1. AQA GCSE Physics 2-3 Work, Energy & Momentum GCSE Physics pages 146 to 159 July 2010

2. AQA GCSE Specification WORK & ENERGY 12.3 What happens to the movement energy when things speed up or slow down? Using skills, knowledge and understanding of how science works: to discuss the transformation of kinetic energy to other forms of energy in particular situations. Skills, knowledge and understanding of how science works set in the context of: When a force causes a body to move through a distance, energy is transferred and work is done. Work done = energy transferred. The amount of work done, force and distance are related by the equation: work done = force applied × distance moved in direction of force Work done against frictional forces is mainly transformed into heat. Elastic potential is the energy stored in an object when work is done on the object to change its shape. The kinetic energy of a body depends on its mass and its speed. HT Calculate the kinetic energy of a body using the equation: kinetic energy = ½ × mass × speed 2 MOMENTUM 12.4 What is momentum? Using skills, knowledge and understanding of how science works: to use the conservation of momentum (in one dimension) to calculate the mass, velocity or momentum of a body involved in a collision or explosion to use the ideas of momentum to explain safety features. Skills, knowledge and understanding of how science works set in the context of: Momentum, mass and velocity are related by the equation: momentum = mass × velocity Momentum has both magnitude and direction. When a force acts on a body that is moving, or able to move, a change in momentum occurs. Momentum is conserved in any collision/explosion provided no external forces act on the colliding/exploding bodies. HT Force, change in momentum and time taken for the change are related by the equation: force = change in momentum / time taken for the change

3. Work When a force causes a body to move through a distance, energy is transferred and work is done. Work done = energy transferred. Both work and energy are measured in joules (J).

4. Work and friction Work done against frictional forces is mainly transformed into heat. Rubbing hands together causes them to become warm. Brakes pads become hot if they are applied for too long. In this case some of the car’s energy may also be transferred to sound in the form of a ‘squeal’

5. The work equation The amount of work done, force and distance are related by the equation: work done = force applied × distance moved in the direction of the force Work is measured in joules (J) Force is measured in newtons (N) Distance is measured in metres (m)

6. also: force = work done ÷ distance moved and: distance = work done ÷ force force distance work

7. Question 1 Calculate the work done when a force of 5 newtons moves through a distance of 3 metres. work = force x distance = 5N x 3m work = 15 joules

8. Question 2 Calculate the work done when a force of 6 newtons moves through a distance of 40 centimetres. work = force x distance = 6 N x 40 cm = 6 N x 0.40 m work = 2.4 joules

9. Question 3 Calculate the value of the force required to do 600 joules of work over a distance of 50 metres. work = force x distance becomes: force = work done ÷ distance = 600 J ÷ 50 m force = 12 newtons

10. Question 4 Calculate the distance moved by a force of 8 newtons when it does 72 joules of work. work = force x distance becomes: distance = work done ÷ force = 72 J ÷ 8 N distance moved = 9 metres

11. Question 5 Calculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cm. work = force x distance The child must exert an upward force equal to its own weight. Therefore: force = 300N This force is exerted upwards and so the distance must also be measured upwards. = (12 x 20cm) = 2.4m therefore: work = 300 N x 2.4 m work = 720 J

12. Question 6 Calculate the work done by a person of mass 80kg who climbs up a set of stairs consisting of 25 steps each of height 10cm. work = force x distance the person must exert an upward force equal their weight the person’s weight = (80kg x 10N/kg) = 800N the distance moved upwards equals (10 x 25cm) = 2.5m work = 800 N x 2.5 m work = 2000 J

13. Complete workforcedistance J50 N3 m 800 J N20 m 500 J250 N m kJ4000 N2 m 2 MJ3.03 N5 km 150 40 2 80 400 Answers

14. Choose appropriate words to fill in the gaps below: Work is done when a _______ moves through a distance. The amount of _______ transferred is also equal to the work done. When a car brakes energy is transformed to ______. Work done is ______ to the force _________ by the distance moved in the __________ of the force. The work done is measured in ______ if the force is measured in newtons and the _________ in metres. multiplieddistancejoulesequalforceenergydirection WORD SELECTION: heat multiplied distance joules equal force energy direction heat

15. Energy and work Notes questions from pages 146 & 147 1.What is meant by ‘work’? 2.Copy both of the equations for work on page 146 along with the units used. 3.Copy and answer questions (a) and (b) on pages 146 and 147. 4.Explain two ways in which the force of friction causes energy to be transformed into heat. 5.Copy the Key Points on page 147. 6.Answer the summary questions on page 147.

16. Energy and work ANSWERS In text questions: (a)To the surroundings as heat energy and sound energy (b)300 J Summary questions: 1.(a) 96 J (b) 96 J 2.(a) (i) 90 J (ii) 4500 J (b) 0.60 m

17. Potential energy Elastic potential energy is the energy stored in an object when work is done on an object to change its shape. An elastic object regains its shape after being stretched or squashed. Elastic potential energy is stored in the bow string when it is pulled by the archer.

18. Gravitational potential energy is the energy stored in an object when work is done in moving the object upwards. The potential energy stored is equal to the weight of the object multiplied by the height lifted. The weightlifter stores gravitational potential energy when he lifts the weights.

19. Kinetic energy Kinetic energy is the energy possessed by a body because of its speed and mass. kinetic energy = ½ x mass x (speed) 2 kinetic energy is measured in joules (J) mass is measured in kilograms (kg) speed is measured in metres per second (m/s)

20. Question 1 Calculate the kinetic energy of a car of mass 1000kg moving at 5 m/s. kinetic energy = ½ x mass x (speed) 2 kinetic energy = ½ x 1000kg x (5m/s) 2 kinetic energy = ½ x 1000 x 25 kinetic energy = 500 x 25 kinetic energy = 12 500 joules

21. Question 2 Calculate the kinetic energy of a child of mass 60kg moving at 3 m/s. kinetic energy = ½ x mass x (speed) 2 k.e. = ½ x 60kg x (3m/s) 2 k.e. = ½ x 60 x 9 k.e. = 30 x 9 kinetic energy = 270 J

22. Question 3 Calculate the kinetic energy of a apple of mass 200g moving at 12m/s. kinetic energy = ½ x mass x (speed) 2 k.e. = ½ x 200g x (12m/s) 2 k.e. = ½ x 0.200kg x 144 k.e. = 0.100 x 144 kinetic energy = 14.4 J

23. Question 4 Calculate the mass of a train if its kinetic energy is 2MJ when it is travelling at 4m/s. kinetic energy = ½ x mass x (speed) 2 2MJ = ½ x mass x (4m/s) 2 2 000 000J = ½ x mass x 16 2 000 000 = 8 x mass 2 000 000 ÷ 8 = mass mass = 250 000 kg

24. Question 5 Calculate the speed of a car of mass 1200kg if its kinetic energy is 15 000J. kinetic energy = ½ x mass x (speed) 2 15 000J = ½ x 1200kg x (speed) 2 15 000 = 600 x (speed) 2 15 000 ÷ 600 = (speed) 2 25 = (speed) 2 speed =  25 speed = 5 m/s

25. Question 6 Calculate the speed of a ball of mass 400g if its kinetic energy is 20J. kinetic energy = ½ x mass x (speed) 2 20J = ½ x 400g x (speed) 2 20 = ½ x 0.400kg x (speed) 2 20 = 0.200 x (speed) 2 20 ÷ 0.200 = (speed) 2 100 = (speed) 2 speed =  100 speed = 10 m/s

26. Complete kinetic energymassspeed J4 kg2 m/s 27 J kg3 m/s 1000 J80 kg m/s kJ200 kg8 m/s 3.2 J3.03g4 m/s 8 6 5 6.4 400 Answers

27. Choose appropriate words to fill in the gaps below: Elastic ________ energy is the energy stored when an object is stretched or ________. This energy is released when the object ________ to its original shape. Kinetic energy is the energy possessed by an object due to its _______ and mass. If the mass of an object is ________ its kinetic energy doubles. If the speed is doubled the kinetic energy will increase by ______ times. When a __________ elastic band is released elastic potential energy is converted into _________ energy. doubledkineticstretchedreturnspotentialsquashedfour WORD SELECTION: speed doubled kinetic stretched returns potential squashed four speed

28. Kinetic energy Notes questions from pages 148 & 149 1.How can gravitational potential energy be calculated? (see the practical on page 148) 2.Copy the equation for kinetic energy at the top of page 149 along with the units used. 3.Repeat the calculation below the kinetic energy equation but this time with a mass of 400 kg moving at a speed of 8 m/s. 4.What does ‘elastic’ mean? 5.What is elastic potential energy? 6.Copy and answer question (b) on page 149. 7.Copy the Key Points on page 149. 8.Answer the summary questions on page 149.

29. Kinetic energy ANSWERS In text question: (b) Heat energy transferred to the surroundings, the foot and the shoe; also sound energy. Summary questions: 1. (a) (i) Chemical energy from the loader is transferred into elastic potential energy of the catapult and some is wasted as heat energy. (ii) Elastic potential energy in the catapult is transformed into kinetic energy of the object and the rubber band and heat energy (plus a little sound energy). (b) (i) 10 J (ii) 10 J 2. (a) 3800 N (b) Friction due to the brakes transforms it from kinetic energy of the car to heat energy in the brakes. (c) 800 kg

30. Momentum momentum = mass x velocity mass is measured in kilograms (kg) velocity is measured in metres per second (m/s) momentum is measured in: kilogram metres per second (kg m/s)

31. Momentum has both magnitude and direction. Its direction is the same as the velocity. The greater the mass of a rugby player the greater is his momentum

32. Question 1 Calculate the momentum of a rugby player, mass 120kg moving at 3m/s. momentum = mass x velocity = 120kg x 3m/s momentum = 360 kg m/s

33. Question 2 Calculate the mass of a car that when moving at 25m/s has a momentum of 20 000 kg m/s. momentum = mass x velocity becomes: mass = momentum ÷ velocity = 20000 kg m/s ÷ 25 m/s mass = 800 kg

34. Complete momentummassvelocity kg m/s50 kg3 m/s 160 kg m/s kg20 m/s 1500 kg m/s250 kg m/s kg m/s500 g8 m/s 3 kg m/skg50 cm/s 150 8 6 4 6 Answers

35. Momentum conservation Momentum is conserved in any collision or explosion provided no external forces act on the colliding or exploding bodies. The initial momentum of the yellow car has been conserved and transferred to the red car

36. Question 1 A truck of mass 0.5kg moving at 1.2m/s collides and remains attached to another, initially stationary truck of mass 1.5kg. Calculate the velocity of the trucks after the collision.

37. total momentum before collision momentum = mass x velocity 0.5 kg truck: = 0.5 kg x 1.2 m/s = 0.6 kg m/s 1.5 kg truck: = 1.5 kg x 0 m/s = 0 kg m/s total initial momentum = 0.6 kg m/s Momentum is conserved in the collision so total momentum after collision = 0.6 kg m/s total momentum = total mass x velocity 0.6 kg m/s = 2.0 kg x velocity 0.6 ÷ 2.0 = velocity velocity = 0.3 m/s

38. Question 2 A train wagon of mass 800 kg moving at 4 m/s collides and remains attached to another wagon of mass 1200 kg that is moving in the same direction at 2 m/s. Calculate the velocity of the wagons after the collision.

39. total momentum before collision momentum = mass x velocity 800 kg wagon: = 800 kg x 4 m/s = 3200 kg m/s 1200 kg truck: = 1200 kg x 2 m/s = 2400 kg m/s total initial momentum = 5600 kg m/s Momentum is conserved in the collision so total momentum after collision = 5600 kg m/s total momentum = total mass x velocity 5600 kg m/s = 2000 kg x velocity 5600 ÷ 2000 = velocity velocity = 2.8 m/s

40. Choose appropriate words to fill in the gaps below: The momentum of an object is equal to its ______ multiplied by its velocity. Momentum has _________, the same as the velocity, and is measured in kilogram _______ per second. In any interaction of bodies, where no external _______ act on the bodies, __________ is conserved. In snooker, a head-on collision of a white ball with a red ball can result in the red ball moving off with the ______ initial velocity of the white ball. This is an example of momentum ____________. momentum forces metresmass directionsame WORD SELECTION: conservation momentum forces metres mass direction same conservation

41. Momentum Notes questions from pages 214 & 215 1.What is ‘momentum’? 2.Copy out the equation at the top of page 214. State the units for each quantity in the equation. 3.Copy and answer question (a) on page 214. 4.Under a heading “Conservation of momentum” copy out the statement in bold at the bottom of page 214. 5.Copy out the worked example on page 215. 6.Copy and answer question (b) on page 215. 7.Copy the Key Points on page 215. 8.Answer the summary questions on page 215.

42. Momentum ANSWERS In text questions: (a)240 kg m/s (b)0.48 m/s Summary questions: 1.(a) mass, velocity (b) momentum, force 2. (a) 5000 kg m/s (b) velocity = momentum / mass = 5000 / 2500 = 2.0 m/s

43. Head-on collisions In this case bodies are moving in opposite directions. Momentum has direction. One direction is treated as positive, the other as negative. In calculations the velocity of one of the colliding bodies must be entered as a NEGATIVE number. + ve velocity - ve velocity NEGATIVEPOSITIVE DIRECTION OF MOTION

44. Question 1 A car of mass 1000 kg moving at 20 m/s makes a head-on collision with a lorry of mass 2000 kg moving at 16 m/s. Calculate their common velocity after the collision if they remain attached to each other. NEGATIVEPOSITIVE DIRECTION OF MOTION car, mass 1000kg lorry, mass 2000kg 16 m/s20 m/s

45. total momentum before collision momentum = mass x velocity car: = 1000 kg x +20 m/s = +20000 kg m/s lorry: = 2000 kg x -16 m/s = -32000 kg m/s total initial momentum = -12000 kg m/s Momentum is conserved in the collision so total momentum after collision = -12000 kg m/s total momentum = total mass x velocity -12000 kg m/s = 3000 kg x velocity -12000 ÷ 3000 = velocity common velocity = - 4 m/s The lorry/car combination will move in the negative direction (to the left in this case) with a common velocity of 4 m/s.

46. Question 2 A car of mass 1000 kg moving at 30 m/s makes a head-on collision with a lorry of mass 2000 kg moving at 15 m/s. Calculate their common velocity after the collision if they remain attached to each other. NEGATIVEPOSITIVE DIRECTION OF MOTION car, mass 1000kg lorry, mass 2000kg 15 m/s30 m/s

47. total momentum before collision momentum = mass x velocity car: = 1000 kg x +30 m/s = +30000 kg m/s lorry: = 2000 kg x -15 m/s = -30000 kg m/s total initial momentum = 0 kg m/s Momentum is conserved in the collision so total momentum after collision = 0 kg m/s The lorry/car combination will not move after the collision.

48. Explosions Before an explosion the total momentum is zero. As momentum is conserved, the total momentum afterwards must also be zero. This means that the different parts of the exploding body must move off in different directions.

49. Question 1 An artillery gun of mass 1500kg fires a shell of mass 20kg at a velocity of 150m/s. Calculate the recoil velocity of the gun. NEGATIVEPOSITIVE DIRECTION OF MOTION shell, mass 20kg 150 m/s recoil artillery gun, mass 1500kg

50. The total momentum before and after the explosion is ZERO momentum = mass x velocity shell: = 20 kg x +150 m/s = +3000 kg m/s This must cancel the momentum of the gun. Therefore the gun’s momentum must be -3000 kg m/s gun: = 1500 kg x recoil velocity = -3000 kg m/s recoil velocity = - 3000 ÷ 1500 = - 2m/s The gun will recoil (move to the left) with a velocity of 2 m/s.

51. Question 2 A girl of mass 60kg throws a boy, mass 90kg out off a swimming pool at a velocity of 2m/s. What is the girl’s recoil velocity? NEGATIVEPOSITIVE DIRECTION OF MOTION recoil girl, mass 60kg 2 m/s boy, mass 90kg recoil girl, mass 60kg 2 m/s boy, mass 90kg

52. The total momentum before and after throwing the boy is ZERO momentum = mass x velocity boy: = 90 kg x +2 m/s = +180 kg m/s This must cancel the momentum of the girl. Therefore the girl’s momentum must be -180 kg m/s gun: = 60 kg x recoil velocity = -180 kg m/s recoil velocity = - 180 ÷ 60 = - 3m/s The girl will recoil (move to the left) with a velocity of 3 m/s.

53. More on collisions and explosions Notes questions from pages 152 & 153 1.Apart from size what other property does momentum have? 2.Copy and answer question (a) on page 152. 3.Explain how conservation of momentum applies in an explosion. 4.Why do guns recoil? 5.Copy and answer question (b) on page 153. 6.Copy the Key Points on page 153. 7.Answer the summary questions on page 153.

54. More on collisions and explosions ANSWERS In text questions: (a)The boat and the person who jumps off move away with equal and opposite amounts of momentum. (b)25 m/s Summary questions: 1.(a) momentum (b) velocity (c) force 2.(a) 60 kg m/s (b) 1.5 m/s

55. Force and momentum A force will cause the velocity of an object to change and therefore also its momentum. The greater the force the faster the momentum changes.

56. force is measured in newtons (N) change in momentum is measured in: kilogram metres per second (kg m/s) time is measured in seconds (s) force = time taken for the change change in momentum

57. Equation proof acceleration = velocity change ÷ time taken multiplying both sides of this equation by ‘mass’ gives: (mass x acceleration) = (mass x velocity) change ÷ time but: (mass x acceleration) = force and: (mass x velocity) = momentum therefore: force = momentum change ÷ time taken

58. Question 1 Calculate the force required to change the momentum of a car by 24000 kgm/s over a 6 second period. force = momentum change ÷ time taken = 24000 kgm/s ÷ 6 s force = 4000N

59. Question 2 Calculate the time taken for a force of 6000N to cause the momentum of truck to change by 42000 kgm/s. force = momentum change ÷ time taken becomes: time taken = momentum change ÷ force = 42000 kgm/s ÷ 6000 N force = 7 seconds

60. Complete forcemomentum change time taken N8000 kgm/s40 s 25 N kgm/s20 s 500 N3000 kgm/s s N8000 kgm/s10 s 4 N kgm/s2 minutes 200 500 6 800 480 Answers

61. Car safety features

62. Crumple zones, air bags and a collapsible steering wheel are designed to increase the time taken for a driver or passenger to change momentum to zero during a crash. The equation: force = momentum change ÷ time taken shows that if the time taken is increased for the same momentum change the force exerted is decreased so is the injury to the driver or passenger.

63. Playground flooring question The picture shows rubber tiles used for playground flooring. Explain how these can reduce injury to children. ANSWER: When a child falls to the floor its momentum changes from a high value to zero. The rubber flooring tiles increase the time taken for this change. force = change in momentum ÷ time taken for the change Therefore the force on the child is reduced and so is the potential injury.

64. Choose appropriate words to fill in the gaps below: The force exerted on an object is equal to the __________ change caused divided by the ______ taken for the change. An airbag activates during a car _______. The inflated airbag _________ the time taken for a driver’s or passenger’s ________ to fall to zero. The time taken for their momentum to fall to ______ is also increased. Therefore the _______ exerted on the driver or passenger is __________ and so is the potential ________ caused. velocity increases zero crash momentumtimeforce WORD SELECTION: decreasedinjury velocity increases zero crash momentum time force decreasedinjury

65. Changing momentum Notes questions from pages 154 & 155 1.What is the purpose of a car’s crumple zones? 2.Copy the key points on page 155. 3.Copy and answer questions (a), (b) and (c) on pages 154 and 155. 4.Copy the Key Points on page 155. 5.Answer the summary questions on page 155.

66. Changing momentum ANSWERS In text questions: (a)If a child falls off a swing, the rubber mat reduces the impact force by increasing the impact time when the child hits the ground. (b)The force is bigger (c)1800 N Summary questions: 1.(a) stays the same (b) increases (c) decreases 2. (a) 24 000 kg m/s (b) (i) 2000 N (ii) 800 N

67. Virtual Physics Laboratory Simulations NOTE: Links work only in school Roller Coaster.exeRoller Coaster.exe - Energy considerations

68. Online Simulations WorkWork (GCSE) - Powerpoint presentation by KT Kinetic EnergyKinetic Energy (GCSE) - Powerpoint presentation by KT Gravitational Potential Energy Gravitational Potential Energy (GCSE) - Powerpoint presentation by KT Energy Skate ParkEnergy Skate Park - Colorado - Learn about conservation of energy with a skater dude! Build tracks, ramps and jumps for the skater and view the kinetic energy, potential energy and friction as he moves. You can also take the skater to different planets or even space! Rollercoaster DemoRollercoaster Demo - Funderstanding Energy conservation with falling particlesEnergy conservation with falling particles - NTNU Ball rolling up a slope Ball rolling up a slope - NTNU Pulley SystemPulley System - Fendt BBC AQA GCSE Bitesize Revision: Work, force and distance Potential and kinetic energy Kinetic energy equation

69. Forces for safety Notes questions from pages 156 & 157 1.Answer questions 1 and 2 on pages 156 and 157.

70. Forces for safety ANSWERS 1.The air bag increases the time taken to stop the person it acts on. This reduces the force of the impact. Also, the force is spread out across the chest by the air bag so its effect is lessened again. 2.(a) 26 100 kg m/s (b) 34.8 m/s (c) Yes.

71. How Science Works ANSWERS (a)No. The upright position is slightly greater although there is no significant difference between the three sets of readings. (b)The upright position propels the ball further. This prediction is even stronger if the 2 nd go/front measurement is considered to be an anomaly. (c)No. The measurements have a wide range within each set. There is even overlap of results. (d)Position of the releae point. (e)Categoric. (f)By measuring the angle of the spoon to the upright. (g)More information can be obtained. A graph can also be drawn and a pattern discerned (or not).