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Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
Newtons laws
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Newtons laws
Newtons laws
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Newtons laws

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  • 1. Topic 2 Newton's Laws
  • 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. 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. Newton’s Laws <ul><li>He lived during the 1600’s. </li></ul><ul><li>His work was published in 1687. </li></ul>
  • 5. Newton’s Laws
  • 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. Newton’s Laws
  • 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. Newton’s Laws
  • 10. Newton’s Laws
  • 11. Newton’s First Law Overview
  • 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. 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 21. Newton’s Laws <ul><li>Statue sitting on a table </li></ul>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 43. Momentum
  • 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>
  • 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>
  • 46. Arabic guy gun demo - momentum
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 52. Momentum <ul><li>Re-arranging so all quantities before the collision are on one side of the equation: </li></ul>
  • 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>
  • 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>
  • 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>
  • 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>
  • 57. Momentum <ul><li>Watch as a 2 kg brick is dropped on a 3 kg loaded cart. </li></ul>
  • 58. Momentum <ul><li>What changes when the cart is only 1 kg and the brick is 2 kg? </li></ul>
  • 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>
  • 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>
  • 61. Solution Before After v neutron v neutron v nitrogen m neutron m neutron m nitrogen
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 67. Solution 53.1 o truck car
  • 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>
  • 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>
  • 70. Solution
  • 71. Solution <ul><li>u c = 20.02 m s -1 </li></ul>
  • 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>
  • 73. Solution – Part (b) <ul><li>= 9.09 m s -1 </li></ul><ul><li>32.7 km h -1 </li></ul>
  • 74. Impulse <ul><li>Application of Newton I I </li></ul><ul><li>During collisions, objects are deformed . </li></ul>
  • 75. Impulse <ul><li>F = m a </li></ul><ul><li>For constant acceleration : </li></ul>
  • 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>
  • 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>
  • 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>
  • 79. Newton’s 2 nd Law Revisited
  • 80. Newton’s 2 nd Law Revisited <ul><li>F net = m a </li></ul><ul><li>Impulse Question </li></ul>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 102. Example <ul><li>Show that the collision between the neutron and the nitrogen nucleus in the earlier example is elastic. </li></ul>
  • 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>
  • 104. Solution
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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>
  • 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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