(a) (i) There is a lift inside the building. The lift travels at a mean velocity of 10 m/s.

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1 Q. The world s tallest building is in Dubai. By Nicolas Lannuzel [CC BY-SA 2.0], via Flickr (a) (i) There is a lift inside the building. The lift travels at a mean velocity of 0 m/s. When the lift is carrying a full load of passengers, the mass of the lift and passengers is 600 kg. Calculate the mean kinetic energy of the lift. Give the unit. Use the correct equation from the Physics Equations Sheet. Show clearly how you work out your answer. Kinetic energy =... (3) Give one reason why your answer to part (a)(i) is the mean kinetic energy. () Page of 20

2 (b) The graph below shows how the velocity of another lift in the building varies with time between the ground floor and an observation deck. Use the graph to calculate the total distance travelled by the lift. Total distance travelled =... m (2) (Total 6 marks) Q2. 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 2 of 20

3 (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) Speed... m/s (3) (Total 7 marks) Page 3 of 20

4 Q3. The diagram below shows a plank being used as a simple machine. The crate is slid up the plank into the back of the lorry. (i) The mass of the crate is 70kg. Calculate the weight of the crate Weight... N (2) Calculate the work done when the crate is lifted a vertical distance of 0.5m Work done... (4) (Total 6 marks) Q4. The outline diagram below shows a tidal power generating system. Gates in the barrage are open when the tide is coming in and the basin is filling to the high tide level. The gates are then closed as the tide begins to fall. Once the tide outside the barrage has dropped the water can flow through large turbines in the barrage which drive generators to produce electrical energy. In one second kg of water flows through the turbines at a speed of 20 m/s. Page 4 of 20

5 (a) Calculate the total kinetic energy of the water which passes through the turbines each second (3) (b) As the height of water in the basin falls, the water speed through the turbines halves. (i) What mass of water will now pass through the turbines each second?.. By how much will the power available to the generators decrease? (5) (Total 8 marks) Q5. (a) The stopping distance of a vehicle is made up of two parts, the thinking distance and the braking distance. (i) What is meant by thinking distance? () State two factors that affect thinking distance. 2 (2) Page 5 of 20

6 (b) A car is travelling at a speed of 20 m/s when the driver applies the brakes. The car decelerates at a constant rate and stops. (i) The mass of the car and driver is 600 kg. Calculate the kinetic energy of the car and driver before the brakes are applied. Use the correct equation from the Physics Equations Sheet. Kinetic energy =... J (2) How much work is done by the braking force to stop the car and driver? Work done =... J () (iii) The braking force used to stop the car and driver was 8000 N. Calculate the braking distance of the car. Use the correct equation from the Physics Equations Sheet. Braking distance =... m (2) (iv) The braking distance of a car depends on the speed of the car and the braking force applied. State one other factor that affects braking distance. () Page 6 of 20

7 (v) Applying the brakes of the car causes the temperature of the brakes to increase. Explain why. (2) (c) Hybrid cars have an electric engine and a petrol engine. This type of car is often fitted with a regenerative braking system. A regenerative braking system not only slows a car down but at the same time causes a generator to charge the car s battery. State and explain the benefit of a hybrid car being fitted with a regenerative braking system. (3) (Total 4 marks) Q6. A machine is used to lift materials on a building site. (a) (i) Write down the equation that links change in gravitational potential energy, change in vertical height and weight.... () Page 7 of 20

8 A 25 kg bag of cement is lifted from the ground to the top of the building. Calculate the gain in the gravitational potential energy of the bag of cement. (On Earth a kg mass has a weight of 0 N.) Change in gravitational potential energy =... joules (2) (b) The conveyor belt delivers six bags of cement each minute to the top of the building. (i) Calculate the useful energy transferred by the machine each second Useful energy transfer each second =... J () The machine is 40% efficient. Use the following equation to calculate the total energy supplied to the machine each second. Show how you work out your answer Total energy supplied each second =... J (2) (Total 6 marks) Q7. The roads were very icy. An accident was recorded by a security camera. Page 8 of 20

9 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.... () 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 =... (3) (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 (3) (Total 7 marks) Q8. The diagram shows the horizontal forces acting on a car of mass 200 kg. Page 9 of 20

10 (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) (b) Explain why the car reaches a top speed even though the thrust force remains constant at 3500 N. (3) Page 0 of 20

11 (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) Q9. 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. Page of 20

12 (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 (3) (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... () Calculate the force used to accelerate the car. Show clearly how you work out your final answer..... Force =... newtons (2) (Total marks) Page 2 of 20

13 M. (a) (i) correct answer with or without working gains 2 marks 2 J allow mark for (providing no subsequent working) accept joule / Joule accept kj if answer is 80 do not accept j mean velocity is used accept velocity / speed varies allow average for mean (b) 456 correct answer with or without working gains 2 marks allow mark for ( ) +(25 2) + ( ) or equivalent or mark for recognition that area under the line is the total distance travelled 2 [6] M2. (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 (b) k.e. = p.e. gains mark but (speed)² = / 65 or 82.5 = ½ 65 (speed)² ecf gains 2 marks Page 3 of 20

14 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] M3. (i) 700 or 686 gets 2 Else mg or 70 0 or () gets 2 350J Else 350 gets 4 gets 3 Else gets 2 Else W = F.d. gets Any answer with unit J may score, 2 or 3 (see general instructions) 4 [6] M4. (a) k = /2mv 2 k = / k = for one mark each 3 Page 4 of 20

15 (b) (i) mass halved speed halved (speed)2 quartered ke and/or power cut to one eight for mark each 5 [8] M5. (a) (i) distance vehicle travels during driver s reaction time accept distance vehicle travels while driver reacts any two from: tiredness (drinking) alcohol (taking) drugs speed age accept as an alternative factor distractions, eg using a mobile phone 2 (b) (i) allow mark for correct substitution, ie subsequent step shown provided no or their (b)(i) (iii) 40 or correctly calculated allow mark for statement work done = KE lost or allow mark for correct substitution, ie 8000 distance = or their (b) 2 Page 5 of 20

16 (iv) any one from: icy / wet roads accept weather conditions (worn) tyres road surface mass (of car and passengers) accept number of passengers (efficiency / condition of the) brakes (v) (work done by) friction (between brakes and wheel) do not accept friction between road and tyres / wheels (causes) decrease in KE and increase in thermal energy accept heat for thermal energy accept KE transferred to thermal energy (c) the battery needs recharging less often accept car for battery or increases the range of the car accept less demand for other fuels or lower emissions or lower fuel costs environmentally friendly is insufficient as the efficiency of the car is increased accept it is energy efficient the decrease in (kinetic) energy / work done charges the battery (up) accept because not all work done / (kinetic) energy is wasted [4] M6. (a) (i) gpe = weight height accept Ep = mgh accept pe= mgh 200 accept values using 9.8 () allow mark for correct substitution 2 Page 6 of 20

17 (b) (i) 20 accept 300 allow b(i) 0.4 for both marks allow mark for correct transformation 2 [6] M7. (i) momentum (change in) = mass velocity (change in) accept... speed 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 = () or error carried forward from part examples 5 (m/s) if 6000 offered in (3) 2.5(m/s) if 5000 offered in (3) 3 [7] Page 7 of 20

18 M8. (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) 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 (c) 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 [] M9. (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 Page 8 of 20

19 (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 (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 2250 credit their (c) 900 for 2 marks credit mark for correct substitution 2 [] Page 9 of 20

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