges. You want the value of g that maximizes the functionf (q) - q(Q - il. Set the derivative df ldq equal to zero. This yields A - 2q:0, or q: Q12,7Assume the spheres are far apart. Then the charge distribution on each of them is sphericallysymmetric and Coulombs law can be used. Let q1 and q2 be the original charges and choose thecoordinate system so the force on ez is positive if it is repelled by qt Take the distance betweenthe charges to be r. Then the force on ez is 1 $Qz Fo -- 4nes 12The negative sign indicates that the spheres attract each other.After the wire is connected, the spheres, being identical, have the same charge. Since chargeis conserved, the total charge is the same as it was originally. This means the charge on eachsphere is (q + q) f 2. The force is now one of repulsion and is given by r;1 I (q+q)2 b r 4" r, 4rz128 Chapter 2l
133.
Solve the two force equations simultaneously for h and e2. The first gives :- QtQz w -4treor| Fo g.gg x 10eN. nf lcand the second gives 0.0360 N et+qz-2r 4nesF6 : 2(0.500 m) - 2.00 x 10-6 C. 8.99 x 10eN . m2 lCThus -(3.00 x 10-tz gzS Qz: Qtand substitution into the second equation gives -3.00 x 10-12 g2 Qr+ Qr - 2.00 x 10-6 C.Multiply by qt to obtain the quadratic equation q? - e.oo x lo-u c)q, - 3.oo x lo-12 g2:0.The solutions are 2.00 x 10-6C+ (-2.00 x 10-e C)2 + 4(3.00 x 10-12 C2) Qr:If the positive sign is used, $:3.00 x 10-e g and if the negative sign is used, $_ -1.00 x10-o g. IJse Qz-1.00 x 10-o g and rf qtidentical, the solutions are essentially the same: one sphere originally had charge -1.00 x 10-e gand the other had charge +3.00 x 10 -6 C.19If the system of three particles is to be in equilibrium, the forceon each parlicle must be zero. Let the charge on the third particleb. qo. The third particle must lie on the r axis since otherwise thetwo forces on it would not be along the same line and could notsum to zero. Thus the A coordinate of the particle must be zero. <_ f __+<_ L_f ___+The third particle must lie between the other two since otherwisethe forces acting on it would be in the same direction and would eo 4.00qnot sum to zero. Suppose the third particle is a distance r fromthe particle with charge e, as shown on the diagram to the right.The force acting on it is then given by F^- 1 lqqo ro 4"rrl7 a.ooqqof_n @)-uwhere the positive direction was taken to be toward the right. Solve this equation for Canceling r.conlmonfactorsyields Ll*- 4.00 lQ-n)2 andtakingthe squarerootyields -2.00 r). Il* l(L-The solution is r :0.333L. Chapter 21 129
134.
The force on qis q- I lry* F-:4"r,LA* looq1 - n Lz ):uSolve for Q0: -4.00qr2 1fz: -0.444q, where eo: r -- 0.333L was used.The force on the particle with charge 4.00q is nD: I 14.00q2 4.00qqol 1 14.00q2 4.00(0.444)q 1l 4q 4resL L2 e-02) 4resL L2 (0.444)L2 r 4res L L2 L2 IWith eo: -0.444q and r :0.333L, all three charges are in equilibrium.25(a) The magnitude of the force between the ions is given by r-., q2 -r @,where g is the charge on either of them and r is the distance between them. Solve for the charge: 3.7 x 10-e 5 q : r {4?t€oF - (5.0 x 1o-to -) :3.2 x 10-1eC. 8.99 x 10eN . m2 lC(b) Let lf be the number of electrons missing from each ion. Then IY e : q and 17 q .// -rv 3.2x10-leC tl - -L. ;3s(a) Every cesium ion at a corner of the cube exerts a force of the same magnitude on the chlorineion at the cube center. Each force is a force of attraction and is directed toward the cesium ionthat exerts rt, along the body diagonal of the cube. We can pair every cesium ion with another,diametrically positioned at the opposite corner of the cube. Since the two ions in such a pairexert forces that have the same magnitude but are oppositely directed, the two forces sum to zeroand, since every cesium ion can be paired in this way, the total force on the chlorine ion is zeto.(b) Rather than remove a cesium ion, superpose charge -e at the position of one cesium ion.This neutralizes the ion and, as far as the electrical force on the chlorine ion is concerned, it isequivalent to removing the ion. The forces of the eight cesium ions at the cube corners sum tozeto, so the only force on the chlorine ion is the force of the added charge.The length of a body diagonal of a cube is l8o, where a is the length of a cube edge. Thus thedistance from the center of the cube to a corner is d - (tftlD".The force has magnitude r I e2 1 e2 D_ 4"r, d "q@ (g.gg x 10eN.m2/c2x1.6-0 x 10-lec)2 - _ L.gx 10_eN. (3 l4)(0.40x 10-g m)2130 Chapter 2l
135.
Since both the added charge and the chlorine ion are negative, the force 1s one of repulsion. Thechlorine ion is pulled away from the site of the missing cesium ion.37None of the reactions given include a beta decay, so the number of protons, the number ofneutrons, and the number of electrons ate each conserved. Atomic numbers (numbers of protonsand numbers of electrons) and molar masses (combined numbers of protons and neutrons) canbe found in Appendix F of the text.(a) tg has 1 proton, I electron, and 0 neutrons and eBe has 4 protons, 4 electronse and94neutrorls. One of the neutrons is freed in the reactioll. X must be boron with a molar mass of5 elmol + 4 elmol - 9 gf mol: eB.(b) tzc has 6 protons, 6 electrons, and 12 - 6:6 neutrons and lH has 1 proton, 1 electron, and0 neutrons, so X has 6 + 1nitrogen with a molar mass of 7 glffiol + 6 glmol - 13 gf mol: 13N.(c) 15N has 7 protons, 7 electrons, and 15 7-8 neutrons; tg has 1 proton, I electron, and 0neutrons; and aHe has 2 protons, 2 electrons, and 4 2protons, 6 electrons, and 8+0 2-6 neutrons. Itmustbe carbonwith a molarmass of6elmol+ 6el^ol - 12glmoI: t2C.39The magnitude of the force of particle I on particle 2 is 1 F - 4nes lqtllqrl a?,+ a?r The signs of the charges are the same, so the particles repel each other along the line that runsthroughthem.Thislinemakesaf|angle0withtheraxissuchthatcoS0-d,zl|m,SothelL component of the force is F* 24(1 .60 x 10-ro 97z(6.00 x 1o--) - (g .gg x 10e c/N .m2) [(2.00 x 10-t *) + (6.00 x 10-z m)2]3/z -1.31 x 10-22N50The magnitude of the gravitational force on a proton near Earths surface is mg, where m isthe mass of the proton (I .67 x 10-27 kg from Appendix B). The electrostatic force between twoprotons is F - Q lanro)(" I d,), where d is their separation. Equate these forces to each otherand solve for d. The result is (t--,1 t ez (8.99 x 10-eY .m2 lC) ( 1 .60 x 10- te C)2 - 0.119m. V 4Tes mg (I .67 x 10- 27 kg)(9.8 ^lt) Chapter 2 I 131
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60The magnitude of the force of particle 1 on particle 4is17 I lqllqolr1 4"rrT - (8 .99 x roe N .m2 lcr(320 x lo-tn cX3.2o x lo-tq c) (0.0300 m)2The charges have opposite signs, so the particles attract each other and the vector force is Fr: -(1 .02-24 N)(cos 35.0o) i - (r.02 x 10-24 N)(sin 35.0") j - -(g .36 x 1o-" N) i - (5.g5 x 1o-" N) j.Particles 2 and 3 repel each other. The force of particle 2 on particle 4 is I lqrllqol ? F2 zn) J 4neg d; x lo-le cX3.2o x 1o-tn c) x loeN . m2 lcrr(3.20 -l (0.0200 m)2 " - -(2.30 x l0-to $ jParticles 3 and 4 repel each other and the force of particle 3 on particle 4 is F3 1 l2qtllqol , 4Tes di x 10-1e-qx3.20-x 10-1e c) _ x -l t0eN. m2 lcr)(6.40 -(4.60 x 10-24 C) ? . (0.0200 m)2The net force is the vector sum of the three forces. The r component is F"4.60N - -5.44 x 10-24N and the A component is Fa: -5.85 x 10-2sN - 2.30 x I0-24NI:-2.89 x 10-24N. The magnitude of the force is F3 + F?a (-5 .44 x 10-24 N)2 + ( -2.89 x L0-24N)2 - 6.T6 x 1024 N.The tangent of the angle 0 between the net force and the positive r axis is tan? : F, I F" :(-2.89 x 10-24 N/(-5 .44 x 10-z+N) - 0.531 and the angle is either 28" or 208". The laterangle is associated with a vector that has negative r and A components and so is the correctangle.69The net force on particle 3 is the vector sum of the forces of particles 1 and 2 and for this tobe zero the two forces must be along the same line. Since electrostatic forces are along the linesthat join the particles, particle 3 must be on the r axis. Its A coordinate is zero.Particle 3 is repelled by one of the other charges and attracted by the other. As a result, particle3 cannot be between the other two particles and must be either to the left of particle 1 or to theright of particle 2. Since the magnitude of gr is greater than the magnitude of qz, partrcle 3 must132 Chapter 2l
137.
be closer to particle 2 than to particle I and so must be to the right of particle 2. Let r be thecoordinate of particle 3. The the r component of the force on it is r-, Ln a-r:l-r-l I I %et ezez I 4Tes L 12 @_ DlIf F" - 0 the solution for :r is -erlq, -(-5.00q) lQ.00q) L 2.72L - . -Qr I q, -(-5.00q) lQ.00q) - Chapter 21 133
138.
Chapter 223Since the magnitude of the electric field produced by a point particle with charge q is givenby E- lqlla"eor2, where r is the distance from the particle to the point where the field hasmagnitude E, the magnitude of the charge is lql -4neor2|- :5.6x10-llc.5Since the charge is uniformly distributed throughout a sphere, the electric field at the surface isexactly the same as it would be if the charge were all at the center. That is, the magnitude ofthe field is 7,-.,- q n 4Tr€oHwhere q is the magnitude of the total charge and R is the sphere radius. The magnitude of thetotal charge is Z e, so E _ z" - (g.gg x 10eN .T2/c2x94x1.q0 x 10-1e c) _ 3.07 x t02].{/c. 4TesRz (6.64 x 10-15 m)2The field is normal to the surface and since the charge is positive it points outward from thesurface.7At points between the partrcles, the individual electric fields arein the same direction and do not cancel. Charge ez has a greater dPmagnitude than charge er, so a point of zero field must be closer Qz Qrto % than to e2. It must be to the right of qr on the diagram.Put the origin at the particle with charge ez and let r be the coordinate of P, the point wherethe field vanishes. Then the total electric field at P is given by E-*lp #lwhere % and Qz are the magnitudes of the charges. If the field is to vanish, Qz: Qt rz (r - d)2L34 Chapter 22
139.
Take the square root of both sides to obtain Viii/x = vfiil / (x - d). The solution for x is x-( Viii Viii - vfiil )d-( V yI4.Oq1vfiil )d 4 .O - ql = ( 2.0 ) d = 2.0d = (2.0)(50 cm) = 100 cm. 2.0 - 1.0The point is 50 cm to the right of ql .2Choose the coordinate axes as shown on the diagram ,Y , / :r: /to the right. At the center of the square, the electric q, a ~2qfields produced by the particles at the lower left and ,, / /upper right comers are both along the x axis and each , /points away from the center and toward the particle ,d , ,/ / / /that produces it. Since each particle is a distance d = a /, ,V2a/2 = a/ Vi away from the center, the net field due / / ,,to these two particles is / ,d / / ,, / -g ,, 2q / / , = _1_ _ q_ = (8.99 x 109 N· m2/C 2)(1.0 x 10- 8 C) = 7.19 x 104 N/C. 41rEo a2 /2 (0.050 m)2/2At the center of the square, the field produced by the particles at the upper left and lower rightcomers are both along the y axis and each points away from the particle that produces it. Thenet field produced at the center by these particles is Ey = 1 [2q q 1 4 - 2/2 - ~/ = -4- q 1 4 2/2 = 7.19 x 10 N/C. KEO a a 2 KEO aThe magnitude of the net field isand the angle it makes with the x axis is E () = tan- I ~ = tan- l (l) = 45° . ExIt is upward in the diagram, from the center of the square toward the center of the upper side.21Think of the quadrupole as composed of two dipoles, each with dipole moment of magnitudep = qd. The moments point in opposite directions and produce fields in opposite directions at Chapter 22 135
140.
points on the quadrupole axis. Consider the point P on the axis, a distance z to the right of thequadrupole center and take a rightward pointing field to be positive. Then the field producedby the right dipole of the pa.r is qdlLnro(z d,l2)t and the field produced by the left dipoleis -qdlTTes(z * d,lD3. Use the binomial expansions (z dlZ-z r z-3 3z-a(-dlz) and(z + dlz)-t N z-3 - 32-4@12) to obtain E- qd l-1 3d1 I 3d1 6qdz 2nes le* 2r^ 7 2ro l 4n esz4Let Q :2qd2. Then 3Q E- 4Tesza27(a) The line ar charge density ) is the charge per unit length of rod. Since the charge is uniformlydistributed on the rod, l- -qlL -(4.23 x 10-tsC) 1Q.0815m) -5.19 x 10-14Cfm. - -(b) and (c) Position the origin at the left endof the rod, as shown in the dtagtarct Let dr drP obe an infinitesimal length of rod at r. The tlllcharge in this segment is dq - ), dr. Since the 0 n L L+asegment may be taken to be a point partrcle, theelectric field it produces at point P has only ann component and this component is given by ), dr dE"- 4Tes(L+a-r)zThe total electric field produced at P by the whole rod is the integral E,: dr llt hlr" (L + a, - r)2 4Tes L+"-*lo :^ 4"r, t1 lA ll L + a) l: ) 4nes a,(L + a) LWhen - q I L is substituted for the result is ^ E* 1 q (8.99 x 10eN. m2 lCX4 .23 x 10-ts C) 1.57 x 1o- N/C . 4nes a(L + a) (0.120 m)(0.0815 m + 0.120 m)The negative sign indicates that the field is toward the rod and makes an angle of 180" with the rpositive direction.(d) Now E*-_ 4res a(L q a)-_(S.qqx (50 mX0.0815 m + 50 m) 1 + tOeX.lt2/C]X+.ZJ_xtO-t5C)__l .szx 10-*N/C.136 Chapter 22
141.
(e) The field of a point particle at the origin is : q (8.99 x loeN . m2 lCX4 .23 x lo-ts C) E* 4Tesa2 (50 m)2 - -1.52 x 10-tN/C.35At a point on the axis of a uniformly charged disk a distance z above the center of the disk, themagnitude of the electric field iswhere R is the radius of the disk and a is the surface charge density on the disk. See Eq. 22-26.The magnitude of the field at the center of the disk (zfor the value of z such that ElE.- Il2. This means E ,21 E" Jz2+R2 2or z2+R2 2Square both sides, then multiply them by z2 + R2 to obtatn z2 - (22 l4)+(R2 lq. Thus z2 - R13and z- Rlrn -(0.600m)lrn -0.346m.39The magnitude of the force acting on the electron is F - €8, where E is the magnitude of theelectric field at its locatiorl. The acceleration of the electron is given by Newtons second law: ! : eE - (1.60 x 10-1e CX2.0O x l04N/C) _ a_ 3.51 x 101 *^, . m m 9.11 x 10-gt kg43(a) The magnitude of the force on the particle is given by F - qE, where q is the magnitude ofthe charge carried by the particle and E is the magnitude of the electric field at the location ofthe particle. Thus E-!_3.ox1o-l N - 1F x 103 N/C. q 2.0 x 10-eC l5The force points downward and the charge is negative, so the field points upward.(b) The magnitude of the electrostatic force on a proton is F": eE - (1.60 x 10-tn cxl.5 x 103 N/c) : 2.4 x l0-16 N .(c) A proton is positively charged, so the force is in the same direction as the field, upward. Chapter 22 137
142.
(d) The magnitude of the gravitational force on the proton is Fs : mg - (1.67 x 1o-" kg)(9.8 mlst) - 1,.64 x ro-26N.The force is downward.(e) The ratio of the force magnitudes is F"- t,ot:,T Fs 0, " I.64x10-26N - 1.5 Lv x roro45(a) The magnitude of the force acting on the proton is F - e E, where E is the magnitudeof the electric field. According to Newtons second laW the acceleration of the proton isa, - F l* - eE lm) where m is the mass of the proton. Thus (1.60 x 10-te CX2.00 x 104 N/C) _ a* L67 x 10-zt kE l.gZ x 101 , ^l(b) Assume the proton starts from rest and use the kinematic equation u2 ,E + Zar (or elser, - iot and n : at) to show that ?): r - l@grx 1or, : r.g6x losm/s.57(a) If q is the positive charge in the dipole and d, is the separation of the charged particles, themagnitude of the dipole moment is p: qd: (1.50 x 10-n CX6.20x 10-u*) - 9.30 x 10-15 C.ril.(b) If the initial angle between the dipole rnoment and the electric field is 0s and the final angleis 0, then the change in the potentral energy as the dipole swings from 0 : 0 to e - 1 80o is L(J : -pU(cos 0 - cos 0o): -(9.30 x 10-15 C.mX1100Nic)(cos l80o - cos0) - 2.05 x 10-11 J.79(a) and (b) Since the field at the point on the r axis with coordinate tr :2.0 cm is in the positiver direction you know that the charged particle is on the r axis. The line through the pointwith coordinates r - 3.0cm and y - 3.0cm and parallel to the field at that point must passthrough the position of the particle. Such a line has slope (3.0)l(4.0) - 0.75 and its equation isy:0.57 + (0.75)r. The solution for A - 0 is tr: -1.0cm, so the particle is located at the pointwith coordinates r: -1.0cm and A - 0.138 Chapter 22
143.
(c) The magnitude of the field at the point on the r axis with coordinate tr : 2.0 cm is given byE - Qlanrilql(2.0cm - r)2, So q:4nesr2E -(0.020m+0.010m)2(100-N/c) _1.0 x 10-n c. 8.99x10eN.nflC81(a) The potential energy of an electric dipole with dipole moment f in an electric field E is : _L lo:-l::,; Tffi ;]f;:T;l :11 :, :::1 ;1;THere we used d,.6: (trnb** eabai arb, to evaluate the scalar product.(b) The torque is F-FxE-(p*i*prilx (tr*i) - -patr*k - -(4.00Xr.24 x 10-30C.mX4000N/C) - -(1.98 x 10-26N.m)t.(c) The work done by the agent is equal to the change in the potential energy of the dipole. Theinitial potential energy is Ut, - -1.49 x 10-26 J, as computed in part (a). The final potentialenergy is u7:: ,1o,ll1,1f .*h1;Hil:1,:IT :l-,0-26 ITheworkdonebythe agent is W: (1.98 x L0-26D-(-1.49 x I0-26D:3.47 x 10-26J. Chapter 22 139
144.
Chapter 231The vector area A and the electric field E are shown on the dia-gram to the right. The angle 0 between them is 180o 35o : L45" ,so the electric flux through the area is O -- E.A- EAcosd -(1800N/C)(3.2 x 10-t cos l45o- -1.5 x l0-2 N. m2 f C. ^)2Let A be the area of one face of the cube, Eu be the magnitude of the electric field at the upperface, and E2 be the magnitude of the field at the lower face. Since the field is downward, theflux through the upper face is negative and the flux through the lower face is positive. The fluxthrough the other faces is zero, so the total flux through the cube surface is OThe net charge inside the cube is given by Gauss law: n:ll rr x 10-t2c2lN m2x100m)2(I00N/c - 60 oN/c) .f:L:?:^l5t9(a) The charge on the surface of the sphere is the product of the surface charge density o andthe surface area of the sphere (4nr2, where r is the radius). Thus q:4nr2o:4n e)t,r., x 10-6 c l^) : 3.7 x r0-5 c. #(b) Choose a Gaussian surface in the form a sphere, concentric with the conducting sphere andwith a slightly larger radius. The flux through the surface is given by Gauss law: q- o- €e :4.rx1o6N.mzfc. 8.85 x 10-t2 Cz/N .m/23The magnitude of the electric field produced by a uniformly charged infinite line is E :where l is the linear charge density and r is the distance from the line to the point where the ^127r€sr,field is measured. See Eq. 23-12. Thus A - )TegEr - 2n(8.85 x l0-12 gz/N .mzx4.5 x l04N/c)(2.0m):5.0 x 10-u cl*. _F140 Chapter 23
145.
27Assume the charge density of both the conducting rod and the shell are uniform. Neglect fringing.Symmetry can be used to show that the electric field is radial, both between the rod and theshell and outside the shell. It is zero, of course, inside the rod and inside the shell since they areconductors.(a) and (b) Take the Gaussian surface to be a cylinder of length L and radius r, concentricwith the conducting rod and shell and with its curved surface outside the shell. The area of thecurved surface is 2rr L The field is normal to the curyed portion of the surface and has uniformmagnitude over it, so the flux through this portion of the surface is O - 2rrLE, where E is themagnitude of the field at the Gaussian surface. The flux through the ends is zero. The chargeenclosed by the Gaussian surface is Qr - 2.00Q1 : -Qr. Gauss law yields 2TresLE: -Qr,so x .ml/q2x3.10_x 10-12 c) E_ _g- (8.99 10e_+r - _0 .2r4N/c. ZresLr (11.00 m)(26.0 x 10-3 *)The magnitude of the field is 0 .2L4N/C. The negative sign indicates that the field points inward.(c) and (d) Take the Gaussian surface to be a cylinder of length L and radius r, concentric withthe conducting rod and shell and with its curved surface between the conducting rod and theshell. As in (a), the flux through the curved portion of the surface is O : 2nr LE, where E isthe magnitude of the field at the Gaussian surface, and the flux through the ends is zero. Thecharge enclosed by the Gaussian surface is only the charge Qr on the conducting rod. Gausslaw yields ZTesrLE : Qr, so r 2(8.99 x 10e NI . nf lCX3 .40 x l0- rz C) E- Q ffi- -+0.SssN/C.The positive sign indicates that the field points outward.(e) Consider a Gaussian surface in the form of a cylinder of length L with the curved portionof its surface completely within the shell. The electric field is zero at all points on the curvedsurface and is parallel to the ends, so the total electric flux through the Gaussian surface is zeroand the net charge within it is zero. Since the conducting rod, which is inside,,the Gaussiancylinder, has charge Q1 , the inner surface of the shell must have charge -At : -3.40 x 10-12 C.(0 Since the shell has total charge -2.00Q1 and has charge -Ar on its inner surface, it musthave charge -At : -3.40 x 10-12 C on its outer surface.35(a) To calculate the electric field at a point very close to the center of a large, uniformly chargedconducting plate, we may replace the finite plate with an infinite plate with the same area chargedensity and take the magnitude of the field to be E - o leo, where o is the area charge densityfor the surface just under the point. The charge is distributed uniformly over both sides of theoriginal plate, with half being on the side near the field point. Thus q 6.0x10-og a. o:il-ffi -_4.69 x lo-4cl^ Chapter 2 3 l4l
146.
The magnitude of the field is _ (j _ 4.69 X 10- 4 C/m2 _ 7 E- EO - 8.85 x 1O-12C 2 /N.m 2 -5.3 x 10 N/C.The field is normal to the plate and since the charge on the plate is positive, it points away fromthe plate.(b) At a point far away from the plate, the electric field is nearly that of a point particle withcharge equal to the total charge on the plate. The magnitude of the field is E = qI 411"Eor2, wherer is the distance from the plate. Thus E = (8.99 X 109 N . m 2 IC 2 )(6.0 x 10- 6 C) = I (30m)2 60N C.41The forces on the ball are shown in the diagram to the right. The grav- Titational force has magnitude mg, where m is the mass of the ball; theelectrical force has magnitude qE, where q is the charge on the ball andE is the electric field at the position of the ball; and the tension in the qEthread is denoted by T. The electric field produced by the plate is normalto the plate and points to the right. Since the ball is positively charged, mgthe electric force on it also points to the right. The tension in the threadmakes the angle () (= 30°) with the vertical.Since the ball is in equilibrium the net force on it vanishes. The sum of the horizontal componentsyields qE - T sin () = 0 and the sum of the vertical components yields T cos () - mg = O. Theexpression T = qE I sin (), from the first equation, is substituted into the second to obtainqE = mgtan().The electric field produced by a large uniform plane of charge is given by E = (j 12Eo, where (jis the surface charge density. Thus q(j - = mgtan() 2Eoand 2Eomgtan () (j = -----=----- q 2(8.85 10- 12 C 2IN· m2)(1.0 x 10- 6 kg)(9.8 m/s2 ) tan 30° X 2.0 x 10- 8 C = 5.0 X 10- 9 C/m2 .45Charge is distributed uniformly over the surface of the sphere and the electric field it produces atpoints outside the sphere is like the field of a point particle with charge equal to the net chargeon the sphere. That is, the magnitude of the field is given by E = qI411"Eor2, where q is the142 Chapter 23
147.
magnitude of the charge on the sphere and r is the distance from the center of the sphere to thepoint where the field is measured. Thus : m)2(3.0x 103 N/c) : q 4nesr2E _(0.15 7 .5x l0- e c . 8.99 x 10eN. m2 lCThe field points inward, toward the sphere center, so the charge is negative: - 7 .5 x 10-e C.49To find an expression for the electric field inside the shell in terms of A and the distance fromthe center of the shell, select A so the field does not depend on the distance.I-Jsea Gaussian surface in the form of a sphere with radius r s, concentric with the sphericalshell and within it (a <distance r s from the shell center.The charge that is both in the shell and within the Gaussian sphere is given by the integralQensdistribution has spherical symmetry, we may take dV to be the volume of a spherical shell withradius r and infinitesimal thickness dr: dV - 4rr2 dr. Thus eenc - 4tr I_* prz dr - on I_* | * dr:4nA I_ r dr:2nArr? - a2).The total charge inside the Gaussian surface is q + eenc : q + 2r A(r? - a2).The electric field is radial, so the flux through the Gaussian surface is O - 4rrln, where tr isthe magnitude of the field. Gauss law yields 4r esUr? : q + 2n Afr? - a2) .Solve for E: E-+l++2nA-ryI 4nes l,ri rznFor the field to be uniform, the first and last terms in the brackets must cancel. They do ifq-2nAa2 - 0 or A- qfLra2: (45.0 x 10-ts C)l2r(2.00 x I0-2m)2: l.7g x 10-tt Cl^2.s9(a) The magnitude tr1 of the electric field produced by the charge q on the spheri cal shell isEt: ql4T€oRZ, where Ro is the radius of the outer surface of the shell. Thus (450 N/CXo.2o m)2_ q- reohR - : 2.0 x 10- s C 8.99 x 10eN . nP lC(b) Since the field at P is outward and is reduced in magnitude the field of Q must be inward. Ais a negative charge and the magnitude of its field at P is E2:450N/C - 180N/C :270N/C.The value of Q is (?70 N/cX020 m)2 e : 4nesEzRZ:- 8.99 x . - - - L/ 10-e c 1.2x . 10eN. m2 lqz Chapter 23 143
148.
(c) Gauss law tells us that since the electric field is zero inside a conductor the net charge insidea spheri cal surface with a radtus that is slightly larger than the inside radius of the shell must bezero. Thus the charge on the inside surface of the shell is +L2 x 10-e C.(d) The remaining charge on the shell must be on its outer surface and this is 2.0 x lQ-e g1.2 x 10-rg:*0.8 x 10-eC.69(a) Draw a spheri cal Gaussian surface with radius r, concentric with the shells. The electricfield, if it exists, is radtal and so is norrnal to the surface. The integral in Gauss law isf E . d,A - 4nrE, where E is the radial component of the field. For r <is zero. Gauss law gives 4nr2E:0, so E:0.(b) For a <4nrz E - qo I eo and E - qo l4lreor2 .(c) For r >and E - (qo + qd l4nesr2 .(d) Consider first a spherical Gaussian with radius just slightly greater than a. The electric fieldis zero everywhere on this surface, so according to Gauss law it encloses zero net charge. Sincethere is no charge in the cavity the charge on the inner surface of the smaller shell is zero. Thetotal charge on the smaller shell is eo and this must reside on the outer surface. Now considera spherical Gaussian surface with radius slight larger than the inner radius of the larger shell.This surface also encloses zero net charge, which is the sum of the charge on the outer surfaceof the smaller shell and the charge on the inner surface of the larger shell. Thus the charge onthe inner surface of the larger shell is -eo. The net charge on the larger shell is Qa, with -Qoon its inner surface, so the charge on its outer surface must be eu - ( -eo) - eu + ea.76(a) The magnitude of the electric field due to a large uniformly charged plate is given by o l2ro,where o is the surface charge density. In the region between the oppositely charged plates thefields of the plates are in the same direction, so the net field has magnitude Eelectrical force on an electron has magnitude eE - eo f es and the gravitational force on it isffig, where m is it mass. If these forces arc to balance, they must have the same magnitude, somg:eofesand m9€o (9.11 x 10-" kg)(9.8 mls2x8.85 x 10-12 gz/N.m2) o: : 4.9 x Io-22 C lrrl e 1.60 x 10-le C(b) The gravitational force is downward, so the electrical force must be upward. Since an electronis negatively charged the electrical force on it is opposite to the electric field, so the electric fieldmust be downward.79(a) Let A be the net charge on the shell, % be the charge on its inner surface and eo be thecharge on its outer surface. Then Q - Qt * eo and et - Q - eo - (-10lrC) - (-1a pC) - *4 p,C.144 Chapter 2 3
149.
(b) Let q be the charge on the particle. Gauss law tells us that since the electric field is zeroinside the conducting shell the net charge inside any spherical surface that entirely within theshell is zero. Thus the sum of the charge on the particle and on the inner surface of the shell iszero, so q + qi = 0 and q = -qi = -4 fLC. Chapter 23 145
150.
Chapter 243(a) An ampere is a coulomb per second, so / 84A.h- (roch (roooi) :30x 1o5c T)(b) The change in potential energy is Ltl : qLV - (3.0 x 10sq(12V):3.6 x 106J.-5ifr. electric field produced by an infinite sheet of charge has magnitude E - of Zes, where o isthe surface charge density. The field is normal to the sheet and is uniform. Place the origin of acoordinate system at the sheet and take the r axis to be parallel to the field and positive in thedirection of the field. Then the electric potentral is v:v, [" rd,r:vr-Er, t-Jv Jowhere V, is the potentral at the sheet. The equipotential surfaces are surfaces of constant r; thatis, they are planes that are parallel to the plane of charge. If two surfaces are separated by Lrthen their potentials differ in magnitude by LV : ELr - (ol2eo)A". Thus 2(8.85 x 10-12c2/N.m:X50v) Lr:2eoLv- o - g.g x 10_3m. 0.10 x 10-6clm2t9(a) The electric potential V at the surface of the drop, the charge q on the drop, and the radius -R of the drop are related by V qf 4res.R. Thus R-#,(b) After the drops combine the total volume is twice the volume of an original drop, so theradius R of the combined drop is given by (n)3 - 2R3 and R - 21 /3 R. The charge is twice,the charge of original drop: q : 2q. Thus q: 2q _22/tV_22/t(5oov):1gov. V,:14"r, R -_L t "qffi-29The disk is uniformly charged. This means that when the full disk is present each quadrantcontributes equally to the electric potentral at P, so the potential at P due to a single quadrantis one-fourth the potentral due to the entire disk. First find an expression for the potential at Pdue to the entire disk.146 Chapter 24
151.
Consider a ring of charge with radius r dr. Its area is Znr dr and it contains charge and widthd,q:2nord,r.Allthechargeinitisadistance!mEfromP,SothepotentialitproduceSatPis a,7 I 2rrordr : or dr d,v: 4nes lmE 2eo r2+D2The total potential at P is o 6 R2+DzThe potential Vsq at P due to a single quadrant is %q:I:*1ffi-Dl .73 x 10-rs l^ 7 8(8.85 x 10-12 C2/N.m2) C - 0 .25q *lI - 4.71 x 10-5 V.39Take the negatives of the partial derivatives of the electric potential with respect to the coordinatesand evaluate the results for r- : 3.00 m, A -2.00 m, and z 4.00 m. This yields - E* Ea: -X E": -y - -2(2.00 Y l^o)raz : Ll -2(2.00 Y l^ox3.00mX -2.00mX4.00m) : 96.0v/m. 0zThe magnitude of the electric field is E- : :1.50x 102Y1*.4tThe work required is equal to the potential energy of the system, rclative to a potential energy ofzero for infinite separation. Number the particles l, 2, 3, and 4, in clockwise order starting withthe particle in the upper left corner of the affangement. The potential energy of the interactionof particles I and 2 is r r QtQz (g.gg x 10eN. m2 lc)(2.30 x 10-t, cx -2.30 x 10-rz c) L./a)-- L- a - 4Tega 0.640 m - -7.43 x 10-14 J .The distance between particles I and 3 is ,Do and both these particles are positively charged,so the potential energy of the interaction between particles I and 3 is Un: -Urrlfr Chapter 24 147
152.
+5.25 x 10- 14 I. The potentral energy of the interaction between particles 1 and 4 is (Ju(Jn : -7.43 x 10-14 J. The potentral energy of the interaction between particles 2 and 3 is(Jzz _ (Jn: -7 .43 x 10- J. The potential energy of the interaction between particles 2 and 4 t4is Uz+is Ut+The total potentral energy of the system is Ut - [Jn + Ur3 + Uru + [Jzz * Uz+ i Uz+ -7.43 x lo-LaJ+5.2s x 10-147- -1.92 x 10-13J.This is equal to the work that must be done to assemble the system from infinite separation.59(a) Use conservation of mechanical energy. The potential energy when the moving particle is atany coordinate A is qV , where V is the electric potential produced at that place by the two fixedparticles. That is, 2Q (l:q 4Tes *a2where r is the coordinate and a is the charge of either one of the fixed particles. The factor2 appears since the two fixed particles produce the same potential at points on the A axis.Conservation of mechanical energy yields 2Q 2Q Ky-K +q : K,+2qQ " 4reg 4nes 4nes +A?where K is the kinetic energy of the movin g particle, the subscrrpt i refers to the initial positionof the movin g particle, and the subscript f refers to the final positioll. Numerically Ky 2(-r5 x 10-6 C;1SO x 10-u C) 1l 4n(8.85 x 10-12 Cz/N .m2) (3.0m)2 + (4.0m) @l - 3.0J.(b) Now K1- 0 and we solve the energy conservation equation for Uf .Conservation of energyfirst yields tJ y : K,i, + Ut The initial potenttal energy is rT - zqQ 2(-15 x 10-6 CX50 x 10-6 C) I. 4nesffi - -/L /---._: 2.7 4r$.85 x 10-12cz/N .m2)Thus Ky: l.zJ- 2.7 J- -1.5J.Now 2qQ Uy 4nes +y?148 Chapter 24
153.
63If the electric potential is zero at infinity, then the electric potential at the surface of the sphereis given by V - qf 4Tesr, where q is the charge on the sphere and r is its radius. Thus V) -: (0 15 mX1500 q : 4nesrv - 2.5 x 10-8 C . / 8 .gg x 10eN . nP lC65(a) The electric potential is the sum of the contributions of the individual spheres. Let h bethe charge on one, ez be the charge on the other, and d be their separation. The point halfwaybetween them is the same distance d,l2 (: 1.0m) from the center of each sphere, So the potentialat the halfway point is $*Qz V: 4nesd,f 2 1.0 m(b) The distance from the center of one sphere to the surface of the other is d, - R, where R isthe radius of either sphere. The potential of either one of the spheres is due to the charge on thatsphere and the charge on the other sphere. The potentral at the surface of sphere 1 is ,i:*[*.h] 1.0 xg 10-a 3.0 x 10-s g =(g.ggx loeN.nf lc) [ 0.030m - 2.0m 0.030m :2.9 x 103 V.(c) The potential at the surface of sphere 2 is Vz I n, *gZ1 ld,- R RJ 1.0 x 10-8 g 3.0 x 10-s g - (g .gg x loeN. m2 lc) L 2.0m - 0.030m 0.030m - -8.9 x 103 V.75The initial potential energy of the three-particle system is Ur:2(q2 f 4reoL) * flrr*"d, where q isthe charge on each particle, L is the length of a triangle side, and Urr*"d is the potenttal energyassociated with the interaction of the two fixed particles. The factor 2 appears since the potential Chapter 24 149
154.
energy is the same for the interaction of the movable particle and each of the fixed particles. Thefinal potential energy is uy 2lqf 4nes(Ll2)l + Un*ed, and the change in the potential energy is - 2q2 4nesLThis is the work that is done by the external agent. If P is the rate with energy is supplied bythe agent and t is the time for the move, then Pt : LU, and t- LIJ 2q , p-ffi- z(B.gg x 10e N . m2 lcXo .rzc)2 1 o, 1 . :1.83x10s.This is 2.1 d.77(a) {Jse Gauss law to find an expression for the electric field. The Gaussian surface is acylindrical surface that is concentric with the cylinder and has a radius r that is greater thanthe radius of the cylinder. The electric field is normal to the Gaussian surface and has uniformmagnitude on tt, so the integral in Gauss law is f E . dA:2rrEL, where L is the length ofthe Gaussian surface. The charge enclosed is where is the charge per unit length on thecylinder. Thus ZrrRLE : ),Lf es and E - ),f 2nesr. ^L, ^Let E n be the magnitude of the field at B and r p be the distance from the central axis to B. LetEs be the magnitude of the field at C and rs be the distance from the central axis to C. SinceE is inversely proportional to the distance from the central axis, trs: ryEp: #(160N/c) :64N/c. Tg u 5.0 cm(b) The magnitude of the field a distance r from the central axis is E - (, e liU n so thepotenttal difference of points B and C is ve vs: l,: TEp d,r: -rBEptn(P rc / = -(o a2omx16oN/c)ln (ffi) -z.ev.(c) The cylinder is conducting, so alI points inside have the same potential, namely VB, soVt Vn: 0.8sConsider a point on the z axis that has coordinate z. All points on the ring are the same distancefrom the point. The distance is r- lW, where R is the radius of the ring. If the electricpotentral is taken to be zero at points that are infinitely far from the ring, then the potential atthe point is V- a 4Tes150 Chapter 24
155.
where A is the charge on the ring. Thus 1l _l nl - (8 .gg x loeN. m2 lcX16.0 x lo-u c) (0.0300 m)2 1(0.0400 *)2 -o3oo{93(a) Forr> r7 1a v- 4"r"; where the zero of potential was taken to be at infinity.(b) To find the potentral in the region 11 <for the electric field, then integrate along a radial path from 12 to r. The Gaussian surface isa sphere of radius r ) concentric with the shell. The field is radial and therefore normal to thesurface. Its magnitude is uniform over the surface, so the flux through the surface is O - 4nr2 E.The volume of the shell is (nl3)(rt, - r?), so the charge density is 3Q p: ar@) - r?)and the charge enclosed by the Gaussian surface is / 4r q: (+) (,Gauss law yields #) /n3-rJ 4TesrzU-QI rand the magnitude of the electric field is D- A r3-3, t) 4"rrffiIf V, is the electric potential at the outer surface of the shell (r : r) then the potential a distancer from the center is given by Chapter 24 151
156.
The potential at the outer surface is found by placing r : 12 in the expression found in part (a).It is V, - Q f 4Tesr2. Make this substitution and collect like terms to find r/_ a 1 (v3 t "i 4"ofirz T 7)Since p : 3Q lan@] - ,3r) this can also be written v-*(+ + +)(c) The electric field vanishes in the cavrty, so the potential is everywhere the same inside andhas the same value as at a point on the inside surface of the shell. Put r - 11 in the result ofpart (b). After collecting terms the result is rr_ A 3(r3-r?) 4?t€o zez - f)or in terms of the charge density v- +?3-?) zeo(d) The solutions agree at r : rr and at r : 12.95The electric potential of a dipole at a point a distance r away is given by Eq. 24-30: rr_ I pcos? v- 4"r, 12 where p is the magnitude of the dipole moment and e is the angle between the dipole momentand the position vector of the point. The potential at infinity was taken to be zero. Take the zaxis to be the dipole axis and consider a point with z positive (on the positive side of the dipole).For this point r : z and 0 : 0. The z component of the electric field is E=--ry:-a ( o - 27reo* - A" - 0, a"rrrz ) pThis is the only nonvanishing component at a point on the dipole axis.For a point with a negative value of z, r: -z and cos 0 - -1, So P E=- -Lz r--p^ -- - d ) +ere orz = 2T esz3103-(a) The electric potentral at the surface of the sphere is given by V : qf 4TeoR, where q is thecharge on the sphere and R is the sphere radius. The charge on the sphere when the potentialreaches 1000 V is : 4nesrV : (0010 mX1000 v) :1.11 q - - x 10-e C . 8.99 x 10eN. m2 lCI52 Chapter 24
157.
The number of electrons that enter the sphere is jV - qle - (1.11 x 10-e ql(l.60 x 10-1e g; -6.95 x 10e. Let.R be the decay rate and tbethe time for the potential to reach it final value. Since -half the resulting electrons enter the sphere jV - (Pl2)t and t zl.VlP - 2(6.95 x 10e) 1Q.70 x108 s-r) : 38 s.(b) The increase in temperature is LT : l/A E lC , where E is the energy deposited by a singleelectron and C is the heat capacity of the sphere. Since iV - (Plz)t, this is LT - (Plz)tLElCand t -zg Y- 2(14J/KX5"0K) :2.4x 107 s. H P LE (3.70 x 108 s-rx100 x 103 eVXl.60 x 10- re J leY)This is about 280 d. Chapter 24 153
158.
Chapter 25-5ftl The capacitance of a parallel-plate capaeitor is given by e - esA I d", where A is the areaofeach plate and d is the plate separation. Since the plates are circulag the plate area is A - T R2 ,where R is the radius of a plate. Thus e_4 d 1.30 x 10-3 m(b) The charge on the positive plate is given by q - CV, where V is the potential differenceacross the plates. Thus q: (I .44 x 10-tofXI20V): 1.73 x 10-8 C - 17.3nC.15The charge initially on the charged capacitor is given by q- CrVo, where Cr (: 100pF) is thecapacitance and Vo (: 50 V) is the initial potential difference. After the battery is disconnectedand the second capaeitor wired in parallel to the first, the charge on the first capacitor iser : CrV, where u (: 35 V) is the new potential difference. Since charge is conserved in theprocess, the charge on the second capacitor is Qz: q er, where ez is the capacitance of thesecond capacitor. Substitute CtVo for q and CrV for gt to obtain q2: Cr(Vo-V). The potentialdifference across the second capacitor is also V , so the capacitance is (1 : ez: L,Z-V- vo v tt: n 5ov -#(roopF) :43pF. , g5V t9(a) After the switches are closed, the potentral differences across the capacitors are the same andthe two capacitors are in parallel. The potenttal difference from a to b is given by Vou - Q lC"r,where A is the net charge on the combination and Ceq is the equivalent eapacitance.The equivalent capacitance is Ceq: Ct+Cz - 4.0 x 10-6F. The total charge on the combinationis the net charge on either pair of connected plates. The charge on capacitor I is er: CtV - (1.0 x 10-u pX100V) - 1.0 x 10-4 Cand the charge on capacrtor 2 is Qz : - (3.0 x 10-u pX100 v) : CzV 3.0 x 10-4 C ,so the net charge on the combination is 3.0 x 10-4 C 1.0 x 10-4 C - 2.0 x l0-4 C. Thepotential difference is x 10-4C- Voa -2.0 x 10-6 F 50v. 4.0(b) The charge on capacitor 1 is now Qt:CrVob-(1.0 x 10-u FXSO V) : 5.0 x 10-5 C.154 Chapter 2 5
159.
(c) The charge on capacitor 2 is now ez: CzVou - (3.0 x 10-upXS0V; - 1.5 x 10-4C.29The total energy is the sum of the energies stored in the individual capacitors. Since they areconnected in parallel, the potenttal difference V across the capacitors is the same and the total :energy is (J: *(ct+C)172: *Q.o x 10-6F+ 4.0 x lO-upX:00D2 - 0.27 J.35(a) Let q be the charge on the positive plate. Since the capacitance of a parallel-plate capacitoris given by esAld. the charge is q :separation is dt and the potential difference is V . Then q esAV ld, and dt v, - ,rAq: d eoA -v-ftv:t6b#(6.oov): - d -- 8.oo mr T6.ov. rrA d(b) The initial energy stored in the eapacitor is (Ji 2- 2d 2(3.00 x 10-3 mm)and the final energy stored isuy: lc,(v,)r: ;# (v,)r: : r.zox ro-ro J.(c) The work done to pull the plates apart is the difference in the energy: W - tI y Ut -1.20 x 10-10J -4.51 x 10-11 J -7.49 x 10-11 J.43The capacitance of a cylindrical capacitor is given by 2" *oL ._,e - c - K(1^- In(W,where Co is the capacitance without the dielectric, K is the dielectric constant, L is the length, ais the inner radius, and b is the outer radius. See Eq. 25-14. The capacitance per unit length ofthe cable is C :: 2n nes zn(2.6x8.85 x 10- t p l^) n 4 , L tn(bla) ln [(o.60mm)le. 10**]]--81 x 10-"Ffm- 81pFfm 1 145The capacitance is given by Cdielectric, K is the dielectric constant, A is the plate area, and d, is the plate separation. The Chapter 2 5 155
160.
electric field between the plates is given by E : V ld, where V is the potential difference betweentheplates. Thus d-Vltr and C : K€yAE lV. Solve for A: CV A- K€gEFor the area to be a minimum, the electric field must be the greatest it can be without breakdownoccurritr.g. That is, (7.0 x 10-8 F114.0 x 103V) 0.63 nl 2.8(8.85 x 10-t2F lmxlS x 106 V lm)51(a) The electric field in the region between the plates is given by E - V I d, where V is thepotenttal difference between the plates and d is the plate separation. The capacitance is given byC - neyAf d,, where A is the plate area and K is the dielectric constant, so d - nesAf e and VC (50 VX100 x 10-tz F) E- x 1o4Yl^. KesA 5.4(8.85 x 10-t2F lmxl00 x 10-4 m2) -1.o =:(b) The free charge on the plates is qf : CV - (100 x 10-tfXS0V) : 5.0 x 10-e C.(c) The electric field is produced by both the free and induced charge. Since the field of a largeuniform layer of charge is q f 2esA, the field between the plates is ,;r_ qf , Qf Qt Qt rJ 2r"A= vr4 zrrA zroAwhere the first term is due to the positive free charge on one plate, the second is due to thenegative free charge on the other plate, the third is due to the positive induced charge on onedielectric surface, and the fourth is due to the negative induced charge on the other dielectricsurface. Note that the field due to the induced charge is opposite the field due to the free charge,so the fields tend to cancel. The induced charge is therefore Qt: Qf- €e AE : 5.0 x 10-e c- (8.85 x 10-12 F lm)(100 x 10-4 m2xl.0 x 104 v l^) :4.I x l0-eC-4. 1nC.6rCapacitors 3 and 4 are in parallel and may be replaced by a capacitor with capacitance Ctq :e3 + e+- 30pF. Capacitors l, 2, and the equivalent capaeitor that replaced 3 and 4 are allin series, so the sum of their potential differences must equal the potential difference across thebattery. Since all of these capacitors have the same capacitance the potential difference acrosseach of them is one-third the battery potential difference or 3.0V. The potential difference acrosscapac rtor 4 is the same as the potential difference across the equivalent capacitor that replaced 3 and,4, so the charge on capacrtor 4 is e+: CqVq - (15 x 10-upX:.0V) :45 x 10-6 C.156 Chapter 25
161.
69(a) and (b) The capacitors have the same plate separation d and the same potenttal difference Vacross their plates, so the electric field are the same within them. The magnitude of the field ineither one is E -Vld, - (600V)l(3.00 x 10--) -2.00 x 105 Yl^.(c) Let A be the area of a plate. Then the surface charge density on the positive plate isoA105 Vlm)- 1,.77 x 10-u Cl^, where CV was substituted for q and the expression esAld, forthe capacitance of a parallel-plate capacitor was substituted for C.(d) Now the capacitance is rces Ald, where rc is the dielectric constant. The surface charge densityon the positive plate is oB: K€08 - KoA- (2.60XI.77 x 10-uCl^) - 4.60 x 10-u Cl^.(e) The electric field in B is produced by the charge on the plates and the induced chargetogether while the field in A is produced by the charge on the plates alone. since the fields atethe same og * oindu_ced : oA, so oinduced: 04 - op: I.77 x 10-o gl^ - 4.60 x 10-u Cl^--2.g3 x 10- Cl^ u .73The electric field in the lower region is due to the charge on both plates and the charge inducedon the upper and lower surfaces of the dielectric in the region. The charge induced on thedielectric surfaces of the upper region has the same magnitude but opposite sign on the twosurfaces and so produces a net field of zerc in the lower region. Simtlarly, the electric field inthe upper region is due to the charge on the plates and the charge induced on the upper and lowersurfaces of dielectric in that region. Thus the electric field in the upper region has magnitudeE oo., : QK,,ryper€sA and the potential difference across that region is l/uppr, - f,upp"rd, where d isthe thickness of the region. The electric field in the lower region is -E1o*", : eKto*er€oA and thepotential difference across that region is I4o*., - Eu,*"rd. The sum of the potential differencesmust equal the potentral difference V across the entire eapacitor, so v -Eupp., d,+Ero*., d- #L#. *]The solution for q is q: KrpperKlower eOA x 10-12N .m2 lc)(z.oo x ro-2m2) (7.00 V v) ffiupp.. * Klo*.. d 3.00 + 4.00 2.00 x l0-3 m 1.06 x 10-e C. Chapter 2 5 157
162.
Chapter 257_(a) The magnitude of the current density is given by Jparticles per unit volume, q is the charge on each particle, and u4 is the drift speed of theparticles. The particle concentration is n- 2.0 x 108 cm-3q:2e:2(L60 x 10-tsC) -3.20 x 10-" C, and the drift speed is 1.0 x 105m/s. Thus J: (2 x 10r4m-t)(3.2 x 10-tnCXl.0 x 105 mls): 6.4A1^ .(b) Since the particles are positively charged, the cuffent density is in the same direction as theirmotion, to the north.(c) The current cannot be calculated unless the cross-sectional area of the beam is known, Theni - JA can be used.t7The resistance of the wire is given by R- pL lA, where p is the resistivity of the material, Lis the length of the wire, and A is the cross-sectional area of the wire. The cross-sectional areais A- nr2radius of the wire. Thus RA p: t: (50 x 10-OXz.85 x 10-t m2) A , :2.0 x lo-t g rr. 2n^t9The resistance of the coil is given by R - pL lA, where L is the length of the wire, p is theresistivity of copper, and A is the cross-sectional area of the wire. Since each turn o1 wire haslength 2nr, where r is the radius of the coil, L - (250)2nr - (250)(2n)(0.12m) - 188.5m. lf r*is the radius of the wire, its cross-sectional areais A - Tr2* - ?T(0.65x 10-*)tAccording to Table 26-I, the resistivity of copper is I.69 x 10-a 9.ril. Thus IL-_ n pL (l .69 x 10-s 9.mX188.5m) _ A t_, ^ - L.+trr. -A 1.33 x 10-6 m22tSince the mass and density of the material do not change, the volume remains the same. If Lsis the original length, L is the new length, Ao is the original cross-sectional are1 and A is thenew cross-sectional area, then LoAo: LA and A- LoAolL: LoAol3fo: Aol3. The newresistance is R-+: P3 Ao13 Lo -s+: Ag epo.158 Chapter 26
163.
where Ro is the original resistance. Thus R - 9(6.0 Q) : 54 O.23The resistance of conductor A is given by RA:#,where ra is the radius of the conductor. If ro is the outside radius of conductor B and ri is itsinside radius, then its cross-sectional area is r(r2" - ,?) and its resistance is Rn- .uD .lL ,?) =" n@2" -The ratio is Ra:rrr_r? Jv R" T39(a) Electrical energy is transferred to internal energy at a rate given by y2 t. D- Ewhere V is the potential difference across the heater and R is the resistance of the heater. Thus P -(1?9p r4r)(b) The cost is given by c- (1.0kwx5.0hx$0.050 /kw .h) - $0.25 .43(a) Let P be the rate of energy dissipation, i be the current in the heater, and V be the potentialdifference across the heater. They are related by P - iV. Solve for i: P 1250w-10.9A. ?- v- rrsv(b) According to the definition of resistance V - dR, where R is the resistance of the heater.Solve for R: R-Y:1ry- i A 10.9 r0.6E2.(c) The thermal energy t produced by the heater in time t (: 1.0 h - 3600 s) is E - Pt : (1250 W)(3600 s) : 4.5 x 106 J . Chapter 26 159
164.
s3(a) and (b) Calculate the electrical resistances of the wires. Let pc be the resistivity of wire C,rs be its radius, and Lc be its length. Then the resistance of this wire is Ln 1.0m Rc:pc +-(2.ox1o-og.m) x 10-t -2.540. Trb 7r(0.50 ^)Let p n be the resistivity of wire D, rp be its radius, and L n be its length. Then the resistanceof this wire is 1.0m Rn: pD 4- TrD (t.o x 10-6e.m) 7r(0 .25 x 10-*)t -5.09f).If i is the current in the wire, the potentral difference between points I and 2 is LVz - iRs - (2.0 AX2.54Q) : 5. 1 Vand the potential difference between points 2 and 3 is LVzt:iRp - (2.0AX5.09 O) - 10V.(c) and (d) The rate of enerry dissipation between points 1 and 2 is Prz : i2 Rg - (2.0 A)2 (2.54o) : 10 wand the rate of energy dissipation between points 2 and 3 is Pzz : i2 Rn - (2.0 A)2(5.09 o) : 20 w .--55(a) The charge that strikes the surface in time Lt is given by Lq - i Lt, where i is the current.Since each particle carries charge 2e, the number of particles that strike the surface is L.q- x 10-: A)-(1.0 s) :2.3 - 2e ry- (0.25 iv x 1012 . 2e 2(1 .6 x 10- tq C)(b) Now let t/ be the number of particles in a length L of the beam. They will all pass throughthe beam cross section at one end in time t - L lr, where u is the particle speed. The current isthe charge that moves through the cross section per unit time. That is, i- 2elt{lt - 2el{uf L.Thus, lV - iLf 2eu.Now find the particle speed. The kinetic energy of a particle is K - 20MeV - QA x 106eD(l.60 x 10-tllev) :3.2 x 10-t2 J.SinceK-}^u,,U:vw.ThemaSSofanalphaparticleisfourtimesthemaSSofaprotonor m:4(1 .67 x 10-27 kg) - 6.68 x 10-" kg, so x 10- tz I) :3.1 2(3 .2 x 107 ml s 6.68 x 1O-zt kE160 Chapter 26
165.
and (Ozs x x t9-,im) : 5.0 x 103 iv - Zeu 2(I.60 x 10- le CX3. I x ?L - 19,6=t)(?o 107 m/s) v .(c) Use conservation of energy. The initial kinetic energy is zero, the final kinetic energy is20 MeV - 3.2 x 10- " J, the initial potential energy is qV - 2eV, and the final potential energyis zero. Here I/ is the electric potential through which the partrcles are accelerated. Conservationof energy leads to K y - LLi, -- 2eV , so v-#- _10x1o6v.59Let Rn be the resistance at the higher temperature (800" C) and let Rr be the resistanceat the lower temperature (200" C). Since the potential difference is the same for the twotemperatures, the rate of energy dissipation at the lower temperature is Pr _ yz I R6 and therate of energy dissipation at the higher temperature is Ps - 172 I Rn, so P7 - (Rn I Rr)Pn.Now Rr : Ru + aRn LT, where LT is the temperature difference Ty - Ts Rs Ps s00w Pr Ps : Rn * aRn LT 1*c-LT 1+(4.0 x 10-4 l.C)(-600"C)75If the resistivity ir po at temperatureTo, then the resistivity at temperatureT is p: po+apy! -70), where a is the temperature coefficient of resistivity. The solution for T is p po* apoTo T_ - aPoSubstitute p : Zpo to obtain I T - To + a:20.0"C + :250oC . 4.3 x 10-3 K-lThe value of a was obtained from Table 26-I. Chapter 26 16l
166.
Chapter 277(a) Let i be the current in the circuit and take it to be positive if it is to the left in R1. I-JseKirchhoffs loop rule: & - iRz - iRr - tz:0. Solve for i: - ? - t2y - 6.oy _ o.5oA. i _ !, Rr+Rz 4.0O+8.0OA positive value was obtained, so the current is counterclockwise around the circuit.(b) and (c) If i is the current in a resistor with resistance R, then the power dissipated by thatresistor is given by P - z2 R. For Rr the power dissipated is P1 - (0.50 A)(4.0 O) : 1.0wand for Rz the power dissipated is P2 - (0.50 A)(8.0 O) : 2.0W .(d) and (e) If i is the cuffent in a battery with emf t, then the battery supplies energy at the rateP - it provided the current and emf are in the same direction. The battery absorbs energy atthe rate P : i,t if the current and emf are in opposite directions. For battery 1 the power is P1 -(0.50AXLZV)-6.0Wand for battery 2 it is P2 - (0.50 AX6.0 V) - 3.0W .(0 and (g) In battery l, the current is in the same direction as the emf so this battery suppliesenergy to the circuit. The battery is discharging. The cuffent in battery 2 is opposite the directionof the emf, so this battery absorbs energy from the circuit. It is charging.13(a) If i is the current and LV is the potential difference, then the power absorbed is given byP - i, LV. Thus Lv: I- i 5ow 1.0A :50vSince energy is absorbed, point A is at a higher potential than point B; that is, Ve - Vn : 50 V.(b) The end-to-end potentral difference is given by Va Vn - +iR + t, where t is the emf ofelement C and is taken to be positive if it is to the left in the diagram. Thus t - Ve - Vs - oR :50 v - (1.0 A)(2.0 o) - 48 V.162 Chapter 27
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(c) A positive value was obtained for t, so it is toward the left. The negative terminal is at B.2t(a) and (b) The circuit is shown in the diagram tothe right. The current is taken to be positive if itis clockwise. The potential difference across battery1 is given by Vthe current must be igives 2E irr irz iR - 0. Substitute i - t lrtand solve for R. You should get R _ 11 12 _0.0160 - 0.012Q - 0.004Q.Now assume that the potential difference across bat- JLtery 2 is zero and carry out the same analysis. Youshould find R: rz - 11. Since 11 >be positive, this situation is not possible. Only the potenttal difference across the battery withthe larger internal resistance can be made to vanish with the proper choice of .R.29Let r be the resistance of each of the thin wires. Since they are in parallel, the resistance R ofthe composite can be determined from 19 R r)or R-r19. Now rr, : 4P( rd,2and oll=. R- rDz )where p is the resistivity of copper. Here rd2 l4 was used for the cross-sectional area of any oneof the original wires and n D2 14 was used for the cross-sectional arca of the replacement wire.Here d and D are diameters. Since the replacement wire is to have the same resistance as thecomposite, 4p( 4p( nD2 - 9rd2Solve for D and obtain D - 3d.33Replace the two resistors on the left with their equivalent resistor. They are in parallel, so theequivalent resistance is ftee - 1.0 O. The circuit now consists of the two emf devices and fourresistors. Take the current to be upward in the right-hand emf device. Then the loop rule givestz - iR"q - 3iR - ty where R - 2.0 O. The current is . L: tz-tr rzv - 5.0v -1.0A. &q+3u l.0o+3(2.0o) Chapter 27 163
168.
To find the potential at point I take a path from ground, through the equivalent resistor andtz, to the point. The result is Vthe potential at point 2 continue the path through the lowest resistor on the digram. It isVz: V + iR: -11 V + (1.0 A)(2.0 O) : -9.0V.47(a) and (b) The copper wire and the aluminum jacket are connected in parallel, so the potentialdifference is the same for them. Since the potenttal difference is the product of the current andthe resistance, ic Rc - i,tRt, where i,g is the current in the copper, i, t is the current in thealuminum, Rc is the resistance of the copper, and Rt is the resistance of the aluminum. Theresistance of either component is given by ft - pL lA, where p is the resistivity, L is the length,and A is the cross-sectional area. The resistance of the copper wire is Rgand the resistance of the aluminum jacket is PNL P n,gSubstitute these expressions into icRc: itRt and cancel the common factors L and 7r toobtain pc ic iePa A2 - b2_a2 Solve this equation simultaneously with L - ic + t A, where i is the total current. You should get Pci o2 is: (b2 a2)pc * pe - a2and -(b2 : a2) Pci- ,i e - (b2 - a2)pc * a2 pt The denominators are the same and each has the value (b2 - a2)pc + a2pt: [(0.380 x 10-"tt)- (0.250 x 10-*)] (1.69 x 10-a g.m) + (0 .250 x l0-*)t (2.75 x 10-8 O .m) -3. 10x 10-15o.m3.Thus (0.250 x I0-t x 10-sg . m)(2.00A) ?,9 - ^)(2.75 3.10 X 10-1s O.m3and [(0.380 x 10- rr)t - (0 .250 x l0- *)t] tt .69 x ro-a g . mX2.00 A) xg: 3.10x l0-l5e.m3 - 0.893 A.164 Chapter 27
169.
(c) Consider the copper wire. If V is the potential difference, then the current is given byV - icRc:icpcLf ra2, so L- razV-- zr(0 .250 x 10-t ^)(12.0.m) -126m. r V) 1A icpc (1.11AXL.69 x 10-8 Os7During charging the charge on the positive plate of the capacitor is given by Eq. 27 -33, withRC : r. That is, q: Ct fr - e-/1 L J/ ,where e is the capacitance, t is applied emf, and r is the time constant. You want the time forwhich q:0.990Ct, so 0.990:1-e-tlThus e-t / o.o 10 .Take the natural logarithm of both sides to obtain tlr: - ln0.010: 4.61 and t- 4.61r.65 (a), (b), and (c) At t : 0, the capaaitor is completely uncharged and the current in the capaeitorbranch is as it would be if the capacitor were replaced by a wire . Let i1 be the current in ^Rr andtake it to be positive if it is to the right. Let i,2 be the current in Rz and take it to be positive if it is downward. Let i,z be the current in Rt and take it to be positive if it is downward. Thejunction rule produces ir : iz * iz, the loop rule applied to the left-hand loop produces t- irRr - izRz: 0,and the loop rule applied to the right-hand loop produces izRz - hRt:0.Since the resistances are all the same, you ean simplify the mathematics by replacing Rr, Rz,and R3 with R. The solution to the three simultaneous equations is it- 2.5-: 1.1 X 10-3 A 3R 3(0.73 x 106 Q) -and !-- ,i2: ,i3- t * t?1y=. : 5.5 x ro-4 A. 3R 3(0.73 v x 106 O)(d), (e), and (0 At t - oo, the capacitor is fully charged and the current in the capaeitor branchis zero. Then : iz and the loop rule yields t - hRt - irRz: 0. Chapter 27 165
170.
The solution is r2 103 Y ,it: ,i2: L_ " 2R 2(0.73x tT"O- 82 x lo-4A(g) and (h) The potentral difference across resistor 2 is Vz: i2R2. At f : 0 it is V2 - (5.5 x 10-4 AXO .73 x 106 O) : 4.0 x 102 Vand at f : m it is V2 - (8.2 x 10-4 AXO .73 x 106 0) : 6.0 x I02 V.(i) The graph of Vz versus t is shown to the right. Vz sl2 tl3 e l673As the capacitor discharges the potential difference across its plates at time t is given byV - Voe-tl, where Vo is the potenttal difference at time t- 0 and r is the capacitive timeconstant. This equation is solved for the time constant, with result t r: rn(vlwSince the time constant is r: RC, where RR is the resistance and C is the capacitance, t rL- ctn(vlw n-For the smaller time interval too * to-1t R -- - : L") o. )48 (o.zzox to-6F)ln fg) / 5.00vand for the larger time interval 6oo xs 1o-3 R -- - r.4g x ro4 e . (0.220x 10-6F)ln fg) 5.00v )75(a) Let i be the cuffent, which is the same in both wires, and t be the applied potential difference.Then the loop equation gives t - iR,q - iRn - 0 and the current is i- t - 0.127 6loY = :70. 1A. Ra+ Rn Q + 0.729n166 Chapter 27
171.
The current density in wire A is i A 70.1 Jt ::: 1.32 x I07 Al^ Trr2o 7r(1.30 x 10-t J-t ^)(b) The potentral difference across wire A is Vt : iRt - (70. 1 AX0.127 A) - 8.90 V.(c) The resistance is Rt : p AL I A, where p is the resistivity, A is the cross-sectional area, andL is the length. The resistivity of wire A is Pa- 44 L 40.0mAccording to Table 26-l the material is copper.(d) Since wire B has the same diameter and length as wire A and carries the same current, thecurrent density in it is the same, I.32 x 107 Al^ .(e) The potential difference across wire B is Vs:iRs - (70. 1AX0.729 O):51.1V.(f) The resistivity of wire B is ReA .fii. Pe - L - 9.68 x 10-8 O 40.0mAccording to Table 26-I the material is iron.77The three circuit elements are in series, so the current is the same in all of them. Since thebattery is discharging, the potential difference across its terminals is Vaut _ t - ir, where t isits emf and r is its internal resistance. Thus t lzv - ll .4v _ - -v - i sOA - 0.012Q tThis is less than 0.0200 Q, so the battery is not defective.The resistance of the cable is Rcable0.040 O. The cable is defective.The potential difference across the motor is Tmotor- II .4V - 3.0V-8.4V and its resistanceis R*oto. - Vmotorli- (8.4V)l(50A) - 0.17 {1, which is less than 0.200O. The motor is notdefective.85Let ftso be the resistance of the silicon resistor at 20" and Rrc be the resistance of the iron resistorat that temperature. At some other temperature T the resistance of the silicon resistor is ftsftso + o.s Rso(T - 20"C) and the resistance of the iron resistor is Rr : Rra * ar Rrc(T - 20" C).Here a s and a 1 are the temperature coefficients of resistivity, The resistors are series so theresistance of the combination is R - ftso + Rrc+ (asRso a arRroXT - 20oC) .We want Rso * Bro to be 10000 and asRso * arRrc to be zero. Then the resistance of thecombination will be independent of the temperature. Chapter 27 167
172.
The second equation gives Rn- -(rs lrt)Rso and when this is used to substitute for Rn inthe first equation the result is Rso - (as l*r)Rso:1000O. The solution for Bso is 1000 f) 1000 (-, ftSO- T_ as -85O ag , I _70 x lo-3 K-l ) a1 1where values for the temperature coefficients of resistivity were obtained from Table 26-1. Theresistance of the iron resistor is Rro: 1000 O 85 O - 91 5 O.9sWhen the capacitor is fully charged the potential difference across its plates is t and the energystored in it is (l - +C t .(a) The current is given as a function of time by i _ (t lB)e-t/, where r (- RC) is thecapacitive time constant. The rate with which the emf device supplies energy is P6 : it and theenergy supplied in fully charging the capacitor is Es: lr* ,rdt -f Ir* "-tt, d,t:+: T2RC : Ct2 RThis is twice the energy stored in the capacitor.(b) The rate with which energy is dissipated in the resistor is Pp : iZR and the energy dissipatedas the capacitor is fully charged is Ep: Ppd,t: d,t: S! : tzRC _ lo* +1,* "-2t/r 2R 2R ice97(a) Immediately after the switch is closed the capacitor is uncharged and since the charge onthe capaaitor is given by q - CVc, the potential difference across its plates is zero. Applythe loop rule to the right-hand loop to find that the potential difference across Rz must also bezero. Now apply the loop rule to the left-hand loop to find that t - irRr: 0 and h - t I Rt-(30 v)lQ0 x 103 o) - 1.5 x 10-3 A.(b) Since the potential difference across Rz is zero and this potential difference is given byVnz : i2Rz, i2: 0.(c) A long time Iater, when the eapacitor is fully charged, the current is zero in the capacitorbranch and the cuffent is the same in the two resistors. The loop rule applied to the left-hand loopgives €-iRr-iRz:0, soi -tl@t+R) -(30Y)lQ0x 103O+10x 1030):1.0x 10-3A.99(a) Rz and Rz are in parallel, with an equivalent resistance of RzRt l@, + Rz), and thiscombination is in series with Rt, so the circuit can be reduced to a single loop with an emf tand a resistance Req: Rr+ Rzful@2* R:) : (ftr Rz+ RrRz+ RzR:,)l(Rr + Rz).The curent is t (Rz+ h)t Req RrRz + Br Rz * RzRt168 Chapter 27
173.
The rate with which the battery supplies energy is (Rz+ Rz)t P-,it: RrRz+ ftr Rz + RzRz The derivative with respect to Rt is d,P gz (Rza RtXRr * Rz)tz srRT dRt RrRz+ RrR: + RzRt (ftr Rz+ RrRz+ RzR)z (ftr Rz+ RrRz+ RzRz)z )where the last form was obtained with a little algebra. The derivative is negative for all (positive)values of the resistances, so P has its maximum value for Rz: 0.(b) Substitute Rt - 0 in the expression for P to obtain p - rt : t: - RrRz Rr gJl: r4.4w. 10.0 O101If the batteries are connected in series the total emfin the circuit is IV t and the equivalentresistance is B+ frr, so the current is ,i - Ntl@+l[r). If R:r, then ,i - Ntl(N + 1)r.If the batteries are connected in parallel then the emf in the circuit is t and the equivalentresistanceisE+rlIV, sothecurrentis i- Sl@*rlIV)- IVtl(Xn*r). If RIV E l(lv + l)r, the same as when they are connected in series. Chapter 27 169
174.
Chapter 283(a) The magnitude of the magnetic force on the proton is given by Fs eu B sin @, where u is -the speed of the proton, B is the magnitude of the magnetic field, and 0 is the angle betweenthe particle velocity and the field when they are drawn with their tails at the same point. Thus t): eB sin - 650x10-17N-- - 4.00x105m/s. {I , / (1.60 x 10-te CX2.60 x 10-3 T) sin 23,0"(b) The kinetic energy of the proton is K - **r: *O.67 x 10-" kg)(4.00 x 105 mf s;2- r.34 x 10-16J.This is (r.34 x l0-16 Dl(l.60 x 10-tJlrv): 835 ev.t7(a) Sinse the kinetic energy is given by K - **r, where m is the mass of the electron and uis its speed, - 2.05 x 107 ml s.(b) The magnitude of the magnetic force is given by eu B and the acceleration of the electron isgiven by ,f r, where r is the radius of the orbit. Newtons second law is euB - mulro so (q.tt x to- 31-kgx2.05 x 197T/s) B-mu: (1.60 x 10-te CX25.0 x t0-2 m) - 4.6g x 10-4T- 46gtrtT. er(c) The frequency f is the number of times the electron goes around per unit time, so u 205x107.m/s f : 2nr- 2r(25.0 x t0-2 m) -1.31 x107Hz:13.1MH2.(d) The period is the reciprocal of the frequency: r 1 - r_ I _ 7.63x10-8s-76.3ns. 7:13ffi-29(a) If u is the speed of the positrotr, then u stn S is the component of its velocity in the plane thatis peqpendicular to the magnetic field. Here 0 is the angle between the velocity and the field(89o). Newtons second law yields eBu sin d: m(u sin ilz f r, where r is the radius of the orbit. :Thus r (*, l"B) sin /. The period is given by 2rr 2nm 2r(9.11 x 10-" kg) v v u stnf eB (1.60 x 10-te CX0.100 T)The expression for r was substituted to obtain the second expression for T.170 Chapter 28
175.
(b) The pitch p is the distance traveled along the line of the magnetic field in a time interval ofone period, Thus p : uT cos /. Use the kinetic energy to find the speed: K - i*r yields 2K u- m -^ - Z.65Ix L07m/s . VThus p: (2.651 x 107 mlsx3.58 x 10-to r) cos 89.0o - I.66 x 10-4 m.(c) The orbit radius is T: ma smp rnd (9.11 x 10-" kg)(2.651 x 107 mls)sin89.0o 1.51 x 10-3m. eB (1.60 x 10-te CX0.l00 T) -.=--:4l :(a) The magnitude of the magnetic force on the wire is given by Fn iLB sin /, where i is thecurrent in the wire, L is the length of the wire, B is the magnitude of the magnetic field, and 0is the angle between the current and the field. In this case Q 70o. Thus : Fn - (5000 A)(100 m)(60.0 x 10-u t) sin 70o - 28.2N .(b) Apply the right-hand rule to the vector product FB - iE x E to show that the force is to thewgst.47 Ut I z,The situation is shown in the left diagram above. The g axis is along the hinge and the magneticfield is in the positive r direction. A torque around the hinge is associated with the wire oppositethe hinge and not with the other wires. The force on this wire is in the positive z direction andhas magnitude I7 - IV ibB, where ,Af is the number of turns.The right diagram shows the view from above. The magnitude of the torque is given by r- Facos 0 - IVibBa cos I : 20(0.10 AX0.10 mX0.50 x l0- fXO.050 m) cos 30o :4.3 x l0-3N.rn. Chapter 28 l7l
176.
IJse the right-hand rule to show that the torque is directed downward, in the negatrve A direction.Thus i--(4.3 x 10-3 11 .m)-i.--I)(t) The magnitude of the magnetic dipole moment is given by p : IYI,A, where ^Af is the numberof turns, i is the current in each turn, and A is the area of a loop. In this case the loops arecircular, so A - nr2, where r is the radius of a turn. Thus F 23oAm2 r. i- lhrrz =- - rLtAF.A(b) The maximum torque occurs when the dipole moment is pe{pendicular to the field (or theplane of the loop is parallel to the field). It is given by , - pB - (2.30A.m2X35.0 x 10-T) -8.05 x l0-2 N.m.s9The magnitude of a magnetic dipole moment of a current loop is given by LLthe current in the loop and A is the area of the loop. Each of these loops is a circle and itsarea is given by A : T R2 , where R is the radius. Thus the dipole moment of the inner loop hasa magnitude of ptouter loop has a magnitude of po : irr2, - (7.00 Azr(0.300 m)z - I.979 A . m2.(a) Both currents are clockwise in Fig. 28-51 so, according to the right-hand rule, both dipolemoments are directed into the page. The magnitude of the net dipole moment is the sum of themagnitudes of the individual momentsi Fr"t: lti* Ito - 0.880A.m2+ 1 .9794.m2 -2.86Arrf .The net dipole moment is directed into the page.(b) Now the dipole moment of the inner loop is directed out of the page. The moments arein opposite directions, so the magnitude of the net moment is Fnet- Fo - ltr - 1.979 A . m20.880A .r#- 1.10A.m2. The net dipole moment is again into the page.63If .,Arr closed loops are formed from the wire of length L, the circumference of each loop is LIIV,the radius of each loop is R - L l2rltr, and the area of each loop is A : r RzL2 l4r 7gz. For maximum torgue, orient the plane of the loops parallel to the magnetic field, sothe dipole moment is pe{pendicular to the field. The magnitude of the torque is then B r: rviAl-(rtr?) (#) B - iLz 4rlYTo maximize the torque, take .,Af to have the smallest possible value, 1. Then T:TB iL2 (4.51 x l0- exo .250 -)t (s .7 | x l0- r) 165 -(a) the magnetic potential energy is given by U -1,.8, where E is the magnetic dipole momentof the coil and B is the magnetic field. The magnitude of the magnetic moment is lr: IV|A,172 Chapter 28
177.
where i is the current in the coil, A is the area of the coil, and .A/ is the number of turns. Themoment is in the negative A direction, as you can tell by wrapping the fingers of your right handaround the coil in the direction of the current. Your thumb is then in the negative A direction.Thus F: -(3.00X2.00AX4.00 x 10-*t)i - -Q.40 x l0-2A.*2)i. The magnetic potentialenergy is : _ly)},: .,olfit ::f:-l i?,r) : _7 2ox Io _, r,where j .i : 0, j .i - 1, and j .k - 0 were used.(b) The magnetic torque on the coil is i-fi,x8-Qtrilx (B*i+Brj+ B,k;- FaB,i- ttaB*k - (-2.40 x l0-2A. m2X-4.00 x 10-r)i - (-2.40 x t0-2 A.m2)(2.00 x l0-r)t - (9.6 x 10-5 N . m) i + 1+.80 x lo-s N . m)t,where j x i - -k, j x j : 0, and j x t - i were used.73The net force on the electron is given by F -e(E + 6 x 87, where E is the electric field, E -is the magnetic field, and 6 is the electrons velocity. Since the electron moves with constantvelocity you know that the net force must vanish. Thus E-- -ix E - -(ri) x (Bk)- -uBj- -(100m/sxs.00r)j-(500vlm)j75(a) and (b) Suppose the particles are accelerated from rest through an electric potential differenceV. Since energy is conserved the kinetic energy of a particle is Kparticles charge. The ratio of the protons kinetic energy to the alpha particles kinetic energy isKellf * - "le - el2e- 0.50. 0.50. The ratio of the deuterons kinetic energy to the alpha particles kineticenergy is KalKr-(c) The magnitude of the magnetic force on a particle is quB and, according to Newtons secondlaw, this must equal mu2 f R, where u is its speed and R is the radius of its orbit. Sincert)-m-Nn, : mtu{ R- mu qB m mlW t2* qB l, V qB - Vm tl tlcThe ratio of the radius of the deuterons path to the radius of the protons path is Ra: ReSince the radius of the protons path is 10cm, the radius of the deuterons path is (1.4X10cm):14cm. Chapter 28 173
178.
(d) The ratio of the radius of the alpha particles path to the radius of the protons path is Ro l4nlr E R" -Vr^or1-- r.4 .Since the radius of the protons path is 10 cm, the radius of the deuterons path is (l.4Xl0 cm) -14cm.77Take the velocity of the particle to be 6 _ u*i * ,rj and the magnetic field to be Bi. Themagnetic force on the particle is then F - q6 x E - e(u*i+urj1 x (B i) - -QUaB k,where q is the charge of the particle. We used i x i * oandjxi--t. The charge is F 0.48N Lr.: ^- rr: 10-2 c . -uab -(4.0 x 103 mls)(sin 37"X5.0 x 10-r T)81(a) If K is the kinetic energy of the electron and m is its mass, then its speed is 2(12 x 103 eV)(l.60 x 10-to I leVl u- x 10-31 - 6.49 x 107 m/s. 9.11Since the electron is traveling along a line that is parallel to the horizontal component of Earthsmagnetic field, that component does not enter into the calculation of the magnetic force on theelectrorl. The magnitude of the force on the electron is eu B and since F - ma) where a, is themagnitude of its acceleration , eu B - ma, and & _ euB: (t.oo x to-te cX6.49 x 101T/sX55.0 x 10-6 T) :6.3 x l0r4 mlrr. m 9.11 x 10-3t kg(b) If the electron does not get far from the r axis we may neglect the influence of the horizontalcomponent of Earths field and assume the electron follows a circular path. Its acceleration isgiven by o : u2 f R, where R is the radius of the path. Thus ) t_ /r At (6.49 x ro7 mf t->,) l-n7 sY R- - 6.72m. a 6.27 x 101 a ml s2The solid curve on the dragram is the path. Suppose it subtendsthe angle 0 at its center. d (- 0 .200 m) is the distance traveledalong the r axis and (. is the deflection. The right triangleyields d : .R sin g, so sin 0 - d,l R and cos g@. The triangle also gives (,- R Rcos9, sot- R- R@. Substitute R- 6.i2m and d-0.2mto obtain L - 0.0030 m.17 4 Chapter 28
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Chapter 291(a) The field due to the wire, at a point 8.0 cm from the wire, must be 3 9 pT and must be directedtoward due south. Since B - Foi,f Znr, 2rrB _ 2zr(0.080mX39 x 10-6T) - lto 4n x 10-7 7 .ml A - 16A.(b) The cuffent must be from west to east to produce a field to the south at points above it.7(a) If the currents are parallel, the two magnetic fields arc in opposite directions in the regionbetween the wires. Since the currents are the same, the net field is zero along the line that runshalfway between the wires. There is no possible current for which the field does not vanish. Ifthere is to be a field on the bisecting line the currents must be in opposite directions. Then thefields are in the same direction in the region between the wires.(b) At a point halfway between the wires, the fields have the same magnitude, Foi f 2rr. Thusthe net field at the midpoint has magnitude B - Foi, ln, and . nrB : ?,- 7r(0.040 m)(300 x 10-6 T) -= 30A lto 4; x tO- f . ,r, lA15Sum the fields of the two straight wires and the circular arc. Look at the derivation of theexpression for the field of a long straight wire, leading to Eq. 29-6. Since the wires we arcconsidering are infinite in only one direction, the field of either of them is half the field of aninfinite wire. That is, the magnitude is Foi lar R, where R is the distance from the end of thewire to the center of the arc. It is the radius of the are. The fields of both wires are out of thepage at the center of the arc.Now find an expression for the field of the ara at its center. Divide the arc into infinitesimalsegments. Each segment produces a field in the same direction. If ds is the length of a segment,the magnitude of the field it produces at the arc center is (ltoi, lar R) dt If 0 is the anglesubtended by the arc in radians, then R0 is the length of the arc and the net field of the arc isttoi? lan R. For the ara of the diagram, the field is into the page. The net field at the center, dueto the wires and arc together, is B-H+ 4nR 4rR ry(z- 4nR l t:rcT: 4rR - r7). Chapter 29 175
180.
For this to vanish, 0 must be exactly 2 radians.19Each wire produces a field with magnitude given by B : pyi f 2nr, where r is the distance fromthe corner of the square to the center. According to the Pythagorean theorem, the diagonal ofthe square has length .,,0.o, so r: and B - Foi,f tD,na. The fields due to the wires at the "ltnupper left and lower right corners both point toward the upper right corner of the square. Thefields due to the wires at the upper right and lower left corners both point toward the upper leftcorner. The horizontal components cancel and the vertical components sum to 2Poi Bnet - 4 ry lDno cos 45o- -ttvL 7I CL 2(47r x l0-7 T ml A)(20 A) . g.0 - - x 10_5 T . zr(0.20 m)In the calculation cos 45" was replaced with Lltn In unit vector notation E - (8.0 x 10-t t)j.2tFollow the same steps as in the solution of Problem 17 above but change the lower limit ofintegration to -L, and the upper limit to 0. The magnitude of the net field is B_rtoi,R fo = 4*==,= FoiR rr I r lo ttoi, L 4n l-t(*2+R2)3/z 4n HWI-, 4*R JTTW 4r x 10-t 7 .ml AX0.693 A) 0.136m 1.32 x 10-7 T. 4r(0.251 m) (0. 136 m)2 a (0.251 m)231The current per unit width of the strip is i l* and the current through a width d"r is (i, lu dr.Treat this as a long straight wire. The magnitude of the field it produces at a point that is adistance d from the edge of the strip is dB : (pa 12")(i I @ dr f r and the net field is B ltoi [o*- d,r - 2nwJa r 2nw d (42- x 10-7 T .m/AX4.61 x 10-6 A) 0.0216m + 0.0491 m ,n 2r(0.0491 m) 0.0216 m 2.23 x 10-11 T.35The magnitude of the force of wire 1 on wire 2 is givenby Foili2f 2rr, where h is the current in wire I, i2 is thecurrent in wire 2, and r is the separation of the wires.The distance between the wires is r: Since the lffir.cuffents are in opposite directions the wires repel each otherso the force on wire 2 is along the line that j oins the wiresand is away from wire 1.17 6 Chapter 29
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To find the r of the force, multiply the magnitude of the force by the cosine of componentthe angle 0 that the force makes with the r axis. This is cos 0 - dzlcomponent.of the force is tw, Thus the r n Jn lloiti,z A dz /,rr d?+ d7 (2n x l0-7 7 .ml AX4.00 x 10-axe .80 x l0-R) 0.0500 m 2n (0.024m)z + (5.00 m)2 : 8.84 x 10-ll T .43(a) Two of the currents are out of the page and one is into the page, so the net current enclosedby the path is 2.0 A, out of the page. Since the path is traversed in the clockwise sense, a cuffentinto the page is positive and a cuffent out of the page is negative, as indicated by the right-handrule associated with Amperes law. Thus i"n": -i and f 6 E.d,i:- Fa,i:-(4n x 10-7T.ml A)(2.0A): -2.5 x 10-6T.rn. --+ J(b) The net current enclosed by the path is zero (two currents are out of the page and two areinto the page), so f E . d,g: Foirn, : 0.s3(a) Assume that the point is inside the solenoid. The field of the solenoid at the point is parallelto the solenoid axis and the field of the wire is perpendicular to the solenoid axis. The net fieldmakes an angle of 45" with the axis if these two fields have equal magnitudes.The magnitude of the magnetic field produced by a solenoid at a point inside is given byBsor - Foisotn, where n is the number of turns per unit length and isol is the current in thesolenoid. The magnitude of the magnetic field produced by a long straight wire at a point adistance r away is given by B*ire : l.L1iwire f 2nr, where i*ir" is the current in the wire. We wantFonisol : Foiwir"f 2nr. The solution for r is i*ir" 6.00 A 4.77 x l0-2 m: 4.77 cm.This distance is less than the radius of the solenoid, so the point is indeed inside as we assumed.(b) The magnitude of the either field at the point isBror: Bwire: Fonirol: (n x 10-7 T .ml AX10.0 x I02 m-1;120.0 x 10-,,t) -2.51 x 10-5T.Each of the two fields is a vector component of the net field, so the magnitude of the net fieldis the square root of the sum of the squares of the individual fields: B - JZ(2.51 x 10-s T)2 -3.55 x 10-5 T. Chapter 29 177
182.
57.The magnitude of the dipole moment is given by p- IVdA, where tr is the number of turns, iis the current, and A is the area. Use A - n R2, where R is the radius. Thus l.L: ItinR2 - (200X0.30A)zr(0.050 m)2 - 0.47 A. m2 .59 :(a) The magnitude of the dipole moment is given by p IY|A, where ,Af is the number of turns,i is the current, and A is the area. Use A - r R2, where R is the radius. Thus p : IYirR2 - (300X4.0 A)n(0.025 m)2 - 2.4 A. nf(b) The magnetic field on the axis of a magnetic dipole, a distance z awa,y, is given by Eq . 29 -27: B_lto P_ 2r 23Solve for zi ry:llto lt1r/t- t 4r x 10-77.m/ A 2 .36,4.. fif l/t : P LznBl 46cm. L 21 5.0 x 10-6 1 I7lUse the Biot-Savart law in the form B lto iLi x r - 4r 13Take Adto be Arj,and r-tobe ri*yj+zk. Then As*xi- Asjx @i+yj*tt<; - Ls(zi-rt<;,wherejXi:-t,ixj--0,andjXk:iwereuSed.Inad.ditiofl,r:ffi.TheBiot-Savart equation becomes tto iLs(zi- zk) B- 4n (*2 * y2 1 ,213 /2(a) For fr:0, A:0, and z - 5.0m, 4r x r0-7 T .mf A (2.0AX0.030mX5.0m)i _ E_ (2.4x 10-ro T) i. 4n (5.0m)3(b) For fr:0, A:6.0m, and z:0, E:0.(c) For r - 7.0m, A :7.0m, and z :0, 4r x l0-7 T .mf A (2.0AX0.030mX-7.0m)k : (4.3 x 10_,, T)t. E_ 4r lQ.0 m)2 + (7.0 m)2l3 /z178 Chapter 29
183.
(d) For r - -3.0m, A: -4.0m, and z:0, E 4r x 10-t7-ml A (2.0AX0.030m)(3.0m)k (1 .4 x 1o-to T) t . 4rr t(-3.0 m)2 + ( -4.0 m)213/z77First consider the finite wire segment shown on the right. aIt extends from A - -d to A : a, d, where a is the Ilength of the segment, and it carries current i in the Ipositive A direction. Let dy be an infinitesimal lengthof wire at coordinate A. According to the Biot-Savartlaw the magnitude of the magnetic field at P due to thisinfinitesimal length is dB : (po laz)(e sin 0 lr) da. Now ,1 rr2 - a2 + R2 and sin 0 - Rl, - Rf ffi, so ds cJB:lto-iR dA 4n (A2 +and the field of the entire segment is B-#iRI:: ffida-pola"h a-d d1 R2+(a-d,)2 ffi)where integral 19 of Appendix E was used.All four sides of the square produce magnetic fields that are into the page at P, so we sum theirmagnitudes. To calculate the field of the left side of the square put d - 3a la and R - a,14. Theresult is Bbrt: tro 4i rI 1 l- 3l tto 4i (1.66). , _l 4r a l{Z ,ml -- = - rT r 57r aThe field of the upper side of the square is the same. To calcul ate the field of the right side ofthe square put d - a,l4 and R - 3a la. The result is tto4i t3 1l po 4i Bright : 4n 3a l.fr r- IIl:l - 3r _ (0.341). Jro]r - a,The field of the bottom side is the same. The total field at P is B - Brcn* Bupp., * Brieht+ Bro*er : fI .66 + I.66+ 0 .341+ 0 .341) #! 4r x 10-7 7 .ml A 4(10 A) x 1o-4T. 4n ffi(4.00):2.0 Chapter 29 179
184.
79(a) Suppose the field is not parallel to the sheet, as shown Iin the upper dragram. Reverse the direction of the current. .] IAccording to the Biot-Savart law, the field reverses, so itwill be as in the second diagram. Now rotate the sheetby 180" about a line that is perpendicular to the sheet. Thefield, of course, will rotate with it and end up in the direction I Ishown in the third diagram. The cuffent distribution is now exactly as it was originally, so the field must also be as itwas originally. But it is not. Only if the field is parallelto the sheet will be final direction of the field be the sameas the original direction. If the current is out of the page, I Iany infinitesimal portion of the sheet in the form of a long t/, Istraight wire produces a field that is to the left above thesheet and to the right below the sheet. The field must be asdrawn in Fig. 29-85.(b) Integrate the tangenttal component of the magnetic fieldaround the rectan galar loop shown with dotted lines. Theupper and lower edges are the same distance from the current ttsheet and each has length L. This means the field has thesame magnitude along these edges. It points to the left along L :=-E- lthe upper edge and to the right along the lower. +_LIf the integration is carried out in the counterclockwise sense, the contribution of the upper edgeis B L, the contribution of the lower edge is also B L, and the contribution of each of the sidesis zero because the field is pe{pendicular to the sides. Thus f A . dg:28L. The total currentthrough the loop is Amperes law yields 2BL: p1^L, so B - po12. ^L81(a) IJse a circular Amperian path that has radius r and is concentric with the cylindrical shellas shown by the dotted circle on Fig. 29-86. The magnetic field is tangent to the path andhas uniform magnitude on tt, so the integral on the left side of the Amperes law equation isf E . d,d : 2rr B. The current through the Amperian path is the current through the regionoutside the circle of radius b and inside the circle of radius r. Since the current is uniformlydistributed through a cross section of the sheltr, the enclosed current is i(r - b) l(o - b). Thus 2nrB: rz 6z , -ri CL OL -and B- ltai, r2 - b2 2n(a2 - b2) r(b) When r to B- 2nr, which is the correct expression for the Foi,ffield of a long straight wire. When r: b it reduces to B : 0, which is correct since there is no180 Chapter 29
185.
field inside the shell. When b - 0 it reduces to B- poirl2naz, which is correct for the fieldinside a cylindrical conductor.(c) The graph is shown below. B (T) 10 x 10-4 8 x 10-4 6 x 10-4 4 x 10-4 2 x 10-4 0.01 0.02 0.03 0.04 0.05 0.06 r (m)89The result of Problem I 1 is used four times, once for each of Pthe sides of the square loop. A point on the axis of the loopis also on a pe{pendicular bisector of each of the loop sides. ,D=*The diagram shows the field due to one of the loop sides, Ethe one on the left. In the expression found in Problem 11, R/ r / /replace L with a, and R with @- +m. /The field due to the side is therefore Foza B_ 4rz * a2 + 2a2The field is in the plane of the dotted triangle shown and iis perpendicular to the line from the midpoint of the loopside to the point P. Therefore it makes the angle 0 with thevertical.When the fields of the four sides are summed vectorially the horizontal components add to zero.The vertical components are all the same, so the total field is given by Bn ur: 48 I _ 4Ba: 4Ba cos 2RffiThus 4 Pgia2 Brorut- r(4r2 + a2) + 2a2For fr : 0, the expression reduces to 4 p,s,ia2 ZtD poi Brorut- na2tQo 7fa Chapter 29 181
186.
in agreement with the result of Problem 12.9l[Jse Amperes law: f E.d,g: pyi"n, where the integral is around aclosed loop and i"n" is thenet current through the loop. For the dashed loop shown on the diagram iintegral t U dd is zero along the bottom, right, and top sides of the loop as it would be if thefield lines are as shown on the diagram. Along the right side the field is zero and along thetop and bottom sides the field is pe{pendicular to dg. If (, is the length of the left edge, thendirect integration yields f E .d,i: Bl, where B is the magnitude of the field at the left sideof the loop. Since neither B nor (. is zero, Amperes law is contradicted. We conclude that thegeometry shown for the magnetic field lines is in error. The lines actually bulge outward andtheir density decreases gradu ally, not precipitously as shown.182 Chapter 29
187.
Chapter 30-5ifr. magnitude of the magnetic field inside the solenoid is B- Fonir, where nis the numberof turns per unit length and i, is the current. The field is parallel to the solenoid axis, so theflux through a cross section of the solenoid is Q6 : ArB: pgrr2"nir, where A, (: nr?) is thecross-sectional area of the solenoid. Since the magnetic field is zero outside the solenoid, this isalso the flux through the coil. The emf in the coil has magnitude c- IYdQs : ponrlxn# L-. dtand the current in the coil is .t 0cR ,, - poTr?lvn di, R dtlwhere lf is the number of turns in the coil and R is the resistance of the coil. The currentchanges linearly by 3.0A in 50ms, so d,i,ldt - (3.0A)l(50 x 10-3 r):60A/r. Thus 0C (4n x t0-7 t . m/n)zr(0.016 m)2(120X220 x 102 m-t)(60 A^) : 3.0 x r0-2 A. 5.3 02t(a) In the region of the smaller loop, the magnetic field produced by the larger loop may be takento be uniform and equal to its value at the center of the smaller loop, on the axis. Eq. 29-26,with z - r and much greater than R, gives B- Foi,R2 213for the magnitude. The field is upward in the diagram. The magnetic flux through the smallerloop is the product of this field and the area (nr) of the smaller loop: n p,sir2 R2 Os- 213(b) The emf is given by Faradays law: t:-#: (ry)#,(*) 3T p,si,rz 2ra Rzu(c) The field of the larger loop is upward and decreases with distance away from the loop. Asthe smaller loop moves away, the flux through it decreases. The induced current is directed soas to produce a magnetic field that is upward through the smaller loop, in the same direction as Chapter 30 183
188.
the field of the larger loop. It is counterclockwise as viewed from above, in the same directionas the current in the larger loop.29Thermal energy is generated at the rute € I R, where t is the emf in the wire and R is theresistance of the wire. The resistance is given by E - pL I A, where p is the resistivity of copper,L is the length of the wire, and A is the cross-sectional area of the wire. The resistivity can befound in Table 26-1. Thus R - pr: (1.69 x 10-8 o px0.500m) _ T.076x r0-2 o. A n-(0.500 x 10-t ^)Faradays liw is used to find the emf. If B is the magnitude of the magnetic field through theloop, then EE AdBldt, where A is the area of the loop. The radius r of the loop is r - LlLnand its area is rr2 : n L2 l4n : L2 lan Thus ( : e 1zdB (0.500q)(10.0 x 10- rr ls): l.ggg x 10-4v. c dt 4?lThe rate of thermal energy generation is D_ gz (1.989 x lo-4 v)2 A - r-E -3.68x10-6W.37(a) The field point is inside the solenoid, so Eq. 30-25 applies. The magnitude of the inducedelectric field is I d,B E_ ; r : : 7 .15 x 10-t v l*. I x 10-t r lsx0 .0220m) E ){e.s(b) Now the field point is outside the solenoid and Eq. 30-27 applies. The magnitude of theinduced field is E -i2 dF Rz:1(u., x 10-trlrr(0o6oomf - r.43x t0-4 yl^. dt r 2 - -" (0.0820 m)51Starting with zero current when the switch is closed, &t time t - 0, the current in an RL seriescircuit at a later time t is given by ,i - 1(r - e-t/L) ,where T7 is the inductive time constant, t is the emf, and R is the resistance. You want tocalcul ate the time t for which i - 0.999At f R. This means o.sssr*: 1(r - e-t/"I) ,184 Chapter 30
189.
SO 0.ggg0-1-_e-t/r-orrake the narural logarithm or both ,io.l r::*: T: r;: ln(0.0010) - -6.e1. rhat is, 6.srinductive time constants must elapse.55(a) If the battery is switched into the circuit at time t* 0, then the current at a later time t isgiven by i-L(, -e-t/n/ E-- q )where r7 - L I R. You want to find the time for which i - 0.800 t I R. This means 0.800 : 1 - e-tlror e-tl"r-o.2ao.Take the natural logarithm of both sides to obtain -(t lrr) - ln(0.200) : - I .609. Thus 1"6019L- 1609(630 x-10:6H) t - l.609rr- R 1.20 x 103 O - 8 .45x l0-e s.(b) At t - 1.012 the current in the circuit is i-*(r -e-o) : (#) (r - e-o) -7.37x r0-3A.59(a) Assume i is from left to right through the closed switch. Let fu be the current in the resistorand take it to be downward. Let i,z be the cuffent in the inductor and also take it to be downward.The junction rule gives i - fi+i2 and the loop rule gives hR- L(dizldt):0. Since dildt:0,the junction rule yields (di,t I dt) - -(diz I dt). Substitute into the loop equation to obtain L+ + irR:0. dtThis equation is similar to Eq. 30-44, and its solution is the function given as Eq. 30-45: ) i,t : ioe-Rt/L ,where io is the current through the resistor at t - 0, just after the switch is closed. Now, justafter the switch is closed, the inductor prevents the rapid build-up of current in its branch, so atthat time, i,z - 0 and it - i. Thus io : i, so i1 : ie- Rt / L Chapter 30 185
190.
and . = . . = . [1 1,2 1, - 1,1 1, - e -Rt/L](b) When i2 = ii, e- Rt / L = 1 _ e- Rt / L ,so e- Rt / L = ~ 2Take the natural logarithm of both sides and use In(1/2) = -In 2 to obtain (Rt / L) = In 2 or L t = R In2.63(a) If the battery is applied at time t = 0, the current is given by i = ~ (1 - e- t / TL ) ,where £ is the emf of the battery, R is the resistance, and TL is the inductive time constant. Interms of R and the inductance L, TL = L / R. Solve the current equation for the time constant.First obtain e- t / TL = 1 - iR £ then take the natural logarithm of both sides to obtain -7£ =In [ 1 - £ t iR]Since In [1 _ iR] = In [1 _ (2.00 x 10- 3 A)(10.0 X 103 Sl)] = -0.5108 £ 50.0V the inductive time constant is TL = t/0.5108 = (5.00 x 10- 3 s)/(0.5108) = 9.79 x 10- 3 s and theinductance is L = TLR = (9.79 x 10- 3 s)(10.0 X 103 0) = 97.9 H.(b) The energy stored in the coil is UB = ~ Li 2 = ~(97.9 H)(2.00 x 10- 3 Ai = 1.96 x 10-4 J. 2 269(a) At any point, the magnetic energy density is given by UB = B2/2f-LO, where B is the magnitudeof the magnetic field at that point. Inside a solenoid, B = f-Loni, where n is the number of turns186 Chapter 30
191.
per unit length and i is the current. For the solenoid of this problem, n- (950)l(0.850m) -1.118 x 103 m-1. The magnetic energy density is us: )uonr: )fo" x 10-7 T .ml AXl.1t8 x 103m (0.60 A)_ 34.zJl^t.(b) Since the magnetic field is uniform inside an ideal solenoid, the total energy stored in thefield is LI a - u nV , where V is the volume of the solenoid. V is calculated as the product ofthe cross-sectional area and the length. Thus (-In - (34.2JfmXt7.0 x 10-4m2X0.850m) - 4.94 x 10-2J.73(a) The mutual inductance M is given by tt- M#,where & is the emf in coil 1 due to the changing current iz in coil 2. Thus M_ffi:ffi(b) The flux linkage in coil 2 is Iy2pzr: M,il : (1 .67 x 10-UX:.60A) : 6.01 x 10-3 Wb.75(a) Assume the cuffent is changittg at the rate di I dt and calcul ate the total emf across both coils.First consider the left-hand coil. The magnetic field due to the current in that coil points to theleft. So does the magnetic field due to the current in coil 2. When the current increases, bothfields increase and both changes in flux contribute emf s in the same direction. Thus the emf incoil 1 is tr: -Qr + ID + af,The magnetic field in coil 2 due to the current in that coil points to the left, as does the field incoil 2 due to the current in coil 1. The two sources of emf are agarn in the same direction andthe emf in coil 2 is tz: - (Lz+ M) + af, .The total emf across both coils is t:tr+tz: -Q1 * Lz+2Mdi. vr / dtThis is exactly the emf that would be produced if the coils were replaced by a single coil withinductance Leq - Ll + Lz * 2M . Chapter 30 187
192.
(b) Reverse the leads of coil 2 so the current enters at the back of the coil rather than the frontas pictured in the diagram. Then the field produced by coil 2 at the site of coil 1 is opposite thefield produced by coil 1 itself. The fluxes have opposite signs. An increasing current in coil 1tends to increase the flux in that coil but an increasing current in coil 2 tends to decrease it. Theemf across coil 1 is tr 0,f,Similarly the emf across coil 2 is di tz: - (Lz M) dtThe total emf across both coils is di t--(Lr*Lz-2M) dtThis the same as the emf that would be produced by a single coil with inductance LeqLraLz-2M.79(a) The electric field lines are circles that are concentric with the cylindrical region and themagnitude of the field is uniform around any circle. Thus the emf around a circle of radius ris€According to Faradays law 2nrE - -rr2(d,B ldt) and E _ *dB - -+(0.050mX-10 x -12 dt 10-3 T/s) :2.5 x 10-4 y l^.Since the normal used to compute the flux was taken to be into the paBe, in the direction of themagnetic field, the positive direction for the electric is clockwise. The calculated value of E ispositive, so the electric field at point ais toward the left and E -- -(2.5 x 10-4V/m)i.The force on the electron is 7i - -"8 and, according to Newtons second law, its acceleration is F _ I eE eD (1.60 x cX 1o-tn -,/-L.J x 1o-4 v/m)i _(4.4xl0,mls)i 2(I/ V/LIL)L i- mm A -_ I.UU2(IU 9.Il x -2.5 kg 10-:t /^A tn7-^_t^2.,?The mass and charge of an electron can be found in Appendix B.(b) The electric field at r - 0 is zero, so the force and acceleration of an election placed at pointb arc zeto.(c) The electric field at point c has the same magnitude as the field at pqinl a but now the fieldis to the right. That is E - (2.5 x 10-4V/m)i and d- -(4.4 x 107 mlrt)i.81(a) The magnetic flux through the loop is Os : B A, where B is the magnitude of the magneticfield and A is the area of the loop. The magnitude of the average emf is given by Faradays law; tave - B LAI Lt, where LA is the change in the area in time Lt Since the final area is zero,the change in area is the initial area and tave: BAILt - (2.0TXO.20m)2 lQ.20s):0.40V.188 Chapter 30
193.
(b) The average current in the loop is the emf divided by the resistance of the loop f iav -tuunl l? - (0.40 v) lQ0 x 10- o) : 20 A.85(a), (b), (c), (d), and (e) Just after the switch is closed the cuffent iz through the inductor is zero.The loop rule applied to the left loop gives t - IrRr:0, so er : tlRt - (10V) 16.0Q) -2.0A.The junction rule gives i, -- i1 - 2.0 A. Since i,z - 0, the potential difference across Rz isVz : izRz - 0. The potential differences across the inductor and resistor must sum to t and, sinceVz: 0, Vt: t :10V. The rate of change of i,2 is dizldt - VrlL - (10V)l(5.0H) : 2.0A/*.(g), (h), (i), 0), (k), and (1) After the switch has been closed for a long time the current i2reaches a constant value. Since its derivative is zero the potential difference across the inductoris Vr : 0. The potenttal differences across both Rr and Rz are equal to the emf of the batteryso z1 - tlRt: (10Y)l(5.0O) - 2.0A and i2: tlRr: (10V)l(10O) - 1.0A. The junctionrule gives i, - it * iz - 3.0 A.9s(a) Because the inductor is in series with the battery the current in the circuit builds slowly andjust after the switch is closed it is zero.(b) Since all currents are zerojust after the switch is closed the emf of the inductor must match theemf of the battery in magnitude. Thus L(diburldt) - t and diaur- tlL - (40V)|60 x 10-H) -8.0 x 102 A/r.(c) Replace the two resistors in parallel with their equivalent resistor. The equivalent resistanceis RrRz (2ok[)X2okq) R"n- Rt.1 - 1 - loko. Rz 20ko + 2okoThe current as a function of time is given by ibut [r - e-/"f ,where 4 is the inductive time constant. Its value rs r7 - LfReq - (50 x 10-3H)l(10 x 103 O) -5.0 x 10-6 s. At t: 3.0 x 10-u r, tlr" - (3 .0)l(5.0) - 0.60 and ibut: 40 v- [t - e-o60] - l8 x 1o-3 A 10 x 103 o(d) Differentiate the expression foliaut to obtain diaut e-tlrr: !"-t/rrq dt L")where 17 - LfReq was used to obtain the last form. At t - 3.0 x 10-6 s diaut 40 V -l dt ffie-o60 - 44 x I02A/t(e) A long time after the switch is closed the currents are constant and the emf of the inductoris zero. The current in the battery is iaut - t lRrq- (40V)/(10 x 103 O) : 4.0 x 10-3 A. Chapter 30 189
194.
(0 The currents are constant and diaurld,t: 0.97(a) and (b) Take clockwise current to be positive and counterclockwise current to be negative.Then according to the right-hand rule we must take the nonnal to the loop to be into the page,so the flux is negative if the magnetic field is out of the page and positive if it is into the page.Assume the field in region 1 is out of the page. We will obtain a negative result for the fieldif the assumption is incorrect. Let r be the distance that the front edge of the loop is intoregion 1. Then while the loop is entering this region flux is -BtHr and, according to Faradayslaw, the emf induced around the loop is t- BrH(drldt): B1Hu. The current in the loop isi-tlR-BrHufR,so Bt: g- (3 0 x 10-6AX0020o) - r.0 x 10-5T. Hu (0.0150mXO.40m/s)The field is positive and therefore out of the page.(c) and (d) Assume that the field 82 of region 2 is out of the page. Let r now be thedistance the front end of the loop is into region 2 as the loop enters that region. The flux is-B1H(D r) B2Hr, the emf is ti - (Bz - Br)Hu f R. The field of region 2 is (-2.0 x 10-u e(0.020 Q) :3.3 x 10-6T. Bz: Br. 1.0 x 10-t T+ #- (0.015 m)(0.40 m/s)The field is positive, indicating that it is out of the page.190 Chapter 30
195.
Chapter 3L7(a) The mass nl, cotresponds to the inductance, so m : I.25 kg.(b) The spring constant k coffesponds to the reciprocal of the capacitance. Since the totalenergy is given by (Ji - QlzC, where A is the maximum charge on the capacitor and C is thecapacitance, t r-,-Q (tlsxlo-uc) -2.69x10-3F u- A, Wand k- _ 372N/-.(c) The maximum displacement nrn coffesponds to the maximum charge, so r,n: 1, .7 5 x 10-4 m .(d) The maximum speed u?,,.1 corresponds to the maximum cuffent. The maximum current is J-ew-&- vv- -3.ozxlo-3A. rcThus utn-3.02 x 10-*/r.15(a) Since the frequency of oscillation f is related to the inductance and capacitance L C byf: I l2n/ LC, the smaller value of C gives the larger value of Hence, f^u* _ f. Il2nm, I - IlZn/TC* and ,mJmm o -c J max f.: : Jmm JC^in(b) You want to choose the additional capacitance C so the ratio of the frequencies is 1.60MHz r: - 2.96 . 0.54MHzSince the additional capacitor is in parallel with the tuning capacitor, its capacitance adds to thatof the tuning capacitor. If C is in picofarads, then re - 2.96 . Chapter 3 t l9l
196.
The solution for e is C- (36s pF) - (2.eq2(10 pF) - 36pF. (2.96)2 1(c) Solve ff : 0.54 MHz. Thus 11 L- 1 :2.2 x 10-4H. @Ctr27Let t be a time at which the eapaeitor is fully charged in some cycle and let q*ax I be the chargeon the eapacitor then. The energy in the eapaeitor at that time is u(t) - @: 2e" 2C A- e-RttL )where llmaxl : Q e-Rt/zLwas used. Here A is the charge at t : 0. One cycle lateg the maximum charge is Qmaxl - Q e-R(t+r)l2Land the energy is q e-R(t+TlL u(t+T) -@- 2C" 2C )where T is the period of oscillation. The fractional loss in energy is LLr U(t) - U(t + T) e-RtlL - e-R(t+r)/L tf U@ e-HwAssume that Rf I L is small compared to 1 (the resistance is small) and use the Maclaurin seriesto expand the exponential. The first two terms are: ,-RT/LE1 y. LReplace T with 2n f u, where u is the angular frequency of oscillation. Thus LLr / 87 RT 2rR, u ^ =1 [t L)- L- oL"33(a) The generator emf is amaximumwhen sin(oat--Tl4) - I or a6t-nl4 ^ (nl2)*2nn, wheren is an integer, including zero. The first time this occurs after f : 0 is wherT a4t - n14 - T12 or 3T . - t-3r: 4ua 4(350 s- ) t 6.73x10-3s.(b) The cuffent is amaximum when sin(a,at-3nf 4) - I, or a6t - 3nl4 - n12*2nn. The firsttime this occurs after t : 0 is when 5n t- 4ua tn - 4(350 s- . 1)(c) The current lags the inductor by n 12 rad, so the circuit element must be an inductor.192 Chapter 3l
197.
(d) The current amplitude I is related to the voltage amplitude Vr by Vnthe inductive rcactance, given by X p - uaL. Furthermore, since there is only one element inthe circuit, the amplitude of the potential difference across the element must be the same as theamplitude of the generator emf: Vr : trn Thus trn - IwyL and r_ trn L_+- 30.0 v _0.138H. Iua rc20 x 10-3 AX350 rad ls)39(a) The capacitive reactance is Xs - 37 9 {l aaC 2r f 6C 2n(60.0 Hz)(70.0 x 10-o F) The inductive reactance is Xr: u)d,L - Lnf aL:2r(60.0H2)(230 x 10-H) - 86.7 Q.The impedance is Z- R2+(Xr-Xd2 (200 o)2 + (37.9 o - 96.7 Q)2 :206Q .(b) The phase angle is Q_tan- (ry)_tan-( 86.7Q-37.9Q ): r) 200 1,3.7"(c) The current amplitude is T C Qyy1, 36.0 V -0.r75A. t- I Z 206Q(d) The voltage amplitudes are vp: IR - (0.t75AX200O) - 35.0V, v7: IXr - (0.i75 AX86.7 O) - 15.2V, Vrand Vc : I Xc - (0 .17 5AX3 7 .g O) : 6,63V .Note that Xy >diagram is drawn to scale on the right.45(a) For a given amplitude trn of the generator emf, the current amplitude is given by c c T I ,- Lyy1, cyyy Z Chapter 31 193
198.
where R is the resistance, L is the inductance, C is the capacitance, and Wd is the angularfrequency. To find the maximum, set the derivative with respect to Wd equal to zero and solvefor Wd. The derivative is dI dWd = -Em [R2 + (Wd L - l/wd c iJ -3/2 [Wd L - _1_] [L + _1_]. WdC w~CThe only factor that can equal zero is WdL - (l/WdC) and it does for Wd = l/VLC. For thegiven circuit, 1 1 Wd = VLC = J (1.00 H)(20.0 X 10-6 F) = 224 rad/ s .(b) For this value of the angular frequency, the impedance is Z = R and the current amplitude is 1= Em = 30.0V = 6.00A. R s.oon(c) and (d) You want to find the values of Wd for which 1= Em /2R. This means Em _ Em JR2 + (WdL - 1/wdC )2 2RCancel the factors Em that appear on both sides, square both sides, and set the reciprocals of thetwo sides equal to each other to obtain R2 + (Wd L _ _ Wd 1_)2 = 4R2. CThus ( Wd L __ 1_) WdC 2 = 3R2 •Now take the square root of both sides and multiply by wdC to obtain w~(LC)±Wd(v3CR) -1=0,where the symbol ± indicates the two possible signs for the square root. The last equation is aquadratic equation for Wd. Its solutions are ±V3CR ± V3C2 R2 + 4LC Wd = 2LCYou want the two positive solutions. The smaller of these is -V3CR+ V3C2R2 +4LC W2 = 2LC -V3(20.0 x 10- 6 F)(S.OO n) 2(1.00 H)(20.0 x 10- 6 F) J3(20.0 X 10- 6 F)2(S.00 n)2 + 4(1.00 H)(20.0 x 10- 6 F) +~~--------------~------~----------- 2(1.00 H)(20.0 x 10- 6 F) = 219 rad/s194 Chapter 31
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and the larger is +t/icft + a1 2LC +/1(zo.o x lo-u pxs.oo o) 2(1.00 HX20.0 x 10-6 F) 3(20.0 x 10- e p)2(5.00 Q)2 + 4(1 .00 Hx20.0 x 10-6 F) 2(1.00 HX20.0 x 10-o F) : 228 rad/s .(e) The fractional width is at-az 228radls - 2lgradls _ ag - 224 radf s 0 .04 .49[Jse the expressions found in Problem 31- 45: +t6cR+ A1 2LCand t,J2 -fjcR+ 2LCAlso use tFLCThus Lra et - Loz: ZtECntR: 3C - 2LC R T55(a) The impedance is given by z-{Rr+(xr-xs)z,where R is the resistance, X 1 is the inductive reactance, and Xs is the capacitive reactance.Thus z - {(12.0o)2+(1.30o -0)2 - l2.lQ.(b) The average rate at which energy is supplied to the aff conditioner is given by s2 Puus: cos @, t Chapter 31 195
200.
where cos / is the power factor. Now cos Q:2: ffi:o.ssz,SO puun: g [(tO,?tl (0.gg2):1.1g x 103w. Ln.to jvl57(a) The power factor is cos /, where 0 is the phase angle when the current is written i/sin(aat - il. Thus Q: -42.0" and cos d: cos(-42.0o) - 0.743.(b) Since 0 <(c) The phase angle is given by tan$- (Xr XdlR, where Xr is the inductive reactance,Xs is the capacitive reactance, and R is the resistance. Now tan Q : tan( -42.0o)negative number. This means X 7 Xs is negative, or Xs >predominantly capacitive.(d) If the circuit is in resonance, X7 is the same as Xs, tanS is zero, and 0 would be zero.Since d is not zero, we conclude the circuit is not in resonance.(e), (0, and (g) Since tan $ is negative and finite, neither the capacitive reactance nor theresistance is zero. This means the box must contain a capacitor and a resistor. The inductivercactance may be zero, so there need not be an inductor. If there is an inductor, its reactancemust be less than that of the capacitor at the operating frequency.(h) The average power is T) 1 Purs: c Tnnna- cos d : 1 5.0 VXI.20 AXO .7 43) - 33.4 W . tE^-I tQ(i) The answers above depend on the frequency only through the phase angle O:,which is given.If values are given for R, L, and C , then the value of the frequency would also be needed tocompute the power factor.63(a) If Ire is the number of primary turns and lf" is the number of secondary turns, then v,: ftu,: (#) lzov)- z4v(b) and (c) The current in the secondary is given by Ohms law: I, R, 15O--o164The current in the primary is I,p- I ffi - (#) (0.16A) :3.2x 1o-3 A.196 Chapter 3I
201.
67use the trigonometric identity, found in AppendiX E, sin a-sin p--2sin (+)cos w),where a and B are any two angles. ThusW -Vz- Asin(c..at)-Asin(aat- I20"): zAsin(I20")cos(aat- 60o): ,trlcos(c,;6t-_60o),where sin( 120") : ,n p was used. Similarly,V-Vt - Asin(c,at)-A sin(o,at_240o) : 2Asin(240) cos(a.at-120") : -ttrlcos(cu 6t-120o) ,where sin(240o) : -E 12 was used, and Vz Vz - Asin(a,at - I20") - Asin(aat - 240o): zAsin(I20")cos(r,,at - 180) - frA cos (wat - l Soo) .All of these are sinusoidal functions of aa and all have amplitudes of tfrA.7l(a) Let Vc be the maximum potential difference across the capacitor, Vr be the maximumpotential difference across the inductor, and Vp be the maximum potential difference across theresistor. Then the phase constant O is tan-,(Vt V":tan-, (ry Vn tan-t(1.00) :45.0o. vp / ,= )-(b) Since the maximum emf is related to the current amplitude by t* _ I Z, where Z is theimpedanceand R-Zcos$, R-ry: :70.,1 {1 .73(a) The frequency of oscillation of an LC circuit is f - I l2nre, where L is the inductanceand C is the capacitance. Thus L-#r* 1 -6 H.(b) The total energy is tl : ifP, where I is the current amplitude. Thus U - LtO.89 x10- tt)(l .20 x lo-s A)2 - l.7g x 10-1r J. Chapter 3l I97
202.
(c) The total energy is also given by U - l2C, where A is the Q charge amplitude. ThusQ-tffi- 2(I .79 x 10-11 JX340 x 10-aF): 1.10 x 10-7 C.83(a) The total energy Lf of the circuit is the sum of the energy IJ n stored in the capacitor andthe energy LIa stored in the inductor at the same time. Since LIs - 2.00[Jp, the total energy istl:3.00t-In Now U- Qlze and Un: q2 lzc, where a is the maximum charge, q is thecharge when the magnetic energy is twice the electrical energy, and C is the cryaeitance. ThusQlzc - 3.ooqlzC and q: A Itm - o .577e.(b) If the capacitor has maximum charge at time t - 0, then qangular frequency of oscillation. This means at - cos- t (0 .577)where T is the period, t-ryr 2r -0.1 s3T8s(a) The energy stored in a eapacitor is given by Un - q2 lzc, where q is the charge and e is thecapaeitance. Now q2 is periodic with a period of T12, where T is the period of the driving emf,so LI n has the same value at the beginning and end of each cycle. Actually IJ n has the samevalue at the beginning and end of each half cycle.(b) The energy stored in an inductor is given by Liz f 2, where i is the current and L is theinductance. The square of the current is periodic with a period of T 12, So it has the same valueat the beginning and end of each cycle.(c) The rate with which the driving emf device supplies energy iswhere ris the currenr ampli tl;:;I o#.Xgglitl:Ho,t, ,!I. angular rrsquency, and a,sa phase constant. The energy supplied over a cycle is trs : ps d,t: It,n sin( aat - ild,t Io Irsin(r,,,af) : I trn [ sin(c,,ar) tsin(a,,a t)cos(il -cos(cr., at)sin(/)] dt , Jowhere the trigonometric identity sin(a - P) - sinocos P-cosasinp was used. Now the integralof sin(rot) over a cycle is T12 and the integral of sin(u,at)cos(wat) over a cycle is zero, soEs -- *t t,"cos @.(d) The rate of energy dissipation in a resistor is given by Pp: i2 R - /2 sin(rot - d)and the energy dissipated over a cycle is Ep: 12 : ]tRT [sin2 (rat - il d,t . Jo(e) Now trn: IZ, where Z is the impedance, and R- Z cos$, so Es - *fTZ cos@ -*tRT: ER.198 Chapter 3l
203.
Chapter 323(a) [Jse Gauss law for magnetism: f E.d,A- 0. Write f E.dA- Or + Qz+Qs, where Qr isthe magnetic flux through the first end mentioned, Q2 is the magnetic flux through the secondend mentioned, and Q g is the magnetic flux through the curved surface. Over the first end, themagnetic field is inward, so the flux is 01field is uniform, normal to the surface, and outward, so the flux is Q2- AB - 7Tr2 B, where Ais the area of the end and r is the radius of the cylinder. Its value is Q2 - z-(0 .I20m)2(1.60 x 10-r) - +7.24 x 10-5 Wb : *72.4 trtWb .Since the three fluxes must sum to zero, Qs(b) The minus sign indicates that the flux is inward through the curved surface.-3Consider a circle of radius r (- 6.0 mm), between the plates and with its center on the axis ofthe capacitor. The cuffent through this circle is zero, so the Ampere-Maxwell law becomes f -+ dQn B.dg: lroroT, fwhere E is the magnetic field at points on the circle and Q p is the electric flux through thecircle. The magnetic field is tangent to the circle at all points on Lt, so f U . d,i : 2rr B. Theelectric flux through the circle is Q6 - r RE, where R (: 3.0 mm) is the radius of a capacitorplate. When these substitutions are made, the Ampere-Maxwell law becomes 2nrB: ltoesTRz#Thus dE 2rB 2(6.A x 10-3 m)(2.0 x 10-t f) :2.4x10lrvl^.s. dt - FoeoRz - (4n x 10-, Hl^X8.85 x 10-12 FmX3.0 x 10-t nr)-13The displacement current is given by . d,tr ia: eoA dt , Chapter 32 199
204.
where A is the area of a plate and tr is the magnitude of the electric field between the plates.The field between the plates is uniform, So E - V ld, where V is the potential difference acrossthe plates and d is the plate separation. Thus . Ld: esA dV d dtNow esAld is the capacitance C of a parallel-plate capacitor without a dielectric, so ia-C#2t(a) For a parallel-plate capacitor, the charge q on the positive plate is given by qwhere A is the plate areU d is the plate separation, and V is the potential difference between theplates. In terms of the electric field E between the plates, V : Ed, So q: eoAtr : €6Oe, whereQ n is the total electric flux through the region between the plates. The true current into thepositive plate is i - dqldt - es dQ t ldt - id" totut, where id, torut is the total displacement currentbetween the plates. Thus id, totur - 2.0 A.(b) Since id, totar: 60 dA t I dt : eyA dE I dt, dE :::i d" ,otul 2.0 A 2.3 x 101V l^. s. dt eoA (8.85 x 10- t2 F lmX1.0 m)2(c) The displacement current is uniformly distributed over the area. If a is the area enclosed bythe dashed lines and A is the area of a plate, then the displacement current through the dashedpath is id""n"- frrdbtar:ffi e.oA): o.5oA.(d) According to Maxwells law of induction, f 6B .di: ltyid,"n, - (4n x 10-tll^X0.50A): 6.3 x 10-7 T.ril. JNotice that the integral is around the dashed path and the displacement current on the rightside of the Maxwells law equation is the displacement current through that path, not the totaldisplacement current.3s(a) The z component of the orbital angular momentum is given by Lorb,z: m(hf 2r, where h isthe Planck constant. Since Tftp: 0, Lorb,, : 0.(b) The z component of the orbital contribution to the magnetic dipole moment is given byForb, z(c) The potentral energy associated with the orbital contribution to the magnetic dipole momentis given by [J[ - -Forb,"Bext, where Be*t is the z component of the externalmagnetic field. Since lJorb,z:0, fJ :0.200 Chapter 3 2
205.
(d) The z component of the spin magnetic dipole moment is either +lrn or -pe, so the potentialenergy is either [Ji - -paBe*t: -e.27 x 10-24 JIT)(35 x 10-T) - _3.2 x 10-25 J.or Ui - +3.2 x 10-25 J.(e) Substitute m4 into the equations given above. The z component of the orbital angularmomentum is (-3X6626 x 10-34Js) Lorb,r: ry- zr 2n - 3a -3.2x r0- J.s.(0 Ihe z component of the orbital contribution to the magnetic dipole moment is Forb, z(g) The potential energy associated with the orbital contribution to the magnetic dipole momentis [] - -Fora, "8"*t: -QJ8 x 10-23 JIT)(35 x l0-T) - -9.7 x 10-25 J.(h) The potential energy associated with spin does not depend on Tftp. It is +3.2 x 10-25 J.39The magnetization is the dipole moment per unit volume, so the dipole moment is given byl.L - MV, where M is the magnetization and V is the volume of the cylinder. Use V: nrzL,where r is the radius of the cylinder and L is its length. Thus Ltr: MrrzL - (5.30 x 103 AIN?r(0.500 x I0-2m)2(5.00 x L0-2 m):2.08 x I0-2 JIT .45(a) The number of atoms per unit volume in states with the dipole moment aligned with themagnetic field is N. - AepB /kr and the number per unit volume in states with the dipolemoment antialigned is lf- - Ae- 1tB lkr, where A is a constant of proportionality. The totalnumber of atoms per unit volume is jV - ff. + l/- - A ("ru /kr * e- p"B /*) . Thus A- et"B=,.= *t e- /kr =,.= p,B/krThe magn ettzatton is the net dipole moment per unit volume. Subtract the magnitude of the totaldipole moment per unit volume of the antialigned moments from the total dipole moment perunit volume of the aligned moments. The result is aE IV peuB /kr - IY pe- FB /kr IY 1t ("ru /kr - e- p"B lkr) ai M: - -lYptanh(p,BlkT). Chapter 32 201
206.
(b) If p,B < - p,B lkT. (See Appendix E forthe power series expansion of the exponential function.) The expression for the magnetizationbecomes IY p, [f t + ptB lkT) - (1 - p,B lkT)] IV pB M= (1 + pB lkT) + (l - p,B lkT) KT(c) If p,B >denominator of the expression for M. Thus m vffi-tVtr. IV M = u"::=t(d) The expression for M predicts that it is linear in B lkT for p,B lkf small and independentof B lkf for p,B lkT large. The figure agrees with these predictions.47(a) The field of a dipole along its axis is given by Eq . 29-27: ts_lto p, L) G7,where LL is the dipole moment and z is the distance from the dipole. Thus the magnitude of themagnetic field is B 2r(10 x 10-n -)(b) The energy of a magnetic dipole with dipole moment I, in a magnetic field E is given byU- -rt. E - -p,Bcos/, where O is the angle between the dipole moment and the field. Theenergy required to turn it end for end (from Q : 0o to 0 - 180") is L(J : -pB(cos 180o - cos0") - 2LtB - 2(1 .5 x L0-23 JIT)(3.0 x 10-uf) : 9.0 x ro-ze J - 5.6 x 1o-lo eV.The mean kinetic energy of translation at room temperature is about 0.04 eV (see Eq. 19-24 orSample Problem 32-3). Thus if dipole-dipole interactions were responsible for aligning dipoles,collisions would easily randomtze the directions of the moments and they would not remainaligned.53(a) If the magn etization of the sphere is saturated, the total dipole moment is Ftotat - IY Lr, wherel/ is the number of iron atoms in the sphere and LL is the dipole moment of an iron atom. Wewish to find the radius of an iron sphere with ,,Af iron atoms. The mass of such a sphere is IVm,where m is the mass of an iron atom. It is also given by 4n pRt 13, where p is the density ofiron and R is the radius of the sphere. Thus IYm:4rpRt 13 and pR3 iv- 4n3m202 Chapter 3 2
207.
Substitute this into Ftotat - Ir, Lt to obtain Ftotal : 4r pR3 p, 3mSolve for R and obtain R - lt*uou1tt L 4rpp I .The mass of an iron atom is m :56 u - (56 uX 1.66 x 10-27 kg/u) - 9.30 x 10-26 kg .So x 10-26 kg)(8.0 x 3(9.30 1 022 I lT) R-L 4r(14 x 103 kelmt)(z] x 10-23 J lT)]t":1.8x105m.(b) The volume of the sphere rs v,- !nt- +(1.82 x 1o5m)3 -2.53 x 1016m3and the volume of Earth is V"- x lo6m)3- 1.08 x Lo2*, +(6.37so the fraction of Earths volume that is occupied by the sphere is 2.53 x r rr-r 1016 m3 : 2.3 x 10-5 . 1.08 x 1021 m3The radius of Earth was obtained from Appendix C.--))(r) The horizontal and vertical directions arepe{pendicular to each other, so the magnitude ofthe field is B-ffi:# lto lt fLUfL l+3sin2 *,,where the trigonometric identity cos rn: I - sin2 was used.(b) The tangent of the inclination angle is ^rn tanQ6- ";^: (ffi) (ffi :#) - 2 tan*,where tan (sin l(cos ,-) was used. ^*: ^*)61(a) The z component of the orbital angular momentum can have the values Lorb," : TrL2h f 2r,where m4 can take on any integer value from -3 to +3, inclusive. There ate seven such values(-3, -2, -1, 0, *1, *2, and +3). Chapter 32 203
208.
(b) The z component of the orbital magnetic moment is given by plorb,z _ -Trl,peh f 4rffi, wherem is the electron mass. Since there is a different value for each possible value of m6 there areseven different values in all.(c) The greatest possible value of Lorb,z occurs rf m,p - +3 is 3hf 2n.(d) The greatest value of L;orb, is 3eh lanm. "(e) Add the orbital and spin angular momentai Lnet,z: Lorb,rt Lr,r: (mthl2r)*(mrhl2n Toobtainthemaximumvalue, setm2equalto+3 and TTLs equalto+j. Theresult LS L1sqs - 3.5h12n.(0 Write Lnesr: Mhl2n, where M is half an odd integer. M can take on all such values from-3.5 to +3.5. There are eight of these: -3.5, -2.5, -1.5, -0.5, +0.5, +1.5, +2.5, and +3.5.204 Chapter 32
209.
Chapter 33IfIf f is the frequency and I is the wavelength of an electromagnetic wave, then f c. The ^_frequency is the same as the frequency of oscillation of the current in the LC circuit of thegenerator. That is, f : llLr{Lc, where e is the capaeitance and L is the inductance. Thus :=11 Lt. ZnJLCThe solution for L is 1z x 10-^) (550 u- r, 5.00 x 10-21 H. 4rzCc2 412(17 x 10-12 FX3.00 x 108 mls)zThis is exceedingly small.2TThe plasma completely reflects all the energy incident on Lt, so the rcdtation pressure is givenbypr:Llfc,where I istheintensity.Theintensityis I-PlA,where P isthepowerand Ais the area intercepted by the radration. Thus 2P P,23Let fbe the fraction of the incident beam intensity that is reflected. The fraction absorbed isI- f . The reflected portion exerts a radiation pressure of p, - (2f Idlc and the absorbed portionexerts a radration pressure of po -- (1 f)Iaf c, where Is is the incident intensity. The factor 2enters the first expression because the momentum of the reflected portion is reversed. The totalrudration pressure is the sum of the two contributions: ,_ 2flo+(1 f)Io J (l + f)Io Ptotal:PrrPo: "To relate the intensity and energy density, consider a tube with length (. and cross-sectionalarea A, lying with its axis along the propagation direction of an electromagnetic wave. Theelectromagnetic energy inside is LI - uA(., where u is the energy density. A11 this energy willpass through the end in time t - ( I , so the intensity is r-Y At uA(.c A( : uc.Thus u density are inherently positive, regardless of thepropagation direction. Chapter 33 205
210.
For the partial|y reflected and parttally absorbed wave, the intensity just outside the surface isI - /o+ f Io - (1 + f)Io, where the first term is associated with the incident beam and the secondis associated with the reflected beam. The energy density is, therefore, ,t.t, : I- (1 + /)/o . cc)the same as radtation pressure.25(a) Since c: where ) is the wavelength and f is the frequency of the wave, ^f, 3oo x m/s f : 9- -108 3.0m - 1.0 x 108 Hz .(b) The angular frequency is ^ u) :2nf :2r(1 .0 x 108 H4: 6.3 x 108 rudf s.(c) The angular wave number is -2n2r k-T-rr_-2. lradlm.(d) The magnetic field amplitude is Brr: En: ?ooY13 c 3.00 x 108 m/s ,(e) E -,rst be in the positi ve z direction when d ir in the positi ve A direction in order for E Eto be in the positive r direction (the direction of propagation). "(0 The time-averaged rate of energy flow or intensity of the wave is r _ E, I-ffi (3oo Y l^), :I.2xI02Wl^.(g) Since the sheet is perfectly absorbitg, the rate per unit area with which momentum is deliveredto it is Il", so d,p- IA: (l|gWlm(2.0m- , - _8.0x10-N. it c 3^0(h) The rcdration pressure is p,- ry-#:4.ox1o-7pa.27If the beam carries energy U away from the spaceship, then it also carries momentum p - U l"away. Since the total momentum of the spaceship and light is conserved, this is the magnitude of206 Chapter 3 3
211.
the momentum acquired by the spaceship . If P is the power of the laser, then the energy carriedawayintime t is (Ji - Pt Thusp - Ptf c and, rf m is mass of the spaceship, its speed is p- Pt: (10 I103yxldx864 x 104s/d)- r.gx 10-r^l- r.gmm/s. J- m mc (1.5 x 103 kgX3.00 x 108 m/s)3sLet Is be in the intensity of the unpol artzed light that is incident on the first polari zing sheet.Then the transmitted intensity is Ir - Lto and the direction of polanzation of the transmittedlight is 0r (: 40") counterclockwise from the A axis in the diagram.The polarizrng direction of the second sheet is 0z (:20") clockwise from the A axis so the anglebetween the direction of polarrzation of the light that is incident on that sheet and the polarizingdirection of the of the sheet is 40" + 20" : 60o. The transmitted intensity is 12: Ircos2 60o - 60o )trcos2and the direction of pol arrzation of the transmitted light is 20o clockwise from the A axis.The polariztng direction of the third sheet is fu (:40) counterclockwise from the A axis so theangle between the direction of polanzation of the light incident on that sheet and the polarizrngdirection of the sheet is 20" + 40o : 60o. The transmitted intensity is : Izcos2 60o _ lrrcos4 60o : 3. I x I0_2 .3.Ioh of the lights initial intensity is transmitted.43(a) The rotation cannot be done with a single sheet. If a sheet is placed with its polarizrngdirection at an angle of 90" to the direction of pol arrzation of the incident radiation, oo radrationis transmitted.It can be done with two sheets. Place the first sheet with its polarizrng direction at some angle0 , between 0 and 90o, to the direction of pol arrzation of the incident radiation. Place the secondsheet with its polari zing direction at 90o to the polari zation direction of the incident radration.The transmitted radiation is then polarized at 90o to the incident polarization direction. Theintensity is /s cos? 0 cost(gO" - 0): /0 cos2 0 srnz 0, where I0 is the incident radiation. If 0 is not0 or 90", the transmitted intensity is not zero.(b) Consider n sheets, with the polariztngdirection of the first sheet making an angle of 0 - 90" lnwith the direction of polarrzation of the incident radration and with the polariztng direction ofeach successive sheet rotated 90" f n in the same direction from the polari zing direction of theprevious sheet. The transmitted rcdration is polari zed with its direction of polanzation makingan angle of 90o with the direction of polarrzation of the incident radration. The intensity isI - .Is cosn (90 ln). You want the smallest integer value of n for which this is greater than0.60,Is. Chapter i 3 207
212.
Start with n : 2 and calcul ate cos2n(90" l.r,). If the result is greater than 0.60, you have obtainedthe solutiotl. If it is less, increase n by 1 and try again. Repeat this process, increasing n by Ieach time, until you have a value for which cos2(90" ld is greater than 0.60. The first one willbe n: 5.51Consider a ruy that grazes the top of the pole, as shown inthe diagram to the right. Here 0r:35o, lt: 0.50m, and(,2- 1.50m. The length of the shadow is tr* L. r is givenbyrthe law of refraction , TL2 stn 02 : rL1 sin 91 . Take T _ 1 and waterTL2 gr t ( sin 35g) ): sin 0z: sin-, ( rr sin- - 25.55o . - )-Dr l33 /-/-L is given by shador,V L - (.ztat0z - (1 .50 m) tan 25.55o - 0 .72 m . LrThe length of the shadow is 0.35 m + 0 .72m - 1.07 m.--33Look at the diagram on the right. The two angleslabeled a have the same value. 0z is the angle ofrefraction. Because the dotted lines are pe{pendicularto the prism surface 0z * a - 90o and aBecause the interior angles of a triangle sum to 1 80o,180o - 202* d: 180 and 0z - 012.Now look at the next dragram and consider the triangleformed by the two normals and the ruy in the interior.The two equal interior angles each have the value 0 -02.Because the exterior angle of a triangle is equal to thesum of the two opposite interior angles, ,lt - 2(e - 0z) jand 0 - ?z+rb 12. Upon substitution for 02thrs becomes -t /.0 - (d + ,1,) 12.According to the law of refraction the index of refraction of the prism material is :: sin 0 sin(/ + rD12 stn 02 sin 012208 Chapter 3 3
213.
05(a) No refraction occurs at the surface eb, so the angle of incidence at surface o,c is 90" O.For total internal reflection at the second surface, ne stn(90" il must be greater than TL,.Here ne is the index of refraction for the glass and TLa is the index of refraction for air. Sincesin(90" il - cos @, you want the largest value of O for which nn cos $ >cos / decreases as 0 increases from zero. When 6 has the largest value for which total internalreflection occurs, then rre cos 0 - tra, or , _ cos-r f _L : 48 e" Q: cos e) r -1 - t 152lThe index of refraction for air was taken to be unity.(b) Replace the air with water. If n* (: 1.33) is the index of refraction for water, then the largestvalue of 0 for which total internal reflection occurs is ry) e) _ cos-l f , Q: cos -l r _ 2e.oo 1.52 /69The angle of incidence 0 s for which reflected light is fully polarized is given by Eq. 33-49of the text. If nr is the index of refraction for the medium of incidence and TL2 is the index ofrefraction for the second medium, then 0e - tan-t (n, l"t) - tarrt (t.53 f I.33) :63.8o.73Let 01 (- 45") be the angle of incidence at the first surface and 0z be the angle of refractionthere. Let fu be the angle of incidence at the second surface. The condition for total internalreflection at the second surface is n srn 03 >of refraction n for which this inequality holds.The law of refraction, applied to the first surface, yields nstn02- sin01 . Consideration of thetriangle formed by the surface of the slab and the ray in the slab tells us that 0zThus the condition for total internal reflection becomes 1 <this equation and use srnz e 2 + cos2 02 - 1 to obtain 1 <sin d2 - ( | I sin 01 to obtain ") 1< *3) : 12 srnz o1 rL /The largest value of n for which this equation is true is the value for which 1Solve for n: sin2 45" : 1.22 Chapter 3 3 209
214.
75Let 0 be the angle of incidence and 0z be the angleof refraction at the left face of the plate. Let nbe the index of refraction of the glass. Then, thelaw of refraction yields sind - nsrn?2. The angleof incidence at the right face is also 02. If fu isthe angle of emergence there, then n srn 02 - sin 93.Thus sin 93 - sin 0 and fu - 0. The emerging ray isparallel to the incident ray.You wish to derive an expression for r in termsof 0. If D is the length of the ray in the glass,then Dcos?z:t and D:tf cos02. Theangle ain the diagram equals 0 0z and r - lJ sin a -D sin(O - 0). Thus M Jr - t sin(0 - 0z) cos 02If all the angles e, 02, 02, and 0 0z are small and measured in radians, then sin 0 N 0,srn 02 = 02, sin(0 0) = 0 02, and cos 02 = 1. Thus r E t(0 0). The law of refractionapplied to the point of incidence at the left face of the plate is now 0 E n?z, So 02 E 0 l, and r= t(r- (n - r)t0 t:77The time for light to travel a distance d in free space is t- dlr, where c is the speed of light(3.00 x 108 m/s).(a) Take d to be 150 km : 150 x 103 m. Then, +_d 150x103m f,:;- - 5.00 x 10-4s.(b) At full moon, the Moon and Sun are on opposite sides of Earth, so the distance traveled bythe light is d,- (1.5 x 108km) + 2(3.8 x 105km) - 1.51 x 108km- 1.51 x 10llm. The timetaken by light to travel this distance is d 1.51 x 1011 m J__ l- c 3.00 x 108 m/s 500 s - 8.4 min .The distances are given in the problem.(c) Take d to be 2(l .3 x 10e km) : 2.6 x 1012 m. Then, u d 2.6 x 1012 m J__ I- 8.7x103s-2.4h c 3.00 x 108 m/s210 Chapter 3 3
215.
(d) Take d to be 6500 ly and the speed of light to be 1.00 IV lV. Then, t-l:#ffi:65ooyThe explosion took place in the year 1054 - 6500: -5446 or B.C.5446.79(a) The amplitude of the magnetic field is B10-B 1. According to the argument of the trigonometric function in the expression for the electricfield, the wave is moving in the negative z direction and the electric field is parallel to the Aaxis. In order for E x E to b. in the negative z direction, E must be in the positive r directionwhen E is in the positive A direction. Thus B* - ( 1,.67 x 10-* r) sinf(l .00 x 106 m- t), + wt]is the only nonvanishing component of the magnetic field.The angular wave number is k-1.00 x 106m-l so the angular frequency is u- kc: (1.00 x106 m-txf .00 x 108 mls) : 3.00 x 1014 s-l and B* - (r.67 x 10-t r) sin[(1.00 x 106m-), + (3.00 x 10r4 s-t)t] .(b) The wavelength is ) - 2nlk -2nl(1.00 x 106m-t) - 6.28 x 10-6m.(c) The period is T :2nf u:2nl(3.00 x 1014 s-1) : 2.09 x 10-14 s.(d)Theintensityofthiswaveis/ - Ez,.l2poc- (5.00V1rr1llZ(+rxl0-ttl^X3.00x108 mls-0.0332W l*. (0 A wavelength of 6.28 x 10-6m places this wave in the infrared portion of theelectromagnetic spectrum. See Fig. 33-1.83(a) The power is the same through any hemisphere centered at the source. The arca of ahemisphere of radius r is A - 2nr2. In this case r is the distance from the source to the aircraft.Thus the intensity at the atrcraft is I - PIA: Pf 2nr2 - (180 x 103 Wl2r(90 x 103 m)2 _3.5 x 10-uW l^.(b) The power of the reflection is the product of the intenstty at the arcc.r:aft and the cross sectionof the akqaft: P, - (3.5 x l0-uw l^Xo .22m2) : 7.8 x 10-7w.(c) The intensity at the detector is P, f 2nr2 -- (7.8 x 10-t W) l2r(90 x 103 m)2 - 1.5 x10-17 w l^.(d) Since the intensity is given by I- E2,.l2por,Ern 2 p,scl(e) The rrns value of the magnetic field is Brrn, : Ernlfr"- (1.1 x 10-Vl^)l(t[Dp.O0 x108 mls) : 2.5 x 10-16 T. Chapter 33 2ll
216.
9tThe critical angle for total internal reflection is given by 0"_ sin-t(tln). For n- 1.456 thisangle is 0":43.38o and for TL: I.470 it is 0.:42.86o.(a) An incidence angle of 42.00o is less than the critical angle for both red and blue light. Therefracted light is white.(b) An incidence angle of 43. 10o is less than the critical angle for red light and greater than thecritical angle for blue light. Red light is refracted but blue light is not. The refracted light isreddish.(c) An incidence angle of 44.00o is greater than the critical angle for both red and blue light.Neither is refracted.103(a) Take the derivative of the functions given for tr and B, then substitute them into a2E c-,o2E ancl ozg c- . - ,o2B M - A., 0t, Arz. The derivatives of tr are AE l0t2 - -a2 E,n srn(kr - .,lt) and AE l0r2 - -pz En sin(kr - olt),so the wave equation for the electric field yields u2 - c2 k2. Since e - ck the function satisfiesthe wave equatiorl. Similarly, the derivatives of B are A2 B lt - -a2 Brnsin(kr .,lt) andA2Bl0*- -TtzBrnsrn(kr - u)t) and the wave equation for the magnetic field yields w2 : c2k2.Since w - ck the function satisfies the wave equation.(b) Let LL : kr * wt and consider f to be a function of u, which in turn is a function of n andt. Then the chain rule of the calculus gives a2tr d,2f (ou d,2f at, - d,r, e/ : a,u*- )and a2 012 tr :#l;) :#k2 f ( ou f d,2 d,Substitullgn into the wave equation agarn yields w2 - c2 k2, So the function obeys the waveequatioll. A similar analysis shows that the function for B also satisfies the wave equation.212 Chapter 3 3
217.
Chapter 345The light bulb is labeled O and its image islabeled I on the digram to the right. Considerthe two rays shown on the dragram to the right.One enters the water at A and is reflected fromthe mirror at B. This ray is petpendicular to thewater line and mirror. The second ruy leavesthe lightbulb at the angle 0, enters the water atC, where it is refracted. It is reflected from themirror at D and leaves the water at E. At C the mlfforangle of incidence is I and the angle of refractionts 0. At D the angles of incidence and reflection d3are both 0. At E the angle of incidence is g/ andthe angle of refraction is 0. The dotted lines that Imeet at I represent extensions of the emerging * Irays. Light appears to come from I. We want to Icompute d3.Consideration of the triangle OBE tells us that the distance d2+ fu is Ltan(90o - 0) - Lf tan?,where L is the distance between A and E. Consideration of the triangle OBC tells us thatthe distance between A and C is d1 tan 0 and consideration of the triangle CDE tells us thatthe distance between C and E is 2dztan?t, so L - drtan? + 2dztan?t, d2 + dzZdztan 0) I tan 0, and , d1 tan 0 + 2dztan 0l tuApply the law of refraction at point C: sin dwater. Since the angles 0 and 0 are small we may approximate their sines by their tangents andwrite tan 0 - ntan?t. Us this to substitute for tan? in the expression for fu to obtain , ndt + 2dz , (133)(250 cm) + 2(20@ rr3l n -42 - 200cm - 350 cm,where the index of refraction of water was taken to be 1.33.9(a) The radius of curvature r and focal length f are positive for a conca/e mirror and are relatedby f :rl2,sor-2(+1 8cm):+36cm. Chapter 34 213
218.
(b) Since (tlD+(lli) - llf , where i is the image distance, (18cm)(12cm) ,i 4- - p- f 12cm- 18cm - -36cm.(c) The magnification is m : -i lp - -(-36 cm) l(12 cm: 3.0.(d) The value obtained for i is negative, so the image is virtual.(e) The value obtained for the magnification is positive, so the image is not inverted.(0 Real images are formed by mirrors on the same side as the object and virtual images areformed on the opposite side. Since the image here is virtual it is on the opposite side of themir:ror from the object.11(a) The radius of curvature r and focal length f are positive for a concave mirror and are relatedby f -r12, so r:2(+l2cm)- +24cm.(b) Since (tlD + (Ili) - Ilf , where i is the image distance, &- (12cmX18cm) - 36cm. i- p- f 18cm- I}cm(c) The magnification is m : -i lp - -(36 cm) 108 cm - -2.0.(d) The value obtained for i, is positive, so the image is real.(e) The value obtained for the magnification is negative, so the image is inverted.(0 Real images are formed by mirrors on the same side as the object. Since the image here isreal it is on the same side of the mirror as the object.15(a) The radius of curvature r and focal length f are negative for a convex mirror and are relatedby f : r 12, so r : 2(-10 cm) : -20cm.(b) Since (tld+(Lli,) - Ilf, where i is the image distance, ,i- - -4.44cm. h-(c) The magnification is m - -ilp - -(-4.44cm)l(8 cm - +0.56.(d) The value obtained for i is negative, so the image is virtual.(e) The value obtained for the magnification is positive, so the image is not inverted.(0 Real images are formed by mirrors on the same side as the object and virtual images areformed on the opposite side. Since the image here is virtual it is on the opposite side of themirror from the object27Since the mirror is convex the radius of curvature is negative. The focal length is f- r 12 -(-40cm)fT--20cm.214 Chapter 34
219.
Since (r lil+ (I li)- (1 I il, if Y: . ?,T"This yields p- +5.0cm if ipositive we select iThe magnification is m - -i lpnegative the image is virtual and on the opposite side of the mirror from the object. Since themagnification is positive the image is not inverted.29Since the magnification m is mdistance, i - -mp. lJse this to substitute for i in (t lD + (lli)length. The solve for p. The result is p-r( :)-(+3ocm)( #) :+1 2ocmSince p must be positive we must use the lower sign. Thus the focal length is -30 cm andthe radius of curvature is rnegative the mirror is convex.The object distance is 1,.2m and the image distance is i - -mp: -(0.20)(l20cm): -24cm.Since the image distance is negative the image is virtual and on the opposite side of the mirrorfrom the object. Since the magnification is positive the image is not inverted.3sSolve r + ?- TLz - TLr p?,7for r. the result is r_ i,p(nz- nt) - (-13 cmX+10 cm) nti * nzp ((1.0X- 13 cm) + (1.5)(+10 cm) - 43 cm.Since the image distance is negative the image is virtual and appears on the same side of thesurface as the obj ect.37Solve r TL rrL TLt TL2 - TL1 -t- pifor r. the result is ,i _, nrrp (1.0X+30 cmX+70 cm) - (nz - nt)p - wrr (1.0 - 1.5)(+70 cm) - (1.5X+30 cm) Chapter 34 215
220.
Since the image distance is negative the image is virtual and appears on the same side of thesurface as the obj ect.4tIJse the lens makers equation, Eq. 34-10: i:(n-1) (* ;) ,where f is the focal length, n is the index of refraction, rr is the radius of curvature of the firstsurface encountered by the light and 12 is the radius of curvature of the second surface. Sinceone surface has twice the radius of the other and since one surface is convex to the incominglight while the other is concave, set 12: -2rt to obtain 3(n - 1) i-(n-1)(*.+): 2rrSolve for 11 : 11- 3(n - r)f - 3(15 - lX6omm) aF :zl)mm. 2 2The radii are 45 mm and 90 mm.47The object distance p and image distance i obey (tld + (Lli,)- (1 1il, where f is the focallength. In addition, p + i - L, where L (: 44 cm) is the distance from the slide to the screen.Use i-L-pto substitute fori inthefirstequationand obtatnp2-pL+Lf :0. Thesolutionis p:L+@- (4cm)+ :22cm. 2251The lens is divergirg, so the focal length is negative. Solve (Ild + (Ili,)- Olfi for i. Theresult is i- - -4.gcm. *-The magnification is m- -ilp: -(-4.8 cm)l(+8.0cm)-0.60. Since the image distance isnegative the image is virtual and appears on the same side of the lens as the object. Since themagnification is positive the image is not inverted.--55The lens is converging, so the focal length is positive. Solve (I lil + (I li,)- (T I fi for i. Theresult is pf (+45 cm)(+20 cm) , n, ?- -(45c@:tJocm p-f216 Chapter 34
221.
The magnification is m - -i lp _ -(36 cm) l@S cm) : -0.80. Since the image distance ispositive the image is real and appears on the opposite side of the lens from the object. Since themagnification is negative the image is inverted.61The focal length is :,e_ rtrz (+30 cm)(-42cm) ft _ r;a,-d: -*3 18cmSolve (t lil + (l li) - (l I fl for i. the result is ,i- -s5cm. *-The magnification is m: -ilp - -(55 cm) IQS cm) : -0.73.Since the image distance is positive the image is real and on the opposite side of the lends fromthe object. Since the magnification is negative the image is inverted.75Since m - -i lp, i - -mp: -(+1.25X+16 cm)The result is p p? (+16 cmx _ zocm) f - p + i - (+16 cm) + ( _ zocm) : *8o cm Since fis positive the lens is a converging lens. Since the image distance is negative the imageis virtual and appears on the same side of the lens as the object. Since the magnification ispositive the image is not inverted.79The image is on the same side of the lens as the object. This means that the image is virtual andthe image distance is negative. Solve (tlD + (Ili) - (llfl for i. The result is .pf ?: p- fand the magnification is i m:-p-- p-f f Since the magnification is less than 1.0, f must be negative and the lens must be a diverginglens. The image distance is i: _ (+5.0 cm)(- 10 cm)and the magnification is m : -i lp - -(- 3.3 cm) l(5.0 cm) - 0 .66 cm. Chapter 34 217
222.
Since the magnification is positive the image is not inverted.81Lens I is converging and so has a positive focal length. Solve (Ilpr)+ Qli)_ (Ilil for theimage distance h associated with the image produced by this lens. The result is i1 : hh.- (2ocmX+9ocm) - r6.4cm. pt h QA cm) - (9.0 cm)This image is the object for lens 2. The object distance is d-p, - (8.0cm)-(16.4cm) : -8.4effi.The negative sign indicates that the image is behind the second lens. The lens equation is stillvalid. The second lens has a positive focal length and the image distance for the image it formsis i2: -+3.rcm. #r-The overall magnification is the product of the individual magnifications: / it / ir / l6.4cm / 3.1cm n7, - t71,1t7l2 Since the final image distance is positive the final image is real and on the opposite side of lens2 from the object. Since the magnification is negative the image is inverted.89(a) IfL is the distance between the lenses, then according to Fig. 34-20, the tube length iss: L- foa f"r:25.0cm -4.00cm- 8.00crrr:13.0cm.(b) Solve (tlD + (Ili) - (Ilf.,) for p. The image distance is L - foa* s - 4.00cm* 13.0cm -17.0 cffi, so p : 4- (17o cmX4oo cm) - 5 .z3cm i f oa 1.7 .0 cm - 4.00 cm .(c) The magnification of the objective is m:-;- -m--3 2s(d) The angul ar magnification of the eyepiece is ,Ls: 25 cm 25 cm f"-g^oor*-3.13.(.) The overall magnification of the microscope is M : TTLrrLo - (--3.25X3.13) - -10.2.93(a) When the eye is relaxed, its lens focuses far-away obj ects on the retrna, a distance i behindthe lens. Set p - cc in the thin lens equation to obtain lli - llf , where f is the focal length of218 Chapter 34
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the relaxed effective lens. Thus iimage distance i remains the same but the object distance and focal length change. If p is thenew object distance and f is the new focal length, then 11 -+ x: f 1 pSubstitute i -f and solve for f . You should obtain f:4 p f + 40.0 cm + 2.50 cm(b) Consider the lensmakers equation 1 /1 1 j:(n- (e d 1) ,where 11 and 12 are the radii of curvature of the two surfaces of the lens and n is the index ofrefraction of the lens matertal. For the lens pictured in Fig. 34-46, 11 and 12 have about the samemagnitude, 11 is positive, and 12 is negative. Since the focal length decreases, the combination0 lrr ) - 0 ld must increase" This can be accomplished by decreasing the magnitudes of eitheror both radii.103For a thin lens, (t ld+ (1 ltl - ( I I il, where p is the object distance, i is the image distance, andf is the focal length. Solve for z: fp p- fLet p- f +*, where:r is positive if the object is outside the focal point and negative if it isinside. Then f(f + r) rNow let i - f + n , where r is positive if the image is outside the focal point and negative if itis inside. Then r,-i f:f(f+r) rr f:fand m - f2.105Place an object far away from the composite lens and find the image distance ,i. Since the imageis at a focal point, i : f , the effective focal length of the composite. The final image is producedby two lenses, with the image of the first lens being the object for the second. For the first lens,Olpt)+ Olir): (1 1il, where h is the focal length of this lens and,fi is the image distance forthe image it forms. Sinc e pl : cc, i1 : f r. Chapter 34 219
224.
The thin lens equation, applied to the second lens, is (l lpz) + 0 li) _ (I I f), where pz is theobject distance, i2 is the image distance, and fz is the focal length. If the thicknesses of thelenses can be ignored, the object distance for the second lens is pz - -ir. The negative signmust be used since the image formed by the first lens is beyond the second lens if ir is positive.This means the object for the second lens is virtual and the object distance is negative. If it isnegative, the image formed by the first lens is in front of the second lens and p2 is positive. Inthe thin lens equation, replac e pz with - fi and iz with f to obtain 11 l- 1 hf f,The solution for f is P ,- hfz a fi+ fzt07(a) and (b) Since the height of the image is twice the height of the fly and since the fly and itsimage have the same orientation the magnification of the lens is nL: +2.0. Since TrL - -ilp,wherep is the object distance andi is the image distance, i - -2p. Now lp+il: d, so l-pl: d,and p : d - 20 cm. The image distance is -40 cm.Solve (t ld + (I li) - (l I f) for f . the result is (20 cmx-40 cm) J p.C:(20 -/ I- p?, ::---: :*40Cm. :(c) and (d) Now m: +0.5 and i- -0.5y). Since lp+il: d,0.5p - d and p:2d:40cm. Theimage distance is -20 cm and the focal length is ,. pi I: p+i a ,. (40 cmx -20 cm) A, (40cm)+(-20cm)220 Chapter 34
225.
Chapter 35-5(a) Take the phases of both waves to be zero at the front surfaces of the layers. The phase ofthe first wave at the back surface of the glass is given bV dl : hL - at, where fu (: 2n I is ^t)the angular wave number and )1 is the wavelength in glass. Simrlarly, the phase of the secondwave at the back surface of the plastic is given by dz - kzL - at, where kz (: 2nl )z) is theangular wave number and ),2 is the wavelength in plastic. The angular frequencies are the samesince the waves have the same wavelength in at and the frequency of a wave does not changewhen the wave enters another medium. The phase difference is / 1- 1) fu-dz-(kl-k)L-tn-t/- tt - t , L -r ,1 )Now,1glass. Similarly, ),2: Iui. lrr, where rL2 is the index of refraction of the plastic. This means thatthe phase difference is h- Qz - (2nlu)(u-rr)L.The value of L that makes this 5.65rad is r -(il - d)^* - 2n4tt rd l 2"(1.60 - 1i0) - 360 x 10-6m -(b) 5.65 rad is less than Zrrad (: 6.28rad), the phase difference for completely constructiveinterference, and greater than r rad (: 3. 1 4 rad), the phase difference for completely destructiveinterference. The interference is therefore intermediate, neither completely constructive norcompletely destructive. It is, however, closer to completely constructive than to completelydestructive.15Interference maxima occur at angles 0 such that d stn 0sources, , is the wavelength, and m is an integer. Since d - 2.0m andthat sin g :. 0.25m. You want all values of m (positive and negative) for which l0.25ml < ^These are -4, -3, -2, -1,0, *1, *2, *3, and +4. For each of these except -4 and a4, thereare two different values for 0. A single value of 0 (-90) is associated with msingle value (-90") is associated with m- +4. There are sixteen different angles in all andtherefore sixteen maxima.t7The angular positions of the maxima of a two-slit interference pattern are given by d sin 0 - ffi,where d is the slit separation, , is the wavelength, and m ts an integer. If 0 is small, sin 0 may beapproximated by 0 in radians. Then d0 : m),. The angular separation of two adlacent maxima Chapter 3 5 221
226.
is L0 - ) f d,. Let )be the wavelength for which the angular separation is 10.0% greater. Then1.10^ f d- Ald, or 1.10):1.10(589nm): 648nm. ^/-t9 -The condition for a maximum in the two-slit interference pattern is d sin 0 mA, where d is theslit separation, . is the wavelength, m is an integer, and 0 is the angle made by the interferingrays with the forward direction. If 0 is small, sin 0 may be approximated by g in radians. Thend0: m), and the angular separation of adjacent maxima, one associated with the integer m andthe other associated with the integer m+ 1, is given by L0 - separation on a screen a :distance D away is given by Ly - D L0 ),D ld. Thus ^ld.The ^ (500 x 10-e *X5.40 m)-2.25 x 10-m-2.25mm. LA: A^- .,29The phasor diagram is shown to the right. Here E1 - 1.00, Ez - 2.00,and d:60o. The resultant amplitude E* is given by the trigonometriclaw of cosines: fi-, - tr? + $- z4r4zcos(l80o - O), Ern: (1.00)2 1 (2.00)2 - 2(1 .00X2.00) cos 120" - 2.6539For complete destructive interference, you want the waves reflected from the front and back ofthe coating to differ in phase by an odd multiple of n rad. Each wave is incident on a mediumof higher index of refraction from a medium of lower index, so both suffer phase changes ofr rud on reflection . If L is the thickness of the coating, the wave reflected from the back surfacetravels a distance 2L farther than the wave reflected from the front. The phase difference is2L(2n l), where )" is the wavelength in the coating. If n is the index of refraction of thecoating, I": ln, where ) is the wavelength in vacuum, and the phase difference rs2nL(2rl).Solve 2nL (T) : (2m + I)nfor L. Here m is an integer. The result i S (2m+ 1)) L- u4nTo find the least thickness for which destructive interference occurs, take m : 0. Then r-/ I - L_ 600 x 10-em - 1 Zx 10_7 m. r" 4n 4U,222 Chapter 35
227.
4tSince nr is greater than n2 there is no change in phase on reflection from the first surface. SincerL2 is less than T there is a change in phase of n rad on reflection from the second surface.One wave travels a distance 2L fuither than the other, so the difference in the phases of the twowaves is anLlz* n, where ),2 is the wavelength in medium 2. Since interference produces aminimum the phase difference must be an odd multiple of r. Thus 4n L I , * n - (2m + I)n,where nL is an integer or zero. Replace ),2 with lry where ) is the wavelength in arc, andsolve for l. The result is 4Lnz 2(380 nmxl. 1.34) 1018 nm 2mmFor nt: I,, -- 1018nm and for m- 2,, - (1018nm)12: 509 llm. Other wavelengths arcshorter. Only l - 509 nm is in the visible range.47There is a phase shift on reflection of r for both waves and one wave travels a distance 2Lfunher than the otheE so the phase difference of the reflected waves is ar L I 2, where ),2 is thewavelength in medium 2. Since the result of the interference is a minimum of intensity the phasedifference must be an odd multiple of n. Thus 4r L I ),2 - (2m + l)n, where m is an integer orzero. Replace )2 with lny where ) is the wavelength in arc, and solve for.. The result is 4Lnz 4(210nmXl.46) 2m+1- 2m+1 2m*1For massociated with wavelengths that are not in the visible range.53(a) Oil has a greater index of refraction than at and water has a still greater index of refraction.There is a change of phase of n rud at each reflection. One wave travels a distance 2L fuitherthan the other, where L is the thickness of the oil. The phase difference of the two reflectedwaves is anLllo, where , is the wavelength in oil, and this must be equal to a multiple of 2rfor a bright reflection. Thus 4nLf )."-Zmn, where m is an integer. Use TLoo, where nois the index of refraction for oil, to find the wavelength in air. The result is ^- ZnoL 2(I .20)(460 nm) 1104 nmFor m- I,, : 1l04nm; for m- 2, : (1104nm)12: 552nm; and for m- 3, I -(1104nm)13 - 368nm. Other wavelengths^ shorter. Only ) - 552nm is in the visible range. ate(b) A maximum in transmission occurs for wavelengths for which the reflection is a minimlrm.The phases of the two reflected waves then differ by an odd multiple of n rad. This means4nL l^" : (2m + I)n and , A- 4noL: 4(I .20)(460 nm) : 2208nm 2m* I 2m* I h+l Chapter 3 5 223
228.
For m(2208 nm)/5 _ 442nm. Other wavelengths are shorter. Only l 442nm falls in the visiblerange.63One wave travels a distance 2L further than the other. This wave is reflected twice, once fromthe back surface and once from the front surface. Since TL2 is greater than T there is no changein phase at the back-surface reflection. Since TL1 is greater than rL2 there is a phase change of nat the front-surface reflection. Thus the phase difference of the two waves as they exit material 2is anLl),2*n, where ),2 is the wavelength in material 2. For a maximum in intensity the phasedifference is a multiple of 2n. Thus an L I ),2 * nfor ),2 is , /., : 4L : 4(41 5 nm) 1660 nm L 2m-1 2m-1 2m-lThe wavelength in aff ls (1.59X1660nm) 2639nm - nzz - 2m-1 - 2m-1 ^For m- 1, ) - m- 2, -- 880nm; for m- 3, - 528nm; and for m 2639nm; forl- ^ 377llm. Other wavelengths are shorter. Only l - 528 nm is in^the visible range.7lConsider the interference of waves reflected from the top and bottom surfaces of the air film.The wave reflected from the upper surface does not change phase on reflection but the wavereflected from the bottom surface changes phase by n rad. At a place where the thickness ofthe aff film is L, the condition for fully constructive interference is 2L - (m + *).f, where ,(- 683 nm) is the wavelength and ?rL is an integer. This is satisfied for m - I40: r.- I (m++)) (140.5X693 x lo-em) . - rr- c 2 - 2 -4.80x10-m-0.048mm.At the thin end of the aLr film, there is a bright fringe. It is associated with mtherefore, 140 bright fringes in all.75Consider the interference pattern formed by waves reflected from the upper and lower surfacesof the air wedge. The wave reflected from the lower surface undergoes a r-rad phase changewhile the wave reflected from the upper surface does not. At a place where the thickness ofthe wedge is d, the condition for a maximum in intensity is 2dwavelength in aff and m is an integer. Thus d - (2m+ 1)) f 4. As the geometry of Fig. 35-47shows, d: R- IF4, where R is the radius of curvature of the lens and r is the radius ofaNewton,Sring.Thus(2m+l)^l4-R-|w.So1veforT,FirstfeaffangethetermsSothe equation becomes (2m+ 1)) -R-224 Chapter 3 5
229.
Now square both sides and solve for 12. When you take the square root, you should get r: V +zl)B) lQm (2m + I)zSz 16 If /? is much larger than a wavelength, the first term dominates the second and81Let Ql be the phase difference of the waves in the two arrns when the tube has air in it and Iet $2be the phase difference when the tube is evacuated. These ate different because the wavelengthin arr is different from the wavelength in vacuum. If , is the wavelength in vacuum, then thewavelength in air is lr, where n is the index of refraction of air. This means 4r(n - I)L Qr_dz:2Ll+ +l:where L is the length of the tube. The factor 2 arises because the light traverses the tube twice,once on the way to a mirror and once after reflection from the mirror.Each shift by one fringe coffesponds to a change in phase of 2r rad, so if the interference patternshifts by t/ fringes as the tube is evacuated, 4n(n - l)L :2lYnand l/, 60(500 x 10-e -) - 1.00030 tL- 1+ n 2(5.0 x 10-2 . ^)87Suppose the wave that goes directly to the receiver travels a distance L1 and the reflected wavetravels a distance L2. Since the index of reftaction of water is greater than that of aff thislast wave suffers a phase change on reflection of half a wavelength. To obtain constructiveinterference at the receiver the difference L2 L2 in the distances traveled must be an oddmultiple of a half wavelength.Look at the diagram on the right. The right triangleon the left, formed by the vertical line from the waterto the transmitter T, the ray incident on the water,and the water line, gives D otriangle on the right, formed by the vertical line from athe water to the receiver R, the reflected r?y, and the Iwater line gives D6: rf tan?. Since Do+ Du: D, r al r I I tan? - D (_ Do D6 --+ Chapter 3 5 225
230.
[Jse the identity stn20- tanz0l0+lrrnz0) to show that sing_ (a+ r)l Dz + (a + r)2. Thismeans D2 + @+ r)z Lzo sin 0 a* rand Lzu r D2+(a+r)2 sin 0 a* rSO L2: L2o* Lza D2+(a**). a* rUse the binomial theorem, with pz large and a2 * 12 small, to approximate this expression:LzB D+(a+r)2 l2D.The distance traveled by the direct wave is L1 : J D+ (a - r)2. IJse the binomial theorem toapproximate this expression: L1 E D + (a r)2 lzD. Thus - a2+2ar+12 r a2 -2ar*12 2ar L2-L1ED+ tt :- 2D 2DDSet this equal to (m + *)X, where m is zero or a positive integer. Solve for r. The result isr-(m+))(Dfa)),.89Bright fringes occur rtan angle 0 suchthat dsin0 -ffi, where d is the slit separation, is the ^wavelength in the medium of propagation, and m $ an integer. Near the center of the pattern theangles are small and sin 0 can be approximated by 0 in radians. Thus 0 - m^ld and the angularseparation of two adjacent bright fringes is L0 - When the arrangement is immersed in ^1d,. : )- f d, where .- is the wavelength inwater the angular sep aration of the fringes becomes L0water. Since,- - ln-, where TLra is the index of refraction of water, L?t - ln*d - @qlTL,y1.Since the units of the angles cancel from this equation we may substitute the angles in degrees ^and obtain L?, - 0.30o f L.33 - 0 .23o .93(a) For wavelength A dark bands occur where the path difference is an odd multiple of ^12.That is, where the path difference is (2m+ I)^12, where m is an integer. The fourth dark bandfrom the central bright fringe is associated with m - 3 and is 7^12:7(500nm)12 - 1750nm.(b) The angular position 0 of the first bright band on either side of the central band is given bysin 0 - xld, where d is the slit separation. The distance on the screen is given by LA - Dtan?,where D is the distance from the slits to the screen. Because 0 is small its sine and tangent arevery nearly equal and Ly - D sin 0 - D^f d.Dark bands have angular positions that are given by sin 0 - (m + |)A I d, and, for the fourth darkband, m - 3 and srn2a- Q l2)^f d. Its distance on the screen from the central fringe is Lyo-Dtan7a: Dsrn2a- 7D^fzd. This means that D^ld: 2Lyal7 - 2(L68 cm)17 :0.48cm.Note that this is Ly.226 Chapter 3 5
231.
97(a) If 1 is the incident intensity then the radration pressure for total absorption is I 1.4x 103 W l^ x 10- 6 Pa . rrr c rn 3.00 x 10s mls 4.67(b) The ratio is 457 x lo-6Pa :4.7 x ro-ll ratio - 1.0 x 105 Pa99IVlinima occur at angles 0 for which sin 0 - (m+ ))^1d,, where , is the wavelength, d, is the slitseparation, and m is an integer. For the first minimum, m:0 and sin01 : For the tenthminimum, m:9 and sin Orc ^12d.The distance on the screen from the central fringe to a minimum is A: Dtan?, where D is thedistance from the slits to the screen. Since the angle is small we may approximate its tangentwith its sine and write Aminima is Ar: D (19^ a d, 2 and d La A-6: (0. 150 x 10-3 *X 18.0 x 10- m) r nn n 1 .7 -6.oox1o-m.103The difference in the path lengths of the two beams is 2r, so their difference in phase whenthey reach the detector is d : 4nr f where l is the wavelength. Assume their amplitudes arcthe same. According to Eq. 35 -22 the^, intensity associated with the addition of two waves isproportional to the square of the cosine of half their phase difference. Thus the intensity ofthe light observed in the interferometer is proportional to cos2 (2rr I D. Since the intensity ismaximum when rmaximum intensity I,n and I - I,ncos2 (2nr l^). Chapter 3 5 227
232.
Chapter 352The condition for a minimum of intensity in a single-slit diffraction pattern is a sin 0 : ffiA,where a is the slit width, is the wavelength, and m is an integer. To find the angular positionof the first minimum to one side of the central maximum, set m : 1: ^ (]) - sin- ( iJO I 10- m ) - 5.89 x 1o-4 rad. 5 89 1 0-e m ,,r- sin- X /If D is the distance from the slit to the screen, the distance on the screen from the center of thepattern to the minimum is Ar - Dtan01- (3.00m)tan(5.89 x 10-4rad): 1.J67 x 10-3m.To find the second minimum, set m - 2: g-r m 0z: ,in-t 2 a 89 I 1 )l I 1. 00 X i0- mThe distance from the pattern center to the minimum is az: Dtan?2: (3.00m)tan(1.178 x10-3rad)- 3.534 x 10-3m. The separation of the two minima is Ly: A2 - hI.7 67 nun - 1,.77 mm.T7(a) The intensity for a single-slit diffraction pattern is given by r_r sin2o t-tr_ AZ, )where a- (nol,)sin9, a, is the slit width and is the wavelength. The angle 0 is measuredfrom the forward direction. You want J - Irnf 2, ^ so sin2a-1 1 1a(b) Evaluate srn2 * and *12 for a- 1.39rad, and compare the results. To be sure that 1.39radis closer to the coffect value for a than any other value with three significant digits, you shouldalso try I .385 rad and 1 .3 95 rud.(c) Since e : (nol )) sin 0, o_sin-(*)Now alr:1.391n -0.442, so o -sin- (0442 rr ) a )228 Chapter 36
233.
The angular separation of the two points of half intensity, one on either side of the center of thediffiaction pattern, is Lo - 20 -2 sin- (0442^ ) a )(d) For al^ - 1.0, L0 - zsin- t (O .442 f I.0) : 0.9 16 rad: 52.5o(e) For a,lx: 5.0, L0 - 2sin- t (O .442 15.0) - 0 .177 rad : 10. 1"(0 For al: 10, t L0 - zsin- (o .442 I I0) - 0.08 84 rad - 5.06o2t(a) Use the Rayleigh criteria. To resolve two point sources, the central maximum of the diffractionpattern of one must lie at or beyond the first minimum of the diffraction pattern of the other.This means the angular separation of the sources must be at least 0 a I .22^ f d, where , is the :wavelength and d is the diameter of the aperture. For the headlights of this problem, x lo-e m) - o LtJ 5.0 x n:12?(550 - 1.3 x 1o-4 rad . 10-3 m(b) If D is the distance from the headlights to the eye when the headlights arc just resolvableand !, is the separation of the headlights, then (.- Dtan?p = D?n where the small angleapproximation tan? p = 0 p was made. This is valid if 0 n is measured in radians. ThusD - ll0n- (1.4m)l(I .34 x 10-4rad): 1.0 x 104m - 10km.25(a) [Jse the Rayleigh criteria: two objects can be resolved if their angular separation at theobserver is greater than 0 pdiameter of the aperture (the eye or mirror) . If D is the distance from the observer to the objects,then the smallest separation (. they can have and still be resolvable is (. - D tan? n t D0 a,where 0p is measured in radians. The small angle approximation tan?p E 0p was made. Thus ( _T.22Dx: 1.22(8.0 x 1010mX550 x 10-em) - 1.1 x I07m: 1.1 x 104km. d 5.0 x 10-3 mThis distance is gteater than the diameter of Mars. One part of the planets surface cannot beresolved from another part.(b) Now d - 5.1m and (.- 1.22(8.0 x l01o m)(550 x 10-n *) -1.1 x104m-l1km. n_ 5lr" Chapter 36 229
234.
29(a) The first minimum in the diffraction pattern is at an angular position 0, measured from thecenter of the pattern, such that sin 0 - 1 .22^ I d,, where is the wavelength and d is the diameter ^of the antenna. If f is the frequency, then the wavelength is x 191 m/s - 9- 3oo f 220 x 10e Hz - r.36x 1o-3 m. ^Thus x 10--) 3 .02 x 10-3 rad . 10-2 m ):The angular width of the central maximum is twice this, or 6.04 x 10-3 rad (0.346").(b) Now )- I.6 cm and d - 2.3 m, so 1.22(I .6 x 10-2 m) o _sin- ( - 8.5 x 10-3 rad . 2.3mThe angular width of the central maximum is 1.7 x I0-2 rad (0.97").39(a) The angular positions 0 of the bright interference fringes are given by d sin 0 - ffi, where d isthe slit separation, . is the wavelength , and m ts an integer. The first diffraction minimum occursat the angle 0t given by asrn?1 - where a is the slit width. The diffraction peak extendsfrom -0t to +0t, So you want to count the number of values of m for which -0r ( 0 < ^,what is the same, the number of values of m for which - sin d1 <-lIo<5.00, so the values of m are m- -4, -3, -2, -1, 0, *1, *2, *3, and +4. There are ninefringes.(b) The intens rty at the screen is given by I - I,n (cos2 P) f lry) a ) ,where aFor the third bright interference fringe , dsin g - 3), so P - 3r rad and cosz pe, - 3naf d - 3n 15.00 - 0.600n rad and 2 sin o6oozr fry) - ( - o .2ss a ) o.6oon ) .The intensity ratio is I I Ir. - 0 .255.45 :The ruling separation is d I lg00 mm- t) 2.5 x 10-3 mm. Diffraction lines occur at angles e -such that d sin 0 - ffi, where is the wavelength and m is afi integer. Notice that for a given ^230 Chapter 36
235.
order, the line associated with a long wavelength is produced at a greater angle than the lineassociated with a shorter wavelength. Take ) to be the longest wavelength in the visible spectrum(700 nm) and find the greatest integer value of m such that 0 is less than 90o. That is, find thegreatest integer value of m for which m), <3.57 , that value is m : 3. There are three complete orders on each side of the m : 0 order. Thesecond and third orders overlap.51(a) Maxima of a diffraction gratingpattern occur at angles d given by dsin 0 - ffi, where d is theslit separation, , is the wavelength , and m rs an integer. The two lines are adj acent, so their ordernumbers differ by unity. Let m be the order number for the line with sin e - 0.2 and mr I be theorder number for the line with sin 0 - 0.3. Then 0.2d: m), and 0.3 d - (m+1)^. Subtractthe firstequation fromthe secondto obtain0.ld- or d- ) l0.l - (600x 10-e^)10.1 -6.0x 10-6m. ^,(b) Minima of the single-slit diffraction pattern occur at angles 0 given by o sin e - m, where &is the slit width. Since the fourth-order interference maximum is missing, it must fall at one ofthese angles. If a is the smallest slit width for which this order is missing, the angle must be givenbyosin0:),. Itisalsogivenbydsin0_ 4^,soa:d,14-(6.0x10-u*)f4-1.5x10-6m.(c) First, set 0 - 90" and find the largest value of m for which m),, <order that is diffracted toward the screen. The condition is the same as m <d,l^ - (6.0 x 10-u*) 1rc00 x 10-n-) - 10.0, the highest order seen is the m:9 order.(d) and (e) The fourth and eighth orders are missing so the observable orders are m - 0, 1,2,3,5, 6, 7, and 9. The second highest order is the m- 7 order and the third highest order is them - 6 order.6tIf a grating just resolves two wavelengths whose average is lave and whose separation is A^,then its resolving power is defined by R - )ave I L^. The text shows this is IY m, where .Ai isthe number of rulings in the grating and rn is the order of the lines. Thus )ave lL^ - IYm and r 7 )uun 656.3 nm // -- o 3 .65 x 103 rulings -tv m L), (1X0.18nm) .73We want the reflections to obey the Bragg condition 2d srn 0 ffi, where is the angle between - Ithe incoming rays and the reflecting planes, is the wavelength, and m is an integer. Solve for0: ^ o_sin_,1#] : ,irr-t (0.1 25 X l g-r m)m 25 , 10-e*) - sin-t(o .24tom) I 2(0 X ]For m: 1 this gives 0 - 14.4o The crystal should be turned 45o 14.4o - 30.6o clockwise.For m:2 it gives 0 - 29.7o. The crystal should be turned 45" - 29.7" - 15.3" clockwise.For m:3 it gives 0 - 48.1o. The crystal should be turned 48.1o - 45" - 3. 1o counterclockwise. Chapter 36 231
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For m- 4rtgives e -82.8o. Thecrystalshouldbeturned 82.80-45o -37.8o counterclockwise.There are no intensity maxima for m >1 for m greater than 4. For clockwise turns the smaller value is 15.3" and the larger value is30.6o. For counterclockwise turns the smaller value is 3.1o and the larger value is 37.8o.77Intensity maxima occur at angles 0 such that d sin 0 - ffi, where d is the separation of adjacentrulings and ) is the wavelength. Here the ruling separation is IlQ00mm-t) - 5.00 x 16-r mnl -5.00 x 10-6 m. Thus d stn0 (5.00 x 10-u rn) sin 30.0o 2.50 x 10-6 mFor m- 1., l - 2500nm; for m- 2,, - l250nm; for m 3r,I: 625 nm; for m- 5, ) - 500offi, and for nL: 6, - 4I7 rlm. Only the last three are in ^the visible range, So the longest wavelength in the visible range is 625ilil, the next longest is500 trffi, and the third longest is 4L7 nm.79Suppose mo is the order of the minimum for orange light, with wavelength Ao, and mbe isthe order of the minimum for blue-green light, with wavelength ),as. Then a sin 0 _ rno).o and astn? : mbe),ug. Thus ffiooThe smallest two integers with this ratio are mbe - 6 and TrLo - 5. The slit width is TTtoo 5(600 x 10-9 m :3.0X10-m. ^) n ", ,.-1 0,:#-Other values for TrLo and mbe are possible but these are associated with a wider slit.81(a) Since the first minimum of the diffraction pattern occurs at the angle 0 such that sin e - lo,where is the wavelength and a, is the slit width, the central maximum extends from 0r - ^- sin-t(Xlo) to 0z: *sin-t(^lo). Maxima of the two-slit interference pattern are at angles 0such that sin d - mAf d, where d is the slit separation and m is an integer. We wish to knowthe number of values of m such that sin-t1*X1a7 hes between - sin-t(Xlq and +sin(^lo) or,what is the same, the number of values of m such that rn I d, lies between -l I o and +l f a. Thegreatest m can be is the greatest integer that is smaller than dlo: (I4 prn)l(z.0 pm):7. (Them : 7 maximum does not appear since it coincides with a minimum of the diffraction pattern.)There are 13 such valuesi 0, + 1 , +2, *3; *4; +5, and *6. Thus 13 interference maximaappear in the central diffraction envelope.(b) The first diffraction envelope extends from 0t: sin-t(Xlo) to 0z: sin-t(ZLlil. Thus wewish to know the number of values of m such that m I d, is greater than L I o and less than 2l o.Since d - 7.0a, m can be 8, 9, 10, 11, 12, or 13. That is, there are 6 interference maxima in thefirst diffraction envelope.232 Chapter 36
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93If you divide the original slit into l/ strips and represent the light from each strip, when itreaches the screen, by a phasor, then at the central maximum in the diffraction pattern you addl/ phasors, all in the same direction and each with the same amplitude. The intensity thereis proportional to 1Jz. If you double the slit width, you need 2lV phasors if they are each tohave the amplitude of the phasors you used for the naffow slit. The intensity at the centralmaximum is proportional to (2LD2 and is, therefore, four times the intensity for the naffow slit.The energy reaching the screen per unit time, however, is only twice the energy reaching it perunit time when the naffow slit is in place. The energy is simply redistributed. For example, thecentral peak is now half as wide and the integral of the intensity over the peak is only twice theanalogous integral for the naffow slit.95(a) Since the resolving power of a grating is given by R - )/A^ and by IY nt , the range ofwavelengths that can just be resolved in order m is A) : ),lIVm. Here l/ is the numberof rulings in the grating and ) is the average wavelength. The frequency f is related to thewavelengthby f^:c,where cisthespeedof light. Thismeans f L,+^AJ:0, so A.: -+ Lf : -^2 Lf . JCwhere f - cl was used. The negative sign means that an increase in frequency coffesponds to wavelength. We may interpret Lf as the range of frequencies that can be resolveda decrease in ^and take it to be positive. Then t2 ^ a.f : l{* c ,and Lf: C l{m),(b) The difference in travel time for waves traveling along the two extreme rays is Lt : LL f c,where LL is the difference in path length. The waves originate at slits that are separated by(,Af - I)d,, where d is the slit separation and tf is the number of slits, so the path difference isLL - (ltr - l)d, sin 0 and the time difference is A,- (ff - Dd, sin 0If t/ is large, this may be approximated by Lt - (lV d lc) sin 9. The lens does not affect thetravel time.(c) Substitute the expressions you derived for Lt and Lf to obtain ^f Lt:(#) (ry) : dsrn? -1 mThe condition d sin 0 - m), for a diffraction line was used to obtain the last result. Chapter 3 6 233
238.
101The dispersion of a grating is given by D = dO / dA, where 0 is the angular position of a lineassociated with wavelength A. The angular position and wavelength are related by R. sin 0 = mA,where R. is the slit separation and m is an integer. Differentiate this with respect to 0 to obtain(dO / dA) R. cos 0 = m or D = R.O = m R.A R. cos 0Now m = (R./ A) sin 0, so D = R. sin 0 __ tan_O R. Acos 0 AThe trigonometric identity tan 0 = sin 0/ cos 0 was used.234 Chapter 36
239.
Chapter 37l1 (: l30m) of the spaceship and its length L as measured by the timing(a) The rest length LsstationarerelatedbyL:Loll:L0m,where1:IlFryandp:ulc.ThusL- (130m)@ -87.4m.(b) The time interval for the passage of the spaceship is . Lt_ L r-J- 87.4m x 1o-7s. u (0140)Q.sg7s x 108 mA) -3.9419The proper time is not measured by clocks in either frame ,S or frame S since a single clockat rest in either frame cannot be present at the origin and at the event. The full Lorcntztransformation must be used: r,:1lr-utl t: jlt- grlrl ,where p - ulc- 0.950 and 1: tlm - rl@ - 3.2026. Thus tr- (3.2026) [tOO x l03m - (0.950)(2.9979 x 108 mls)(200 x 10-u r] - 1.38 x 10s m - 138 kmand (0.950x 100 x 103m) t - x 10-4 s - -3 74 (3 .2026) froo x 10-6 s - 2.9979 x 108 m/s t: -3 .74 p,s .29IJse Eq. 37 -29 with It - 0 .40c and ?J :0.60c. Then + .60c 0.40c 0 ?t: 1 + (0 .a0Q(0.60c) - -0.81c. I "33Calculate the speed of the micrometeorite relative to the spaceship. Let ,S be the reference framefor which the data is given and attach frame ,S to the spaceship. Suppose the micrometeorite isgoing in the positive n direction and the spaceship is going in the negative r direction, both asviewed from S . Then, in Eq. 37 -29, ?.,0 - A.82c and u Chapter 37 235
240.
is the velocity of S relative to S. Thus the velocity of the micrometeorite in the frame of thespaceship is u + v 0.82e + 0.82e 09 u= = = . 806e. 1 + uv I e2 1 + (0.82e)(0.82e)1 e2The time for the micrometeorite to pass the spaceship is L 350m tlt = - = = 1.19 X 1O- 6 s. u (0.9806)(2.9979 X 108 m/s)37The spaceship is moving away from Earth, so the frequency received is given by f = fay H 1+73where fa is the frequency in the frame of the spaceship, {3 = v I e, and v is the speed of thespaceship relative to Earth. See Eq. 37 - 31. Thus f = (100 MHz) 1 - 0.9000 = 22.9 MHz. 1 + 0.900039The spaceship is moving away from Earth, so the frequency received is given by f = fay H 1+73where fa is the frequency in the frame of the spaceship, {3 = vie, and v is the speed of thespaceship relative to Earth. See Eq. 37- 31. The frequency f and wavelength A are related byfA = e, so if AO is the wavelength of the light as seen on the spaceship and A is the wavelengthdetected on Earth, then A = AO ~ = (450nm) +{3 -- 1-{3 1 + 0.20 1 - 0.2 0 = 550nm.This is in the yellow-green portion of the visible spectrum.43Use the two expressions for the total energy: E = me2 + K and E = ,me2 , where m is the mass Jof an electron, K is the kinetic energy, and, = 1I 1 - {32. Thus me2 + K = ,m2 and K (100.000 x 106 eV)(1.602176462J/eV) 9 69 ,=1+-=1+ =16 5 me2 (9.10938188 x 10- 31 kg)(2.99792458 x 10 8 m/s)2 ..236 Chapter 37
241.
Now y2 - p), so 1 1 @ - 0 .999 987 .53The energy equivalent of one tablet rs mcT - (320x 10-ut gX2.gg79 x 108 mls)2 2.88 x 10r3 J. :This provides the same energy as (2.88 x 1013 Dl(3.65 x L07 JIL)_ 7.89 x 105 L of gasoline.The distance the car can go is d - (7.89 x 105 LXI2.75km lL) 1.01 x 107km. -7lThe energyofthee1ectronisgivenbyE_mc2lre,whichyields lH c: -0.gggg9994cxcfor the speed u of the electron. In the rest frame of Earth the trip took time t_ 26y. A clocktraveling with the electron records the proper time of the trip, so the trip in the rest frame of theelectron took time tt : t l^y. Now : r_- 1533 MevXl .602 x 10_ J/x,fev) : 3.0 x 103 13 .y mc2 (9.11 x 10-rt kg)(2.9979 x 108 m/s)and t - (26y)/(3.0 x 103) : 8.7 x 10-y. The distance traveled is 8.7 x 10-ly.73 :Start with (pr)z K2 +2Kmc2, where p is the momentum of the particle, /( is its kinetic energy,and m is its mass. For an electron mc2 0.511MeV, so - pc: - (2.00 MeV)z 1 2(2.00 MeV)(0.5 1 1 MeV) : 2.46MeV .Thus p:2.46MeV/c.75The work required is the increased in the energy of the proton. The energy is given by trmcT llt @ ldz} Let u1 be the initial speed and u2 be the final speed. Then the work is W mc2 mc2 938 MeV 938 MeV 189 MeV, (, l r) (o.gg60)2 1 - (o.gg5o)2where mcT : 938 MeV was used.77(a) Let u be the speed of either satellite, relative to Earth. According to the Galilean velocitytransformation equation the relative speed is urer : 2u : 2(2.7 x 104 km/h : 5 .4 x 104 krnf h. Chapter 37 237
242.
(b) The correct relativistic transformation equation is 2v Vre1 = 2 . 1+~ c2The fractional error is 2v - Vre1 1 fract err = = 1- ---0- 2 . 2v 1+ ~ c2The speed of light is 1.08 x 109 km/h, so 1 fract err = = 6.3 X 10- 10 . 1 + (2.7 X 104 km/hi (1.08 x 109km/h)2238 Chapter 37
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Chapter 3I7(a) Let ^R be the rate of photon emission (number of photons emitted per unit time) and let tr bethe energy of a single photon. Then the power output of a lamp is given by P - Rtr if all thepower goes into photon production. Now E - hf - hcf ),, where h is the Planck constant, f isthe frequency of the light emitted, and is the wavelength. Thus P - Rhcl ), and R - lhc. ^ ^PThe lamp emitting light with the longer wavelength (the 700-nm lamp) emits more photons perunit time. The energy of each photon is less so it must emit photons at a greater rate.(b) Let ,R be the rate of photon production for the 700 nm lamp Then za )P Lw H- (700 x 10-e m)(400J/s) I41 x photon/s hc (6.626x 10- 34 J.sX2 .ggigx 108 -/t - L021 t7The energy of an incident photon is tr - h f hc f ),, where h is the Planck constant, f is thefrequency of the electromagnetic rudration, and , is its wavelength. The kinetic energy of themost energetic electron emitted is Krn - E - O - (hcl D - O, where O is the work function forsodium. The stopping potential Vo is related to the maximum kinetic energy by eVo: K,n soeVo-(hclD- O and h, x J. .9979 x m/s) I, - a: (6.626 19-34 sX2 108 1 n r n_j "Vr.Here eVo: 5.0 eV was used.2l(a) The kinetic energy Krn of the fastest electron emitted is given by Krn: hf -O - (hclD-O,where O is the work function of aluminum, f is the frequency of the incident radtation, and ,is its wavelength. The relationship f : cl was used to obtain the second form. Thus ^ K,n (200 x 10-e mX | .602 x 10- ts J lev)(b) The slowest electron just breaks free of the surface and so has zero kinetic energy.(c) The stopping potential Vo is given by Krn: €Vo, so Vo: Krnl": (2.00eV) l" - 2.00V.(d) The value of the cutoff wavelength is such that Krn - 0. Thus h"l, - O or I-U O (4.2 eVX | .602 x 10- re J I eY) Chapter 38 239
244.
Ifthe wavelength is longer, the photon energy is less and a photon does not have sufficientenergy to knock even the most energetic electron out of the aluminum sample.29(a) When a photon scatters from an electron initially at rest, the change in wavelength is givenby A, - (hl*cxl - cos 0), where m is the mass of an electron and O is the scattering angle.Now hf mc-2.43 x 10-12m-2.43pffi, so Al - (2.43pmx1 - cos30o):0.326pm. The finalwavelength is ) : ) + A, - 2.4pm+ 0.326pm - 2.73pm.(b) Now A. - (2.43pmxl - cos 120"): 3.645pm and X -2.4pm+3.645pm - 6.05pm.43Since the kinetic energy K and momentum p are related by K - p2 lzm, the momentum of theelectronisp-MandthewavelengthofitsmatterwaVeis^-hlp-hfffi.ReplaceK with €V, where V is the accelerating potential and e is the fundamental charge, to obtain h x 10-34J.s 6.626 {2meV 2(9. 109 x 10-:tkg)(1 "602 x 10-teCX25.0 x 103V) -7.75 x 10-12m- 7.7Spm.47(a) The kinetic energy acquired is K - qV, where q is the charge on an ion and V is theaccelerating potential. Thus K - (I .602 x 10-1e CX300V) - 4.80 x l0-r7 J. The mass ofa single sodium atom is, from Appendix F, m- (22.9898g1mo1)l(6.02 x 1023 afimfmol)3.819 x 10-23 g - 3.819 x 10-26 kg. Thus the momentum of an ion is p:17: 2(3.819 x 10-20 kg)(4.80 x 10-17 J) - 1.91 x 10-2t kg.m/s.(b) The de Broglie wavelength is , h 6.63x10-34J.s", u: )- p t ="r""-^^"-,. .m/s -3.47 x 10-13m. 1.91 x I0-2t kg49 K and momentum p ate related by K - p2 lzm, the momentum of theSince the kinetic energyelectronisp-ffiandthewavelengthofitsmatterwaVeis^-hlp-hf,ffi.Thus I 6.626 x 10-34 J . s) K- (*) 10ru 2m 2(9.11 x 590 x 10-e m) ) - 6.92 x l0 -2s J - 4.33 x 10-6 eV.59The angular wave number k is related to the wavelength l by k :2rfAandthe wavelength isrelated to the particle momentum p by , - hlp, so k:2rpf h. Now the kinetic energy K and240 Chapter 38
245.
themomentumarere1atedbyK-p2l2m,wheremisthemaSSofthepartic1e.Thusp:ffiand k- ZnrffiR61For Ut - (Jo, Schrodingers equation becomes , d,zr! 8r2m r 1.,+ n ltr- Lb]qr-0. o,tr- ,",Substitute * - rho"ik* . The second derivative is d,2rb ld* - -tz{o"ik" - -kzr!. The result is -k2q)+ryW -uoltb:0. -/ 7^r2Solve for k and obtain67(a) If rn is the mass of the particle and tr is its energy, then the transmission coefficient for abarrier of height U and width L is given by T _ e-2kLwhereIf the change LU in U is small (as it is), the change in the transmission coefficient is given by Lr:# L(-r: -zLr # L(.r .Now dk : 8n2m(U d(I 2yffi 2(u - E) 2(u - tr)Thus LT: -LTK AU (J-EFor the data of Sample Problem 38-7,ZkL:10.0, so kL:5.0 and LT kL*- (oo1oX68ev) T- Lr - E- -(lu)6.gev-5.1ev - -o .zo. -(5.g;There is a 20% decrease in the transmission coefficient. Chapter 38 241
246.
(b) The change in the transmission coefficient is given by : LT # LL: -2ke-zkL LL - -zkr LLand LT : -2k LL: - 2(6.67 x 10e m-tX0.010X750 x l0-" *) : -0.10. TThere is a I0% decrease in the transmission coefficient.(c) The change in the transmission coefficient is given by LT: #LE: -ZLe-2L#LE: -2LT#LE.Now d,kf dtr - -dkld,U - -klz(U - E), so +: kL#- (5 s;(ooloxsltv) - 0.r5.There is a 15% increase in the transmission coefficient.79The uncertaintyin the momentum is Lp: mLu - (0.50kg)(1.0mls):0.50kg.ffi/s, where Auis the uncertainty in the velocity. Solve the uncertainty relationship Lr Lp > n for the minimumuncertainty in the coordinate r: Lr -nf Lp - (0.60J.s) l2n(0.50kg.mls):0.19m.242 Chapter 38
247.
Chapter 3913The probability that the electron is found in any interval is given by P _ f lVl d*, where theintegral is over the intenral. If the interval width Lr is small, the probability can be approximatedby P : lrfl L*, where the wave function is evaluated for the center of the interval, say. For anelectron trapped in an infinite well of width L, the ground state probability density is lrhl: lrrn(T) ,so P-ff)sin2 (T)(a) Take L:100pffi, r - 25pm, and Lr - 5.0pm. Then P-[ffi] sin2L*#] -ooso(b) Take L - 100pffi, r- 50pm, and Lr - 5.0pm. Then p lzfs0 Pm)1 fzr(so Pm)l - o. to .(c) Take L:100pffi, r:90pm, and Ln - 5.0pm. Then P-rry] L loopn sin2[#ffi] -oooe525The energy levels are given by En*nv: #ln.frlwhere the substitutions L*- L and La _ 2L were made. In units of hl8mLz, the energylevels are given by nI* f 4. The lowest five levels are 81,1 : I.25, Et,2 - 2.00, EtJ - 3.25,E2; : 4.25, and Ez,z: Er,4 - 5.00. A little thought should convince you that there ate no other "1possible values for the energy less than 5.The frequency of the light emitted or absorbed when the electron goes from an initial state z to afinal state f is f - (Et - Ei,)lh and in units of hl8mLz is simply the difference in the values ofn?.+nll+ for the two states. The possible frequencies are 0.75 (L,2--+ 1,1), 2.00 (1,3 -+ 1,1),3.00 (2,1 ------+ 1,1), 3.75 (2,2 ----+ 1,1), 1.25 (1,3 + 1,2),2.25 (2,1 ----J I,2),3.00 (2,2 ----J 1,2),1.00 (2,I 1,3), I.75 (2,2---+ 1,3), 0.75 (2,2-----+ 2,1), all in units of hl8mL2. -> Chapter 39 243
248.
There are 8 different frequencies in all. In units of h l8mL2 the lowest is 0.7 5, the second lowestis 1 .00, and the third lowest is I .25. The highest is 3.7 5, the second highest is 3.00, and thethird highest is 2.25.33If kinetic energy is not conserved some of the neutrons initial kinetic energy is used to excitethe hydrogen atom. The least energy that the hydrogen atom can accept is the differencebetween the first excited state (n : 2) and the ground state (na state with principal quantum number n is -(13.6eY)f n2, the smallest excitation energy is13.6eV-(13.6eV)lQ)2- l0.2eV. The neutron does not have sufficient kinetic energy to excitethe hydrogen atom, so the hydrogen atom is left in its ground state and all the initial kineticenergy of the neutron ends up as the final kinetic energies of the neutron and atom. The collisionmust be elastic.37The energy E of the photon emitted when a hydrogen atom jumps from a state with principalquantum numb er u to a state with principal quantum numb er (, is given by E:A(; #)where A - 13.6eV. The frequency f of the electromagnetic wave is given by f : Elh and thewavelength is given by - cl f. Thus ^ 1- T- E- A (L 1 c hc hcZt ur) ^The shortest wavelength occurs at the series limit, for which u - co. For the Balmer series, : 2 and the shortest wavelength is .s : 4lr,c I A. For the L5rman series , t : 1 and the shortest(.wavelength is ),r- hclA. The ratio is ),n I t, : 4.43The proposed wave function is ,r!,:+ t/ e_ rlo /2" 1 ftotwhere a, is the Bohr radius. Substitute this into the right side of Schrodingers equation andshow that the result is zero. The derivative is drh:- Io-r/a d, - fiTas/2e )SO *, drb - r2 ^-, / o r- dr:244 Chapter 39
249.
and 1 7 #(*#): #ll.;] e-"::ll.;]Now the energy of the ground state is given by tr - -me4 l8elhz and the Bohr radius is given ,bby o: h2rof nme2, so E - -e2 f 8nesa. The potenttal energy is given by (Ji - -e2 f 4nesr, so ryw- ut,h - ry |h. hll, 1: ry h[+ .?],t, t_1 z1 I l- - n+o l-an;l - ; L-; *;i trmeT z1 Yrs bThe two terms in Schrodingers equation obviously cancel and the proposed function ,lt satisfiesthat equation.47The radial probability function for the ground state of hydrogen is P(r) - 7412 I ot)e-2r /o" , wherea, is the Bohr radius. (See Eq. 39-44.) You want to evaluate the integral /r"" P(r)d,r. Eq. 15in the integral table of Appendix E is an integral of this form. Set n- 2 and replace a in thegiven formula with 2lo and r with r. Then P(r) dr : 4f* e-2, /o d,r 4 l,* *J, T2 a3 (2 I a)349(a) th21o is real. Simply square it to obtain the probability density: It^ol_ Le-/o 32naj cos2 o.(b) Each of the other functions is multiplied by its complex conjugate, obtained by replacing iwith -i in the function. Since eif e-if : e0 - I, the result is the square of the function withoutthe exponential factor: lrhr,*, lt - #e-/o srn2 o lrhr,-,lt _ #e-/o srn2 o .The last two functions lead to the same probability density.(c) For IrL4is greatest along the z axis, and for a given distance from the nucleus decreases in proportion tocos2 d for points away from the z axis. This is consistent with the dot plot of Fig. 39-24 (a).For TrLp : + 1 the radial probability density decreases strongly with distance from the nucleus, isgreatest in the fr,, a plane, and for a given distance from the nucleus decreases in proportion to srn2 0 for points away from that plane. Thus it is consistent with the dot plot of Fig . 3g-24(b). Chapter 39 245
250.
(d) The total probability density for the three states is the sum: l,hz,ol,+l,bzr*r|2+l,b,I-|2:#e_,/o[,o,,0+|,,n,0+ 1,,"rl r2 : 32na)The trigonometric identity cos2 0 + stn2 0 -e-r/a.depend on 0 or 0. It is spherically symmetric.57The wave function is { : tE "-kr. Substitute this function into Schrodingers equation,Since d,2rh ld*- tFCtte-k* - k2th, the result is h2 k2 ffirb+Uorh - E1h.The solution for k is ry(to_ E).Thus the function given for th is a solution to Schrodingers equation provided k has the valuecalculated from the expression given above.246 Chapter 39
251.
Chapter 402Since L2 = L; -+ L; -+ L;, J Li + L~ = JL2 - L;. Replace L2 with IV, + 1)n2 and Lz withmjlnto obtain JLi+L~= JR(R+l)-m~n.For a given value of R, the greatest that mjl can be is R, so the smallest that J L~ + L~ can 2be is jR(R + 1) - R n = V£n. The smallest possible magnitude ofmjl is zero, so the largestJ Li + L~ can be is JR(!! + l)n. Thus11(a) For R = 3, the magnitude of the orbital angular momentum is L = JR(R -+ l)n = J3(3 + l)n =JI2Ti, = 3.46Ti,.(b) The magnitude of the orbital dipole moment is /--Lorb = JR(R + 1)/--LB = VU/--LB = 3.46 /--LB·(c) The largest possible value ofmjl is R, which is +3.Cd) The corresponding value of the z component of the angular momentum is L z = fn = +3n.(e) The direction of the orbital magnetic dipole moment is opposite that of the orbital angularmomentum, so the corresponding value of the z component of the orbital dipole moment is/--Lorb, z = -3/--LB.(f) The angle () between i and the z axis is Lz 3n () = cos- 1 - = cos- 1 - - = 30.0 0 . L 3.46n(g) The second largest value ofmjl is mp = R - 1 = 2 and the angle is e = cos- 1 -Lz 1 = cos- - - 2n = 0 54.7 . L 3.46n(h) The most negative value of mp is -3 and the angle is L -3n e = cos- 1 ~ = cos- l -- = 150 0 • L 3.46n Chapter 40 247
252.
15The acceleration 1s F : 0.t cos 0) (dB ldr) a,- M Mwhere M is the mass of a silver atom, p is its magnetic dipole moment, B is the magnetic field,and 0 is the angle between the dipole moment and the magnetic field. Take the moment and thefield to be parallel (cos 0 - 1) and use the data given in Sample Problem 40-1 to obtain a- (9.27 x 10-ollrxl.4x 103T1*) -7.21 x 104*lrt25In terms of the quantum nurnbers frr, fra, and fr2, the single-particle energy levels are given by En*,rlsrlz: (n* * r?r , ,2"). #The lowest single-particle level coffesponds to nr: 1, na: 1, and nz: 1 and is ErJ1 -3(h2 f 8mL2). There are two electrons with this energy, one with spin up and one with spindown.The next lowest single-particle level is three-fold degenerate in the three integer quantum numbers.The energy is 8t1,2: Er,2,t: E2J,r - 6(hl8*L. Each of these states can be occupied by aspin up and a spin down electron, so six electrons in all can occupy the states. This completesthe assignment of the eight electrons to single-particle states. The ground state energy of thesystem is Es (2)(r(h2 lSmL) * (6X6) (h lSmL) (42)(h, lB*L,). - -31(a) All states with principal quantum numb er n - I are filled. The next lowest states have rL : 2.The orbital quantum number can have the values (. - 0 or 1 and of these, the t - 0 states have thelowest energy. The magnetic quantum number must be m4 - 0 since this is the only possibilityrf ( : 0. The spin quantum number can have either of the values TTLs: -+ or +|. Since thereis no external magnetic field, the energies of these two states are the same. Thus, in the groundstate, the quantum numbers of the third electron are either n- 2, (,- 0, wlptu : 2, | : 0, TfL1: 0, TTLs - +t.(b)Thenextloweststateinenergyis ann-2,(: l state. A11 n:3 statesarehigherinenergy.The magnetic quantum number can be TTLy: -1, 0, or +1; the spin quanfum number can beTTLs: -+ or +|. If both external and internal magnetic fields can be neglected, all these stateshave the same energy. The possible states are (2, I, l, +Il2), (2, t, 1, -1 l2), (2, 1, 0, +ll2),(2, 1, 0, -1 l2), (2, I, -1 , +ll2), and (2, l, -l , Il2).37(a) The cut-off wavelength )*in is charucteristic of the incident electrons, not of the targetmaterial. This wavelength is the wavelength of a photon with energy equal to the kinetic energy248 Chapter 40
253.
of an incident electron. Thus (6;926 x 1=0 ) - Y- LE _11 l, :l(300 x, (35 x 103 eVX1.60 x 10-1eJ/eV) l98 T1s) - 3.ss x 10-r m - 3s.s pm.(b) A K* photon results when an electron in atarget atom jumps fromthe l-shell to the K-shell.The energy of this photon is 25.5 1 keV - 3 .56 keV - 21 .95 keV and its wavelength is _ hcA: Lt (6.626 x 10-34 J. sX3.00 x 108 m/s)(c) A K p photon results when an electron in a target atom jumps from the M-shell to theK-shell. The energy of this photon is 25.5 1 keV - 0.53 keV : 24.98 keV and its wavelength is , hc )- +- (6.626 x 10-34 J . sX3.00 x 108 m/s) x 10-rrm:49.6pm. LE (24.98 x 103 eVXl.60 x 10-te JleY) -4.964tSince the frequency of an x-ray emission is proportional to (Z I)2, where Z is the atomicnumber of the target atom, the ratio of the wavelength Axu for the K o line of niobium to thewavelength lcu for the Ko line of gallium is given by )Nalcu: (Zca- DlGr*- I), whereZNa is the atomic number of niobium (41) and the Zcu is the atomic number of gallium (3 1).Thus )Nb I cu - (3 0) (4q2 - 9 f 16.49The number of atoms in a state with energy tr is proportional to e- E lr , where T is thetemperature on the Kelvin scale and k is the Boltzmann constant. Thus the ratio of the numberof atoms in the thirteenth excited state to the nurnber in the eleventh excited state is TLtz n_ LE /kr - ut-e )where LE is the difference in the energies: LE- En En: 2(L.2eY): 2.4eV. For thegiven temperature , kT - (8 .62 x l0-2 eV/KX2000 K) : 0.1 724eV. Hence, TLtz -1 Tt tt6s(a) The intensity at the target is given by I - P lA, where P is the power output of the sourceand A is the area of the beam at the target You want to compute I and comp are the result with108 wl^.The beam spreads because diffraction occurs at the aperture of the laser. Consider the part ofthe beam that is within the central diffiaction maximum. The angular position of the edge isgiven by sin d - 1.22^ld, where l is the wavelength and d is the diameter of the aperture (see Chapter 40 249
254.
Problem 50). At the target, a distance D away, the radius of the beam is r - D tan?. Since e issmall, we may approximate both sin0 and tan? by 0, in radians. Then r: D0:1.22D^ld, and I- I,, P Pdz (5.0 x 106W(4.0ttt)2 rr ?r(I .22D^)2 Tr 1t.2213000 x 103 m)(3.0 x 10-u *)] :2.1 x 105 W l^ ,not great enough to destroy the missile.(b) Solve for the wavelength in terms of the intensity and substitute f- 1.0 x 108 W f m2: d,E t- nzD _ I 4.0m 5.0 x 106 W 1 .22(3000 x 103 m) zr(1.0x108W1*) "I - L.4 x 10-7 m- I40 nm.7l(a) The length of the pulse is I : c Lt, where Lt is its duratioll. Thus L - (3.00 x 10* -/sxl0 x10-15 s - 3.0 x l0-6m. The number of wavelengths in the pulse is ,Afl0- 6 l(500 x 10-e -) : 6.0. ^)(b) Solve for X: x-#-r.ox 1or4s.Since I year contains 356 days, each day contains 24hours, and each hour contains 3600 seconds,the value of X in years is 1.0 x 10la s :3.2 x 106y. (365 .2 d)(24h1dX3600 s/h)250 Chapter 40
255.
Chapter 4L1(a) At absolute temperature T - 0, the probability is zero that any state with energy above theFermi energy is occupied.(b) The probability that a state with energy E is occupied at temperature T is given by P(E) - ,where k istheBoltzmannconstantand trp istheFermienergy. Now, tr-Ep -0.062eVand(E - Ep) lkf - (0 .062 eV)/(8.62 x 10-t ,Y lKX320 K) - 2.248, so P(E)-+ I I -0.0956. "2.248See Appendix B for the value of k.11The Fermi-Dirac occupation probability is given by Pro - I I ("^u lkr + 1) and the Boltzmannoccupation probability is given by Ps : e- Ltr /kr . Let f be the fractional difference. Then e LE /kr -W ^- t 1 $_Ps-Pro J P" e- LE /kr-[Jsing a common denominator and a little algebra yields t- e-LE/kr e-LElkr + 1 The solution for e- LE lkr is e- r.E/kr: L 1fTake the natural logarithm of both sides and solve for T. The result is rm.Ltr - ry1 ,UIII I f )(a) Put f equal to 0.01 and evaluate the expression for T: q-1 _ (1.00eVXl.60x 10-relleVl _ 25ox1o3K f (1.38x lo-23 Chapter 41 251
256.
(b) Put f equal to 0.10 and evaluate the expression for ?: (1 .00 eVXl .60 x 10- te I T- 1eV| - 5.30 x 103 K. (r.38 x LI-L3J/K) ln (#ffi)t7(a) According to Appendix F the molar mass of silver is 107.870 glmol and the density isp : 10. a9 glc*3. The mass of a silver atom is 107.870 x 10-3 kg/mol M_ 6.022 x 1023 mol- IThe number of atoms per unit volume is p 10.49 x 103 kgl*t F (1 z. no n-M - 1Since silver is monovalent this is the same as the number density of conduction electrons.(b) The Fermi energy is (0. 1 2I)(6.626 x 10-34 J. s)2 Ep (5.86 x 1028m r)2/3 rn 9.109 x 10-3t kg - 8.80 x 10-le ; - 5.49 eV .(c) Since Ee: *,*ur, up: tl M- 2(8.80x 10-1r J) Vm 9.109 x 10-:t kg(d) The de Broglie wavelength is hh x 10-34J.s 6.626 tr^a : : : - /1 e H- x 10-lo m. p p TnLU p (9. 109 x 10-31 kg)( I .39 x 106 m/s)31(a) Since the electron jumps from the conduction band to the valence band, the energy ofthe photon equals the energy gap between those two bands. The photon energy is given byhf : hcf A, where f is the frequency of the electromagnetic wave and. is its wavelength. ThusEs - hcf ), and A_ hc- , (6.63 x 10-34 J.sX3.00 x 108m/s) _ __ 226 x 11 m ^ ^r 10- 1 : 226 nm Es (5.50 ev)(l.60 x lo-re J/ev)Photons from other transitions have a greater energy, so their waves have shorter wavelengths.252 Chapter 4I
257.
(b) These photons are in the ultraviolet portion of the electromagnetic spectrum.37Sample Problem 4I -6 gives the fraction of silicon atoms that must be replaced by phosphorusatoms. Find the number the silicon atoms in 1.0 g, then the number that must be replacod, andfinally the mass of the replacement phosphorus atoms. The molar mass of silicon ts 28.08 6 gf mol,so the mass of one silicon atom is (28.086g1^01)l(6.022 x 1023 mol-l) - 4.66 x 10-23 gand the nurnber of atoms in 1.0g is (1.0 g)l(4.66 x t0-23 g) - 2.I4 x 1022. According toSample Problem 41 -6 one of every 5 x 106 silicon atoms is replaced with a phosphorus atom.This means there will be (2.I4 x 1022)16 x 106) - 4.29 x 1015 phosphorus atoms in 1.0g ofsilicon. The molar mass of phosphorus is 30.97 58 g/mol so the mass of a phosphorus atom is(30.9758g/-ol)l$.022 x 10-23 mol-t) - 5.14 x 10-" g.The mass of phosphorus that must beadded to 1.0g of silicon is (4.29 x l0ltxS .14 x IA-23 g): 2.2 x 10-7 o39The energy received by each electron is exactly the difference in energy between the bottom ofthe conduction band and the top of the valence band (1.1 eV). The number of electrons that canbe excited across the gap by a single 662-keV photon is l/ - (662x 103 eYl(1.1eV) - 6.0 x 105.Since each electron that jumps the gap leaves a hole behind, this is also the number of electron-hole pairs that can be created.4e(a) According to Eq. 4I-6 P(E) - . e@-Ep)lkr + 1Its derivative is d,P(e) -1 e@-EP)/kT dE l"(t-EF)/kr + Il kT For E - Ee, e(tr-trF)/kr : e0: L, so the derivative at E + Ep is -1 f 4kT.(b) Represent the tangent line by P - A+ BE, where A and B are constants. We want P - ll2and dPldU: -1l4kf for E- Ep. This means A+ BEe: Il2 and B- -1l4kf. Thesolution for A is A- (Il2)+ (EelLkT). Thus P- (Il2) (E Ee)lkT. The intercept isfound by setting P equal to zero and solving for E. The result is E - E p + 2kT . Chapter 41 253
258.
Chapter 4213(a) The de Broglie wavelength is given by - h lp, where p is the magnitude of the momentum.The kinetic energy K and momentum ate related by Eq. 37 *54, which yields ^ pc- - (200MeV)2 a2(200MeV)(0.511MeV) - 200.5 MeV.Thus t- hc 200J.106.V-(6. 18 x 10-6 nm - 6. 18 fm 1240 eV.Ilm - 1!--" " ^^^^- - . W(b) The diameter of a copper nucleus, for example, is about 8.6 fm, just a little larger than the deBroglie wavelength of a 200-MeV electroll. To resolve detail, the wavelength should be smallerthan the target, ideally a tenth of the diameter or less. 200-MeV electrons are perhaps at thelower limit in energy for useful probes.t7The binding energy is given by AEt.number (number of protons), A is the mass number (number of nucleons), TL17 is the massof a hydrogen atom, mn is the mass of a neutron, and Mpu is the mass of a llp" atom. Inprincipal, nuclear masses should have been used, but the mass of the Z electrons includedin Z M ru is canceled by the mass of the Z electrons included in Mpu, so the result is thesame. First, calcul ate the mass difference in atomic mass units: Lm(239 94)(1 .00867 u) (239.A5216 u) - L.94101 u. Since I u is equivalent to 93I.5 MeV, LEa" - (l .94101 u)(93 1.5 MeV/u) - 1808 MeV. Since there are 239 nucleons, the bindingenergy per nucleon is LEa"n: EIA- (1808MeV) lQ39): 7.56MeV.t9If protons and tf neutroos, its binding energy is Atr. - (ZmH + IVmn a nucleus contains Zm)c2, where TTL7 is the mass of a hydrogen atom, mn is the mass of a neutron, and m isthe mass of the atom containing the nucleus of interest. If the masses are given in atomicmass units, then mass excesses are defined by L11 - (*n l)r2, Ln _ (mn l)r2, andAE -(ZLs* NA", A) +(Z +t/ - A)"- ZLs*l[A,r, - A, where A- Z+ IV was used.For t?!X", Z : 79 and iV - I97 - 79 : 118. Hence AEu. -- (79)(1 .29 MeV) + (118X8.07 MeV) - (-3l.zMeV) : 1560MeV.This means the binding energy per nucleon is LEu"n: (1550MeV)1Q97) -7.92MeV.254 Chapter 42
259.
27(a) The half-life Tr /z and the disintegration constant , are related by Tr lz - (ln 2) I soTt/z - (ln 2)1Q.0108 h-t) - 64.2h. ^,(b) At time t, the number of undecayed nuclei remaining is given by jV - ffo e-t - l/o s-(ln2)t/Trf2 .Substitute t- 3Tr /z to obtain ,Ar --3tn2_0.I25. l/r -eIn each half-life, the number of undecayed nuclei is reduced by half. At the end of one half-life, Niv-AIo/g-0.125Ah.(c) Use iV - Afo e-t10.0d is 240h., so )t - (0.01086-t)Q40h):2.592 and lr _2.ss2 2J- lrr:e -0074935(a) Assume that the chlorine in the sample had the naturally occurring isotopic mixture, so theaverage mass number was 35.453, as given in Appendix F. Then the mass o1 226Ra was m: 226 g) : 76l x 1o- s 226 + zlzts lsY(o10The mass of a 226pu nucleus is (226uX 1.66I x 10-24 glu)226pu nuclei present was iV (76.I x 10-3 gl(3.75 x 10-22 - g): 2.03 x 1020.(b) The decay rate is given by R- N^ - (lf hlTr/2, where is the disintegration constant, ^Tr/z is the half-life, and tf is the number of nuclei. The relationship l - (ln2)lTrp was used.Thus R- - 2.7gx 10es-l .43If ,Af is the number of undecayed nuclei present at time t, then ry - dt R_)N,where R is the rate of production by the cyclotron andis the disintegration constant. The ^second term gives the rate of decay. Rearrange the equation slightly and integrate: dtY [* R-A/ - [ n,, uui J*o l, Chapter 42 255
260.
where Afg is the number of undecayed nuclei present at time f, : 0. This yields r tn*-xvo:t R- ) ^rrSolve for l/: / R rv-R (.^o x) e i* ^tAfter many half-lives, the exponential is small and the second term can be neglected. Then,,Af - Rl^, regardless of the initial value fig. At times that are long compared to the half-life,the rate of production equals the rate of dec ay and .,Af is a constant.49The fraction of undecayed nuclei remaining after time t is given by ,,Ar : tf, - e-t e-(lnDtl 12where is the disintegratton constant and Trlz (- (ln 2)lD is the half-life. The time for half the ^original 238IJ nuclei to decay is 4.5 x 10ey. For 2aapu atthat time (h2X45 x loey) ry- Tt 8.2 x 107 y -3g.0 /zand tr : : e-38o 31 x 10-17 F; For za9g^ at that time (In2)t (ln 2X4.5 x 10e y) _ 9n0 - 3.4 x y Tt /z 105and { : e-st7o - 3.31 x 10-3e83 . lroFor any reasonably sized sample this is less than one nucleus and may be taken to be zero. Yourcalculator probably cannot evaluate e-er70 directly. Treat it as (e-el70;100.--35Let Mc, be the mass of one atom of t33Cr and Msu be the mass of one atom of t3Znu. Toobtain the nuclear masses we must subtract the mass of 55 electrons from Mc" and the massof 56 electrons from Msu. The energy released is A- l(Mc, - 55*) - (Msa- 56m) -mf c2,where m is the mass of an electron. Once cancellations have been made, Q : (Mc, - Msu)r isobtained. Thus Q - [136.9071 u - 136.9058u] c2 - (0.00 13u)c2 - (0.0013u)(g3zMeV/u): 1.21 MeV.256 Chapter 42
261.
59Since the electron has the maximum possible kinetic energy no neutrino is emitted. Sincemomentum is conserved, the momentum of the electron and the momentum of the residual sulfurnucleus arc equal in magnitude and opposite in direction. If p" is the momentum of the electronand ps is the momentum of the sulfur nucleus, then ps: -pu. The kinetic energy Ks ofthe sulfur nucleus is Ks - pLlLtWt - p?l2Ms, where Ms is the mass of the sulfur nucleus.Now the electrons kinetic energy K" is related to its momentum by the relativistic equation(prc)z - K?+2K"fficT, where m is the mass of an electron. See Eq. 37-54. Thus (P"c)z _ KZ + 2K"mc2 (1 .71 MeV) + 2(1 .71MeV)(0.511 MeV) Kg - 2Mscz 2Msc2 - 2(32u)(93 1.5 MeV l") - 7.83 x 10-s MeV - 78.3 eVwhere mcz - 0.511 MeV was used.67The decay rute R is related to the number of nuclei l[ by R - ,LAf , where is the disintegration Iconstant. The disintegration constant is related to the half-life Tt/z by (ln 2)lTt/2, so -.Ar - : RIA RTr/zlln2. Since 1 Ci :3.7 x 1010 disintegrations/s, ^ Ar _ Q50 CiX3 .7 x 1010 s- lct)(2.7 dX8 .64 x 104 s/d) ]VThe mass of a le84n atom is M _ (198uXI.661 x 10-24 glu)- 3.29 x 10-22 g so the massrequired is M: (3.11 x lOltxf .2g x 10-tg) - 1.02 x 10-g- l.02mg.73A generaltzed formation reaction can be written X + x -+ Y, where X is the target nucleus, x isthe incident light particle, and Y is the excited compound nucleus (oN.). Assume X is initiallyat rest. Then conservation of energy yields ITLyc2 + mrr2 + K*: TTLyr2 + Ky i Ev ,where ffLy, ffir, and rft,y are masses, K, and Ky are kinetic energies, and Ey is the excitationenergy of Y. Conservation of momentum yields P*: PYNow Ky - p+12^t - p?*12*" : (m*l*")K*, so TTLyc2 + mnc2 + K*: TTLyc2 + (m*l*t)K, + Evand K* TTLy - ffir Chapter 42 257
262.
(a) Let r rcpresent the alpha particle and X represent the 160 nucleus. Then (mv -wLy -mn)r :(19.99244u - 15.99491 u - 4.00260u)(931.5 MeV/n) - -4.722MeV and 19.99244u Ko: (-4.722MeV + 25.0MeV) : 25.35 MeV. 19.99244u - 4.00260u(b) Let n represent the proton and X represent the leF nucleus. Then (mv TLy TTL)" -(19 .99244u - 18.99841 u - 1.00783 u)(931.5 MeV/,r) - - 12.85 MeV and 19.99244u Ko 19.99244u- 1.00783u (-12.85 MeV + 25.0MeV) - I2.80MeV.(c) Let r represent the photon and X represent the 20Ne nucleus. Since the mass of the photonis zero, we must rewrite the conservation of energy equation: if E.y is the energy of the photon,then E^,t *mxc2 : TTLyc2 + Ky + Ev. Since ?TLy: TTLr, this equation becomes E^t: Ky + Ev.Since the momentum and energy of a photon are related by pt - E^t f c, the conservation ofmomentum equation becomes E^, l" - pv. The kinetic energy of the compound nucleus isKv - pTl2^t- 8?1 f2mycz. Substitute this result into the conservation of energy equation toobtain E^-3 EY ^Y zmva* This quadratic equation has the solutions E^r : TTl,y c2 + (mvr2)2 2mvc2EvIfthe problem is solved using the relativistic relationship between the energy and momentumof the compound nucleus, only one solution would be obtained, the one corresponding to thenegative sign above. Since my c2 - ( L9 .99244u)(93 1 .5 MeV/u) : I .862 x 104 MeV, E^t - (I.862 x 104 MeV) - (I .862 x 104 MeV)2 - 2(I .862 x 104 MeV) (25.0 MeV) :25.0MeV.The kinetic energy of the compound nucleus is very small; essentially all of the photon energygoes to excite the nucleus.75Let A be the area over which fallout occurs and a be the area that produces a count rate of R-74000counts/s. The count rate is R- where, is the disintegration constant and l/ isthe number of radioactive nuclei in area a. ^ltr, number of atoms in the entire fallout is M l*, Thewhere M is the mass of e0Sr produced and m is the mass of a single nucleus of e0Sr. Thus thecount rate for the area a is R- The half-lifeT1 p is related to the disintegration ^(Mlm)(alA).constant by - (Inz) lTr /2, so R - (ln 2lT, p)(Att ld@lA). Solve for a,: ^ a-AR(#)(#)258 Chapter 42
263.
The molar mass of e0 Sr is 90 glmol, so the mass of a single e0 Sr nucleus is (90-l kgl^ol)/(6 .02 x1023 mol-t): 1.50 x l0-2skg. The half-life is (z9yx365dlfle4hld)(3600s/h) -9. l5 x 108s.Therefore &-(2000X106mQ4000counts/s)()(#):7.3X10-2m2.85The number of undecayed nuclei at time t is given by iV - l[oe-t, where Ah is the number atf : 0 and A is the disintegration constant. The rate of decay is R - -d,IV ld,t :,Ah e-t - ,Arand the rate att:) is fto: Ah. Thus RlRo: Nllr{o- e-lt. The solution for t is 1R t- -,tr*0.The disintegration constant is related to the half-life Tr/z by ) - (ln2)lTt/2, so r - +r,r*: -{+h(o .ozo):3.2x looy. L--lt87Let fiX represent the unknown nuclide. The reaction equation is fx+fn*_!e+2tRr.Conservation of charge yields Z+0- -1 +4 or Z-3. Conservation of mass number yieldsA+ I :0+8 or A: 7. According to the periodic table in Appendix E, lithium has atomicnumber 3, so the nuclide must be lti. Chapter 42 259
264.
Chapter 4313(a) If X represents the unknown fragment, then the reaction can be written ltu+fn-!ic.+fX,where A is the mass number and Z is the atomic number of the fragment. Conservation ofchargeyields 92+0- 32+2, so Z - 60. Conservationof massnumberyields 235+1_83+,4,so A_ 153. Look in Appendix F or G for nuclides with Zunknown fragment is t;ANd.(b) and (c) Ignore the small kinetic energy and momentum carried by the neutron that triggers thefission event. Then A - Kc" * KNo, where Kc" is the kinetic energy of the gefinanium nucleusand /fxa is the kinetic energy of the neodymium nucleus. Conservation of momentum yieldspce + prva : 0, where pce is the momentum of the gennanium nucleus and ft.-ro is the momentumof the neodymium nucleus. Since pNd : -pce, the kinetic energy of the neodymium nucleus is lfxd : Piro PL" Mc" Kc" 2MNd 2Mxd MxoThus the energy equation becomes e:Kc".mKc"-WKc"and Kc.: # Q:#(l7oMev): lloMevSimilarly, : -#(170 Mev) : 6o Mev lrNd ffie .The mass conversion factor can be found in Appendix C.(d) The initial speed of the gennanium nucleus is -, -1.60x l07m/s. V(e) The initial speed of the neodymium nucleus is : 2(60 x 106 eVX1.60 x 10-le J/eV) :8.69 uNd x 106 m/s. (153 uX 1.661 x 10-27 kg/,r)260 Chapter 43
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15(a) The energy yield of the bomb is E - (66 x 10--rgaton)(2.6 x l028MeV/megaton)I.72 x 1027 MeV. (The energy conversion factor is given in Problem 16.) At 200MeV perfission event, (l .72 x 1027 MeV) lQ}AMeV) - 8.58 x L024 fission events take place. Sinceonly 4.0oh of the 23sg nuclei originally present undergo fission, there must have been (8.58 x rc2l(0.040) : 2.L4 x 1026 nuclei originally present. The mass og 23s1; originally present was(2.I4 x 1026Xn5uXI.66l x 10-27 kg/.r) - 83.7kg. The mass conversion factor can be foundin Appendix C.(b) Two fragments are produced in each fission event, so the total number of fragments is2(8.58 x 10241 - !.72 x 102s.(c) One neutron produced in a fission event is used to trigger the next fission event, so theaverage number of neutrons released to the environment in each event is 1.5. The total numberreleased is (8.58 x 1024X1.5): L.29 x 1025.23(a) Let un,i be the initial velocity of the neutron, unf be its final velocity, and u y be the finalvelocity of the target nucleus. Then, since the target nucleus is initially at rest, conservationof momentum yields nLnuni, - TTLnunf + rnu f and conservation of energy yields **rult -*,**ult + i*rr. Solve these two equations simultaneously for r) y. This can be done, forexample, by using the conservation of momentum equation to obtain an expression for unf interms of u y and substituting the expression into the conservation of energy equatioll. Solve theresultirg equation for ?)y. You should obtain uy: 2mrnni,l(*+ mn). The energy lost by theneutron is the same as the energy gained by the target nucleus, soThe initial kinetic energy of the neutron is K - *.*rzrt, so AK 4mrm K (m+ mn)2(b) The mass of a neutron is 1.0 u and the mass of a hydrogen atom is also 1.0 u. (Atomicmasses can be found in Appendix G.) Thus (A/O lK - 4(1 .0uxl.0u)l(1.0u+ 1.0u)2(c) The mass of a deuterium atom is 2.0v, so (A/O lK - 4(1 .0uX2.0u)/(2.0u* 1.0 u)2 - 0.89.(d) The mass of a carbon atom is l2u, so (A/{) lK - 4(1 .0uXI2u)lQzu* 1.0u)2 - 0.28.(e) The mass of a lead atom is 207u, so (A/O lK - 4(1 .0u)(207u)l(207 u* 1.0u)2 - 0.019.(0 During each collisior, the energy of the neutron is reduced by the factor I - 0.89 - 0.11. IfEi is the initial energy, then the energy after n collisions is given by E - (0.1L)n Ei. Take thenatural logarithm of both sides and solve for n. The result is tn(E I trr,) ln(0 .025 eY l1 .00 eV) tt: ln 0.1I ln 0.11 -7.9. Chapter 43 261
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The energy first falls below 0.025 eV on the eighth collision.25Let Po be the initial power output, P be the final power output, k be the multiplication factor, tbe the time for the power reduction, and tgen be the neutron generation time. Then according tothe result of Problem 18, P = Po kt/tgen .Divide by Po, then take the natural logarithm of both sides of the equation and solve for In k.You should obtain In k = tgen In P . t PoHencewhere _ tgen P _ 1.3 X 10- 3 s 350.00MW _ -4 a - -t- In Po - 2.6000s In 1200.0MW - -6.161 x 10 .This yields k = .99938.29Let t be the present time and t = 0 be the time when the ratio of 235U to 238U was 3.0%. LetN 235 be the number of 235U nuclei present in a sample now and N 235 , 0 be the number presentat t = O. Let N 238 be the number of 238U nuclei present in the sample now and N238, 0 be thenumber present at t = O. The law of radioactive decay holds for each specie, so N 235 = N 235,0 e -.>..tand N 238 = N 238 , 0 e -.>..t .Divide the first equation by the second to obtain r = ro e-(>"->")t ,where r = N235/N238 (= 0.0072) and ro = N 235 , O/N238 , 0 (= 0.030). Solve for t: 1 r t=- In- . ..235 - ..238 roNow use ..235 = (In 2)/T235 and ..238 = (In 2)/T23 8, where T 235 and T 238 are the half-lives, toobtain _ T235T238 I r_ (7.0 x 108 y)(4.5x 109 y) 10.0072_17 109 t-- n--- n -.Ix y. (T238 - T 235 ) In 2 ro (4.5 x 109 y - 7.0 X 108 y) In 2 0.030262 Chapter 43
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31The height of the Coulomb barrier is taken to be the value of the kinetic energy K eachdeuteron must initially have if they are to come to rest when their surfaces touch (see SampleProblem 43-4). If r is the radius of a deuteror, conservation of energy yields 1 t. 2K:4rreg 2r -LL )SO 4: K - -]- 4r (8 ,ggx 10e m/F) - 2.74 x 10-14 J . 4nesThis is l7}keV.43(a) The mass of a carbon atom is (I2.0 uX I.66I x 10-27 kg/,r)number of carbon atoms in 1.00kg of carbon is x 10-ukg) (1.00kg) 1Q.99(The mass conversion factor can be found in Appendix C.) The heat of combustion per atom is(3.3 x 107 JlkglO.Oz x 1025 alrrmfkg) - 6.58 x 10-tJlutom. This is 4.lIeYlatom.(b) In each combustion event, two oxygen atoms combine with one carbon atom, so the totalmass involved is 2(16.0u)+(12.0u) - 44v. This is (44uXL.661 x 10-27 kglu) -7.31x 10-26 kg.Each combustion event produces 6.58 x 10-le J so the energy produced per unit mass of reactantsis (6.58 x 10-te DlQ.3r x L0-26 kg) -9.00 x 106 Jlkg.(c) If the Sun were composed of the appropriate mixture of carbon and oxygen, the number ofcombustion events that could occur before the Sun burns out would be (2.0 x 1030 kgl(7.31 x L0-26 kg) - 2.74x 1055. The total energy released would be E - (2.74x 10ssx6.58 x 10-tnJ) -1 .80 x 1037 J. If P is the power output of the Sun, the burn time would be t(1.80 x 1037 DlQ9 x 1026 w) - 4.62 x 1010 s. This is 1460y. Chapter 43 263
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Chapter 4411(a) The conservation laws considered so far are associated with energy, momentum, angularmomentum, charge, baryon number, and the three lepton numbers. The rest energy of the muonis 105.7MeV, the rest energy of the electron is 0.511MeV, and the rest energy of the neutrino iszero. Thus the total rest energy before the decay is greater than the total rest energy after. Theexcess energy can be carried away as the kinetic energies of the decay products and energy canbe conserved. Momentum is conserved if the electron and neutrino move away from the decayin opposite directions with equal magnitudes of momenta. Since the orbital angular momentumis zero, we consider only spin angular momentum. All the particles have spin nlz. The totalangular momentum after the decay must be either h, (if the spins are aligned) or zero (if the spinsare antialigned). Since the spin before the decay ish, 12, angular momentum cannot be conseryed.The muon has charge -e) the electron has charge -e) and the neutrino has charge zeto, so thetotal charge before the decay is -e and the total charge after is -e. Charge is conserved. All theparticles have baryon number zero, so baryon number is conserved. The muon lepton numberof the muon is 11, the muon lepton number of the muon neutrino is *1, and the muon leptonnumber of the electron is 0. Muon lepton number is conseryed. The electron lepton numbers ofthe muon and muon neutrino are 0 and the electron lepton number of the electron is +1. Electronlepton number is not conserved. The laws of consenration of angular momentum and electronlepton number are not obeyed and this decay does not occur..(b) Analyze the decay in the same way. You should find that only charge is not conserved.(c) Here you should find that energy and muon lepton number cannot be conserved.29(a) Look at Table 44-5. Since the particle is abaryon, it must consist of three quarks. To obtaina strangeness of -2, two of them must be s quarks. Each of these has a charge of -e13, so thesum of their charges is -2e 13. To obtain a total charge of e, the charge on the third quark mustbe 5ef3. There is no quark with this charge, so the particle cannot be constructed. In fact, sucha particle has never been observed.(b) Again the particle consists of three quarks (and no antiquarks). To obtain a strangeness ofzero, none of them may be s quarks. We must find a combination of three u and d quarks witha total charge of 2e. The only such combination consists of three u quarks.4t(a) The mass M wtthin Earths orbit is used to calculate the gravitational force on Earth. If r is264 Chapter 44
269.
the radius of the orbit, R is the radius of the new Sun, and M s is the mass of the Sun, then tvr /Yrr . *)t r.ssx M - (L)rtr*: (t ?9 19ll 1030ke) -3.27 x rr2skg. n/ s.lo x lol ) ^The gravitational force on Earth is given by GMmlr, where m is the mass of Earth and G isthe universal gravitational constant. Since the centripetal acceleration is given by , lr, where uis the speed of Earth, GMmlr -- mr)l, and IGM x 10-ll m3 lt.kgX3 .27 x t)-l; (6.67 1.50 x l0ll m 102s kg) - l.2l x 102 m/s .(b) The period of revolution is T* 2nr : 2n(1 .50 x 101t ttt) - 7.82 x 109 s. u I.2l x I02 m/sThis is 248y.45The energy released would be twice the rest energy of Earth, or E- 2mc2 - 2(5.98 x1024 kgX3.00 x 108 ml s)zC.47(a) Since ,S-1 f3. To obtain a meson with charge quantum number *1, the s quark must be combined withan antiquark with strangeness 0 and charge quantum number *4f3. There is no such antiquark.(b) Now S- +1, so the meson contains an S quark, which has a charge quantum number of+l 13. To obtain a charge quantum number of - 1 it must also contain a quark with chargequantum number -4 12. There is no such quark.51(a) After time At the distance between the galaxy and Earth is r * ra Lt, where r is thedistance when the light is emiffed. The distance when the light reaches Earth must be cLt, socLt - r + raLt and Lt - r I @ - rcr).(b) The detected waveleneth is longer than I by )c Lt, so L^l^ - a Lt - ar l@ - crr).(c) Since c > , :;[r AA ar r -: ar1-r ar I- / czr r art2 ...] T] T[+ ff)+ (T)+ - T.ff) *ff) *(d) If only the first term is retained L^l, - ar f c. Chapter 44 265
270.
(e) If u - Hr,where H is the Hubble constant, then A^/^_ ulc: Hrf c. Comparison withL^l, - ar f c gives e - H - 0.0218 m/s . ly.(f) Solve A) f - ra l@ rc-) for r. The result is - ^ (300 x m/sX0 :(411)) - (0.02 lSmls . ly)(l + 0.050) : 6.6x 108 ly. - a(L + A 91% 108 ^lD(g) According to the result of part (a) Lt-c-rra - 3.00 x (6;6 110tlI),(?16*, 108 mls - (0.02 l8m/s.lyX6.6 x 10ltT11I) ,=, :z.zx1016s. 108ly) --This is 6.9 x 108 y.(h) The time is Lt - rlc: x 108 l9l(1.00 lVlg:6.6 x 108y. (:6.6(i) The distance is cLt - (1.00 IVlg(6.9 x 108 y) : 6.9 x 108 ly.0) [Jse the equation developed in part (0: r:ffi:(k) The result of part (a) gives r At- (c-rc-) 3.00 x 108 mls - (I .02 x 10e1VX0.021 Sm/s.ly) v"This is 1.1 x 10ey.(1) Galaxy B emits light Lt- 1.1 x 10ey - 6.9 x 108yDuring that time the universe expands, so that the distance of galaxy B from Earth at the time Aemits is 0.0218mls . ly ,n(L+ oA t): (1,.02x tOe ly) + (4.1x 108 yX3. 16 x 107 r/vl] [t 9.46 x 101t *fiy -1.05 x 10ely.The separation of the galaxies at the time A emits is 1.05 x 10ely- 6.6 x 108 ly: 3.9 x 108 ly.266 Chapter 44
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