topic: energy, power and momentum specification...
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Topic: Energy, Power and Momentum Specification reference: 3.4.1.6; 3.4.1.7; 3.4.1.8
Marks available: 111 Time allowed (minutes): 133 Examination questions from AQA. Don’t forget your data sheet! Mark scheme begins on page 23
Q1. The diagram shows a cue striking a stationary snooker ball of mass 140 g. The contact
time of the cue with the ball is 12 ms. The ball leaves the cue with a velocity v of 0.40 m
s−1
(a) Show that there is an impulse of about 6 × 10−2 N s when the cue is in contact with the snooker ball.
(2)
(b) Calculate the average force exerted by the cue on the snooker ball when they are in contact.
average force ____________________ N
(2)
(Total 4 marks)
Q2. Figure 1 shows a jet engine.
Figure 1
Air enters the engine at A and is heated before leaving B at a much higher speed.
(a) State what happens to the momentum of the air as it passes through the engine.
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(1)
(b) Explain, using appropriate laws of motion, why the air exerts a force on the engine in the forward direction.
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(3)
(c) In one second a mass of 210 kg of air enters at A. The speed of this mass of air increases by 570 m s−1 as it passes through the engine.
Calculate the force that the air exerts on the engine.
force = ____________________ N
(1)
(d) When an aircraft lands, its jet engines exert a decelerating force on the aircraft by making use of deflector plates. These cause the air leaving the engines to be deflected at an angle to the direction the aircraft is travelling as shown in Figure 2.
Figure 2
The speed of the air leaving B is the same as the speed of the deflected air.
Explain why the momentum of the air changes.
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(2)
(e) The total horizontal decelerating force exerted on the deflector plates of the jet engines is 190 kN.
Calculate the deceleration of the aircraft when it has a mass of 7.0 × 104 kg.
deceleration = ____________________ m s−2
(1)
(f) The aircraft lands on the runway travelling at a speed of 68 m s−1 with the deflector plates acting.
Calculate the distance the aircraft travels along the runway until it comes to rest. You may assume that the decelerating force acting on the jet engines remains constant.
distance = ____________________ m
(2)
(g) Suggest why in practice the decelerating force provided by the deflector plates may not remain constant.
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(2)
(Total 12 marks)
Q3. A fully-loaded lorry transporting water starts from rest and travels along a straight road. The figure is a graph showing how the speed of the lorry varies with time. The driving force on the lorry remains constant.
The total resistive force acting on the lorry increases with both speed and mass of the lorry. A large proportion of the mass of the lorry is due to the water which it is carrying.
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A similar lorry, also loaded with water, has the same initial mass. However, at the instant it begins to move, a large leak develops and all the water leaks out during the time covered by the graph.
Discuss how the speed–time graph will be different from that shown in the figure.
Your answer should include an explanation of:
• the shape of the graph in the figure • the effect of water loss on the initial gradient of the graph • the effect of water loss on the final speed of the lorry.
You may draw on the figure to help you with your answer.
The quality of your written communication will be assessed in your answer.
(Total 6 marks)
Q4. (a) State the law of conservation of energy.
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(2)
(b) The diagram shows a block on a horizontal table top initially held against a spring so that the spring is compressed. The other end of the spring is fixed to a wall. When released the block is pushed away by the spring. When the spring reaches its natural length the block leaves the spring and then slides along the table top. A constant frictional force acting between the moving block and the table top eventually brings the block to rest.
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(i) When the block leaves the spring, the block has a kinetic energy of 2.2 J. The mass of the block is 0.40 kg. Calculate the maximum velocity of the block.
maximum velocity = ____________________ m s−1
(1)
(ii) The block travels 1.2 m after leaving the spring before coming to rest. Show that the frictional force between the block and the table top is about 1.8 N.
(1)
(iii) The spring was initially compressed through 0.20 m. The constant frictional force acts on the block whenever it is moving. Calculate the elastic potential energy in the spring when in its initial compressed position. Assume the spring has negligible mass. State an appropriate unit for your answer.
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elastic potential energy = __________ unit = _________
(3)
(iv) The force exerted on the block by the spring is proportional to the compression of the spring. Calculate the maximum force exerted on the block by the spring.
maximum force = ____________________ N
(1)
(Total 8 marks)
Q5. The speed of an air rifle pellet is measured by firing it into a wooden block suspended from a rigid support. The wooden block can swing freely at the end of a light inextensible string as shown in the figure below.
initial position of block
A pellet of mass 8.80 g strikes a stationary wooden block and is completely embedded in it. The centre of mass of the block rises by 0.63 m. The wooden block has a mass of 450 g.
(a) Determine the speed of the pellet when it strikes the wooden block.
speed = ____________________ m s–1
(4)
(b) The wooden block is replaced by a steel block of the same mass. The experiment is repeated with the steel block and an identical pellet. The pellet rebounds after striking the block.
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Discuss how the height the steel block reaches compares with the height of 0.63 m reached by the wooden block. In your answer compare the energy and momentum changes that occur in the two experiments.
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(4)
(c) Discuss which experiment is likely to give the more accurate value for the velocity of the pellet.
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(2)
(Total 10 marks)
Q6. The diagram shows a girl bouncing vertically on a trampoline. The highest point that she reaches is H.
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Describe the energy changes involved as the girl bounces from position H and back to the same position shown in the diagram.
You should consider the energy losses that occur during this motion.
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(Total 3 marks)
Q7. Spectacle lenses can be tested by dropping a small steel ball onto the lens, as shown in the figure below, and then checking the lens for damage.
A test requires the following specifications: diameter of ball = 16 mm mass of ball = 16 g height of drop = 1.27 m
(a) Calculate the density of the steel used for the ball.
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density = ____________________kg m–3
(3)
(b) In a test the ball bounced back to a height of 0.85 m.
Calculate the speed of the ball just before impact.
speed = ____________________m s–1
(2)
(c) Calculate the speed of the ball just after impact.
speed = m ____________________m s–1
(2)
(d) Calculate the change in momentum of the ball due to the impact.
momentum = m ____________________ kg m s–1
(2)
(e) The time of contact was 40 ms. Calculate the average force of the ball on the lens during the impact.
average force = ____________________ N
(2)
(f) Explain, with reference to momentum, why the test should also specify the material of the plinth the lens sits on.
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(2)
(Total 13 marks)
Q8. An electric wheelchair, powered by a battery, allows the user to move around independently.
One type of electric wheelchair has a mass of 55 kg. The maximum distance it can travel on level ground is 12 km when carrying a user of mass 65 kg and travelling at its maximum speed of 1.5 m s−1.
The battery used has an emf of 12 V and can deliver 7.2 × 104 C as it discharges fully.
(a) Show that the average power output of the battery during the journey is about 100 W.
(3)
(b) During the journey, forces due to friction and air resistance act on the wheelchair and its user.
Assume that all the energy available in the battery is used to move the wheelchair and its user during the journey.
Calculate the total mean resistive force that acts on the wheelchair and its user.
total mean resistive force = ____________________ N
(2)
The diagram below shows the wheelchair and its user travelling up a hill. The hill makes an angle of 4.5° to the horizontal.
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(c) Calculate the force that gravity exerts on the wheelchair and its user parallel to the slope.
force parallel to the slope = ____________________ N
(1)
(d) Calculate the maximum speed of the wheelchair and its user when travelling up this hill when the power output of the battery is 100 W.
Assume that the resistive forces due to friction and air resistance are the same as in part (b).
maximum speed = ____________________ m s−1
(2)
(e) Explain how and why the maximum range of the wheelchair on level ground is affected by
• the mass of the user
• the speed at which the wheelchair travels.
Effect of mass _______________________________________________________
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Effect of speed ______________________________________________________
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(4)
(Total 12 marks)
Q9.
Figure 1 shows the variation of velocity v with time t for a Formula 1 car during a test
drive along a straight, horizontal track.
The total mass of the car and driver is 640 kg. The car engine provides a constant driving force of 5800 N.
Figure 1
(a) (i) Determine the distance travelled by the car during the first 10 s.
distance ____________________ m
(3)
(ii) Show that the instantaneous acceleration is about 4 m s−2 when t is 10 s.
(2)
(iii) Calculate the magnitude of the resistive forces on the car when t is 10 s.
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resistive forces ____________________ N
(3)
(iv) Calculate the power, in kW, of the car at the maximum speed during the test drive.
power ____________________ kW
(2)
(b) Figure 2 shows the aerofoil that is fitted to a Formula 1 car to increase its speed around corners.
Figure 2
However, the aerofoil exerts an unwanted drag force on the car when it is travelling in a straight line so a Drag Reduction System (DRS) is fitted. This system enables the driver to change the angle of the aerofoil to reduce the drag.
The graph in Figure 1 is for a test drive along a straight, horizontal track. Under the conditions for this test drive, the DRS was not in use and the engine produced a constant driving force.
Explain why the velocity varies in the way shown in the graph.
Go on to explain how the graph will be different when the DRS is in use and the driving force is the same.
The quality of written communication will be assessed in your answer.
(6)
(Total 16 marks)
Q10. Figure 1 shows a model of a system being designed to move concrete building blocks from an upper to a lower level.
Figure 1
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The model consists of two identical trolleys of mass M on a ramp which is at 35° to the horizontal. The trolleys are connected by a wire that passes around a pulley of negligible mass at the top of the ramp.
Two concrete blocks each of mass m are loaded onto trolley A at the top of the ramp. The trolley is released and accelerates to the bottom of the ramp where it is stopped by a flexible buffer. The blocks are unloaded from trolley A and two blocks are loaded onto trolley B that is now at the top of the ramp. The trolleys are released and the process is repeated.
Figure 2 shows the side view of trolley A when it is moving down the ramp.
Figure 2
(a) The tension in the wire when the trolleys are moving is T.
Draw and label arrows on Figure 2 to represent the magnitudes and directions of any forces and components of forces that act on trolley A parallel to the ramp as it travels down the ramp.
(1)
(b) Assume that no friction acts at the axle of the pulley or at the axles of the trolleys and that air resistance is negligible.
Show that the acceleration a of trolley B along the ramp is given by
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(2)
(c) Compare the momentum of loaded trolley A as it moves downwards with the momentum of loaded trolley B.
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(2)
(d) In practice, for safety reasons there is a friction brake in the pulley that provides a resistive force to reduce the acceleration to 25% of the maximum possible acceleration.
The distance travelled for each journey down the ramp is 9.0 m.
The following data apply to the arrangement.
Mass of a trolley M = 95 kg
Mass of a concrete block m = 30 kg
Calculate the time taken for a loaded trolley to travel down the ramp.
time = ____________________ s
(3)
(e) It takes 12 s to remove the blocks from the lower trolley and reload the upper trolley.
Calculate the number of blocks that can be transferred to the lower level in 30 minutes.
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number = ____________________
(2)
(Total 10 marks)
Q11. What is represented by the area under a force–displacement graph?
A rate of change of kinetic energy
B change in momentum
C work done
D acceleration
(Total 1 mark)
Q12. In a test a 500 kg car travelling at 10 m s–1 hits a wall. The front 0.30 m of the car crumples as the car is brought to rest.
What is the average force on the car during the impact?
A 830 N
B 7500 N
C 8300 N
D 83 000 N
(Total 1 mark)
Q13. Which line, A to D, in the table correctly shows what is conserved in an elastic collision?
Mass Momentum Kinetic energy
Total energy
A conserved not conserved conserved conserved
B not conserved conserved conserved not conserved
C conserved conserved not conserved conserved
D conserved conserved conserved conserved
(Total 1 mark)
Q14. The graph shows how the resultant force applied to an object of mass 2.0 kg, initially at rest, varies with time.
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What is the speed of the object after 1.0 s?
A 2.5 m s−1
B 5.0 m s−1
C 7.5 m s−1
D 10 m s−1
(Total 1 mark)
Q15. Which of the following is not a unit of power?
A N m s−1
B J s
C W
D kg m2 s−3
(Total 1 mark)
Q16. A roller coaster car is raised to a height of 65 m and released from rest.
What is the maximum possible speed of the car?
A 11 m s−1
B 25 m s−1
C 36 m s−1
D 130 m s−1
(Total 1 mark)
Q17. A firework rocket is fired vertically into the air and explodes at its highest point. What are the changes to the total kinetic energy of the rocket and the total momentum of the rocket as a result of the explosion?
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total kinetic energy
of rocket
total momentum of rocket
A unchanged unchanged
B unchanged increased
C increased unchanged
D increased increased
(Total 1 mark)
Q18. Which of the following is not a unit of power?
A N m s–1
B kg m2 s–3
C J s–1
D kg m–1 s–1
(Total 1 mark)
Q19. A car exerts a driving force of 500 N when travelling at a constant speed of 72 km h–1 on a level track. What is the work done in 5 minutes?
A 3.0 × 106 J
B 2.0 × 106 J
C 2.0 × 105 J
D 1.1 × 105 J
(Total 1 mark)
Q20. An electric motor of input power 100 W raises a mass of 10 kg vertically at a steady speed of 0.5 m s–1. What is the efficiency of the system?
A 5%
B 12%
C 50%
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D 100%
(Total 1 mark)
Q21. An object is accelerated from rest by a constant force F for a time t. Which graphs represent the variation of time with the change in the kinetic energy and the change in momentum of the object?
Kinetic energy Momentum
A
B
C
D
A
B
C
D
(Total 1 mark)
Q22. A body falls freely, with negligible air resistance. What quantity of the body is its rate of change of momentum?
A mass
B power
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C kinetic energy
D weight
(Total 1 mark)
Q23. A bullet of mass 10 g is fired with a velocity of 100 m s–1 from a stationary rifle of mass 4.0 kg. Consider the rifle and bullet to be an isolated system.
What are the recoil velocity of the rifle and the total momentum of the rifle and bullet just after firing?
Recoil velocity /
m s–1
Total momentum / kg m s–1
A 0.25 0
B 0.25 1.0
C 0.40 0
D 0.40 1.0
(Total 1 mark)
Q24.
The graph shows how the force F applied to an object varies with time t.
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What is the momentum gained by the object from t = 0 to t = 10 s?
A 18 kg m s–1
B 32 kg m s–1
C 40 kg m s–1
D 58 kg m s–1
(Total 1 mark)
Q25. Two bodies of different masses undergo an elastic collision in the absence of any external force.
Which row gives the effect on the total kinetic energy of the masses and the magnitudes of the forces exerted on the masses during the collision?
Total kinetic energy Magnitudes of forces
A remains unchanged same on both masses
B remains unchanged greater on the smaller
mass
C decreases same on both masses
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D decreases greater on the smaller
mass
(Total 1 mark)
Q26. A mass of 2.5 kg is released from rest at X and slides down a ramp, of height 3.0 m, to point Y as shown.
When the mass reaches Y at the bottom of the ramp it has a velocity of 5.0 m s–1.
What is the average frictional force between the mass and the ramp?
A 8.5 N
B 10.6 N
C 14.7 N
D 24.5 N
(Total 1 mark)
Q27. A car of mass 580 kg collides with the rear of a stationary van of mass 1200 kg.
Following the collision, the van moves with a velocity of 6.20 m s–1 and the car recoils in the opposite direction with a velocity of 1.60 m s–1.
What is the initial speed of the car?
A 5.43 m s–1
B 11.2 m s–1
C 12.8 m s–1
D 14.4 m s–1
(Total 1 mark)
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Mark schemes
Q1. (a) F∆t = ∆p (or stated in words)
∆p = 0.14 × 0.4 or 0.056
Accept use of F = ma with / in F∆t
Accept 4.67 × 12 × 10-3 = 0.056 2
(b) 0.056 / 12 × 10-3 or 0.06 / 12 × 10-3
4.7 (4.67) (N)
Condone power of 10 error in t for C1 mark
Allow 1 sf answer of 5 (N) for 0.06 (kg m s-1) 2
[4]
Q2.
(a) (momentum of air) increases ✔
words implying increase 1
(b) (rate of change of momentum so) force acting on the air (Newton 2) ✔ 1
it/air exerts force (on engine) of the same/equal magnitude/size ✔ 1
but opposite in direction (Newton 3) ✔
allow backwards and forwards to indicate opposite 1
(c) (use of F = ∆mv/t)
F = 210 × 570 = 120 000 (N) (119 700) ✔ 1
(d) momentum/velocity is a vector OR momentum/velocity has direction ✔ 1
there is a change (in the air’s) direction ✔ 1
(e) (use of F = ma)
a = (−) 190 000/7.0 × 104 = 2.7 (2.71) (m s−2) ✔ 1
(f) (use of v2 = u2 +2as)
CE from 01.5
accept range 850 − 860
if forget to square u or double a score 1 mark 1
0 = 682 − 2 × 2.7 × s ✔
s = 682/(2 × 2.7) = 860 (m) (856)
accept alternatives using s = ut +1/2at2 OR average speed – first mark for time calculation AND correct substitution
1
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(g) rate of intake of air decreases (as plane slows) OR volume/mass/amount of air (passing through engine) per
second decreases ✔
allow argument in terms of (air) resistance
(air) resistance decreases as speed of aircraft decreases for 1 mark
1
(as) smaller rate of change of momentum OR momentum change ✔
NOT FRICTION 1
[12]
Q3. High Level – Good to Excellent The relationship between the acceleration and the gradient of the graph or how it changes should be given including the reason why the initial acceleration would remain the same. Also a reference to terminal velocity should be made and an explanation of why the terminal speed is greater. The information presented as a whole should be well organised using appropriate specialist vocabulary. There should only be one or two spelling or grammatical errors for this mark.
6 marks = 6 points given from the descriptor list but at least one must come from each Group.
5 marks = 5 points given from the descriptor list but at least one must come from each Group.
5 - 6
Intermediate Level – Modest to Adequate The relationship between the acceleration and the gradient of the graph or how it changes should be given. Also something should be said about the initial acceleration being the same and/or the terminal velocity being larger. With this restriction marks can come from any marking point that is clearly given. The grammar and spelling may have a few shortcomings but the ideas must be clear.
4 marks = 4 points from the descriptor list but at least one must come from two Groups.
3 marks = 3 points from the descriptor list but at least one must come from two Groups.
3 - 4
Low Level – Poor to Limited Any two valid statements that cover any of the parts given below. There may be many grammatical and spelling errors and the information may be poorly organised.
2 marks = any 2 points from the descriptor list with no restriction on which Group they come from
1 mark = any point from the descriptor list 1 - 2
The description expected in a competent answer should include: First Group: 1. the acceleration is the gradient of the graph (or the link between acceleration and
gradient is clearly made) 2. the acceleration is greatest initially / (continuously) decreases 3. a constant final velocity / terminal velocity is shown by the graph becoming
horizontal / flat.
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Second Group: 4. initial acceleration or the initial gradient is the same or only slightly greater (as
initially the mass is the same) 5. a = F/m (in words or symbols) Third Group: 6. at the end / eventually / at terminal velocity resistance force = drive force 7. resistance force is now less at the same speed (because mass is less) 8. speed will be higher or graph has higher line (this may come from the figure) 9. in the middle of the graph the acceleration will be greater.
[6]
Q4.
(a) energy cannot be created or destroyed ✔
it can only be transferred / changed / converted from one form to another ✔
‘Transformed’ can be taken to mean transferred from one form to another.
2
(b) (i) (using Ek = ½ mv2 ) 2.2 = ½ × 0.40 × v2
v = 3.3 (ms-1) ✔
Ignore errors in 3 sig fig.
Answer only can gain mark. 1
(ii) (using work done = F × s ) 2.2 = F × 1.2 ✔ ( F = 1.83 N ) or
(using a = (v2 – u2) / 2 s) a = (02 – 3.322) / 2 × 1.2 = (-) 4.59 (m s-1)
(F = ma) = 0.4 × 4.59 ✔ = (1.84 N)
A substitution of numbers are necessary for the mark 1
(iii) (work done in moving 0.2 m) = 1.8 × 0.2 (J) ✔ (= 0.36 J)
(allow ecf (bii) × 0.2)
total work done = 2.2 + 0.36 = 2.6 ✔ (same answer is achieved if F = 2N)
J or joule ✔ 3
(iv) (use of energy = ½ F x) 2.6 = ½ Fmax 0.2
Fmax = 26 N ✔
(allow ecf 10 × (biii))
Allow mark for answer only even for ecf. 1
[8]
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Q5.
(a) Max GPE of block = Mgh = 0.46 × 9.81 × 0.63 = 2.84 J ✓
The first mark is for working out the GPE of the block 1
Initial KE of block = ½ Mv2 = 2.84 J
Initial speed of block v2 = (2 × 2.84) / 0.46
v = 3.51 ms–1 ✓
The second mark is for working out the speed of the block initially
1
momentum lost by pellet = momentum gained by block
= Mv = 0.46 × 3.51 = 1.61 kg m s–1 ✓
The third mark is for working out the momentum of the block (and therefore pellet)
1
Speed of pellet = 1.58 / m = 1.58 / 8.8 × 10−3 = 180 ms–1 (183) ✓
The final mark is for the speed of the pellet 1
At each step the mark is for the method rather than the calculated answer
Allow one consequential error in the final answer
(b) As pellet rebounds, change in momentum of pellet greater and therefore the
change in momentum of the block is greater ✓
Ignore any discussion of air resistance 1
Initial speed of block is greater ✓ 1
(Mass stays the same)
Initial KE of block greater ✓ 1
Therefore height reached by steel block is greater than with wooden block ✓ 1
(c) Calculation of steel method will need to assume that collision is elastic so that
change of momentum can be calculated ✓ 1
This is unlikely due to deformation of bullet, production of sound etc. ✓ 1
And therefore steel method unlikely to produce accurate results.
[10]
Q6.
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GPE to KE and / or elastic potential (and reverse)
Energy is lost due to work done on air / trampoline
Work done by child (on trampoline) makes up for energy losses
No B1 mark for an incorrect energy change
Accept air resistance for an energy loss
[3]
Q7.
(a) m = 16 g = 0.016 kg r = 0.008 m
Use of V = 4 / 3 π r3 to give V = 4 / 3 π (0.008)3
= 2.1 × 10 − 6 m3✓
The first mark is for calculating the volume 1
Use of density = m / V to give density = 0.016 / 2.1 × 10−6✓
The second mark is for substituting into the density equation using the correct units
1
Density = 7.4 × 103 kg m−3 ✓
The final mark is for the answer. 1
(b) Use of v2 = u2 + 2as to give v2 = 2 (9.81) (1.27) ✓
(allow use of mgΔh = ½ mv2)
v2 = 25 (24.9)
The first mark is for using the equation 1
v = 5.0 (m s-1) ✓
The second for the final answer 1
(c) Use of v2 = u2 + 2as to give 0 = u2 + 2 (-9.81) (0.85) ✓
The first mark is for using the equation 1
u2 = 17 (16.7)
u = 4.1 m s-1 ✓
The second for the final answer 1
(d) Change in momentum = mv + mu = 0.016 × 5 + 0.016 × 4.1✓
The first mark is for using the equation 1
= 0.15 (0.146) kg m s-1 ✓
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The second for the final answer 1
(e) Use of Force = change in momentum / time taken
= 0.15 / 40 × 10-3✓
The first mark is for using the equation 1
= 3.6 N ✓
The second for the final answer 1
(f) Impact time can be increased if the plinth material is not stiff✓
Alternative
A softer plinth would decrease the change in momentum of
the ball (or reduce the height of rebound) ✓ 1
Increased impact time would reduce the force of the impact. ✓
Smaller change in momentum would reduce the force of
impact ✓ 1
[13]
Q8. (a) Calculation of energy = 12 × 7.2 × 104 = 8.64 × 105 J
Or time = 12000 / 1.5 = 8000 s ✔
Calculation of other quantity and substitution in power =
useful energy / time taken ✔
Power = 110 (108 W) ✔
Or
Time = 8000 s ✔
Allow ecf for current or time
Current = charge / time = 9 A ✔
Power = VI = 108 (W) ✔ 3
(b) Attempt to use Power / velocity ✔
Allow use of 100W for P
73 N ✔
Ignore inclusion of KE in calculation
If 108 used then answer is 72 N
If 100 used then answer is 67 N
or
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work done = F × 12000 ✔
equates to 110 × 8000 so F = 73 N ✔
allow ecf from 3.1 2
(c) Force parallel to slope = 120 × 9.81 × sin 4.5 = 92 N ✔ 1
(d) Total resistive force = ans to (c) + ans to (b) (= 165 N) ✔
Allow ecf for incorrect F
Allow 0.66 / 0.67 if 108 W or 110 W used 2
(e) Increasing the mass
Reward discussion of compression of tyres
Reduces the range ✔
increases the friction on the bearings/tyres
OR More energy / power is used accelerating the user to the final speed
OR user and wheelchair have higher KE/ more energy to
move ✔
Increasing the speed
Reduces the range ✔
Air resistance increases with speed ✔
Treat as independent parts
If not explicit about increasing / decreasing lose the first mark in each part
Within each part, second mark is dependent on the first
Allow opposite answers for decreasing mass / speed 4
[12]
Q9. (a) (i) Attempt to determine area under curve
Number of “large” squares (22 ± 1) or distance per square = 20 (large) or 0.8 (small) 435 ± 10 (m)
Zero marks for use of suvat 3
(ii) Tangent to curve seen drawn at 10 s Gradient of tangent = 4.2 (allow 3.5 – 4.5) (m s-2)
Zero marks for use of suvat 2
GetThoseGrades
(iii) 640 × 4 or 2560 or 640 × their 7aii 5800-2560 or 5800-(640 × their 7aii) 3200 (3240) (N)
3
(iv) 89 (read off from graph) cnao 520 (516) (kW)
If answer is in W, then unit must be given 2
(b) The marking scheme for this question includes an overall assessment for the quality of written communication (QWC). There are no discrete marks for the assessment of QWC but the candidate’s QWC in this answer will be one of the criteria used to assign a level and award the marks for this question.
Descriptor – an answer will be expected to meet most of the criteria in the level descriptor.
Level 3 – good -claims supported by an appropriate range of evidence; -good use of information or ideas about physics, going beyond those given in the question; -argument is well structured with minimal repetition or irrelevant points; -accurate and clear expression of ideas with only minor errors of grammar, punctuation and spelling. Level 2 – modest -claims partly supported by evidence; -good use of information or ideas about physics given in the question but limited beyond this; -the argument shows some attempt at structure; -the ideas are expressed with reasonable clarity but with a few errors of grammar, punctuation and spelling. Level 1 – limited -valid points but not clearly linked to an argument structure; -limited use of information about physics; -unstructured; -errors in spelling, punctuation and grammar or lack of fluency. Level 0 -incorrect, inappropriate or no response.
Allow named resistive forces e.g. air resistance, drag, friction.
Allow thrust for driving force.
Explanation of velocity variation:
Car accelerates / Velocity increases because driving force is larger than resistive forces
Maximum acceleration at start
Resistive forces are zero at start
Resistive forces increase with increasing velocity
Drag proportional to velocity-squared
Resultant force decreases with increasing velocity
Acceleration decreases with increasing velocity
Terminal velocity reached when resultant force is zero / resistive forces balance the driving force
Explanation of change with DRS in use:
GetThoseGrades
Resistive forces reduced (because air resistance / drag) is reduced
Resultant force is non-zero for longer time
Acceleration occurs over longer time
Terminal velocity reached in longer time
Higher terminal velocity achieved
Higher maximum velocity on graph
Same initial acceleration / gradient on graph 6
[16]
Q10. (a) arrow parallel to slope labelled (M+2m)gsin35 and label
parallel to slope labelled tension OR T ✔
Ignore arrows not parallel to ground e.g. weight
Ignore friction
W not acceptable for (M +2m)g 1
(b) T – Mgsin35 = Ma
AND (M+2m)gsin35 – T = (M+2m)a ✔
add two equations
(M + 2m)gsin35 – Mg sin35 = Ma + (M +2m)a ✔
HENCE (a= mgsin35 / (M+m))
OR
(M +2m)gsin35 – Mgsin35 ✔ (= (2M+2m)a)
a = 2mgsin35 /(2M +2m) ✔
HENCE (a= mgsin35 / (M+m))
1
1
(c) SECOND MARK CONDITIONAL ON FIRST
mass / impulse / acceleration (of trollies) is the same ✔
momenta (trolley A and B) the same
SECOND MARK CONDITIONAL ON FIRST
both have same speed / magnitude of velocity but different
masses ✔
GetThoseGrades
(hence) momentum of A is greater / momenta in opposite
directions ✔ 1
1
(d) ✔
(use of v2 =2as)
v = √(2 × 0.338 × 9.0) = 2.47 ✔
✔
OR (use of s =1/2at2)
9 = ½ × 0.338 × t2 ✔
t = 7.3 s ✔
CE from acceleration calculation
If used g for acceleration then no marks awarded 1
1
1
(e) number of journeys = (1800/(12 + 7.3) = 93 or 94 ✔
number of blocks = 2 × 93 = 186 or 2 × 94 = 188 ✔
Allow CE from 06.4
Allow between 93 to 94
Allow CE from incorrect number of journeys
Allow 186 to 188 1
1
[10]
Q11. C
[1]
Q12. D
[1]
Q13. D
[1]
Q14. B
[1]
GetThoseGrades
Q15. B
[1]
Q16. C
[1]
Q17. C
[1]
Q18. D
[1]
Q19. A
[1]
Q20. C
[1]
Q21. A
[1]
Q22. D
[1]
Q23. A
[1]
Q24. D
[1]
Q25. A
[1]
