Arrows X and Y show two forces acting on the sky-diver as he falls. Y... (2) (1) Which of the forces will be bigger?... (1)......

Transcription

1 Q. A sky-diver jumps from a plane. The sky-diver is shown in the diagram below. (a) Arrows X and Y show two forces acting on the sky-diver as he falls. (i) Name the forces X and Y. X... Y... (2) (ii) Explain why force X acts in an upward direction. () (iii) At first forces X and Y are unbalanced. Which of the forces will be bigger?... () (iv) How does this unbalanced force affect the sky-diver? (2) Page of 66

2 (b) After some time the sky-diver pulls the rip cord and the parachute opens. The sky-diver and parachute are shown in the diagram below. After a while forces X and Y are balanced. Underline the correct answer in each line below. Force X has increased / stayed the same / decreased. Force Y has increased / stayed the same / decreased. The speed of the sky-diver will increase / stay the same / decrease. Page 2 of 66

3 (c) The graph below shows how the height of the sky-diver changes with time. (i) Which part of the graph, AB, BC or CD shows the sky-diver falling at a constant speed?... () (ii) What distance does the sky-diver fall at a constant speed? Distance... m () (iii) How long does he fall at this speed? Time... s () Page 3 of 66

4 (iv) Calculate this speed. Speed... m/s (2) (Total 4 marks) Q2. Five forces, A, B, C, D and E act on the van. (a) Complete the following sentences by choosing the correct forces from A to E. Force... is the forward force from the engine. Force... is the force resisting the van s motion. () (b) The size of forces A and E can change. Complete the table to show how big force A is compared to force E for each motion of the van. Do this by placing a tick in the correct box. The first one has been done for you. MOTION OF VAN FORCE A SMALLER THAN FORCE E FORCE A EQUAL TO FORCE E FORCE A BIGGER THAN FORCE E Not moving Speeding up Constant speed Slowing down Page 4 of 66

5 (c) When is force E zero?... () (d) The van has a fault and leaks one drop of oil every second. The diagram below shows the oil drops left on the road as the van moves from W to Z. Describe the motion of the van as it moves from: W to X... X to Y... Y to Z... (e) The driver and passengers wear seatbelts. Seatbelts reduce the risk of injury if the van stops suddenly. backwards downwards force forwards mass weight Complete the following sentences, using words from the list above, to explain why the risk of injury is reduced if the van stops suddenly. A large... is needed to stop the van suddenly. The driver and passengers would continue to move.... The seatbelts supply a... force to keep the driver and passengers in their seats. (Total marks) Q3. The brick shown in the diagram is being pushed but it is not moving. Page 5 of 66

6 (a) The pushing force does not make the brick move. Explain why.. () (b) The weight of the brick does not make it move downwards. Explain why. () (c) A bigger pushing force does make the brick slide across the table. Write down one thing that the sliding brick will do to the surface of the table.... () (Total 3 marks) Q4. A car travelling along a straight road has to stop and wait at red traffic lights. The graph shows how the velocity of the car changes after the traffic lights turn green. (a) Between the traffic lights changing to green and the car starting to move there is a time delay. This is called the reaction time. Write down one factor that could affect the driver s reaction time. () (b) Calculate the distance the car travels while accelerating. Show clearly how you work out your answer. Distance =...metres Page 6 of 66

7 (c) Calculate the acceleration of the car. Show clearly how you work out your final answer and give the units. Acceleration =... (4) (d) The mass of the car is 900 kg. (i) Write down the equation that links acceleration, force and mass.... () (ii) Calculate the force used to accelerate the car. Show clearly how you work out your final answer Force =... newtons (2) (Total marks) Page 7 of 66

8 Q5. The distance-time graph represents the motion of a car during a race. (a) Describe the motion of the car between point A and point D. You should not carry out any calculations. To gain full marks in this question you should write your ideas in good English. Put them into a sensible order and use the correct scientific words. Page 8 of 66

9 (b) Calculate the gradient of the graph between point B and point C. Show clearly how you get your answer. gradient =... (Total 6 marks) Q6. The diagram shows the horizontal forces acting on a car of mass 200 kg. (a) Calculate the acceleration of the car at the instant shown in the diagram. Write down the equation you use, and then show clearly how you work out your answer and give the unit Acceleration =... (4) Page 9 of 66

10 (b) Explain why the car reaches a top speed even though the thrust force remains constant at 3500 N (c) The diagram shows a car and a van. The two vehicles have the same mass and identical engines. Explain why the top speed of the car is higher than the top speed of the van (4) (Total marks) Page 0 of 66

11 Q7. (a) The diagram shows an aircraft and the horizontal forces acting on it as it moves along a runway. The resultant force on the aircraft is zero. (i) What is meant by the term resultant force? () (ii) Describe the movement of the aircraft when the resultant force is zero. () (b) The aircraft has a take-off mass of kg. Each of the 4 engines can produce a maximum force of 240 kn. Use the equation in the box to calculate the maximum acceleration of the aircraft. resultant force = mass acceleration Show clearly how you work out your answer and give the unit. Acceleration =... Page of 66

12 (c) As the aircraft moves along the runway to take off, its acceleration decreases even though the force from the engines is constant. Explain why. (2) (Total 7 marks) Q8. (a) A car driver takes a short time to react to an emergency before applying the brakes. The distance the car will travel during this time is called the thinking distance. The graph shows how the thinking distance of a driver depends on the speed of the car. (i) What is the connection between thinking distance and speed?... () (ii) Many people drive while they are tired. Draw a new line on the graph to show how thinking distance changes with speed for a tired driver. () Page 2 of 66

13 (iii) The graph was drawn using data given in the Highway Code. Do you think that the data given in the Highway Code is likely to be reliable? Draw a ring around your answer. Yes No Maybe Give a reason for your answer () (b) The distance a car travels once the brakes are applied is called the braking distance. (i) What is the relationship between thinking distance, braking distance and stopping distance?... () (ii) State two factors that could increase the braking distance of a car at a speed of 5 m/s (2) (Total 6 marks) Page 3 of 66

14 Q9. A car driver sees a dog on the road ahead and has to make an emergency stop. The graph shows how the speed of the car changes with time after the driver first sees the dog. (a) Which part of the graph represents the reaction time or thinking time of the driver? (b) (i) What is the thinking time of the driver? Time... seconds () () (ii) Calculate the distance travelled by the car in this thinking time. Distance... m Page 4 of 66

15 (c) Calculate the acceleration of the car after the brakes are applied. Acceleration... (4) (d) Calculate the distance travelled by the car during braking. Distance... m (e) The mass of the car is 800 kg. Calculate the braking force. Braking force... N (Total 5 marks) Page 5 of 66

16 ## The table shows the braking distances for a car at different speeds and kinetic energy. The braking distance is how far the car travels once the brakes have been applied. Braking distance in m Speed of car in m/s Kinetic energy of car in kj (a) A student suggests, the braking distance is directly proportional to the kinetic energy. (i) Draw a line graph to test this suggestion. (ii) Does the graph show that the student s suggestion was correct or incorrect? Give a reason for your answer () Page 6 of 66

17 (iii) Use your graph and the following equation to predict a braking distance for a speed of 35 metres per second (m/s). The mass of the car is 800 kilograms (kg). Show clearly how you obtain your answer. kinetic energy = ½ mv Braking distance =... m (2) (iv) State one factor, apart from speed, which would increase the car s braking distance.... () (b) The diagram shows a car before and during a crash test. The car hits the wall at 4 metres per second (m/s) and takes 0.25 seconds (s) to stop. (i) Write down the equation which links acceleration, change in velocity and time taken.... () (ii) Calculate the deceleration of the car.... Deceleration =... m/s 2 () (iii) In an accident the crumple zone at the front of a car collapses progressively. This increases the time it takes the car to stop. In a front end collision the injury to the car passengers should be reduced. Explain why. The answer has been started for you. By increasing the time it takes for the car to stop, the (2) (Total marks) Page 7 of 66

18 Q. The Highway Code gives tables of the shortest stopping distances for cars travelling at various speeds. An extract from the Highway Code is given below. thinking distance + braking distance = total stopping distance (a) A driver s reaction time is 0.7 s. (i) Write down two factors which could increase a driver s reaction time (2) (ii) What effect does an increase in reaction time have on: A thinking distance;... B braking distance;... C total stopping distance?... (b) Explain why the braking distance would change on a wet road (2) Page 8 of 66

19 (c) A car was travelling at 30 m/s. The driver braked. The graph below is a velocity-time graph showing the velocity of the car during braking. Calculate: (i) the rate at which the velocity decreases (deceleration); Rate... m/s² (2) (ii) the braking force, if the mass of the car is 900 kg; Braking force... N (2) (iii) the braking distance. Braking distance... m (2) (Total 3 marks) Page 9 of 66

20 Q2. (a) The diagram shows a steel ball-bearing falling through a tube of oil. The forces, L and M, act on the ball-bearing. What causes force L? () (b) The distance time graph represents the motion of the ball-bearing as it falls through the oil. Page 20 of 66

21 (i) Explain, in terms of the forces, L and M, why the ball-bearing accelerates at first but then falls at constant speed. (ii) What name is given to the constant speed reached by the falling ball-bearing? () (iii) Calculate the constant speed reached by the ball-bearing. Show clearly how you use the graph to work out your answer. Speed =... m/s (2) (Total 7 marks) Q3. A sky-diver steps out of an aeroplane. After 0 seconds she is falling at a steady speed of 50m/s. She then opens her parachute. After another 5 seconds she is once again falling at a steady speed. This speed is now only 0m/s. Page 2 of 66

22 (a) Calculate the sky-diver s average acceleration during the time from when she opens her parachute until she reaches her slower steady speed. (Show your working.) (b) Explain, as fully as you can: (i) why the sky-diver eventually reaches a steady speed (with or without her parachute). (ii) why the sky-diver s steady speed is lower when her parachute is open. () (c) The sky-diver and her equipment have a total mass of 75kg. Calculate the gravitational force acting on this mass. (Show your working.) Answer... N () (Total 8 marks) Page 22 of 66

23 Q4. A sky-diver jumps from a plane. The sky-diver is shown in the diagram below. (a) Arrows X and Y show two forces acting on the sky-diver as he falls. (i) Name the forces X and Y. X... Y... (2) (ii) Explain why force X acts in an upward direction. () (iii) At first forces X and Y are unbalanced. Which of the forces will be bigger?... () (iv) How does this unbalanced force affect the sky-diver? (2) Page 23 of 66

24 (b) After some time the sky-diver pulls the rip cord and the parachute opens. The sky-diver and parachute are shown in the diagram below. After a while forces X and Y are balanced. Underline the correct answer in each line below. Force X has increased / stayed the same / decreased. Force Y has increased / stayed the same / decreased. The speed of the sky-diver will increase / stay the same / decrease. Page 24 of 66

25 (c) The graph below shows how the height of the sky-diver changes with time. (i) Which part of the graph, AB, BC or CD shows the sky-diver falling at a constant speed?... () (ii) What distance does the sky-diver fall at a constant speed? Distance... m () (iii) How long does he fall at this speed? Time... s () Page 25 of 66

26 (iv) Calculate this speed. Speed... m/s (2) (Total 4 marks) Q5. When a bungee-jump is made the jumper steps off a high platform. An elastic cord from the platform is tied to the jumper. The diagram below shows different stages in a bungee-jump. Forces A, B and C are forces acting on the jumper at each stage. moving down moving down moving down large acceleration small acceleration slowing acceleration diagram X diagram Y diagram Z (a) Name force A.... () Page 26 of 66

27 (b) The motion of the jumper is shown in the diagrams. By comparing forces A, B and C, state how the motion is caused in: (i) diagram X;. (ii) diagram Y;. (iii) diagram Z.. (c) The table gives results for a bungee cord when it is being stretched. STRETCHING FORCE (N) LENGTH OF CORD (m) (i) Plot a graph of these results on the graph paper. Page 27 of 66

28 (ii) Use the graph to find the length of the cord before it was stretched. Length... m () (Total 8 marks) Q6. In bungee jumping, a fixed rubber cord is fastened to the jumper s ankles. The graph shows how the bungee jumper s velocity changes during part of the jump. (a) Calculate the acceleration of the bungee jumper between 2 and 4 seconds. Show your working. Acceleration =... m/s 2 Page 28 of 66

29 (b) Describe, in as much detail as you can, what happens to the bungee jumper after 4 seconds. (Total 6 marks) Q7. A racing driver is driving his car along a straight and level road as shown in the diagram below. (a) The driver pushes the accelerator pedal as far down as possible. The car does not accelerate above a certain maximum speed. Explain the reasons for this in terms of the forces acting on the car. (4) Page 29 of 66

30 (b) The racing car has a mass of 250 kg. When the brake pedal is pushed down a constant braking force of N is exerted on the car. (i) Calculate the acceleration of the car. (ii) Calculate the kinetic energy of the car when it is travelling at a speed of 48 m/s. (iii) When the brakes are applied with a constant force of N the car travels a distance of 44 m before it stops. Calculate the work done in stopping the car. (2) (Total 6 marks) Q8. The diagram shows a high jumper. In order to jump over the bar, the high jumper must raise his mass by.25 m. The high jumper has a mass of 65 kg. The gravitational field strength is 0 N/kg. Page 30 of 66

31 (a) The high jumper just clears the bar. Use the following equations to calculate the gain in his gravitational potential energy. weight = mass gravitational field strength (newton, N) (kilogram, kg) (newton/kilogram, N/kg) change in gravitational potential energy = weight change in vertical height (joule, J) (Newton, N) (metre, m) Gain in gravitational potential energy... J (4 (b) Use the following equation to calculate the minimum speed the high jumper must reach for take-off in order to jump over the bar. kinetic energy = mass [speed] 2 (joule, J) (kilogram, kg) [(metre/second) 2, (m/s) 2 Speed... m/s (Total 7 marks) Page 3 of 66

32 Q9. A car is driven along a straight, snow covered, road. The graph shows how the velocity of the car changes from the moment the driver sees a very slow moving queue of traffic ahead. (a) Use the graph to calculate the distance the car travels while it is slowing down. Show clearly how you work out your answer Distance =... m (b) The car has a mass of 200 kg. Calculate the kinetic of the car when it travels at a speed of 2 m/s. Write down the equation you use, and then show clearly how you work out your answer Kinetic energy =... J (2) (Total 5 marks) Page 32 of 66

33 Q20. A crane is used to lift a steel girder to the top of a high building. When it is lifted by the crane: the girder accelerates from rest to a speed of 0.6 m/s in the first 3 seconds; it then rises at a steady speed. (a) Calculate the acceleration of the girder. (Show your working.) (b) (i) What is the weight of the steel girder? Answer... N () (ii) Calculate the power of the crane motor as it lifts the girder at a steady speed of 0.6 m/s. (Show your working. You can ignore the weight of the cable and hook which is small compared to the weight of the girder.) Answer... W (2) Page 33 of 66

34 (c) A new motor is fitted to the crane. This motor accelerates the girder at 0.3 m/s 2. Calculate the force which the crane applies to the girder to produce this acceleration. (Show your working.) Answer... N (Total 9 marks) Q2. The diagram shows a child on a playground swing. The playground has a rubber safety surface. (a) The child, with a mass of 35 kg, falls off the swing and hits the ground at a speed of 6 m/s. (i) Use the equation in the box to calculate the momentum of the child as it hits the ground. momentum = mass velocity Show clearly how you work out your answer and give the unit. Momentum =... Page 34 of 66

35 (ii) After hitting the ground, the child slows down and stops in 0.25 s. Use the equation in the box to calculate the force exerted by the ground on the child. force = Show clearly how you work out your answer. Force =... N (2) (b) The diagram shows the type of rubber tile used to cover the playground surface. Explain how the rubber tiles reduce the risk of children being seriously injured when they fall off the playground equipment. Page 35 of 66

36 (c) The critical fall height is the height that a child can fall and not be expected to sustain a life-threatening head injury. A new type of tile, made in a range of different thicknesses, was tested in a laboratory using test dummies and the critical fall height measured. Only one test was completed on each tile. The results are shown in the graph. The critical fall height for playground equipment varies from 0.5 m to 3.0 m. Suggest two reasons why more tests are needed before this new type of tile can be used in a playground (2) (d) Developments in technology allow manufacturers to make rubber tiles from scrap car tyres. Suggest why this process may benefit the environment. () (Total marks) Page 36 of 66

37 Q22. The roads were very icy. An accident was recorded by a security camera. Car A was waiting at a road junction. Car B, travelling at 0 m/s, went into the back of car A. This reduced car B s speed to 4 m/s and caused car A to move forward. The total mass of car A was 200 kg and the total mass of car B was 500 kg. (i) Write down the equation, in words, which you need to use to calculate momentum. () (ii) Calculate the change in momentum of car B in this accident. Show clearly how you work out your final answer and give the unit. Change in momentum =... (iii) Use your knowledge of the conservation of momentum to calculate the speed, in m/s, of car A when it was moved forward in this accident. Show clearly how you work out your final answer. Speed =... m/s (Total 7 marks) Page 37 of 66

38 Q23. In an experiment at an accident research laboratory, a car driven by remote control was crashed into the back of an identical stationary car. On impact the two cars joined together and moved in a straight line. (a) The graph shows how the velocity of the remote-controlled car changed during the experiment. (i) How is the velocity of a car different from the speed of a car? () (ii) Use the graph to calculate the distance travelled by the remote-controlled car before the collision. Show clearly how you work out your answer. Distance =... m (2) Page 38 of 66

39 (iii) Draw, on the grid below, a graph to show how the velocity of the second car changed during the experiment. (2) (iv) The total momentum of the two cars was not conserved. What does this statement mean? () Page 39 of 66

40 (b) The graph line shows how the force from a seat belt on a car driver changes during a collision. Scientists at the accident research laboratory want to develop a seat belt that produces a constant force throughout a collision. Use the idea of momentum to explain why this type of seat belt would be better for a car driver. (2) (Total 8 marks) Q24. (a) When two objects collide, and no other forces act, then conservation of momentum applies. (i) What does the term conservation of momentum mean? (2) (ii) Apart from collisions and similar events, give another type of event in which conservation of momentum applies. () (iii) Write, in words, the equation which you need to use to calculate momentum. () Page 40 of 66

41 (iv) The diagram shows a straight and horizontal runway and two trolleys, X and Y, which can move on the runway. X has a mass of 0.2 kg and its velocity is.2 m/s to the right. Y has a mass of 0. kg and is stationary. When X collides with Y they stick together. Calculate the velocity of the trolleys after the collision. Show clearly how you work out your answer and give the unit and direction. Velocity of the trolleys =... (5) (v) What assumption did you make in order to calculate your answer to part (a)(iv)? () (b) Just before it hits a target, a bullet has a momentum of 5 kg m/s. It takes s for the target to stop the bullet. Calculate the force, in newtons, needed to do this. Write, in words, the equation that you will need to use and show clearly how you work out your answer. Force =... newtons (Total 3 marks) Page 4 of 66

42 Q25. (a) The picture shows two ice hockey players skating towards the puck. The players, travelling in opposite directions, collide, fall over and stop. (i) Use the following equation and the data given in the box to calculate the momentum of player number 3 before the collision. Show clearly how you work out your answer and give the unit. momentum = mass velocity Momentum of player 3 =... (ii) What is the momentum of player 4 just before the collision?... () (iii) The collision between the two players is not elastic. What is meant by an elastic collision? () Page 42 of 66

43 (b) The pictures show what happened when someone tried to jump from a stationary rowing boat to a jetty. Use the idea of momentum to explain why this happened (2) (c) The diagram shows one type of padded body protector which may be worn by a horse rider. Page 43 of 66

44 If the rider falls off the horse, the body protector reduces the chance of the rider being injured. Use the idea of momentum to explain why (Total 0 marks) Q26. The drawing below shows two railway trucks A and B, moving in the same direction. Truck A, of mass 500 kg, is initially moving at a speed of 8 m/s. Truck B, of mass 2000 kg, is initially moving at a speed of m/s. Truck A catches up and collides with truck B. The two trucks become coupled together as shown in the diagram. (a) Calculate: (i) the initial momentum of truck A.... momentum... kg m/s (ii) the initial momentum of truck B.... momentum... kg m/s Page 44 of 66

45 (iii) the total momentum of the trucks before the collision.... total momentum... kg m/s (6) (b) Calculate the speed of the coupled trucks after the collision. (5) (c) (i) How is the total kinetic energy of the trucks changed as a result of the collision? A calculated answer is not needed for full marks. (ii) State an energy transfer which accounts for part of the change in the total kinetic energy of the trucks during the collision. (iii) What would have been the effect on the change of total kinetic energy of the trucks if the collision had been more elastic? (Total 4 marks) Page 45 of 66

46 M. (a) (i) air resistance/drag/friction (or upthrust) weight/gravitational pull/gravity for mark each (ii) (iii) (iv) air resistance/friction acts in opposite direction to motion Y the sky-diver accelerates/his speed increases in downward direction/towards the Earth/falls for mark each 2 (b) force X has increased force Y has stayed the same the speed of the sky-diver will stay the same for mark each 3 (c) (i) CD 3 (iv) 0 (but apply e.c.f. from (ii) and (iii)) gets 2 marks or 500/50 or d/t gets mark 2 [4] M2. (a) A then E for one mark (b) A > E A = E A < E in this order for mark each 3 (c) when van stops / is stationary / is parked for one mark Page 46 of 66

47 (d) WX slowing down (owtte) XY constant speed (owtte) YZ speeding up (owtte) for mark each 3 (e).. force. forwards. backwards for mark each 3 [] M3. (a) idea that balanced by friction force* / pushing force equals friction force (*note balanced by unspecified force) or specification of relevant force but no reference to balancing in both (a) and (b) gains mark overall for mark (b) balanced by upwards force of table* for mark (c) makes it (slightly) warm / hot or wears it away (slightly) / damages surface for mark [3] M4. (a) concentration / tiredness / drugs / alcohol accept any reasonable factor that could affect a driver s reactions do not accept speed or any physical condition unrelated to the driver (b) 3.25 credit for mark correct attempt to calculate the area under the slope or for using the equation distance = average velocity (speed) time credit for mark use of correct velocity change (2.5) and correct time (5) or answer of Page 47 of 66

48 (c) 2.5 credit for mark triangle drawn on slope or correct equation or two correct pairs of coordinates credit for mark use of correct velocity change (2.5) and correct time (5) accept time = between 4.8 and 5.2 if used in (b) do not accept an attempt using one pair of coordinates taken from the slope 3 metres / second / second or metres / second / squared or m/s 2 or ms 2 (d) (i) force = mass acceleration accept correct transformation accept F = m a accept provided subsequent use of Δ is correct do not accept an equation in units (ii) 2250 credit their (c) 900 for 2 marks credit mark for correct substitution 2 [] M5. (a) Quality of written communication A B constant speed for correct use of term speed in all correct examples Q Q describes all 3 sections correctly for 2 marks describes 2 or section correctly for mark do not accept pace for speed B C (has accelerated) to a higher (constant) speed C D goes back to original / lower (constant) speed allow for mark, initial and final (constant) speeds are the same accept velocity for speed ignore reference to direction max 2 Page 48 of 66

49 (b) 62.5 allow answer to 2 s.f. allow mark for drawing a correct triangle or for using two correct pairs of coordinates allow mark for correct use of y/x ignore units 3 [6] M6. (a).25 allow mark for correct resultant force ie 500N allow 2 marks for correct transformation and substitution ie allow mark for a correct transformation but clearly substituting an incorrect value for force eg = 3 m/s 2 (b) (c) as speed increases so does the size of the drag force accept frictional force / resistive force / air resistance for drag eventually the drag force becomes equal to the thrust the resultant force is now equal to zero and therefore there is no further acceleration the car and van will reach top speed when the forward force equals the drag force accept air resistance / frictional / resistive force for drag force the drag force at any speed is smaller for the car than for the van as the car is more streamlined therefore the car s drag force will equal the forward force at a higher speed allow converse throughout [] Page 49 of 66

50 M7. (a) (i) a single force that has the same effect as all the forces combined accept all the forces added / the sum of the forces / overall force (ii) constant speed (in a straight line) do not accept stationary or constant velocity (b) 3 allow mark for correct substitution into transformed equation accept answer gains mark answer = 0.75 gains mark 2 m/s 2 (c) as speed increases air resistance increases accept drag / friction for air resistance reducing the resultant force [7] M8. (a) (i) as one goes up so does the other or (directly) proportional accept change by the same ratio (ii) steeper straight line through the origin judge by eye (iii) Yes with reason eg data would have been checked / repeated accept produced by a reliable/ official/ government source do not accept it needs to be reliable or No with reason eg does not apply to all conditions / cars / drivers or are only average values or Maybe with a suitable reason eg cannot tell due to insufficient information Page 50 of 66

51 (b) (i) stopping distance = thinking distance + braking distance (ii) any two from: factors must be to do with increasing braking distance smooth road / loose surface rain / snow / ice accept wet road/ petrol spills do not accept condition of road unless suitably qualified badly maintained brakes accept worn brakes accept bad/ worn/ rusty brakes do not accept old brakes worn tyres accept bald tyres accept lack of grip on tyres do not accept old tyres downhill slope/gradient heavily loaded car 2 [6] M9. (a) AB for mark (b) (i) 0.7 for mark each (ii) 6.8 gains 2 marks 2 but correct working (d = v.t, d = , or in terms of area under graph) gains mark (c) a = (v-u)/t = 24/4 = 6 m/s 2 (see marking of calculations) (can work in terms of graph gradient) 4 Page 5 of 66

52 (d) d = v.t = 24/2 4 = 48 (see marking of calculations) (can work in terms of area under graph) 3 (e) F = ma = = 4800 (see marking of calculations) 3 [5] M0. (a) (i) linear scales used do not credit if less than half paper used points plotted correctly all of paper used (straight) line of best fit drawn allow a tolerance of half square (ii) correct and straight line through origin all needed e.c.f. if their (a)(i) is straight but not through the origin - incorrect because line does not go through origin credit a calculation that shows proportionality (iii) 62 ± 0.5 (m) credit mark for KE = or 490kJ credit mark for correct use of graph clearly shown 2 Page 52 of 66

53 (iv) any one from: wet or icy or worn or smooth road accept slippery slope brakes worn accept faulty brakes car heavily loaded worn tyres downhill slope do not accept anything to do with thinking distance e.g. driver tired or drunk (b) (i) acceleration = accept correct transformation accept accept m/s = 2 do not accept acceleration = (ii) 56 accept 56 (iii) deceleration is reduced accept deceleration is slower accept acceleration force on car and or passengers is reduced accept an answer in terms of change in momentum for full credit [] M. (a) (i) tiredness / boredom drugs alcohol distraction any two for mark each 2 Page 53 of 66

54 (ii) A greater / longer B no effect C greater / longer each for mark 3 (b) on a wet road: there is less friction / grip for mark braking distance is greater / takes longer to stop or car skids / slides forward for mark 2 (c) (i) deceleration = gradient or 30 / 4.8 each for mark (ii) force = mass acceleration or each for mark 2 2 (iii) distance = area under graph or or average speed time or Accept answer in terms of change in k.e. = work done if incorrect unit given (eg 72km) then no mark each for mark 2 [3] M2. (a) gravity accept weight do not accept mass accept gravitational pull (b) (i) Initially force L greater than force M accept there is a resultant force downwards (as speed increases) force M increases accept the resultant force decreases when M = L, (speed is constant) accept resultant force is 0 accept gravity/weighty for L accept drag/ upthrust/resistance/friction for M do not accept air resistance for M but penalise only once Page 54 of 66

55 (ii) terminal velocity (iii) 0.5 accept an answer between an answer of 0. gains no credit allow mark for showing correct use of the graph 2 [7] ## (a) evidence of or gains mark (credit 50/0 or 5 with mark) NOT 40/0 or 50/5 but 8 [N.B. negative not required] gains 2 marks units metres per second per second or (metres per second squared or m/s²) for mark 3 (b) (i) idea that accelerates at first due to gravity air/wind resistance friction/resistance/drag with air increases with speed eventually gravity and friction cancel balance or (no net/accelerating force) [NOT terminal velocity] each for mark 3 (ii) idea a bigger resistance/friction/drag at any given speed (credit a bigger drag (factor)) for mark (c) evidence of 0 / 9.8 / 9.8 or 750/735(75) for mark [8] M4. (a) (i) air resistance/drag/friction (or upthrust) weight/gravitational pull/gravity for mark each Page 55 of 66

56 (ii) (iii) (iv) air resistance/friction acts in opposite direction to motion Y the sky-diver accelerates/his speed increases in downward direction/towards the Earth/falls for mark each 2 (b) force X has increased force Y has stayed the same the speed of the sky-diver will stay the same for mark each 3 (c) (i) CD 3 (iv) 0 (but apply e.c.f. from (ii) and (iii)) gets 2 marks or 500/50 or d/t gets mark 2 [4] M5. (a) weight or gravity or gravitational for mark (b) (i) only force A acts / force A > air resistance / gravity / weight for mark (ii) (iii) force A > force B for mark force C > force A for mark (Forces A, B and C need not be used, description of forces are OK) Page 56 of 66

57 (c) (i) graph points all correct ± little square gains 2 marks one point wrong gains mark 2+ points wrong gains 0 mark appropriate line good freehand OK gains mark Bar chart gets 0, but if points clear can get 2 3 (ii) 6 or candidates own intercept should be 6 m in range -9 if no kinks on graph line for mark [8] M6. (a) 7.5 correct answer with no working = 3 if incorrect allow mark for (change in velocity from graph =) 5 mark for 2 marks for N.B. correct answer from the incorrectly recalled relationship (b) = 2 marks (4 5 seconds) the bungee jumper slows down (decelerates) (the rubber cord) stops the fall 3 (5 6 seconds) the bungee jumper starts moving (accelerating) upwards (in the opposite direction) max 2 marks if no correct indication of time [6] Page 57 of 66

58 M7. (a) there is a (maximum) forward force drag/friction/resistance (opposes motion) (not pressure) increases with speed till forward and backward forces equal so no net force/acceleration any 4 for mark each (b) (i) F = ma = 250a a = 8 m/s 2 for mark each 4 4 (ii) ke = /2 mv 2 ke = / ke = J for mark each 4 (iii) W = Fd W = W = J for mark each 4 [6] M8. (a) W = 65 0 (allow a maximum of 3 marks if candidate uses g = 9.8N / Kg (as ecf)) gains mark but W = 650 (N) (allow use of p.e.= m g h) gains 2 marks but PE change = or gains 3 marks but PE change = 82.5 (J) (allow 83J or 82J) gains 4 marks 4 Page 58 of 66

59 (b) k.e. = p.e. gains mark but (speed)² = / 65 or 82.5 = ½ 65 (speed)² ecf gains 2 marks but speed = 5 (m/s) (allow ) (if answer = 25mls check working: 82.5 = ½ m v gains mark for KE = PE) (but if 82.5 = ½m v² = ½ 65 v 2 or v 2 = gains 2 marks) 25, with no working shown gains 0 marks gains 3 marks 3 [7] M9. (a) 35 (m) allow mark for indicating the correct area allow mark for obtaining correct figures from the graph allow mark for calculating area of triangle (25) but omitting the rectangle underneath (2 x 5) 3 (b) allow mark for correct substitution into the correct equation ie / [5] Page 59 of 66

60 M20. (a) *evidence of acceleration = or gains mark but 0.2 gains 2 marks units m/s 2 for mark 3 (b) (i) 2000 or 960 for mark (ii) evidence of power = or weight speed (credit figures)/ (iii) gains mark but 200/76 or figure consistent with (b)(i) gains 2 marks 2 (c) evidence of force = mass acceleration or gains mark but 60 gains 2 marks but 60 + weight of girder (2060/2020*) (or figure consistent with (b)(i)) gains 3 marks 3 [9] M2. (a) (i) 20 allow mark for correct substitution i.e kg m/s or Ns do not accept n for N accept g m/s for 3 marks 2 (ii) 840 if answer given is not 840 accept their (a)(i) in kg m/s 0.25 correctly calculated for both marks allow mark for correct substitution i.e or their (a)(i) Page 60 of 66

61 (b) increases the time to stop accept increases impact time do not accept any references to slowing down time decreases rate of change in momentum accept reduces acceleration/deceleration reduces momentum is insufficient reduces the force (on the child) (c) any two from: insufficient range of tests/thicknesses for required cfh accept need data for thicknesses above 80 mm/ cfh 2.7 m not enough tests is insufficient (seems to be) some anomalous data (repeats) needed to improve reliability (of data) accept data/ results are unreliable do not accept maybe systematic/random error do not accept reference to precision need to test greater range/variety of dummies accept children for dummies accept specific factor such as weight/height/size 2 (d) Tyres do not need to be dumped/burned/ less land-fill/ saves on raw materials accept less waste do not accept recycling on its own [] M22. (i) momentum (change in) = mass velocity (change in) accept... speed (ii) for mark but not from incorrect equation kilogram metre(s) per second or kg m/s 2 (iii) either 7.5 (m/s) or change in momentum of car B change in momentum of car A () 9000 = 200 v () or v = () Page 6 of 66

62 or error carried forward from part (ii) examples 5 (m/s) if 6000 offered in (ii) 2.5(m/s) if 5000 offered in (ii) 3 [7] M23. (a) (i) velocity includes direction accept velocity is a vector (ii) 64 allow mark for obtaining values of 6 and 4 from the graph or marking correct area or correct attempt to calculate an area 2 (iii) any two from: velocity zero from 0 to 4 seconds increasing in 0.2 s (or very rapidly) to 8 m/s decreasing to zero over the next 8 seconds 2 (iv) momentum before does not equal momentum after ignore reference to energy or total momentum changes or an external force was applied (b) to reduce the momentum of the driver a smaller (constant) force would be needed do not accept reduces the impact / impulse on the driver [8] Page 62 of 66

63 M24. (a) (i) either the momentum in a particular direction after (the collision) is the same as the momentum in that direction before (the collision) accept momentum before equals momentum after for mark (ii) or total momentum after (the collision) equals the total momentum before (the collision) (2) accept momentum before equals momentum after for mark explosion(s) or (action of a) rocket (motor(s)) or (action of a) jet (engine) or firing a gun accept any other activity in which things move apart as a result of the release of internal energy eg throwing a ball 2 (iii) momentum = mass velocity or any correctly transposed version accept momentum = mass speed accept p = mv do not accept momentum = ms or M = mv (iv) 0.8 if answer 0.8 not given, any two for () each: momentum of X = = momentum of X and Y after impact = 0.3 v or = ( ) v 3 m/s to the right (v) any one from: conservation of momentum (applies) no external forces do not accept just no (other) forces act friction is negligible / insignificant no friction no air resistance Page 63 of 66

64 (b) force = (change in) momentum time or any correctly transposed version 4000 or 4 kilonewtons dependent on correct or no equation force = gains mark 2 [3] M25. (a) (i) direction indicated accept to right or + or or arrow drawn on diagram 300 kg m/s or Ns (ii) 300 (kg m/s) (iii) there is no change in the total KE or total KE is constant (b) momentum of person towards jetty = momentum of boat away from jetty or total momentum is constant so as person goes one way boat goes the other mark is for the idea of momentum conservation is for direction 2 (c) time of collision increases do not accept momentum is conserved so a smaller force is exerted do not accept designed to absorb energy or momentum to produce the same change of momentum or impulse force do not accept cushions fall [0] Page 64 of 66

65 M26. (a) Throughout the question the equation M = mv is credited once only. This is the first time it appears. The mark scheme below assumes it will appear in (i). (i) M = mv m v sufficient not m s, mass speed = = (see marking of calculations) 3 (ii) M = mv M = 2000 = 2000 (see marking of calculations) 2 (iii) must be sum of (i) and (ii) for mark (b) total mass = 3500 momentum = (conserved) M = mv or v = 4 000/3500 v = 4 m/s (c) (i) it reduces for mark 5 (ii) ke to sound/heat for mark (iii) change smaller for mark [4] Page 65 of 66

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