Newtons laws

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Newtons laws

  1. 1. Topic 2 Newton's Laws
  2. 2. Contents <ul><li>Newtons Laws </li></ul><ul><li>Inertial Mass </li></ul><ul><li>Gravitational Mass </li></ul><ul><li>Weight </li></ul><ul><li>Difference Between Mass and Weight </li></ul><ul><li>Linear Momentum </li></ul><ul><li>Application of Newtons Second Law </li></ul><ul><li>Newtons 2 nd Law Revisited </li></ul><ul><li>Energy </li></ul>
  3. 3. Newton’s Laws <ul><li>Newton is an English Scientist who worked in this area. </li></ul><ul><li>He is the most famous scientist in the world today: </li></ul><ul><ul><li>Remember the apple falling on his head? </li></ul></ul><ul><ul><li>Besides Albert Einstein. </li></ul></ul>
  4. 4. Newton’s Laws <ul><li>He lived during the 1600’s. </li></ul><ul><li>His work was published in 1687. </li></ul>
  5. 5. Newton’s Laws
  6. 6. Newton’s Laws <ul><li>Science is not exact, mistakes are made. </li></ul><ul><li>Newton found mistakes in his own work after it was published. </li></ul><ul><li>He then corrected the mistakes in his personal copy. </li></ul>
  7. 7. Newton’s Laws
  8. 8. Newton’s Laws <ul><li>Ist Law: </li></ul><ul><li>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) </li></ul>
  9. 9. Newton’s Laws
  10. 10. Newton’s Laws
  11. 11. Newton’s First Law Overview
  12. 12. Newton’s Laws <ul><li>If the forces on an object are balanced, </li></ul><ul><ul><li>the object moves with , </li></ul></ul><ul><ul><li>a constant velocity. </li></ul></ul><ul><li>If however they are not balanced, </li></ul><ul><ul><li>the velocity will change </li></ul></ul><ul><ul><li>ie accelerate. </li></ul></ul>
  13. 13. Newton’s Laws <ul><li>There is a relationship between : </li></ul><ul><ul><li>force and acceleration. </li></ul></ul><ul><li>This will be explored in : </li></ul><ul><ul><li>a practical on Newton’s Second Law. </li></ul></ul>
  14. 15. Newton’s Laws <ul><li>2nd Law </li></ul><ul><li>The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. </li></ul><ul><li>The direction of the acceleration is in the direction of the applied net force. ( F = m a ) </li></ul>
  15. 16. Newton’s Laws <ul><li>3rd Law </li></ul><ul><li>Whenever one object exerts a force on a second object, the second exerts an equal and opposite on the first . </li></ul>
  16. 17. Newton’s Laws <ul><li>Forces occur in pairs in every interaction. </li></ul><ul><li>Use your foot to push against the floor to move, </li></ul><ul><ul><li>The floor pushes back. </li></ul></ul><ul><li>You push backwards against the water in a pool, </li></ul><ul><ul><li>The water pushes back and moves you forward. </li></ul></ul>
  17. 18. Newton’s Laws <ul><li>What happens if the force pair is not obvious? </li></ul><ul><li>E.g., a rock falling down a cliff? </li></ul><ul><ul><li>Earth pulls rock </li></ul></ul><ul><ul><li>Rock pulls earth. </li></ul></ul>
  18. 19. Newton’s Laws <ul><li>E.g., Car accelerating from a stop light? </li></ul><ul><ul><li>Tyre pushes on road </li></ul></ul><ul><ul><li>Road pushes on tyre </li></ul></ul>
  19. 20. Newton’s Laws <ul><li>Rifle and bullet </li></ul><ul><ul><li>Rifle pushes bullet out of barrel. </li></ul></ul><ul><ul><li>Bullet pushes rifle back into the shoulder, </li></ul></ul><ul><ul><ul><li>Kickback. </li></ul></ul></ul>
  20. 21. Newton’s Laws <ul><li>Statue sitting on a table </li></ul>
  21. 22. Newton’s Laws <ul><li>Be careful choosing the action/reaction pairs. </li></ul><ul><li>F N ’ is the reaction force to F N </li></ul><ul><li>F N ’ = Force exerted on the table by the statue </li></ul><ul><li>The reaction force to F g is not shown. </li></ul>
  22. 23. Newton’s Laws <ul><li>Why don’t forces cancel? </li></ul><ul><li>Kick football: </li></ul><ul><ul><li>it accelerates away from you. </li></ul></ul><ul><li>There is only one force: </li></ul><ul><ul><li>By NII, if force applied, object accelerates. </li></ul></ul><ul><li>What about equal and opposite force? </li></ul><ul><ul><li>It doesn’t act on the ball: </li></ul></ul><ul><ul><ul><li>It acts on your foot. </li></ul></ul></ul>
  23. 24. Newton’s Laws <ul><li>Two people kicking the same soccer ball with equal but opposite force? </li></ul><ul><li>Two forces on the ball. </li></ul><ul><ul><li>Cancel each other out. </li></ul></ul>
  24. 25. Inertial Mass <ul><li>Kick an empty tin can and: </li></ul><ul><ul><li>it moves. </li></ul></ul><ul><li>Kick a can full of sand and it: </li></ul><ul><ul><li>moves a bit </li></ul></ul><ul><li>Kick a can full of lead and: </li></ul><ul><ul><li>it hurts, </li></ul></ul><ul><ul><li>but doesn’t move. </li></ul></ul>
  25. 26. Inertial Mass <ul><li>The lead filled can has more inertia than the sand filled can. </li></ul><ul><li>The can with the most matter has the greatest inertia. </li></ul><ul><li>The amount of inertia an object has depends on its mass. </li></ul><ul><ul><li>The amount of matter in the object. </li></ul></ul>
  26. 27. Inertial Mass <ul><li>The more mass an object has: </li></ul><ul><ul><li>The more for it takes to change its state of motion. </li></ul></ul><ul><li>As F = ma : </li></ul><ul><ul><li>m = F/a </li></ul></ul><ul><li>Inertial mass is: </li></ul><ul><ul><li>Ratio of resultant force to acceleration </li></ul></ul>
  27. 28. Inertial Mass <ul><li>An object with 3 times the mass needs: </li></ul><ul><ul><li>3 times the force. </li></ul></ul><ul><li>If your mass is 60 kg, you have: </li></ul><ul><ul><li>60 times the inertia than a 1 kg bag of sugar. </li></ul></ul><ul><li>If you have a mass of 90 kg and: </li></ul><ul><ul><li>an opponent on the football field has, </li></ul></ul><ul><ul><li>a mass of 60 kg </li></ul></ul>
  28. 29. Inertial Mass <ul><li>You have 1.5 times the inertia of the opponent. </li></ul><ul><li>When you collide: </li></ul><ul><ul><li>Travelling at the same speed, </li></ul></ul><ul><ul><li>He will get knocked over. </li></ul></ul>
  29. 30. Gravitational Mass <ul><li>The amount of gravitational mass an object has can be determined by experiment. </li></ul><ul><li>Accelerate objects with: </li></ul><ul><ul><li>different masses in, </li></ul></ul><ul><ul><li>a gravitational field. </li></ul></ul>
  30. 31. Gravitational Mass <ul><li>If we conduct the experiment in a vacuum: </li></ul><ul><ul><li>No frictional forces. </li></ul></ul><ul><li>Drop a light and heavy object at the same time: </li></ul><ul><ul><li>They reach the ground at the same time. </li></ul></ul><ul><li>They must have the same acceleration. </li></ul>
  31. 32. Gravitational Mass <ul><li>The force acting on both objects is: </li></ul><ul><ul><li>The weight of the object </li></ul></ul><ul><ul><li>F = W </li></ul></ul><ul><li>Mass is a measure of gravitational force of attraction of one body on another. </li></ul><ul><li>One body will attract another body with the same force as the second body attracts the first. </li></ul>
  32. 33. Gravitational Mass <ul><li>No experiments have shown that inertial mass and gravitational mass are not equal. </li></ul><ul><li>This is called the equivalence principal: </li></ul><ul><li>Gravitational mass is equivalent to inertial mass. </li></ul>
  33. 34. Gravitational Mass <ul><li>Physicists have thought for a long time that: </li></ul><ul><ul><li>Mass is the source of gravitation, </li></ul></ul><ul><ul><li>Has the inertial property of resistance. </li></ul></ul><ul><li>This overlooks how different the two properties are. </li></ul><ul><li>It took Einstein to conclude that Inertia and mass are equivalent. </li></ul>
  34. 35. Gravitational Mass <ul><li>The values of the two are numerically equal: </li></ul><ul><ul><li>a g = 9.81 m s -2 </li></ul></ul><ul><ul><li>Gravitational Field Strength = 9.81 N kg -1 </li></ul></ul>
  35. 36. Weight <ul><li>Commonly, weight is the measure of: </li></ul><ul><ul><li>The force of gravity on an object. </li></ul></ul><ul><li>You can use a spring balance to: </li></ul><ul><ul><li>measure your weight in Panama. </li></ul></ul><ul><li>Repeat the procedure in Fairbanks Alaska and: </li></ul><ul><ul><li>the scale will read ½ % more. </li></ul></ul>
  36. 37. Weight <ul><li>Fly to Mars and the reading would be: </li></ul><ul><ul><li>1/3 of the original reading. </li></ul></ul><ul><li>The problem occurs because: </li></ul><ul><ul><li>we are measuring the force, </li></ul></ul><ul><ul><li>trying to stretch the spring. </li></ul></ul>
  37. 38. Weight <ul><li>The force resisting the stretching of the spring is: </li></ul><ul><ul><li>The same while, </li></ul></ul><ul><ul><li>The force of gravity is not. </li></ul></ul><ul><li>This is a non-equilibrium situation. </li></ul><ul><li>Use a Beam Balance and the reading will be different. </li></ul>
  38. 39. Weight <ul><li>If two objects have the same weight in Panama: </li></ul><ul><ul><li>The scale will come to a balance position. </li></ul></ul><ul><li>Moving to Fairbanks: </li></ul><ul><ul><li>Both objects will weigh less and, </li></ul></ul><ul><ul><li>Still come to a balance position. </li></ul></ul><ul><li>It also works on Mars. </li></ul>
  39. 40. Weight <ul><li>If we use F = m a to determine the weight: </li></ul><ul><ul><li>We get a particular value. </li></ul></ul><ul><li>If we use a scale on a balance: </li></ul><ul><ul><li>We get a different value. </li></ul></ul><ul><li>The term weight: </li></ul><ul><ul><li>Can be ambiguous </li></ul></ul><ul><li>Make sure you read the ‘weight’ in an equilibrium situation </li></ul>
  40. 41. Difference Between Mass & Weight <ul><li>Weight is a measure of the gravitational attraction between bodies. </li></ul><ul><li>Mass is the amount of matter in that same body. </li></ul>
  41. 42. Difference Between Mass & Weight <ul><li>Mass does not vary: </li></ul><ul><ul><li>It has the same number of atoms in it. </li></ul></ul><ul><li>Weight does vary: </li></ul><ul><ul><li>It depends on how strong the gravitational force of attraction is. </li></ul></ul><ul><li>Weight can be related to mass by the formula: </li></ul><ul><ul><li>W = ma </li></ul></ul>
  42. 43. Momentum
  43. 44. Momentum <ul><li>We have previously considered single particles. </li></ul><ul><li>We now need to consider the interaction of two particles. </li></ul>
  44. 45. Momentum <ul><li>Newton’s 3 rd law describes how : </li></ul><ul><ul><li>there is an equal and opposite force , </li></ul></ul><ul><ul><li>in every interaction. </li></ul></ul><ul><li>T he force which the earth pulls the moon around in a circular orbit : </li></ul><ul><ul><li>is matched by an equal and opposite gravitational force , </li></ul></ul><ul><ul><li>exerted by the moon on the earth. </li></ul></ul>
  45. 46. Arabic guy gun demo - momentum
  46. 47. Momentum <ul><li>To investigate the force acting in an interaction between particles : </li></ul><ul><ul><li>we must make sure the interaction is , </li></ul></ul><ul><ul><li>isolated from any external influences. </li></ul></ul><ul><li>Examples of isolated systems are: </li></ul>
  47. 48. Momentum <ul><li>The contents of a thermos are : </li></ul><ul><ul><li>thermally isolated from the atmosphere. </li></ul></ul><ul><li>A sound proofed room is : </li></ul><ul><ul><li>acoustically isolated. </li></ul></ul>
  48. 49. Momentum <ul><li>Consider two gliders colliding on an air track. </li></ul><ul><li>If the force applied on glider 2 by glider 1 is F 1 , then : </li></ul><ul><ul><li>the force on glider 1 by glider 2 is, </li></ul></ul><ul><ul><li>according to Newton’s 3 rd law; </li></ul></ul><ul><ul><li>- F 2 . </li></ul></ul>
  49. 50. Momentum <ul><li>From Newton’s 2 nd law; </li></ul><ul><li>m 1 a 1 = -m 2 a 2 </li></ul><ul><li>and as acceleration is defined as : </li></ul><ul><li>and acceleration by glider 2 is; </li></ul>
  50. 51. Momentum <ul><li>We have assumed that : </li></ul><ul><ul><li>the time taken for the collision is , </li></ul></ul><ul><ul><li>the same for both gliders. </li></ul></ul><ul><li>The equation now becomes: </li></ul><ul><li>Multiplying both sides by t : </li></ul>
  51. 52. Momentum <ul><li>Re-arranging so all quantities before the collision are on one side of the equation: </li></ul>
  52. 53. Momentum <ul><li>This tells us that if we add the vectors before the collision : </li></ul><ul><ul><li>it will equal the vectors after the collision. </li></ul></ul><ul><li>The quantity m v is conserved : </li></ul><ul><ul><li>When no external forces act. </li></ul></ul>
  53. 54. Momentum <ul><li>To simplify matters, we call the quantity m v : </li></ul><ul><li>Momentum . </li></ul><ul><li>The momentum p of an object is defined as : </li></ul><ul><ul><li>the product of its mass and velocity , </li></ul></ul><ul><ul><li>p = m v . </li></ul></ul><ul><li>The units of momentum are: kg m s -1 . </li></ul>
  54. 55. Momentum <ul><li>We stated previously, </li></ul><ul><ul><li>the quantity m v , or momentum, </li></ul></ul><ul><ul><li>is conserved. </li></ul></ul><ul><li>This law can be stated: </li></ul><ul><li>The total momentum of an isolated system is constant and : </li></ul><ul><ul><li>is unaffected by the interaction of it’s parts . </li></ul></ul>
  55. 56. Cons of Momentum Law <ul><li>Provided no external forces act, the total momentum of a system of interacting particles remains constant, despite the interaction of its parts. </li></ul>
  56. 57. Momentum <ul><li>Watch as a 2 kg brick is dropped on a 3 kg loaded cart. </li></ul>
  57. 58. Momentum <ul><li>What changes when the cart is only 1 kg and the brick is 2 kg? </li></ul>
  58. 59. Example <ul><li>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 10 7 ms -1 and it rebounds after the collision with a velocity of 1.30 x 10 7 ms -1 in the opposite direction. What is the final velocity of the nitrogen nucleus? </li></ul>
  59. 60. Solution <ul><li>m neutron = 1.67 x 10 -27 kg </li></ul><ul><li>m nitrogen = 2.31 x 10 -26 kg </li></ul><ul><li>u neutron = 1.50 x 10 7 m s -1 </li></ul><ul><li>v neutron = -1.30 x 10 7 m s -1 </li></ul><ul><li>u nitrogen = 0 ms -1 </li></ul><ul><li>v nitrogen = ? ms -1 </li></ul>
  60. 61. Solution Before After v neutron v neutron v nitrogen m neutron m neutron m nitrogen
  61. 62. Solution <ul><li>By the law of conservation of momentum </li></ul><ul><li>p i = p f </li></ul><ul><li>p i = m neutron u neutron + m nitrogen u nitrogen </li></ul><ul><li>p i = (1.67 x 10 -27 ) x (1.50 x 10 7 ) + 0 </li></ul><ul><li>p i = 2.505 x 10 -20 kg m s -1 </li></ul>
  62. 63. Solution <ul><li>p f = m neutron v neutron + m nitrogen v nitrogen </li></ul><ul><li>p f = ( 1.67 x 10 -27 ) x (-1.30 x 10 7 ) +(2.31 x 10 -26 ) x v nitrogen </li></ul><ul><li>p f = (2.31 x 10 -26 v nitrogen ) - 2.171 x 10 -20 </li></ul><ul><li>As p i = p f </li></ul>
  63. 64. Solution <ul><li>2.505 x 10 -20 = (2.31 x 10 -26 v nitrogen ) - 2.171 x 10 -20 </li></ul><ul><li>2.505 x 10 -20 +2.171 x 10 -20 = 2.31 x 10 -26 v nitrogen </li></ul><ul><li>4.676 x 10 -20 = v nitrogen </li></ul><ul><li>2.31 x 10 -26 </li></ul><ul><li>v nitrogen = 2.02 x 10 6 m s -1 </li></ul><ul><ul><li>in the initial direction of the neutron </li></ul></ul>
  64. 65. Example <ul><li>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 N53.1 o E. </li></ul>
  65. 66. Example <ul><li>Neglect friction in answering the following questions. </li></ul><ul><li>a)    Was the car exceeding the speed limit of 60 km h -1 ? </li></ul><ul><li>b)    What was the common speed of the car and the truck after the collision? </li></ul>
  66. 67. Solution 53.1 o truck car
  67. 68. Solution <ul><li>m c = 750 kg </li></ul><ul><li>u c = ? north </li></ul><ul><li>m T = 2000 kg </li></ul><ul><li>u T = 36 km h -1 east </li></ul><ul><li>u T = 10 m s -1 east </li></ul><ul><li>u c+T = ? N53.1 o E </li></ul>
  68. 69. Solution <ul><li>By conservation of momentum, </li></ul><ul><li>m c u c + m T u T = m c+T v c+T </li></ul><ul><li>750 u c N + 2000 x 10 E = 2750 x v c+T N53.1 o E </li></ul><ul><li>Diagrammatically : </li></ul>
  69. 70. Solution
  70. 71. Solution <ul><li>u c = 20.02 m s -1 </li></ul>
  71. 72. Solution – Part (a) <ul><li>Speed of car = 20.0 ms -1 </li></ul><ul><li>= 72 kmh -1 </li></ul><ul><li>The car was exceeding the speed limit. </li></ul>
  72. 73. Solution – Part (b) <ul><li>= 9.09 m s -1 </li></ul><ul><li>32.7 km h -1 </li></ul>
  73. 74. Impulse <ul><li>Application of Newton I I </li></ul><ul><li>During collisions, objects are deformed . </li></ul>
  74. 75. Impulse <ul><li>F = m a </li></ul><ul><li>For constant acceleration : </li></ul>
  75. 76. Impulse <ul><li>F t = m v – m u </li></ul><ul><li>F t =  p </li></ul><ul><li>F t is called the : </li></ul><ul><ul><li>impulse of the force . </li></ul></ul><ul><li>This impulse causes the : </li></ul><ul><ul><li>momentum to change. </li></ul></ul><ul><li>An impulse is a short duration force : </li></ul><ul><ul><li>usually of non-constant magnitude. </li></ul></ul>
  76. 77. Impulse <ul><li>Units are the same as those for momentum : </li></ul><ul><ul><li>kg m s -1 or s N </li></ul></ul><ul><li>Defined as : </li></ul><ul><ul><li>the product of the force and the time , </li></ul></ul><ul><ul><li>over which the force acts . </li></ul></ul><ul><li>During collisions : </li></ul><ul><ul><li> t is often very small so , </li></ul></ul><ul><ul><li>the F av is often very large. </li></ul></ul>
  77. 78. Newton’s 2 nd Law Revisited <ul><li>We can now redefine the term force. </li></ul><ul><li>Force is the rate of change of : </li></ul><ul><ul><li>the linear momentum of a body. </li></ul></ul><ul><li>From this we can derive Newton’s 2 nd Law. </li></ul>
  78. 79. Newton’s 2 nd Law Revisited
  79. 80. Newton’s 2 nd Law Revisited <ul><li>F net = m a </li></ul><ul><li>Impulse Question </li></ul>
  80. 81. Momentum and Energy <ul><li>The total energy in an isolated system is conserved but : </li></ul><ul><ul><li>energy can be transferred from one object to another and , </li></ul></ul><ul><ul><li>can be converted from one form to another. </li></ul></ul>
  81. 82. Momentum and Energy <ul><li>The units are Joules (J) and : </li></ul><ul><ul><li>it is a scalar quantity and , </li></ul></ul><ul><ul><li>does not have a direction. </li></ul></ul><ul><li>In collisions : </li></ul><ul><ul><li>total energy is always conserved. </li></ul></ul>
  82. 83. Momentum and Energy <ul><li>The kinetic energy will not always remain constant but : </li></ul><ul><ul><li>may be converted to other forms. </li></ul></ul><ul><li>This could be : </li></ul><ul><ul><li>rotational kinetic energy, </li></ul></ul><ul><ul><li>sound or , </li></ul></ul><ul><ul><li>heat. </li></ul></ul><ul><li>There are three types of collisions: </li></ul>
  83. 84. Momentum and Energy <ul><li>1. Elastic : </li></ul><ul><li>Momentum is conserved and : </li></ul><ul><ul><li>no kinetic energy is lost. </li></ul></ul><ul><li>This occurs on the microscopic scale : </li></ul><ul><ul><li>such as between nuclei. </li></ul></ul>
  84. 85. Momentum and Energy <ul><li>What happens when a light car rear-ends a heavy truck and collides elastically? </li></ul><ul><li>Calculate the kinetic energy. </li></ul>
  85. 86. Momentum and Energy <ul><li>Before </li></ul><ul><li>E k = ½ mv 2 </li></ul><ul><li>E k = ½ x 1000 x 20 2 </li></ul><ul><li>E k = 2.0 x 10 5 J </li></ul>
  86. 87. Momentum and Energy <ul><li>After </li></ul><ul><li>E k = ½mv 2 + ½mv 2 </li></ul><ul><li>E k = (½x 1000 x 10 2 ) + (½x 3000 x 10 2 ) </li></ul><ul><li>E k = (50000) + (150000) </li></ul><ul><li>E k = 200000 </li></ul><ul><li>E k = 2.0 x 10 5 J </li></ul><ul><li>E k is conserved </li></ul>
  87. 88. Momentum and Energy <ul><li>What happens when the heavy truck rear-ends the light car elastically? </li></ul><ul><li>Calculate the kinetic energy. </li></ul>
  88. 89. Momentum and Energy <ul><li>Before </li></ul><ul><li>E k = ½ mv 2 </li></ul><ul><li>E k = ½ x 3000 x 20 2 </li></ul><ul><li>E k = 600000 J </li></ul><ul><li>E k = 6.0 x 10 5 J </li></ul>
  89. 90. Momentum and Energy <ul><li>After </li></ul><ul><li>E k = ½mv 2 + ½mv 2 </li></ul><ul><li>E k = (½x 3000 x 10 2 ) + (½x 1000 x 30 2 ) </li></ul><ul><li>E k = 150000 + 450000 </li></ul><ul><li>E k = 600000 J </li></ul><ul><li>E k = 6.0 x 10 5 J </li></ul><ul><li>E k is conserved </li></ul>
  90. 91. Momentum and Energy <ul><li>2. Inelastic : </li></ul><ul><li>Momentum is conserved but : </li></ul><ul><ul><li>kinetic energy is lost. </li></ul></ul><ul><li>All macroscopic collisions are inelastic. </li></ul><ul><li>Some collisions are almost elastic </li></ul><ul><ul><li>Billiard balls and , </li></ul></ul><ul><ul><li>air track/table gliders. </li></ul></ul>
  91. 92. Momentum and Energy <ul><li>What happens in a head-on where the collision is inelastic? </li></ul><ul><li>Calculate the kinetic energy. </li></ul>
  92. 93. Momentum and Energy <ul><li>Before </li></ul><ul><li>E k = ½mv 2 + ½mv 2 </li></ul><ul><li>E k = (½x 1000 x 20 2 ) + (½x 3000 x 20 2 ) </li></ul><ul><li>E k = 20000o + 600000 </li></ul><ul><li>E k = 800000 J </li></ul><ul><li>E k = 8.0 x 10 5 J </li></ul>
  93. 94. Momentum and Energy <ul><li>After </li></ul><ul><li>E k = ½mv 2 + ½mv 2 </li></ul><ul><li>E k = (½x 1000 x 10 2 ) + (½x 3000 x 10 2 ) </li></ul><ul><li>E k = 50000 + 150000 </li></ul><ul><li>E k = 200000 J </li></ul><ul><li>E k = 2.0 x 10 5 J </li></ul><ul><li>E k is not conserved </li></ul>
  94. 95. Momentum and Energy <ul><li>3. Perfectly inelastic : </li></ul><ul><li>Momentum is conserved but : </li></ul><ul><ul><li>kinetic energy is lost and , </li></ul></ul><ul><ul><li>the bodies stick together , </li></ul></ul><ul><ul><li>after the collision. </li></ul></ul><ul><li>Elastic & Inelastic Collisions </li></ul>
  95. 96. Momentum and Energy <ul><li>What happens when a light car rear-ends a heavy truck and collides and sticks together? </li></ul><ul><li>Determine the kinetic energy. </li></ul>
  96. 97. Momentum and Energy <ul><li>Before </li></ul><ul><li>E k = ½ mv 2 </li></ul><ul><li>E k = ½ x 1000 x 20 2 </li></ul><ul><li>E k = 200000 J </li></ul><ul><li>E k = 2.0 x 10 5 J </li></ul>
  97. 98. Momentum and Energy <ul><li>After </li></ul><ul><li>E k = ½mv 2 + ½mv 2 </li></ul><ul><li>E k = (½x 3000 x 5 2 ) + (½x 1000 x 5 2 ) </li></ul><ul><li>E k = 37500 + 12500 </li></ul><ul><li>E k = 50000 J </li></ul><ul><li>E k = 5.0 x 10 4 J </li></ul><ul><li>E k is not conserved </li></ul>
  98. 99. Momentum and Energy <ul><li>What happens when the truck rear ends the car and they stick together? </li></ul><ul><li>Calculate the kinetic energy. </li></ul>
  99. 100. Momentum and Energy <ul><li>Before </li></ul><ul><li>E k = ½ mv 2 </li></ul><ul><li>E k = ½ x 3000 x 20 2 </li></ul><ul><li>E k = 600000 J </li></ul><ul><li>E k = 6.0 x 10 5 J </li></ul>
  100. 101. Momentum and Energy <ul><li>After </li></ul><ul><li>E k = ½mv 2 + ½mv 2 </li></ul><ul><li>E k = (½x 3000 x 15 2 ) + (½x 1000 x 15 2 ) </li></ul><ul><li>E k = 37500 + 12500 </li></ul><ul><li>E k = 337500 J </li></ul><ul><li>E k = 3.4 x 10 5 J </li></ul><ul><li>E k is not conserved </li></ul>
  101. 102. Example <ul><li>Show that the collision between the neutron and the nitrogen nucleus in the earlier example is elastic. </li></ul>
  102. 103. Solution <ul><li>m nitrogen = 2.31 x 10 -26 kg </li></ul><ul><li>u neutron = 1.50 x 10 7 m s -1 </li></ul><ul><li>u nitrogen = 0 ms -1 </li></ul><ul><li>v neutron = -1.30 x 10 7 m s -1 </li></ul><ul><li>v nitrogen = 2.02 x 10 6 m s -1 </li></ul><ul><li>E kinitial = ½ m neutron u 2 neutron + ½ m nitrogen u 2 nitrogen </li></ul>
  103. 104. Solution
  104. 105. Solution <ul><li>= ½(1.67 x 10 -27 ) x (1.50 x 10 7 ) 2 + ½(2.31 x 10 -26 ) x 0 2 </li></ul><ul><li>=1.88 x 10 -13 J </li></ul><ul><li>E kfinal = ½ m neutron v 2 neutron + ½ m nitrogen v 2 nitrogen </li></ul><ul><li>= ½(1.67 x 10 -27 ) x (1.30 x 10 7 ) 2 + ½(2.31 x 10 -26 ) x(2.02 x 10 6 ) 2 </li></ul><ul><li>=1.88 x 10 -13 J </li></ul><ul><li>Since E kinitial = E kfinal , the collision is elastic. </li></ul>
  105. 106. Example <ul><li>An atom of helium gas moving with a speed of 3.0 x 10 2 ms -1 collides elastically with a stationary atom of the same gas. After the collision the first atom moves off at an angle of 30 o to its initial direction. </li></ul><ul><li>a) Find the speed of the incident atom after the collision. </li></ul><ul><li>b) Find the velocity of the struck atom after collision. </li></ul>
  106. 107. Solution – Part (a) <ul><li>The masses of the incident atom, and the struck atom, m s , are equal since : </li></ul><ul><ul><li>they are both helium atoms. </li></ul></ul><ul><li>m i = m s = m </li></ul><ul><li>The collision is isolated, and so : </li></ul>
  107. 108. Solution – Part (a) <ul><li>p i = p f </li></ul><ul><li>m i u i = m i v i + m s v s </li></ul><ul><li>i.e. m u i = m v i + m v s </li></ul><ul><li>Thus u i = v i + v s </li></ul>
  108. 109. Solution – Part (a) <ul><li>The collision is elastic, so </li></ul><ul><li>E ki = E kf </li></ul><ul><li>½ m i u i 2 = ½ m i v i 2 + ½ m s v s 2 </li></ul><ul><li>i.e. ½ mu i 2 = ½ mv i 2 + ½ mv s 2 </li></ul><ul><li>Thus u i 2 = v i 2 + v s 2 </li></ul>
  109. 110. Solution – Part (a) <ul><li>Therefore u i is a hypotenuse and  = 90 o in the vector-addition triangle, by Pythagoras’s theorem. </li></ul><ul><li>By trigonometry, </li></ul>
  110. 111. Solution – Part (a) <ul><li>v i = u i cos30 o = 300 x  3/2 </li></ul><ul><li>= 2.6 x 10 2 ms -1 </li></ul><ul><li>Notice that we will always get  = 90 o just from the masses being equal and the collision elastic. </li></ul>
  111. 112. Solution – Part (b) <ul><li>By trigonometry in the vector diagram, </li></ul><ul><li>v s = u i sin30 o = 300 x ½ </li></ul><ul><li>=150 m s -1 </li></ul><ul><li>v s = 1.5 x 10 2 m s -1 at 90o to v i </li></ul>

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