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# Newtons laws 2012

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### Newtons laws 2012

1. 1. Newtons Laws
2. 2. Contents Newtons LawsInertial MassGravitational MassWeightDifference Between Mass and WeightLinear MomentumApplication of Newtons Second LawNewtons 2nd Law RevisitedEnergy
3. 3. Newton’s Laws  Newton is an English Scientist who worked in this area. He is the most famous scientist in the world today:  Remember the apple falling on his head?  Besides Albert Einstein.
4. 4. Newton’s Laws  He lived during the 1600’s. His work was published in 1687.
5. 5. Newton’s Laws 
6. 6. Newton’s Laws  Science is not exact, mistakes are made. Newton found mistakes in his own work after it was published. He then corrected the mistakes in his personal copy.
7. 7. Newton’s Laws 
8. 8. Newton’s Laws  Ist Law: Every body continues in its state of rest or uniform speed in a straight line unless it is compelled to change that state by a net force acting on it. (Law of inertia)
9. 9. Newton’s Laws 
10. 10. Newton’s Laws 
11. 11. Newton’s First Law Overview 
12. 12. Newton’s Laws  If the forces on an object are balanced,  the object moves with,  a constant velocity. If however they are not balanced,  the velocity will change  ie accelerate.
13. 13. Newton’s Laws  There is a relationship between:  force and acceleration. This will be explored in:  a practical on Newton’s Second Law.
14. 14. Newton’s 2nd Law Overview 
15. 15. Newton’s Laws  2nd Law The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the applied net force. (F = ma)
16. 16. Newton’s Laws  3rd Law Whenever one object exerts a force on a second object, the second exerts an equal and opposite on the first.
17. 17. Newton’s Laws Forces occur in pairs in every interaction.Use your foot to push against the floor to move,  The floor pushes back.You push backwards against the water in a pool,  The water pushes back and moves you forward.
18. 18. Newton’s Laws  What happens if the force pair is not obvious? E.g., a rock falling down a cliff?  Earth pulls rock  Rock pulls earth.
19. 19. Newton’s Laws  E.g., Car accelerating from a stop light?  Tyre pushes on road  Road pushes on tyre
20. 20. Newton’s Laws  Rifle and bullet  Rifle pushes bullet out of barrel.  Bullet pushes rifle back into the shoulder,  Kickback.
21. 21. Newton’s Laws  Statue sitting on a table
22. 22. Newton’s Laws  Be careful choosing the action/reaction pairs. FN’ is the reaction force to FN FN’ = Force exerted on the table by the statue The reaction force to Fg is not shown.
23. 23. Newton’s Laws  Newton’s Second Law also applies Mass of bullet is small  Acceleration is large Mass of rifle is relatively large  Acceleration is large
24. 24. Newton’s Laws  747 Does a 747 move by exhaust gases pushing on the atmosphere? If so, what would it push against in space? Exhaust gases push against the rocket Rocket pushes against the exhaust gas and moves forwards.
25. 25. Newton’s Laws  Waging Tail of Dog Tail wags dog Most noticeable when tail has a large relative mass
26. 26. Newton’s Laws  Why don’t forces cancel? Kick football:  it accelerates away from you. There is only one force:  By NII, if force applied, object accelerates. What about equal and opposite force?  It doesn’t act on the ball:  It acts on your foot.
27. 27. Newton’s Laws  Two people kicking the same soccer ball with equal but opposite force? Two forces on the ball.  Cancel each other out.
28. 28. Newton’s Laws  TOK Newton created many laws. What makes a law? How does the meaning of the word law change from subject to subject? Laws simplifies concepts and makes things easier to understand. What are we losing by simplification?
29. 29. Inertial Mass  Kick an empty tin can and:  it moves. Kick a can full of sand and it:  moves a bit Kick a can full of lead and:  it hurts,  but doesn’t move.
30. 30. Inertial Mass  The lead filled can has more inertia than the sand filled can. The can with the most matter has the greatest inertia. The amount of inertia an object has depends on its mass.  The amount of matter in the object.
31. 31. Inertial Mass  The more mass an object has:  The more for it takes to change its state of motion. As F = ma:  m = F/a Inertial mass is:  Ratio of resultant force to acceleration
32. 32. Inertial Mass  An object with 3 times the mass needs:  3 times the force. If your mass is 60 kg, you have:  60 times the inertia than a 1 kg bag of sugar. If you have a mass of 90 kg and:  an opponent on the football field has,  a mass of 60 kg
33. 33. Inertial Mass  You have 1.5 times the inertia of the opponent. When you collide:  Travelling at the same speed,  He will get knocked over.
34. 34. Gravitational Mass  The amount of gravitational mass an object has can be determined by experiment. Accelerate objects with:  different masses in,  a gravitational field.
35. 35. Gravitational Mass  If we conduct the experiment in a vacuum:  No frictional forces. Drop a light and heavy object at the same time:  They reach the ground at the same time. They must have the same acceleration.
36. 36. Gravitational Mass  The force acting on both objects is:  The weight of the object F=W Mass is a measure of gravitational force of attraction of one body on another. One body will attract another body with the same force as the second body attracts the first.
37. 37. Gravitational Mass  No experiments have shown that inertial mass and gravitational mass are not equal. This is called the equivalence principal: Gravitational mass is equivalent to inertial mass.
38. 38. Gravitational Mass  Physicists have thought for a long time that:  Mass is the source of gravitation,  Has the inertial property of resistance. This overlooks how different the two properties are. It took Einstein to conclude that Inertia and mass are equivalent.
39. 39. Gravitational Mass  The values of the two are numerically equal:  ag = 9.81 m s-2  Gravitational Field Strength = 9.81 N kg-1
40. 40. Weight  Commonly, weight is the measure of:  The force of gravity on an object. You can use a spring balance to:  measure your weight in Panama. Repeat the procedure in Fairbanks Alaska and:  the scale will read ½ % more.
41. 41. Weight  Fly to Mars and the reading would be:  1/3 of the original reading. The problem occurs because:  we are measuring the force,  trying to stretch the spring.
42. 42. Weight  The force resisting the stretching of the spring is:  The same while,  The force of gravity is not. This is a non-equilibrium situation. Use a Beam Balance and the reading will be different.
43. 43. Weight  If two objects have the same weight in Panama:  The scale will come to a balance position. Moving to Fairbanks:  Both objects will weigh less and,  Still come to a balance position. It also works on Mars.
44. 44. Weight If we use F = ma to determine the weight:  We get a particular value.If we use a scale on a balance:  We get a different value.The term weight:  Can be ambiguousMake sure you read the ‘weight’ in an equilibrium situation
45. 45. Difference Between Mass & Weight  Weight is a measure of the gravitational attraction between bodies. Mass is the amount of matter in that same body.
46. 46. Difference Between Mass & Weight  Mass does not vary:  It has the same number of atoms in it. Weight does vary:  It depends on how strong the gravitational force of attraction is. Weight can be related to mass by the formula:  W = ma
47. 47. Momentum 
48. 48. Momentum  We have previously considered single particles. We now need to consider the interaction of two particles.
49. 49. Momentum  Newton’s 3rd law describes how:  there is an equal and opposite force,  in every interaction. The force which the earth pulls the moon around in a circular orbit:  is matched by an equal and opposite gravitational force,  exerted by the Moon on the Earth.
50. 50. What is Momentum?  What we will study
51. 51. Momentum  To investigate the force acting in an interaction between particles:  we must make sure the interaction is,  isolated from any external influences. Examples of isolated systems are:
52. 52. Momentum  The contents of a thermos are:  thermally isolated from the atmosphere. A sound proofed room is:  acoustically isolated.
53. 53. Momentum  on an air track. Consider two gliders colliding If the force applied on glider 2 by glider 1 is F1, then:  the force on glider 1 by glider 2 is,  according to Newton’s 3rd law;  -F2.
54. 54. Momentum  law;From Newton’s 2 ndm1a1 = -m2a2and as acceleration is defined as : v v1 u1 a1 t tand acceleration by glider 2 is; v2 u 2 a2 t
55. 55. Momentum  that:We have assumed the time taken for the collision is, the same for both gliders.The equation now becomes: v1 u1 v2 u2 m1 m2 t tMultiplying both sides by t:
56. 56. Momentum  m1 (v1 u1 ) m2 (v2 u2 ) m1v1 m1u1 m2v2 m2u2Re-arranging so all quantities before the collision are on one side of the equation: m1u1 m2u2 m1v1 m2v2
57. 57. Momentum This tells us that if we add the vectors before the collision: it will equal the vectors after the collision.The quantity mv is conserved: When no external forces act.
58. 58. Momentum  To simplify matters, we call the quantity mv: Momentum. The momentum p of an object is defined as:  the product of its mass and velocity,  p = mv. The units of momentum are: kg m s-1.
59. 59. Momentum  We stated previously,  the quantity mv, or momentum,  is conserved. This law can be stated: The total momentum of an isolated system is constant and:  is unaffected by the interaction of it’s parts.
60. 60. Cons of Momentum Law  Provided no external forces act, the total momentum of a system of interacting particles remains constant, despite the interaction of its parts.
61. 61. Momentum  Watch as a 2 kg brick is dropped on a 3 kg loaded cart.
62. 62. Momentum  What changes when the cart is only 1 kg and the brick is 2 kg?
63. 63. Example  A neutron of mass 1.67 x 10-27 kg collides head on with a stationary nitrogen nucleus of mass 23.1 x 10- 27 kg. The initial velocity of the neutron is 1.50 x 107 ms-1 and it rebounds after the collision with a velocity of 1.30 x 107 ms-1 in the opposite direction. What is the final velocity of the nitrogen nucleus?
64. 64. Solution  mneutron= 1.67 x 10-27 kg mnitrogen = 2.31 x 10-26 kg uneutron = 1.50 x 107 m s-1 vneutron = -1.30 x 107 m s-1 unitrogen = 0 ms-1 vnitrogen = ? ms-1
65. 65. Solution v neutron v neutron v nitrogenm neutron m neutron m nitrogen Before After
66. 66. Solution  By the law of conservation of momentum pi = pf pi = mneutron uneutron + mnitrogen unitrogen pi = (1.67 x 10-27) x (1.50 x 107) + 0 pi = 2.505 x 10-20 kg m s-1
67. 67. Solution  pf = mneutron vneutron + mnitrogen vnitrogen pf = (1.67 x 10-27) x (-1.30 x 107) +(2.31 x 10-26) x vnitrogen pf = (2.31 x 10-26 vnitrogen) - 2.171 x 10-20 As pi = pf
68. 68. Solution  2.505 x 10-20 = (2.31 x 10-26 vnitrogen) - 2.171 x 10-20 2.505 x 10-20+2.171 x 10-20 = 2.31 x 10-26 vnitrogen 4.676 x 10-20 = vnitrogen 2.31 x 10-26 vnitrogen = 2.02 x 106 m s-1  in the initial direction of the neutron
69. 69. Example A car of mass 750 kg was travelling due north when it collided with a truck of mass 2000 kg which was travelling due east. The police accident investigation squad determined from witnesses that the truck was travelling at a speed of 36 km h-1 and that, after the collision, the car and truck stuck together and moved on the direction N53.1oE.
70. 70. Example  Neglect friction in answering the following questions. a) Was the car exceeding the speed limit of 60 km h- 1? b) What was the common speed of the car and the truck after the collision?
71. 71. Solution  53.1otruck car
72. 72. Solution  mc = 750 kg uc = ? north mT = 2000 kg uT = 36 km h-1 east uT = 10 m s-1 east uc+T = ? N53.1oE
73. 73. Solution  By conservation of momentum, mcuc + mTuT = mc+Tvc+T 750 uc N + 2000 x 10 E = 2750 x vc+T N53.1oE Diagrammatically:
74. 74. Solution  20 000750uc o 2750vc+T 53.1
75. 75. Solution 20000 tan 53.1 o 750 uc 20000 uc 750 x tan 53.1o uc = 20.02 m s-1
76. 76. Solution – Part (a)  Speed of car = 20.0 ms-1 20 x 3600 1000 = 72 kmh-1 The car was exceeding the speed limit.
77. 77. Solution – Part (b)  o 20000 sin 53 .1 2750 x vc T 20000 vc T 2750 x sin 53.1o = 9.09 m s-1 32.7 km h-1
78. 78. Impulse  Application of Newton II During collisions, objects are deformed. F t
79. 79. Impulse  F = ma For constant acceleration: v F m t (v u) F m t
80. 80. Impulse  Ft = mv – mu Ft = p Ft is called the:  impulse of the force. This impulse causes the:  momentum to change. An impulse is a short duration force:  usually of non-constant magnitude.
81. 81. Impulse  Units are the same as those for momentum:  kg m s-1 or s N Defined as: the product of the force and the time, over which the force acts. During collisions:  t is often very small so,  the Fav is often very large.
82. 82. Newton’s Law 2nd Revisited  We can now redefine the term force. Force is the rate of change of:  the linear momentum of a body. From this we can derive Newton’s 2nd Law.
83. 83. Newton’s Law 2nd Revisited p p  p f iFnet t t mv muFnet t m(v u)Fnet t
84. 84. Newton’s Law 2nd Revisited m v  Fnet t Fnet = ma Impulse Question
85. 85. Momentum and Energy  The total energy in an isolated system is conserved but:  energy can be transferred from one object to another and,  can be converted from one form to another.
86. 86. Momentum and Energy  The units are Joules (J) and:  it is a scalar quantity and,  does not have a direction. In collisions:  total energy is always conserved.
87. 87. Momentum and Energy  The kinetic energy will not always remain constant but:  may be converted to other forms. This could be:  rotational kinetic energy,  sound or,  heat. There are three types of collisions:
88. 88. Momentum and Energy 1. Elastic: Momentum is conserved and:  no kinetic energy is lost. This occurs on the microscopic scale:  such as between nuclei.
89. 89. Momentum and Energy  What happens when a light car rear-ends a heavy truck and collides elastically? Calculate the kinetic energy.
90. 90. Momentum and Energy  Before Ek = ½ mv2 Ek = ½ x 1000 x 202 Ek = 2.0 x 105 J
91. 91. Momentum and Energy  After Ek = ½mv2 + ½mv2 Ek = (½x 1000 x 102) + (½x 3000 x 102) Ek = (50000) + (150000) Ek = 200000 Ek = 2.0 x 105 J Ek is conserved
92. 92. Momentum and Energy What happens when the heavy truck rear-ends the light car elastically?Calculate the kinetic energy.
93. 93. Momentum and Energy  Before Ek = ½ mv2 Ek = ½ x 3000 x 202 Ek = 600000 J Ek = 6.0 x 105 J
94. 94. Momentum and Energy  After Ek = ½mv2 + ½mv2 Ek = (½x 3000 x 102) + (½x 1000 x 302) Ek = 150000 + 450000 Ek = 600000 J Ek = 6.0 x 105 J Ek is conserved
95. 95. Momentum and Energy 2. Inelastic: Momentum is conserved but:  kinetic energy is lost. All macroscopic collisions are inelastic. Some collisions are almost elastic  Billiard balls and,  air track/table gliders.
96. 96. Momentum and Energy  What happens in a head-on where the collision is inelastic? Calculate the kinetic energy.
97. 97. Momentum and Energy  Before Ek = ½mv2 + ½mv2 Ek = (½x 1000 x 202) + (½x 3000 x 202) Ek = 20000o + 600000 Ek = 800000 J Ek = 8.0 x 105 J
98. 98. Momentum and Energy  After Ek = ½mv2 + ½mv2 Ek = (½x 1000 x 102) + (½x 3000 x 102) Ek = 50000 + 150000 Ek = 200000 J Ek = 2.0 x 105 J Ek is not conserved
99. 99. Momentum and Energy 3. Perfectly inelastic: Momentum is conserved but:  kinetic energy is lost and,  the bodies stick together,  after the collision. Elastic & Inelastic Collisions
100. 100. Momentum and Energy  What happens when a light car rear-ends a heavy truck and collides and sticks together? Determine the kinetic energy.
101. 101. Momentum and Energy  Before Ek = ½ mv2 Ek = ½ x 1000 x 202 Ek = 200000 J Ek = 2.0 x 105 J
102. 102. Momentum and Energy  After Ek = ½mv2 + ½mv2 Ek = (½x 3000 x 52) + (½x 1000 x 52) Ek = 37500 + 12500 Ek = 50000 J Ek = 5.0 x 104 J Ek is not conserved
103. 103. Momentum and Energy  What happens when the truck rear ends the car and they stick together? Calculate the kinetic energy.
104. 104. Momentum and Energy  Before Ek = ½ mv2 Ek = ½ x 3000 x 202 Ek = 600000 J Ek = 6.0 x 105 J
105. 105. Momentum and Energy  After Ek = ½mv2 + ½mv2 Ek = (½x 3000 x 152) + (½x 1000 x 152) Ek = 37500 + 12500 Ek = 337500 J Ek = 3.4 x 105 J Ek is not conserved
106. 106. Example  Show that the collision between the neutron and the nitrogen nucleus in the earlier example is elastic.
107. 107. Solution  mnitrogen = 2.31 x 10-26 kg uneutron = 1.50 x 107 m s-1 unitrogen = 0 ms-1 vneutron = -1.30 x 107 m s-1 vnitrogen = 2.02 x 106 m s-1 Ekinitial = ½mneutronu2neutron + ½mnitrogenu2nitrogen
108. 108. Solution 
109. 109. Solution  = ½(1.67 x 10-27) x (1.50 x 107)2 + ½(2.31 x 10-26) x 02=1.88 x 10-13 JEkfinal = ½mneutronv2neutron + ½mnitrogenv2nitrogen= ½(1.67 x 10-27) x (1.30 x 107)2 + ½(2.31 x 10-26) x(2.02 x 106)2=1.88 x 10-13 JSince Ekinitial = Ekfinal , the collision is elastic.
110. 110. Example An atom of helium gas moving with a speed of 3.0 x 102 ms-1 collides elastically with a stationary atom of the same gas. After the collision the first atom moves off at an angle of 30o to its initial direction.a) Find the speed of the incident atom after the collision. b) Find the velocity of the struck atom after collision.
111. 111. Solution – Part (a)  The masses of the incident atom, and the struck atom, ms, are equal since:  they are both helium atoms. mi = ms = m The collision is isolated, and so:
112. 112. Solution – Part (a)p =p i f  miui = mivi + msvs i.e. mui = mvi + mvs Thus ui = vi + vs
113. 113. Solution – Part (a)  The collision is elastic, so Eki = Ekf ½miui2 = ½mivi2 + ½msvs2 i.e. ½mui2 = ½mvi2 + ½mvs2 Thus ui2 = vi2 + vs2
114. 114. Solution – Part (a)  Therefore ui is a hypotenuse and = 90o in the vector- addition triangle, by Pythagoras’s theorem. By trigonometry,
115. 115. Solution – Part (a)  vi = uicos30o = 300 x 3/2 = 2.6 x 102 ms-1 Notice that we will always get = 90o just from the masses being equal and the collision elastic.
116. 116. Solution – Part (b)  By trigonometry in the vector diagram, vs = uisin30o = 300 x ½ =150 m s-1 vs= 1.5 x 102 m s-1 at 90o to vi