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# Sm chapter9

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### Sm chapter9

1. 1. 9 Linear Momentum and Collisions CHAPTER OUTLINE 9.1 Linear Momentum and Its Conservation 9.2 Impulse and Momentum 9.3 Collisions in One Dimension 9.4 Two-Dimensional Collisions 9.5 The Center of Mass 9.6 Motion of a System of Particles 9.7 Deformable Systems 9.8 Rocket Propulsion ANSWERS TO QUESTIONS *Q9.1 (a) No. Impulse, F∆t, depends on the force and the time for which it is applied. (b) No. Work depends on the force and on the distance over which it acts. *Q9.2 The momentum magnitude is proportional to the speed and the kinetic energy is proportional to the speed squared. (i) The speed of the constant-mass object becomes 4 times larger and the kinetic energy 16 times larger. Answer (a). (ii) The speed and the momentum become two times larger. Answer (d). *Q9.3 (i) answer (c). For example, if one particle has 5 times larger mass, it will have 5 times smaller speed and 5 times smaller kinetic energy. (ii) answer (d). Momentum is a vector. *Q9.4 (i) Equal net work inputs imply equal kinetic energies. Answer (c). (ii) Imagine one particle has four times more mass. For equal kinetic energy it must have half the speed. Then this more massive particle has 4(1ր2) = 2 times more momentum. Answer (a). Q9.5 (a) It does not carry force, for if it did, it could accelerate itself. (b) It cannot deliver more kinetic energy than it possesses. This would violate the law of energy conservation. (c) It can deliver more momentum in a collision than it possesses in its ﬂight, by bouncing from the object it strikes. *Q9.6 Mutual gravitation brings the ball and the Earth together. As the ball moves downward, the Earth moves upward, although with an acceleration on the order of 1025 times smaller than that of the ball. The two objects meet, rebound, and separate. Momentum of the ball-Earth system is conserved. Answer (d). Q9.7 (a) Linear momentum is conserved since there are no external forces acting on the system. The fragments go off in different directions and their vector momenta add to zero. (b) Kinetic energy is not conserved because the chemical potential energy initially in the explosive is converted into kinetic energy of the pieces of the bomb. 209 13794_09_ch09_p209-244 copy.indd209 20913794_09_ch09_p209-244 copy.indd209 209 1/8/07 5:47:49 PM1/8/07 5:47:49 PM
3. 3. Linear Momentum and Collisions 211 Q9.17 The center of mass of the balls is in free fall, moving up and then down with the acceleration due to gravity, during the 40% of the time when the juggler’s hands are empty. During the 60% of the time when the juggler is engaged in catching and tossing, the center of mass must accelerate up with a somewhat smaller average acceleration. The center of mass moves around in a little closed loop with a parabolic top and likely a circular bottom, making three revolutions for every one revolution that one ball makes. Letting T represent the time for one cycle and Fg the weight of one ball, we have F T F TJ g0 60 3. = and F FJ g= 5 . The average force exerted by the juggler is ﬁve times the weight of one ball. Q9.18 In empty space, the center of mass of a rocket-plus-fuel system does not accelerate during a burn, because no outside force acts on this system. The rocket body does accelerate as it blows exhaust containing momentum out the back. According to the text’s ‘basic expression for rocket propulsion,’ the change in speed of the rocket body will be larger than the speed of the exhaust relative to the rocket, if the ﬁnal mass is less than 37% of the original mass. Q9.19 To generalize broadly, around 1740 the English favored position (a), the Germans position (b), and the French position (c). But in France Emilie de Chatelet translated Newton’s Principia and argued for a more inclusive view. A Frenchman, Jean D’Alembert, is most responsible for showing that each theory is consistent with the others. All the theories are equally correct. Each is useful for giving a mathematically simple and conceptually clear solution for some problems. There is another comprehensive mechanical theory, the angular impulse—angular momentum theorem, which we will glimpse in Chapter 11. It identiﬁes the product of the torque of a force and the time it acts as the cause of a change in motion, and change in angular momentum as the effect. We have here an example of how scientiﬁc theories are different from what people call a theory in everyday life. People who think that different theories are mutually exclusive should bring their thinking up to date to around 1750. SOLUTIONS TO PROBLEMS Section 9.1 Linear Momentum and Its Conservation P9.1 m = 3 00. kg, v i j= −( )3 00 4 00. ˆ . ˆ m s (a) p v i j= = −( ) ⋅m 9 00 12 0. ˆ . ˆ kg m s Thus, px = ⋅9 00. kg m s and py = − ⋅12 0. kg m s (b) p p px y= + = ( ) + ( ) = ⋅2 2 2 2 9 00 12 0 15 0. . . kg m s θ = ⎛ ⎝⎜ ⎞ ⎠⎟ = −( ) =− − tan tan .1 1 1 33 307 p p y x ° *P9.2 (a) Whomever we consider the aggressor, brother and sister exert equal-magnitude oppositely- directed forces on each other, to give each other equal magnitudes of momentum. We take the eastward component of the equation total original momentum = total ﬁnal momentum for the two-sibling system 0 = 65 kg (−2.9 mրs) + 40 kg v v = 4.71 mրs, meaning she moves at 4.71 mրs east continued on next page 13794_09_ch09_p209-244 copy.indd211 21113794_09_ch09_p209-244 copy.indd211 211 1/8/07 5:47:50 PM1/8/07 5:47:50 PM
4. 4. 212 Chapter 9 (b) original chemical energy in girl’s body = total ﬁnal kinetic energy Uchemical = (1ր2)(65 kg)(2.9 mրs)2 + (1ր2)(40 kg)(4.71 mրs)2 = 717 J (c) System momentum is conserved with the value zero. The net forces on the two siblings are of equal magnitude in opposite directions. Their impulses add to zero. Their ﬁnal momenta are of equal magnitude in opposite directions, to add as vectors to zero. P9.3 I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground with speed given by v vf i f ia x x2 2 2− = −( ): 0 2 9 80 0 2502 − = −( )( )vi . .m s m2 vi = 2 20. m s Total momentum of the system of the Earth and me is conserved as I push the planet down and myself up: 0 5 98 10 85 0 2 2024 = ×( ) − + ( )( ). ( ) . . ~ kg kg m sv v e e 110 23− m s *P9.4 (a) For the system of two blocks ∆p = 0 , or p pi f= Therefore, 0 3 2 00= + ( )( )M Mmv . m s Solving gives vm = −6 00. m s (motion toward the left). (b) 1 2 1 2 1 2 3 8 402 2 3 2 kx M MM M= + ( ) =v v . J (c) The original energy is in the spring. A force had to be exerted over a distance to com- press the spring, transferring energy into it by work. The cord exerts force, but over no distance. (d) System momentum is conserved with the value zero. The forces on the two blocks are of equal magnitude in opposite directions. Their impulses add to zero. The ﬁnal momenta of the two blocks are of equal magnitude in opposite directions. P9.5 (a) The momentum is p m= v, so v = p m and the kinetic energy is K m m p m p m = = ⎛ ⎝ ⎞ ⎠ = 1 2 1 2 2 2 2 2 v (b) K m= 1 2 2 v implies v = 2K m , so p m m K m mK= = =v 2 2 FIG. P9.4 13794_09_ch09_p209-244 copy.indd212 21213794_09_ch09_p209-244 copy.indd212 212 1/8/07 5:47:51 PM1/8/07 5:47:51 PM
5. 5. Linear Momentum and Collisions 213 Section 9.2 Impulse and Momentum *P9.6 From the impulse-momentum theorem, F t p m mf i∆ ∆( )= = −v v , the average force required to hold onto the child is F m t f i = −( ) ( ) = ( ) −( ) − v v ∆ 12 0 60 0 050 0 1kg mi h s. mm s mi h N 2 237 6 44 103 . . ⎛ ⎝⎜ ⎞ ⎠⎟ = − × In trying to hang onto the child, he would have to exert a force of 6.44 kN (over 1400 lb) toward the back of the car, to slow down the child’s forward motion. He is not strong enough to exert so large a force. If he were belted in and his arms were ﬁrmly tied around the child, the child would exert this size force on him toward the front of the car. A person cannot safely exert or feel a force of this magnitude and a safety device should be used. P9.7 (a) I Fdt= =∫ area under curve I = ×( )( )= ⋅−1 2 1 50 10 18 000 13 53 . .s N N s (b) F = ⋅ × =− 13 5 9 003 . . N s 1.50 10 s kN (c) From the graph, we see that Fmax .= 18 0 kN P9.8 The impact speed is given by 1 2 1 2 1m mgyv = . The rebound speed is given by mgy m2 2 21 2 = v . The impulse the ﬂoor imparts to the ball is the change in the ball’s momentum, m m m m gh gh v v v v2 1 2 1 2 12 2 up down up up − = +( ) = +( ) == ( ) +( ) = 0 15 2 9 8 0 960 1 25 1 39 . . . . . kg m s m m up2 kg m s upward⋅ P9.9 ∆ ∆ ∆ p F= = −( )= ( )− t p m m my fy iyv v v vcos . cos60 0 60° .. sin . sin . sin 0 0 60 0 60 0 2 60 ° ° ° = = − −( ) = −∆p m mx v v v .. . . . . 0 2 3 00 10 0 0 866 52 0 ° = − ( )( )( ) = − ⋅ kg m s kg m ss kg m s s NavgF p t x = = − ⋅ = − ∆ ∆ 52 0 0 200 260 . . P9.10 Assume the initial direction of the ball in the −x direction. (a) Impulse, I p p p i= = − = ( )( ) − ( )∆ f i 0 060 0 40 0 0 060 0 50. . ˆ . .00 5 40( ) −( )= ⋅ˆ . ˆi i N s (b) Work = − = ( ) ( ) − ( )⎡⎣ ⎤⎦ = −K Kf i 1 2 0 060 0 40 0 50 0 27 0 2 2 . . . . JJ FIG. P9.7 FIG. P9.9 13794_09_ch09_p209-244 copy.indd213 21313794_09_ch09_p209-244 copy.indd213 213 1/8/07 5:47:52 PM1/8/07 5:47:52 PM
6. 6. 214 Chapter 9 *P9.11 (a) The impulse is to the right and equal to the area under the F-t graph: I = [(0 + 4 N)ր2](2 s − 0) + (4 N)(3 s − 2 s) + (2 N)(2 s) = 12.0 N ⋅ s ˆi (b) m t mi fv F v+ = (2.5 kg)(0) + 12 ˆi N ⋅ s = (2.5 kg) v f v f = 4.80 ˆi mրs (c) From the same equation, (2.5 kg)(−2 ˆi mրs) + 12 ˆi N ⋅ s = (2.5 kg) v f v f = 2.80 ˆi mրs (d) F i Favg avgt∆ = ⋅12 0. ˆ N s = (5 s) = NF iavg 2 40. ˆ *P9.12 (a) A graph of the expression for force shows a parabola opening down, with the value zero at the beginning and end of the 0.8 s interval. I Fdt t t dt= = −( )∫0 0 8 2 0 0 9200 11500 . s 2 N/s N/s ..8s ∫ = N/s N/s2 9200 2 115002 3 ( ) − ( )t t/ /33 9200 0 8 2 0 0 8 2 ⎡⎣ ⎤⎦ ( ) . ( . / s = N/s s) −− ( )11500 0 8 33 N/s s) = 2944 N 2 ( . / ⋅⋅ − ⋅ = ⋅s N s 981 N s1963 The athlete imparts downward impulse to the platform, so the platform imparts 981 N⋅s of upward impulse to her. (b) We could ﬁnd her impact speed as a free-fall calculation, but we choose to write it as a conservation-of energy calculation: mgytop = (1ր2)mvimpact 2 vimpact = (2gytop )1ր2 = [2(9.8 mրs2 )0.6 m]1ր2 = 3.43 mրs down (c) Gravity, as well as the platform, imparts impulse to her during the interaction with the platform. mvi + Iplatform + mgt = mvf (65 kg)(−3.43 mրs) + 981 N ⋅ s − (65 kg)(9.8 mրs2 )(0.8 s) = 65 kg vf −223 N ⋅ s + 981 N ⋅ s − 510 N ⋅ s = 65 kg vf vf = 249 N ⋅ sր65 kg = 3.83 mրs up Note that the athlete is putting a lot of effort into jumping and does not exert any force “on herself.” The usefulness of the force platform is to measure her effort by showing the force she exerts on the ﬂoor. (d) Again energy is conserved in upward ﬂight. (1ր2)mvtakeoff 2 = mgytop ytop = vtakeoff 2 ր2g = (3.83 mրs)2 ր2(9.8 mրs2 ) = 0.748 m P9.13 (a) Energy is conserved for the spring-mass system: K U K Ui si f sf+ = + : 0 1 2 1 2 02 2 + = +kx mv v = x k m (b) From the equation, a smaller value of m makes v = x k m larger. (c) I m mx k m x kmf i f= − = − = =p p v 0 (d) From the equation, a larger value of m makes I x km= larger. (e) For the glider, W K K m kxf i= − = − = 1 2 0 1 2 2 2 v The mass makes no difference to the work. 13794_09_ch09_p209-244 copy.indd214 21413794_09_ch09_p209-244 copy.indd214 214 1/8/07 5:47:52 PM1/8/07 5:47:52 PM
7. 7. Linear Momentum and Collisions 215 *P9.14 After 3 s of pouring, the bucket contains (3s)(0.25 Lրs) = 0.75 liter of water, with mass 0.75 L(1 kgր1 L) = 0.75 kg, and feeling gravitational force 0.75 kg(9.8 mրs2 ) = 7.35 N. The scale through the bucket must exert 7.35 N upward on this stationary water to support its weight. The scale must exert another 7.35 N to support the 0.75-kg bucket itself. Water is entering the bucket with speed given by mgytop = (1ր2)mvimpact 2 vimpact = (2gytop )1ր2 = [2(9.8 mրs2 )2.6 m]1ր2 = 7.14 mրs downward The scale exerts an extra upward force to stop the downward motion of this additional water, as described by mvimpact + Fextra t = mvf The rate of change of momentum is the force itself: (dmրdt) vimpact + Fextra = 0 Fextra = −(dmրdt) vimpact = −(0.25 kgրs)(−7.14 mրs) = +1.78 N Altogether the scale must exert 7.35 N + 7.35 N + 1.78 N = 16.5 N Section 9.3 Collisions in One Dimension P9.15 Momentum is conserved for the bullet-block system 10 0 10 5 01 0 600 301 3 . . .×( ) = ( )( ) = − kg kg m s m s v v P9.16 (a) m m mi i fv v v1 23 4+ = where m = ×2 50 104 . kg vf = + ( ) = 4 00 3 2 00 4 2 50 . . . m s (b) K K m m mf i f i i− = ( ) − + ( )⎡ ⎣⎢ ⎤ ⎦⎥ = 1 2 4 1 2 1 2 3 22 1 2 2 2 v v v .550 10 12 5 8 00 6 00 3 75 104 4 ×( ) − −( ) = − ×. . . . J Ki = Kf + ∆Eint ∆Eint = +37.5 kJ P9.17 (a) The internal forces exerted by the actor do not change the total momentum of the system of the four cars and the movie actor 4 3 2 00 4 00 6 00 4 m m mi i ( ) = ( )( )+ ( ) = + v v . . . m s m s m s .. . 00 4 2 50 m s m s= (b) W K K m mf iactor m s m s= − = ( )( ) + ( )⎡ ⎣ 1 2 3 2 00 4 00 2 2 . . ⎤⎤ ⎦ − ( )( ) 1 2 4 2 50 2 m m s. Wactor kg m s= ×( ) + −( )( ) = 2 50 10 2 12 0 16 0 25 0 4 2. . . . 337 5. kJ (c) The event considered here is the time reversal of the perfectly inelastic collision in the previous problem. The same momentum conservation equation describes both processes. FIG. P9.17 13794_09_ch09_p209-244 copy.indd215 21513794_09_ch09_p209-244 copy.indd215 215 1/8/07 5:47:54 PM1/8/07 5:47:54 PM
8. 8. 216 Chapter 9 P9.18 Energy is conserved for the bob-Earth system between bottom and top of swing. At the top the stiff rod is in compression and the bob nearly at rest. K U K Ui i f f+ = + : 1 2 0 0 22 M Mgbv + = + ᐉ vb g2 4= ᐉ so vb g= 2 ᐉ Momentum of the bob-bullet system is conserved in the collision: m m M gv v = + ( )2 2 ᐉ v = 4M m gᐉ P9.19 First we ﬁnd v1, the speed of m1 at B before collision. 1 2 2 9 80 5 00 9 90 1 1 2 1 1 m m ghv v = = ( )( ) =. . . m s Now we use the text’s analysis of one-dimensional elastic collisions to ﬁnd v1 f , the speed of m1 at B just after collision. v v1 1 2 1 2 1 1 3 9 90 3 30f m m m m = − + = − ( ) = −. .m s m s Now the 5-kg block bounces back up to its highest point after collision according to m gh m1 1 21 2 3 30max .= −( ) hmax . . .= −( ) ( ) = 3 30 2 9 80 0 556 2 m s m s m2 *P9.20 (a) We assume that energy is conserved in the fall of the basketball and the tennis ball. Each reaches its lowest point with a speed given by K U K U mgy m g g i b b +( ) = +( ) + = + release bottom 0 1 2 02 v v == = ( )( ) =2 2 9 8 1 20 4 85gyi . . .m s m m s2 FIG. P9.19 continued on next page v/2 FIG. P9.18 13794_09_ch09_p209-244 copy.indd216 21613794_09_ch09_p209-244 copy.indd216 216 1/8/07 5:47:55 PM1/8/07 5:47:55 PM
9. 9. Linear Momentum and Collisions 217 (b) The two balls exert no forces on each other as they move down. They collide with each other after the basketball has its velocity reversed by the ﬂoor. We choose upward as positive. Momentum conservation: 57 4 85 590 4 85 57 2g m s g m s g( ) −( )+ ( )( )= ( ). . v f ++ ( )590 1g v f To describe the elastic character of the collision, we use the relative velocity equation 4 85 4 85 2 1. .m s m s− −( )= −v vf f we solve by substitution v v v v 1 2 2 2 9 70 2 580 57 590 f f f = − = ( ) + ( ) . m s gm s g g ff f f −( ) = ( ) + ( ) − 9 70 57 590 5 7202 2 . m s g g gm sv v v22 8 310 647 12 8f = = m s m s. Now the tennis ball-Earth system keeps constant energy as the ball rises: 1 2 57 12 8 57 9 8 165 2 g m s g m s2 ( )( ) = ( )( ) = . . y y f f mm s m s m 2 2 2 2 9 8 8 41 . . ( ) = P9.21 (a), (b) Let vg and vp be the x-components of velocity of the girl and the plank relative to the ice surface. Then we may say that v vg p− is the velocity of the girl relative to the plank, so that v vg p− = 1 50. (1) But also we must have m mg g p pv v+ = 0, since total momentum of the girl-plank system is zero relative to the ice surface. Therefore 45 0 150 0. v vg p+ = , or v vg p= −3 33. Putting this into the equation (1) above gives − − =3 33 1 50. .v vp p or vp = −0 346. ˆ m si (answer b) Then vg = − −( ) =3 33 0 346 1 15. . . ˆ m si (answer a) FIG. P9.20(b) FIG. P9.21 4.85 m/s 4.85 m/s v f2 v f1 13794_09_ch09_p209-244 copy.indd217 21713794_09_ch09_p209-244 copy.indd217 217 1/8/07 5:47:56 PM1/8/07 5:47:56 PM
10. 10. 218 Chapter 9 P9.22 We assume equal ﬁring speeds v and equal forces F required for the two bullets to push wood ﬁbers apart. These equal forces act backward on the two bullets. For the ﬁrst, K E Ki f+ =∆ mech 1 2 7 00 10 8 00 10 03 2 2 . .×( ) − ×( )=− − kg mv F For the second, p pi f= 7 00 10 1 0143 . .×( ) = ( )− kg kgv vf v v f = ×( )− 7 00 10 1 014 3 . . Again, K E Ki f+ =∆ mech : 1 2 7 00 10 1 2 1 0143 2 2 . .×( ) − = ( )− kg kgv vFd f Substituting for vf , 1 2 7 00 10 1 2 1 014 7 00 103 2 3 . . . ×( ) − = ( ) ×− − kg kgv Fd vv 1 014 2 . ⎛ ⎝⎜ ⎞ ⎠⎟ Fd = ×( ) − ×( )− − 1 2 7 00 10 1 2 7 00 10 1 014 3 2 3 2 2 . . . v v Substituting for v, Fd F= ×( ) − ×⎛ ⎝⎜ ⎞ ⎠⎟ − − 8 00 10 1 7 00 10 1 014 2 3 . . . m d = 7 94. cm P9.23 (a) From the text’s analysis of a one-dimensional elastic collision with an originally stationary target, the x-component of the neutron’s velocity changes from vi to v1f = (1 − 12)vi ր13 = −11vi ր13. The x-component of the target nucleus velocity is v2 f = 2vi ր13. The neutron started with kinetic energy (1ր2) m1 vi 2 The target nucleus ends up with kinetic energy (1ր2) (12 m1 )( 2vi ր13)2 Then the fraction transferred is 1 2 1 2 1 2 1 2 12 2 13 48 169 0 284 m m i i ( / ) . v v = = Because the collision is elastic, the other 71.6% of the original energy stays with the neutron. The carbon is functioning as a moderator in the reactor, slowing down neutrons to make them more likely to produce reactions in the fuel. (b) Kn = ( ) ×( )= ×− − 0 716 1 6 10 1 15 1013 13 . . .J J KC = ( ) ×( )= ×− − 0 284 1 6 10 4 54 1013 14 . . .J J 13794_09_ch09_p209-244 copy.indd218 21813794_09_ch09_p209-244 copy.indd218 218 1/8/07 5:47:57 PM1/8/07 5:47:57 PM
11. 11. Linear Momentum and Collisions 219 P9.24 (a) Using conservation of momentum, p p∑ ∑( ) = ( )before after , gives 4 0 5 0 10 3 0 3 0 4. . . .kg m s kg m s kg( )( )+ ( )( )+ ( ) − .. . .0 4 0 10 3 0m s kg( )= + +( )[ ]v Therefore, v = +2 24. m s, or 2 24. m s toward the right (b) No. For example, if the 10-kg and 3.0-kg mass were to stick together ﬁrst, they would move with a speed given by solving 13 10 3 0 3 0 4 01kg kg m s kg m s( ) = ( )( )+ ( ) −( )v . . . , or v1 1 38= + . m s Then when this 13 kg combined mass collides with the 4.0 kg mass, we have 17 13 1 38 4 0 5 0kg kg m s kg m s( ) = ( )( )+ ( )( )v . . . , and v = +2 24. m s just as in part (a). Coupling order makes no difference to the ﬁnal velocity. P9.25 During impact, momentum of the clay-block system is conserved: m m mv v1 1 2 2= +( ) During sliding, the change in kinetic energy of the clay-block-surface system is equal to the increase in internal energy: 1 2 1 2 0 112 1 2 2 2 1 2 2 2 m m f d m m gdf+( ) = = +( ) ( ) = v v µ . kg 00 650 0 112 9 80 7 50. . . .kg m s m2 ( )( )( ) v2 2 95 6= . m s2 2 v2 9 77= . m s 12 0 10 0 112 9 773 1. . .×( ) = ( )( )− kg kg m sv v1 91 2= . m s Section 9.4 Two-Dimensional Collisions *P9.26 (a) Over a very short time interval, outside forces have no time to impart signiﬁcant impulse—thus the interaction is a collision. The opponent grabs the fullback and does not let go, so the two players move together at the end of their interaction—thus the collision is completely inelastic. FIG. P9.25 continued on next page 13794_09_ch09_p209-244 copy.indd219 21913794_09_ch09_p209-244 copy.indd219 219 1/8/07 5:47:57 PM1/8/07 5:47:57 PM
12. 12. 220 Chapter 9 (b) First, we conserve momentum for the system of two football players in the x direction (the direction of travel of the fullback). 90 0 5 00 0 185. . coskg m s kg( )( )+ = ( )V θ where θ is the angle between the direction of the ﬁnal velocity V and the x axis. We ﬁnd V cos .θ = 2 43 m s (1) Now consider conservation of momentum of the system in the y direction (the direction of travel of the opponent). 95 0 3 00 0 185. . sinkg m s kg( )( )+ = ( )( )V θ which gives V sin .θ = 1 54 m s (2) Divide equation (2) by (1) tan . . .θ = = 1 54 2 43 0 633 From which θ = °32 3. Then, either (1) or (2) gives V = 2 88. m s (c) Ki = ( )( ) + ( ) 1 2 90 0 5 00 1 2 95 0 3 00 2 . . . .kg m s kg m s(( ) = × 2 3 1 55 10. J Kf = ( )( ) = × 1 2 185 2 88 7 67 10 2 2 kg m s J. . Thus, the kinetic energy lost is 786 J into internal energy . P9.27 By conservation of momentum for the system of the two billiard balls (with all masses equal), in the x and y directions separately, 5 00 0 4 33 30 0 1 25 2 2 . . cos . . m s m s m + = ( ) + = ° v v fx fx ss m s m s 0 4 33 30 0 2 16 2 2 2 = ( ) + = − . sin . . ° v v fy fy fv == −2 50 60 0. .m s at ° Note that we did not need to explicitly use the fact that the collision is perfectly elastic. FIG. P9.27 13794_09_ch09_p209-244 copy.indd220 22013794_09_ch09_p209-244 copy.indd220 220 1/8/07 5:47:58 PM1/8/07 5:47:58 PM
13. 13. Linear Momentum and Collisions 221 P9.28 We use conservation of momentum for the system of two vehicles for both northward and eastward components, to ﬁnd the original speed of car number 2. For the eastward direction: M MVf13 0 2 55 0. cos .m s( )= ° For the northward direction: M MVi fv2 2 55 0= °sin . Divide the northward equation by the eastward equation to ﬁnd: v2 13 0 55 0 18 6 41 5i = ( ) = =. tan . . .m s m s mi h° Thus, the driver of the north bound car was untruthful. His original speed was more than 35 miրh. P9.29 p pxf xi= m m m v vO Y m s cos . cos . . 37 0 53 0 5 00 ° °+ = ( ) 0 799 0 602 5 00. . .v vO Y m s+ = (1) p p m m yf yi= − =v vO Ysin . sin .37 0 53 0 0° ° 0 602 0 799. .v vO Y= (2) Solving (1) and (2) simultaneously, vO m s= 3 99. and vY m s= 3 01. P9.30 p pxf xi= : m m m iv v vO Ycos cos .θ θ+ −( ) =90 0° v v vO Ycos sinθ θ+ = i (1) p pyf yi= : m mv vO Ysin sin .θ θ− −( ) =90 0 0° v vO Ysin cosθ θ= (2) From equation (2), v vO Y= ⎛ ⎝ ⎞ ⎠ cos sin θ θ (3) Substituting into equation (1), v v vY Y cos sin sin 2 θ θ θ ⎛ ⎝⎜ ⎞ ⎠⎟ + = i so v vY cos sin sin2 2 θ θ θ+( )= i , and v vY = i sinθ Then, from equation (3), v vO i= cosθ . We did not need to write down an equation expressing conservation of mechanical energy. In the prob- lem situation, the requirement of perpendicular ﬁnal velocities is equivalent to the condition of elasticity. FIG. P9.29 FIG. P9.30 FIG. P9.28 13794_09_ch09_p209-244 copy.indd221 22113794_09_ch09_p209-244 copy.indd221 221 1/8/07 5:47:59 PM1/8/07 5:47:59 PM
14. 14. 222 Chapter 9 P9.31 m m m mi i f1 1 2 2 1 2v v v+ = +( ) : 3 00 5 00 6 00 5 00. . ˆ . ˆ .( ) − =i j v v i j= −( )3 00 1 20. ˆ . ˆ m s P9.32 x-component of momentum for the system of the two objects: p p p pix ix fx fx1 2 1 2+ = + : − + = +m m mi i xv v v3 0 3 2 y-component of momentum of the system: 0 0 31 2+ = − +m my yv v by conservation of energy of the system: + + = + +( )1 2 1 2 3 1 2 1 2 32 2 1 2 2 2 2 2 m m m mi i y x yv v v v v we have v v 2 2 3 x i = also v v1 23y y= So the energy equation becomes 4 9 4 3 32 2 2 2 2 2 v v v vi y i y= + + 8 3 12 2 2 2v vi y= or v v 2 2 3 y i = (a) The object of mass m has ﬁnal speed v v v1 23 2y y i = = and the object of mass 3 m moves at v v v v 2 2 2 2 2 2 4 9 2 9 x y i i + = + v v v2 2 2 2 2 3 x y i+ = (b) θ = ⎛ ⎝⎜ ⎞ ⎠⎟ − tan 1 2 2 v v y x θ = ⎛ ⎝⎜ ⎞ ⎠⎟ = °− tan .1 2 3 3 2 35 3 v v i i P9.33 m0 27 17 0 10= × − . kg vi = 0 (the parent nucleus) m1 27 5 00 10= × − . kg v j1 6 6 00 10= ×. ˆ m s m2 27 8 40 10= × − . kg v i2 6 4 00 10= ×. ˆ m s (a) m m m1 1 2 2 3 3 0v v v+ + = where m m m m3 0 1 2 27 3 60 10= − − = × − . kg 5 00 10 6 00 10 8 40 10 4 00 127 6 27 . . ˆ . .×( ) ×( )+ ×( ) ×− − j 00 3 60 10 0 9 33 10 8 6 27 3 3 6 ˆ . . ˆ . i v v i ( )+ ×( ) = = − × − − 333 106 ×( )ˆj m s FIG. P9.33 continued on next page 13794_09_ch09_p209-244 copy.indd222 22213794_09_ch09_p209-244 copy.indd222 222 1/8/07 5:48:01 PM1/8/07 5:48:01 PM
15. 15. Linear Momentum and Collisions 223 (b) E m m m= + + 1 2 1 2 1 2 1 1 2 2 2 2 3 3 2 v v v E = ×( ) ×( ) + ×( )− −1 2 5 00 10 6 00 10 8 40 10 4 027 6 2 27 . . . . 00 10 3 60 10 12 5 10 4 39 6 2 27 6 2 ×( )⎡ ⎣ + ×( ) ×( ) ⎤ ⎦ = − . . .E ×× − 10 13 J P9.34 The initial momentum of the system is 0. Thus, 1 20 10 0. .m mBi( ) = ( )v m s and vBi = 8 33. m s K m m mi = ( ) + ( )( ) = 1 2 10 0 1 2 1 20 8 33 1 2 183 2 2 . . .m s m s m s2 2 ( ) = ( ) + ( )( ) =K m m mf G B 1 2 1 2 1 20 1 2 1 2 183 2 2 v v. m s2 2 ( )⎛ ⎝ ⎞ ⎠ or v vG B 2 2 1 20 91 7+ =. . m s2 2 (1) From conservation of momentum, m mG Bv v= ( )1 20. or v vG B= 1 20. (2) Solving (1) and (2) simultaneously, we ﬁnd ( . ) . . (1 20 1 20 91 7 912 2 v v vB B B+ = =m s2 2 .. / . ) / 7 2 64 1 2 m s2 2 vB = 5 89. m s (speed of blue puck after collision) and vG = 7 07. m s (speed of green puck after collision) Section 9.5 The Center of Mass P9.35 The x-coordinate of the center of mass is x m x m i i i CM kg kg kg = = + + + + + ∑ ∑ 0 0 0 0 2 00 3 00 2 50. . . ++( ) = 4 00 0 . kg CMx and the y-coordinate of the center of mass is y m y m i i i CM kg m kg = = ( )( )+ ( )∑ ∑ 2 00 3 00 3 00 2 5. . . . 00 2 50 0 4 00 0 500 2 00 m kg kg m( )+ ( )( )+ ( ) −( ). . . . kg kg kg kg mCM + + + = 3 00 2 50 4 00 1 00 . . . .y 13794_09_ch09_p209-244 copy.indd223 22313794_09_ch09_p209-244 copy.indd223 223 1/8/07 5:48:02 PM1/8/07 5:48:02 PM
16. 16. 224 Chapter 9 P9.36 Let the x axis start at the Earth’s center and point toward the Moon. x m x m x m m CM kg 0 = + + = × + ×1 1 2 2 1 2 24 22 5 98 10 7 36 10. . kg m kg 3 84 10 6 05 10 8 24 . . ×( ) × = 4.67 × 106 m from the Earth’s center The center of mass is within the Earth, which has radius 6 37 106 . × m. It is 1.7 Mm below the point on the Earth’s surface where the Moon is straight overhead. P9.37 Let A1 represent the area of the bottom row of squares, A2 the middle square, and A3 the top pair. A A A A M M M M M A M A = + + = + + = 1 2 3 1 2 3 1 1 A1 300= cm2 , A2 100= cm2 , A3 200= cm2 , A = 600 cm2 M M A A M M M M A A 1 1 2 2 300 600 2 1 = ⎛ ⎝ ⎞ ⎠ = = = ⎛ ⎝ ⎞ ⎠ = cm cm 2 2 000 600 6 200 600 3 3 cm cm cm cm 2 2 2 M M M M A A = = ⎛ ⎝ ⎞ ⎠ = 22 M M = 3 x x M x M x M M M M CM cm cm = + + = ( )+1 1 2 2 3 3 1 2 1 615 0 5 00. . (( )+ ( ) = = ( 10 0 11 7 5 00 1 3 1 2 . . . cm cm cm CM CM M M x y M ))+ ( )+ ( )( ) = = 1 6 1 315 0 25 0 13 3 M M M y . . . cm cm cm CM 113 3. cm P9.38 (a) Represent the height of a particle of mass dm within the object as y. Its contribution to the gravitational energy of the object-Earth system is dm gy( ) . The total gravitational energy is U gydm g ydmg = =∫ ∫all mass . For the center of mass we have y M ydmCM = ∫ 1 , so U gMyg = CM (b) The volume of the ramp is 1 2 3 6 15 7 64 8 1 83 103 . . . .m m m m3 ( )( )( ) = × . Its mass is ρV = ( ) ×( )= ×3800 1 83 10 6 96 103 6 kg m m kg3 3 . . . As shown in the chapter, its center of mass is above its base by one-third of its height, yCM m m= = 1 3 15 7 5 23. . . Then U Mgyg = = × ( ) = ×CM 2 kg m s m6 96 10 9 8 5 23 3 57 106 . . . . 88 J FIG. P9.37 13794_09_ch09_p209-244 copy.indd224 22413794_09_ch09_p209-244 copy.indd224 224 1/8/07 5:48:02 PM1/8/07 5:48:02 PM
17. 17. Linear Momentum and Collisions 225 P9.39 This object can be made by wrapping tape around a light stiff uniform rod. (a) M dx x dx= = +⎡⎣ ⎤⎦∫ λ 0 0 300 0 0 30 50 0 20 0 . . . . m 2 g m g m 00 m ∫ M x x= +⎡⎣ ⎤⎦ =50 0 10 0 15 92 0 0 300 . . . . g m g m g2 m (b) x xdm M M xdx M xCM all mass m g= = = ∫ ∫ 1 1 50 0 0 0 300 λ . . mm g m2 m +⎡⎣ ⎤⎦∫ 20 0 2 0 0 300 . . x dx x x x CM 2 g g m g m = + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 15 9 25 0 20 3 2 3 0 0 300 . . . mm m= 0 153. P9.40 Take the origin at the center of curvature. We have L r= 1 4 2π , r L = 2 π . An incremental bit of the rod at angle θ from the x axis has mass given by dm rd M Lθ = , dm Mr L d= θ where we have used the deﬁnition of radian measure. Now y M ydm M r Mr L d r CM all mass = = =∫ ∫= ° ° 1 1 45 135 sinθ θ θ 22 45 135 2 45 135 2 1 L d L L sin cos θ θ π θ ° ° ° ° ∫ = ⎛ ⎝ ⎞ ⎠ −( ) = 44 1 2 1 2 4 2 2 2 L L π π + ⎛ ⎝⎜ ⎞ ⎠⎟ = The top of the bar is above the origin by r L = 2 π , so the center of mass is below the middle of the bar by 2 4 2 2 1 2 2 0 063 52 L L L L π π π π − = − ⎛ ⎝⎜ ⎞ ⎠⎟ = . . Section 9.6 Motion of a System of Particles P9.41 (a) v v v v i CM kg m = = + = ( ) ∑m M m m M i i 1 1 2 2 2 00 2 00. . ˆ ss m s kg m s m s−( )+ ( ) +3 00 3 00 1 00 6 00. ˆ . . ˆ . ˆj i j(( ) 5 00. kg v i jCM m s= +( )1 40 2 40. ˆ . ˆ (b) p v i j i= = ( ) +( ) =M CM kg m s5 00 1 40 2 40 7 00. . ˆ . ˆ . ˆ ++( ) ⋅12 0. ˆj kg m s FIG. P9.40 x y θ 13794_09_ch09_p209-244 copy.indd225 22513794_09_ch09_p209-244 copy.indd225 225 1/8/07 5:48:03 PM1/8/07 5:48:03 PM
18. 18. 226 Chapter 9 *P9.42 r r r i j j CM m m m m t t = + + = + +⎡ ⎣1 1 2 2 1 2 2 3 5 3 3 2. ( ˆ ˆ) ˆ ⎤⎤ ⎦ + − −⎡ ⎣ ⎤ ⎦ + 5 5 3 2 6 3 5 5 5 1 8 2 . ˆ ˆ ˆ . . . i i jt t = 33 1 17 1 22 2 5 0 7782 2 + −( ) + − +( ). . ˆ . . ˆt t t ti j (a) At t = 2.5 s, r iCM = + ⋅ − ⋅( ) + − ⋅ +1 83 1 17 2 1 22 6 25 2 5 2 5 0. . . . ˆ . ..5 .. . ˆ778 6 25⋅( )j = (−2.89 ˆi − 1.39ˆj)cm We can conveniently do part (c) on the way to part (b): v r i jCM CMd dt t t= = −( ) + − +( )1 17 2 44 2 5 1 56. . ˆ . . ˆ att s,t CM= = − ⋅( ) + − +2 5 1 17 2 44 2 5 2 5 1 5. . . . ˆ . .v i 66 2 5⋅( ). ˆj = (−4.94ˆi + 1.39ˆj) cmրs (b) Now the total linear momentum is the total mass times the velocity of the center of mass: (9 g) (−4.94 ˆi + 1.39 ˆj) cmրs = (−44.5ˆi + 12.5ˆj) g⋅cmրs (d) Differentiating again, a v i jCM CMd dt = = −( ) +2 44 1 56. ˆ . ˆ The center of mass acceleration is (−2.44 ˆi + 1.56 ˆj) cmրs2 at t = 2.5 s and at all times. (e) The net force on the system is equal to the total mass times the acceleration of the center of mass: (9 g) (−2.44 ˆi + 1.56 ˆj) cmրs2 = (−220 ˆi + 140 ˆj) µN P9.43 Let x = distance from shore to center of boat ᐉ = length of boat ′ =x distance boat moves as Juliet moves toward Romeo The center of mass stays ﬁxed. Before: x M x M x M x M M M b J R B J R CM = + −( )+ +( )⎡⎣ ⎤⎦ + +( ) ᐉ ᐉ 2 2 After: x M x x M x x M x x M B J R B CM = − ′( )+ + − ′( )+ + − ′( )⎡⎣ ⎤⎦ + ᐉ ᐉ 2 2 MM MJ R+( ) ᐉ ᐉ − +⎛ ⎝ ⎞ ⎠ = ′ − − −( )+ 55 0 2 77 0 2 80 0 55 0 77 0 2 55 . . . . . .x 00 77 0 55 0 212 55 0 2 70 212 0 700 +( ) ′ = = ( ) = . . . . .x ᐉ m P9.44 (a) Conservation of momentum for the two-ball system gives us: 0 200 1 50 0 300 0 400 0 200. . . . .kg m s kg m s( )+ −( )= kg kgv v1 20 300f f+ . Relative velocity equation: v v2 1 1 90f f− = . m s Then 0 300 0 120 0 200 0 300 1 901 1. . . . .− = + +( )v vf f v1 0 780f = − . m s v2 1 12f = . m s v i1 0 780f = − . ˆ m s v i2 1 12f = . ˆ m s FIG. P9.43 continued on next page 13794_09_ch09_p209-244 copy.indd226 22613794_09_ch09_p209-244 copy.indd226 226 1/8/07 5:48:04 PM1/8/07 5:48:04 PM
19. 19. Linear Momentum and Collisions 227 (b) Before, v i CM kg m s kg = ( )( ) + ( ) −0 200 1 50 0 300 0 400. . ˆ . . m s kg ( )ˆ . i 0 500 v iCM m s= ( )0 360. ˆ Afterwards, the center of mass must move at the same velocity, because the momentum of the system is conserved. Section 9.7 Deformable Systems *P9.45 (a) Yes. The only horizontal force on the vehicle is the frictional force exerted by the ﬂoor, so it gives the vehicle all of its ﬁnal momentum, (6 kg)(3 ˆi mրs) = 18.0 ˆi kg⋅mրs (b) No. The friction force exerted by the ﬂoor on each stationary bit of caterpillar tread acts over no distance, so it does zero work. (c) Yes, we could say that the ﬁnal momentum of the cart came from the ﬂoor or from the planet through the ﬂoor, because the ﬂoor imparts impulse. (d) No. The ﬂoor does no work. The ﬁnal kinetic energy came from the original gravitational energy of the elevated load , in amount (1ր2)(6 kg)(3 mրs)2 = 27.0 J. (e) Yes. The acceleration is caused by the static friction force exerted by the ﬂoor that pre- vents the caterpillar tracks from slipping backward. *P9.46 (a) Yes. The ﬂoor exerts a force, larger than the person’s weight over time as he is taking off. (b) No. The work by the ﬂoor on the person is zero because the force exerted by the ﬂoor acts over zero distance. (c) He leaves the ﬂoor with a speed given by (1ր2)mv2 = mgyf v = [2(9.8 mրs2 )0.15 m]1ր2 = 1.71 mրs, so his momentum immediately after he leaves the ﬂoor is (60 kg)(1.71 mրs up) = 103 kg⋅mրs up (d) Yes. You could say that it came from the planet, that gained momentum 103 kg⋅mրs down, but it came through the force exerted by the ﬂoor over a time interval on the per- son, so it came through the ﬂoor or from the ﬂoor through direct contact. (e) (1ր2)(60 kg)(1.71 mրs)2 = 88.2 J (f) No. The energy came from chemical energy in the person’s leg muscles. The ﬂoor did no work on the person. 13794_09_ch09_p209-244 copy.indd227 22713794_09_ch09_p209-244 copy.indd227 227 1/8/07 5:48:05 PM1/8/07 5:48:05 PM
20. 20. 228 Chapter 9 *P9.47 (a) When the cart hits the bumper it immediately stops, and the hanging particle keeps moving with its original speed vi . The particle swings up as a pendulum on a ﬁxed pivot, keeping constant energy. Measure elevations from the pivot: (1ր2)mvi 2 + mg(−L) = 0 + mg(−L cos θ) Then vi = [2gL(1 − cosθ)]1ր2 (b) vi = [2gL(1 − cosθ)]1ր2 = [2(9.8 mրs2 )(1.2 m)(1 − cos 35°)]1ր2 = 2.06 mրs (c) Yes. The bumper must provide the horizontal force to the left to slow down the swing of the particle to the right, to reverse its rightward motion, and to make it speed up to the left. When the particle passes its straight-down position moving to the left, the bumper stops exerting force. It is at this moment that the cart-particle system momentarily has zero horizontal accelera- tion for its center of mass. *P9.48 Depending on the length of the cord and the time interval ∆t for which the force is applied, the sphere may have moved very little when the force is removed, or we may have x1 and x2 nearly equal, or the sphere may have swung back, or it may have swung back and forth several times. Our solution applies equally to all of these cases. (a) The applied force is constant, so the center of mass of the glider-sphere system moves with constant acceleration. It starts, we deﬁne, from x = 0 and moves to (x1 + x2 )ր2. Let v1 and v2 represent the horizontal components of velocity of glider and sphere at the moment the force stops. Then the velocity of the center of mass is vCM = (v1 + v2 )ր2 and because the acceleration is constant we have (x1 + x2 )ր2 = [(v1 + v2 )ր2]∆tր2 ∆t = 2(x1 + x2 )ր(v1 + v2 ) The impulse-momentum theorem for the glider-sphere system is F∆t = mv1 + mv2 F2(x1 + x2 )ր(v1 + v2 ) = m(v1 + v2 ) F2(x1 + x2 )րm = (v1 + v2 )2 F2(x1 + x2 )ր4m = (v1 + v2 )2 ր4 = vCM 2 Then vCM = [F(x1 + x2 )ր2m]1ր2 (b) The applied force does work that becomes, after the force is removed, kinetic energy of the constant-velocity center-of-mass motion plus kinetic energy of the vibration of the glider and sphere relative to their center of mass. The applied force acts only on the glider, so the work-energy theorem for the pushing process is Fx1 = (1ր2)(2 m) vCM 2 + Evib Substitution gives Fx1 = (1ր2)(2 m)F(x1 + x2 )ր2m + Evib = Fx1 ր2 + Fx2 ր2 + Evib Then Evib = Fx1 ր2 − Fx2 ր2 When the cord makes its largest angle with the vertical, the vibrational motion is turning around. No kinetic energy is associated with the vibration at this moment, but only gravita- tional energy: mgL(1 − cos θ) = F(x1 − x2 )ր2 Solving gives θ = cos−1 [1 − F(x1 − x2 )ր2mgL] P9.49 A picture one second later differs by showing ﬁve extra kilograms of sand moving on the belt. (a) ∆ ∆ p t x = ( )( ) = 5 00 0 750 1 00 3 75 . . . . kg m s s N (b) The only horizontal force on the sand is belt friction, so from p f t pxi xf+ =∆ this is f p t x = = ∆ ∆ 3 75. N (c) The belt is in equilibrium: F max x∑ = : + − =F fext 0 and Fext N= 3 75. continued on next page 13794_09_ch09_p209-244 copy.indd228 22813794_09_ch09_p209-244 copy.indd228 228 1/8/07 5:48:06 PM1/8/07 5:48:06 PM
21. 21. Linear Momentum and Collisions 229 (d) W F r= = ( ) =∆ cos . cos .θ 3 75 0 2 81N 0.750 m J° (e) 1 2 1 2 5 00 0 750 1 412 2 ∆m( ) = ( ) =v . . .kg m s J (f) One-half of the work input becomes kinetic energy of the moving sand and the other half becomes additional internal energy. The internal energy appears when the sand does not elastically bounce under the hopper, but has friction eliminate its horizontal motion relative to the belt. By contrast, all of the impulse input becomes momentum of the moving sand. Section 9.8 Rocket Propulsion P9.50 (a) The fuel burns at a rate dM dt = = × −12 7 6 68 10 3. . g 1.90 s kg s Thrust = ve dM dt : 5 26 6 68 10 3 . .N kg s= ×( )− ve ve = 787 m s (b) v v vf i e i f M M − = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ln : vf − = ( ) + + − 0 787 53 5 25 5 25 5 1 m s g g 53.5 g g ln . . . 22 7. g ⎛ ⎝⎜ ⎞ ⎠⎟ vf = 138 m s P9.51 (a) Thrust = ve dM dt Thrust = ×( ) ×( )= ×2 60 10 1 50 10 3 90 103 4 7 . . .m s kg s N (b) F Mg May∑ = − =Thrust : 3 90 10 3 00 10 9 80 3 00 107 6 6 . . . .× − ×( )( ) = ×( )a a = 3 20. m s2 P9.52 (a) From the equation for rocket propulsion in the text, v v v− = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − ⎛ ⎝⎜ ⎞ ⎠⎟0 e i f e f i M M M M ln ln Now, M M ktf i= − , so v v v= − −⎛ ⎝⎜ ⎞ ⎠⎟ = − − ⎛ ⎝⎜ ⎞ ⎠⎟e i i e i M kt M k M tln ln 1 With the deﬁnition T M k p i ≡ , this becomes v vt t T e p ( ) = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ln 1 (b) With ve = 1500 m s, andTp = 144 s, v = −( ) −⎛ ⎝ ⎞ ⎠ 1500 1 144 m s s ln t t s( ) ( )v m s 0 0 20 224 40 488 60 808 80 1220 100 1780 120 26690 132 3730 FIG. P9.52(b) v (m/s) 2500 0 20 40 60 80 100 120 140 2000 1500 1000 500 0 3000 3500 4000 t (s) continued on next page 13794_09_ch09_p209-244 copy.indd229 22913794_09_ch09_p209-244 copy.indd229 229 1/8/07 5:48:07 PM1/8/07 5:48:07 PM
22. 22. 230 Chapter 9 (c) a t d dt d dt e t T e t T p p ( ) = = − −( )⎡ ⎣ ⎤ ⎦ = − − ⎛ ⎝ ⎜ ⎞v v v ln 1 1 1 ⎠⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 1 1T Tp e p t Tp v , or a t T t e p ( ) = − v (d) With ve = 1500 m s, and Tp = 144 s, a t = − 1500 144 m s s t s a( ) ( )m s 0 10.4 20 12.1 40 14.4 60 17.9 80 23.4 100 3 2 44.1 120 62.5 132 125 (e) x t dt t T dt t e p t ( ) = + = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ∫ ∫0 1 0 0 v v ln == − ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟∫ve p p p t T t T dt T ln 1 0 x t T t T t T t T e p p p p ( ) = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − ⎛ ⎝ ⎜ ⎞ v 1 1 1ln ⎠⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ( ) = −( ) − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + 0 1 t e p p ex t T t t T tv vln (f) With ve = =1500 1 50m s km s. , and Tp = 144 s, x t t t= −( ) −⎛ ⎝ ⎞ ⎠ +1 50 144 1 144 1 50. ln . t xs km 0 20 40 60 80 100 1 ( ) ( ) 0 2 19 9 23 22 1 42 2 71 7 . . . . . 220 132 115 153 FIG. P9.52(f) x (km) 100 0 20 40 60 80 100 120 140 80 60 40 20 0 120 140 160 t (s) a (m/s2 ) 100 0 20 40 60 80 100 120 140 80 60 40 20 0 120 140 t (s) FIG. P9.52(d) 13794_09_ch09_p209-244 copy.indd230 23013794_09_ch09_p209-244 copy.indd230 230 1/8/07 5:48:08 PM1/8/07 5:48:08 PM
23. 23. Linear Momentum and Collisions 231 P9.53 In v v= e i f M M ln we solve for Mi . (a) M e Mi f e = v v M ei = ×( )= ×5 3 5 3 00 10 4 45 10. .kg kg The mass of fuel and oxidizer is ∆M M Mi f= − = −( )× =445 3 00 10 4423 . kg metric tons (b) ∆M e= ( )− =2 3 00 3 00 19. .metric tons metric tons ..2 metric tons This is much less than the suggested value of 442ր2.5. Mathematically, the logarithm in the rocket propulsion equation is not a linear function. Physically, a higher exhaust speed has an extra-large cumulative effect on the rocket body’s ﬁnal speed, by counting again and again in the speed the body attains second after second during its burn. Because of the exponential, a relatively small increase in engine efﬁciency causes a large change in the amount of fuel and oxidizer required. Additional Problems P9.54 (a) When the spring is fully compressed, each cart moves with same velocity v. Apply conser- vation of momentum for the system of two gliders p pi f= : m m m m1 1 2 2 1 2v v v+ = +( ) v v v = + + m m m m 1 1 2 2 1 2 (b) Only conservative forces act; therefore ∆E = 0. 1 2 1 2 1 2 1 2 1 1 2 2 2 2 1 2 2 2 m m m m kxmv v v+ = +( ) + Substitute for v from (a) and solve for xm. x m m m m m m m m m 2 1 2 1 1 2 1 2 2 2 2 1 1 2 2 2 = +( ) + +( ) − ( ) − (v v v v )) − +( ) = + −( ) 2 1 2 1 2 1 2 1 2 1 2 2 2 1 2 2 2 m m k m m x m m m v v v v v v kk m m m m k m m1 2 1 2 1 2 1 2+( ) = −( ) +( ) v v continued on next page 13794_09_ch09_p209-244 copy.indd231 23113794_09_ch09_p209-244 copy.indd231 231 1/8/07 5:48:09 PM1/8/07 5:48:09 PM
24. 24. 232 Chapter 9 (c) m m m mf f1 1 2 2 1 1 2 2v v v v+ = + Conservation of momentum: m mf f1 1 1 2 2 2v v v v−( )= −( ) (1) Conservation of energy: 1 2 1 2 1 2 1 2 1 1 2 2 2 2 1 1 2 2 2 2 m m m mf fv v v v+ = + which simpliﬁes to: m mf f1 1 2 1 2 2 2 2 2 2 v v v v−( )= −( ) Factoring gives m mf f f f1 1 1 1 1 2 2 2 2v v v v v v v−( )⋅ +( )= −( )⋅ + vv2( ) and with the use of the momentum equation (equation (1)), this reduces to v v v v1 1 2 2+( )= +( )f f or v v v v1 2 2 1f f= + − (2) Substituting equation (2) into equation (1) and simplifying yields: v v v2 1 1 2 1 2 1 1 2 2 2 f m m m m m m m = + ⎛ ⎝⎜ ⎞ ⎠⎟ + − + ⎛ ⎝⎜ ⎞ ⎠⎟ Upon substitution of this expression for v2 f into equation 2, one ﬁnds v v v1 1 2 1 2 1 2 1 2 2 2 f m m m m m m m = − + ⎛ ⎝⎜ ⎞ ⎠⎟ + + ⎛ ⎝⎜ ⎞ ⎠⎟ Observe that these results are the same as two equations given in the chapter text for the situation of a perfectly elastic collision in one dimension. Whatever the details of how the spring behaves, this collision ends up being just such a perfectly elastic collision in one dimension. P9.55 We hope the momentum of the wrench provides enough recoil so that the astronaut can reach the ship before he loses life support! We might expect the elapsed time to be on the order of several minutes based on the description of the situation. No external force acts on the system (astronaut plus wrench), so the total momentum is constant. Since the ﬁnal momentum (wrench plus astronaut) must be zero, we have ﬁnal momentum = initial momentum = 0. m mwrench wrench astronaut astronautv v+ = 0 Thus v v astronaut wrench wrench astronaut = − = − m m 0 500. kg m s kg m s ( )( ) = − 20 0 80 0 0 125 . . . At this speed, the time to travel to the ship is t = = = 30 0 240 4 00 . . m 0.125 m s s minutes The astronaut is fortunate that the wrench gave him sufﬁcient momentum to return to the ship in a reasonable amount of time! In this problem, we did not think of the astronaut as drifting away from the ship when he threw the wrench. However slowly, he must be drifting away since he did not encounter an external force that would reduce his velocity away from the ship. There is no air friction beyond earth’s atmosphere. In a real-life situation, the astronaut would have to throw the wrench hard enough to overcome his momentum caused by his original push away from the ship. 13794_09_ch09_p209-244 copy.indd232 23213794_09_ch09_p209-244 copy.indd232 232 1/8/07 5:48:10 PM1/8/07 5:48:10 PM
25. 25. Linear Momentum and Collisions 233 *P9.56 Proceeding step by step, we ﬁnd the real actor’s speed just before collision, using energy conser- vation in the swing-down process: ma gyi = (1ր2) ma vi 2 [2(9.8 mրs2 )(1.8 m)]1ր2 = vi = 5.94 mրs Now for the elastic collision with a stationary target we use the specialized equation from the chapter text v2 f = (2 m1 v1i )ր(m1 + m2 ) = 2(80 kg)(5.94 mրs)ր(80 kg + m) = (950 kg⋅mրs)ր(80 kg + m) The time for the clone’s fall into the ocean is given by ∆y = vyi t + (1ր2)ay t2 − 36 m = 0 + (1ր2)(−9.8 mրs2 )t2 t = 2.71 s so his horizontal range is R = v2 f t = (2.71 s )(950 kg⋅mրs)ր(80 kg + m) = 2.58 × 103 kg⋅mր(80 kg + m) (b) By substitution, 2576 kg⋅m (80 kg + 79 kg)−1 = 16.2 m (c) A little heavier and he does not go so far: 2576 kg⋅m (80 kg + 81 kg)−1 = 16.0 m (d) We solve 30 m = 2580 kg⋅m (80 kg + m)−1 80 kg + m = 85.87 kg m = 5.87 kg (e) The maximum value for R is 2576ր80 = 32.2 m , obtained in the limit as (f) we make m go to zero . (g) The minimum value of R is approaching zero , obtained in the limit as (h) we make m go to inﬁnity . (i) Yes, mechanical energy is conserved until the clone splashes down. This principle is not sufﬁcient to solve the problem. We need also conservation of momentum in the collision. (j) Yes, but it is not useful to include the planet in the analysis of momentum. We use instead momentum conservation for the actor-clone system while they are in contact. (k) In symbols we have vi = [2 g (1.8 m)]1ր2 v2f = 2(80 kg) [2 g (1.8 m)]1ր2 ր(80 kg + m) t = [2(36 m)րg]1ր2 and R = [2(36 m)րg]1ր2 2(80 kg) [2 g (1.8 m)]1ր2 ր(80 kg + m) Here g divides out. At a location with weaker gravity, the actor would be moving more slowly before the collision, but the clone would follow the same trajectory, moving more slowly over a longer time interval. P9.57 Using conservation of momentum from just before to just after the impact of the bullet with the block: m M mi fv v= +( ) or v vi f M m m = +⎛ ⎝ ⎞ ⎠ (1) The speed of the block and embedded bullet just after impact may be found using kinematic equations: d tf= v and h gt= 1 2 2 Thus, t h g = 2 and vf d t d g h gd h = = = 2 2 2 Substituting into (1) from above gives vi M m m gd h = +⎛ ⎝ ⎞ ⎠ 2 2 . FIG. P9.57 M vi m h d 13794_09_ch09_p209-244 copy.indd233 23313794_09_ch09_p209-244 copy.indd233 233 1/8/07 5:48:11 PM1/8/07 5:48:11 PM
26. 26. 234 Chapter 9 P9.58 (a) The initial momentum of the system is zero, which remains constant throughout the motion. Therefore, when m1 leaves the wedge, we must have m m2 1 0v vwedge block+ = or 3 00 0 500 4 00 0. . .kg kg m swedge( ) + ( ) +( )=v so vwedge m s= −0 667. (b) Using conservation of energy for the block- wedge-Earth system as the block slides down the smooth (frictionless) wedge, we have K U K K U i iblock system wedge block sys+⎡⎣ ⎤⎦ + ⎡⎣ ⎤⎦ = + ttem wedge ⎡⎣ ⎤⎦ + ⎡⎣ ⎤⎦f f K or 0 0 1 2 4 00 0 1 2 0 6671 1 2 2+[ ]+ = ( ) +⎡ ⎣⎢ ⎤ ⎦⎥ + −( )m gh m m. . 22 which gives h = 0 952. m P9.59 (a) Conservation of momentum: 0 5 2 3 1 1 5 1 2 3. ˆ ˆ ˆ . ˆ ˆ ˆkg m s kgi j k i j k− +( ) + − + −( ) m s kg m s kg= − + −( ) +0 5 1 3 8 1 5 2 2 . ˆ ˆ ˆ .i j k v v f ff = − + −( ) ⋅ + − +0 5 1 5 4 0 5 1 5 4. ˆ . ˆ ˆ . ˆ . ˆ ˆi j k i j kkg m s (( ) ⋅ = kg m s kg1 5 0 . The original kinetic energy is 1 2 0 5 2 3 1 1 2 1 5 1 2 32 2 2 2 2 2 . .kg m s kg2 2 + +( ) + + +( ) mm s J2 2 = 14 0. The ﬁnal kinetic energy is 1 2 0 5 1 3 8 0 18 52 2 2 . .kg m s J2 2 + +( ) + = different from the original energy so the collision is inelastic . (b) We follow the same steps as in part (a): − + −( ) ⋅ = − +0 5 1 5 4 0 5 0 25 0 75. ˆ . ˆ ˆ . . ˆ .i j k ikg m s kg ˆˆ ˆ . . ˆ . ˆ j k v v i j −( ) + = − + − 2 1 5 0 5 1 5 4 2 2 m s kg f f ˆˆ . ˆ . ˆ ˆ . k i j k( ) ⋅ + − +( ) ⋅kg m s kg m s0 125 0 375 1 1 5 kkg m s= − + −( )0 250 0 750 2 00. ˆ . ˆ . ˆi j k We see v v2 1f f= , so the collision is perfectly inelastic . continued on next page FIG. P9.58 v = 4.00 m/sblock vwedge +x 13794_09_ch09_p209-244 copy.indd234 23413794_09_ch09_p209-244 copy.indd234 234 1/8/07 5:48:12 PM1/8/07 5:48:12 PM
27. 27. Linear Momentum and Collisions 235 (c) Conservation of momentum: − + −( ) ⋅ = − + +0 5 1 5 4 0 5 1 3. ˆ . ˆ ˆ . ˆ ˆ ˆi j k i j kkg m s kg a(( ) + = − + −( ) m s kg k 1 5 0 5 1 5 4 2 2 . . ˆ . ˆ ˆ v v i j k f f gg m s kg m s kg ⋅ + − −( ) ⋅ = − 0 5 1 5 0 5 1 5 2 6 . ˆ . ˆ . ˆ . . i j ka 77 0 333−( ). ˆa k m s Conservation of energy: 14 0 1 2 0 5 1 3 1 2 1 5 2 672 2 2 . . . .J kg m s kg2 2 = + +( ) +a ++( ) = + + + + 0 333 2 5 0 25 5 33 1 33 2 2 . . . . . a a a m s J J 2 2 00 083 3 2 . a 0 0 333 1 33 6 167 1 33 1 33 4 0 333 2 2 = + − = − ± − ( ) . . . . . . a a a −−( ) = − 6 167 0 667 2 74 6 74 . . . .a or . Either value is possible. with a = 2 74. , v k k2 2 67 0 333 2 74 3 58f = − − ( )( ) = −. . . ˆ . ˆm s m s with a = −6 74. , v k k2 2 67 0 333 6 74 0 419f = − − −( )( ) = −. . . ˆ . ˆm s m s P9.60 Consider the motion of the ﬁreﬁghter during the three intervals: (1) before, (2) during, and (3) after collision with the platform. (a) While falling a height of 4.00 m, her speed changes from vi = 0 to v1 as found from ∆E K U K Uf f i i= +( )− −( ), or K E U K Uf f i i= − + +∆ When the initial position of the platform is taken as the zero level of gravitational potential, we have 1 2 180 0 01 2 m fh mghv = ( )− + +cos ° Solving for v1 gives v1 2 2 300 4 00 75 0 9 80 4 00 = − +( ) = − ( )+ ( )( )fh mgh m . . . . 775 0 6 81 . .= m s FIG. P9.60 v1 v2 continued on next page 13794_09_ch09_p209-244 copy.indd235 23513794_09_ch09_p209-244 copy.indd235 235 1/8/07 5:48:13 PM1/8/07 5:48:13 PM
28. 28. 236 Chapter 9 (b) During the inelastic collision, momentum is conserved; and if v2 is the speed of the ﬁre- ﬁghter and platform just after collision, we have m m Mv v1 2= +( ) or v v 2 1 1 75 0 6 81 75 0 20 0 5 38= + = ( ) + = m m M . . . . . m s Following the collision and again solving for the work done by non-conservative forces, using the distances as labeled in the ﬁgure, we have (with the zero level of gravitational potential at the initial position of the platform): ∆E K U U K U Uf fg fs i ig is= + + − − − , or − = + +( ) −( )+ − +( ) − −fs m M g s ks m M0 1 2 1 2 0 02 2 v This results in a quadratic equation in s: 2 000 931 300 1375 02 s s s− ( ) + − = or s = 1 00. m P9.61 (a) Each primate swings down according to mgR m= 1 2 1 2 v MgR M= 1 2 1 2 v v1 2= gR The collision: − + = + +( )m M m Mv v v1 1 2 v v2 1= − + M m M m Swinging up: 1 2 1 352 2 M m M m gR+( ) = +( ) −( )v cos ° v2 2 1 35= −( )gR cos ° 2 1 35 2 0 425 0 425 gR M m M m gR M m M −( ) +( ) = −( ) + = − cos . . ° mm m M m M 1 425 0 575 0 403 . . . = = (b) No change is required if the force is different. The nature of the forces within the system of colliding objects does not affect the total momentum of the system. With strong magnetic attraction, the heavier object will be moving somewhat faster and the lighter object faster still. Their extra kinetic energy will all be immediately converted into extra internal energy when the objects latch together. Momentum conservation guarantees that none of the extra kinetic energy remains after the objects join to make them swing higher. 13794_09_ch09_p209-244 copy.indd236 23613794_09_ch09_p209-244 copy.indd236 236 1/8/07 5:48:14 PM1/8/07 5:48:14 PM
29. 29. Linear Momentum and Collisions 237 P9.62 (a) Utilizing conservation of momentum, m m m m m m gh A B A A 1 1 1 2 1 1 2 1 1 2 6 29 v v v v = +( ) = + ≅ . m s (b) Utilizing the two equations 1 2 2 gt y= and x tA= v1 we combine them to ﬁnd v1 2 2 A x y g x g y = = / From the data, v1 6 16A = . m s Most of the 2% difference between the values for speed is accounted for by the uncertainty in the data, estimated as 0 01 8 68 0 1 68 8 1 263 1 257 0 1 85 3 1 1 . . . . . . . %+ + + + = . *P9.63 (a) In the same symbols as in the text’s Example, the original kinetic energy is KA = (1ր2)m1 v1A 2 . The example shows that the kinetic energy immediately after latching together is KB = (1ր2) m1 2 v1A 2 ր(m1 + m2 ) so the fraction of kinetic energy remaining as kinetic energy is KB րKA = m1 ր(m1 + m2 ) (b) KB րKA = 9.6 kgր(9.6 kg + 214 kg) = 0.0429 (c) Momentum is conserved in the collision so momentum after divided by momentum before is 1.00 . (d) Energy is an entirely different thing from momentum. A comparison: When a photographer’s single-use ﬂashbulb ﬂashes, a magnesium ﬁlament oxidizes. Chemical energy disappears. (Internal energy appears and light carries some energy away.) The measured mass of the ﬂashbulb is the same before and after. It can be the same in spite of the 100% energy conver- sion, because energy and mass are totally different things in classical physics. In the ballistic pendulum, conversion of energy from mechanical into internal does not upset conservation of mass or conservation of momentum. FIG. P9.62 x y v1i 13794_09_ch09_p209-244 copy.indd237 23713794_09_ch09_p209-244 copy.indd237 237 1/8/07 5:48:15 PM1/8/07 5:48:15 PM
30. 30. 238 Chapter 9 *P9.64 (a) The mass of the sleigh plus you is 270 kg. Your velocity is 7.50 m/s in the x direction. You unbolt a 15.0-kg seat and throw it back at the ravening wolves, giving it a speed of 8.00 m/s relative to you. Find the velocity of the seat relative to the ground after your action, and the velocity of the sleigh. (b) We substitute v v1 28f f= −m s 270 15 8 2552 2kg 7.5 m s kg m s kg( )= − +( )+ ( )v vf f 22 025 120 270 2145 2 2 kg m s kg m s kg⋅ = − ⋅ + ( ) = v v f f mm s m s m s m s m s 270 7 94 8 7 94 0 055 61 = = − = . . .v f The ﬁnal velocity of the seat is −0 055 6. ˆm s i. That of the sleigh is 7 94. ˆm s i. (c) You do work on both the sleigh and the seat, to change their kinetic energy according to K W K K W i f f+ = + ( )( ) + = ( ) 1 2 21 2 270 7 5 1 2 15kg m s kg. 00 055 6 1 2 255 7 94 7 594 2 2 . .m s kg m s J ( ) + ( )( ) + =W 00 0231 8 047 453 . J J J + =W 13794_09_ch09_p209-244 copy.indd238 23813794_09_ch09_p209-244 copy.indd238 238 1/8/07 5:48:16 PM1/8/07 5:48:16 PM
31. 31. Linear Momentum and Collisions 239 *P9.65 The force exerted by the spring on each block is in magnitude F kxs = = ( )( ) =3 85 0 08 0 308. . .N m m N. (a) With no friction, the elastic energy in the spring becomes kinetic energy of the blocks, which have momenta of equal magnitude in opposite directions. The blocks move with constant speed after they leave the spring. K U K U kx m m i f f f +( ) = +( ) = + 1 2 1 2 1 2 1 2 3 85 2 1 1 2 2 2 2 v v . NN m m kg kg( )( ) = ( ) + ( )0 08 1 2 0 25 1 2 0 50 2 1 2 . . .v vf 22 2 1 1 2 2 1 1 2 2 0 0 25 f i i f fm m m mv v v v+ = + = ( ). kg v11 2 1 2 0 50 2 0 012 3 1 2 0 f f f f −( )+ ( ) = = ˆ . ˆ . i ikg J v v v .. . .25 2 1 2 0 50 1 2 1 52 2 2 2 kg kg kg( )( ) + ( ) = ( )v vf f vv v 2 2 2 1 2 2 0 123 0 128 f f = ⎛ ⎝⎜ ⎞ ⎠⎟ = . . J 0.75 kg m s v ff f f = = ( )= = 0 128 2 0 128 0 256 01 1 . ˆ . . m s m s m s i vv .. ˆ256 m s −( )i (b) For the lighter block, F may y∑ = , n − ( )=0 25 0. kg 9.8 m s2 , n = 2 45. N, f nk k= = ( ) =µ 0 1 2 45 0 245. . .N N. We assume that the maximum force of static friction is a similar size. Since 0.308 N is larger than 0.245 N, this block moves. For the heavier block, the normal force and the frictional force are twice as large: fk = 0 490. N. Since 0.308 N is less than this, the heavier block stands still. In this case, the frictional forces exerted by the ﬂoor change the momentum of the two-block system. The lighter block will gain speed as long as the spring force is larger than the friction force: that is until the spring compression becomes xf given by F kxs = , 0 245 3 85. .N N m= ( )xf , 0 063 6. m = xf . Now for the energy of the lighter block as it moves to this maximum-speed point we have K U f d K Ui i k f f+ − = + + − −0 0 012 3 0 245 0 08 0 063. . . .J N 66 1 2 0 25 1 2 3 85 0 0632 m kg N m 6 m( )= ( ) + ( )( ). . .vf 22 2 0 012 3 0 004 01 1 2 0 25 0 007 80. . . .J J kg− = ( ) +vf J J kg m 2 0 000 515 0 25 0 064 2 1 2 . . . ( )⎛ ⎝⎜ ⎞ ⎠⎟ = =vf ss Thus for the heavier block the maximum velocity is 0 and for the lighter 0 064 2. ˆm s −( )i . (c) For the lighter block, fk = ( ) =0 462 2 45 1 13. . .N N. The force of static friction must be at least as large. The 0.308-N spring force is too small to produce motion of either block. Each has 0 maximum speed. 13794_09_ch09_p209-244 copy.indd239 23913794_09_ch09_p209-244 copy.indd239 239 1/8/07 5:48:16 PM1/8/07 5:48:16 PM
32. 32. 240 Chapter 9 P9.66 The orbital speed of the Earth is vE 7 m 3.156 10 s = = × × = × 2 2 1 496 10 2 98 10 11 4π πr T . . mm s In six months the Earth reverses its direction, to undergo momentum change m mE E E E kg m s∆v = = ×( ) ×( )=2 2 5 98 10 2 98 10 324 4 v . . ..56 1029 × ⋅kg m s Relative to the center of mass, the sun always has momentum of the same magnitude in the opposite direction. Its 6-month momentum change is the same size, mS S kg m s∆v = × ⋅3 56 1029 . . Then ∆vS kg m s kg m s= × ⋅ × = 3 56 10 1 991 10 0 179 29 30 . . . P9.67 (a) Find the speed when the bullet emerges from the block by using momentum conservation: m MV mi iv v= + The block moves a distance of 5.00 cm. Assume for an approximation that the block quickly reaches its maximum velocity, Vi , and the bullet kept going with a constant velocity, v. The block then compresses the spring and stops. 1 2 1 2 900 5 00 10 1 00 2 2 2 2 MV kx V i i = = ( ) ×( )− N m m k . . gg m s kg m s = = − = ×( )( )− 1 50 5 00 10 4003 . . v vm MV m i i −− ( )( ) × = − 1 00 1 50 5 00 10 100 3 . . . kg m s kg m sv (b) ∆ ∆ ∆E K U= + = ×( )( ) − ×−1 2 5 00 10 100 1 2 5 00 13 2 . .kg m s 00 400 1 2 900 5 00 10 3 2 2 2 − − ( )( ) + ( ) ×( ) kg m s N m m. ∆E = −374 J, or there is a mechanical energy loss of 374 J . FIG. P9.66 S CM E FIG. P9.67 400 m/s 5.00 cm v 13794_09_ch09_p209-244 copy.indd240 24013794_09_ch09_p209-244 copy.indd240 240 1/8/07 5:48:17 PM1/8/07 5:48:17 PM
33. 33. Linear Momentum and Collisions 241 P9.68 (a) p F pi ft+ = : 3 00 7 00 12 0 5 00 3 0. . ˆ . ˆ . .kg m s N s( )( ) + ( )( ) =j i 00 kg( )v f v i jf = +( )20 0 7 00. ˆ . ˆ m s (b) a v v = −f i t : a i j j i= + −( ) = 20 0 7 00 7 00 5 00 4 00 . ˆ . ˆ . ˆ . . ˆ m s s m ss2 (c) a F = ∑ m : a i i= = 12 0 4 00 . ˆ . ˆN 3.00 kg m s2 (d) ∆r v a= +it t 1 2 2 : ∆r j i= ( )( )+ ( )7 00 5 00 1 2 4 00 5 00. ˆ . . ˆ .m s s m s2 ss( )2 ∆r i j= +( )50 0 35 0. ˆ . ˆ m (e) W = ⋅F r∆ : W = ( )⋅ +( )=12 0 50 0 35 0 600. ˆ . ˆ . ˆN m m Ji i j (f) 1 2 1 2 3 00 20 0 7 00 20 0 72 m fv = ( ) +( )⋅ +. . ˆ . ˆ . ˆ .kg i j i 000ˆj( ) m s2 2 1 2 1 50 449 6742 m fv = ( )( )=. kg m s J2 2 (g) 1 2 1 2 3 00 7 00 600 6742 2 m Wiv + = ( )( ) + =. .kg m s J J (h) The accelerations computed in different ways agree. The kinetic energies computed in different ways agree. The three theories are consistent. 13794_09_ch09_p209-244 copy.indd241 24113794_09_ch09_p209-244 copy.indd241 241 1/8/07 5:48:18 PM1/8/07 5:48:18 PM
34. 34. 242 Chapter 9 P9.69 The force exerted by the table is equal to the change in momentum of each of the links in the chain. By the calculus chain rule of derivatives, F dp dt d m dt dm dt m d dt 1 = = ( ) = + v v v We choose to account for the change in momentum of each link by having it pass from our area of interest just before it hits the table, so that v dm dt ≠ 0 and m d dt v = 0 Since the mass per unit length is uniform, we can express each link of length dx as having a mass dm: dm M L dx= The magnitude of the force on the falling chain is the force that will be necessary to stop each of the elements dm. F dm dt M L dx dt M L 1 2 = = ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ v v v After falling a distance x, the square of the velocity of each link v2 2= gx (from kinematics), hence F Mgx L 1 2 = The links already on the table have a total length x, and their weight is supported by a force F2: F Mgx L 2 = Hence, the total force on the chain is F F F Mgx L total = + =1 2 3 That is, the total force is three times the weight of the chain on the table at that instant. ANSWERS TO EVEN PROBLEMS P9.2 (a) She moves at 4.71 mրs east. (b) 717 J (c) System momentum is conserved with the value zero. The forces on the two siblings are of equal magnitude in opposite directions. Their impulses add to zero. Their ﬁnal momenta are of equal magnitude in opposite directions. P9.4 (a) 6.00 (−ˆi) mրs (b) 8.40 J (c) The original energy is in the spring. A force had to be exerted over a distance to compress the spring, transferring energy into it by work. The cord exerts force, but over no distance. (d) System momentum is conserved with the value zero. The forces on the two blocks are of equal magnitude in opposite directions. Their impulses add to zero. The ﬁnal momenta of the two blocks are of equal magnitude in opposite directions. P9.6 In trying to hang onto the child, he would have to exert a force of 6.44 kN toward the back of the car, to slow down the child’s forward motion. He is not strong enough to exert so large a force. If he were solidly belted in and tied to the child, the child would exert this size force on him toward the front of the car. FIG. P9.69 13794_09_ch09_p209-244 copy.indd242 24213794_09_ch09_p209-244 copy.indd242 242 1/8/07 5:48:19 PM1/8/07 5:48:19 PM
35. 35. Linear Momentum and Collisions 243 P9.8 1 39. kg m s upward⋅ P9.10 (a) 5 40. N s⋅ toward the net (b) −27 0. J P9.12 (a) 981 N⋅s up (b) 3.43 mրs (c) 3.83 mրs (d) 0.748 m P9.14 16.5 N P9.16 (a) 2 50. m s (b) 3 75 104 . × J P9.18 v = 4M m gᐉ P9.20 (a) 4.85 mրs (b) 8.41 m P9.22 7.94 cm P2.24 (a) 2 24. m s toward the right (b) No, coupling order makes no difference P9.26 (a) Over a very short time interval, outside forces have no time to impart signiﬁcant impulse— thus the interaction is a collision. The opponent grabs the fullback and does not let go, so the two players move together at the end of their interaction—thus the collision is completely inelastic. (b) 2.88 mրs at 32.3° (c) 783 J becomes internal energy. P9.28 No; his speed was 41 5. mi h P9.30 v vY = i sinθ; v vO i= cosθ P9.32 (a) 2vi; 2 3 vi (b) 35.3° P9.34 vBlue = 5.89 mրs and vGreen = 7.07 mրs P9.36 4.67 × 106 m from the Earth’s center P9.38 (a) see the solution (b) 3 57 108 . × J P9.40 0.063 5L P9.42 (a) (−2.89 ˆi −1.39 ˆj)cm (b) (−44.5 ˆi + 12.5 ˆj) g⋅cmրs (c) (−4.94 ˆi + 1.39ˆj) cmրs (d) (−2.44 ˆi + 1.56 ˆj) cmրs2 (e) (−220 ˆi + 140 ˆj) µN P9.44 (a) −0 780. ˆi m s; 1.12ˆi mրs (b) 0 360. ˆi m s P9.46 (a) Yes (b) No. The work by the ﬂoor on the person is zero. (c) 103 kg⋅mրs up (d) Yes. You could say that it came from the planet, that gained momentum 103 kg⋅mրs down, but it came through the force exerted by the ﬂoor over a time interval on the person, so it came through the ﬂoor or from the ﬂoor through direct contact. (e) 88.2 J (f) No. The energy came from chemical energy in the person’s leg muscles. The ﬂoor did no work on the person. P9.48 (a) [F(x1 + x2 )ր2m]1ր2 (b) cos−1 [1 − F(x1 − x2 )ր2mgL] P9.50 (a) 787 m s (b) 138 m s P9.52 see the solution 13794_09_ch09_p209-244 copy.indd243 24313794_09_ch09_p209-244 copy.indd243 243 1/8/07 5:48:19 PM1/8/07 5:48:19 PM
36. 36. 244 Chapter 9 P9.54 (a) m m m m 1 1 2 2 1 2 v v+ + (b) v v1 2 1 2 1 2 −( ) +( ) m m k m m (c) v v v1 1 2 1 2 1 2 1 2 2 2 f m m m m m m m = − + ⎛ ⎝⎜ ⎞ ⎠⎟ + + ⎛ ⎝⎜ ⎞ ⎠⎟ ; v v v2 1 1 2 1 2 1 1 2 2 2 f m m m m m m m = + ⎛ ⎝⎜ ⎞ ⎠⎟ + − + ⎛ ⎝⎜ ⎞ ⎠⎟ P9.56 (a) R = 2580 kg ⋅ m (80 kg + m)−1 (b) 16.2 m (c) 16.0 m (d) 5.87 kg (e) 32.2 m (f) m→0 (g) 0 (h) m→∞ (i) Yes, until the clone splashes down. No; we need also conservation of momentum in the collision. (j) Yes, but it is not useful to include the planet in the analysis of momentum. We use instead momentum conservation for the actor-clone system while they are in contact. (k) At a location with weaker gravity, the actor would be moving more slowly before the collision, but the clone would follow the same trajectory, moving more slowly over a longer time interval. P9.58 (a) −0 667. m s (b) 0.952 m P9.60 (a) 6 81. m s (b) 1.00 m P9.62 (a) 6.29 mրs (b) 6.16 mրs (c) Most of the 2% difference between the values for speed is accounted for by the uncertainty in the data, estimated as 0 01 8 68 0 1 68 8 1 263 1 257 0 1 85 3 1 1 . . . . . . . %+ + + + = P9.64 (a) The mass of the sleigh plus you is 270 kg and your velocity is 7.50 mրs in the x direction. You unbolt a 15.0-kg seat and throw it back at the wolves, giving it a speed of 8.00 mրs relative to you. Find the velocity of the seat relative to the ground afterward, and the velocity of the sleigh afterward. (b) 0.055 6 mրs in the −x direction; 7.94 mրs in the +x direction (c) 453 J P9.66 0 179. m s P9.68 (a) 20 0 7 00. ˆ . ˆi j+( ) m s (b) 4 00. ˆi m s2 (c) 4 00. ˆi m s2 (d) 50 0 35 0. ˆ . ˆi j+( ) m (e) 600 J (f) 674 J (g) 674 J (h) The accelerations computed in different ways agree. The kinetic energies computed in different ways agree. The three theories are consistent. 13794_09_ch09_p209-244 copy.indd244 24413794_09_ch09_p209-244 copy.indd244 244 1/8/07 5:48:21 PM1/8/07 5:48:21 PM