2022 May/June · Paper 31 · Q2 (d)1 markMotion (kinematics)
2022 May/June · Paper 42 · Q1 (c(ii))1 markMass, weight & density
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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Energy, work & power · Core · Paper 2 style · 6 marks
(a)An electric hoist in a Lahore warehouse lifts a crate of mass 45 kg vertically through 6.0 m at a steady speed in 12 s. Take g = 9.8 N/kg.Calculate the gravitational potential energy gained by the crate.[2]
(b)Calculate the useful power output of the hoist.[2]
(c)The electrical power input to the hoist is 400 W. Calculate its efficiency.[2]
Q2Energy, work & power · Core · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) B 30 W
Work = F × d = 60 × 4.0 = 240 J; P = 240 ÷ 8.0 = 30 W. A (7.5 W) divides force by time and ignores distance.
Q3Energy, work & power · Standard · Paper 2 style · 8 marks
(a)A cyclist and her bicycle have a total mass of 80 kg. She is travelling at 9.0 m/s at the bottom of a hill and stops pedalling.Calculate the kinetic energy of the cyclist and bicycle.[2]
(b)Assuming no energy is dissipated, calculate the greatest vertical height she could rise up the hill before stopping. Take g = 9.8 N/kg.[2]
(c)In fact she rises only 3.1 m vertically. Calculate the energy dissipated.[2]
(d)State where this dissipated energy ends up and what causes the transfer.[2]
Show answer
(a) 3240 J
Eₖ = ½mv² M1; = ½ × 80 × 9.0² = 3240 J A1
(b) 4.13 m
mgh = ½mv² so h = 3240 ÷ (80 × 9.8) M1; = 4.13 m A1
(d) As thermal (internal) energy of the surroundings / tyres / air, caused by friction and air resistance
B1 thermal energy in surroundings; B1 work done against friction / air resistance (drag)
Q4Energy, work & power · Standard · Paper 2 style · 7 marks
(a)A car of mass 1200 kg travels at 20 m/s on the Motorway. The driver brakes and the car stops in a distance of 40 m.Calculate the kinetic energy of the car before braking.[2]
(b)Calculate the average braking force, assuming all the kinetic energy is removed by the brakes.[2]
(c)State the main energy transfer that takes place in the brakes.[1]
(d)The car now travels at 40 m/s. Using the same braking force, state the factor by which the braking distance increases. Explain your answer.[2]
Show answer
(a) 240000 J (2.4 × 10⁵ J)
Eₖ = ½mv² = ½ × 1200 × 20² M1; = 240 000 J A1
(b) 6000 N
work done = F × d so F = 240 000 ÷ 40 M1; = 6000 N A1
(c) Kinetic energy → thermal (internal) energy of brakes and surroundings
B1
(d) 4 times
B1 factor 4; B1 kinetic energy ∝ v², so doubling speed gives 4 × the energy and work = F × d needs 4 × the distance
Q5Energy, work & power · Challenge · Paper 2 style · 7 marks
(a)A solar-powered pump on a farm in Tharparkar lifts 1200 kg of water every hour from a well through a vertical height of 18 m. Take g = 9.8 N/kg.Calculate the gravitational potential energy gained by the water in one hour.[2]
(b)Calculate the useful power output of the pump.[1]
(c)The solar panel has an area of 1.5 m² and receives 600 W of sunlight on each square metre. The panel converts 20 % of this into electrical power for the pump.Calculate (i) the electrical power supplied to the pump and (ii) the efficiency of the pump.[3]
(d)Calculate the overall efficiency of the system, from sunlight to gravitational potential energy of water.[1]
Q6Motion (kinematics) · Core · Paper 2 style · 7 marks
(a)A bus in Karachi starts from rest and accelerates uniformly to 12 m/s in 8.0 s. It then travels at 12 m/s for 20 s and finally decelerates uniformly to rest in 6.0 s.Calculate the acceleration of the bus during the first 8.0 s.[2]
(b)Calculate the total distance travelled by the bus. (Use the area under the speed–time graph.)[3]
(c)Calculate the average speed of the bus for the whole journey.[2]
Show answer
(a) 1.5 m/s²
a = Δv/t = 12 ÷ 8.0 M1; = 1.5 m/s² A1
(b) 324 m
½ × 8.0 × 12 = 48 m; 20 × 12 = 240 m; ½ × 6.0 × 12 = 36 m C2 (any two areas); total = 324 m A1
(c) 9.53 m/s
average speed = total distance ÷ total time = 324 ÷ 34 M1; = 9.53 m/s A1
Q7Motion (kinematics) · Standard · Paper 1 style · 1 mark
(a)The velocity of a trolley increases uniformly from 4.0 m/s to 10 m/s in 3.0 s. What distance does it travel in this time?[1]
Show answer
(a) C 21 m
Distance = area under velocity–time graph = ½ × (4.0 + 10) × 3.0 = 21 m. D (30 m) uses the final velocity for the whole time.
Q8Motion (kinematics) · Standard · Paper 2 style · 6 marks
(a)A stone is dropped from rest from a bridge over the River Ravi and hits the water 2.4 s later. Ignore air resistance and take the acceleration of free fall as 9.8 m/s².Calculate the speed of the stone just before it hits the water.[2]
(b)The speed–time graph of the stone is a straight line from the origin. Use the area under this graph to calculate the height of the bridge above the water.[2]
(c)State how the speed–time graph would be different if air resistance were not negligible.[2]
Show answer
(a) 23.5 m/s
v = a × t = 9.8 × 2.4 M1; = 23.52 ≈ 23.5 m/s A1
(b) 28.2 m
area = ½ × 2.4 × 23.52 M1; = 28.2 m A1
(c) The line would curve, with its gradient decreasing, so the final speed would be lower
B1 gradient (acceleration) decreases with time / graph curves towards the time axis; B1 lower final speed (or graph lies below the straight line)
Q9Motion (kinematics) · Challenge · Paper 2 style · 9 marks
(a)A skydiver of mass 75 kg jumps from a hovering helicopter. Take g = 9.8 N/kg.State the acceleration of the skydiver at the instant she leaves the helicopter. Give a reason.[2]
(b)Explain why her acceleration decreases as she falls.[2]
(c)She reaches a terminal velocity of 55 m/s. Calculate the air resistance acting on her at terminal velocity.[2]
(d)She then opens her parachute. Describe and explain what happens to her speed until she reaches the ground.[3]
Show answer
(a) 9.8 m/s²
B1 9.8 m/s² (acceleration of free fall); B1 air resistance is zero because speed is zero, so the only force is her weight
(b) Air resistance increases with speed, so the resultant (downward) force decreases, so acceleration decreases
B1 air resistance increases as speed increases; B1 resultant force = weight − air resistance gets smaller (a = F/m)
(c) 735 N
at terminal velocity resultant force = 0 so air resistance = weight M1; = 75 × 9.8 = 735 N A1
(d) Speed decreases (she decelerates) because air resistance becomes greater than weight; as speed falls air resistance falls until it equals weight; she then falls at a new, lower terminal velocity
B1 air resistance > weight / resultant force upwards; B1 speed decreases; B1 reaches new lower constant (terminal) velocity when forces balance again
Q10Forces & Newton's laws · Core · Paper 2 style · 4 marks
(a)A tractor pulling a loaded trailer has a forward driving force of 3600 N. The total resistive force on the tractor and trailer is 1400 N.Calculate the resultant force.[1]
(b)The total mass of the tractor and trailer is 2750 kg. Calculate the acceleration.[2]
(c)Later the driving force becomes equal to the resistive force. Describe the motion of the tractor.[1]
Show answer
(a) 2200 N forwards
3600 − 1400 = 2200 N B1
(b) 0.80 m/s²
a = F/m = 2200 ÷ 2750 M1; = 0.80 m/s² A1
(c) It moves at constant speed in a straight line (constant velocity)
B1 constant velocity / zero acceleration (not 'stops')
Q11Forces & Newton's laws · Core · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) B The resultant force on the satellite is towards the centre of the circle.
Motion in a circle needs a force perpendicular to the velocity, towards the centre (here gravity). A is wrong: the direction of velocity keeps changing, so there must be a resultant force.
Q12Forces & Newton's laws · Standard · Paper 2 style · 6 marks
(a)A lift of total mass 600 kg hangs from a steel cable. At one moment the tension in the cable is 6500 N. Take g = 9.8 N/kg.Calculate the weight of the lift.[1]
(b)Calculate the resultant force and the acceleration of the lift, giving the direction of each.[3]
(c)The lift is moving downwards at this moment. Describe its motion.[1]
(d)State the force that, according to Newton's third law, pairs with the weight of the lift.[1]
(c) It is slowing down (decelerating) while moving downwards
B1
(d) The gravitational pull of the lift on the Earth (upwards, same size)
B1 force of lift on Earth; not the cable tension
Q13Forces & Newton's laws · Challenge · Paper 2 style · 7 marks
(a)A car is travelling at 25 m/s on a dry road. The driver sees a hazard. His reaction time is 0.7 s, after which the brakes produce a constant deceleration of 6.25 m/s².Calculate the thinking distance.[1]
(b)Calculate the time taken for the brakes to stop the car.[1]
(c)Calculate the braking distance and the total stopping distance.[3]
(d)State two factors, other than speed, that would increase the braking distance.[2]
Show answer
(a) 17.5 m
25 × 0.70 = 17.5 m B1
(b) 4.0 s
t = Δv/a = 25 ÷ 6.25 = 4.0 s B1
(c) braking 50 m; stopping 67.5 m
braking distance = area under graph = ½ × 25 × 4.0 M1 = 50 m A1; stopping = 17.5 + 50 = 67.5 m A1
(d) e.g. wet or icy road; worn tyres; worn brakes; greater mass of car (load)
1 mark each for any two; not 'tiredness' (that affects thinking distance)
Q14Pressure · Core · Paper 2 style · 6 marks
(a)A woman of mass 60 kg stands on a wooden floor. Take g = 9.8 N/kg.Calculate her weight.[1]
(b)Wearing flat sandals, the total area of contact with the floor is 0.03 m². Calculate the pressure on the floor.[2]
(c)Wearing narrow high heels, at one moment all her weight rests on the heels, which have a total area of 2.0 cm². Calculate the pressure in Pa.[2]
(d)Explain why the heels may dent the floor but the sandals do not.[1]
Show answer
(a) 588 N
60 × 9.8 = 588 N B1
(b) 19600 Pa
p = F/A = 588 ÷ 0.030 M1; = 19 600 Pa A1
(c) 2.94 × 10⁶ Pa
2.0 cm² = 2.0 × 10⁻⁴ m² C1; p = 588 ÷ 2.0 × 10⁻⁴ = 2.94 × 10⁶ Pa A1
(d) Same force on a much smaller area gives a much larger pressure
B1
Q15Pressure · Standard · Paper 2 style · 6 marks
(a)A diver is 12 m below the surface of the Arabian Sea. The density of sea water is 1030 kg/m³ and g = 9.8 N/kg.Calculate the pressure due to the water at this depth.[2]
(b)Atmospheric pressure is 1.01 × 10⁵ Pa. Calculate the total pressure on the diver.[1]
(c)The glass of the diver's mask has an area of 0.015 m². Calculate the force on the outside of the glass.[2]
(d)State the direction in which the water pressure acts on the glass.[1]
(d) At right angles (perpendicular) to the surface of the glass
B1
Q16Pressure · Standard · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) C 24 500 Pa
p = ρgh = 1000 × 9.8 × 2.5 = 24 500 Pa. D is ten times too large (a power-of-ten slip).
Q17Mass, weight & density · Core · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) C 2.5 g/cm³
Volume = 146.0 − 50.0 = 96.0 cm³; density = 240 ÷ 96.0 = 2.5 g/cm³. B divides by the final reading (146 cm³) instead of the change in volume.
Q18Mass, weight & density · Standard · Paper 2 style · 7 marks
(a)An astronaut's spacesuit and equipment have a mass of 85 kg. The gravitational field strength is 9.8 N/kg on Earth and 1.6 N/kg on the Moon.Calculate the weight of the equipment on Earth and on the Moon.[2]
(b)State the mass of the equipment on the Moon and explain your answer.[2]
(c)Define gravitational field strength.[1]
(d)Suggest why it is easier for the astronaut to lift the equipment on the Moon but just as hard to start it moving sideways quickly.[2]
Show answer
(a) Earth 833 N; Moon 136 N
W = mg: 85 × 9.8 = 833 N B1; 85 × 1.6 = 136 N B1
(b) 85 kg
B1 85 kg; B1 mass is the amount of matter (resistance to change in motion), which does not depend on location
(c) Force per unit mass (on a mass in the field)
B1 g = W/m, force per unit mass
(d) Lifting depends on weight, which is smaller on the Moon; sideways acceleration depends on mass (F = ma), which is unchanged
B1 weight smaller; B1 same mass/inertia so same force needed for same acceleration
Q19Mass, weight & density · Challenge · Paper 2 style · 7 marks
(a)A student finds the density of cooking oil. An empty measuring cylinder has a mass of 112.4 g. With 80.0 cm³ of oil in it, the mass is 185.2 g.Calculate the mass of the oil.[1]
(b)Calculate the density of the oil in g/cm³ and in kg/m³.[3]
(c)Oil and water do not mix. The density of water is 1.00 g/cm³. State which liquid forms the upper layer and explain why.[1]
(d)Calculate the mass in kg of oil in a full 2.5 litre bottle. (1 litre = 1000 cm³)[2]
m = ρV = 0.91 × 2500 M1; = 2275 g = 2.275 kg ≈ 2.3 kg A1
Q20Momentum & impulse · Core · Paper 1 style · 1 mark
(a)An airbag inflates in a car crash. Why does it reduce the force on the passenger?[1]
Show answer
(a) B It increases the time taken for the passenger's momentum to change.
Force = change in momentum ÷ time. The change in momentum is fixed by the passenger's mass and speed, so a longer stopping time gives a smaller force. A is wrong: the passenger still stops, so the momentum change is the same.
Q21Momentum & impulse · Standard · Paper 2 style · 6 marks
(a)A cricket ball of mass 0.16 kg travels towards a batter at 30 m/s.Calculate the momentum of the ball.[2]
(b)The batter hits the ball straight back at 25 m/s. Calculate the change in momentum of the ball.[2]
(c)The ball is in contact with the bat for 0.002 s. Calculate the average force on the ball.[2]
Show answer
(a) 4.8 kg m/s
p = mv = 0.16 × 30 M1; = 4.8 kg m/s A1
(b) 8.8 kg m/s
Δp = 0.16 × 25 − (−0.16 × 30) M1 (signs/directions); = 4.0 + 4.8 = 8.8 kg m/s A1
(c) 4400 N
F = Δp/Δt = 8.8 ÷ 0.0020 M1; = 4400 N A1
Q22Momentum & impulse · Challenge · Paper 2 style · 8 marks
(a)In a railway yard a wagon of mass 12000 kg moving at 3.0 m/s collides with a stationary wagon of mass 8000 kg. They couple together and move off as one.State the principle of conservation of momentum.[1]
(b)Calculate the speed of the coupled wagons just after the collision.[3]
(c)Calculate the kinetic energy lost in the collision.[3]
(d)Suggest what happens to the kinetic energy that is lost.[1]
Show answer
(a) The total momentum of a system stays constant if no external resultant force acts
B1
(b) 1.8 m/s
momentum before = 12 000 × 3.0 = 36 000 kg m/s C1; 36 000 = 20 000 × v M1; v = 1.8 m/s A1
(c) 21600 J
before ½ × 12 000 × 3.0² = 54 000 J C1; after ½ × 20 000 × 1.8² = 32 400 J C1; lost 21 600 J A1
(d) Transferred to thermal energy (and sound) in the couplings and surroundings
B1
Q23Turning effect & centre of gravity · Core · Paper 2 style · 6 marks
(a)A girl of weight 350 N sits 1.6 m from the pivot of a playground see-saw.Calculate the moment of her weight about the pivot.[2]
(b)State the principle of moments.[2]
(c)Her brother of weight 400 N sits on the other side so the see-saw balances. Calculate his distance from the pivot.[2]
Show answer
(a) 560 N m
moment = F × perpendicular distance = 350 × 1.6 M1; = 560 N m A1
(b) For an object in equilibrium, the sum of clockwise moments about a point equals the sum of anticlockwise moments about the same point
B1 equilibrium / balanced; B1 clockwise = anticlockwise about the same point
(c) 1.4 m
400 × d = 560 M1; d = 1.4 m A1
Q24Turning effect & centre of gravity · Standard · Paper 3 style · 6 marks
(a)A uniform metre rule is balanced on a knife-edge at the 40 cm mark when a load of weight 0.3 N hangs from the 15 cm mark. The centre of gravity of the rule is at the 50 cm mark.Calculate the weight of the metre rule.[3]
(b)Calculate the mass of the rule. Take g = 9.8 N/kg.[1]
(c)Calculate the upward force exerted by the knife-edge on the rule.[1]
(d)Suggest why the knife-edge is used rather than a broad support.[1]
Show answer
(a) 0.75 N
load moment = 0.30 × (40 − 15) = 7.5 N cm C1; W × (50 − 40) = 7.5 M1; W = 0.75 N A1
(b) 0.077 kg (≈ 77 g)
0.75 ÷ 9.8 = 0.0765 kg B1
(c) 1.05 N
sum of downward forces = 0.75 + 0.30 = 1.05 N B1
(d) So the distances to the pivot can be measured accurately from a single, well-defined point
B1
Q25Turning effect & centre of gravity · Challenge · Paper 2 style · 8 marks
(a)A labourer lifts the handles of a loaded wheelbarrow so that its legs just leave the ground. The wheel axle acts as the pivot. The load of weight 600 N has its centre of gravity 0.4 m from the axle (measured horizontally). The barrow itself weighs 150 N with its centre of gravity 0.6 m from the axle. The labourer's vertical upward force acts at the handles, 1.5 m from the axle.Calculate the upward force the labourer must apply.[3]
(b)Calculate the upward force of the ground on the wheel.[2]
(c)Suggest why the load should be placed as close to the wheel as possible.[1]
(d)A bus manufacturer wants a double-decker bus to be stable. State two design features that help.[2]
Show answer
(a) 220 N
moments of weights = 600 × 0.40 + 150 × 0.60 = 240 + 90 = 330 N m C1; F × 1.5 = 330 M1; F = 220 N A1
(b) 530 N
upward forces = downward forces: 220 + N = 750 M1; N = 530 N A1
(c) Smaller moment of the load about the axle, so a smaller lifting force is needed
B1
(d) Low centre of gravity (heavy parts such as engine low down); wide wheelbase / wide base
B1 each
Q26Measurement & physical quantities · Core · Paper 1 style · 1 mark
(a)Which quantity is a vector?[1]
Show answer
(a) C acceleration
Acceleration has both size and direction. Speed (B) is a scalar: it has size only; velocity is the vector.
Q27Measurement & physical quantities · Standard · Paper 2 style · 6 marks
(a)State the difference between a scalar quantity and a vector quantity.[1]
(b)A boat on the River Indus is steered straight across the river at 3.0 m/s relative to the water. The river current flows at 1.6 m/s at right angles to this. Calculate the magnitude of the resultant velocity of the boat.[2]
(c)Calculate the angle between the resultant velocity and the direction in which the boat is steered.[2]
(d)State one other vector quantity and one other scalar quantity.[1]
Show answer
(a) A scalar has magnitude only; a vector has magnitude and direction
B1
(b) 3.4 m/s
v = √(3.0² + 1.6²) M1; = 3.4 m/s A1
(c) 28°
tan θ = 1.6 ÷ 3.0 M1; θ = 28.1° ≈ 28° A1
(d) e.g. vector: force / momentum / displacement; scalar: mass / time / energy / distance
B1 both correct
Q28Springs & deformation · Core · Paper 2 style · 5 marks
(a)A load of 2.0 N stretches a spring by 8.0 cm. Calculate the spring constant of the spring in N/m.[2]
(b)Assuming the limit of proportionality is not reached, calculate the extension when the load is 3.5 N.[2]
(c)A much larger load is added and removed. The spring is now longer than at the start. State what has happened to the spring.[1]
Show answer
(a) 25 N/m
k = F/x = 2.0 ÷ 0.080 M1; = 25 N/m A1
(b) 14 cm
x = F/k = 3.5 ÷ 25 = 0.14 m M1; = 14 cm A1
(c) It has been stretched beyond its elastic limit (permanently deformed)
B1
Q29Springs & deformation · Standard · Paper 3 style · 6 marks
A student hangs loads on a spring and measures its length.Load / N: 0, 1.0, 2.0, 3.0, 4.0, 5.0Length / cm: 4.0, 5.6, 7.2, 8.8, 10.6, 12.8
(a)State the extension of the spring when the load is 3.0 N.[1]
(b)Calculate the spring constant in N/m, using the readings up to 3.0 N.[2]
(c)State between which two loads the limit of proportionality is passed. Explain how the readings show this.[2]
(d)Calculate the load that would give an extension of 2.4 cm.[1]
Show answer
(a) 4.8 cm
8.8 − 4.0 = 4.8 cm B1
(b) 62.5 N/m (0.625 N/cm)
k = F/x = 3.0 ÷ 0.048 m M1; = 62.5 N/m A1
(c) Between 3.0 N and 4.0 N
B1 3 N–4 N; B1 up to 3 N each 1 N adds 1.6 cm; from 3 N to 4 N extension increases by 1.8 cm (not equal increments)
(d) 1.5 N
2.4 ÷ 1.6 = 1.5 N B1
Q30Energy resources · Standard · Paper 2 style · 8 marks
(a)Describe the energy transfers in a hydroelectric power station such as the one at Tarbela Dam.[2]
(b)At one turbine, 2000 kg of water flows each second through a vertical drop of 100 m. The station converts 85 % of the energy lost by the water into electrical energy. Take g = 9.8 N/kg. Calculate the electrical power output of this turbine.[3]
(c)Give one advantage and one disadvantage of hydroelectric power compared with a coal-fired power station.[2]
(d)State the original source of the energy in hydroelectric power.[1]
Show answer
(a) Gravitational potential energy of stored water → kinetic energy of water/turbine → electrical energy from the generator
Eₚ lost per second = 2000 × 9.8 × 100 = 1.96 × 10⁶ W C1; × 0.85 M1; = 1.67 × 10⁶ W A1
(c) Advantage: renewable / no CO₂ or air pollution when operating / reliable, quick start. Disadvantage: floods land and habitats / displaces people / high building cost / depends on rainfall
B1 each
(d) The Sun (it evaporates water that later falls as rain at height)
B1
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
Energy, work & powerBanker72 questions · 9/9 sessions
