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IB Physics Standard Level · Topic A.2

A.2 Forces and momentum

212 free, high-quality past-paper-style questions on Forces and momentum for IB Physics SL, each with a full mark scheme. Covers: Newton's laws & force pairs; Contact & field forces; Drag & buoyancy; and more. For when you have finished the past papers, or want to practise one topic at a time.A.2 at Higher Level →Read the A.2 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,083 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1A-1A-05Newton's laws & force pairs Paper 1AEasy1 mark

A book rests on a horizontal table.

Which force is the Newton's third law pair to the gravitational force exerted on the book by the Earth?

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Marking pointMarkNotes
Step 1A third-law pair is the same type of force, acts on two different bodies, and is equal and opposite.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The stated force is gravitational, exerted by the Earth on the book, so its pair is gravitational and exerted by the book on the Earth.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: a third-law pair is the same kind of force, acts on two different bodies, and is equal in magnitude and opposite in direction.
  • BThis balances the weight, but it is not its third-law pair: both forces act on the same body, and they are different kinds of force.
  • CThat is a force between the Earth and the table, a different interaction altogether.
  • DThis is a contact force, and is the third-law pair to the normal force rather than to the weight.
2A-1A-06Contact & field forces Paper 1AEasy1 mark

Forces can be classified as contact forces or field forces.

Which row gives a contact force and a field force?

Contact forceField force
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Marking pointMarkNotes
Step 1Buoyancy is exerted by the surrounding fluid on the surface of the body, so it is a contact force; gravitational, electric and magnetic forces act at a distance through a field.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: buoyancy is the push of the displaced fluid on the body's surface (contact), and the magnetic force acts through a field.
  • BWeight is the gravitational force on the body, which is a field force, not a contact force.
  • CViscous drag is exerted by the fluid in contact with the moving body, so it is a contact force, not a field force.
  • DThese are the right kinds of force in the wrong columns: the magnetic force is the field force and the normal force is the contact force.
3A-1A-22Newton's laws & force pairs Paper 1AEasy1 mark

A parachutist is descending vertically at a constant speed.

Which statement about the forces on her is correct?

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Marking pointMarkNotes
Step 1Newton's first law: a body moving at constant velocity has zero resultant force, so the upward drag must be equal in magnitude to the downward weight.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AIf the drag exceeded the weight there would be a net upward force and she would be slowing down, not moving at a constant speed.
  • BThat would leave a net downward force, so she would still be accelerating downwards.
  • CBoth her weight and the drag are still acting. It is their resultant that is zero, not the individual forces.
  • DCorrect: constant velocity means zero resultant force, so the upward drag exactly balances the downward weight.
4A-1A-25Drag & buoyancy Paper 1AEasy1 mark

A wooden block floats at rest on water, partly submerged.

Which statement about the buoyancy force on the block is correct?

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Marking pointMarkNotes
Step 1The block is at rest, so by Newton's first law the buoyancy force balances the weight.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Fb = ρVg with ρ the density of the water and V the volume displaced, which is the weight of the water displaced.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AIf the buoyancy force exceeded the weight, the resultant force would be upwards and the block would accelerate upwards. A floating body is in equilibrium.
  • BCorrect: the block is in equilibrium, so Fb = weight; and Fb = ρVg is the weight of the water displaced.
  • CV in Fb = ρVg is the volume of fluid displaced, which is only the submerged part of the block.
  • DThe density in Fb = ρVg is that of the fluid. The density of the wood only decides how much of the block is submerged.
5A-1A-27Hooke's law & springs Paper 1AEasy1 mark

A spring extends by 4.0 cm when a force of 12 N is applied to it. The spring obeys Hooke's law.

What is the spring constant?

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Step 1Hooke's law gives k = F/x. Convert the extension to metres first: 4.0 cm = 0.040 m, so k = 12 / 0.040 = 300 N m⁻¹.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AThis divides by 4.0 without converting centimetres to metres — the answer is 100 times too small.
  • BThis converts 4.0 cm to 0.40 m (dividing by 10 instead of 100), giving 12/0.40 = 30 N m−1.
  • CThis multiplies the force by the extension instead of dividing. k = F/x.
  • DCorrect: k = F/x = 12 / 0.040 = 300 N m⁻¹.
6A-1A-29Explosions & recoil Paper 1AEasy1 mark

An object at rest explodes into two fragments that move off in opposite directions. No external force acts.

Which row describes the total momentum and the total kinetic energy of the two fragments just after the explosion?

Total momentumTotal kinetic energy
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Step 1Momentum is conserved and was zero, so the fragments' momenta are equal and opposite and the total is zero; kinetic energy is a scalar, each fragment has some, and it has come from the chemical energy released.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AThis treats kinetic energy as conserved like momentum. The fragments are both moving, and energy that was stored in the object has become kinetic energy.
  • BThis adds the magnitudes of the momenta. Momentum is a vector: equal momenta in opposite directions have a total of zero, as it was before the explosion.
  • CThis swaps the behaviour of the two quantities: it is the vector, momentum, that sums to zero, while the scalar kinetic energies are both positive and add.
  • DCorrect: the total momentum stays zero because no external force acts, while the total kinetic energy has increased from zero.
7A-1A-42Friction & inclined planes Paper 1AEasy1 mark

A block slides down a rough slope. In each diagram the slope descends to the right and the block is moving down the slope.

Which diagram is the free-body diagram of the block?

Four free-body diagrams, A to D, for a block sliding down a rough slope that descends to the right.A.weightnormalfrictionB.weightnormalfrictionC.weightnormalfrictionD.weightnormalfrictionmotion
Diagram NOT accurately drawn
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Step 1Three forces act on the block: its weight vertically downwards, the normal force perpendicular to the surface of the slope, and friction parallel to the slope opposing the sliding, i.e. up the slope.✓ 1Answer C

Answer: C  ·  1 stage of work, one mark

Every option, and why

  • AThe friction arrow points down the slope, in the direction of motion. Friction opposes the relative motion, so it acts up the slope.
  • BThe normal force is drawn vertically. The normal force is perpendicular to the surface that exerts it, which here is the slope.
  • CCorrect: weight vertically down, normal force perpendicular to the slope, friction up the slope, and no other forces.
  • DThis adds a separate 'force of motion' down the slope. The pull down the slope is the component of the weight, which is already drawn; a body needs no force along its motion to keep moving.
8A-1A-53Newton's laws & force pairs Paper 1AEasy1 mark

A van of mass 3600 kg and a car of mass 1200 kg collide head-on.

During the collision, what are (magnitude of force on car) ÷ (magnitude of force on van) and (magnitude of acceleration of car) ÷ (magnitude of acceleration of van)?

Force on car ÷ force on vanAcceleration of car ÷ acceleration of van
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Marking pointMarkNotes
Step 1The force of the van on the car and the force of the car on the van are a Newton's third law pair, so they are equal in magnitude: ratio 1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2a = F/m, so the car, with one third of the mass, has three times the acceleration: ratio 3600/1200 = 3.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis assumes that the heavier van pushes three times as hard on the car as the car pushes on the van. The two forces are a third-law pair and are always equal in magnitude.
  • BCorrect: the forces are a third-law pair (ratio 1), and the car has a third of the mass, so three times the acceleration.
  • CEqual forces do not give equal accelerations: a = F/m, so the less massive car has the larger acceleration.
  • DThis inverts a = F/m, as if acceleration were proportional to mass.
9A-1A-56Momentum & impulse Paper 1AEasy1 mark

A gymnast lands vertically with the same speed on a hard floor and on a thick crash mat, and is brought to rest each time. The time taken to stop is 0.040 s on the floor and 0.32 s on the mat.

What are (change of momentum of the gymnast on the mat) ÷ (change of momentum of the gymnast on the floor) and (average resultant force on the gymnast on the mat) ÷ (average resultant force on the gymnast on the floor)?

Δp on mat ÷ Δp on floorForce on mat ÷ force on floor
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Marking pointMarkNotes
Step 1The gymnast goes from the same landing speed to rest each time, so the change of momentum is the same: ratio 1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Average resultant force = Δp/Δt, so the ratio is 0.040/0.32 = ⅛.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis assumes that the change of momentum grows with the stopping time while the force stays the same. The change of momentum is fixed by the initial and final speeds.
  • BThis inverts F = Δp/Δt: a longer stopping time gives a smaller, not a larger, average force.
  • CCorrect: the same change of momentum spread over eight times the time needs one eighth of the average resultant force.
  • DThe mat does not reduce the change of momentum: the gymnast still goes from the same speed to rest. Only the force is reduced.
10A-1A-59Circular motion Paper 1AEasy1 mark

A car travels around a horizontal circular bend at constant speed.

Which statement about the resultant force on the car is correct?

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Step 1The direction of the velocity changes, so the car accelerates towards the centre even at constant speed. On a level road the only horizontal force from outside the car that can point towards the centre is the friction of the road on the tyres.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: the centripetal force is provided by sideways friction between the tyres and the road.
  • BConstant speed is not constant velocity: the direction of motion changes, so there is an acceleration and a resultant force.
  • CThere is no outward force on the car. Without enough friction the car would continue in a straight line, along the tangent, which only appears to be outwards.
  • DA resultant force along the velocity would change the speed. The resultant force is perpendicular to the velocity, so it changes only the direction.
11A-1A-86Newton's laws & force pairs Paper 1AEasy1 mark

A passenger is standing, without holding on, in a bus that is moving at constant velocity along a straight, level road. The bus brakes suddenly and the passenger starts to fall forwards relative to the bus.

Which statement best explains this?

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Step 1Newton's first law: a body keeps a constant velocity unless a resultant force acts on it. The floor slows the feet by friction, but no horizontal force acts on the upper body, so it carries on at the original velocity while the bus slows down beneath it.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: with no horizontal force on it, the upper body keeps its velocity (first law) while the bus and the feet decelerate, so the passenger tips forwards relative to the bus.
  • BThis assumes that a force is needed in the direction of motion. The only horizontal force on the passenger, friction on the feet, acts backwards.
  • CInertia is the tendency of a body to keep its velocity; it is not a force exerted by any other body.
  • DThe passenger does not speed up relative to the road; the passenger only fails to slow down as quickly as the bus.
12A-1A-87Contact & field forces Paper 1AEasy1 mark

A robotic rover has a mass of 1025 kg on the Earth. The gravitational field strength at the surface of Mars is 3.71 N kg−1.

Which row gives the mass and the weight of the rover on the surface of Mars?

MassWeight
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Step 1Mass does not depend on location, so it is still 1025 kg. The weight is the gravitational force Fg = mg = 1025 × 3.71 = 3803 N ≈ 3.80 kN.✓ 1Answer C

Answer: C  ·  1 stage of work, one mark

Every option, and why

  • AThis scales the mass by 3.71/9.81 and then multiplies by the field strength on Mars again, so the weight is reduced twice.
  • BThe weight is right, but the mass has been scaled by 3.71/9.81. Mass is a property of the rover and is the same everywhere.
  • CCorrect: the mass is unchanged and Fg = 1025 × 3.71 = 3.80 kN.
  • DThis is the weight on the Earth, 1025 × 9.81. On Mars the field strength is 3.71 N kg−1.
13A-1A-89Hooke's law & springs Paper 1AEasy1 mark

One end of a spring is fixed. The other end is moved along the axis of the spring so that its displacement from the natural length is x, where x is positive for an extension and negative for a compression. The spring obeys Hooke's law for both.

Which graph shows how the elastic restoring force F exerted by the spring varies with x?

Four sketch graphs of restoring force against displacementA.xFB.xFC.xFD.xF
Sketch graphs, not to scale
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Step 1FH = −kx: the force is proportional to the displacement and always opposite to it, so the graph is a straight line through the origin with a negative gradient, for compressions as well as extensions.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AThis leaves out the minus sign: the restoring force acts towards the natural length, opposite to the displacement, not along it.
  • BThis makes the force negative for compressions too. A compressed spring pushes outwards, in the positive direction.
  • CThis assumes that a spring exerts a force only when stretched. A compressed spring that obeys Hooke's law pushes back with a force kx.
  • DCorrect: FH = −kx is a straight line through the origin with gradient −k on both sides.
14A-1A-93Drag & buoyancy Paper 1AEasy1 mark

The viscous drag force on a small sphere moving through a fluid is Fd = 6πηrv.

What is the unit of the viscosity η in SI base units?

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Step 1η = Fd/(6πrv), and 6π has no unit.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Unit = (kg m s−2)/(m × m s−1) = kg m−1 s−1.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis divides the force by the speed only and leaves out the radius.
  • BCorrect: (kg m s−2) ÷ (m × m s−1) = kg m−1 s−1.
  • CThis drops the s−1 from the unit of speed, dividing by m × m instead of by m × m s−1.
  • DThis multiplies the units of F, r and v instead of dividing the force by r and v.
15A-1A-102Circular motion Paper 1AEasy1 mark

Which row correctly identifies the force that provides the centripetal force?

Body moving in a circleForce that provides the centripetal force
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Step 1The centripetal force is the resultant force towards the centre, and it is always provided by real forces: here the electric attraction between the negative electron and the positive proton.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: the electric force on the electron is directed towards the proton, at the centre of the orbit.
  • BOn a flat road the normal force is vertical. The horizontal force towards the centre of the bend is friction between the tyres and the road.
  • CThe friction from the wall is vertical and supports the rider's weight. The normal force from the wall points towards the axis and provides the centripetal force.
  • DThe weight is vertical, so it cannot point towards the centre of a horizontal circle. The horizontal component of the tension in the wire does.
16IBP2S-003-04-02Connected bodies Paper 1AEasy1 mark

A warehouse robot pushes horizontally with force F on trolley P, of mass 2m. Trolley P is coupled to trolley Q, of mass m, which is in front of it. Friction is negligible. What is the magnitude of the force in the coupling between P and Q?

P 2mQ mFcouplingmotion
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Both trolleys share the acceleration a = F/(3m). The coupling is the only horizontal force on Q, so the coupling force = m × F/(3m) = F/3. Option B is the force needed to accelerate P alone. Option C assumes the force is shared equally between the trolleys. Option D assumes the whole applied force is passed on to Q.✓ 1Answer A

Answer: A

Common error: Do not assume the full applied force reaches the last body. A coupling only needs to supply the force that accelerates the bodies beyond it.

17IBP2S-003-04-03Collisions & energy Paper 1AEasy1 mark

A curling stone A slides with speed v and makes a head-on elastic collision with an identical stone B, which is at rest. Friction during the collision is negligible. What happens immediately after the collision?

AvB (at rest)
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For equal masses, conserving both momentum and kinetic energy requires the stones to exchange velocities. A therefore stops and B moves off at v. Option B conserves momentum but loses half of the kinetic energy, so it describes an inelastic collision. Option C treats B like an immovable wall and does not conserve momentum. Option D conserves neither momentum nor kinetic energy.✓ 1Answer A

Answer: A

Exam tip: An elastic collision must satisfy both conservation laws. Test each option against momentum first, then against kinetic energy.

18IBP2S-003-04-06Explosions & recoil Paper 1AEasy1 mark

An astronaut floats at rest outside a space station. Her mass, including her suit but not the tool she is holding, is 90 kg. She throws a 3.0 kg tool away from her. The tool moves at 6.0 m s−1. What is her recoil speed?

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The system starts at rest, so its total momentum stays zero. Her momentum must equal the tool's momentum in size: 90v = 3.0 × 6.0 = 18 kg m s−1. This gives v = 18/90 = 0.20 m s−1.
Option A (18/93) wrongly counts the tool as part of her mass after the throw.
Option C sets the two kinetic energies equal instead of the two momenta.
Option D assumes both bodies move at the same speed.
✓ 1Answer B

Answer: B

Common error: In a recoil from rest, the momenta are equal and opposite, not the kinetic energies. Once the tool has left her hand, use only the astronaut's mass for her recoil.

19IBP2S-003-04-09Momentum & impulse Paper 1AEasy1 mark

A gymnast lands from the same height either on a thick foam mat or on a hard floor. In both cases she comes to rest without bouncing. Compared with landing on the floor, landing on the mat gives

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Her change in momentum is the same in both landings, so the impulse is the same. The mat compresses, which increases the stopping time. Because F = Δp/Δt, the mean force is smaller. Option B wrongly reduces the impulse. Option C has the time change in the wrong direction. Option D contradicts FΔt = Δp.✓ 1Answer A

Answer: A

Common error: A cushion does not reduce the impulse. It spreads the same impulse over a longer time, which reduces the mean force.

20IBP2S-003-03-01Momentum & impulse Paper 1AEasy1 mark

A raindrop and a hailstone have the same mass and both strike a horizontal car roof vertically at the same speed v. The raindrop comes to rest on the roof. The hailstone rebounds vertically with speed v/2.
What is the ratio \(\frac{\text{impulse exerted on roof by hailstone}}{\text{impulse exerted on roof by raindrop}}\)?

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Impulse equals change in momentum. Raindrop: Δp = mv. Hailstone: the velocity reverses, so Δp = mv + m(v/2) = 1.5mv. The ratio is 1.5. Option A subtracts the rebound momentum instead of adding it. Option B ignores the rebound. Option D assumes the hailstone rebounds at full speed.✓ 1Answer C

Answer: C

Common error: Momentum is a vector. When an object reverses direction, the magnitudes of the initial and final momenta add to give the change in momentum.

21IBP2S-003-03-02Newton's laws & force pairs Paper 1AEasy1 mark

A communications satellite is in a circular orbit around the Earth. Earth exerts a gravitational force on the satellite.
According to Newton's third law, what is the other force in this pair?

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Newton's third-law pairs act on different bodies. They are the same type of force and have equal magnitude and opposite direction. If Earth pulls the satellite gravitationally, the satellite pulls Earth gravitationally with a force of equal magnitude in the opposite direction. Option A is not a separate force: in this case the centripetal force is the gravitational force itself. Option C is not a real force in an inertial frame. Option D does not exist, because an orbit is a path, not an object.✓ 1Answer B

Answer: B

Exam tip: To check a third-law pair, ask: do the two forces act on two different objects, and are they the same type of force? Both must be yes.

22IBP2S-003-03-05Friction & inclined planes Paper 1AEasy1 mark

At an airport, a suitcase is placed at rest on a horizontal conveyor belt that is moving forward at constant speed. For a short time the suitcase slips on the belt until it reaches the belt's speed.
While the suitcase is slipping, what is the direction of the frictional force exerted by the belt on the suitcase?

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Relative to the belt, the suitcase slides backward. Friction opposes this relative motion, so it acts forward on the suitcase. This forward friction is the resultant force that accelerates the suitcase up to the belt's speed. Option B assumes friction always opposes the direction of motion relative to the ground. Option C describes the normal force. Option D confuses the belt's constant speed with the suitcase's changing speed.✓ 1Answer A

Answer: A

Common error: Friction opposes relative motion (or attempted relative motion) between the surfaces. It does not necessarily oppose the object's motion relative to the ground.

23IBP2S-003-03-10Explosions & recoil Paper 1AEasy1 mark

An astronaut of total mass 80 kg floats at rest just outside a space station. She throws a 2.0 kg wrench directly away from the station at 4.0 m s−1.
What is the velocity of the astronaut immediately afterwards?

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The total momentum stays zero. The wrench gains 2.0 × 4.0 = 8.0 kg m s−1 away from the station, so the astronaut gains 8.0 kg m s−1 towards the station. Her speed is 8.0/80 = 0.10 m s−1 towards the station. Option A confuses conservation of total momentum with each object staying at rest. Option B has the wrong direction. Option D assumes the two objects have equal speeds rather than equal momenta.✓ 1Answer C

Answer: C

Exam tip: In a recoil problem, the two objects have momenta of equal magnitude in opposite directions. Their speeds are inversely proportional to their masses.

24IBP2S-003-00-01Angular velocity & period Paper 1AEasy1 mark

A blood sample tube in a laboratory centrifuge moves in a horizontal circle of fixed radius. At one rotation rate, the centripetal acceleration of the tube is 2000 m s−2. The angular speed of the rotor is then doubled while the radius stays the same. What is the new centripetal acceleration of the tube?

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Centripetal acceleration is a = ω2r. With r fixed, doubling ω multiplies a by 22 = 4, so a = 4 × 2000 = 8000 m s−2. 4000 m s−2 assumes a ∝ ω. 2800 m s−2 assumes a ∝ √ω. 1000 m s−2 treats ω as if it were the period.✓ 1Answer D

Answer: D

Common error: Centripetal acceleration depends on the square of the angular speed, and also on the square of the linear speed in a = v²/r. Check which variable is squared before scaling.

25IBP2S-003-00-02Hooke's law & springs Paper 1AEasy1 mark

A bird feeder hangs from two identical light springs joined end to end, one below the other. Each spring has spring constant k. What is the spring constant of the combination?

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Each spring carries the full weight F of the feeder, so each spring extends by F/k. The total extension is 2F/k, so the effective constant is F ÷ (2F/k) = k/2. 2k is the result for springs side by side (in parallel). k ignores the fact that the extensions add. k/4 wrongly squares the factor of 2.✓ 1Answer B

Answer: B

Exam tip: In series, every spring carries the same force and the extensions add. In parallel, every spring has the same extension and the forces add.

26IBP2S-003-00-07Hooke's law & springs Paper 1AEasy1 mark

The spring in a luggage balance has a natural length of 20.0 cm. When a 4.0 N load is hung from it, its length becomes 25.0 cm. The spring obeys Hooke's law. What is the length of the spring when a 10.0 N load is hung from it?

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The extension is 5.0 cm for 4.0 N, so k = 0.80 N cm−1. For 10.0 N the extension is 10.0/0.80 = 12.5 cm, so the length is 20.0 + 12.5 = 32.5 cm. 12.5 cm is the extension, not the length. 37.5 cm adds the new extension to the stretched length instead of the natural length. 62.5 cm assumes the total length is proportional to the load.✓ 1Answer A

Answer: A

Common error: Hooke's law relates force to extension, not to total length. Subtract the natural length first.

27IBP2S-003-00-09Angular velocity & period Paper 1AEasy1 mark

A point on the platter of a computer hard-disk drive moves in a circle at a constant rate of rotation. Which quantity for this point changes as it moves?

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Momentum p = mv is a vector. Its direction is always tangential and turns continuously, so the momentum changes even though its magnitude is constant. The speed, the magnitude of the centripetal acceleration (v2/r) and the angular speed all stay constant in uniform circular motion.✓ 1Answer A

Answer: A

Common error: A vector changes if its direction changes, even when its magnitude stays the same. Velocity, momentum and the acceleration vector all change in uniform circular motion.

28IBP2S-003-05-01Contact & field forces Paper 1AEasy1 mark

A delivery drone hovers at rest above a garden. The Earth exerts a gravitational force on the drone. According to Newton's third law, what is the partner of this force?

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A Newton's third law pair acts on two different bodies and is of the same type. The Earth pulls the drone gravitationally, so the partner is the drone pulling the Earth gravitationally, upward, with equal magnitude. The upward force of the air on the drone balances the weight, but it acts on the same body and is a contact force, so it is a first-law balance and not a third-law pair.✓ 1Answer C

Answer: C

Common error: Two forces that are equal and opposite on the SAME object are balanced forces, not a third-law pair. Third-law partners always act on different objects and are the same type of force.

29IBP2S-003-05-02Circular motion Paper 1AEasy1 mark

An astronaut of mass 80 kg floats inside the International Space Station (mass 4.2 × 105 kg). She pushes gently on a wall and moves away from it. During the push, which statement is correct?

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By Newton's third law, the force of the astronaut on the wall and the force of the wall on the astronaut are equal in magnitude at every instant. By Newton's second law, a = F/m, so the same force produces a far larger acceleration of the 80 kg astronaut than of the 4.2 × 105 kg station.✓ 1Answer A

Answer: A

Exam tip: Equal forces do not mean equal accelerations. Apply the third law to compare the forces, then apply F = ma separately to each body.

30IBP2S-003-05-07Variable mass & F = Δp/Δt Paper 1AEasy1 mark

A space probe is far from any star or planet, so gravitational forces on it are negligible. It is moving at 15 km s−1 when its engine is switched off. What happens to the probe afterwards?

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By Newton's first law, a body with no resultant force on it keeps a constant velocity. With the engine off and negligible gravity, the resultant force is zero, so the probe continues at 15 km s−1 in a straight line. No force is needed to keep it moving.✓ 1Answer B

Answer: B

Common error: A moving object does not 'carry' a force that gets used up. A force is needed to change velocity, not to keep it.

31IBP2S-003-05-10Contact & field forces Paper 1AEasy1 mark

A magnet is held at rest on the vertical steel door of a refrigerator. Which force prevents the magnet from sliding down?

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The weight of the magnet acts vertically downward. The magnetic attraction and the normal reaction are both horizontal, perpendicular to the door, so neither can balance a vertical force. The upward static friction force from the door balances the weight. The magnetic force matters only because it increases the normal force, which raises the maximum friction available.✓ 1Answer C

Answer: C

Exam tip: When describing a free-body diagram in words, check the direction of every force. Only a force with a vertical component can balance a weight.

32IBP2S-003-02-01Friction & inclined planes Paper 1AEasy1 mark

A smartphone of mass m rests without sliding on the rough surface of a stand tilted at angle θ to the horizontal. The coefficient of static friction between the phone and the stand is μ. What is the magnitude of the frictional force on the phone?

phone, mass mθ
Show mark scheme
Marking pointMarkNotes
The phone is in equilibrium, so the net force along the slope is zero. Static friction acts up the slope and exactly balances the component of weight down the slope, mg sin θ. μmg cos θ is only the maximum possible static friction, reached when the phone is just about to slip. μmg uses the weight instead of the normal force, and mg cos θ is the normal force.✓ 1Answer A

Answer: A

Common error: F = μN gives the limiting value of static friction. Below that limit, static friction is only as large as it needs to be to prevent sliding.

33IBP2S-003-02-02Explosions & recoil Paper 1AEasy1 mark

A spring-loaded toy at rest on a frictionless track splits into two parts, X and Y. X has mass m and Y has mass 3m. What is \(\frac{\text{kinetic energy of X}}{\text{kinetic energy of Y}}\) immediately after the split?

Show mark scheme
Marking pointMarkNotes
Total momentum stays zero, so X and Y have equal and opposite momenta of magnitude p. Since Ek = p2/2m, kinetic energy is inversely proportional to mass for equal p. The ratio is therefore 3m/m = 3. 1/3 comes from assuming kinetic energy is proportional to mass. 1 comes from assuming the energy is shared equally. 9 comes from squaring the velocity ratio without multiplying by the mass ratio.✓ 1Answer C

Answer: C

Exam tip: For any two-body split from rest, use Ek = p²/2m: the lighter fragment always carries the larger share of the kinetic energy.

34IBP2S-003-02-07Friction & inclined planes Paper 1AEasy1 mark

A 50 kg crate rests on a level warehouse floor. The coefficient of static friction between the crate and the floor is 0.40. A worker pushes horizontally on the crate with a force of 150 N and the crate does not move. What is the frictional force on the crate?

Show mark scheme
Marking pointMarkNotes
The maximum static friction is μmg = 0.40 × 50 × 9.8 = 196 N. The 150 N push is below this limit, so the crate stays in equilibrium and static friction exactly balances the push: 150 N. 196 N is the limiting value, not the actual value. 346 N adds the push to the limit. 0 N assumes friction acts only when the crate moves.✓ 1Answer B

Answer: B

Common error: Always compare the applied force with μsN first. If the applied force is smaller, the friction force simply equals it.

35IBP2S-003-06-01Momentum & impulse Paper 1AEasy1 mark

A tennis ball machine launches balls of mass 58 g horizontally from rest to a speed of 30 m s−1. It fires 2 balls every second. What is the average horizontal force exerted by the machine on the balls?

Show mark scheme
Marking pointMarkNotes
Average force = rate of change of momentum = (number per second) × (momentum per ball) = 2 × 0.058 × 30 = 3.5 N. 1.7 N ignores the firing rate. 7.0 N wrongly doubles the momentum change as if the ball bounced. 3.5 × 103 N uses the mass in grams instead of kilograms.✓ 1Answer B

Answer: B

Exam tip: For a stream of objects, multiply the momentum change of one object by the number per second to get the force.

36IBP2S-003-06-02Variable mass & F = Δp/Δt Paper 1AEasy1 mark

A rocket engine on a test stand ejects exhaust gas at a rate of 40 kg s−1 with a speed of 2500 m s−1 relative to the engine. What is the thrust on the engine?

Show mark scheme
Marking pointMarkNotes
Thrust = (Δm/Δt) × v = 40 × 2500 = 1.0 × 105 N, because this is the rate at which momentum is given to the gas, and by Newton's third law the gas pushes back on the engine equally. 62.5 N divides instead of multiplies. 5.0 × 104 N introduces a factor of ½ from kinetic-energy thinking. 2.0 × 105 N wrongly doubles the momentum change.✓ 1Answer C

Answer: C

Exam tip: Check units: kg s⁻¹ × m s⁻¹ = kg m s⁻² = N, confirming that thrust is a mass flow rate times a speed.

37IBP2S-003-06-03Variable mass & F = Δp/Δt Paper 1AEasy1 mark

A probe in deep space, far from any gravitational influence, fires its engine. The thrust and the rate at which fuel is ejected are both constant. How does the acceleration of the probe change while the engine is firing?

Show mark scheme
Marking pointMarkNotes
Acceleration = thrust ÷ current mass. The thrust is constant while the mass falls steadily as fuel is ejected, so the acceleration increases. A constant thrust does not mean a constant acceleration when the mass changes. Using up fuel lowers the mass; it does not weaken a constant thrust. The thrust depends on the exhaust speed relative to the probe, so it does not vanish when the probe reaches any particular speed.✓ 1Answer B

Answer: B

Common error: Using the initial mass throughout a burn. With variable mass, a = F/m must use the mass at that instant.

38IBP2S-003-01-01Drag & buoyancy Paper 1AEasy1 mark

A remotely operated vehicle (ROV) used to inspect an offshore wind-turbine foundation is fully submerged in sea water. It descends from a depth of 20 m to a depth of 200 m. The volume of the ROV and the density of the sea water stay constant. What happens to the buoyant force on the ROV during the descent?

Show mark scheme
Marking pointMarkNotes
The buoyant force equals the weight of fluid displaced, ρVg. The ROV's volume, the water density and g are all unchanged, so the buoyant force is unchanged. The pressure does rise with depth, but it rises equally on the top and bottom surfaces, so the pressure difference that produces the buoyant force stays the same. Buoyancy does not depend on the speed of the object.✓ 1Answer C

Answer: C

Common error: Buoyant force depends on the pressure difference between the lower and upper surfaces, not on the absolute pressure. For a fully submerged rigid object in a fluid of uniform density, it does not change with depth.

39IBP2S-003-01-02Drag & buoyancy Paper 1AEasy1 mark

In an environmental-science laboratory, a spherical microplastic bead of radius r moves through still water at speed v and experiences a viscous drag force F given by Stokes's law. A second bead of radius 2r moves through the same water at speed v/2. What is the drag force on the second bead?

Show mark scheme
Marking pointMarkNotes
Stokes's law gives F = 6πηrv, so the drag is proportional to the product rv. For the second bead, (2r)(v/2) = rv, so the drag is again F.
The distractors come from these errors:
• 2F accounts only for the doubled radius.
• F/2 accounts only for the halved speed.
• 4F treats drag as proportional to cross-sectional area (r2) and ignores the change in speed.
✓ 1Answer B

Answer: B

Exam tip: Write the formula as a proportionality (F ∝ rv) and multiply the scale factors together. This is quicker and safer than substituting numbers.

40IBP2S-003-01-05Free-body diagrams Paper 1AEasy1 mark

A hospital lift carries a patient on a trolley. The lift is moving downward at a constant speed. What is the resultant force on the trolley and patient?

Show mark scheme
Marking pointMarkNotes
The velocity is constant, so the acceleration is zero. By Newton's second law the resultant force is therefore zero: the upward normal force from the lift floor exactly balances the weight. The direction of motion does not determine the direction of the resultant force.✓ 1Answer A

Answer: A

Common error: Motion does not need a resultant force. A resultant force is needed only to change the velocity.

41A-1A-07Circular motion Paper 1AMedium1 mark

A small ball on a string moves in a vertical circle. Consider the instant when the ball is at the lowest point of the circle.

Which row compares the tension in the string with the weight of the ball, and gives the direction of the resultant force on the ball, at that instant?

Tension in the stringDirection of resultant force
Show mark scheme
Marking pointMarkNotes
Step 1The ball moves in a circle, so its acceleration, and hence the resultant force, is directed towards the centre: at the lowest point this is vertically upwards, perpendicular to the velocity.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Resultant = T − mg = mv²/r > 0, so the tension is greater than the weight.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe ball's speed may be momentarily constant, but its direction is changing, so it is accelerating. A body moving in a circle can never have zero resultant force.
  • BThis treats the resultant as an outward, 'centrifugal' force. The resultant on a body in circular motion points towards the centre, which is above the ball.
  • CCorrect: the resultant force T − mg = mv²/r acts upwards towards the centre, so T must exceed mg.
  • DThis assumes a force is needed along the direction of motion to keep the ball moving. The centripetal force is perpendicular to the velocity.
42A-1A-08Friction & inclined planes Paper 1AMedium1 mark

A block slides down a rough plane inclined at an angle θ to the horizontal. The coefficient of dynamic friction between the block and the plane is μd.

What is the magnitude of the acceleration of the block?

blockθ
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Step 1Perpendicular to the plane there is no acceleration, so FN = mg cos θ and the friction up the plane is μdmg cos θ.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Along the plane: ma = mg sin θ − μdmg cos θ, so a = g(sin θ − μd cos θ).✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis takes the normal force as mg. On an incline only the component mg cos θ is balanced by the normal force.
  • BThis swaps the components: the weight's component along the plane is mg sin θ and its component into the plane is mg cos θ.
  • CThis adds the friction to the driving component. Friction opposes the sliding, so it acts up the plane.
  • DCorrect: FN = mg cos θ, friction = μdmg cos θ up the plane, and a = g(sin θ − μd cos θ).
43A-1A-23Resultant forces Paper 1AMedium1 mark

A lamp of weight W hangs at rest from two identical cables. Each cable makes an angle θ with the vertical, as shown.

What is the tension in each cable?

A lamp hanging from two cables, each at angle theta to the vertical.θθlamp
Diagram NOT accurately drawn
Show mark scheme
Marking pointMarkNotes
Step 1The lamp is in translational equilibrium, so the vertical components of the two tensions balance the weight: 2T cos θ = W (the horizontal components cancel).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2T = W/(2 cos θ).✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis shares the weight between the cables but ignores their angle. Only the vertical component, T cos θ, of each tension supports the lamp.
  • BCorrect: 2T cos θ = W, so T = W/(2 cos θ), which is larger than W/2.
  • CThis resolves with the wrong trigonometric function. The angle is measured from the vertical, so the vertical component is T cos θ.
  • DThis lets one cable support the whole weight. Two cables each provide T cos θ upwards.
44A-1A-28Momentum & impulse Paper 1AMedium1 mark

A ball of mass 0.40 kg is at rest on a smooth horizontal surface. It is struck, and the graph shows how the force on it varies with time.

What is the speed of the ball immediately afterwards?

Force-time graph: a triangular pulse rising to 40 N at 0.15 s and back to zero at 0.30 s.0.00.10.20.30.4Time (s)01020304050Force (N)
Graph drawn to scale
Show mark scheme
Marking pointMarkNotes
Step 1The force is not constant, so use impulse = area under the force–time graph. The shape is a triangle of base 0.30 s and height 40 N.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Impulse = ½ × 0.30 × 40 = 6.0 N s.—
Step 3The impulse equals the change in momentum, and the ball started at rest: mv = 6.0, so v = 6.0 / 0.40 = 15 m s⁻¹.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis is the impulse in N s, not the speed. It still has to be divided by the mass.
  • BCorrect: the impulse is the area under the graph, ½ × 0.30 × 40 = 6.0 N s, and 6.0 / 0.40 = 15 m s⁻¹.
  • CThis uses the peak force × the whole contact time, which is the area of a rectangle. The pulse is triangular, so it is half of that.
  • DThis is the peak force ÷ the mass, giving the peak acceleration in m s⁻². It is not a speed.
45A-1A-30Angular velocity & period Paper 1AMedium1 mark

A disc rotates about its axis at a constant rate. Point X on the disc is a distance r from the axis and point Y is a distance 2r from the axis.

What are the ratios of the speed and of the centripetal acceleration of Y to those of X?

Speed of Y ÷ speed of XAcceleration of Y ÷ acceleration of X
Show mark scheme
Marking pointMarkNotes
Step 1Both points turn through the same angle in the same time, so they have the same angular velocity ω; v = ωr gives a speed ratio of 2.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2a = ω²r with the same ω gives an acceleration ratio of 2 (equivalently v²/r = 2²/2 = 2).✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: same ω, so v = ωr and a = ω²r both double when r doubles.
  • BThis gives both points the same linear speed, confusing speed with angular velocity, and then uses a = v²/r with the larger radius.
  • CThe speed ratio is right, but a = v²/r has been used with only the speed doubled. The radius has doubled too, so 2²/2 = 2.
  • DThe speed ratio is right, but the acceleration has been taken as inversely proportional to r, which is true only at constant speed, not at constant ω.
46A-1A-40Momentum & impulse Paper 1AMedium1 mark

A ball of mass m hits a wall at speed u and rebounds along the same line at speed v. The ball is in contact with the wall for a time Δt.

What is the magnitude of the average force exerted on the ball by the wall?

Show mark scheme
Marking pointMarkNotes
Step 1Taking the rebound direction as positive, the initial momentum is −mu and the final momentum is +mv, so Δp = mv − (−mu) = m(u + v).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2F = Δp/Δt = m(u + v)/Δt.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis subtracts the speeds as though the ball kept moving in the same direction. Its velocity reverses, so the two contributions add.
  • BThis uses only the momentum after the rebound. The wall must first remove the momentum mu before giving the ball momentum mv.
  • CThis halves the change in momentum, as though averaging the two momenta. The impulse is the full change m(u + v).
  • DCorrect: momentum is a vector; the change is m(u + v) and the average force is that impulse divided by Δt.
47A-1A-52Resultant forces Paper 1AMedium1 mark

A gardener pushes a lawnmower of mass m across level grass at constant velocity. She pushes along the handle with a force F directed at an angle θ below the horizontal.

Which row gives the horizontal resistive force on the mower and the normal force exerted on the mower by the ground?

mower, mass mθF
Diagram not to scale
Horizontal resistive forceNormal force
Show mark scheme
Marking pointMarkNotes
Step 1The velocity is constant, so the resultant force is zero (Newton's first law). Horizontally: resistive force = F cos θ.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Vertically the push has a downward component F sin θ, so FN = mg + F sin θ.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe horizontal force is right, but the vertical component of the push has been taken as upwards. A push directed below the horizontal presses the mower into the ground.
  • BThis swaps the components. The angle θ is measured from the horizontal, so the horizontal component is F cos θ and the vertical component is F sin θ.
  • CCorrect: zero resultant force horizontally gives F cos θ; vertically the normal force balances the weight and the downward component of the push.
  • DThis ignores the angle of the handle, as if the whole push were horizontal and the weight were the only vertical force.
48A-1A-54Collisions & energy Paper 1AMedium1 mark

A freight wagon P of mass 2.0 × 104 kg moves east at 3.0 m s−1 along a straight, level track. It collides with a wagon Q of mass 1.0 × 104 kg moving west at 4.0 m s−1, and the wagons couple together. External horizontal forces are negligible.

What are the speed and the direction of the coupled wagons just after the collision?

P 2.0×10⁴ kg3.0 m s⁻¹, eastQ 1.0×10⁴ kg4.0 m s⁻¹, westeast
Diagram not to scale
Speed / m s−1Direction
Show mark scheme
Marking pointMarkNotes
Step 1Take east as positive. Total momentum = (2.0 × 104)(+3.0) + (1.0 × 104)(−4.0) = +2.0 × 104 kg m s−1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2v = 2.0 × 104 ÷ 3.0 × 104 = +0.67 m s−1, so 0.67 m s−1 east.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis averages the two velocities, (3.0 − 4.0)/2, ignoring the masses. Momentum, not velocity, is conserved, and the heavier wagon P dominates.
  • BCorrect: the total momentum is 2.0 × 104 kg m s−1 east and the combined mass is 3.0 × 104 kg, so 0.67 m s−1 east.
  • CThe net momentum has been divided by the mass of P only. After coupling, both wagons move together, so the total mass 3.0 × 104 kg must be used.
  • DThis adds the magnitudes of the two momenta, ignoring the fact that the wagons move in opposite directions.
49A-1A-57Variable mass & F = Δp/Δt Paper 1AMedium1 mark

A rocket of total mass M stands on its launch pad. When its engine fires, exhaust gas of mass R is ejected vertically downwards each second at speed u relative to the rocket.

What is the acceleration of the rocket just after it leaves the pad, while its speed is still very small?

Show mark scheme
Marking pointMarkNotes
Step 1Each second the engine gives momentum Ru to the gas, so by F = Δp/Δt the gas exerts an upward force Ru on the rocket (Newton's third law).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Resultant force = Ru − Mg, so a = (Ru − Mg)/M = Ru/M − g.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe weight has been added to the thrust. Gravity acts downwards on the rocket, against the upward thrust.
  • BThis ignores the weight of the rocket: Ru is the thrust, not the resultant force.
  • CThis uses an average gas speed of u/2. The gas leaves with speed u, so the momentum given to it each second is Ru.
  • DCorrect: the thrust is the rate of change of momentum of the gas, Ru; subtracting the weight and dividing by M gives Ru/M − g.
50A-1A-85Contact & field forces Paper 1AMedium1 mark

A brick of weight W rests, without sliding, on a rough plank that is inclined at an angle θ to the horizontal. Which row gives the magnitude and the direction of the total contact force exerted by the plank on the brick?

brickθ
Diagram not to scale
MagnitudeDirection
Show mark scheme
Marking pointMarkNotes
Step 1Only two forces act on the brick: its weight and the contact force from the plank. The brick is at rest, so the resultant force is zero and the contact force must be equal and opposite to the weight.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The contact force is therefore W vertically upwards. Its component perpendicular to the plank is the normal force W cos θ and its component along the plank is the static friction W sin θ.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis is only the normal force, the component of the contact force perpendicular to the surface. The plank also exerts static friction along its surface, and the contact force is the vector sum of the two.
  • BThis is only the frictional component of the contact force. The normal component, W cos θ, must be added to it as a vector.
  • CThe magnitude is right, but the contact force is perpendicular to the surface only when there is no friction. Here static friction acts up the plank, so the contact force is tilted away from the normal.
  • DCorrect: the contact force is the only force that can balance the weight, so it is W vertically upwards, whatever the angle θ.
51A-1A-88Hooke's law & springs Paper 1AMedium1 mark

A planter of weight W is supported by two identical elastic cords that obey Hooke's law. The masses of the cords are negligible.

When the cords hang side by side, both attached to the planter, the extension of each cord is xP. When the same cords are joined end to end and the planter hangs from the lower end, the total extension of the two cords is xS.

What is xS/xP?

Two identical cords supporting a planter side by side and end to endplanterplanterside by sideend to end
Diagram NOT accurately drawn
Show mark scheme
Marking pointMarkNotes
Step 1Side by side, each cord supports W/2, so xP = W/2k.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2End to end, each cord supports the whole weight W and extends by W/k, so xS = 2W/k and xS/xP = 4.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis is xP/xS: the ratio has been inverted.
  • BThis assumes that two identical cords have the same stiffness however they are joined, so the extension depends only on the load.
  • CThis treats each side-by-side cord as carrying the whole weight, so xP = W/k; in fact each carries only W/2.
  • DCorrect: xP = W/2k and xS = 2W/k, so the ratio is 4.
52A-1A-90Connected bodies Paper 1AMedium1 mark

An electric tug pulls two baggage carts in a straight line along a level floor. Cart P, of mass 2m, is coupled to the tug, and cart Q, of mass m, is coupled behind P. The tug exerts a horizontal force F on P. Resistive forces on the carts are negligible.

What is the magnitude of the force exerted by P on Q?

Tug pulling cart P, with cart Q coupled behind PQmass mPmass 2mtugFforce of tug on P
Diagram NOT accurately drawn
Show mark scheme
Marking pointMarkNotes
Step 1Treating P and Q as one system of mass 3m, the only horizontal external force is F, so a = F/3m.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The only horizontal force on Q is the force from P, so it equals ma = F/3.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the carts share the acceleration F/3m, and Q needs a resultant force m × F/3m = F/3.
  • BThis shares the force equally between the two carts, ignoring that P has twice the mass of Q.
  • CThis is the resultant force on P, 2m × F/3m, not the force that P exerts on Q.
  • DThis assumes the whole of F is passed on to Q. Part of F is needed to accelerate P itself.
53A-1A-92Drag & buoyancy Paper 1AMedium1 mark

A small bubble of gas rises vertically at its terminal speed through a glass of sparkling water. The weight of the bubble is W and the buoyancy force on it is Fb.

Which row gives the direction of the viscous drag on the bubble and compares Fb with W?

Direction of viscous dragFb compared with W
Show mark scheme
Marking pointMarkNotes
Step 1The viscous drag opposes the motion of the bubble through the water, so it acts downwards.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2At terminal speed the resultant force is zero: Fb = W + Fd, so Fb is greater than W.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis draws the drag upwards, as for a sphere falling through a fluid. The bubble moves up, so the drag acts down; with an upward drag the balance Fb + Fd = W would follow.
  • BThis draws the drag upwards and then balances the buoyancy against the weight alone, as for a body floating at rest.
  • CCorrect: drag opposes the upward motion, and zero resultant force needs Fb = W + Fd > W.
  • DThe drag direction is right, but the drag has been left out of the force balance, as for a body floating at rest. The downward drag must also be balanced.
54A-1A-95Momentum & impulse Paper 1AMedium1 mark

The graph shows how the momentum p of a trolley moving along a straight track varies with time t.

What is the magnitude of the resultant force on the trolley at t = 3.0 s?

Momentum of a trolley against time012345678t / s02468101214p / kg m s⁻¹
Graph drawn to scale
Show mark scheme
Marking pointMarkNotes
Step 1The resultant force is the rate of change of momentum, F = Δp/Δt, which is the gradient of the p–t graph.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Between t = 2.0 s and t = 4.5 s the momentum is constant, so the gradient, and the resultant force, at t = 3.0 s are zero.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the momentum is constant around t = 3.0 s, so Δp/Δt = 0 and no resultant force acts.
  • BThis divides the momentum at 3.0 s by the time, 12/3.0, instead of finding the gradient at that instant.
  • CThis is the gradient of the first section of the graph, (12 − 2)/2.0, which applies before t = 2.0 s.
  • DThis reads the value of the momentum as if it were the force.
55A-1A-96Momentum & impulse Paper 1AMedium1 mark

A ball is released from rest and falls freely towards the Earth. Air resistance is negligible.

Which statement about momentum during the fall is correct?

Show mark scheme
Marking pointMarkNotes
Step 1The ball and the Earth attract each other with gravitational forces that form a Newton's third law pair: equal in magnitude, opposite in direction and acting for the same time.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2So their impulses, and the changes in momentum of the ball and the Earth, are equal and opposite; the total momentum of the ball–Earth system stays constant.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe weight of the ball is a resultant external force on the ball alone, so the ball's momentum increases. Momentum is conserved only for a system on which no resultant external force acts.
  • BDoing work changes the kinetic energy, not the total momentum. The gravitational forces between the ball and the Earth are internal to that system, so its total momentum is constant.
  • CThe ball pulls the Earth with a force equal in size to the ball's weight. The Earth's change in velocity is negligible because its mass is huge, but its change in momentum is as large as the ball's.
  • DCorrect: equal and opposite forces act for the same time, so the two changes in momentum are equal and opposite.
56A-1A-97Explosions & recoil Paper 1AMedium1 mark

Far from any planet, a capsule of mass 1200 kg and a service module of mass 3600 kg are joined together and at rest. A spring mechanism pushes them apart along a straight line. No external force acts.

What is (kinetic energy of the capsule) ÷ (kinetic energy of the service module) after the separation?

Show mark scheme
Marking pointMarkNotes
Step 1The total momentum stays zero, so the two momenta are equal in magnitude: the capsule's speed is 3600/1200 = 3 times that of the module.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2With equal momenta, Ek = p²/2m ∝ 1/m, so the ratio is 3600/1200 = 3.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis inverts the ratio, as though kinetic energy were proportional to mass at equal momentum.
  • BThis assumes the spring gives equal shares of energy to the two bodies, because it pushes on them with equal forces. The lighter body moves further during the push, so more work is done on it.
  • CCorrect: equal momenta and Ek = p²/2m give Ec/Em = mm/mc = 3.
  • DThis squares the speed ratio, 3² = 9, but leaves out the mass ratio in ½mv².
57A-1A-99Angular velocity & period Paper 1AMedium1 mark

The blades of a wind turbine are 60 m long and rotate at a steady 15 revolutions per minute.

Which row gives the angular velocity of the blades and the speed of a blade tip?

Angular velocity / rad s−1Tip speed / m s−1
Show mark scheme
Marking pointMarkNotes
Step 1ω = 2π × 15/60 = 1.571 rad s−1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2v = ωr = 1.571 × 60 = 94.2 m s−1 ≈ 94 m s−1.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis gives the rate in revolutions per second, 15/60, and uses it as an angular velocity without the factor 2π.
  • BCorrect: ω = 2π × 15/60 = 1.6 rad s−1 and v = ωr = 94 m s−1.
  • CThe angular velocity is right, but the speed has been found with the diameter of the rotor, 120 m, instead of the radius.
  • DThis multiplies 15 revolutions by 2π without converting minutes to seconds, so both values are 60 times too large.
58A-1A-100Circular motion Paper 1AMedium1 mark

A sample tube in a laboratory centrifuge moves in a horizontal circle of radius r at a steady N revolutions per minute. The tube is moved to a radius 2r and the rate is reduced to N/2 revolutions per minute.

What is (new centripetal acceleration of the tube) ÷ (original centripetal acceleration)?

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Marking pointMarkNotes
Step 1Halving the rate doubles the period T. With a = 4π²r/T², doubling r doubles a and doubling T divides a by 4.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2New/original = 2 × ¼ = ½.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis uses a = v²/r with the speed proportional to the rate only, (½)²/2. The speed also doubles because the radius doubles, so v is unchanged.
  • BCorrect: a = 4π²r/T² with r × 2 and T × 2 gives a factor of 2/4 = ½.
  • CThis takes a to be proportional to the rate rather than to its square, so the two changes appear to cancel.
  • DThis allows for the change of radius but not for the reduced rate of rotation.
59A-1A-104Variable mass & F = Δp/Δt Paper 1AMedium1 mark

An aircraft flies horizontally at a steady speed of 250 m s−1. Each of its engines takes in 180 kg of air every second and ejects it backwards at 600 m s−1 relative to the aircraft. The mass of fuel added to the air is negligible.

What is the thrust of one engine?

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Marking pointMarkNotes
Step 1Relative to the aircraft, the air enters at 250 m s−1 and leaves at 600 m s−1, so each kilogram gains 350 m s−1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2F = Δp/Δt = 180 × 350 = 63 000 N ≈ 63 kN.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis halves the change in velocity of the air, as though using its average speed. F = Δp/Δt uses the full change in velocity.
  • BCorrect: the momentum given to the air each second is 180 × (600 − 250) = 6.3 × 104 N.
  • CThis treats the incoming air as being at rest relative to the engine, using 180 × 600.
  • DThis adds the two speeds, 180 × (600 + 250), as though the incoming air were moving in the opposite direction to the exhaust.
60IBP2S-003-04-01Collisions & energy Paper 1AMedium1 mark

A supply capsule of mass m approaches a space station of mass 3m with speed v, measured relative to the station. The capsule docks and the two move off together. What fraction of the initial kinetic energy, measured in the station's original rest frame, remains as kinetic energy after docking?

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Marking pointMarkNotes
Momentum is conserved: mv = 4m·vf, so vf = v/4. The final kinetic energy is ½(4m)(v/4)2 = mv2/8. The initial kinetic energy is ½mv2, so the fraction remaining is (1/8)/(1/2) = 1/4. Option B squares the speed ratio but ignores the fourfold increase in mass. Option C is the fraction lost, not the fraction remaining. Option D is the result for two equal masses.✓ 1Answer A

Answer: A

Exam tip: For a perfectly inelastic collision with one body initially at rest, the fraction of kinetic energy kept is mmoving/(mtotal). This is a quick way to check your answer.

61IBP2S-003-04-04Connected bodies Paper 1AMedium1 mark

In a model of a lift, a 3.0 kg mass and a 1.0 kg mass hang from opposite ends of a light string passing over a frictionless pulley. The masses are released from rest. What is the tension in the string while the masses move?

3.0 kg1.0 kglight string overfrictionless pulley
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Marking pointMarkNotes
The acceleration is a = (3.0 − 1.0)g/(3.0 + 1.0) = g/2 = 4.9 m s−2. For the 1.0 kg mass, which accelerates upwards: T − 9.8 = 1.0 × 4.9, so T = 14.7 N. Option A is the weight of the lighter mass, which would apply only in equilibrium. Option C is the difference between the two weights. Option D is the weight of the heavier mass.✓ 1Answer B

Answer: B

Common error: While the masses accelerate, the tension lies between the two weights. It is never equal to either weight.

62IBP2S-003-04-05Explosions & recoil Paper 1AMedium1 mark

A firework shell at rest explodes into two fragments of masses m and 3m, which move in opposite directions. What is the ratio \(\frac{\text{kinetic energy of lighter fragment}}{\text{kinetic energy of heavier fragment}}\)?

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Marking pointMarkNotes
Total momentum stays zero, so both fragments have the same magnitude of momentum p. Since Ek = p2/(2m), equal momenta mean kinetic energy is inversely proportional to mass, giving a ratio of 3 : 1. Option A inverts this relationship. Option C uses the square of the speed ratio but ignores the masses. Option D wrongly assumes the energy is shared equally.✓ 1Answer B

Answer: B

Exam tip: When a system splits from rest, use Ek = p²/2m. The equal momenta make the energy comparison a single step.

63IBP2S-003-04-07Connected bodies Paper 1AMedium1 mark

A block on a rough horizontal bench is connected by a light string, passing over a frictionless pulley at the bench edge, to a hanging mass m. The hanging mass accelerates downwards. Which statement about the tension T in the string is correct?

rough benchblockm
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Marking pointMarkNotes
The hanging mass accelerates downwards, so the resultant force on it is downward: mg − T = ma > 0. Therefore T < mg. Option A would be true only if the mass were in equilibrium. Option C would make the mass accelerate upwards. Option D would mean zero resultant force on the block, which contradicts the fact that it accelerates.✓ 1Answer B

Answer: B

Exam tip: Compare a force with a weight by reading the direction of acceleration. The larger force points the way the body accelerates.

64IBP2S-003-04-10Hooke's law & springs Paper 1AMedium1 mark

A locomotive pulls five identical carriages along a straight track with a constant acceleration. The resistive force on each carriage is the same. What is the ratio \(\frac{\text{tension in coupling between locomotive and carriage 1}}{\text{tension in coupling between carriage 4 and carriage 5}}\)?

locomotive12345accelerationcarriages
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Marking pointMarkNotes
Each carriage needs the same force, ma + f. The first coupling must supply this force for all five carriages. The last coupling supplies it only for carriage 5. The ratio is therefore 5. Option B counts the couplings instead of the carriages pulled. Option C confuses the ratio with a ratio of carriage numbers. Option D assumes the tension is the same throughout the train.✓ 1Answer A

Answer: A

Exam tip: The tension in any coupling equals the total force needed by everything behind it. Count the bodies being pulled.

65IBP2S-003-03-03Momentum & impulse Paper 1AMedium1 mark

A golf ball of mass 0.040 kg is initially at rest on a tee. When the club strikes it, the force on the ball rises uniformly from zero to a maximum of 800 N. The force then falls uniformly back to zero. The total contact time is 8.0 ms.
What is the speed of the ball as it leaves the club?

01234567890100200300400500600700800900t / msF / N
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Marking pointMarkNotes
The force–time variation is a triangle, so the impulse = ½ × 800 × 8.0×10−3 = 3.2 N s. This equals the change in momentum, so v = 3.2 / 0.040 = 80 m s−1. Option A gives the impulse rather than the speed. Option B halves the impulse a second time. Option D uses peak force × time, treating the force as constant at its maximum.✓ 1Answer C

Answer: C

Exam tip: Impulse is the area under a force–time relationship. For a force that rises and falls linearly, use the area of a triangle, not peak force × time.

66IBP2S-003-03-04Variable mass & F = Δp/Δt Paper 1AMedium1 mark

An ion thruster on a deep-space probe ejects xenon ions at a rate of 2.0×10−6 kg s−1. The ions leave at a speed of 4.0×104 m s−1 relative to the probe.
What is the thrust on the probe?

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Marking pointMarkNotes
Thrust is the rate of change of momentum of the ejected ions: F = (Δm/Δt)v = 2.0×10−6 × 4.0×104 = 8.0×10−2 N. Option A divides instead of multiplying. Option B wrongly includes a factor of ½, as in kinetic energy. Option D doubles the result, as if the ions reversed direction.✓ 1Answer C

Answer: C

Common error: Do not mix up the rate of change of momentum, (Δm/Δt)v, with the rate of transfer of kinetic energy, which is ½(Δm/Δt)v².

67IBP2S-003-03-06Free-body diagrams Paper 1AMedium1 mark

A person of mass 70 kg stands on a set of bathroom scales in a lift in a tall office building. The scales read 600 N.
Which describes a possible motion of the lift?

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Marking pointMarkNotes
The person's weight is 70 × 9.8 ≈ 686 N. The normal force (the scale reading) is 600 N, which is less than the weight. The resultant force on the person is therefore downward, so the acceleration is downward. A lift moving upward and slowing down has a downward acceleration. Option A would give a reading of 686 N. Options B and D both involve an upward acceleration, which would give a reading greater than 686 N.✓ 1Answer C

Answer: C

Exam tip: A scale reading tells you the direction of the acceleration, not the direction of the velocity. Work out the acceleration first, then list the motions that are consistent with it.

68IBP2S-003-03-07Collisions & energy Paper 1AMedium1 mark

In a railway freight yard, a wagon moving at speed v collides with an identical stationary wagon on a straight horizontal track. The two wagons couple together and move off as one.
What fraction of the initial kinetic energy is transferred to other forms in the collision?

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Marking pointMarkNotes
Momentum is conserved: mv = 2mv′, so v′ = v/2. The final kinetic energy is ½(2m)(v/2)² = ¼mv², which is half of the initial ½mv². Half of the kinetic energy is therefore transferred to other forms. Option A assumes kinetic energy is conserved, as it would be in an elastic collision. Option B confuses the fraction remaining with the fraction lost incorrectly. Option D comes from squaring the speed ratio but forgetting that the mass doubles.✓ 1Answer C

Answer: C

Exam tip: In any collision where the objects stick together, momentum is conserved but kinetic energy is not. Always find the final velocity from momentum first.

69IBP2S-003-00-03Circular motion Paper 1AMedium1 mark

A bobsleigh travels around a banked bend at exactly the speed for which no sideways friction is needed. Air resistance can be ignored. Which statement about the bobsleigh at this instant is correct?

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Marking pointMarkNotes
Only two forces act: the weight and the normal reaction. The normal reaction is perpendicular to the banked surface, so it has a horizontal component pointing towards the centre of the bend. That component is the resultant force and provides the centripetal force. The vertical component balances the weight, so N cos θ = mg and N > mg. No outward 'centrifugal' force acts in an inertial frame. The velocity is changing direction, so the resultant force cannot be zero.✓ 1Answer C

Answer: C

Common error: Constant speed does not mean zero resultant force. In circular motion the direction of the velocity changes, so a resultant force towards the centre must act.

70IBP2S-003-00-04Circular motion Paper 1AMedium1 mark

A stunt pilot of mass m flies a vertical loop of radius r. At the lowest point of the loop the aircraft has speed v. What is the magnitude of the force exerted by the seat on the pilot at this point?

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Marking pointMarkNotes
At the lowest point the centre of the circle is directly above the pilot. The resultant upward force must equal mv2/r, so N − mg = mv2/r, giving N = m(g + v2/r). m(v2/r − g) is the value at the top of the loop. mv2/r ignores the weight. mg ignores the circular motion.✓ 1Answer B

Answer: B

Exam tip: Write 'resultant force towards the centre = mv²/r' first. Then add or subtract the weight depending on whether the centre is above or below the object.

71IBP2S-003-00-05Hooke's law & springs Paper 1AMedium1 mark

An engineer cuts a uniform spring of spring constant k into three equal lengths. She then mounts the three pieces side by side (in parallel) to support a load. What is the spring constant of the new arrangement?

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Marking pointMarkNotes
Under a given force, each third of the original spring extends one third as much, so each piece has constant 3k. In parallel the constants add: 3k + 3k + 3k = 9k. 3k forgets that the shorter pieces are stiffer. k assumes cutting and recombining changes nothing. k/9 inverts the result.✓ 1Answer D

Answer: D

Common error: Cutting a spring does not leave its spring constant unchanged. A shorter piece of the same spring is stiffer, in inverse proportion to its length.

72IBP2S-003-00-06Circular motion Paper 1AMedium1 mark

A design for a rotating space habitat uses a ring of radius 39.2 m. Astronauts standing on the inside of the outer wall should have a centripetal acceleration of 9.8 m s−2. What is the required period of rotation of the ring?

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Marking pointMarkNotes
a = 4π2r/T2, so T = 2π√(r/a) = 2π√(39.2/9.8) = 2π × 2 = 4π ≈ 12.6 s. 2.0 s omits the factor 2π. 25.1 s omits the square root (2π × 4). 19.6 s is √(ar), which is the rim speed in m s−1, not a period.✓ 1Answer B

Answer: B

Exam tip: Rearrange symbolically before substituting. Here r/a = 4 exactly, so the arithmetic needs no calculator.

73IBP2S-003-00-10Circular motion Paper 1AMedium1 mark

An e-scooter rider turns on a flat, horizontal circular path of fixed radius. On a dry surface the maximum speed without skidding is v. On ice, the maximum frictional force available is one quarter of its dry value. What is the maximum speed on ice?

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Marking pointMarkNotes
Friction provides the centripetal force: μmg = mv2/r, so v = √(μgr). If μ becomes μ/4, v becomes √(1/4) × v = v/2. v/4 assumes v ∝ μ. v/16 squares the factor instead of taking its square root. v assumes that because the mass cancels, nothing else changes either.✓ 1Answer C

Answer: C

Exam tip: For proportional-reasoning questions, write the formula for the unknown, here v ∝ √μ, before applying the scale factor.

74IBP2S-003-05-03Resultant forces Paper 1AMedium1 mark

A hand strap hangs freely from the ceiling of a train that is accelerating uniformly along a straight, level track. The strap makes a steady angle θ with the vertical. What is the magnitude of the acceleration of the train?

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Marking pointMarkNotes
Only two forces act on the strap: tension T along the strap and weight mg. Vertically there is no acceleration, so T cos θ = mg. Horizontally the strap accelerates with the train, so T sin θ = ma. Dividing the two equations gives a = g tan θ.✓ 1Answer C

Answer: C

Common error: Choosing g sin θ treats the strap like a block sliding down a slope. Here the acceleration is horizontal, so resolve the forces horizontally and vertically.

75IBP2S-003-05-04Variable mass & F = Δp/Δt Paper 1AMedium1 mark

A spacecraft in deep space fires its engine. Exhaust gas leaves the nozzle at 2500 m s−1 relative to the spacecraft, at a rate of 40 kg s−1. What is the magnitude of the thrust on the spacecraft?

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Marking pointMarkNotes
Each second the engine gives 40 kg of gas a momentum of 40 × 2500 = 1.0 × 105 kg m s−1. The force on the gas is the rate of change of its momentum: F = (Δm/Δt)v = 1.0 × 105 N. By Newton's third law, the gas pushes the spacecraft forward with a force of the same magnitude.✓ 1Answer D

Answer: D

Exam tip: Thrust = (mass flow rate) × (exhaust speed). Check the units: kg s⁻¹ × m s⁻¹ = kg m s⁻² = N. Dividing the two quantities, or using a ½mv² energy method, gives a wrong answer.

76IBP2S-003-05-05Friction & inclined planes Paper 1AMedium1 mark

In a warehouse, a robot pushes horizontally with a force of 24 N on a 6.0 kg box. This box is in contact with a 2.0 kg box in front of it. The floor is frictionless. What is the magnitude of the force that the 6.0 kg box exerts on the 2.0 kg box?

6.0 kg2.0 kgrobot24 N
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Marking pointMarkNotes
Both boxes share the same acceleration: a = 24 / (6.0 + 2.0) = 3.0 m s−2. The only horizontal force on the 2.0 kg box is the contact force, so F = 2.0 × 3.0 = 6.0 N. By the third law, the 2.0 kg box pushes back on the 6.0 kg box with 6.0 N, and that is why the net force on the 6.0 kg box is 18 N.✓ 1Answer A

Answer: A

Common error: The full 24 N is not passed on to the front box. Part of the push is needed to accelerate the rear box. 18 N is the net force on the 6.0 kg box, not the contact force.

77IBP2S-003-05-06Friction & inclined planes Paper 1AMedium1 mark

A crate of medical supplies is dropped from an aircraft. Just after its parachute opens, the crate is still moving downward but is slowing down. Which statement describes the forces on the crate at this moment?

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Marking pointMarkNotes
The crate is moving down but slowing, so its acceleration is upward. By Newton's second law, the resultant force points the same way as the acceleration, which is upward. The upward drag must therefore be greater than the weight. The direction of motion does not set the direction of the resultant force.✓ 1Answer A

Answer: A

Common error: The resultant force is in the direction of the acceleration, not the velocity. 'Moving down' does not mean 'net force down'.

78IBP2S-003-05-09Variable mass & F = Δp/Δt Paper 1AMedium1 mark

Grain falls vertically onto a horizontal conveyor belt at a steady rate of 15 kg s−1. The belt moves at a constant 2.0 m s−1. Ignore friction in the rollers. What horizontal force is needed to keep the belt moving at constant speed?

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Marking pointMarkNotes
Each second, 15 kg of grain must be given a horizontal velocity of 2.0 m s−1. The rate of change of horizontal momentum is 15 × 2.0 = 30 kg m s−2, so the force is F = Δp/Δt = 30 N. 15 N comes from equating the force to the rate of kinetic energy gain divided by speed. That method ignores the energy dissipated as the grain slips before it moves with the belt.✓ 1Answer D

Answer: D

Common error: 'Constant speed' does not mean zero force when mass is being added. Use F = Δp/Δt, not F = ma with a fixed mass.

79IBP2S-003-02-03Friction & inclined planes Paper 1AMedium1 mark

A parcel sits on the flat floor of a delivery van. The van accelerates forward along a level road and the parcel does not slide relative to the floor. What is the direction of the frictional force that the floor exerts on the parcel?

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Marking pointMarkNotes
The parcel accelerates forward with the van. By Newton's second law it needs a forward resultant force, and the only horizontal force on it is static friction from the floor. Friction therefore acts forwards. It opposes the relative motion the parcel would have without friction (sliding backwards relative to the floor), not the motion of the parcel itself. Zero friction would leave the parcel with no acceleration, and the vertical upward force is the normal force, not friction.✓ 1Answer B

Answer: B

Common error: Friction does not always oppose motion. It opposes relative sliding between the surfaces, and it can be the very force that makes an object accelerate.

80IBP2S-003-02-04Friction & inclined planes Paper 1AMedium1 mark

In a ski-wax test, a ski slides down a snow slope inclined at θ to the horizontal at constant velocity. The same ski is then launched up the same slope. While it moves up the slope, what is the magnitude of its deceleration?

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Marking pointMarkNotes
Sliding down at constant velocity means kinetic friction equals the weight component along the slope: F = mg sin θ. Moving up the slope, friction now acts down the slope, in the same direction as the weight component. The resultant force is 2mg sin θ down the slope, so the deceleration is 2g sin θ. g sin θ ignores friction. Zero wrongly assumes that constant velocity carries over to motion in the opposite direction.✓ 1Answer B

Answer: B

Exam tip: When an object reverses direction on a slope, redraw the friction arrow. Kinetic friction always opposes the current direction of sliding.

81IBP2S-003-02-05Explosions & recoil Paper 1AMedium1 mark

An astronaut of mass 90 kg, including her spacesuit but not the wrench she is holding, floats at rest next to a space station. She throws a 1.5 kg wrench, which moves away at 6.0 m s−1 relative to the station. What is her speed relative to the station immediately afterwards?

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Marking pointMarkNotes
Total momentum is zero before and after the throw, so momentum is conserved: 90v = 1.5 × 6.0. This gives v = 9.0/90 = 0.10 m s−1.
The distractors come from these errors:
• 0.098 m s−1: dividing by the combined mass of 91.5 kg, although only the 90 kg astronaut recoils at this speed.
• 0.77 m s−1: wrongly setting the two kinetic energies equal.
• 9.0 m s−1: this is the wrench's momentum in kg m s−1, not a speed.
✓ 1Answer B

Answer: B

Common error: To find the recoil speed, divide the momentum by the mass of the recoiling body only. Do not add the mass of the projectile.

82IBP2S-003-02-06Explosions & recoil Paper 1AMedium1 mark

A stationary nucleus with nucleon number 226 emits an alpha particle (nucleon number 4). The daughter nucleus has nucleon number 222. Treat the masses as proportional to the nucleon numbers. What fraction of the total kinetic energy released is carried by the alpha particle?

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Marking pointMarkNotes
The momenta are equal and opposite, so Ek = p2/2m is inversely proportional to mass. Eα : Edaughter = 222 : 4. The alpha particle's fraction is therefore 222/(222 + 4) = 222/226. 4/226 is the daughter's share, which comes from assuming energy is proportional to mass. 1/2 assumes the energy is shared equally. 4/222 is the ratio of masses rather than a fraction of the total.✓ 1Answer D

Answer: D

Exam tip: In a two-body break-up, each fragment's share of the kinetic energy equals the other fragment's mass divided by the total mass.

83IBP2S-003-02-08Friction & inclined planes Paper 1AMedium1 mark

Two aluminium blocks, P of mass m and Q of mass 2m, are placed side by side on a wooden board. One end of the board is slowly raised. Both blocks have the same coefficient of static friction with the board. Which statement is correct?

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Marking pointMarkNotes
Slipping begins when mg sin θ = μmg cos θ, which gives tan θ = μ. The mass cancels, so both blocks slip at the same angle. Option A considers only the larger weight component. Option B considers only the smaller normal force. Option D considers only the larger maximum friction. In each case the other force also scales with mass, so the effect cancels.✓ 1Answer C

Answer: C

Exam tip: The tilt-until-slip method measures μs = tan θ directly, and the result does not depend on the mass of the block.

84IBP2S-003-02-10Variable mass & F = Δp/Δt Paper 1AMedium1 mark

A satellite's ion thruster ejects xenon ions at a steady rate of 3.0 mg s−1. The ions leave with a speed of 30 km s−1 relative to the satellite. What is the thrust on the satellite?

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Marking pointMarkNotes
Thrust = rate of change of momentum of the ejected ions = (Δm/Δt)v = (3.0×10−6 kg s−1)(3.0×104 m s−1) = 0.090 N. 9.0×10−5 N uses 30 m s−1 instead of 30 km s−1. 90 N treats mg as grams. 0.045 N wrongly includes a factor of ½, as if from a kinetic energy expression.✓ 1Answer B

Answer: B

Exam tip: For any jet or exhaust, F = (Δm/Δt) × v. Convert mg s⁻¹ to kg s⁻¹ (× 10⁻⁶) before multiplying.

85IBP2S-003-06-04Variable mass & F = Δp/Δt Paper 1AMedium1 mark

A horizontal jet of water strikes a flat vertical wall. The water runs down the wall with negligible horizontal velocity, and the force on the wall is F. The same jet is now aimed at a curved surface that sends the water straight back horizontally with its speed unchanged. What is the force on the curved surface?

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Marking pointMarkNotes
For the flat wall, each kilogram of water changes velocity from v to 0, so the momentum change per second is (Δm/Δt)v. When the water is turned back, its velocity changes from +v to −v, a change of 2v. The mass flow rate is the same, so the force doubles to 2F. Answer B assumes that only the speed matters, not the reversal of direction. Answer D wrongly squares the factor of 2.✓ 1Answer C

Answer: C

Exam tip: Momentum is a vector. A reversal of direction gives a change of 2mv, not zero and not mv.

86IBP2S-003-06-05Collisions & energy Paper 1AMedium1 mark

An open-topped railway wagon rolls freely along a straight, horizontal, frictionless track. Rain falls vertically into the wagon and collects inside it. Air resistance is negligible. Which statement describes the wagon and the water it contains?

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Marking pointMarkNotes
There is no external horizontal force, so the horizontal momentum of the wagon plus the collected water is conserved. The raindrops arrive with zero horizontal velocity and must be accelerated up to the wagon's speed. Because the total mass increases while mv stays constant, the speed must decrease. Option C forgets that no external horizontal force acts. Options B and D ignore the need to give horizontal momentum to the rain.✓ 1Answer A

Answer: A

Common error: Thinking that vertical rain cannot affect horizontal motion. The rain has to gain horizontal momentum, and that momentum comes from the wagon.

87IBP2S-003-06-06Variable mass & F = Δp/Δt Paper 1AMedium1 mark

A leaf blower forces air out through a nozzle of fixed cross-sectional area. The air leaves at speed v after starting from rest. The fan is adjusted so that the air leaves the same nozzle at speed 2v. By what factor does the reaction force on the blower increase?

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Marking pointMarkNotes
The mass of air leaving per second is ρAv, and each kilogram gains momentum v, so F = ρAv2. Doubling v doubles both the mass flow rate and the momentum per kilogram, so the force increases by a factor of 4. Answer 2 forgets that the mass flow rate also rises. Answer 8 confuses the force with power, which is proportional to v3.✓ 1Answer C

Answer: C

Common error: Treating the mass flow rate as fixed when the speed changes. For a fixed nozzle area, the mass flow rate is itself proportional to v.

88IBP2S-003-06-07Variable mass & F = Δp/Δt Paper 1AMedium1 mark

A helicopter of mass 1200 kg hovers at rest. Its rotor pushes air that was initially at rest vertically downwards at a speed of 20 m s−1. What is the mass of air pushed downwards each second?

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Marking pointMarkNotes
To hover, the upward force from the air must equal the helicopter's weight: 1200 × 9.8 = 1.18 × 104 N. This force equals the rate of change of momentum of the air, (Δm/Δt) × 20. So Δm/Δt = 1.18 × 104/20 ≈ 590 kg each second. 60 kg uses the mass instead of the weight. 290 kg introduces a spurious factor of ½. 2.4 × 105 kg multiplies the weight by the speed instead of dividing.✓ 1Answer C

Answer: C

Exam tip: For a hovering vehicle, set weight = (Δm/Δt) × v. The equilibrium condition tells you the force before you consider the momentum.

89IBP2S-003-06-08Variable mass & F = Δp/Δt Paper 1AMedium1 mark

An ion thruster on a satellite produces a thrust of 0.090 N. The ions leave the thruster with a speed of 30 km s−1 relative to the satellite. What is the mass of ions ejected per second?

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Marking pointMarkNotes
Thrust = (Δm/Δt)v, so Δm/Δt = 0.090/(3.0 × 104) = 3.0 × 10−6 kg s−1. 3.0 × 10−3 kg s−1 uses the speed in km s−1 instead of m s−1. 2.7 × 103 kg s−1 multiplies the thrust by the speed. 3.3 × 105 kg s−1 divides the speed by the thrust.✓ 1Answer A

Answer: A

Common error: Leaving speeds in km s⁻¹. Convert to m s⁻¹ before using F = (Δm/Δt)v.

90IBP2S-003-01-03Drag & buoyancy Paper 1AMedium1 mark

In a blood sedimentation test, red blood cells are modelled as small spheres falling at terminal speed through plasma, with Stokes's law applying. In some illnesses the cells clump into aggregates. Model each aggregate as a sphere of the same density as a single cell but with 3 times the radius. What is the ratio

terminal speed of an aggregate ÷ terminal speed of a single cell?

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Marking pointMarkNotes
At terminal speed, weight − buoyant force = drag. The weight and the buoyant force are both proportional to volume, so the net downward force is proportional to r3. The Stokes drag is 6πηrv. Setting r3 ∝ rv gives v ∝ r2. Tripling r therefore multiplies the terminal speed by 32 = 9. Answer 27 ignores the fact that drag also grows with r. Answer 3 ignores the fact that weight grows as r3.✓ 1Answer B

Answer: B

Common error: Remember that the radius appears on both sides of the force balance: r³ in the weight and buoyancy, and r in the drag. Leaving out either one gives 27 or 3 instead of 9.

91IBP2S-003-01-04Drag & buoyancy Paper 1AMedium1 mark

A helium weather balloon carrying a radiosonde rises vertically through still air at constant velocity. Which statement about the forces on the balloon and its load is correct?

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Marking pointMarkNotes
Constant velocity means the resultant force is zero (Newton's first law). The balloon moves upward, so air resistance acts downward, in the same direction as the weight. The upward buoyant force must therefore balance weight + drag. Option C describes an accelerating balloon. Option D puts the drag in the wrong direction. Option A ignores the drag.✓ 1Answer B

Answer: B

Exam tip: First decide the direction of motion, then point drag opposite to it. Only after that write the force balance.

92IBP2S-003-01-06Drag & buoyancy Paper 1AMedium1 mark

A floating hydrometer is used to test the electrolyte in a lead–acid battery. It floats upright, first in pure water and then in battery acid, which is denser than water. Compared with floating in water, how do the buoyant force on the hydrometer and the volume of the hydrometer below the surface change when it floats in the acid?

Show mark scheme
Marking pointMarkNotes
A floating object is in equilibrium, so the buoyant force equals its weight in both liquids. The buoyant force is therefore the same. Since buoyant force = ρfluidVsubg is constant, a larger ρfluid requires a smaller submerged volume, so the hydrometer floats higher in the acid.✓ 1Answer B

Answer: B

Exam tip: For any floating object, buoyant force = weight. Use this first, then use ρVg to find what happens to the submerged volume.

93IBP2S-003-01-07Variable mass & F = Δp/Δt Paper 1AMedium1 mark

A camera drone of weight 12 N hovers at a fixed position in a steady horizontal wind. The wind exerts a horizontal drag force of 5.0 N on the drone. What is the magnitude of the thrust that the rotors must provide?

Show mark scheme
Marking pointMarkNotes
The drone is in equilibrium, so the thrust must balance the vector sum of the weight (vertical) and the drag (horizontal). These two forces are perpendicular, so thrust = √(122 + 5.02) = √169 = 13 N. The thrust is therefore tilted into the wind. Answers 17 N and 7.0 N add or subtract the magnitudes as if the forces were in a straight line. Answer 12 N ignores the wind.✓ 1Answer C

Answer: C

Common error: Perpendicular forces combine using Pythagoras, not by simple addition or subtraction.

94IBP2S-003-01-09Newton's laws & force pairs Paper 1AMedium1 mark

A skydiver falls vertically at a terminal speed of 50 m s−1 and then opens her parachute. Immediately after the parachute opens, what are the directions of her velocity and of the resultant force on her?

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Marking pointMarkNotes
Opening the parachute greatly increases the drag, which becomes larger than her weight. The resultant force is therefore upward, so she decelerates. She is still moving downward, however, so her velocity is downward. Velocity and resultant force need not be in the same direction.✓ 1Answer B

Answer: B

Common error: The parachute does not make the skydiver move upward. The upward resultant force only reduces her downward speed until a new, lower terminal speed is reached.

95A-1A-24Friction & inclined planes Paper 1AHard1 mark

A crate rests on a rough horizontal floor. A horizontal force P is applied to the crate and is increased slowly from zero. The coefficient of static friction is greater than the coefficient of dynamic friction.

Which graph shows how the frictional force F on the crate varies with P?

Four sketches of frictional force F against applied force P.A.FP0B.FP0C.FP0D.FP0
Sketch graphs, not to scale
Show mark scheme
Marking pointMarkNotes
Step 1While the crate is at rest the friction is static and balances the applied force exactly, F = P, up to the limit μsFN.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Once P exceeds μsFN the crate slides and the friction drops to the constant dynamic value μdFN, which is smaller.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis rises correctly while the crate is at rest but then stays at the maximum static value. Once the crate slides, the friction is μdFN, which is less than μsFN.
  • BThis assumes the friction always equals the applied force, so the crate would never move. Static friction has a maximum value, μsFN.
  • CCorrect: F = P while the crate is at rest (Ff ≤ μsFN), then a drop to the constant sliding value Ff = μdFN.
  • DThis gives the crate a friction force even when no force is applied. With P = 0 there is nothing for friction to oppose, so F = 0.
96A-1A-26Drag & buoyancy Paper 1AHard1 mark

Two solid spheres X and Y, made of the same steel, fall vertically through the same oil at their terminal speeds. The radius of Y is twice the radius of X. The viscous drag on a sphere is Fd = 6πηrv.

What is (terminal speed of Y)/(terminal speed of X)?

Show mark scheme
Marking pointMarkNotes
Step 1At terminal speed the forces balance: Fd + Fb = weight, so 6πηrv = (ρsteel − ρoil) × (4/3)πr³g.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The right-hand side goes as r³ and the drag as rv, so v ∝ r²: the ratio is 2² = 4.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis assumes all spheres fall at the same speed, as they would in free fall. In a viscous fluid the balance between drag, buoyancy and weight depends on the radius.
  • BThis takes the weight (and buoyancy) to be proportional to r², like an area. They depend on the volume, which goes as r³.
  • CCorrect: weight − buoyancy ∝ r³ and Fd ∝ rv, so v ∝ r² and doubling the radius gives 4 times the terminal speed.
  • DThis scales the terminal speed with the volume, r³, forgetting that the drag 6πηrv also increases with the radius.
97A-1A-31Variable mass & F = Δp/Δt Paper 1AHard1 mark

Sand falls vertically onto a horizontal conveyor belt at a constant rate of R kilograms per second and is carried away by the belt. The belt moves at a constant speed v. Friction in the belt mechanism is negligible.

What are the horizontal force that the motor must exert on the belt and the power the motor must deliver?

ForcePower
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Marking pointMarkNotes
Step 1Each second a mass R of sand gains horizontal velocity v, so the belt must supply a horizontal force F = Δp/Δt = Rv.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The motor pushes the belt with this force at speed v, so P = Fv = Rv². Only half of this, ½Rv², becomes kinetic energy of the sand; the rest is dissipated while the sand slips on the belt.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe belt moves at constant speed, but the mass on it keeps increasing: new sand must be accelerated continuously, so a force is needed (F = Δp/Δt, not F = ma with constant mass).
  • BThis starts from the rate of gain of kinetic energy, ½Rv², and divides by v to get a force. The force comes from the rate of change of momentum, Rv.
  • CThe force is right, but this power is only the rate at which the sand gains kinetic energy. The motor must also supply the energy dissipated as the sand slips before reaching speed v.
  • DCorrect: F = Δp/Δt = Rv and P = Fv = Rv², twice the rate at which the sand gains kinetic energy.
98A-1A-39Collisions & energy Paper 1AHard1 mark

A body of mass m moving with speed u collides with a stationary body of mass M. The two bodies stick together. No external force acts.

What fraction of the initial kinetic energy remains as kinetic energy after the collision?

Show mark scheme
Marking pointMarkNotes
Step 1Momentum is conserved, so the momentum p is the same before and after; only the moving mass changes, from m to m + M.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2With Ek = p²/2m: Eafter/Ebefore = (p²/2(m + M))/(p²/2m) = m/(m + M).✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: at equal momentum Ek ∝ 1/mass, so the kinetic energy falls by the factor m/(m + M).
  • BThis is the fraction of the kinetic energy that is lost (transferred to internal energy), not the fraction that remains.
  • CThis is (V/u)², the square of the speed ratio, which forgets that the moving mass has increased from m to m + M.
  • DThis is the initial kinetic energy divided by the final one: the ratio is inverted, giving a value greater than 1.
99A-1A-41Circular motion Paper 1AHard1 mark

A small bob on a light string moves at constant speed in a horizontal circle of radius r. The string makes a constant angle θ with the vertical, as shown.

What is the speed of the bob?

A conical pendulum: a bob on a string moving in a horizontal circle of radius r, the string at angle theta to the vertical.rθ
Diagram NOT accurately drawn
Show mark scheme
Marking pointMarkNotes
Step 1The bob has no vertical acceleration, so T cos θ = mg; the horizontal component provides the centripetal force, T sin θ = mv²/r.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Dividing: tan θ = v²/gr, so v = √(gr tan θ).✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis takes the tension to be equal to the weight, T = mg. Only the vertical component of the tension, T cos θ, balances the weight.
  • BCorrect: T cos θ = mg and T sin θ = mv²/r give v² = gr tan θ.
  • CThis swaps the components of the tension. With θ measured from the vertical, the horizontal component is T sin θ and the vertical component is T cos θ.
  • DThis sets the centripetal force equal to the weight. The centripetal force is the horizontal component of the tension, mg tan θ.
100A-1A-44Collisions & energy Paper 1AHard1 mark

Trolley X moves towards a stationary trolley Y along a horizontal track. A spring buffer on Y is compressed during the collision, which is elastic. Friction is negligible.

Which statement is correct at the instant when the spring is most compressed?

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Marking pointMarkNotes
Step 1The spring stops being compressed further when the trolleys stop approaching each other, i.e. when they have the same velocity.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Momentum is conserved throughout (no external force), but kinetic energy is temporarily stored as elastic potential energy in the spring, which is at its maximum, so the kinetic energy is at its minimum.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe total momentum was not zero to begin with (X was moving), and it cannot change without an external force, so the trolleys cannot both be at rest.
  • B'Elastic' means the total kinetic energy after the collision equals that before it. During the collision some of it is stored as elastic potential energy in the spring.
  • CCorrect: the trolleys move with a common velocity, the elastic potential energy in the spring is at its maximum, and so the kinetic energy is at its minimum.
  • DMomentum cannot be stored in a spring. With no external force the total momentum is constant at every instant of the collision.
101A-1A-55Collisions & energy Paper 1AHard1 mark

Trolley X of mass 1.0 kg moves at 4.0 m s−1 along a horizontal track towards a stationary trolley Y of mass 3.0 kg. Friction is negligible. Velocities are positive in the original direction of X.

Which row could give the velocities of X and Y after an elastic collision?

X 1.0 kg4.0 m s⁻¹Y 3.0 kgat rest
Diagram not to scale
Velocity of X / m s−1Velocity of Y / m s−1
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Marking pointMarkNotes
Step 1Before: total momentum = 1.0 × 4.0 = 4.0 kg m s−1; total kinetic energy = ½ × 1.0 × 4.0² = 8.0 J. An elastic collision must conserve both.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Row D: momentum = 1.0(−2.0) + 3.0(+2.0) = 4.0 kg m s−1; kinetic energy = 2.0 + 6.0 = 8.0 J. Both are conserved.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis row conserves kinetic energy (2.0 J + 6.0 J = 8.0 J) but not momentum (2.0 + 6.0 = 8.0 kg m s−1, not 4.0), so it is impossible with no external force.
  • BThe trolleys move off together: momentum is conserved (4.0 kg m s−1) but only 2.0 J of the 8.0 J of kinetic energy remains. This is a perfectly inelastic collision.
  • CMomentum is conserved (−0.50 + 4.5 = 4.0 kg m s−1) but the kinetic energy falls to 3.5 J, so the collision is inelastic even though X rebounds.
  • DCorrect: momentum stays 4.0 kg m s−1 and kinetic energy stays 8.0 J, as an elastic collision requires.
102A-1A-58Friction & inclined planes Paper 1AHard1 mark

A student holds a book of mass m at rest against a vertical wall by pushing it horizontally towards the wall with a force P. The coefficient of static friction between the book and the wall is μs. The book does not slide.

What is the magnitude of the frictional force exerted by the wall on the book?

book, mass mPwall
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Step 1The book is in equilibrium, so vertically the upward frictional force balances the weight: Ff = mg.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Here FN = P, and Ff ≤ μsFN gives only the greatest possible static friction; the friction equals μsP only when the book is on the point of sliding.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis is the maximum static friction, μsFN with FN = P. Static friction is only as large as it needs to be, which here is the weight of the book.
  • BCorrect: the book is in equilibrium, so the upward frictional force balances the weight mg, whatever the size of P.
  • CThis uses μsFN with FN = mg, as for a book lying on a table. The normal force here is the horizontal push, and the friction need not be at its maximum value.
  • DThis is the normal force exerted by the wall, which balances the horizontal push. It acts perpendicular to the wall, not along it.
103A-1A-60Circular motion Paper 1AHard1 mark

An aerobatic aircraft flies in a vertical circle of radius r. At the top of the circle the pilot is upside down, and the seat pushes on the pilot with a force equal to three times the pilot's weight.

What is the speed of the aircraft at the top of the circle?

Show mark scheme
Marking pointMarkNotes
Step 1At the top both the weight mg and the push of the seat, 3mg, act vertically downwards, towards the centre of the circle.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2mv²/r = 3mg + mg = 4mg, so v = √(4gr) = 2√(gr).✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis takes the weight as the only centripetal force, which is the condition for the seat to exert no force. Here the seat also pushes towards the centre.
  • BThis treats the weight as acting away from the centre (3mg − mg = mv²/r), as at the bottom of the circle. At the top the weight also points towards the centre.
  • CThis uses only the force of the seat as the centripetal force and leaves out the weight of the pilot.
  • DCorrect: the seat force and the weight both point towards the centre, giving 4mg = mv²/r.
104A-1A-91Resultant forces Paper 1AHard1 mark

A rescue helicopter winches a casualty of mass m vertically upwards on a cable. Near the top, the casualty is still moving upwards but is slowing down with a deceleration of magnitude a, where a < g. Air resistance is negligible.

What is the tension in the cable?

Show mark scheme
Marking pointMarkNotes
Step 1The casualty is slowing down while moving upwards, so the acceleration is downwards, of magnitude a.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Resultant force downwards: mg − T = ma, so T = m(g − a).✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the acceleration is downwards, so the weight must exceed the tension: T = m(g − a).
  • BThis takes the acceleration to be upwards because the casualty is moving upwards. The direction of the acceleration is that of the change in velocity, which is downwards here.
  • CThis assumes the forces balance because the casualty is moving at the instant considered. The velocity is changing, so there is a resultant force.
  • DThis takes the tension as the resultant force and leaves out the weight of the casualty.
105A-1A-94Momentum & impulse Paper 1AHard1 mark

A car of mass m drives round a horizontal semicircular bend of radius r at a constant speed v, so that it leaves the bend travelling in the opposite direction to the one in which it entered. Which row gives the magnitude of the impulse of the resultant force on the car and the magnitude of the average resultant force on the car while it goes round the bend?

Magnitude of impulseMagnitude of average resultant force
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Marking pointMarkNotes
Step 1The velocity reverses, so the change in momentum is mv − (−mv) = 2mv: the impulse of the resultant force is 2mv.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The time taken is Δt = πr/v, so the average resultant force is J/Δt = 2mv/(πr/v) = 2mv2/(πr).✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis takes the momentum to be unchanged because the speed is unchanged. Momentum is a vector, and the velocity has reversed.
  • BThe impulse is right, but the average force has been taken as the magnitude of the centripetal force, mv2/r. The direction of this force keeps changing, so its average over the bend is smaller: 2mv2/(πr).
  • CCorrect: the impulse equals the change in momentum, 2mv, and dividing by the time πr/v gives the average resultant force.
  • DThis multiplies the magnitude of the centripetal force, mv2/r, by the time πr/v. The direction of the force changes all the time, so its magnitude cannot simply be multiplied by the time.
106A-1A-98Collisions & energy Paper 1AHard1 mark

Two bodies collide head-on along a straight line. No external force acts on them.

Which statement is correct?

Show mark scheme
Marking pointMarkNotes
Step 1Momentum is conserved in every collision with no external force; kinetic energy is conserved only in an elastic collision.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2With momenta equal and opposite the total momentum is zero, so bodies that stick together are at rest afterwards and all the kinetic energy has been transferred.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • ABodies that stick together still move with the total momentum they had before. Their kinetic energy falls to zero only if that total momentum is zero.
  • BMomentum is conserved whenever no external force acts, whether the collision is elastic or inelastic. Only kinetic energy is lost in an inelastic collision.
  • CCorrect: zero total momentum means the joined bodies are at rest, so none of the kinetic energy remains.
  • DBodies that bounce apart may still have less total kinetic energy than before. A collision is elastic only if the total kinetic energy is unchanged.
107A-1A-101Circular motion Paper 1AHard1 mark

A ball on a light string moves in a vertical circle with centre O. The diagram shows the ball at the instant when the string is horizontal, as the ball moves upwards. Air resistance is negligible.

Which arrow could show the direction of the resultant force on the ball at this instant?

Ball on a string in a vertical circle at the instant the string is horizontalOvABCD
Diagram NOT accurately drawn
Show mark scheme
Marking pointMarkNotes
Step 1Two forces act on the ball: the tension, horizontally towards O, and the weight, vertically downwards.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Their resultant points below the horizontal and towards the side of O: it has a component towards the centre and a component opposite to the velocity, which slows the ball.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis assumes the resultant always points to the centre. That is true only when the speed is constant; here the weight is along the path and slows the ball.
  • BCorrect: the tension (towards O) plus the weight (downwards) gives a resultant pointing down and towards the centre side.
  • CThis includes only the weight. The string must also pull towards O to give the ball its centripetal acceleration.
  • DThis adds an upward force to keep the ball moving upwards. No force acts along the velocity; the weight acts against it.
108A-1A-103Variable mass & F = Δp/Δt Paper 1AHard1 mark

A cart carrying an open tank of water rolls along a straight, horizontal track. Friction and air resistance are negligible. Water drains out of a hole in the floor of the tank, falling vertically downwards relative to the cart.

Which statement about the speed of the cart is correct?

Show mark scheme
Marking pointMarkNotes
Step 1As it leaves, the water has the same horizontal velocity as the cart: its velocity relative to the cart is vertical.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2So the water and the cart exert no horizontal force on each other, and the cart's velocity does not change: the momentum carried away, (Δm)v, is in the same proportion as the mass carried away.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis applies conservation of momentum to the cart alone. The water that leaves takes its share, mv, of the momentum, so the cart's momentum falls in proportion to its mass.
  • BThe water does carry momentum away, but it also carries away mass in the same proportion, so the cart's velocity is unchanged.
  • CA rocket ejects gas backwards relative to itself. Here the water has no horizontal velocity relative to the cart, so it exerts no horizontal push.
  • DCorrect: with no horizontal force between the water and the cart, the cart's velocity does not change.
109IBP2S-003-04-08Collisions & energy Paper 1AHard1 mark

In a nuclear reactor, a neutron of mass m makes a head-on elastic collision with a stationary carbon nucleus of mass 12m. What fraction of the neutron's kinetic energy is transferred to the carbon nucleus?

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Marking pointMarkNotes
Conserving momentum and kinetic energy for this head-on elastic collision gives the neutron a final velocity of −11v/13 and the carbon nucleus 2v/13. The carbon nucleus's kinetic energy is ½(12m)(2v/13)2 = (48/169) × ½mv2. Option C is the fraction the neutron keeps. Option B treats the collision as perfectly inelastic. Option D shares the energy in proportion to mass.✓ 1Answer A

Answer: A

Exam tip: A light particle hitting a heavy one hands over only a small fraction of its energy. This is why moderators use light nuclei.

110IBP2S-003-03-08Explosions & recoil Paper 1AHard1 mark

In deep space, a crew capsule of mass 300 kg is attached to a service stage of mass 900 kg. The system is initially at rest in a particular inertial frame. A compressed spring then pushes the capsule and the stage apart.
What is \(\frac{\text{kinetic energy of capsule}}{\text{kinetic energy of stage}}\) after separation?

Show mark scheme
Marking pointMarkNotes
The total momentum is zero, so the capsule and the stage have momenta of equal magnitude p. Kinetic energy is Ek = p²/2m, so for equal p it is inversely proportional to mass. The ratio is 900/300 = 3. Option B gives the mass ratio the wrong way round. Option D squares the mass ratio, as if the velocities rather than the momenta were equal in magnitude. Option A combines both of these errors.✓ 1Answer C

Answer: C

Exam tip: When the total momentum is zero, use Ek = p²/2m. The lighter object always carries the larger share of the kinetic energy.

111IBP2S-003-03-09Drag & buoyancy Paper 1AHard1 mark

In a blood sedimentation test, spherical cells sink through plasma at terminal speed. Viscous drag on each cell is given by Stokes' law. Cell A has twice the radius of cell B. Both cells are made of the same material.
What is \(\frac{\text{terminal speed of A}}{\text{terminal speed of B}}\)?

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Marking pointMarkNotes
At terminal speed, weight − buoyancy = 6πηrv. Both weight and buoyancy are proportional to volume, so the net downward force is proportional to r³. Therefore 6πηrv ∝ r³, which gives v ∝ r². Doubling the radius multiplies the terminal speed by 4. Option A assumes the net force is proportional to r. Option C ignores the factor of r in the drag. Option D assumes that greater drag means a slower fall.✓ 1Answer B

Answer: B

Common error: A larger sphere does experience more drag at a given speed. However, its weight grows faster (∝ r³) than its drag (∝ r), so it falls faster.

112IBP2S-003-00-08Circular motion Paper 1AHard1 mark

On a fairground chair-swing ride, identical chains hang from points on a rotating canopy. A heavy adult and a light child sit in two chairs whose chains hang from the canopy at the same distance from the axis. The ride rotates at constant angular speed and the chairs move in horizontal circles. The masses of the chains and chairs are negligible. Which statement is correct?

Show mark scheme
Marking pointMarkNotes
For each rider, T cos θ = mg and T sin θ = mω2r, so tan θ = ω2r/g. The mass cancels. With the same ω, chain length and attachment point, r and θ are the same for both riders. The tension T = mg/cos θ is proportional to m, so it is greater for the adult. Options A and B wrongly assume the mass affects the angle. Option D wrongly assumes it does not affect the tension.✓ 1Answer C

Answer: C

Exam tip: Divide the horizontal equation by the vertical one to eliminate the tension. This often shows which quantities the angle depends on.

113IBP2S-003-05-08Connected bodies Paper 1AHard1 mark

A recovery truck tows a car along a level road using a light rope, and both vehicles are accelerating forward.
I. The force of the rope on the car is equal in magnitude and opposite in direction to the force of the car on the rope.
II. The vehicles accelerate because the truck pulls forward on the car harder than the car pulls backward on the truck.
III. The forward force of the road on the truck's driven wheels is greater than the total resistive force acting on the truck and the car.
Which statements are correct?

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Marking pointMarkNotes
I is correct: these two forces are a third-law pair. II is wrong: through a light rope, the truck's pull on the car and the car's pull on the truck are always equal in magnitude, whatever the motion. III is correct: treat the truck, rope and car as one system. The only external horizontal forces are the forward friction from the road on the driven wheels and the resistive forces. The system accelerates forward, so the forward force must be the larger.✓ 1Answer B

Answer: B

Exam tip: To explain why a system accelerates, consider only the external forces on that system. Internal third-law pairs, such as rope–car, always cancel.

114IBP2S-003-02-09Momentum & impulse Paper 1AHard1 mark

A signal flare moves horizontally with speed v at the highest point of its path. It splits into two halves of equal mass. Immediately after the split, one half has zero velocity. What is the speed of the other half immediately after the split?

Show mark scheme
Marking pointMarkNotes
The internal forces of the split do not change the total momentum, and gravity has no time to act during the split. Before the split the momentum is Mv, and afterwards it is (M/2)(0) + (M/2)u. So u = 2v. Option v assumes each half keeps the original velocity. Option √2 v comes from wrongly conserving kinetic energy, even though the split releases chemical energy. Option 4v comes from a further mass-halving error.✓ 1Answer C

Answer: C

Common error: Explosions conserve momentum but not kinetic energy. Never use energy conservation to find fragment speeds unless the energy released is given.

115IBP2S-003-06-09Variable mass & F = Δp/Δt Paper 1AHard1 mark

A cart loaded with sand rolls at constant speed along a frictionless horizontal track. Sand starts to leak through a small hole in the floor of the cart. Each grain leaves the hole moving horizontally with the same velocity as the cart. What happens to the speed of the cart while the sand leaks?

Show mark scheme
Marking pointMarkNotes
Each grain leaves with the same horizontal velocity as the cart, so no horizontal force acts between the grain and the cart as it leaves. With no horizontal force on the cart, its velocity is unchanged. The total momentum of the cart plus the leaked sand is conserved, but the leaked sand carries its own share of that momentum. The cart's speed therefore does not rise. A rocket accelerates only because its exhaust is pushed backwards relative to it.✓ 1Answer D

Answer: D

Common error: Applying p = mv to the cart alone and concluding that falling mass means rising speed. Ask whether the ejected mass is pushed relative to the body. If it is not, no force acts on the body.

116IBP2S-003-06-10Variable mass & F = Δp/Δt Paper 1AHard1 mark

In a water turbine test rig, a horizontal jet carrying 3.0 kg s−1 of water at 8.0 m s−1 strikes a fixed curved vane. The vane turns the water through 90° in the horizontal plane without changing its speed. What is the magnitude of the force exerted by the water on the vane?

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Marking pointMarkNotes
Each second, the water's momentum changes from 24 kg m s−1 in one direction to 24 kg m s−1 at 90° to it. The magnitude of the vector change is √(242 + 242) = 24√2 ≈ 34 kg m s−1 per second, so the force is 34 N. 0 N assumes that an unchanged speed means no force. 24 N counts only one component. 48 N treats the deflection as a complete reversal.✓ 1Answer B

Answer: B

Exam tip: Draw the initial and final momentum vectors tip-to-tail mentally. The change is a vector subtraction, not a difference of magnitudes.

117IBP2S-003-01-08Drag & buoyancy Paper 1AHard1 mark

A small steel ball is released from rest in a tall tube of oil. The drag on the ball obeys Stokes's law and the buoyant force on it is not negligible. Immediately after release, the acceleration of the ball is a0. What is the acceleration of the ball when its speed is half of its terminal speed?

Show mark scheme
Marking pointMarkNotes
At release v = 0, so there is no drag and ma0 = W − U. At terminal speed the drag equals W − U. Stokes drag is proportional to v, so at half the terminal speed the drag is ½(W − U). The resultant force is then ½(W − U), giving an acceleration of a0/2. Answer 3a0/4 would apply if drag were proportional to v2. Answer g/2 ignores the buoyant force. Answer a0 ignores the drag.✓ 1Answer C

Answer: C

Exam tip: For linear (Stokes) drag, the acceleration falls in proportion to the remaining speed: a = a₀(1 − v/vt).

118IBP2S-003-01-10Newton's laws & force pairs Paper 1AHard1 mark

A beaker of water stands on a top-pan balance, which reads 15.0 N. A metal block of weight 6.0 N hangs from a spring balance. The block is lowered until it is fully submerged in the water without touching the beaker. When fully submerged, the buoyant force on the block is 2.0 N. What are the new readings?

Show mark scheme
Marking pointMarkNotes
The water pushes up on the block with 2.0 N, so the spring balance needs to support only 6.0 − 2.0 = 4.0 N. By Newton's third law, the block pushes down on the water with 2.0 N, so the top-pan reading rises to 15.0 + 2.0 = 17.0 N. As a check, the two readings together (21.0 N) equal the total weight of the beaker, water and block. Option D adds the whole weight of the block to the top pan.✓ 1Answer B

Answer: B

Exam tip: Every buoyant force has a Newton's third law partner: the object pushes down on the fluid with an equal force. Use this to explain changes in balance readings.

119A-1B-11Explosions & recoil Paper 1BEasy9 marks

Two trolleys A and B are held at rest on a level track with a compressed spring between them. When the spring is released the trolleys move apart in opposite directions. Each trolley carries a card of length 50.0 ± 0.5 mm, which passes through a light gate close to the starting position. The light-gate timer has a resolution of 0.1 ms.

TrolleyMass / kgTime for card to pass gate / ms
A0.512 ± 0.00176.7
B1.030 ± 0.001156.6
A 0.512 kgB 1.030 kgcard 50.0 mmcard 50.0 mmlight gatelight gate
Diagram not to scale; trolleys shown before release
(a)

State the total momentum of the two trolleys before the spring is released.

(1)
(b)

Show that the speed of A after the release is about 0.65 m s−1.

(1)
(c)

Determine the momentum of A together with its absolute uncertainty.

(3)
(d)

The momentum of B is 0.329 ± 0.004 kg m s−1. Deduce whether the results are consistent with the conservation of momentum.

(2)
(e)

Calculate the total kinetic energy of the trolleys after the release and identify where this energy came from.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
zero «both trolleys at rest»✓ 1
Part (b)
v = 0.0500 / 0.0767 = 0.652 m s−1✓ 1Must see conversion to m and s OR answer to 3 s.f.
Part (c)
p = 0.512 × 0.652 = 0.334 kg m s−1✓ 1
fractional uncertainty = 0.5/50.0 + 0.1/76.7 + 0.001/0.512 = 0.013 «1.3 %»✓ 1Accept omission of the timing term (0.13 %). Award [0] for adding absolute uncertainties.
p = 0.334 ± 0.004 kg m s−1✓ 1MP3 is for matching precision; unit required. Accept ±0.005.
Part (d)
the momenta are in opposite directions, so the total after release is 0.334 − 0.329 = 0.005 kg m s−1 «should be zero»✓ 1
this difference is less than the combined uncertainty 0.008 kg m s−1 «the ranges overlap», so the results are consistent with conservation of momentum✓ 1Allow ECF from (c).
Part (e)
Ek = ½ × 0.512 × 0.652² + ½ × 1.030 × 0.319² = 0.161 J✓ 1Accept 0.16 J.
the elastic potential energy stored in the compressed spring✓ 1

Answers: (b) 0.652 m s−1  ·  (c) 0.334 ± 0.004 kg m s−1  ·  (e) 0.161 J (the remaining parts are explanations — see the table above)

120A-1B-24Friction & inclined planes Paper 1BEasy8 marks

A wooden block rests on a hinged board. A smartphone inclinometer app fixed to the board displays the angle θ of the board to the horizontal to the nearest 0.1°. The board is raised slowly until the block just begins to slide, and θ is recorded. Masses are fixed to the block to change its total mass m.

For m = 0.400 kg the five readings of θ are:

22.4° 23.6° 21.9° 23.1° 22.7°

The table gives the coefficient of static friction μs found in the same way for the other masses.

m / kgμs
0.2000.41 ± 0.02
0.400
0.6000.43 ± 0.02
0.8000.42 ± 0.01
block, mass mθhinge
Diagram not to scale
(a)

State the resolution of the inclinometer app.

(1)
(b)

For m = 0.400 kg, calculate the mean value of θ and its absolute uncertainty.

(2)
(c)

Show that, when the block is just about to slide, μs = tan θ.

(1)
(d)

Determine μs for m = 0.400 kg, together with its absolute uncertainty.

(2)
(e)

The student's hypothesis is that μs does not depend on the mass of the block. Discuss whether the data support this hypothesis.

(1)
(f)

Suggest one improvement to the method that would reduce the spread of the readings of θ.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
0.1°✓ 1
Part (b)
mean θ = 113.7/5 = 22.74 ≈ 22.7°✓ 1
uncertainty = ½ × (23.6 − 21.9) = ±0.85 ≈ ±0.9°✓ 1Accept ±0.85°. Do not accept ±0.1° (the resolution).
Part (c)
on the point of sliding: Ff = μsFN with Ff = mg sin θ and FN = mg cos θ, so μs = sin θ/cos θ = tan θ✓ 1Both components needed.
Part (d)
μs = tan 22.7° = 0.418✓ 1
tan 23.6° = 0.437 and tan 21.8° = 0.400, so μs = 0.42 ± 0.02✓ 1Allow ECF from (b). Accept ±0.01–0.02. MP2 is for matching precision of value and uncertainty.
Part (e)
the range for 0.400 kg (0.40–0.44) overlaps the ranges for all the other masses, so the data are consistent with μs being independent of mass / support the hypothesis✓ 1Allow ECF from (d). Must refer to the uncertainty ranges.
Part (f)
any one specific improvement, eg: raise the board very slowly and steadily, for example with a screw jack, so it does not overshoot the angle at which sliding starts; clean the surfaces / always place the block on the same part of the board; tap the board gently in the same way each time✓ 1Do not accept 'take more readings' (does not reduce the spread).

Answers: (b) 22.7 ± 0.9°  ·  (d) 0.42 ± 0.02 (the remaining parts are explanations — see the table above)

121A-1B-39Hooke's law & springs Paper 1BEasy10 marks

A student tests a steel spring from a trampoline. The spring is held in a test rig that stretches it in steps. At each step a force sensor gives the tension F in the spring and a digital calliper gives the length L of the spring. The unstretched length L0 is 165.0 mm. Each length is uncertain by ±0.2 mm because the calliper jaws cannot be placed exactly at the ends of the coils, and each force is uncertain by ±10 N. The extension is x = L − L0.

The graph shows F against x with the error bars for F. One point has not been plotted.

L / mmx / mmF / N
177.012.065
189.024.0121
201.036.0192
213.048.0248
225.0307
237.072.0378
249.084.0434
261.096.0463
273.0108.0482
Tension in the spring against extension020406080100120x / mm0100200300400500600F / N
Graph drawn to scale
(a)

State the absolute uncertainty in each value of x.

(1)
(b)

Complete the table and plot the missing point on the graph.

(1)
(c)

Draw, on the graph, the line of best fit for the data in the region where the spring obeys Hooke's law.

(1)
(d)

Estimate the largest tension for which the spring obeys Hooke's law.

(1)
(e)

Determine the spring constant k, in N m−1, using your line.

(2)
(f)

By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in k.

(2)
(g)

The manufacturer states that k = 5.0 kN m−1 ± 3 %. Deduce whether the student's result is consistent with this.

(1)
(h)

When all the force is removed after the last reading, the length of the spring is 166.8 mm. Outline what this shows.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
±0.4 mm «uncertainties in L and L0 add»✓ 1Do not accept ±0.2 mm.
Part (b)
x = 225.0 − 165.0 = 60.0 mm AND point plotted correctly «with its error bar»✓ 1
Part (c)
single straight line through the first 7 points, within all their error bars, passing close to the origin; the last two points are not used✓ 1Do not award a line forced through the two highest points or a curve through all the points.
Part (d)
about 434 N «the last point on the straight line, x = 84 mm»✓ 1Accept 420–460 N. Allow ECF from their line.
Part (e)
gradient from a large triangle on the line, with x converted to metres✓ 1Award [1] max for 5.2 N mm−1 without conversion.
k = 5200 N m−1 «5.2 kN m−1»✓ 1Accept 5.0–5.4 × 103 N m−1.
Part (f)
steepest and shallowest lines drawn through all the error bars of the Hooke's-law points✓ 1
Δk = ½(max − min) ≈ ½(5403 − 4938) ≈ 200 N m−1✓ 1Accept 100–300 N m−1. Allow ECF from their lines.
Part (g)
the result 5200 ± 200 N m−1 «5.0–5.4 kN m−1» overlaps the manufacturer's range 4.85–5.15 kN m−1, so it is consistent✓ 1Allow ECF from (e) and (f). A comparison of ranges is needed.
Part (h)
the spring does not return to its original length: it has been permanently deformed «stretched beyond its elastic limit»✓ 1OWTTE.

Answers: (a) ±0.4 mm  ·  (b) 60.0 mm  ·  (e) 5200 N m−1  ·  (f) ±200 N m−1 (the remaining parts are explanations — see the table above)

122A-1B-45Drag & buoyancy Paper 1BEasy10 marks

A student makes a hydrometer from a flat-bottomed plastic tube of external diameter 18.0 ± 0.1 mm with some lead shot inside it. The tube floats upright in sugar solutions of different concentrations, and the depth h of the bottom of the tube below the liquid surface is read from a scale on the tube to ±1 mm. The densities ρ of the solutions are taken from a data book.

The graph shows h against 1/ρ for five of the solutions.

Sugar concentration / %ρ / kg m−3(1/ρ) / 10−4 m³ kg−1h / mm
099810.02104
1010389.6399
20108196
3011278.8792
4011768.5088
5012308.1384
Depth of immersion against 1/density88.48.89.29.610(1/ρ) / 10⁻⁴ m³ kg⁻¹8084889296100104h / mm
Graph drawn to scale
(a)

Outline why the buoyancy force on the hydrometer is the same in every solution.

(1)
(b)

Show that h = m/(ρA), where m is the mass of the hydrometer and A is its cross-sectional area.

(1)
(c)

Complete the table for the 20 % solution, plot the point and draw the line of best fit.

(2)
(d)

Determine the gradient of the line in kg m−2.

(2)
(e)

Determine m.

(1)
(f)

The uncertainty in the value of m from the graph is ±5 %. The mass of the hydrometer measured on a balance is 26.30 g. Deduce whether the two values agree.

(1)
(g)

In a sports drink the hydrometer floats with h = 90 mm. Determine the density of the drink.

(1)
(h)

Suggest one change to the hydrometer that would make the readings of h more sensitive to changes in density.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
the hydrometer floats at rest in each solution, so the buoyancy force equals its weight, which is the same each time✓ 1
Part (b)
ρ(Ah)g = mg «Fb = ρVg with V = Ah», so h = m/(ρA)✓ 1Must see V = Ah.
Part (c)
1/ρ = 1/1081 = 9.25 × 10−4 m³ kg−1 AND point plotted✓ 1
single straight line with the points scattered within their error bars✓ 1
Part (d)
gradient from a large triangle, with h in metres and 1/ρ in m³ kg−1✓ 1
gradient = 104 kg m−2✓ 1Accept 98–106 kg m−2.
Part (e)
A = π × (9.0 × 10−3)² = 2.545 × 10−4 m² so m = gradient × A = 26.4 g✓ 1Accept 0.025–0.027 kg. Allow ECF from (d).
Part (f)
range 25.1–27.7 g includes 26.30 g, so the values agree✓ 1Allow ECF from (e).
Part (g)
ρ = gradient/h = 104/0.090 = 1150 kg m−3✓ 1Accept a value read from an extended line. Accept 1110–1180 kg m−3. Allow ECF from (d).
Part (h)
use a narrower tube «smaller A», so that h changes by more for the same change in density✓ 1Accept a heavier tube of the same diameter «greater m», provided it still floats.

Answers: (c) 9.25 × 10−4 m³ kg−1  ·  (d) 104 kg m−2  ·  (e) 26.4 g  ·  (g) 1150 kg m−3 (the remaining parts are explanations — see the table above)

123A-1B-46Resultant forces Paper 1BEasy10 marks

A load hangs at rest from the midpoint of a light cord. Each end of the cord is attached to a force sensor, and the two sensors are clamped at the same height. The sensors are moved apart in steps so that the angle θ between each half of the cord and the vertical increases. For each θ, measured with a protractor to ±1°, the mean tension T shown by the two sensors is recorded. Each value of T is uncertain by ±0.10 N.

The graph shows T against 1/cos θ with error bars. The horizontal error bar for θ = 70° has not been drawn.

θ / °1/cos θT / N
151.0354.24
301.1554.63
405.31
501.5566.19
602.0008.05
702.92411.54
Tension against 1/cos θ11.41.82.22.631/cos θ3456789101112T / N
Graph drawn to scale
(a)

Show that T = W/(2 cos θ), where W is the weight of the load.

(1)
(b)

Complete the table for θ = 40° and plot the point. The horizontal error bar is not required.

(1)
(c)

Calculate the absolute uncertainty in 1/cos θ for θ = 70° and draw the horizontal error bar for this point on the graph.

(2)
(d)

Draw the line of best fit.

(1)
(e)

Determine W using the gradient of the line.

(2)
(f)

The uncertainty in the gradient is ±3 %. The mass of the load is 0.800 kg. Deduce whether the value of W from the graph is consistent with this mass.

(1)
(g)

The line does not pass through the origin. Suggest a reason for this and state its effect on your answer to (e).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the load is in equilibrium, so the vertical components of the two tensions balance the weight: 2T cos θ = W «the horizontal components cancel»✓ 1Must see the vertical components.
Part (b)
1/cos 40° = 1.305 AND point plotted correctly✓ 1
Part (c)
½(1/cos 71° − 1/cos 69°) = ±0.14✓ 1Accept ±0.13 to ±0.15. Accept the larger of the two differences from 2.92.
horizontal error bar from about 2.79 to 3.07✓ 1Allow ECF from their uncertainty.
Part (d)
single straight line through all the error bars, not forced through the origin✓ 1
Part (e)
gradient = 3.89 N «= W/2»✓ 1Accept 3.8–4.1 N.
W = 2 × gradient = 7.78 N✓ 1Allow ECF from their gradient.
Part (f)
W = 0.800 × 9.81 = 7.85 N, which lies within 7.55–8.01 N, so it is consistent✓ 1Allow ECF from (e).
Part (g)
the sensors have a zero error «each reads about 0.2 N too high, eg they were not zeroed before use»✓ 1OWTTE.
a constant offset changes the intercept but not the gradient, so W is unaffected✓ 1

Answers: (b) 1.305  ·  (c) ±0.14  ·  (e) 7.78 N (the remaining parts are explanations — see the table above)

124A-1B-03Resultant forces Paper 1BMedium10 marks

A glider on a level air track is connected by a light thread, passing over a pulley at the end of the track, to a hanger. Slotted masses are moved one at a time from the glider to the hanger, so the total mass M of the glider, hanger and slotted masses stays the same. For each hanging mass m the acceleration a of the glider is found using two light gates. Each value of a is uncertain by ±0.02 m s−2.

m / g102030405060
a / m s−20.1640.3420.5550.7410.9501.130
Acceleration of the glider against hanging mass010203040506070m / g00.20.40.60.811.21.4a / m s⁻²
Graph drawn to scale
(a)

Explain why the slotted masses are moved from the glider to the hanger rather than taken from the bench.

(1)
(b)

Draw the line of best fit on the graph.

(1)
(c)

Determine the gradient of the line, in m s−2 kg−1.

(2)
(d)

Ignoring friction, mg = Ma. Use your answer to (c) to determine M.

(2)
(e)

The uncertainty in the gradient is ±3 %. M measured with a balance is 0.500 ± 0.001 kg. Deduce whether the two values of M agree.

(2)
(f)
(i)

The line of best fit meets the m axis at a positive value. Suggest a reason for this.

(1)
(ii)

Estimate the size of the force responsible.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
so the total mass being accelerated stays constant and a depends only on the hanging weight «the accelerating force»✓ 1OWTTE. Do not accept 'fair test' alone.
Part (b)
single straight line through all error bars, not forced through the origin✓ 1
Part (c)
gradient from a large triangle, with m converted to kg✓ 1Award [1] max for 0.0195 without conversion to kg.
19.5 m s−2 kg−1✓ 1Accept 19.0–20.1.
Part (d)
gradient = g/M so M = 9.81/gradient✓ 1
M = 0.502 kg✓ 1Accept 0.488–0.516 kg. Allow ECF from (c).
Part (e)
ΔM = 3 % of 0.502 = 0.015 kg «so M = 0.502 ± 0.015 kg»✓ 1
range 0.487–0.517 kg includes 0.500 kg, so the values agree✓ 1Allow ECF.
Part (f)(i)
a constant frictional force «at the pulley or on the track» must be overcome before the glider accelerates✓ 1Accept track not exactly level «sloping upwards towards the pulley».
Part (f)(ii)
intercept ≈ 1.9 g, so force = 0.0019 × 9.81 ≈ 0.02 N✓ 1Accept 0.01–0.03 N.

Answers: (c) 19.5 m s−2 kg−1  ·  (d) 0.50 kg  ·  (e) ±0.015 kg  ·  (f)(ii) 0.02 N (the remaining parts are explanations — see the table above)

125A-1B-10Friction & inclined planes Paper 1BMedium10 marks

A wooden block is pulled at a slow steady speed along a horizontal varnished-wood bench by a force sensor attached to a horizontal string. Masses are placed on the block to change the total mass m. For each m the mean sensor reading F during the steady part of the pull is recorded. The readings fluctuate, so each value of F is uncertain by ±0.08 N.

m / kgFN / NF / N
0.2502.450.97
0.4501.51
0.6506.382.22
0.8508.342.78
1.0503.50
1.25012.34.05
Mean pulling force against normal force02468101214normal force / N012345F / N
Graph drawn to scale
(a)

Explain why the block must be pulled at a constant speed.

(1)
(b)

Complete the table by calculating the normal force FN for the two missing rows.

(1)
(c)

Draw the line of best fit on the graph.

(1)
(d)

Determine the coefficient of dynamic friction μd between the block and the bench, together with its absolute uncertainty.

(3)
(e)

A data handbook gives μd = 0.40 for wood on wood. Comment on this in the light of your answer to (d).

(1)
(f)
(i)

The line of best fit does not pass through the origin. Identify a systematic error that would cause this.

(1)
(ii)

State the effect of this error on the value of μd.

(1)
(g)

Suggest one change that would reduce the fluctuation in the readings of F.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
at constant velocity the resultant force is zero «Newton's first law», so the pulling force equals the frictional force✓ 1
Part (b)
FN = mg: 4.41 N and 10.3 N✓ 1Both needed.
Part (c)
single straight line through all error bars, not forced through the origin✓ 1
Part (d)
μd = gradient = 0.32✓ 1Accept 0.31–0.33.
maximum and minimum gradient lines drawn through all the error bars✓ 1
Δμd = ½(max − min) ≈ 0.01, so μd = 0.32 ± 0.01✓ 1No unit. Accept Δμ 0.01–0.02. MP3 is for matching precision.
Part (e)
0.40 lies outside the range of the result, so the handbook value does not apply to these surfaces «friction depends on the condition of the surfaces, eg varnish, polish, moisture»✓ 1Allow ECF from (d).
Part (f)(i)
a zero error in the force sensor «it reads about 0.16 N with no force applied»✓ 1Accept a constant extra force such as the string rubbing on the edge of the bench.
Part (f)(ii)
none: a constant offset shifts the line without changing its gradient✓ 1
Part (g)
pull with a motor/winch at constant speed OR take the mean of the sensor data over a longer steady section OR clean the surfaces✓ 1Any one sensible, specific change.

Answers: (b) 4.41 N, 10.3 N  ·  (d) 0.32 ± 0.01 (the remaining parts are explanations — see the table above)

126A-1B-12Momentum & impulse Paper 1BMedium10 marks

A rubber ball of mass 0.160 ± 0.001 kg is dropped onto a force plate connected to a data logger. The graph shows the force F exerted by the plate on the ball during the bounce.

A high-speed video shows that the ball reaches the plate moving downwards at 5.40 ± 0.10 m s−1 and leaves it moving upwards at 3.62 ± 0.10 m s−1.

Force exerted by the plate on the ball against time012345678910t / ms050100150200250300F / N
Graph drawn to scale
(a)

Determine the impulse exerted by the plate on the ball, using the graph.

(2)
(b)

Calculate the magnitude of the change in momentum of the ball, using the video data.

(2)
(c)

Determine the absolute uncertainty in your answer to (b).

(2)
(d)

Deduce whether your answers to (a) and (b) agree.

(1)
(e)

Show that the impulse of the weight of the ball during the contact is less than 1 % of the change in momentum.

(1)
(f)

Determine the average force exerted by the plate on the ball during the contact.

(1)
(g)

The speeds are found from the positions of the ball in the video frames just before and just after the contact. Suggest one improvement to this measurement.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
impulse = area under the graph, by counting squares or an equivalent shape «one large square = 50 N × 1 ms = 0.05 N s»✓ 1
1.46 N s✓ 1Accept 1.38–1.54 N s.
Part (b)
velocities are in opposite directions, so Δv = 5.40 + 3.62 = 9.02 m s−1✓ 1Award [0] for 0.28 N s (speeds subtracted).
Δp = 0.160 × 9.02 = 1.44 N s✓ 1Accept kg m s−1.
Part (c)
fractional uncertainty = 0.20/9.02 + 0.001/0.160 = 0.028 «2.8 %»✓ 1Absolute uncertainties of the two speeds add because they are summed.
Δ(Δp) = 0.04 N s✓ 1Accept 0.03–0.05 N s. Allow ECF from (b).
Part (d)
range 1.40–1.48 N s includes the impulse from the graph, so they agree «within uncertainty»✓ 1Allow ECF from (a), (b), (c).
Part (e)
mgΔt = 0.160 × 9.81 × 0.0080 = 0.0126 N s, which is 0.9 % of 1.44 N s✓ 1Must see the contact time from the graph (8 ms).
Part (f)
F = J/Δt = 1.46/0.0080 = 182 ≈ 180 N✓ 1Accept Δp/Δt = 1.44/0.0080 = 180 N, since the impulse of the weight is negligible (e). Accept 175–190 N.
Part (g)
use a higher frame rate «so the positions used are closer to the moment of contact» OR use light gates immediately above the plate✓ 1OWTTE.

Answers: (a) 1.46 N s  ·  (b) 1.44 N s  ·  (c) ±0.04 N s  ·  (e) 0.013 N s  ·  (f) 180 N (the remaining parts are explanations — see the table above)

127A-1B-13Drag & buoyancy Paper 1BMedium10 marks

A metal cylinder of diameter 38.0 ± 0.5 mm hangs vertically from a force sensor. It is lowered into a salt solution in steps, and the sensor reading F is recorded for each depth h of the bottom of the cylinder below the liquid surface. The graph shows the data and the line of best fit.

h / cm012345678
F / N2.702.592.452.352.222.091.991.851.74
Force-sensor reading against depth of immersion012345678h / cm1.61.822.22.42.62.8F / N
Graph drawn to scale
(a)

The cross-sectional area of the cylinder is A and the density of the solution is ρ. Explain why F = W − ρgAh, where W is the weight of the cylinder.

(2)
(b)

Determine the gradient of the line in N m−1.

(2)
(c)

Determine ρ.

(2)
(d)

The uncertainty in the gradient is ±3 %. Determine the absolute uncertainty in ρ.

(2)
(e)

The solution was prepared to have a density of 1100 kg m−3. Comment on your result.

(1)
(f)

The force sensor was later found to read 0.03 N when nothing was attached to it. State the effect of this on the value of ρ.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
the upthrust equals the weight of liquid displaced: Fb = ρVg with V = Ah✓ 1
the cylinder is in equilibrium, so F + Fb = W, giving F = W − ρgAh✓ 1
Part (b)
uses a large triangle on the line and converts cm to m✓ 1Award [1] max for −0.121 N cm−1.
gradient = −12.1 N m−1✓ 1Accept −11.7 to −12.4 N m−1. Sign not required.
Part (c)
A = π × (0.0190)² = 1.134 × 10−3 m² and ρ = |gradient|/(gA)✓ 1
ρ = 12.1/(9.81 × 1.134 × 10−3) = 1088 ≈ 1090 kg m−3✓ 1Accept 1050–1120 kg m−3. Allow ECF from (b).
Part (d)
fractional uncertainty = 0.03 + 2 × (0.5/38.0) = 0.056✓ 1The area depends on D², so the fractional uncertainty in D is doubled. Award [1] max if the factor of 2 is omitted.
Δρ = 0.056 × 1088 ≈ 61 kg m−3✓ 1Accept 60–70 kg m−3.
Part (e)
range 1027–1149 kg m−3 includes 1100 kg m−3, so the result is consistent✓ 1Allow ECF.
Part (f)
no effect: every reading is shifted by the same amount, so the gradient «and ρ» is unchanged✓ 1

Answers: (b) −12.1 N m−1  ·  (c) 1090 kg m−3  ·  (d) ±61 kg m−3 (the remaining parts are explanations — see the table above)

128A-1B-25Collisions & energy Paper 1BMedium9 marks

Trolley A, of mass 0.500 ± 0.001 kg, is pushed along a level track towards a stationary trolley B, of mass 0.750 ± 0.001 kg. Hook-and-loop pads make the trolleys stick together. A light gate measures the velocity u of A just before the collision, and a second light gate measures the common velocity v just after it. The run is repeated with different values of u.

Each value of v is uncertain by ±0.010 m s−1; the uncertainty in u is negligible.

u / m s−1v / m s−1
0.3120.122
0.4250.160
0.5380.211
0.6610.262
0.8020.312
0.9150.360
Common velocity after the collision against the velocity of A before it00.20.40.60.81u / m s⁻¹00.10.20.30.4v / m s⁻¹
Graph drawn to scale
(a)

Show that, if momentum is conserved, v = 0.400u.

(1)
(b)

Draw the line of best fit on the graph.

(1)
(c)

By drawing lines of maximum and minimum gradient, determine the gradient of the line and its absolute uncertainty.

(2)
(d)

Deduce whether the data are consistent with the conservation of momentum.

(1)
(e)

For the run with u = 0.802 m s−1, calculate the fraction of the initial kinetic energy that is transferred to other forms in the collision.

(2)
(f)

Friction acts on the trolleys between the collision and the second light gate. State and explain its effect on the measured common velocities, and suggest how the experiment could be modified to reduce this effect.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
mAu = (mA + mB)v, so v = 0.500u/1.250 = 0.400u✓ 1Must see substitution.
Part (b)
single straight line through all error bars✓ 1Do not accept a line joined point to point.
Part (c)
gradient of best-fit line = 0.397✓ 1Accept 0.39–0.41.
lines of maximum and minimum gradient through the error bars, giving ±0.03✓ 1Accept ±0.02–0.04. Allow ECF from their lines.
Part (d)
the range 0.37–0.43 includes 0.400, so the data are consistent with the conservation of momentum✓ 1Allow ECF from (c).
Part (e)
initial Ek = ½ × 0.500 × 0.802² = 0.1608 J AND final Ek = ½ × 1.250 × 0.312² = 0.0608 J✓ 1
fraction transferred = 1 − 0.0608/0.1608 = 0.62✓ 1Accept 0.61–0.63. Award [2] for CNA. Award [1] for the fraction remaining, 0.38, given as the answer.
Part (f)
friction decelerates the joined trolleys before they reach the second gate, so each measured common velocity is smaller than the velocity just after the collision «the points lie below the line predicted by conservation of momentum»✓ 1OWTTE. Do not accept “the gradient changes” without the effect on the measured velocities.
tilt the track slightly so that a trolley given a push moves at constant velocity «friction-compensated track» OR place the second gate as close as possible to the collision point✓ 1Do not accept 'use a smoother track' without detail.

Answers: (c) 0.40 ± 0.03  ·  (e) 0.62 (the remaining parts are explanations — see the table above)

129A-1B-26Circular motion Paper 1BMedium11 marks

A tyre manufacturer publishes test results for one of its tyres. A test car is driven around flat, horizontal circles of different radius r on a dry asphalt test area, and the greatest steady speed v at which the car can go round each circle without sliding is recorded. The published speeds are given to the nearest 1 km h−1; take the uncertainty in each speed as ±1 km h−1.

A student uses the published data to find the coefficient of static friction μs between the tyres and the road. The graph shows v² against r with error bars.

r / mv / km h−1v / m s−1v² / m² s−2
154111.4130
255214.4209
406718.6346
6082
809526.4696
10010529.2851
Square of the maximum cornering speed against radius of the circle0102030405060708090100110r / m01002003004005006007008009001000v² / m² s⁻²
Graph drawn to scale
(a)

Complete the table for r = 60 m.

(2)
(b)

Show that, at the greatest speed, v² = μsgr.

(1)
(c)

Draw the line of best fit on the graph.

(1)
(d)

Determine μs using the graph.

(2)
(e)

Calculate the percentage uncertainty in v² for r = 15 m.

(2)
(f)

The uncertainty in the gradient is ±2 %. The manufacturer's advertising states that μs = 0.95 for this tyre on dry asphalt. Deduce whether the published test data support this claim.

(2)
(g)

Outline one advantage of finding μs from the gradient of the graph rather than from the result for a single circle.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = 82/3.6 = 22.8 «m s−1»✓ 1Accept 22.78 m s−1.
v² = 519 «m² s−2»✓ 1Accept 517–521. Allow ECF.
Part (b)
the centripetal force is provided by the greatest static friction: μsmg = mv²/r, so v² = μsgr✓ 1Must see FN = mg and mv²/r.
Part (c)
single straight line through all error bars, passing close to the origin✓ 1
Part (d)
gradient = 8.58 «m s−2»✓ 1Accept 8.44–8.75.
μs = gradient/g = 8.58/9.81 = 0.88✓ 1Accept 0.861–0.892. Allow ECF from their gradient.
Part (e)
percentage uncertainty in v = 1/41 × 100 = 2.4 %✓ 1
percentage uncertainty in v² = 2 × 2.4 = 4.9 %✓ 1Accept 5 %. Award [2] for CNA.
Part (f)
Δμs = 0.02 × 0.88 = ±0.02, so μs = 0.88 ± 0.02, in the range 0.86–0.90✓ 1Allow ECF from (d).
0.95 lies outside this range, so the test data do not support the claim✓ 1
Part (g)
the gradient uses all the data, so random errors in individual results are reduced, OR it shows whether v² is proportional to r / reveals a systematic error as a non-zero intercept✓ 1OWTTE.

Answers: (a) 22.8 m s−1, 519 m² s−2  ·  (d) μs ≈ 0.88  ·  (e) 4.9 %  ·  (f) 0.88 ± 0.02 (the remaining parts are explanations — see the table above)

130A-1B-41Circular motion Paper 1BMedium12 marks

A small bob hangs from a light string attached to a motor fixed to the ceiling. When the motor turns, the bob moves at constant speed in a horizontal circle and the string traces out a cone. For each motor setting, the vertical distance h from the point of support to the bob is measured with a metre rule, and the time for 10 revolutions is measured with a stopwatch. The uncertainty in each time for 10 revolutions is ±0.2 s.

The graph shows four of the values of T², where T is the period, with their error bars.

h / mtime for 10 revolutions / sT / sT² / s²
0.1508.080.808
0.25010.261.0261.053
0.35012.071.2071.457
0.45013.631.363
0.55015.041.5042.262
0.65016.321.6322.663
Square of the period against h for a conical pendulum00.10.20.30.40.50.60.7h / m00.511.522.53T² / s²
Graph drawn to scale
(a)

Explain why the bob is accelerating even though its speed is constant.

(1)
(b)

By considering the forces on the bob, show that T = 2π√(h/g). The radius of the circle is r and the string makes an angle θ with the vertical.

(2)
(c)

Calculate T² for h = 0.150 m together with its absolute uncertainty.

(2)
(d)

Calculate T² for h = 0.450 m, plot the two missing points and draw the line of best fit.

(2)
(e)

Determine g using the gradient of the line.

(2)
(f)

The uncertainty in this value of g is ±3 %. Comment on whether it is consistent with g = 9.81 m s−2.

(1)
(g)

The line does not pass through the origin. Suggest a systematic error in the measurement of h that would cause this, and state its effect on the value of g.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
its velocity changes direction continuously, so it has a «centripetal» acceleration towards the centre of the circle✓ 1Do not accept “the forces are unbalanced” alone.
Part (b)
vertically FT cos θ = mg AND horizontally FT sin θ = mω²r, so tan θ = ω²r/g✓ 1FT is the tension. Accept 4π²r/T² for ω²r.
tan θ = r/h, so ω² = g/h and T = 2π/ω = 2π√(h/g)✓ 1
Part (c)
T² = 0.808² = 0.653 s²✓ 1
fractional uncertainty = 2 × 0.02/0.808 = 0.050, so T² = 0.65 ± 0.03 s²✓ 1Accept ±0.03 to ±0.04 s². MP2 is for matching precision of value and uncertainty.
Part (d)
T² = 1.858 s² AND both points plotted correctly✓ 1Allow ECF from (c) for the first point.
single straight line within all the error bars, not forced through the origin✓ 1
Part (e)
gradient = 4.02 s² m−1✓ 1Accept 3.9–4.1 s² m−1.
g = 4π²/gradient = 9.81 m s−2✓ 1Accept 9.6–10.1 m s−2. Allow ECF from their gradient.
Part (f)
range 9.5–10.1 m s−2 includes 9.81 m s−2, so it is consistent✓ 1Allow ECF from (e).
Part (g)
h was measured to the top of the bob «or to the bottom of the motor's clamp» instead of to the centre of the bob, so every h is too small by the same amount «the line is shifted along the h-axis»✓ 1OWTTE.
the gradient is not affected, so g is unaffected✓ 1MP2 only scores if MP1 scores.

Answers: (c) 0.65 ± 0.03 s²  ·  (d) 1.858 s²  ·  (e) 9.81 m s−2 (the remaining parts are explanations — see the table above)

131A-1B-42Momentum & impulse Paper 1BMedium10 marks

A student stands still on a force plate in a crouched position and then jumps vertically upwards without first dipping down. The force plate is connected to a data logger. The graph is a simplified version of the record of the vertical force F exerted by the plate on the student. The student starts to push at t = 0 and leaves the plate at t = 0.32 s. Air resistance is negligible.

A video of the jump shows that the centre of mass of the student rises 0.25 ± 0.02 m after leaving the plate.

Force exerted by the plate on the student against time−0.1000.100.200.300.40t / s0300600900120015001800F / N
Graph drawn to scale
(a)

Show, using the graph, that the mass of the student is about 70 kg.

(1)
(b)

Outline why the area under the graph between t = 0 and t = 0.32 s is not equal to the momentum of the student at take-off.

(1)
(c)

Determine the impulse of the resultant force on the student between t = 0 and t = 0.32 s.

(3)
(d)

Calculate the speed of the student at take-off.

(1)
(e)

Determine whether your answer to (d) is consistent with the video measurement.

(2)
(f)

Suppose instead that the force plate had a zero error, so that every reading, including the reading with nothing on the plate, was 15 N too large. State and explain the effect of this zero error on the take-off speed calculated by the method of (a) to (d).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
before t = 0 the student is at rest, so F = weight = 690 N and m = 690/9.81 = 70.3 kg✓ 1Must see the weight read from the graph.
Part (b)
the weight of the student also acts «downwards» during the push, so the area is the impulse of the plate's force only, not of the resultant force✓ 1OWTTE.
Part (c)
area under the graph from 0 to 0.32 s = 117 + 165 + 99 = 381 N s✓ 1Accept 370–390 N s, by counting squares or by shapes.
impulse of the weight = 690 × 0.32 = 221 N s✓ 1
impulse of the resultant force = 381 − 221 = 160 N s «upwards»✓ 1Award [3] for CNA. Award [1] max for 381 N s «weight not subtracted».
Part (d)
v = 160/70.3 = 2.28 m s−1✓ 1Allow ECF from (a) and (c).
Part (e)
h = v²/2g = 2.28²/(2 × 9.81) = 0.26 m✓ 1Allow ECF from (d).
0.26 m lies within 0.23–0.27 m, so the results are consistent✓ 1A comparison using the uncertainty range is needed.
Part (f)
every reading, including the reading of the weight, is 15 N too large, so F − W and the impulse of the resultant force are unchanged✓ 1
but the mass is overestimated by 15/9.81 ≈ 1.5 kg «about 2 %», so the calculated speed is slightly too small «2.23 m s−1 instead of 2.28 m s−1»✓ 1Accept “negligible effect” only with this reasoning.

Answers: (a) 70.3 kg  ·  (c) 160 N s  ·  (d) 2.28 m s−1  ·  (e) 0.26 m (the remaining parts are explanations — see the table above)

132A-1B-43Newton's laws & force pairs Paper 1BMedium11 marks

A cart carries a battery-driven fan that pushes air backwards, so that the cart accelerates forwards along a level track. The thrust of the fan is the same in every run. Masses are added to the cart to change its total mass m, and the acceleration a is found from the velocity–time graph recorded by a motion sensor. Each value of a is uncertain by ±0.02 m s−2.

The graph shows a against 1/m, with error bars, for four of the six readings.

m / kg(1/m) / kg−1a / m s−2
0.5002.000.808
0.6001.670.650
0.7500.526
0.9001.110.421
1.1000.351
1.4000.710.254
Acceleration of the fan cart against 1/m00.40.81.21.62(1/m) / kg⁻¹−0.100.10.20.30.40.50.60.70.80.9a / m s⁻²
Graph drawn to scale
(a)

Outline why a is plotted against 1/m rather than against m.

(1)
(b)

Complete the table and plot the two missing points on the graph.

(2)
(c)

Draw the line of best fit, extending it to meet the vertical axis.

(1)
(d)

Determine the gradient of the line, including its unit.

(2)
(e)

By drawing lines of maximum and minimum gradient, determine the uncertainty in the gradient. The thrust measured directly with a force sensor is 0.42 ± 0.01 N. Deduce whether the gradient is consistent with this value.

(2)
(f)
(i)

The resistive force on the cart is μmg, where μ is a constant. Explain why the line has a negative intercept on the a-axis.

(2)
(ii)

Determine μ.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
if F is constant, a = F/m «Newton's second law», so a graph of a against 1/m should be a straight line, which is easier to test and whose gradient gives F✓ 1OWTTE.
Part (b)
1.33 kg−1 AND 0.91 kg−1✓ 1Both needed. Accept 3 s.f.
both points plotted correctly✓ 1Allow ECF from their values.
Part (c)
single straight line through all the error bars, extended to the a-axis below the origin✓ 1
Part (d)
gradient from a large triangle on the line✓ 1
0.42 N «kg m s−2»✓ 1Accept 0.40–0.44 N. Unit required «accept kg m s−2».
Part (e)
steepest and shallowest lines through all the error bars give an uncertainty of about ±0.02 N✓ 1Accept ±0.01 to ±0.03 N. Allow ECF from their lines.
the range 0.40–0.45 N overlaps 0.41–0.43 N, so the gradient is consistent with the measured thrust✓ 1Allow ECF from (d).
Part (f)(i)
resultant force: F − μmg = ma, so a = F × (1/m) − μg✓ 1
so the graph is a straight line of gradient F whose intercept is −μg, a negative constant✓ 1
Part (f)(ii)
intercept ≈ −0.043 m s−2, so μ = 0.043/9.81 = 0.0044✓ 1Accept 0.002–0.006. Allow ECF from their intercept.

Answers: (b) 1.33 kg−1, 0.91 kg−1  ·  (d) 0.42 N  ·  (e) ±0.02 N  ·  (f)(ii) 0.0044 (the remaining parts are explanations — see the table above)

133A-1B-04Circular motion Paper 1BHard12 marks

A smartphone is taped flat to a horizontal platform that is rotated at different constant angular velocities ω by a motor. The phone's gyroscope measures ω and its accelerometer measures the magnitude a of the horizontal acceleration of the sensor chip inside the phone. The position of the chip inside the phone is not known.

The student tests the hypothesis a = kωn, where k and n are constants, by plotting lg a against lg ω.

ω / rad s−1a / m s−2lg(ω / rad s−1)lg(a / m s−2)
2.000.4300.301−0.367
3.000.9260.477−0.033
4.001.710.6020.233
5.503.14
7.005.200.8450.716
9.008.450.9540.927
lg a against lg omega00.20.40.60.81lg(ω / rad s⁻¹)−1.2−0.8−0.400.40.81.2lg(a / m s⁻²)
Graph drawn to scale
(a)

Explain why a graph of lg a against lg ω is used to test the hypothesis.

(2)
(b)

Complete the table.

(1)
(c)

Draw the line of best fit on the graph.

(1)
(d)

Determine n.

(2)
(e)

State whether your value of n supports the relationship a = ω²r for circular motion.

(1)
(f)

Determine the distance r of the sensor chip from the axis of rotation.

(2)
(g)

The distance from the axis to the centre of the phone, measured with a ruler, is 0.125 m. Suggest why this differs from your answer to (f), and explain why the difference does not affect n.

(2)
(h)

The gyroscope readings are uncertain by ±0.05 rad s−1. Calculate the percentage uncertainty in ω² for the smallest value of ω.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
taking logs: lg a = n lg ω + lg k✓ 1
so the graph is a straight line whose gradient is n «and intercept lg k»✓ 1
Part (b)
0.740 and 0.497✓ 1Both needed. Accept 2 or 3 d.p.
Part (c)
single straight line with points scattered evenly, extended to lg ω = 0✓ 1The line must reach the vertical axis for (f).
Part (d)
gradient from a large triangle on the line✓ 1
n = 1.99✓ 1Accept 1.9–2.1.
Part (e)
yes: a = ω²r predicts n = 2, and the value found is 2 within experimental error✓ 1Allow ECF from (d).
Part (f)
intercept = lg r ≈ −0.97✓ 1Accept −1.00 to −0.94.
r = 10−0.97 = 0.107 m✓ 1Accept 0.100–0.115 m. Allow ECF from their intercept.
Part (g)
the sensor chip is not at the centre of the phone «it is nearer the axis», so its radius is smaller✓ 1OWTTE.
r only changes the intercept «lg k» of the log graph; the gradient n is independent of r✓ 1
Part (h)
2 × (0.05/2.00) × 100 = 5 %✓ 1

Answers: (b) 0.740, 0.497  ·  (d) n = 2.0  ·  (f) 0.107 m  ·  (h) 5 % (the remaining parts are explanations — see the table above)

134A-1B-27Variable mass & F = Δp/Δt Paper 1BHard11 marks

Dry sand flows at a constant rate R (in kg s−1) through a small hole in a hopper and falls vertically through 0.800 m onto a tray fixed to a force sensor. The sensor is set to read zero with the empty tray. The time t is measured from the instant the first sand reaches the tray, and the sensor reading F is recorded every 1.0 s. Each reading is uncertain by ±0.02 N.

The sand lands with speed v and comes to rest on the tray without bouncing.

t / sF / N
1.00.691
2.01.168
3.01.668
4.02.146
5.02.652
6.03.133
7.03.631
8.04.111
Force-sensor reading against time while sand falls onto the tray012345678t / s00.511.522.533.544.5F / N
Graph drawn to scale
(a)

Explain why F = Rgt + Rv.

(2)
(b)

Draw the line of best fit on the graph.

(1)
(c)

Determine R using the gradient of the line.

(2)
(d)

Determine v using the intercept of the line and your answer to (c).

(2)
(e)

The uncertainty in the intercept is ±0.02 N and the uncertainty in R is negligible. Deduce whether your value of v is consistent with the sand falling freely from rest through 0.800 m.

(2)
(f)

As sand collects on the tray, the height through which the sand falls decreases slightly. State and explain the effect of this on the value of R found in (c).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
after time t the tray holds sand of mass Rt, whose weight Rtg the sensor supports✓ 1
each second sand of mass R is brought to rest from speed v, so by F = Δp/Δt the tray must exert an extra force Rv «and, by Newton's third law, the sand pushes on the tray with Rv»✓ 1Must link Rv to the rate of change of momentum.
Part (b)
single straight line through all error bars, NOT passing through the origin✓ 1Do not accept a line forced through the origin.
Part (c)
gradient = 0.490 «N s−1»✓ 1Accept 0.485–0.494.
R = gradient/g = 0.490/9.81 = 0.0499 «kg s−1»✓ 1Accept 0.049–0.051 kg s−1. Allow ECF.
Part (d)
intercept = Rv = 0.195 «N»✓ 1Accept 0.18–0.22 N.
v = 0.195/0.0499 = 3.9 «m s−1»✓ 1Allow ECF from (c) and their intercept.
Part (e)
percentage uncertainty in v = 0.02/0.195 ≈ 10 %, so v = 3.9 ± 0.4 m s−1✓ 1Allow ECF.
free fall gives √(2gh) = √(2 × 9.81 × 0.800) = 3.96 m s−1, which lies within this range, so the value is consistent✓ 1Award this mark only with a comparison that uses the uncertainty range.
Part (f)
the impact speed, and so the term Rv, decreases as time goes on, so F increases more slowly than it would otherwise✓ 1
the gradient is reduced, so R «= gradient/g» is underestimated✓ 1MP2 only scores if MP1 scores.

Answers: (c) 0.0499 kg s−1  ·  (d) 3.9 m s−1  ·  (e) 3.9 ± 0.4 m s−1; √(2gh) = 3.96 m s−1 (the remaining parts are explanations — see the table above)

135A-1B-40Drag & buoyancy Paper 1BHard12 marks

Steel ball bearings of different diameters d are released at the surface of glycerol in a wide glass cylinder. Two rubber bands around the cylinder are 0.200 m apart; the upper band is well below the surface. For each ball the time t to fall between the bands is measured three times and the mean is recorded. The diameters are measured with a micrometer.

Density of steel ρs = 7850 kg m−3; density of glycerol ρf = 1260 kg m−3. The viscous drag on a sphere of radius r moving at speed v is Fd = 6πηrv.

The graph shows lg v against lg r, where r is the radius of the ball, for all the balls except one.

d / mmt / sv / m s−1lg( r / m )lg( v / m s−1 )
2.0016.620.0120−3.000−1.920
2.5010.360.0193−2.903−1.714
3.007.370.0271−2.824−1.566
4.004.07
5.002.660.0752−2.602−1.124
6.001.810.110−2.523−0.957
lg v against lg r for ball bearings in glycerol−3.1−3.0−2.9−2.8−2.7−2.6−2.5−2.4lg( r / m )−2.0−1.8−1.6−1.4−1.2−1.0−0.8lg( v / m s⁻¹ )
Graph drawn to scale
(a)

Show that the terminal speed of a ball is v = 2(ρs − ρf)gr²/9η.

(2)
(b)

Complete the table for d = 4.00 mm.

(2)
(c)

Plot the missing point and draw the line of best fit.

(1)
(d)

Determine the gradient of the line and state whether it supports the relationship in (a).

(2)
(e)

Assuming that the gradient is 2, use a point on your line to determine the viscosity η of the glycerol.

(2)
(f)

The uncertainty in η is ±6 %. A data book gives η = 1.41 Pa s for glycerol at 20 °C. Deduce whether the result is consistent with this value.

(1)
(g)

Suggest one reason for the difference, other than random error, and one change to the method that would allow for it.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
at terminal speed the resultant force is zero: weight = buoyancy + drag, ρs(4/3)πr³g = ρf(4/3)πr³g + 6πηrv✓ 1Buoyancy must be included.
rearranging: (ρs − ρf)(4/3)πr³g = 6πηrv, so v = 2(ρs − ρf)gr²/9η✓ 1Must see the cancellation of π and one power of r.
Part (b)
v = 0.200/4.07 = 0.0491 m s−1✓ 1
lg r = lg(0.0020) = −2.699 AND lg v = −1.309✓ 1Both logs needed. Accept 2 or 3 d.p. Award [1] for lg r = lg(0.004) «diameter used».
Part (c)
point plotted correctly AND single straight line with the points scattered evenly about it✓ 1
Part (d)
gradient from a large triangle = 2.01✓ 1Accept 1.9–2.1.
the model predicts v ∝ r², a gradient of 2, so the data support it «within experimental error»✓ 1Allow ECF from their gradient.
Part (e)
eg «−2.700, −1.31»: v/r² = 104.085 = 12200 m−1 s−1✓ 1Any point on their line. Accept the intercept method if the line is extended correctly.
η = 2 × 6590 × 9.81/(9 × 12200) = 1.18 Pa s✓ 1Accept 1.10–1.26 Pa s. Allow ECF.
Part (f)
range 1.11–1.25 Pa s does not include 1.41 Pa s, so the result is not consistent with the data-book value✓ 1Allow ECF from (e). A comparison of the range is needed.
Part (g)
the glycerol was warmer than 20 °C «viscosity falls steeply as the temperature rises» OR the glycerol had absorbed water from the air, reducing its viscosity✓ 1
measure the temperature of the glycerol with a thermometer «and compare with data for that temperature» OR keep the cylinder in a water bath at 20 °C OR use fresh glycerol kept in a sealed container✓ 1The change must match the reason given.

Answers: (b) 0.0491 m s−1; −2.699; −1.309  ·  (d) 2.01  ·  (e) 1.18 Pa s (the remaining parts are explanations — see the table above)

136A-1B-44Collisions & energy Paper 1BHard12 marks

A vehicle-safety test centre publishes braking data for its test road. A test car is driven at a speed u, its wheels are locked, and the distance s it skids before stopping is measured. Each speed is measured by radar to ±1 km h−1; the graph shows u² against s with the corresponding error bars.

In a staged crash on the same road, car X of mass 1480 kg, with its wheels locked, runs into the back of car Y of mass 1120 kg, which is at rest with its wheels locked. The two cars lock together and skid 6.8 ± 0.2 m before stopping.

u / km h−1s / mu / m s−1u² / m² s−2
305.48.3369
409.011.11123
5014.713.89193
6020.5
7028.619.44378
8036.822.22494
Square of the initial speed against skid distance0510152025303540s / m050100150200250300350400450500550u² / m² s⁻²
Graph drawn to scale
(a)

Complete the table for u = 60 km h−1 and plot the point on the graph.

(1)
(b)

Show that, for a car skidding to rest with its wheels locked, u² = 2μdgs, where μd is the coefficient of dynamic friction between the tyres and the road.

(2)
(c)

Draw the line of best fit.

(1)
(d)

Determine μd using the graph.

(2)
(e)

Calculate the speed of the two cars immediately after the collision.

(1)
(f)

Determine the speed of X immediately before the collision, together with its absolute uncertainty. The uncertainty in μd is ±4 % and the uncertainties in the masses are negligible.

(3)
(g)

State one assumption made when applying the conservation of momentum to the collision.

(1)
(h)

The driver of X states that X was travelling below the speed limit of 50 km h−1. Comment on this statement.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
u = 16.67 m s−1 AND u² = 278 m² s−2, plotted correctly✓ 1Accept 277–278 m² s−2.
Part (b)
friction μdmg is the resultant force, so the deceleration is μdg✓ 1Must see FN = mg.
0 = u² − 2μdgs, so u² = 2μdgs✓ 1
Part (c)
single straight line through all the error bars, passing close to the origin✓ 1
Part (d)
gradient = 13.4 m s−2✓ 1Accept 13.0–13.8 m s−2.
μd = gradient/2g = 13.4/19.6 = 0.68✓ 1Accept 0.66–0.70. Allow ECF.
Part (e)
V = √(2 × 0.68 × 9.81 × 6.8) = 9.5 m s−1✓ 1Allow ECF from (d).
Part (f)
mXu = (mX + mY)V: u = 9.54 × 2600/1480 = 16.8 m s−1✓ 1Allow ECF from (e).
fractional uncertainty in V = ½(0.04 + 0.2/6.8) = 0.035✓ 1The factor ½ comes from the square root. Award [1] max for MP2 and MP3 if the ½ is omitted.
u = 16.8 ± 0.6 m s−1 «60 ± 2 km h−1»✓ 1MP3 is for matching precision of value and uncertainty.
Part (g)
the external forces «friction from the road» during the short collision are negligible compared with the forces between the cars, so the total momentum is conserved✓ 1OWTTE.
Part (h)
the lowest speed consistent with the data, 58 km h−1, is above 50 km h−1, so the statement is not supported✓ 1Allow ECF from (f). A comparison using the uncertainty is needed.

Answers: (a) 16.67 m s−1, 278 m² s−2  ·  (d) 0.68  ·  (e) 9.5 m s−1  ·  (f) 16.8 ± 0.6 m s−1 (the remaining parts are explanations — see the table above)

137A-1B-47Variable mass & F = Δp/Δt Paper 1BHard11 marks

A ducted fan for a model aircraft is fixed to a stand on a digital balance, so that the fan blows air vertically upwards. When the fan runs, the thrust F on the fan pushes the stand down and the reading of the balance increases. The rotation rate n of the fan, in revolutions per second, is measured with an optical tachometer. The student tests the hypothesis F = knp, where k and p are constants.

The graph shows lg F against lg n for five of the readings.

n / s−1increase in balance reading / gF / Nlg( n / s−1 )lg( F / N )
1001041.022.0000.009
1502262.222.1760.346
2004134.052.3010.608
2506246.122.3980.787
3009312.477
350123612.12.5441.084
lg F against lg n for a ducted fan1.922.12.22.32.42.52.6lg( n / s⁻¹ )−0.200.20.40.60.811.2lg( F / N )
Graph drawn to scale
(a)

The mass of air pushed through the fan each second and the speed of the air leaving the fan are both proportional to n. The air enters the fan from rest. Explain, using Newton's laws, why this suggests that p = 2.

(2)
(b)

Complete the table for n = 300 s−1.

(2)
(c)

Plot the missing point and draw the line of best fit.

(1)
(d)

Determine p and comment on the hypothesis in (a).

(2)
(e)

A vertical-take-off model of total mass 1.60 kg is lifted by two of these fans. Assuming p = 2, use your line to determine the smallest rotation rate at which the model can take off, together with its uncertainty. The uncertainty in the thrust given by the line is ±4 %.

(3)
(f)

The battery voltage fell during the experiment, so the rotation rate drifted during some readings. Suggest how the student could reduce the effect of this drift on the results.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
the fan exerts a force on the air equal to its rate of change of momentum: F = (Δm/Δt)v «air from rest»; by Newton's third law the air exerts an equal and opposite force «the thrust» on the fan✓ 1
Δm/Δt ∝ n and v ∝ n, so F ∝ n², p = 2✓ 1
Part (b)
F = 931 × 10−3 × 9.81 = 9.13 N✓ 1Award [0] for using the reading in grams as a force.
lg F = 0.961✓ 1Accept 2 or 3 d.p. Allow ECF from their F.
Part (c)
point plotted correctly AND single straight line with the points evenly scattered about it✓ 1
Part (d)
p = gradient = 1.99✓ 1Accept 1.9–2.1.
this is 2 within experimental error, so the data support the hypothesis✓ 1Allow ECF from their gradient.
Part (e)
thrust needed from each fan = 1.60 × 9.81/2 = 7.85 N✓ 1
from the line k = F/n² ≈ 1.00 × 10−4 N s², so n = √(7.85/k) = 280 s−1✓ 1Accept a value of n read from the line at lg F = 0.89. Accept 270–290 s−1.
n ∝ √F, so the fractional uncertainty is ½ × 4 % = 2 %: n = 280 ± 6 s−1✓ 1MP3 is for halving the percentage uncertainty and quoting a matching precision.
Part (f)
for each reading, record n and the balance reading at the same instant «or average both over the same time interval», OR recharge/replace the battery or use a regulated power supply✓ 1OWTTE. Do not accept “repeat the readings” alone.

Answers: (b) 9.13 N; 0.961  ·  (d) 1.99  ·  (e) 280 ± 6 s−1 (the remaining parts are explanations — see the table above)

138A-2-04Friction & inclined planes Paper 2Easy11 marks

A crate of mass 18 kg rests on a loading ramp. The angle θ between the ramp and the horizontal is increased slowly. The crate begins to slide when θ = 24°. Figure 1 shows the crate on the ramp; the dot marks the centre of the crate.

A crate resting on a loading ramp inclined at angle theta to the horizontal; a dot marks the centre of the cratecrateθhorizontal ground
Figure 1 — Diagram NOT accurately drawn
(a)

Draw and label, on Figure 1, the forces acting on the crate just before it begins to slide.

(3)
(b)

Show that the coefficient of static friction between the crate and the ramp is about 0.45.

(1)
(c)

The ramp is lowered to θ = 15° with the crate at rest on it. Calculate the frictional force on the crate.

(2)
(d)

The ramp is returned to 24° and the crate slides down it. The coefficient of dynamic friction is 0.35. Calculate the acceleration of the crate.

(3)
(e)

Once the crate is sliding, the angle of the ramp is reduced until the crate slides down it at constant speed. Calculate this angle.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
weight «mg, Fg» vertically downwards from the dot✓ 1
normal force «FN» perpendicular to the ramp surface, away from it✓ 1
friction «Ff» parallel to the ramp surface, directed up the slope✓ 1Each arrow must be labelled. Do not award a mark for a separate “force down the slope” «award [2 max] if one is added».
Part (b)
μs = Ff / FN = mg sin 24° / mg cos 24° = tan 24° «= 0.445»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)
the crate is in equilibrium, so the friction equals the component of the weight down the slope: Ff = 18 × 9.81 × sin 15°✓ 1
Ff = 45.7 N✓ 1Do not accept μsFN = 75.9 N: that is the maximum possible friction, not the actual friction.
Part (d)
Ff = 0.35 × 18 × 9.81 × cos 24° = 56.5 N✓ 1
resultant down the slope = 71.8 − 56.5 = 15.4 N✓ 1
a = 15.4 / 18 = 0.85 m s−2✓ 1Accept a = g(sin θ − μd cos θ). Award [3] for CNA.
Part (e)
constant velocity so the resultant force is zero: mg sin θ = μd mg cos θ «tan θ = μd»✓ 1
θ = tan−1 0.35 = 19°✓ 1Accept 19.3°. Award [2] for CNA. Do not accept 24°: sliding is opposed by dynamic, not static, friction.

Answers: (b) 0.445  ·  (c) 45.7 N  ·  (d) 0.85 m s−2 · (e) 19° (the remaining parts are explanations — see the table above)

139A-2-16Free-body diagrams Paper 2Easy9 marks

A student of mass 58 kg stands on a set of bathroom scales on the floor of a lift. The scales measure the normal force they exert on the student. Figure 1 shows how the upward velocity v of the lift varies with time t as it travels up from rest.

Velocity of the lift against time: rises to 2.4 m s⁻¹ at 2.0 s, constant to 8.0 s, falls to zero at 11.0 s0123456789101112t / s00.511.522.53v / m s⁻¹
Figure 1 — graph drawn to scale
A dot representing the studentstudent
Figure 2 — for (b)
(a)

Calculate the weight of the student.

(1)
(b)

Draw and label, on Figure 2, the forces acting on the student during the first 2.0 s. The relative lengths of your arrows should be correct.

(2)
(c)

Determine the reading of the scales during the first 2.0 s.

(2)
(d)

State the reading of the scales between t = 2.0 s and t = 8.0 s.

(1)
(e)

Calculate the reading of the scales between t = 8.0 s and t = 11.0 s.

(2)
(f)

Identify the force that forms a Newton’s third law pair with the normal force exerted by the scales on the student.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
W = 58 × 9.81 = 569 N✓ 1
Part (b)
weight vertically downwards AND normal force vertically upwards, from the dot and labelled✓ 1
normal force arrow longer than the weight arrow✓ 1Do not award MP2 if any third force is drawn.
Part (c)
a = 2.4 / 2.0 = 1.2 m s−2 AND FN − mg = ma✓ 1
FN = 58(9.81 + 1.2) = 639 N✓ 1Award [2] for CNA.
Part (d)
569 N «equal to the weight: constant velocity so zero resultant force»✓ 1Allow ECF from (a).
Part (e)
a = −2.4 / 3.0 = −0.80 m s−2✓ 1Accept a deceleration of 0.80 m s−2.
FN = 58(9.81 − 0.80) = 523 N✓ 1
Part (f)
the «downward» force exerted by the student on the scales✓ 1Do not accept the weight of the student or the gravitational force.

Answers: (a) 569 N  ·  (c) 639 N  ·  (d) 569 N  ·  (e) 523 N (the remaining parts are explanations — see the table above)

140A-2-37Circular motion Paper 2Easy10 marks

A salad spinner dries washed lettuce. The lettuce is placed in a circular basket of radius 11 cm whose wall has many small holes. The basket rotates about a vertical axis at a constant 9.0 revolutions per second. The diagram shows the basket from above.

Top view of the circular basket of a salad spinner with small holes in its wall. The basket rotates anticlockwise about its centre O. Point P is a hole at the right-hand side of the basket.OProtation
Diagram NOT accurately drawn
(a)
(i)

Calculate the period of rotation of the basket.

(1)
(ii)

Show that the angular velocity of the basket is about 57 rad s−1.

(1)
(b)
(i)

Calculate the centripetal acceleration of a point on the wall of the basket, and express it as a multiple of g.

(2)
(ii)

A drop of water of mass 5.0 × 10−5 kg rests against the wall. Calculate the resultant force on the drop.

(1)
(c)

A drop of water passes through the hole at P as P passes through the position shown. Draw, on the diagram, an arrow labelled v to show the direction of the velocity of the drop at P, and a line to show the path of the drop after it has left the basket.

(2)
(d)

Explain why drops of water leave the basket through the holes while the lettuce stays inside.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
T = 1/9.0 = 0.11 s✓ 1Accept 0.111 s.
Part (a)(ii)
ω = 2π × 9.0 = 56.5 «rad s−1»✓ 1Accept 2π/T. Must see full substitution OR answer to 3 s.f.
Part (b)(i)
a = ω²r = 56.5² × 0.11 = 3.5 × 102 m s−2✓ 1Accept v²/r.
≈ 36 g✓ 1Accept 35–36 g. Allow ECF.
Part (b)(ii)
F = ma = 5.0 × 10−5 × 352 = 0.018 N «towards the centre»✓ 1Allow ECF from (b)(i).
Part (c)
arrow at P tangent to the circle, pointing up the page «in the sense of rotation»✓ 1Do not award MP1 for a radial arrow.
path after leaving: a straight line continuing along the tangent at P✓ 1Do not award MP2 for a curved or radial path.
Part (d)
the wall pushes the lettuce towards the centre: the normal force provides the centripetal force it needs✓ 1
at a hole there is no wall to push the drop inwards «adhesion to the lettuce is too small to provide the force needed»✓ 1
so the drop continues in a straight line along the tangent «Newton’s first law» and leaves through the hole✓ 1Do not accept “the drops are thrown outwards by a centrifugal force”.

Answers: (a)(i) 0.11 s  ·  (a)(ii) 56.5 rad s−1  ·  (b)(i) 3.5 × 102 m s−2 ≈ 36 g  ·  (b)(ii) 0.018 N (the remaining parts are explanations — see the table above)

141A-2-44Newton's laws & force pairs Paper 2Easy13 marks

Two identical ring magnets, each of mass 0.025 kg, are placed on a vertical wooden rod fixed to a base, which stands on a top-pan balance (Figure 1). The magnets are arranged so that their north faces are towards each other. The upper magnet floats at rest 12 mm above the lower magnet, which rests on the base. Friction between the magnets and the rod is negligible.

Two ring magnets on a vertical wooden rod fixed to a base on a top-pan balance. The lower magnet rests on the base with its N face upwards; the upper magnet floats 12 mm above it with its N face downwards.top-pan balancebaseNSNSSNSNupper magnetlower magnetwooden rod12 mm
Figure 1 — Diagram NOT accurately drawn
Two boxes, each with a dot: one for the upper magnet and one for the lower magnetupper magnetlower magnet
Figure 2 — for (a) and (c)
Side view of the two magnets with their N faces towards each other across the gapSNNSupper magnetlower magnet
Figure 3 — for (g)
(a)

Draw and label, on the dot for the upper magnet in Figure 2, the forces acting on the upper magnet.

(2)
(b)

State the magnitude of the magnetic force on the upper magnet.

(1)
(c)

Draw and label, on the dot for the lower magnet in Figure 2, the forces acting on the lower magnet. The relative lengths of your arrows should be correct.

(2)
(d)

State the force that forms a Newton’s third law pair with the magnetic force on the upper magnet.

(1)
(e)

A student predicts that the reading of the balance will be smaller with the upper magnet floating than with it resting directly on the lower magnet, because the floating magnet “does not touch anything”. Explain why the student is wrong.

(3)
(f)

Over this range of separations, the magnetic force between the magnets may be modelled as F = C/d⁴, where d is the separation and C is a constant. A small brass mass of 0.025 kg is placed on top of the upper magnet. Determine the new separation of the magnets.

(2)
(g)

Sketch, on Figure 3, the magnetic field lines in the gap between the two magnets.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
downward arrow labelled weight «Fg, gravitational force» and upward arrow labelled magnetic force✓ 1Arrows must start on the dot.
arrows of equal length✓ 1Do not award MP2 if any additional force «eg normal force» is drawn.
Part (b)
mg = 0.025 × 9.81 = 0.245 N✓ 1
Part (c)
downward arrows labelled weight and magnetic force «from the upper magnet»; upward arrow labelled normal force «from the base»✓ 1
normal-force arrow about twice the length of each downward arrow✓ 1Allow a single downward arrow of the combined force only if labelled.
Part (d)
the «upward» magnetic force on the upper magnet pairs with the downward magnetic force exerted by the upper magnet on the lower magnet✓ 1Both direction and the body acted on are needed. Do not accept the weight of the upper magnet.
Part (e)
the upper magnet is in equilibrium, so the upward magnetic force on it equals its weight✓ 1
by Newton’s third law the upper magnet exerts an equal downward magnetic force on the lower magnet✓ 1
so the base «and the balance» must push up with a force equal to the total weight of both magnets, rod and base: the reading is unchanged✓ 1OWTTE. Accept an argument treating the whole apparatus as one system with only weight and the normal force acting.
Part (f)
the magnetic force must now be twice as large, so d⁴ halves✓ 1
d = 12/21/4 = 10.1 mm✓ 1Award [2] for CNA. Award [1] max for 6.0 mm «d halved».
Part (g)
lines leave both N faces and curve outwards, away from the axis, without joining the two magnets✓ 1Arrows on the lines must point away from the N faces.
a neutral point «no field lines» midway between the faces on the axis✓ 1

Answers: (b) 0.245 N  ·  (c) normal force 0.491 N  ·  (f) 10.1 mm (the remaining parts are explanations — see the table above)

142A-2-45Momentum & impulse Paper 2Easy13 marks

In a drop test, a phone of mass 0.195 kg in a rigid case is released from rest 1.20 m above a hard floor. It comes to rest on the floor without bouncing, and the impact lasts 2.0 ms. Air resistance is negligible.

Blank axes for force F / N from 0 to 1000 against time t / ms from 0 to 701234567t / ms02004006008001000F / N
Figure 1 — axes for (e)
(a)

Show that the speed of the phone just before it hits the floor is about 4.9 m s−1.

(1)
(b)

Calculate the momentum of the phone just before it hits the floor.

(1)
(c)
(i)

Calculate the magnitude of the average acceleration of the phone during the impact, and express it as a multiple of g.

(2)
(ii)

State the magnitude and direction of the impulse of the resultant force on the phone during the impact.

(1)
(d)

Calculate the magnitude of the average force exerted by the floor on the phone during the impact.

(1)
(e)

Model the force exerted by the floor on the phone during the impact as a triangular pulse. Sketch, on Figure 1, a graph of this force against time. On the same axes, sketch the graph for a second test in which the phone, in a soft rubber case, again comes to rest without bouncing, but the impact lasts 6.0 ms. Label the two graphs.

(2)
(f)

Explain why the soft case reduces the chance of the screen cracking.

(3)
(g)

In another test the phone rebounds vertically at 2.0 m s−1. State and explain whether the impulse of the resultant force on the phone is greater or smaller than in (c)(ii).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = √(2gh) = √(2 × 9.81 × 1.20) = 4.85 «m s−1»✓ 1Must see full substitution OR answer to 3 s.f. Accept use of mgh = ½mv².
Part (b)
p = 0.195 × 4.85 = 0.946 kg m s−1✓ 1Accept N s.
Part (c)(i)
a = Δv/Δt = 4.85/2.0 × 10−3 = 2.4 × 103 m s−2✓ 1Accept 2400–2450 m s−2. Do not award MP1 if 2.0 ms is not converted to seconds.
a/g = 2430/9.81 ≈ 250 «a ≈ 250g»✓ 1Accept 240–250. Allow ECF.
Part (c)(ii)
0.946 N s «equal to the change in momentum», upwards✓ 1Allow ECF from (b). Direction required.
Part (d)
F = Δp/Δt = 0.946/2.0 × 10−3 = 473 N✓ 1Accept 470–480 N; the weight «1.9 N» is negligible. Accept ma from (c)(i).
Part (e)
rigid case: triangle starting at t = 0, lasting 2.0 ms, with a peak of about 950 N «twice the average force»✓ 1Accept a peak of 900–1000 N. Allow ECF from (d).
soft case: triangle lasting 6.0 ms with a peak of about 320 N, so that the two areas are equal✓ 1Accept a peak of 280–360 N. The areas must look equal.
Part (f)
the change in momentum «impulse» is the same in both impacts✓ 1
it takes place over a longer time✓ 1
so the average «and the maximum» force on the phone is smaller, F = Δp/Δt, so the force on the glass is less likely to crack it✓ 1Do not award MP3 for “absorbs the force”.
Part (g)
greater✓ 1No mark for “greater” alone.
the change in momentum is now m(v + vr) = 0.195 × (4.85 + 2.0) = 1.34 N s, larger than 0.946 N s, because the momentum reverses direction✓ 1Accept a correct explanation without the numerical value.

Answers: (a) 4.85 m s−1  ·  (b) 0.946 kg m s−1  ·  (c)(i) 2.4 × 103 m s−2 ≈ 250g  ·  (c)(ii) 0.946 N s upwards  ·  (d) 473 N  ·  (g) 1.34 N s (the remaining parts are explanations — see the table above)

143A-2-71Newton's laws & force pairs Paper 2Easy12 marks

During the first strides of a 100 m race, a sprinter of mass 68 kg accelerates along a straight, level track. Averaged over each stride, her horizontal acceleration is 4.0 m s−2 and her vertical acceleration is zero. Treat the sprinter as a particle and use the average forces over a stride. Air resistance is negligible except in (g).

Figure 1 shows the sprinter, represented by a dot, and the direction in which she moves.

A dot representing the sprinter just above a level track; an arrow shows that she is moving to the righttrackdirection of motion
Figure 1 — for part (a)
(a)

Draw and label, on the dot in Figure 1, the forces acting on the sprinter. The relative lengths of your arrows should be approximately correct.

(2)
(b)
(i)

Calculate the resultant force on the sprinter.

(1)
(ii)

Identify the force that provides this resultant force, and the body that exerts it.

(1)
(c)

Explain, with reference to Newton's third law, how the sprinter is able to accelerate forwards.

(2)
(d)

Calculate the magnitude of the total contact force exerted by the track on the sprinter, and its angle to the vertical.

(2)
(e)

Outline why it is static friction, rather than dynamic friction, that acts on the sprinter's shoe while it is in contact with the track.

(1)
(f)

On an icy path the coefficient of static friction between her shoes and the ice is 0.10. Determine whether she could accelerate at 4.0 m s−2 on the ice.

(2)
(g)

Later, she runs along the track at a constant top speed of 11 m s−1. The air resistance on her is then 30 N. State the average horizontal force that the track exerts on her.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Weight «Fg, mg» vertically downwards AND normal force «FN» vertically upwards, from the dot, labelled and of equal length✓ 1
Frictional force «Ff» horizontal, in the direction of motion, labelled, clearly shorter than the weight «about 0.4 of its length»✓ 1Do not award MP2 if friction is drawn backwards or if a separate 'forward force' is added.
Part (b)(i)
F = ma = 68 × 4.0 = 272 N «forwards»✓ 1Accept 270 N.
Part (b)(ii)
Friction «static frictional force» exerted by the track on her shoes✓ 1Both the force and the body needed.
Part (c)
Her foot pushes backwards on the track «friction exerted by the shoe on the track»✓ 1
By Newton's third law the track pushes forwards on her with a force of equal magnitude; this is the only horizontal force on her, so it is the resultant force and she accelerates forwards✓ 1OWTTE.
Part (d)
F = √(667² + 272²) = 720 N✓ 1Accept 720–721 N. Allow ECF from (b)(i).
angle = tan−1(272/667) = 22° to the vertical, tilted forwards✓ 1Direction «forwards» needed.
Part (e)
The shoe does not slide «there is no relative motion» over the track surface while it is in contact✓ 1OWTTE.
Part (f)
Maximum friction = μsFN = 0.10 × 667 = 67 N «maximum acceleration = μsg = 0.98 m s−2»✓ 1
67 N < 272 N «0.98 m s−2 < 4.0 m s−2», so she cannot: her feet would slip✓ 1A comparison is needed. Allow ECF from (b)(i).
Part (g)
30 N forwards «resultant force zero at constant velocity»✓ 1Direction needed.

Answers: (b)(i) 270 N  ·  (d) 720 N at 22° to the vertical  ·  (f) maximum friction 67 N  ·  (g) 30 N (the remaining parts are explanations — see the table above)

144A-2-75Explosions & recoil Paper 2Easy11 marks

The fruit of the squirting cucumber plant is full of seeds in a liquid under pressure. When a ripe fruit breaks away from its stalk, the seeds and liquid are forced out as a jet through the hole where the stalk was attached, and the empty fruit moves off in the opposite direction.

In a model of one fruit, the total mass is 32 g, of which 6.0 g is seeds and liquid. The fruit is at rest when the ejection starts. The seeds and liquid leave as a jet at 11 m s−1, and the ejection takes 4.0 ms. Treat the motion during the ejection as one-dimensional.

(a)

State the condition for the total momentum of the fruit and its contents to remain constant.

(1)
(b)

Calculate the speed of the empty fruit immediately after the ejection.

(2)
(c)

The weight of the fruit and its contents acts during the ejection. Show that the impulse of this weight during the ejection is less than 2 % of the momentum of the jet.

(1)
(d)
(i)

Show that the total kinetic energy of the jet and the empty fruit just after the ejection is about 0.45 J.

(1)
(ii)

Outline where this kinetic energy has come from.

(1)
(e)

Calculate the average force exerted on the seeds and liquid during the ejection.

(1)
(f)

The jet leaves at 45° above the horizontal from a point at ground level.

(i)

Calculate the horizontal distance from the plant at which the seeds land. Assume that air resistance is negligible.

(3)
(ii)

Seeds are found up to about 6 m from such plants. Suggest one reason for the difference from your answer to (f)(i).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
No resultant external force acts on the system «fruit + contents»✓ 1Do not accept 'no forces act'.
Part (b)
0.0060 × 11 = 0.026 × v «momentum before = 0»✓ 1
v = 2.5 m s−1✓ 1Award [2] for CNA. Award [1] max for 2.1 m s−1 «total mass 0.032 kg used».
Part (c)
Impulse of weight = 0.032 × 9.81 × 4.0 × 10−3 = 1.3 × 10−3 N s, which is 1.9 % of the jet's momentum 0.066 kg m s−1✓ 1Must see both values OR the percentage to 2 s.f.
Part (d)(i)
½ × 0.0060 × 11² + ½ × 0.026 × 2.54² = 0.363 + 0.084 = 0.447 «J»✓ 1Must see full substitution OR answer to 3 s.f. Allow ECF from (b).
Part (d)(ii)
Elastic potential energy stored in the stretched wall of the fruit «work done by the pressure of the liquid»✓ 1Accept energy stored in the compressed liquid/fruit. Do not accept 'from the plant'.
Part (e)
F = Δp/Δt = 0.066/4.0 × 10−3 = 16.5 N✓ 1Accept 16–17 N.
Part (f)(i)
vx = vy = 11 cos 45° = 7.78 m s−1✓ 1
time of flight = 2 × 7.78/9.81 = 1.59 s✓ 1
distance = 7.78 × 1.59 = 12 m✓ 1Award [3] for CNA.
Part (f)(ii)
Air resistance on the small seeds «and drops of liquid» reduces their horizontal speed, so they land closer✓ 1Accept: the jet breaks up / the launch angle varies / the seeds are not all launched at 11 m s−1. OWTTE.

Answers: (b) 2.5 m s−1  ·  (e) 16.5 N  ·  (f)(i) 12 m (the remaining parts are explanations — see the table above)

145A-2-76Angular velocity & period Paper 2Easy11 marks

In an optical-disc drive, data are read by a laser spot that moves slowly outwards from the inner track, 23 mm from the centre of the disc, to the outer track, 58 mm from the centre. The drive changes the rotation rate of the disc so that the track always passes the laser spot at the same linear speed of 1.22 m s−1.

Blank axes for the sketch: angular velocity ω from 0 to 60 rad s−1 against radius r of the track from 20 mm to 60 mm2030405060r / mm0102030405060ω / rad s⁻¹
Figure 1 — for part (c)
(a)

Calculate the angular velocity of the disc when the inner track is being read.

(1)
(b)

Show that the disc makes about 200 revolutions per minute when the outer track is being read.

(1)
(c)

Sketch, on Figure 1, a graph to show how the angular velocity ω of the disc varies with the radius r of the track being read, from r = 23 mm to r = 58 mm.

(2)
(d)

A speck of dust is stuck to the rim of the disc, 60 mm from the centre, while the inner track is being read. Calculate the speed of the speck and its centripetal acceleration. Express the acceleration as a multiple of g.

(3)
(e)

Explain why the speck accelerates even though its speed is constant.

(2)
(f)

When the drive starts, the disc speeds up from rest. Outline why, during this time, the resultant force on the speck is not directed towards the centre of the disc.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
ω = v/r = 1.22/0.023 = 53 rad s−1✓ 1Accept 53.0–53.1 rad s−1.
Part (b)
ω = 1.22/0.058 = 21.0 rad s−1; 21.0/2π × 60 = 201 «rev min−1»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)
A curve that decreases as r increases, falling less steeply at large r «ω ∝ 1/r», not a straight line✓ 1
Passing through «23, 53» and «58, 21»✓ 1Allow ±2 rad s−1. Allow ECF from (a) and (b).
Part (d)
v = ωr = 53.04 × 0.060 = 3.2 m s−1✓ 1Allow ECF from (a).
a = ω²r = 53.04² × 0.060 = 169 m s−2✓ 1Accept v²/r. Allow ECF.
= 17g✓ 1Accept 17g. Award [3] for CNA of the speed and of the acceleration in terms of g.
Part (e)
Velocity is a vector and the direction of the speck's velocity changes continuously✓ 1
So its velocity changes and it has an acceleration «a resultant force acts», directed towards the centre, perpendicular to the velocity✓ 1OWTTE.
Part (f)
The speed of the speck is increasing, so the resultant force has a component along «tangent to» the circle, in the direction of motion✓ 1
It also has a component towards the centre «to change the direction», so the resultant is at an angle to the radius «ahead of the radius»✓ 1OWTTE.

Answers: (a) 53 rad s−1  ·  (d) 3.2 m s−1; 169 m s−2 ≈ 17g (the remaining parts are explanations — see the table above)

146A-2-81Hooke's law & springs Paper 2Easy11 marks

A car of mass 1500 kg is stuck in mud on level ground. A recovery truck pulls it out with a horizontal nylon rope that obeys Hooke's law. Figure 1 shows how the tension T in the rope varies with its extension x.

The mud exerts a horizontal resistive force on the car. The car starts to move only when the tension exceeds 12.0 kN. Once the car is moving, the resistive force is constant at 9.0 kN.

Tension T in the rope against extension x: a straight line through the origin, reaching 36 kN at 2.4 m00.40.81.21.622.4x / m0510152025303540T / kN
Figure 1 — graph drawn to scale
A dot representing the stuck car on level ground; the rope pulls towards the righttowards the truck
Figure 2 — for part (c)
(a)

Determine the spring constant of the rope.

(2)
(b)

Calculate the extension of the rope when the car starts to move.

(1)
(c)

Draw and label, on the dot in Figure 2, the forces acting on the car just before it starts to move.

(2)
(d)

At one instant while the car is moving, the extension of the rope is 1.10 m.

(i)

Calculate the acceleration of the car at this instant.

(2)
(ii)

State the magnitude and direction of the force exerted by the rope on the truck at this instant.

(1)
(iii)

Calculate the elastic potential energy stored in the rope at this instant.

(1)
(e)

The rope breaks when the tension exceeds 45 kN. Calculate the greatest extension the rope can have, assuming that it obeys Hooke's law up to this tension.

(1)
(f)

Suggest one advantage of using a stretchy rope rather than a steel cable, which hardly stretches.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Gradient of the line, with kN converted to N «e.g. 36 × 103/2.4»✓ 1
k = 1.5 × 104 N m−1✓ 1Accept 1.45–1.55 × 104 N m−1. Award [1] max for 15 «N m−1, kN not converted».
Part (b)
x = 12.0 × 103/1.5 × 104 = 0.80 m✓ 1Allow ECF from (a). Accept a value read from the graph.
Part (c)
Weight vertically downwards AND normal force vertically upwards, labelled, of equal length✓ 1
Tension towards the truck AND resistive force «from the mud» in the opposite direction, labelled, of equal length✓ 1Award [1] max if any additional force is drawn.
Part (d)(i)
T = 1.5 × 104 × 1.10 = 16.5 kN✓ 1Accept a value read from the graph.
a = (16.5 − 9.0) × 103/1500 = 5.0 m s−2✓ 1Award [2] for CNA. Award [1] max for 11 m s−2 «resistive force omitted».
Part (d)(ii)
16.5 kN, backwards «towards the car»✓ 1Both magnitude and direction needed.
Part (d)(iii)
EH = ½ × 1.5 × 104 × 1.10² = 9.1 × 103 J✓ 1Allow ECF from (a).
Part (e)
x = 45 × 103/1.5 × 104 = 3.0 m✓ 1Allow ECF from (a).
Part (f)
The rope stretches, so the car is accelerated over a longer time «and distance» and the peak force is smaller «less likely to break or damage the vehicles»✓ 1Accept: energy stored in the rope keeps pulling the car after the truck stops. OWTTE.

Answers: (a) 1.5 × 104 N m−1  ·  (b) 0.80 m  ·  (d)(i) 5.0 m s−2  ·  (d)(ii) 16.5 kN  ·  (d)(iii) 9.1 × 103 J  ·  (e) 3.0 m (the remaining parts are explanations — see the table above)

147A-2-87Hooke's law & springs Paper 2Easy13 marks

Geologists use a simple model of how a fault in the Earth’s crust moves during an earthquake. A brick of mass 2.50 kg rests on a strip of sandpaper fixed to a horizontal bench. One end of a horizontal spring is attached to the brick, and the other end is pulled horizontally away from the brick by a motor at a constant speed of 2.0 mm s−1 (Figure 1). The spring is at its natural length when the motor starts.

The brick stays at rest while the spring stretches. It then suddenly slides forwards, stops, and stays at rest again while the spring stretches further. The spring has negligible mass and obeys Hooke’s law with a spring constant of 180 N m−1. Between the brick and the sandpaper, the coefficient of static friction is 0.62 and the coefficient of dynamic friction is 0.45. Air resistance is negligible.

Brick on a sandpaper strip pulled through a spring by a motorbrickmotorspringpulled at 2.0 mm s⁻¹sandpaper strip fixed to the benchbench
Figure 1 — Diagram NOT accurately drawn
Dot representing the brickbrick
Figure 2 — for part (a)
(a)

Draw and label, on the dot in Figure 2, the forces acting on the brick just before it starts to slide. The relative lengths of your arrows should be approximately correct.

(2)
(b)

For each of the following forces, state the force that forms a Newton’s third law pair with it. Give the body that exerts it, the body it acts on, and its direction.

(i)

The normal force exerted on the brick by the sandpaper.

(1)
(ii)

The frictional force exerted on the brick by the sandpaper.

(1)
(c)

Show that the force exerted by the spring on the brick just before it starts to slide is about 15 N.

(1)
(d)

Calculate the extension of the spring just before the brick starts to slide.

(1)
(e)

Calculate the time from the start of the motor to the instant at which the brick first starts to slide.

(1)
(f)

Calculate the acceleration of the brick just after it starts to slide.

(2)
(g)

Explain why the acceleration of the brick decreases as it slides forwards.

(2)
(h)

The brick is replaced by a brick of the same material with twice the mass. Deduce how the extension of the spring when the brick starts to slide, and the acceleration of the brick just after it starts to slide, compare with your answers to (d) and (f).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
weight «Fg, mg» vertically downwards AND normal force «FN» vertically upwards, from the dot, labelled and of equal length✓ 1
spring force «tension, FH» horizontally towards the motor AND static friction «Ff» horizontally away from the motor, labelled, equal in length to each other and shorter than the weight arrow «about 0.6 of its length»✓ 1Accept 0.4 to 0.8 of the weight arrow. Do not award MP2 if friction is drawn towards the motor or if any other force is added.
Part (b)(i)
the «normal» contact force exerted by the brick on the sandpaper, vertically downwards✓ 1Body acted on and direction needed. Do not accept the weight of the brick.
Part (b)(ii)
the frictional force exerted by the brick on the sandpaper, horizontally towards the motor «forwards»✓ 1Body acted on and direction needed.
Part (c)
the brick is in equilibrium and the static friction has reached its maximum, so F = μsmg = 0.62 × 2.50 × 9.81 = 15.2 «N»✓ 1Must see full substitution OR answer to 3 s.f.
Part (d)
x = F/k = 15.2/180 = 0.084 m «8.4 cm»✓ 1Accept 0.083–0.085 m. Allow ECF from (c).
Part (e)
t = 0.084/2.0 × 10−3 = 42 s✓ 1Accept 41–43 s. Allow ECF from (d).
Part (f)
dynamic friction = μdmg = 0.45 × 2.50 × 9.81 = 11.0 N «the spring force is still 15.2 N»✓ 1
a = (15.2 − 11.0)/2.50 = 1.7 m s−2✓ 1Award [2] for CNA. Award [1] max for 6.1 m s−2 «friction ignored». Allow ECF from (c).
Part (g)
the brick moves towards the motor much faster than the end of the spring is pulled, so the extension decreases and the spring force kx «Hooke’s law» decreases✓ 1
the dynamic friction μdmg stays constant, so the resultant force kx − μdmg, and so the acceleration, decreases «and becomes a deceleration once kx < μdmg»✓ 1OWTTE.
Part (h)
the normal force and so the greatest static friction double, so the extension doubles «to 0.17 m»✓ 1Allow ECF from (d).
the resultant force (μs − μd)mg and the mass both double, so the acceleration is unchanged «a = (μs − μd)g = 1.7 m s−2»✓ 1A reason is needed for each mark.

Answers: (c) 15.2 N  ·  (d) 0.084 m  ·  (e) 42 s  ·  (f) 1.7 m s−2  ·  (h) extension doubles «0.17 m»; acceleration unchanged «1.7 m s−2» (the remaining parts are explanations — see the table above)

148IBP2S-003-04-11Connected bodies Paper 2Easy5 marks

An electric scooter and its rider have a combined mass of 95 kg. The scooter tows a cargo trailer of mass 25 kg using a rigid, horizontal tow bar on level ground. The driving force on the scooter is 180 N. The resistive forces are 30 N on the scooter and 10 N on the trailer.

scooter + rider 95 kgtrailer 25 kgtow bar30 N10 Ndriving force 180 N
Resistive forces shown acting on each part
(a)

Calculate the acceleration of the scooter and trailer.

(2)
(b)

Outline why the tension in the tow bar is smaller than the driving force.

(1)
(c)

Determine the tension in the tow bar.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Resultant force = 180 − 30 − 10 = 140 N✓ 1
a = 140/120 = 1.2 m s−2 (1.17 m s−2)✓ 1
Part (b)
The tow bar only has to accelerate the trailer and overcome the trailer's resistance. The driving force must accelerate the whole system, including the scooter and rider, and overcome all of the resistance✓ 1
Part (c)
For the trailer: T − 10 = 25 × 1.17✓ 1
T = 39 N (accept 39–40 N, e.g. 40 N from a = 1.2 m s−2; allow ECF from (a))✓ 1

Exam tip: Find the acceleration from the whole system, then isolate the trailer to find the internal force. Choosing the smaller body makes the algebra shortest.

149IBP2S-003-03-11Collisions & energy Paper 2Easy6 marks

A curling stone of mass 19.0 kg is released on a horizontal ice sheet with a speed of 2.4 m s−1. It slides in a straight line and comes to rest after travelling 28 m. Assume the frictional force is constant and air resistance is negligible. Take g = 9.8 m s−2.

(a)

State the forces acting on the stone while it slides, and identify each as a contact force or a field force.

(2)
(b)

Calculate the magnitude of the deceleration of the stone.

(2)
(c)

Determine the coefficient of dynamic friction between the stone and the ice.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Weight (gravitational force), acting vertically downward, is a field force . Normal force (upward) and friction (opposite to the motion) are both contact forces .✓ 2
Part (b)
Uses v2 = u2 + 2as with v = 0: 0 = 2.42 + 2a(28) . Magnitude of a = 0.10 m s−2 (0.103 m s−2), directed opposite to the motion .✓ 2
Part (c)
Friction is the resultant force, so μmg = ma, giving μ = a/g . μ = 0.103/9.8 = 0.010 (0.0105). Accept 0.010–0.011 depending on rounding, and allow an error carried forward from the deceleration found in the previous part .✓ 2

Exam tip: The mass cancels when you find μ for an object on a horizontal surface. Write μ = a/g explicitly, and keep unrounded values until the final step so that rounding errors do not change your answer.

150IBP2S-003-00-11Hooke's law & springs Paper 2Easy5 marks

A kitchen scale uses a spring in its load cell. A student adds masses to the spring and measures its extension. The results are listed below.
mass / g: 0, 100, 200, 300, 400, 500
extension / mm: 0.0, 1.6, 3.3, 4.9, 6.5, 9.4

(a)

State Hooke's law.

(1)
(b)

Identify the reading which suggests the limit of proportionality has been exceeded. Explain your answer.

(2)
(c)

Determine the spring constant of the spring in N m−1.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The extension of a spring is directly proportional to the force applied to it, provided the limit of proportionality is not exceeded .✓ 1
Part (b)
The 500 g reading (9.4 mm) .✓ 1
The extension increases by 2.9 mm from 400 g to 500 g, compared with about 1.6 mm for each previous 100 g, so the extension is no longer proportional to the force .✓ 1
Part (c)
Use a point in the linear region, e.g. F = 0.400 × 9.8 = 3.92 N for x = 6.5×10−3 m .✓ 1
k = F/x ≈ 6.0×102 N m−1 (accept 590–610 N m−1; no credit if the 500 g point is used) .✓ 1

Exam tip: Convert grams to kilograms and millimetres to metres before calculating k. Base the spring constant only on data from the linear region.

151IBP2S-003-05-11Newton's laws & force pairs Paper 2Easy6 marks

A sprinter of mass 68 kg starts a race from rest by pushing backward on the starting blocks. The average horizontal force on the sprinter while she is in contact with the blocks is 750 N, and the contact lasts 0.36 s.

(a)

State Newton's third law of motion.

(1)
(b)

Identify the Newton's third law partner of the forward force of the blocks on the sprinter.

(1)
(c)

Calculate the average horizontal acceleration of the sprinter while she is in contact with the blocks.

(2)
(d)

Determine her horizontal speed as she leaves the blocks.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
If body A exerts a force on body B, then B exerts a force on A that is equal in magnitude and opposite in direction✓ 1
Part (b)
The backward (horizontal contact) force of the sprinter on the blocks✓ 1
Part (c)
a = F/m = 750/68✓ 1
a = 11 m s−2 (11.0)✓ 1
Part (d)
v = u + at = 0 + 11.0 × 0.36, or Δp = FΔt = 750 × 0.36 = 270 N s with v = 270/68 (allow error carried forward from the acceleration calculated in the previous part)✓ 1
v = 4.0 m s−1 (3.97)✓ 1

Exam tip: When identifying a force pair, name the type of force, the body exerting it, the body it acts on, and its direction, e.g. 'the contact force of the sprinter pushing backward on the blocks'.

152IBP2S-003-02-11Friction & inclined planes Paper 2Easy6 marks

A rescue sled of mass 12 kg is released from rest on a uniform snow slope inclined at 15° to the horizontal. The coefficient of kinetic friction between the sled and the snow is 0.080. Air resistance is negligible.

sled 12 kg15°motion
(a)

Outline the forces acting on the sled as it slides down the slope.

(2)
(b)

Show that the acceleration of the sled is about 1.8 m s−2.

(2)
(c)

Calculate the speed of the sled after it has travelled 20 m along the slope.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Weight acting vertically downwards and a normal force acting perpendicular to the slope surface . Kinetic friction acting up the slope, parallel to the surface, opposing the motion .✓ 2
Part (b)
Resultant force = mg sin 15° − μmg cos 15°, so a = g(sin 15° − 0.080 cos 15°) . a = 9.8(0.2588 − 0.0773) = 1.78 m s−2 ≈ 1.8 m s−2 .✓ 2
Part (c)
v2 = u2 + 2as = 0 + 2 × 1.78 × 20 . v = 8.4 m s−1 . Accept 8.5 m s−1 if 1.8 m s−2 is used. ECF from (b).✓ 2

Exam tip: On a 'show that' question, give the answer to at least one more significant figure than the value quoted, and use the unrounded value in later parts.

153IBP2S-003-02-12Friction & inclined planes Paper 2Easy6 marks

A footwear designer tests shoe soles. A sole is placed on a flat vinyl board. One end of the board is raised slowly until the sole just begins to slip. The angle θ between the board and the horizontal at that moment is recorded.

Sole material — slipping angle θ / °
rubber — 38
EVA foam — 31
leather — 22
hard plastic — 17

(a)

Show that the coefficient of static friction μs between a sole and the board is equal to tan θ.

(2)
(b)

Calculate μs for the leather sole.

(2)
(c)

A hospital access ramp is surfaced with the same vinyl and is inclined at 20°. Explain, using the data, which sole material(s) would slip if a person stood still on the ramp.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
At the point of slipping, friction has its limiting value μsN and balances the component of weight down the slope: mg sin θ = μsN, with N = mg cos θ . Dividing gives μs = sin θ / cos θ = tan θ .✓ 2
Part (b)
μs = tan 22° . μs = 0.40 .✓ 2
Part (c)
A sole slips if its slipping angle is less than 20°, which is the same as μs < tan 20° = 0.36 . Only the hard plastic sole (17°) would slip. Rubber, EVA foam and leather do not slip, although leather (22°) has only a small margin .✓ 2

Common error: The mass of the shoe is not needed, so do not invent a value. The mass cancels when you divide the friction condition through by mg.

154IBP2S-003-06-11Momentum & impulse Paper 2Easy6 marks

During a hailstorm, hailstones fall vertically onto a horizontal solar panel on a roof. Each hailstone has a mass of 1.5 g and strikes the panel at 12 m s−1. On average, 400 hailstones hit the panel each second. Initially, the hailstones come to rest on impact without bouncing. Ignore the weight of hail that collects on the panel.

(a)

State Newton's second law of motion in terms of momentum.

(1)
(b)

Calculate the magnitude of the change in momentum of one hailstone when it hits the panel.

(1)
(c)

Calculate the average force exerted on the panel by the hail.

(2)
(d)

Later, the hailstones bounce off the panel. Explain how this affects the average force on the panel, assuming the arrival rate and impact speed are unchanged.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The resultant force on a body is equal to its rate of change of momentum (F = Δp/Δt)✓ 1
Part (b)
Δp = 1.5 × 10−3 × 12 = 1.8 × 10−2 kg m s−1✓ 1
Part (c)
F = (number per second) × Δp = 400 × 1.8 × 10−2✓ 1
F = 7.2 N✓ 1
Allow error carried forward from the previous part.—
Part (d)
The velocity of each hailstone now reverses direction, so its change in momentum is greater than before✓ 1
The rate of change of momentum is therefore greater, so the force on the panel increases (up to a maximum of double, 14.4 N, for a perfectly elastic bounce)✓ 1

Exam tip: Convert grams to kilograms before calculating momentum. A factor of 1000 error in the force is easy to miss.

155IBP2S-003-01-11Drag & buoyancy Paper 2Easy5 marks

Hydrostatic weighing is used in sports medicine to estimate body density. A volunteer of mass 72.0 kg is lowered on a harness until fully submerged in a tank of water of density 1000 kg m−3. While fully submerged, the harness supports an apparent weight of 30.0 N. Air remaining in the lungs may be ignored.

(a)

Outline why a body submerged in water experiences an upward buoyant force.

(1)
(b)

Calculate the buoyant force on the volunteer when fully submerged.

(2)
(c)

Determine the average density of the volunteer's body.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Pressure in the water increases with depth, so the upward force on the lower surfaces is greater than the downward force on the upper surfaces, giving a net upward force .✓ 1
Part (b)
Weight = 72.0 × 9.8 = 706 N .✓ 1
Buoyant force = 706 − 30.0 = 676 N .✓ 1
Part (c)
V = Fb/(ρwg) = 676/(1000 × 9.8) = 6.89×10−2 m3 .✓ 1
ρ = m/V = 72.0/6.89×10−2 = 1.04×103 kg m−3 .✓ 1
ECF from (b).—

Exam tip: Apparent weight = true weight − buoyant force. Find the buoyant force first: it gives the displaced volume directly.

156A-2-05Explosions & recoil Paper 2Medium10 marks

A nucleus of polonium-210, 21084Po, is at rest. It decays by emitting an alpha particle, and the daughter nucleus is an isotope of lead, Pb. The energy released in the decay is 5.41 MeV, and all of it becomes kinetic energy of the alpha particle and the lead nucleus.

Take the mass of the alpha particle to be 4.00 u and the mass of the lead nucleus to be 206 u.

(a)

Complete the equation for the decay: 21084Po → ……Pb + ……α

(1)
(b)

Explain why the alpha particle and the lead nucleus move off in opposite directions with momenta of equal magnitude.

(2)
(c)

Show that Eα / EPb = mPb / mα, where E is kinetic energy and m is mass.

(2)
(d)

Calculate, in MeV, the kinetic energy of the alpha particle.

(2)
(e)
(i)

Determine the speed of the alpha particle.

(2)
(ii)

Calculate the speed of the lead nucleus.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
20682Pb AND 42α✓ 1Both needed.
Part (b)
the total momentum before the decay is zero «nucleus at rest»✓ 1
no external force acts, so the total momentum afterwards is also zero, so the two momenta are equal in magnitude and opposite in direction✓ 1OWTTE.
Part (c)
E = p² / 2m for each particle✓ 1Accept ½mv² with mαvα = mPbvPb.
p is the same for both, so Eα/EPb = (p²/2mα) / (p²/2mPb) = mPb/mα✓ 1Must see the momenta cancel.
Part (d)
Eα = 5.41 × 206 / (206 + 4.00)✓ 1Accept a ratio route: Eα = 51.5 EPb with Eα + EPb = 5.41 MeV.
Eα = 5.31 MeV✓ 1Award [2] for CNA. Award [0] for 5.41 MeV.
Part (e)(i)
Eα = 5.31 × 106 × 1.60 × 10−19 = 8.49 × 10−13 J AND mα = 4.00 × 1.661 × 10−27 kg✓ 1Allow ECF from (d).
v = √(2E/m) = 1.6 × 107 m s−1✓ 1
Part (e)(ii)
vPb = 1.60 × 107 × 4.00 / 206 = 3.1 × 105 m s−1✓ 1Allow ECF from (e)(i). Accept use of EPb = 0.10 MeV.

Answers: (d) 5.31 MeV  ·  (e)(i) 1.6 × 107 m s−1  ·  (e)(ii) 3.1 × 105 m s−1 (the remaining parts are explanations — see the table above)

157A-2-06Drag & buoyancy Paper 2Medium12 marks

Clouds consist of very small droplets of water. A spherical droplet of radius 12 μm falls vertically through still air.

Density of water = 1000 kg m−3; density of air = 1.2 kg m−3; viscosity of air η = 1.8 × 10−5 Pa s.

A box containing a single dot that represents the falling water droplet.droplet
Box for (b)
(a)
(i)

Calculate the mass of the droplet.

(1)
(ii)

Show that the buoyancy force on the droplet is about 0.1 % of its weight.

(1)
(b)

Draw and label, on the dot, the forces acting on the droplet when it falls at its terminal speed. The relative lengths of your arrows should be correct.

(2)
(c)

Determine the terminal speed of the droplet.

(2)
(d)

Inside the cloud the air rises vertically at a steady 2.5 cm s−1. State and explain the motion of the droplet relative to the ground.

(2)
(e)
(i)

Calculate the terminal speed that the same model predicts for a raindrop of radius 1.0 mm.

(2)
(ii)

The measured terminal speed of a raindrop of this size is about 6.5 m s−1. Suggest why the model fails for the raindrop.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
m = 1000 × 4/3 × π × (12 × 10−6)³ = 7.2 × 10−12 kg✓ 1Accept 7.24 × 10−12 kg.
Part (a)(ii)
the droplet displaces its own volume of air, so Fb/W = ρairVg/ρwaterVg = 1.2/1000 = 1.2 × 10−3 «0.12 %»✓ 1ALTERNATIVE: W = 7.1 × 10−11 N and Fb = 8.5 × 10−14 N. Must see full substitution OR answer to 3 s.f.
Part (b)
weight vertically downwards AND drag vertically upwards, both labelled✓ 1Do not accept “gravity” for weight without “force”. Do not credit a force in the direction of motion.
drag arrow equal in length to the weight arrow «by eye», buoyancy upwards and very much shorter «or labelled as negligible»✓ 1Award MP2 only for a diagram with zero resultant force.
Part (c)
mg = 6πηrv «buoyancy negligible»✓ 1Allow inclusion of the buoyancy force.
v = 7.1 × 10−11/(6π × 1.8 × 10−5 × 12 × 10−6) = 0.017 m s−1✓ 1Accept 0.0174 m s−1. Award [2] for CNA. Allow ECF from (a)(i).
Part (d)
the drag depends on the velocity of the droplet relative to the air «the droplet still falls at 1.7 cm s−1 relative to the air»✓ 1
so the droplet moves upwards relative to the ground at about 0.8 cm s−1✓ 1Accept 0.76 cm s−1. Allow ECF from (c). MP2 only scores if MP1 scores.
Part (e)(i)
weight ∝ r³ and drag ∝ rv, so the terminal speed ∝ r² OR full recalculation with r = 1.0 × 10−3 m✓ 1
v = 0.0174 × (1.0 × 10−3/12 × 10−6)² = 120 m s−1✓ 1Accept 120–121 m s−1. Allow ECF from (c).
Part (e)(ii)
for a large drop moving fast the drag is no longer given by 6πηrv: it increases much more rapidly with speed «turbulent flow; drag roughly ∝ v²»✓ 1OWTTE.
so the drag becomes equal to the weight at a much lower speed than the model predicts✓ 1Accept: a large drop flattens as it falls, increasing the drag at a given speed.

Answers: (a)(i) 7.2 × 10−12 kg  ·  (c) 0.017 m s−1  ·  (d) upwards at about 0.8 cm s−1  ·  (e)(i) 120 m s−1 (the remaining parts are explanations — see the table above)

158A-2-17Newton's laws & force pairs Paper 2Medium11 marks

A beaker of water stands on a top-pan balance that reads the force on its pan in newtons. The reading is 14.50 N. An aluminium block of mass 0.540 kg and volume 2.00 × 10−4 m3 hangs from a newton-meter. The block is lowered until it is completely under water, without touching the beaker (Figure 1). The density of water is 1000 kg m−3.

A block hanging from a newton-meter, fully submerged in water in a beaker that stands on a top-pan balancenewton-meterwaterblocktop-pan balance
Figure 1 — Diagram NOT accurately drawn
A dot representing the submerged blockblock
Figure 2 — for (b)(i)
(a)
(i)

Calculate the reading on the newton-meter before the block enters the water.

(1)
(ii)

Calculate the buoyancy force on the block when it is completely under water.

(1)
(b)
(i)

Draw and label, on Figure 2, the forces acting on the block when it is completely under water.

(2)
(ii)

Show that the reading on the newton-meter is now about 3.3 N.

(1)
(c)

State and explain the change in the reading on the balance as the block is lowered into the water, and determine the new reading.

(4)
(d)

The string is cut and the block comes to rest on the bottom of the beaker. Determine the reading on the balance.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
0.540 × 9.81 = 5.30 N✓ 1
Part (a)(ii)
Fb = ρVg = 1000 × 2.00 × 10−4 × 9.81 = 1.96 N✓ 1
Part (b)(i)
weight downwards, tension «force from the newton-meter/string» upwards, buoyancy upwards, all labelled✓ 1
the two upward arrows together equal the weight in length AND tension longer than buoyancy✓ 1Award [1 max] if any other force is drawn.
Part (b)(ii)
T = 5.30 − 1.96 «= 3.34 N»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)
the water exerts an upward buoyancy force on the block✓ 1
by Newton’s third law the block exerts an equal downward force on the water✓ 1
this force is transmitted to the pan, so the balance reading increases by the buoyancy force✓ 1
new reading = 14.50 + 1.96 = 16.5 N✓ 1Allow ECF from (a)(ii).
Part (d)
the balance now supports the whole weight of the beaker, the water and the block «the newton-meter supports nothing»✓ 1
14.50 + 5.30 = 19.8 N✓ 1Allow ECF from (a)(i). Award [2] for CNA.

Answers: (a)(i) 5.30 N  ·  (a)(ii) 1.96 N  ·  (b)(ii) 3.34 N  ·  (c) 16.5 N  ·  (d) 19.8 N (the remaining parts are explanations — see the table above)

159A-2-19Variable mass & F = Δp/Δt Paper 2Medium12 marks

A camera drone of mass 1.6 kg hovers at a fixed position. Its rotors take in air that is effectively at rest and push it vertically downwards at a speed of 9.0 m s−1.

The drone is powered by a battery with a terminal potential difference of 14.8 V. While the drone hovers, the current in the battery is 7.5 A.

(a)

Explain, using Newton’s laws of motion, how the rotors produce an upward force on the drone.

(3)
(b)

Show that about 1.7 kg of air is pushed downwards each second.

(2)
(c)
(i)

Calculate the rate at which kinetic energy is given to the air.

(1)
(ii)

Determine the efficiency with which the drone converts electrical energy into kinetic energy of the air.

(2)
(d)

The battery stores 2.7 × 105 J of energy. Estimate the maximum time for which the drone can hover.

(1)
(e)

A designer suggests fitting larger rotors that push a greater mass of air downwards each second, at a lower speed. Explain why this reduces the power needed for the drone to hover.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
the rotors exert a downward force on the air✓ 1
the air gains downward momentum; the force equals its rate of change of momentum «Newton’s second law»✓ 1
by Newton’s third law the air exerts an equal and opposite «upward» force on the rotors✓ 1
Part (b)
upward force = weight = 1.6 × 9.81 = 15.7 N AND F = Δp/Δt = (Δm/Δt)v✓ 1
Δm/Δt = 15.7 / 9.0 «= 1.74 kg s−1»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)(i)
½ × 1.74 × 9.0² = 71 W✓ 1Accept 69–71 W.
Part (c)(ii)
electrical power = 14.8 × 7.5 = 111 W✓ 1
efficiency = 70.6 / 111 = 0.64 «64 %»✓ 1Allow ECF from (c)(i).
Part (d)
2.7 × 105 / 111 = 2.4 × 103 s «≈ 41 min»✓ 1Allow ECF from (c)(ii).
Part (e)
the upward force must still equal the weight: (Δm/Δt)v = mg is fixed✓ 1
the power given to the air is ½(Δm/Δt)v² = ½mgv✓ 1
so a lower air speed v means a smaller power «e.g. half the speed, twice the mass flow, half the power»✓ 1MP3 only scores if the force is recognised as fixed.

Answers: (b) 1.74 kg s−1  ·  (c)(i) 71 W  ·  (c)(ii) 0.64  ·  (d) 2.4 × 103 s (the remaining parts are explanations — see the table above)

160A-2-20Explosions & recoil Paper 2Medium15 marks

A person stands in a small boat that is at rest on a calm lake. The total mass of the boat and the person is 180 kg. The person throws a mooring weight of mass 12 kg horizontally from the back of the boat. The weight leaves the person’s hands with a horizontal velocity of 4.5 m s−1 relative to the water. The throw lasts 0.60 s. The resistive force of the water on the boat is negligible during the throw.

The graph shows how the horizontal momentum of the weight varies with time t from the start of the throw. The direction in which the weight is thrown is positive.

Horizontal momentum of the weight against time during and just after the throw: an S-shaped curve rising from 0 to 54 kg m s⁻¹ at 0.60 s, then constant.00.10.20.30.40.50.60.7t / s−60−40−200204060horizontal momentum / kg m s⁻¹weight
Graph drawn to scale
(a)
(i)

State why the total momentum of the boat, the person and the weight is zero immediately after the throw.

(1)
(ii)

Calculate the speed of the boat immediately after the throw.

(2)
(b)

Draw, on the graph, a line to show how the horizontal momentum of the boat and the person varies with time from t = 0 to t = 0.60 s.

(2)
(c)
(i)

Show that the total kinetic energy of the boat, the person and the weight just after the throw is about 130 J.

(1)
(ii)

Outline where this kinetic energy has come from.

(1)
(iii)

The force exerted by the person on the weight and the force exerted by the weight on the person are equal in magnitude at every instant of the throw. Explain, by considering the work done by these forces, why the weight gains much more kinetic energy than the boat and the person.

(3)
(d)
(i)

After the throw, the water brings the boat to rest in 6.0 s. Calculate the average resistive force exerted by the water on the boat.

(2)
(ii)

The person now stands on a fixed jetty and throws the same weight horizontally, doing the same amount of work as in the throw from the boat. Deduce whether the weight now leaves the person’s hands faster than 4.5 m s−1.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
the total momentum was zero before the throw AND no «resultant» external horizontal force acts on the system during the throw✓ 1Do not accept a bare “momentum is conserved”.
Part (a)(ii)
0 = 12 × 4.5 − 180v OR momentum of boat and person = 54 kg m s−1✓ 1
v = 0.30 m s−1✓ 1Award [2] for CNA. Award [1] max for 0.28 m s−1 «mass of the weight added to the boat and person».
Part (b)
line below the t-axis that is the mirror image of the given curve «equal magnitude, opposite sign at every instant»✓ 1The candidate’s line and the given curve must add to zero at each time, by eye.
line starts at 0 and ends at −54 kg m s−1 at t = 0.60 s✓ 1Allow ± 3 kg m s−1.
Part (c)(i)
½ × 12 × 4.5² + ½ × 180 × 0.30² = 121.5 + 8.1 = 129.6 «J»✓ 1Must see full substitution OR answer of 129.6 J.
Part (c)(ii)
chemical «potential» energy in the person’s muscles, transferred by the work the person does «on the weight and on the boat»✓ 1
Part (c)(iii)
the two forces are equal in magnitude «Newton’s third law» and act for the same time✓ 1
the momenta are equal in magnitude at every instant, so the weight always moves 15 times as fast as the boat and travels 15 times as far «relative to the water» during the throw✓ 1
W = Fs, so 15 times as much work is done on the weight, which therefore gains 15 times as much kinetic energy «121.5 J compared with 8.1 J»✓ 1Allow ECF from (a)(ii). An argument using only Ek = p²/2m with equal momenta scores [1] max.
Part (d)(i)
impulse = change in momentum = 54 N s✓ 1Allow ECF from (a)(ii).
F = 54/6.0 = 9.0 N✓ 1Award [2] for CNA.
Part (d)(ii)
the jetty does not move, so all 129.6 J «≈ 130 J» becomes kinetic energy of the weight✓ 1
v = √(2 × 129.6/12) = 4.6 m s−1✓ 1Accept 4.6–4.7 m s−1. Allow ECF from (c)(i).
4.6 m s−1 > 4.5 m s−1 so the weight leaves faster «by about 3 %»✓ 1MP3 requires a comparison of two speeds. A bare “faster” scores [0].

Answers: (a)(ii) 0.30 m s−1  ·  (c)(i) 129.6 J  ·  (d)(i) 9.0 N  ·  (d)(ii) 4.6 m s−1; faster (the remaining parts are explanations — see the table above)

161A-2-22Circular motion Paper 2Medium13 marks

A skateboarder of mass 62 kg (including the skateboard) starts from rest at the edge of a half-pipe. The cross-section of the half-pipe is a semicircle of radius 3.2 m, and the starting point is level with the centre O of the semicircle. Treat the skateboarder as a point mass. Unless stated otherwise, friction and air resistance are negligible.

Cross-section of a half-pipe: a semicircle of radius 3.2 m. The skateboarder starts at rest at the left lip, level with the centre O. Point P is the lowest point. A separate box with a dot represents the skateboarder at P.O3.2 mstartPskateboarder at P
Diagram NOT accurately drawn
(a)
(i)

Show that the speed of the skateboarder at the lowest point P is about 7.9 m s−1.

(1)
(ii)

State the direction of the resultant force on the skateboarder at P.

(1)
(b)
(i)

Determine the normal force exerted by the half-pipe on the skateboarder at P.

(3)
(ii)

Draw and label, on the dot in the box, the forces acting on the skateboarder at P. The lengths of your arrows should be consistent with your answer to (b)(i).

(2)
(iii)

A friend claims that a half-pipe of larger radius would reduce the normal force on the skateboarder at P, for a start from rest level with O. Deduce whether this claim is correct.

(2)
(c)

In practice, friction acts and the skateboarder reaches P at 7.2 m s−1.

(i)

Calculate the energy transferred by friction as the skateboarder moves from the start to P.

(2)
(ii)

State and explain how the normal force at P compares with your answer to (b)(i).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
½mv² = mgr so v = √(2 × 9.81 × 3.2) = 7.92 «m s−1»✓ 1Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
vertically upwards «towards O / the centre of the circle»✓ 1
Part (b)(i)
FN − mg = mv²/r✓ 1Accept a correct equation in words.
mv²/r = 62 × 7.92²/3.2 = 1.22 × 103 N✓ 1
FN = 1.82 × 103 N✓ 1Accept 1.8 × 103 N. Award [1] max for 1.22 × 103 N «centripetal force given as the answer». Award [2] max for 608 N «weight subtracted».
Part (b)(ii)
weight vertically down AND normal force vertically up, both labelled✓ 1
normal force arrow about three times as long as the weight arrow✓ 1Accept 2.5 to 3.5 times. Allow ECF from (b)(i). Do not award MP2 for arrows of equal length.
Part (b)(iii)
v² = 2gr, so the centripetal force mv²/r = 2mg, which does not depend on r✓ 1
so FN = 3mg for any radius: the claim is incorrect✓ 1MP2 only scores if MP1 scores. OWTTE.
Part (c)(i)
mgr − ½mv² = 62 × 9.81 × 3.2 − ½ × 62 × 7.2²✓ 1
= 340 J✓ 1Accept 339 J. Award [2] for CNA.
Part (c)(ii)
smaller speed so a smaller centripetal force mv²/r is needed✓ 1
FN = mg + mv²/r is smaller «1.61 × 103 N»✓ 1Award [0] for “smaller” with no reasoning.

Answers: (a)(i) 7.92 m s−1  ·  (b)(i) 1.82 × 103 N  ·  (c)(i) 340 J (the remaining parts are explanations — see the table above)

162A-2-23Drag & buoyancy Paper 2Medium13 marks

A weather balloon is filled with helium at ground level, where its volume is 4.0 m³ and the density of the air is 1.2 kg m−3. The balloon fabric and the instrument package have a total mass of 2.5 kg and the helium has a mass of 0.68 kg. When the balloon moves upwards through the air at speed v, the drag force on it is kv², where k = 0.65 kg m−1. The volume of the instrument package is negligible.

A box containing a single dot that represents the weather balloon and its payload.balloon
Diagram for (a)
(a)

The balloon rises at a constant speed. Draw and label, on the dot, the forces acting on the balloon. The relative lengths of your arrows should be approximately correct.

(2)
(b)
(i)

Show that the buoyancy force on the balloon at ground level is about 47 N.

(1)
(ii)

Calculate the acceleration of the balloon immediately after it is released from rest.

(3)
(iii)

Determine the terminal speed of the balloon near the ground.

(2)
(c)

At a height of 5.0 km the air pressure is 54 kPa and the temperature is 255 K. At ground level they were 101 kPa and 288 K. The helium in the balloon stays at the same pressure and temperature as the surrounding air, and none escapes.

(i)

Calculate the volume of the balloon at a height of 5.0 km.

(2)
(ii)

The density of the air at 5.0 km is 0.72 kg m−3. Determine whether the buoyancy force on the balloon is different at this height from its value at ground level.

(2)
(iii)

State one assumption about the helium that you made in (c)(i).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
buoyancy «upthrust, Fb» upwards; weight «mg» AND drag «Fd» downwards; all labelled✓ 1Do not award MP1 if drag is drawn upwards.
buoyancy arrow equal in length to the weight and drag arrows combined✓ 1By eye.
Part (b)(i)
Fb = ρVg = 1.2 × 4.0 × 9.81 = 47.1 «N»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
total mass = 2.5 + 0.68 = 3.18 kg, so weight = 31.2 N✓ 1
resultant force = 47.1 − 31.2 = 15.9 N✓ 1
a = 15.9/3.18 = 5.0 m s−2✓ 1Award [3] for CNA. Award [2] max for 9.0 m s−2 «mass of helium omitted».
Part (b)(iii)
at terminal speed kv² = Fb − mg = 15.9 N✓ 1
v = √(15.9/0.65) = 4.9 m s−1✓ 1Allow ECF from (b)(ii).
Part (c)(i)
V2 = V1 × (P1/P2) × (T2/T1) = 4.0 × (101/54) × (255/288)✓ 1Accept use of PV/T = constant.
= 6.6 m³✓ 1Award [1] max for 7.5 m³ «temperature change ignored».
Part (c)(ii)
Fb = 0.72 × 6.6 × 9.81 = 47 N✓ 1Allow ECF from (c)(i).
«about» the same as the 47 N at ground level: the lower air density is compensated by the larger volume✓ 1MP2 requires a comparison with the value from (b)(i).
Part (c)(iii)
the helium behaves as an ideal gas «so that PV/T = constant applies»✓ 1Do not accept conditions that are given in the stem «no helium escapes; same pressure and temperature as the air».

Answers: (b)(i) 47.1 N  ·  (b)(ii) 5.0 m s−2  ·  (b)(iii) 4.9 m s−1  ·  (c)(i) 6.6 m³  ·  (c)(ii) 47 N; about the same (the remaining parts are explanations — see the table above)

163A-2-30Variable mass & F = Δp/Δt Paper 2Medium14 marks

A space probe of mass 650 kg is driven by an ion thruster. Inside the thruster, xenon ions, each of mass 2.18 × 10−25 kg and charge +1.60 × 10−19 C, are accelerated from rest through a potential difference of 1200 V and leave the probe as a narrow beam. The current carried by the beam is 1.8 A. Electrons of negligible mass are also emitted so that the probe stays uncharged.

The probe is far from any planet, so gravitational forces on it are negligible.

(a)
(i)

Show that the kinetic energy gained by each ion is about 1.9 × 10−16 J.

(1)
(ii)

Calculate the speed of an ion as it leaves the thruster.

(2)
(b)
(i)

Calculate the number of ions leaving the thruster each second.

(1)
(ii)

Determine the magnitude of the force exerted on the probe by the thruster.

(3)
(c)

Explain why the force on the probe would soon decrease if the electrons were not emitted.

(2)
(d)
(i)

Calculate the increase in the speed of the probe after one day of continuous operation.

(2)
(ii)

In (d)(i) the mass of the probe was treated as constant. Explain why this is justified for one day but would not be justified for a mission in which the thruster runs continuously for 4 years.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
W = qV = 1.60 × 10−19 × 1200 = 1.92 × 10−16 «J»✓ 1Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
½ × 2.18 × 10−25 × v² = 1.92 × 10−16✓ 1
v = 4.2 × 104 m s−1✓ 1Award [2] for CNA.
Part (b)(i)
N = I/e = 1.8/1.60 × 10−19 = 1.1 × 1019 «s−1»✓ 1Accept 1.13 × 1019.
Part (b)(ii)
mass ejected per second = 1.13 × 1019 × 2.18 × 10−25 = 2.45 × 10−6 kg s−1✓ 1Allow ECF from (b)(i).
F = Δp/Δt = (Δm/Δt) × v✓ 1Accept momentum given to the ions each second.
F = 2.45 × 10−6 × 4.20 × 104 = 0.10 N✓ 1Allow ECF from (a)(ii). Award [3] for CNA.
Part (c)
each ion carries positive charge away from the probe, so «by conservation of charge» the probe would become increasingly negatively charged✓ 1
the negative probe would attract the positive ions back «opposite charges attract», reducing the momentum carried away by the beam each second and so the force on the probe✓ 1Accept the ions being slowed down or pulled back to the probe.
Part (d)(i)
a = 0.103/650 = 1.6 × 10−4 m s−2✓ 1
Δv = 1.58 × 10−4 × 86 400 = 14 m s−1✓ 1Accept 13–14 m s−1. Allow ECF from (b)(ii).
Part (d)(ii)
mass of xenon ejected in one day ≈ 2.45 × 10−6 × 86 400 = 0.21 kg, negligible compared with 650 kg✓ 1
in 4 years about 3 × 102 kg is ejected, a large fraction of the probe’s mass✓ 1Accept any estimate of order 102 kg.
so the mass of the probe falls significantly and, for the same force, its acceleration increases: F = ma with constant m no longer applies «use F = Δp/Δt»✓ 1

Answers: (a)(i) 1.92 × 10−16 J  ·  (a)(ii) 4.2 × 104 m s−1  ·  (b)(i) 1.1 × 1019 s−1  ·  (b)(ii) 0.10 N  ·  (d)(i) 14 m s−1 (the remaining parts are explanations — see the table above)

164A-2-35Hooke's law & springs Paper 2Medium14 marks

The body of a car is supported by four identical springs, one at each wheel. When passengers of total mass 280 kg get into the car, the body moves down by 3.5 cm. Assume that the load is shared equally between the four springs. The mass of the loaded car body is 1380 kg.

Axes of displacement x against time t, with marks at multiples of the period T and at x₀ and −x₀. A dot marks the starting displacement x₀ at t = 0.0T2T3T4Ttime t−x₀0x₀displacement x
Axes for (d)(i)
(a)
(i)

State Hooke’s law.

(1)
(ii)

Show that the spring constant of each spring is about 2.0 × 104 N m−1.

(2)
(iii)

Outline the significance of the negative sign in FH = −kx.

(1)
(b)

The loaded car body is pushed down and released, and it oscillates vertically on the springs. Determine the natural frequency of this oscillation.

(3)
(c)

The loaded car is driven at a constant speed along a road that has small ridges across it, 9.0 m apart.

(i)

Calculate the speed at which the car body oscillates with the greatest amplitude.

(2)
(ii)

Explain why the amplitude of oscillation is large at this speed.

(2)
(d)

Shock absorbers provide damping. The car body is displaced by x0 and released at t = 0. On the axes, T is the period of the undamped oscillation.

(i)

Sketch, on the axes, graphs to show how the displacement of the car body varies with time when the oscillation is lightly damped and when it is critically damped. Label your graphs L and C.

(2)
(ii)

Suggest why car suspensions are designed to be close to critically damped.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
the force «exerted by a spring» is proportional to its extension or compression «within the elastic limit»✓ 1Accept FH = −kx with symbols explained.
Part (a)(ii)
force on each spring = 280 × 9.81/4 = 687 N✓ 1
k = 687/0.035 = 1.96 × 104 «N m−1»✓ 1Must see full substitution OR answer to 3 s.f.
Part (a)(iii)
the «restoring» force is in the opposite direction to the displacement «extension» of the spring✓ 1
Part (b)
each spring carries 345 kg OR total spring constant 4k = 7.85 × 104 N m−1✓ 1
T = 2π√(345/1.96 × 104) = 0.83 s✓ 1Accept 2π√(m/4k).
f = 1/T = 1.2 Hz✓ 1Award [2] max for 0.60 Hz «one spring carrying the whole mass» or 1.3 Hz «unloaded mass used».
Part (c)(i)
greatest amplitude «resonance» when the time between ridges equals the natural period: v/9.0 = 1.2 Hz✓ 1
v = 11 m s−1✓ 1Accept 10.6–11 m s−1. Allow ECF from (b).
Part (c)(ii)
the driving frequency equals the natural frequency «resonance»✓ 1
energy is transferred to the oscillation most efficiently, so the amplitude builds up until the energy dissipated by damping each cycle equals the energy supplied✓ 1OWTTE.
Part (d)(i)
L: starts at x0, oscillates about zero with decreasing amplitude and «roughly» constant period✓ 1Period should be close to the marked T.
C: falls from x0 to zero without crossing the time axis, reaching «close to» zero within about one period✓ 1
Part (d)(ii)
the body returns to its equilibrium position quickly without repeated bouncing «comfort, tyres stay in contact with the road»✓ 1

Answers: (a)(ii) 1.96 × 104 N m−1  ·  (b) 1.2 Hz  ·  (c)(i) 11 m s−1 (the remaining parts are explanations — see the table above)

165A-2-46Resultant forces Paper 2Medium13 marks

A slackline is a flat strip of nylon webbing tied between two trees. The anchor points are at the same height and 12.0 m apart. A person of mass 68 kg stands at rest at the midpoint P of the line, which sags 0.75 m below the anchors (Figure 1). The mass of the line is negligible and each half of the line is straight.

A slackline tied between two trees 12.0 m apart, sagging 0.75 m at its midpoint P where the person stands, shown as a dotP0.75 m12.0 mtreetree
Figure 1 — Diagram NOT accurately drawn
(a)

Draw and label, on the dot at P in Figure 1, the forces acting on the person.

(2)
(b)

Show that each half of the line makes an angle of about 7° with the horizontal.

(1)
(c)

Calculate the tension in the line.

(2)
(d)

Explain why the tension in the line is much greater than the weight of the person.

(2)
(e)

When nobody stands on it, the line is straight and its length between the anchors is 12.0 m, with negligible tension. The webbing obeys Hooke’s law. Calculate the spring constant of the line.

(2)
(f)

The line is re-tied more tightly, so that the same person makes it sag by only 0.25 m. Each anchor is rated to withstand a force of 10 kN. Determine whether the anchors are safe.

(3)
(g)

Suggest why the anchors need to be rated for a force well above the value you found in (f).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
downward arrow labelled weight «Fg»✓ 1
two arrows labelled tension, directed along each half of the line away from P, of equal length✓ 1Arrows must start on the dot.
Part (b)
tan θ = 0.75/6.0 so θ = 7.13°✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)
2T sin θ = mg so T = 68 × 9.81/(2 sin 7.13°)✓ 1
T = 2700 N✓ 1Accept 2.6–2.8 kN. Award [1] max for T = mg/2 = 334 N.
Part (d)
only the vertical components of the tensions support the weight: 2T sin θ = mg✓ 1
θ is small, so sin θ is small «≈ 0.12» and T must be large «T ≈ 4mg»✓ 1Do not accept “the line is stretched” alone.
Part (e)
length of each half = √(6.0² + 0.75²) = 6.047 m, so extension = 2 × 6.047 − 12.0 = 0.0934 m✓ 1
k = T/x = 2690/0.0934 = 2.9 × 104 N m−1✓ 1Allow ECF from (c). Accept 2.8–3.0 × 104 N m−1.
Part (f)
new angle: tan θ = 0.25/6.0, θ = 2.39°✓ 1
T = 68 × 9.81/(2 sin 2.39°) = 8000 N✓ 1
8.0 kN < 10 kN, so the anchors are safe «but only just»✓ 1MP3 needs a comparison. Allow ECF from MP2.
Part (g)
when the person walks, bounces or falls onto the line, the line must also give an upward acceleration, so the tension is greater than in the static case✓ 1Accept “safety margin for dynamic loads” only with a reason.

Answers: (b) 7.13°  ·  (c) 2700 N  ·  (e) 2.9 × 104 N m−1  ·  (f) 8000 N: safe (the remaining parts are explanations — see the table above)

166A-2-47Collisions & energy Paper 2Medium12 marks

In a game of curling, stone A of mass 19.1 kg is released at 2.20 m s−1 and slides in a straight line across horizontal ice towards an identical stone B, which is at rest 20.0 m away (Figure 1). The coefficient of dynamic friction between a stone and the ice is 0.0080. A hits B head-on, and immediately after the collision A moves forward at 0.10 m s−1 along the same line.

A circular target, the house, has its centre 8.0 m beyond the initial position of B, on the line of motion, and a radius of 1.83 m.

Plan view of the ice: stone A released 20.0 m from stationary stone B; the centre of the house is 8.0 m beyond B and the house has a radius of 1.83 mAreleaseBhouse20.0 m8.0 m1.83 m
Figure 1 — Diagram NOT accurately drawn
(a)

Show that the deceleration of a sliding stone is about 0.08 m s−2.

(1)
(b)

Calculate the speed of A just before it hits B.

(2)
(c)

Calculate the speed of B immediately after the collision.

(2)
(d)

Calculate the kinetic energy transferred to other forms during the collision, and suggest one form of energy into which it is transferred.

(3)
(e)

Determine whether B comes to rest inside the house.

(3)
(f)

Outline why the friction of the ice on the stones can be ignored when applying the conservation of momentum to the collision.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Ff = μdmg so a = μdg = 0.0080 × 9.81 = 0.0785 «m s−2»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)
v² = 2.20² − 2 × 0.0785 × 20.0 = 1.70✓ 1
v = 1.30 m s−1✓ 1Award [2] for CNA.
Part (c)
m × 1.30 = m × 0.10 + m × vB✓ 1Allow ECF from (b).
vB = 1.20 m s−1✓ 1
Part (d)
Ek before = ½ × 19.1 × 1.30² = 16.2 J; after = ½ × 19.1 × (0.10² + 1.20²) = 13.9 J✓ 1Allow ECF from (b) and (c).
energy transferred = 2.3 J✓ 1Accept 2.2–2.4 J.
internal energy «thermal energy» of the stones OR sound✓ 1Do not accept “heat” alone without the stones; do not accept “friction”.
Part (e)
s = vB²/2a = 1.20²/(2 × 0.0785)✓ 1Allow ECF from (c).
s = 9.2 m✓ 1
the house extends from 6.17 m to 9.83 m beyond B, so B stops inside the house✓ 1MP3 needs the comparison with the edges of the house.
Part (f)
the collision lasts a very short time and the collision forces are much larger than friction, so the impulse of friction during the collision is negligible✓ 1OWTTE.

Answers: (a) 0.0785 m s−2  ·  (b) 1.30 m s−1  ·  (c) 1.20 m s−1  ·  (d) 2.3 J  ·  (e) 9.2 m: inside the house (the remaining parts are explanations — see the table above)

167A-2-48Circular motion Paper 2Medium14 marks

A car travels at a constant speed around a flat, horizontal roundabout on a circular path of radius 18 m. A phone of mass 0.17 kg lies on a horizontal rubber mat on the dashboard. The coefficient of static friction between the phone and the mat is 0.40. Figure 1 shows the phone as seen from behind the car.

View from behind the car: a phone, shown with a dot at its centre, lies on a horizontal rubber mat on the dashboard; the centre of the roundabout is to the leftdashboard (rubber mat)phonetowards centre of roundabout
Figure 1 — Diagram NOT accurately drawn
(a)

Draw and label, on the dot in Figure 1, the forces acting on the phone while it moves around the roundabout without sliding.

(2)
(b)

The car travels at 7.5 m s−1.

(i)

Calculate the angular velocity of the car.

(1)
(ii)

Determine whether the phone slides on the mat.

(3)
(c)

Deduce the maximum speed at which the car can travel around the roundabout without the phone sliding, and explain why this speed does not depend on the mass of the phone.

(2)
(d)

At a higher speed the phone slides towards the outside of the curve. The driver says that the phone is “thrown outwards by a centrifugal force”. Explain, with reference to Newton’s laws of motion, why the phone slides towards the outside of the curve.

(4)
(e)

On a wet day the coefficient of static friction between the tyres and the road is 0.35. Deduce whether the phone or the car is the first to slide as the driver gradually increases the speed around the roundabout.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
weight downwards and normal force upwards, of equal length✓ 1
frictional force horizontal, towards the centre of the roundabout «to the left»✓ 1Do not award MP2 if an outward force is also drawn.
Part (b)(i)
ω = v/r = 7.5/18 = 0.42 rad s−1✓ 1
Part (b)(ii)
friction needed = mv²/r = 0.17 × 7.5²/18 = 0.53 N✓ 1
maximum friction = μsmg = 0.40 × 0.17 × 9.81 = 0.67 N✓ 1
the maximum friction is greater than the force needed, so the phone does not slide✓ 1Accept the comparison of accelerations: 3.1 m s−2 < μsg = 3.9 m s−2. MP3 needs a comparison.
Part (c)
μsmg = mv²/r gives v = √(μsgr) = 8.4 m s−1✓ 1
both the maximum friction and the required centripetal force are proportional to m, so m cancels✓ 1
Part (d)
the friction from the mat is now less than the centripetal force mv²/r needed for the phone to follow the car’s circular path✓ 1
by Newton’s first law the phone tends to continue in a straight line, along the tangent to its path✓ 1
friction still acts on it towards the centre but is too small, so the phone moves on a path of larger radius than the car✓ 1
the car turns towards the centre underneath it, so relative to the car the phone moves outwards; no outward force acts on the phone✓ 1Do not award MP4 for “the centrifugal force is greater than friction”.
Part (e)
the car skids when v > √(0.35 × 9.81 × 18) = 7.9 m s−1✓ 1
this is less than 8.4 m s−1, so the car slides first✓ 1Accept the comparison 0.35 < 0.40 with a reason «same r, v ∝ √μ».

Answers: (b)(i) 0.42 rad s−1  ·  (b)(ii) needs 0.53 N < 0.67 N: does not slide  ·  (c) 8.4 m s−1  ·  (e) car first, at 7.9 m s−1 (the remaining parts are explanations — see the table above)

168A-2-72Friction & inclined planes Paper 2Medium13 marks

A window-cleaning robot of mass 1.20 kg works on a vertical glass window. A fan inside the robot removes air from a sealed chamber between the robot and the glass, so that the pressure in the chamber is lower than atmospheric pressure. The chamber is in contact with the glass over an area of 0.040 m2, so the atmosphere pushes the robot towards the glass with a force equal to (pressure difference) × 0.040 m2.

The robot touches the glass only through rubber treads. The coefficient of static friction between the treads and dry glass is 0.50. Figure 1 shows a side view.

Side view: a vertical glass window on the left; the robot is a box pressed against the right-hand face of the glass, with a dot at its centre and a sealed low-pressure chamber between it and the glassglassrobotsealed chamber at low pressurerubber treads
Figure 1 — Diagram NOT accurately drawn
(a)

Draw and label, on the dot in Figure 1, the forces acting on the robot when it is at rest on the dry window. Treat the push of the atmosphere as a single force.

(2)
(b)

Show that the normal force on the robot must be at least about 24 N for the robot to remain at rest on the dry window.

(1)
(c)

Calculate the minimum pressure difference between the atmosphere and the chamber.

(2)
(d)

In normal use the pressure difference is 1.5 kPa. While the robot cleans, the glass is wet and the coefficient of static friction falls to 0.25. Determine whether the robot can remain at rest on the wet window.

(3)
(e)

The robot climbs vertically up the dry window at constant speed, driven by its treads. State and explain the magnitude and direction of the frictional force exerted by the glass on the robot.

(2)
(f)

Calculate the greatest upward acceleration that the robot can have on the wet window when the pressure difference is 1.5 kPa.

(2)
(g)

Suggest why the robot is also attached to the room by a safety cord.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Weight «mg» vertically downwards AND frictional force «Ff» vertically upwards, labelled and of equal length✓ 1Arrows must start on the dot.
Normal force «FN», horizontal and directed away from the glass, AND push of the atmosphere «pressure force», horizontal and directed towards the glass, labelled and of equal length✓ 1Do not award MP2 if the normal force is drawn towards the glass.
Part (b)
Ff = mg = 11.8 N and Ff ≤ μsFN, so FN ≥ 11.8/0.50 = 23.5 «N»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)
No acceleration perpendicular to the glass, so push of the atmosphere = FN = ΔP × A✓ 1
ΔP = 23.5/0.040 = 589 Pa✓ 1Accept 585–600 Pa «600 Pa from 24 N». Award [2] for CNA.
Part (d)
FN = 1.5 × 103 × 0.040 = 60 N✓ 1
Maximum friction = 0.25 × 60 = 15 N✓ 1
15 N > 11.8 N «the weight», so the robot remains at rest✓ 1MP3 needs a comparison. Allow ECF from MP1, MP2.
Part (e)
11.8 N, vertically upwards✓ 1MP1 only scores if MP2 scores.
The velocity is constant, so the resultant force is zero «Newton's first law»; friction is the only upward force, so it balances the weight✓ 1OWTTE.
Part (f)
Ff,max − mg = ma: 15 − 11.8 = 3.23 N✓ 1Allow ECF from (d).
a = 3.23/1.20 = 2.7 m s−2✓ 1Award [2] for CNA. Award [1] max for 12.5 m s−2 «weight ignored».
Part (g)
If the fan stops «power fails», the pressure difference and so the normal force fall to zero, so there is no friction to support the weight and the robot would fall✓ 1OWTTE.

Answers: (c) 590 Pa  ·  (d) maximum friction 15 N  ·  (e) 11.8 N  ·  (f) 2.7 m s−2 (the remaining parts are explanations — see the table above)

169A-2-73Variable mass & F = Δp/Δt Paper 2Medium13 marks

A firefighter holds the nozzle at the end of a fire hose. Water leaves the nozzle at a speed of 32 m s−1. The volume flow rate of the water is 390 litres per minute. The water moves slowly in the wide hose, so treat it as entering the nozzle with negligible speed. The mass of the firefighter, including her equipment, is 85 kg. Ignore the weight of the water in the nozzle and any force exerted by the hose.

Density of water = 1000 kg m−3; 1 litre = 1.0 × 10−3 m3.

The firefighter, a block standing on level ground, holds the nozzle; in (e) the jet leaves the nozzle at 35° above the horizontalfirefighterhosejet35°
Figure 1 — Diagram NOT accurately drawn
(a)

Show that the mass of water leaving the nozzle each second is about 6.5 kg.

(1)
(b)

At first the jet is horizontal. Calculate the magnitude of the force exerted on the water by the nozzle.

(2)
(c)

Explain, using Newton's laws of motion, why the firefighter must push forwards on the nozzle to keep it at rest.

(2)
(d)

The ground is covered with foam, and the coefficient of static friction between her boots and the ground is 0.22. Determine whether the firefighter slides backwards.

(2)
(e)

The firefighter now aims the jet upwards at 35° to the horizontal, as shown in Figure 1. The speed and the flow rate of the water are unchanged. State the direction of the force now exerted on the firefighter by the nozzle, and determine whether she slides backwards.

(4)
(f)
(i)

Calculate the rate at which kinetic energy is given to the water.

(1)
(ii)

Suggest why the pump must supply energy to the water at a greater rate than your answer to (f)(i).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
390 × 1.0 × 10−3 × 1000/60 = 6.50 «kg s−1»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)
F = Δp/Δt = (Δm/Δt)v✓ 1
F = 6.5 × 32 = 208 N✓ 1Award [2] for CNA. Accept 210 N.
Part (c)
By Newton's third law the water exerts an equal and opposite force «208 N, backwards» on the nozzle✓ 1
For the nozzle to stay at rest the resultant force on it must be zero «Newton's first law», so she must push forwards with 208 N✓ 1OWTTE.
Part (d)
Maximum friction = 0.22 × 85 × 9.81 = 183 N✓ 1
183 N < 208 N, so she slides backwards✓ 1MP2 needs a comparison. Allow ECF from (b).
Part (e)
Backwards and downwards, at 35° below the horizontal «opposite to the direction of the jet»✓ 1Accept a correctly described or drawn direction.
Horizontal component of the force = 208 cos 35° = 170 N✓ 1
Normal force = 85 × 9.81 + 208 sin 35° = 953 N✓ 1Award MP3 only if the downward component is added to the weight.
Maximum friction = 0.22 × 953 = 210 N > 170 N, so she does not slide✓ 1MP4 needs a comparison. Allow ECF from (b).
Part (f)(i)
½ × 6.5 × 32² = 3.3 × 103 W✓ 1Accept 3.3 kW.
Part (f)(ii)
Any one of: energy is dissipated by friction «viscous forces» in the hose; the pump is not 100 % efficient; the water may need to be raised to the height of the nozzle✓ 1OWTTE.

Answers: (b) 208 N  ·  (d) maximum friction 183 N  ·  (e) 170 N needed; maximum friction 210 N  ·  (f)(i) 3.3 kW (the remaining parts are explanations — see the table above)

170A-2-74Collisions & energy Paper 2Medium13 marks

At a fairground, dodgem A (a small electric bumper car) moves at 2.4 m s−1 along a straight line and hits dodgem B from behind. B is at rest. The mass of A with its driver is 260 kg and the mass of B with its driver is 200 kg. Each driver is held firmly in the seat. Friction and air resistance are negligible during the collision.

An accelerometer on B records the collision. Figure 1 shows how the force F exerted by A on B varies with time t. Forces in the direction in which A was moving are positive.

Force F exerted by A on B against time t: rises linearly from 0 to 2000 N in 0.08 s, stays at 2000 N until 0.20 s, then falls linearly to zero at 0.30 s. The axis extends to −2500 N00.050.10.150.20.250.30.350.4t / s-2500-2000-1500-1000-50005001000150020002500F / N
Figure 1 — graph drawn to scale
(a)

Determine the impulse exerted on B by A.

(2)
(b)

Calculate the speed of B immediately after the collision.

(1)
(c)

Sketch, on Figure 1, a graph to show how the force exerted by B on A varies with time during the collision.

(2)
(d)

Calculate the velocity of A immediately after the collision.

(2)
(e)

Determine whether the collision is elastic.

(3)
(f)

Outline what happens to the kinetic energy that is not conserved.

(1)
(g)

State and explain which dodgem, with its driver, has the greater acceleration during the collision.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Impulse = area under the F–t graph «½ × 0.08 × 2000 + 0.12 × 2000 + ½ × 0.10 × 2000»✓ 1
= 420 N s✓ 1Accept kg m s−1. Accept 410–430 N s. Award [2] for CNA.
Part (b)
v = 420/200 = 2.1 m s−1✓ 1Allow ECF from (a).
Part (c)
Graph below the time axis, starting at t = 0 and returning to zero at t = 0.30 s, the mirror image of the given graph✓ 1
Constant at −2000 N from 0.08 s to 0.20 s «equal magnitude to the given force at every instant»✓ 1Allow ±100 N by eye.
Part (d)
260 × 2.4 = 260 vA + 420 «OR impulse on A = −420 N s»✓ 1Allow ECF from (a).
vA = 0.78 m s−1 in the original direction of motion✓ 1Direction needed. Accept 0.78–0.79 m s−1.
Part (e)
Ek before = ½ × 260 × 2.4² = 749 J✓ 1
Ek after = ½ × 260 × 0.785² + ½ × 200 × 2.1² = 521 J✓ 1Allow ECF from (b) and (d).
Ek has decreased «by about 228 J», so the collision is not elastic «it is inelastic»✓ 1MP3 needs a comparison of the two values.
Part (f)
It is transferred to internal energy «thermal energy» of the rubber bumpers «and the cars», and to sound✓ 1Do not accept 'lost' or 'friction' alone.
Part (g)
B «and its driver»✓ 1MP1 only scores if MP2 scores.
The forces on A and B are equal in magnitude «Newton's third law» and B has the smaller mass, so a = F/m is larger for B✓ 1OWTTE.

Answers: (a) 420 N s  ·  (b) 2.1 m s−1  ·  (d) 0.78 m s−1  ·  (e) 749 J before, 521 J after (the remaining parts are explanations — see the table above)

171A-2-78Contact & field forces Paper 2Medium13 marks

In a steel stockyard, a crane lifts steel plates with an electromagnet that hangs from the crane cable (Figure 1). A horizontal steel plate of mass 1200 kg is held against the flat underside of the electromagnet only by the magnetic force that the electromagnet exerts on it. With the normal current in the electromagnet, this magnetic force is vertical and has a magnitude of 30.0 kN.

The mass of the electromagnet is 450 kg. The mass of the cable is negligible. Air resistance is negligible.

A crane cable holds an electromagnet, a wide flat block; a horizontal steel plate is held against the flat underside of the electromagnetcableelectromagnetsteel plate
Figure 1 — Diagram NOT accurately drawn
A dot representing the steel plate
Figure 2 — for part (a)
(a)

Draw and label, on the dot in Figure 2, the forces acting on the plate while it hangs at rest under the electromagnet. The relative lengths of your arrows should be approximately correct.

(2)
(b)

Show that the normal force exerted on the plate by the electromagnet is about 18 kN.

(1)
(c)

For each of the following forces, state the force that forms a Newton's third law pair with it. Give the type of force, the body that exerts it, the body it acts on, and its direction.

(i)

The magnetic force exerted on the plate by the electromagnet.

(1)
(ii)

The normal force exerted on the plate by the electromagnet.

(1)
(d)

The crane lifts the electromagnet and the plate with an upward acceleration of 1.5 m s−2. Calculate the tension in the cable.

(2)
(e)

Later, the plate is being lowered. As it approaches the ground it is moving downwards and slowing down at 2.0 m s−2. Deduce whether the normal force on the plate is then greater or smaller than in (b), and calculate it.

(2)
(f)

During a fault, the magnetic force on the plate falls to 13.0 kN while the crane is lifting with an upward acceleration of 1.5 m s−2. Determine whether the plate stays attached to the electromagnet.

(3)
(g)

Suggest why such electromagnets are designed so that the magnetic force is two to three times the weight of the heaviest plate to be lifted.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Magnetic force vertically upwards AND weight «mg» vertically downwards, from the dot, labelled✓ 1
Normal force «FN» vertically downwards, labelled, with the magnetic-force arrow equal in length to weight + normal force «normal force about 1.5 times the weight»✓ 1Do not award MP2 if the normal force is drawn upwards. Accept a normal-force arrow 1.2 to 1.8 times the weight arrow.
Part (b)
FN = 30.0 × 103 − 1200 × 9.81 = 1.82 × 104 «N»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)(i)
Magnetic force exerted on the electromagnet by the plate, vertically downwards✓ 1All of type, bodies and direction needed.
Part (c)(ii)
Normal «contact» force exerted on the electromagnet by the plate, vertically upwards✓ 1All of type, bodies and direction needed. Do not accept the weight of the plate.
Part (d)
T − (1200 + 450) × 9.81 = (1200 + 450) × 1.5✓ 1Accept T = (m1 + m2)(g + a) with correct substitution.
T = 1.87 × 104 N✓ 1Award [2] for CNA. Award [1] max for 1.62 × 104 N «acceleration ignored» or 1.36 × 104 N «electromagnet omitted».
Part (e)
The plate slows down while moving downwards, so its acceleration «and the resultant force» is upwards; Fm − mg − FN = ma with a = 2.0 m s−2 upwards✓ 1
FN = 30.0 × 103 − 11 772 − 1200 × 2.0 = 1.58 × 104 N, smaller than in (b)✓ 1MP2 only scores if the acceleration is taken as upwards. Award [1] max for 2.06 × 104 N «acceleration taken downwards».
Part (f)
To give the plate an upward acceleration of 1.5 m s−2 with FN ≥ 0 the magnetic force must be at least m(g + a)✓ 1
= 1200 × (9.81 + 1.5) = 1.36 × 104 N✓ 1
13.0 kN < 13.6 kN «the normal force would have to be negative», so the plate does not stay attached «it falls away»✓ 1MP3 needs a comparison. Do not award MP3 for a comparison with the weight only «13.0 kN > 11.8 kN». ALTERNATIVE: largest acceleration with the plate attached = (13.0 × 103 − 11 772)/1200 = 1.0 m s−2 [2] < 1.5 m s−2 [1].
Part (g)
The magnetic force must also provide the extra upward force needed when the load is accelerated upwards or slowed while moving down «and allow for a reduced force from a gap, rust or a fault»✓ 1OWTTE.

Answers: (b) 1.82 × 104 N  ·  (d) 1.87 × 104 N  ·  (e) 1.58 × 104 N, smaller  ·  (f) 13.6 kN needed; the plate falls away (the remaining parts are explanations — see the table above)

172A-2-80Drag & buoyancy Paper 2Medium14 marks

A salvage diver raises a steel anchor of mass 180 kg from the seabed using a lift bag: an open-bottomed bag that the diver fills with air from a cylinder. The bag is tied to the anchor by a short rope (Figure 1). The mass of the empty bag and the rope together is 4.0 kg and their volume is negligible. The mass of the air is negligible.

Density of steel = 7850 kg m−3; density of sea water = 1025 kg m−3. Ignore the drag force of the water.

An open-bottomed lift bag filled with air, tied by a short rope to a steel anchor resting on the seabedairlift bagropeanchorseabed
Figure 1 — Diagram NOT accurately drawn
A dot representing the anchor
Figure 2 — for part (b)
(a)

Show that the buoyancy force on the anchor is about 230 N.

(1)
(b)

Draw and label, on the dot in Figure 2, the forces acting on the anchor while it is raised at a constant speed. The relative lengths of your arrows should be approximately correct.

(2)
(c)

The diver first puts 0.12 m3 of air into the bag. Determine whether the anchor is lifted off the seabed.

(3)
(d)

The diver adds air until the bag holds 0.20 m3 of air, and the anchor starts to rise from rest.

(i)

Calculate the initial acceleration of the anchor.

(2)
(ii)

Calculate the tension in the rope as the anchor starts to rise.

(2)
(iii)

Outline why this tension is greater than the tension when the anchor rises at constant speed.

(1)
(e)

The anchor starts to rise where the pressure of the water is 400 kPa. The temperature of the air in the bag stays constant and no air escapes. Calculate the volume of the air in the bag when the pressure has fallen to 200 kPa, and hence explain why lift bags are fitted with a valve that releases air during the ascent.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
V = 180/7850 = 0.0229 m3; Fb = 1025 × 0.0229 × 9.81 = 231 «N»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)
Weight vertically downwards; tension AND buoyancy vertically upwards; all labelled✓ 1Do not award MP1 if any other force is drawn.
Tension + buoyancy equal in length to the weight, with the buoyancy arrow much shorter «about one eighth of the weight»✓ 1By eye.
Part (c)
Upward forces = 1025 × 0.12 × 9.81 + 231 = 1437 N✓ 1
Total weight = 184 × 9.81 = 1805 N✓ 1Award MP2 if the bag's 4.0 kg is omitted «1766 N».
1437 N < 1805 N, so the anchor is not lifted✓ 1ALTERNATIVE: minimum volume = 0.157 m3 [2] > 0.12 m3 [1]. MP3 needs a comparison.
Part (d)(i)
Resultant force = 1025 × 0.2229 × 9.81 − 1805 = 437 N✓ 1
a = 437/184 = 2.4 m s−2✓ 1Award [2] for CNA. Award [1] max for 2.6 m s−2 «bag and rope omitted».
Part (d)(ii)
For the anchor: T + 231 − 1766 = 180 × 2.37✓ 1ALTERNATIVE: for the bag: 2011 − T − 39.2 = 4.0 × 2.37.
T = 1962 N✓ 1Accept 1960 N. Allow ECF from (d)(i).
Part (d)(iii)
The anchor accelerates upwards, so the resultant force on it must be upwards; at constant speed the resultant is zero and T = 1535 N✓ 1OWTTE.
Part (e)
PV = constant: V = 0.20 × 400/200 = 0.40 m3✓ 1
As the bag rises the pressure decreases, so the air expands and the buoyancy force on the bag increases✓ 1
The resultant upward force and the acceleration increase, giving an uncontrolled «dangerously fast» ascent; releasing air keeps the volume and the buoyancy roughly constant✓ 1OWTTE.

Answers: (c) 1440 N upwards, 1810 N downwards  ·  (d)(i) 2.4 m s−2  ·  (d)(ii) 1960 N  ·  (e) 0.40 m3 (the remaining parts are explanations — see the table above)

173A-2-84Newton's laws & force pairs Paper 2Medium13 marks

During a spacewalk, an astronaut is joined to a small satellite by a light tether. The astronaut and the satellite are at rest relative to each other, 12.0 m apart (Figure 1). The mass of the astronaut, including her spacesuit, is 140 kg and the mass of the satellite is 420 kg.

The astronaut pulls the tether in, hand over hand, so that the tension in it is a constant 21 N. Ignore all other forces on the astronaut and the satellite.

The astronaut on the left and the satellite on the right, 12.0 m apart, joined by a straight tetherastronaut140 kgsatellite420 kgtether12.0 m
Figure 1 — Diagram NOT accurately drawn
(a)

State and explain how the force exerted by the tether on the satellite compares with the force exerted by the tether on the astronaut.

(2)
(b)

Calculate the magnitude of the acceleration of the astronaut and of the satellite.

(2)
(c)

Show that the astronaut moves 9.0 m before she reaches the satellite.

(2)
(d)

Calculate the time taken for the astronaut to reach the satellite.

(1)
(e)

Show that the total momentum of the astronaut and the satellite is zero just before they meet.

(2)
(f)

The astronaut grabs the satellite and holds on to it. State their velocity after this.

(1)
(g)

Outline the energy transfers from the start of the pull until the astronaut is holding the satellite.

(2)
(h)

Suggest why it would be safer for the astronaut to pull with a smaller tension.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Equal in magnitude «21 N» and opposite in direction «each towards the other»✓ 1MP1 only scores if MP2 scores.
The tether has negligible mass, so the resultant force on it must be zero and it pulls equally on both ends «the two forces are not a Newton's third law pair»✓ 1OWTTE. Accept an argument that the tension is the same throughout a light tether.
Part (b)
Astronaut: 21/140 = 0.15 m s−2✓ 1
Satellite: 21/420 = 0.050 m s−2✓ 1
Part (c)
They move for the same time from rest, so s = ½at² gives distances in the ratio of the accelerations, 3 : 1✓ 1
Astronaut's distance = 12.0 × 3/4 = 9.0 «m»✓ 1Must see the ratio OR an equivalent calculation. ALTERNATIVE: ½ × 0.15t² + ½ × 0.050t² = 12.0.
Part (d)
t = √(2 × 9.0/0.15) = 11 s✓ 1Accept 11.0 s.
Part (e)
vA = 0.15 × 10.95 = 1.64 m s−1 and vS = 0.050 × 10.95 = 0.548 m s−1✓ 1Allow ECF from (d).
pA = 140 × 1.64 = 230 kg m s−1 and pS = 420 × 0.548 = 230 kg m s−1 in opposite directions, so the total is zero✓ 1Must see both momenta and their opposite directions.
Part (f)
Zero «they are at rest»✓ 1
Part (g)
Chemical energy in her muscles is transferred, by the work she does, to kinetic energy of the astronaut and the satellite «252 J»✓ 1
When she grabs the satellite all of this kinetic energy is transferred to internal «thermal» energy «a perfectly inelastic collision»✓ 1OWTTE.
Part (h)
They would meet at lower speeds, so there is less kinetic energy to absorb and smaller forces when she grabs the satellite✓ 1OWTTE.

Answers: (b) 0.15 m s−2 and 0.050 m s−2  ·  (d) 11 s  ·  (e) 230 kg m s−1 each (the remaining parts are explanations — see the table above)

174A-2-85Friction & inclined planes Paper 2Medium13 marks

A snowboarder of mass 75 kg is pulled up a straight slope inclined at 14° to the horizontal by a tow lift. The tow rope makes an angle of 30° with the surface of the slope and lies in the vertical plane that contains the line of the slope (Figure 1). The snowboarder moves up the slope at a constant speed of 2.5 m s−1. The coefficient of dynamic friction between the snowboard and the snow is 0.10. Air resistance is negligible. Treat the snowboarder as a particle.

A slope rising to the right at 14° to the horizontal; a dot represents the snowboarder on the slope; the tow rope runs up the slope from the dot at 30° to the slope surface14°tow rope30°direction of motion
Figure 1 — Diagram NOT accurately drawn
(a)

Draw and label, on the dot in Figure 1, the forces acting on the snowboarder.

(3)
(b)

State the resultant force on the snowboarder.

(1)
(c)

Show that the normal force on the snowboarder is FN = mg cos 14° − T sin 30°, where T is the tension in the rope.

(1)
(d)

Determine T.

(3)
(e)

Calculate the frictional force on the snowboarder.

(1)
(f)

Calculate the rate at which the rope does work on the snowboarder.

(2)
(g)

A rider suggests that the tension would be smaller if the rope were parallel to the slope. Deduce whether this suggestion is correct.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Weight «mg» vertically downwards, labelled✓ 1
Normal force «FN» perpendicular to the slope, away from it, labelled✓ 1
Friction «Ff» parallel to the slope, down the slope, AND tension along the rope, both labelled✓ 1Award [2] max if any additional force «eg a separate 'force up the slope'» is drawn.
Part (b)
Zero «the velocity is constant»✓ 1
Part (c)
No acceleration perpendicular to the slope: FN + T sin 30° = mg cos 14°, which rearranges to the given expression✓ 1Must see the components perpendicular to the slope balanced.
Part (d)
Along the slope: T cos 30° = mg sin 14° + 0.10 FN✓ 1
T (cos 30° + 0.10 sin 30°) = mg (sin 14° + 0.10 cos 14°)✓ 1Accept a correct substitution of the expression from (c).
T = 272 N✓ 1Award [3] for CNA. Award [2] max for 288 N «FN taken as mg cos 14°» or 290 N «FN taken as mg».
Part (e)
Ff = 0.10 × (714 − 136) = 58 N✓ 1Allow ECF from (d).
Part (f)
Only the component of the tension along the velocity does work: P = T cos 30° × v✓ 1
P = 272 × cos 30° × 2.5 = 589 W✓ 1Award [1] max for 681 W «Tv». Allow ECF from (d).
Part (g)
With a parallel rope, FN = mg cos 14° and T = mg(sin 14° + 0.10 cos 14°) = 249 N✓ 1
249 N < 272 N, so the suggestion is correct «the reduction in friction with the angled rope is smaller than the loss of the tension component along the slope»✓ 1MP2 needs a comparison. Allow ECF from (d).

Answers: (d) 272 N  ·  (e) 58 N  ·  (f) 590 W  ·  (g) 249 N (the remaining parts are explanations — see the table above)

175A-2-86Free-body diagrams Paper 2Medium13 marks

A paraglider and its pilot have a total mass of 105 kg. In still air they glide at a constant velocity of 10.0 m s−1 along a straight path inclined at 8.0° below the horizontal. The air exerts two forces on the paraglider: a drag force D opposite to the velocity, and a lift force L perpendicular to the velocity. Treat the paraglider and pilot as a single particle. Figure 1 shows the direction of the velocity.

A dot representing the paraglider; a dashed line through it shows the flight path, descending to the right at 8.0° below the horizontal, with an arrow giving the direction of the velocity8.0°velocity
Figure 1 — Diagram NOT accurately drawn
(a)

State the resultant force on the paraglider, giving a reason.

(1)
(b)

Draw and label, on the dot in Figure 1, the forces acting on the paraglider. The relative lengths of your arrows should be approximately correct.

(3)
(c)

Calculate D and L.

(2)
(d)

State the magnitude and the direction of the total force exerted by the air on the paraglider.

(1)
(e)

Calculate the rate at which the paraglider loses height.

(1)
(f)

Show that the rate at which the paraglider loses gravitational potential energy is equal to the rate at which work is done against the drag force.

(2)
(g)

The pilot launches from a hilltop 650 m above a landing field that is 4.0 km away horizontally. Determine whether the paraglider can reach the field in still air.

(2)
(h)

Suggest one reason why the paraglider might not reach the field on a real day.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Zero, because the velocity is constant «Newton's first law»✓ 1
Part (b)
Weight vertically downwards, labelled✓ 1
Drag along the flight path, opposite to the velocity «backwards and 8° above the horizontal», labelled✓ 1
Lift perpendicular to the flight path «tilted 8° forwards from the vertical», labelled, about as long as the weight, with the drag arrow much shorter «about one seventh»✓ 1Award [2] max if any additional force is drawn.
Part (c)
D = mg sin 8.0° = 143 N✓ 1
L = mg cos 8.0° = 1020 N✓ 1Award [1] max if sin and cos are interchanged.
Part (d)
1030 N vertically upwards «equal and opposite to the weight»✓ 1Both needed.
Part (e)
10.0 × sin 8.0° = 1.39 m s−1✓ 1
Part (f)
Rate of loss of Ep = mg × 1.392 = 1434 W✓ 1
Rate of work against drag = Dv = 143.4 × 10.0 = 1434 W, the same✓ 1Must see both values.
Part (g)
Horizontal distance = 650/tan 8.0° = 4.6 km✓ 1Accept a ratio argument: 7.1 m forward for each metre of height.
4.6 km > 4.0 km, so yes✓ 1MP2 needs a comparison.
Part (h)
Any one of: a headwind; sinking air; turns needed on the way; a change of glide angle✓ 1OWTTE.

Answers: (c) D = 143 N, L = 1020 N  ·  (d) 1030 N  ·  (e) 1.39 m s−1  ·  (g) 4.6 km (the remaining parts are explanations — see the table above)

176IBP2S-003-04-12Momentum & impulse Paper 2Medium7 marks

A tennis ball of mass 0.058 kg is fired vertically down onto three different court surfaces. Each time it rebounds vertically. The table shows the data.

Surface | impact speed / m s−1 | rebound speed / m s−1 | contact time / ms
Grass | 6.0 | 3.0 | 8.0
Clay | 6.0 | 3.9 | 5.0
Hard court | 6.0 | 4.5 | 4.0

Neglect the weight of the ball during contact.

(a)

Calculate the magnitude of the impulse on the ball from the clay surface.

(2)
(b)

Determine which surface exerts the largest mean force on the ball.

(2)
(c)

Outline, using the data, whether these collisions are elastic.

(1)
(d)

Calculate the fraction of the ball's kinetic energy lost in the impact with grass.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δp = m(v + u), taking account of the reversal of direction, = 0.058 × (6.0 + 3.9)✓ 1
= 0.57 N s (0.574 N s)✓ 1
Part (b)
Use F = Δp/Δt for each surface, e.g. grass: 0.058 × 9.0/0.0080 = 65 N; clay: 115 N✓ 1
Hard court: 0.058 × 10.5/0.0040 = 150 N (152 N), which is the largest✓ 1
ECF from (a) for the clay value—
Part (c)
They are inelastic, because the rebound speed is less than the impact speed, so kinetic energy is lost (transferred to internal energy and sound)✓ 1
Part (d)
Ek is proportional to v2, so the fraction remaining = (3.0/6.0)2 = 0.25✓ 1
Fraction lost = 0.75✓ 1

Common error: When a ball rebounds, the change in velocity is the sum of the two speeds, not their difference. Using the difference makes the impulse far too small.

177IBP2S-003-04-14Friction & inclined planes Paper 2Medium7 marks

In a loading bay, a crate of mass 40 kg rests on a ramp inclined at 15° to the horizontal. A light cable runs from the crate, parallel to the ramp, over a frictionless pulley at the top of the ramp, to a hanging counterweight of mass 25 kg. The coefficient of dynamic friction between the crate and the ramp is 0.20. The counterweight is released, and the crate moves up the ramp. Take g = 9.8 m s−2.

crate 40 kg25 kg15°
Diagram not to scale. The counterweight hangs vertically.
(a)

Outline, in words, the forces acting on the crate as it moves up the ramp.

(2)
(b)

Calculate the friction force on the crate.

(2)
(c)

Determine the acceleration of the crate.

(2)
(d)

Calculate the tension in the cable.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Weight acting vertically downwards, and normal reaction acting perpendicular to the ramp surface✓ 1
Tension acting up the ramp, parallel to the cable, and friction acting down the ramp, opposing the motion✓ 1
Part (b)
N = 40 × 9.8 × cos 15° = 379 N✓ 1
F = 0.20 × 379 = 76 N✓ 1
Part (c)
Resultant force on the system = 25 × 9.8 − 40 × 9.8 × sin 15° − 76 = 245 − 101 − 76 = 68 N✓ 1
a = 68/65 = 1.04 m s−2 (accept 1.0 m s−2)✓ 1
Allow ECF from (b). Writing separate equations for the crate and the counterweight and solving them simultaneously gives the same result and earns full credit.—
Part (d)
T = 25 × (9.8 − 1.04) = 219 N (accept 219–220 N; the crate equation T = 40 × 1.04 + 101 + 76 gives the same result)✓ 1
Allow ECF from (c).—

Common error: Calculate the normal reaction on a slope using mg cos θ, not mg. Using the full weight overestimates the friction.

178IBP2S-003-04-16Collisions & energy Paper 2Medium6 marks

In a collision reconstruction, car A of mass 1200 kg ran into the back of car B of mass 1500 kg, which was stationary at traffic lights with its brakes applied. The cars locked together and skidded 4.5 m in a straight line before stopping. The coefficient of dynamic friction between the tyres and the road is 0.70. Assume all wheels skidded throughout.

BeforeA 1200 kgB 1500 kgstationary, brakes onAfterA and B locked togetherskids 4.5 m
Diagram not to scale
(a)

Calculate the deceleration of the wreckage.

(1)
(b)

Calculate the speed of the wreckage immediately after the collision.

(2)
(c)

Determine the speed of car A just before the collision.

(2)
(d)

State one assumption made in (c).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
a = μg = 0.70 × 9.8 = 6.9 m s−2 (6.86 m s−2)✓ 1
Part (b)
v2 = 2as✓ 1
v = √(2 × 6.86 × 4.5) = 7.9 m s−1✓ 1
ECF from (a)—
Part (c)
1200u = 2700 × 7.86✓ 1
u = 18 m s−1 (17.7 m s−1)✓ 1
ECF from (b)—
Part (d)
Momentum is conserved during the impact: the collision time is so short that the impulse from road friction is negligible✓ 1

Exam tip: Work backwards in stages. Kinematics gives the speed just after impact, then conservation of momentum gives the speed just before it.

179IBP2S-003-03-13Collisions & energy Paper 2Medium6 marks

An uncrewed cargo spacecraft of mass 7200 kg docks with an orbiting space station of mass 4.2×105 kg. Just before docking, the spacecraft approaches at 0.10 m s−1 relative to the station, along the line joining their centres. Treat the station as initially at rest in this frame. The two remain joined after docking.

(a)

State the condition required for the total momentum of the spacecraft–station system to be conserved during docking.

(1)
(b)

Calculate the change in velocity of the space station as a result of docking.

(2)
(c)

Determine the kinetic energy transferred to other forms during docking.

(2)
(d)

Outline where the energy transferred in (c) goes.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
There must be no resultant external force acting on the system, i.e. the system is isolated .✓ 1
Part (b)
Momentum conservation: 7200 × 0.10 = (7200 + 4.2×105)v .✓ 1
v = 1.7×10−3 m s−1 in the original direction of motion of the spacecraft .✓ 1
Part (c)
Initial Ek = ½ × 7200 × 0.102 = 36 J and final Ek = ½ × 4.272×105 × (1.69×10−3)2 ≈ 0.61 J .✓ 1
Energy transferred = 36 − 0.61 ≈ 35 J (accept 35.4 J; allow ECF from (b)) .✓ 1
Part (d)
It becomes internal (thermal) energy, sound or vibration, or work done deforming the docking mechanism, shock absorbers or latches .✓ 1

Exam tip: In a perfectly inelastic collision where a light body hits a much heavier one at rest, almost all of the initial kinetic energy is transferred. A result close to the initial 36 J is a quick check that your answer is reasonable.

180IBP2S-003-03-14Newton's laws & force pairs Paper 2Medium7 marks

A sprinter of mass 68 kg starts a race from rest using starting blocks. She is in contact with the blocks and the track for 0.32 s. She leaves them with a horizontal velocity of 3.4 m s−1. During the push, the vertical acceleration of her centre of mass may be taken as negligible.

(a)

Calculate the average horizontal force exerted on the sprinter during the push.

(2)
(b)

Explain why the average vertical component of the contact force on the sprinter is approximately equal to her weight.

(2)
(c)

Determine the magnitude and direction of the average total contact force exerted on the sprinter by the blocks and track.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δp = 68 × 3.4 = 231 kg m s−1 . F = 231/0.32 = 720 N (722.5 N) .✓ 2
Part (b)
The vertical acceleration is negligible, so the resultant vertical force is approximately zero (Newton's first/second law) . The only other vertical force is her weight, so the upward contact force ≈ mg = 670 N (666 N) .✓ 2
Part (c)
Combines the components: F = √(722.5² + 666.4²) . F ≈ 980 N (983 N) . Direction: tan−1(666.4/722.5) ≈ 43° above the horizontal, in the forward direction. ECF from (a) and (b) .✓ 3

Exam tip: A contact force from the ground usually has two components: friction parallel to the surface and a normal force perpendicular to it. Treat them separately, then combine them using vectors.

181IBP2S-003-03-15Collisions & energy Paper 2Medium7 marks

A freight wagon of mass 2.0×104 kg rolls at 1.5 m s−1 into the spring buffers at the end of a siding. The buffers behave as a single ideal spring of spring constant 8.0×105 N m−1. Friction is negligible.

wagon 2.0×10⁴ kg1.5 m s⁻¹buffers
Buffers modelled as a single spring; diagram not to scale
(a)

Calculate the momentum of the wagon before it reaches the buffers.

(1)
(b)

Determine the maximum compression of the buffers.

(2)
(c)

Calculate the maximum deceleration of the wagon.

(2)
(d)

The wagon rebounds from the buffers at 1.5 m s−1. Determine the magnitude of the total impulse exerted on the wagon by the buffers.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
p = mv = 2.0×104 × 1.5 = 3.0×104 kg m s−1 .✓ 1
Part (b)
Use energy conservation: ½mv² = ½kx², so x = v√(m/k) .✓ 1
x = 1.5 × √(2.0×104 / 8.0×105) = 0.24 m .✓ 1
Part (c)
Maximum force: Fmax = kx = 8.0×105 × 0.237 = 1.9×105 N .✓ 1
Then a = F/m = 9.5 m s−2 .✓ 1
Accept 9.5–9.6 m s−2; using the rounded value 0.24 m gives 9.6 m s−2. Allow error carried forward from (b).—
Part (d)
Take the rebound direction as positive, so the initial velocity is −1.5 m s−1.—
Impulse = Δp = m(1.5 − (−1.5)), or 2 × 3.0×104 .✓ 1
Magnitude of impulse = 6.0×104 N s .✓ 1
Allow error carried forward from (a).—

Common error: Using only the stopping phase gives an impulse of 3.0×10⁴ N s. The impulse over the whole contact also includes the rebound, so the momentum change is doubled.

182IBP2S-003-00-12Angular velocity & period Paper 2Medium8 marks

A medical centrifuge separates blood cells from plasma. A sample tube is held 9.0 cm from the axis of rotation and moves in a horizontal circle. The centrifuge is set so that the tube's centripetal acceleration is 1500 times the gravitational field strength, where g = 9.8 m s−2. Gravitational effects on the tube can be ignored.

(a)

Calculate the angular speed of the rotor.

(2)
(b)

Determine the rotation rate of the rotor in revolutions per minute.

(2)
(c)

The tube and its contents have a total mass of 4.0 g. Calculate the magnitude of the resultant force on the tube.

(2)
(d)

Explain the direction of the force that the holder exerts on the tube.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
a = 1500 × 9.8 = 1.47×104 m s−2 .✓ 1
ω = √(a/r) = √(1.47×104/0.090) ≈ 4.0×102 rad s−1 (404 rad s−1) .✓ 1
Part (b)
f = ω/2π = 404/2π ≈ 64.3 s−1 .✓ 1
Rate = 64.3 × 60 ≈ 3.9×103 revolutions per minute . ECF from (a).✓ 1
Part (c)
F = ma = 0.0040 × 1.47×104 .✓ 1
F ≈ 59 N . ECF from the value of a used in (a).✓ 1
Part (d)
The tube moves in a circle, so a resultant force directed towards the centre (the axis) is needed .✓ 1
The holder pushes the tube inward. No outward force acts on the tube: the tube's tendency to continue along a tangent presses it against the holder, so by Newton's third law the tube pushes outward on the holder .✓ 1

Common error: Do not draw or describe a 'centrifugal force' acting on the tube. The only horizontal force on the tube is the inward push of the holder.

183IBP2S-003-00-14Circular motion Paper 2Medium8 marks

In the hammer throw, an athlete swings a hammer of mass 7.26 kg on a wire. Just before release, the hammer head moves in an approximately horizontal circle of radius 1.8 m at a speed of 28 m s−1. The mass of the wire can be ignored.

(a)

Calculate the centripetal force on the hammer just before release.

(2)
(b)

Explain why, when released, the hammer moves off along a tangent to the circle.

(2)
(c)

Outline why the wire cannot be exactly horizontal while the hammer moves in a horizontal circle.

(2)
(d)

Estimate the angle between the wire and the horizontal.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
F = mv2/r = 7.26 × 282/1.8 .✓ 1
F ≈ 3.2×103 N .✓ 1
Part (b)
Once the wire is released, there is no longer a horizontal force on the hammer .✓ 1
By Newton's first law, it continues with the velocity it had at the moment of release, and that velocity is tangential to the circle .✓ 1
Part (c)
The weight of the hammer acts vertically downwards and must be balanced, because there is no vertical acceleration .✓ 1
Only the tension can supply an upward component, so the wire must slope downwards from the athlete's hands to the hammer head .✓ 1
Part (d)
tan θ = mg/(mv2/r) = gr/v2 = 9.8 × 1.8/282 ≈ 0.023 .✓ 1
θ ≈ 1.3° . ECF from (a).✓ 1

Exam tip: When a tension has two roles, resolve it into components. The horizontal component provides the centripetal force and the vertical component balances the weight.

184IBP2S-003-00-16Hooke's law & springs Paper 2Medium7 marks

A simple accelerometer for use in a lift consists of a mass of 0.040 kg hanging from a light spring of spring constant 25 N m−1. The spring is attached to the ceiling of the lift.

(a)

Calculate the extension of the spring when the lift is at rest.

(2)
(b)

At one moment the extension of the spring is 1.90 cm and the mass is at rest relative to the lift. Determine the magnitude and direction of the acceleration of the lift.

(3)
(c)

The lift cable breaks and the lift falls freely. The mass is at rest relative to the lift. State and explain the extension of the spring.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
At equilibrium, kx = mg .✓ 1
x = 0.040 × 9.8 / 25 = 1.6×10−2 m (1.6 cm) .✓ 1
Part (b)
Spring force = 25 × 0.0190 = 0.475 N .✓ 1
Resultant force = 0.475 − 0.392 = 0.083 N upwards .✓ 1
a = 0.083 / 0.040 ≈ 2.1 m s−2, directed upwards .✓ 1
Part (c)
The extension is zero, so the spring is at its natural length .✓ 1
The mass and the lift both accelerate at g, so the weight alone provides the acceleration of the mass and no spring force is needed .✓ 1

Common error: An upward acceleration does not tell you that the lift is moving upwards. It could be moving down and slowing. State the direction of the acceleration, not the direction of the motion.

185IBP2S-003-05-13Variable mass & F = Δp/Δt Paper 2Medium7 marks

A deep-space probe of mass 540 kg is propelled by an ion thruster. The thruster expels xenon ions backward at 30 km s−1 relative to the probe and produces a constant thrust of 0.090 N. Gravitational forces on the probe may be neglected.

(a)

Explain, with reference to Newton's third law, how the thruster produces a forward force on the probe.

(2)
(b)

Calculate the acceleration of the probe.

(1)
(c)

Determine the time needed for the probe to increase its speed by 1.0 km s−1. Assume the mass of the probe stays constant.

(2)
(d)

Calculate the mass of xenon expelled per second.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The thruster exerts a backward force on the ions to accelerate them out of the probe✓ 1
By Newton's third law, the ions exert an equal and opposite (forward) force on the thruster and therefore on the probe✓ 1
Part (b)
a = F/m = 0.090/540 = 1.7 × 10−4 m s−2✓ 1
Part (c)
t = Δv/a = 1000/(1.67 × 10−4), allowing error carried forward from the acceleration calculated above✓ 1
t = 6.0 × 106 s, about 69 days; accept 5.9 × 106 s if the rounded acceleration 1.7 × 10−4 m s−2 is used✓ 1
Part (d)
F = (Δm/Δt)v, so Δm/Δt = F/v✓ 1
Δm/Δt = 0.090/(3.0 × 104) = 3.0 × 10−6 kg s−1✓ 1

Exam tip: Convert km s⁻¹ to m s⁻¹ before substituting. A missed factor of 1000 is a common source of lost marks in thrust calculations.

186IBP2S-003-05-14Connected bodies Paper 2Medium6 marks

At an airport, a baggage tug pulls two carts in a line along a level apron. The front cart A has mass 400 kg and the rear cart B has mass 300 kg. The coupling between the tug and cart A has a tension of 1050 N. Rolling resistance on cart A is 60 N and on cart B is 45 N. The couplings are horizontal and have negligible mass.

tugcart A 400 kgcart B 300 kg1050 N60 N45 N
Tension in the tug coupling and rolling resistances shown
(a)

Describe the horizontal forces acting on cart B.

(2)
(b)

Calculate the acceleration of the carts.

(2)
(c)

Determine the tension in the coupling between cart A and cart B.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A forward tension from the coupling between A and B✓ 1
A backward rolling resistance of 45 N✓ 1
Part (b)
Resultant force on A and B together = 1050 − 60 − 45 = 945 N✓ 1
a = 945/700 = 1.35 m s−2 (accept 1.4 m s−2)✓ 1
Part (c)
For B: T − 45 = 300 × 1.35, or for A: 1050 − 60 − T = 400 × 1.35✓ 1
T = 450 N (allow error carried forward from the acceleration calculated earlier)✓ 1

Exam tip: Find the acceleration by treating the carts as one system, then isolate the rear cart to find the internal tension. The rear cart has fewer forces on it, which makes the algebra simpler.

187IBP2S-003-02-13Explosions & recoil Paper 2Medium7 marks

A spacecraft is at rest relative to a nearby space station. The spacecraft has total mass 1200 kg, which includes a 50 kg CubeSat. A compressed spring releases the CubeSat, which leaves at 1.8 m s−1 relative to the station. The spring is in contact with the CubeSat for 0.15 s.

(a)

State the principle used to find the velocity of the spacecraft after the release.

(1)
(b)

Calculate the recoil speed of the spacecraft after the release.

(2)
(c)

Determine the elastic potential energy that was stored in the spring.

(2)
(d)

Calculate the average force exerted by the spring on the CubeSat.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Conservation of linear momentum: the total momentum of the system stays constant (here zero) when no external force acts .✓ 1
Part (b)
0 = 50 × 1.8 − 1150v, using the remaining mass of 1150 kg . v = 0.078 m s−1 .✓ 2
Part (c)
E = ½ × 50 × 1.82 + ½ × 1150 × 0.0782 = 81 + 3.5 . E = 85 J . ECF from (b).✓ 2
Part (d)
F = Δp/Δt = (50 × 1.8)/0.15 . F = 600 N .✓ 2

Common error: After the release, the spacecraft's mass is 1200 − 50 = 1150 kg. Using 1200 kg gives a slightly wrong recoil speed.

188IBP2S-003-02-14Friction & inclined planes Paper 2Medium9 marks

A 40 kg equipment crate is pushed up a loading ramp inclined at 25° to the horizontal. The push is parallel to the ramp. The coefficient of kinetic friction between the crate and the ramp is 0.30 and the coefficient of static friction is 0.45.

crate 40 kg25°push
(a)

The crate moves up the ramp at constant speed. Describe the free-body diagram for the crate, stating the direction of each force.

(2)
(b)

Calculate the magnitude of the push.

(2)
(c)

The push is removed while the crate is moving up the ramp at 1.5 m s−1. Determine the further distance the crate travels up the ramp.

(3)
(d)

Determine whether the crate remains at rest after it stops.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Weight acts vertically downwards, and the normal force acts perpendicular to the ramp, away from the surface . The push acts parallel to the ramp, up the slope, and kinetic friction acts parallel to the ramp, down the slope .✓ 2
Part (b)
F = mg sin 25° + μkmg cos 25° . F = 392 × (0.423 + 0.30 × 0.906) ≈ 270 N (accept 272 N) .✓ 2
Part (c)
Friction still acts down the slope while the crate moves up, so the deceleration is a = g(sin 25° + 0.30 cos 25°) . a ≈ 6.8 m s−2 . s = u2/2a = 2.25/13.6 ≈ 0.17 m (accept 0.165 m) .✓ 3
Part (d)
Compare tan 25° = 0.47 with μs = 0.45. Alternatively, compare the component of weight down the slope, mg sin 25° ≈ 166 N, with the maximum static friction, μsmg cos 25° ≈ 160 N . Since tan 25° > μs (or 166 N > 160 N), static friction cannot hold the crate, so it slides back down the ramp .✓ 2

Exam tip: An object that stops on a slope does not necessarily stay there. To decide, compare μs with tan θ, or compare mg sin θ with the maximum static friction.

189IBP2S-003-06-13Variable mass & F = Δp/Δt Paper 2Medium7 marks

A student investigates the force exerted by a water jet. Water leaves a narrow tube vertically downwards at speed v. It strikes a flat plate resting on a top-pan balance just below the end of the tube. The water spreads out horizontally across the plate. The plate is so close to the tube that any change in the speed or cross-section of the jet before it reaches the plate is negligible. The tube has a constant internal cross-sectional area A. The student changes v and records the extra force F on the balance, after allowing for the water collecting on the plate. The density of water is 1000 kg m−3.

v / m s−1: 2.0 | 3.0 | 4.0 | 5.0
F / N: 0.079 | 0.18 | 0.33 | 0.50

tube, area Avplatetop-pan balance
(a)

Show that F = ρAv2, where ρ is the density of water.

(2)
(b)

Outline why a graph of F against v2 is expected to be a straight line through the origin.

(1)
(c)

Use all the data to determine A.

(2)
(d)

In a repeat run, some water splashes upwards off the plate. Suggest how this affects the calculated value of A.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The mass of water hitting the plate per second is ρAv✓ 1
All of the water's vertical velocity v is lost, so F = (Δm/Δt)v = (ρAv)v = ρAv2✓ 1
Part (b)
ρ and A are constant, so F is directly proportional to v2, with gradient ρA✓ 1
Part (c)
Gradient of the F against v2 line (or the mean of F/v2) = ρA ≈ 0.020 N s2 m−2✓ 1
A = 0.020/1000 = 2.0 × 10−5 m2 (accept 1.9–2.1 × 10−5 m2)✓ 1
Part (d)
The splashing water leaves with an upward velocity component, so its change in momentum per second, and hence F, is larger for the same v✓ 1
The gradient is therefore larger, and A is overestimated✓ 1

Exam tip: When you test a derived relationship, use the gradient of a best-fit line (or several data points) rather than a single reading. This reduces the effect of random error.

190IBP2S-003-06-15Variable mass & F = Δp/Δt Paper 2Medium7 marks

In a grain store, wheat falls vertically from a hopper onto a horizontal conveyor belt at a rate of 25 kg s−1. The belt moves at a constant speed of 1.6 m s−1. Ignore friction in the belt's rollers.

1.6 m s⁻¹hopperwheat 25 kg s⁻¹
(a)

Explain why a horizontal force must act on the belt to keep it moving at constant speed.

(2)
(b)

Calculate the horizontal force that the motor must supply.

(1)
(c)

Calculate the power that the motor must supply to provide this force.

(1)
(d)

Calculate the rate at which the wheat gains kinetic energy.

(1)
(e)

Explain why the power supplied by the motor differs from the rate at which the wheat gains kinetic energy.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The wheat lands with zero horizontal velocity, so the belt must exert a forward force on it to give it horizontal momentum✓ 1
By Newton's third law, the wheat exerts an equal backward force on the belt, so the motor must supply an equal forward force to keep the belt's velocity constant✓ 1
Part (b)
F = (Δm/Δt)v = 25 × 1.6 = 40 N✓ 1
Part (c)
P = Fv = 40 × 1.6 = 64 W✓ 1
Allow a follow-through from the force calculated previously.—
Part (d)
Rate = ½(Δm/Δt)v2 = ½ × 25 × 1.62 = 32 W✓ 1
Part (e)
The remaining 32 W (half of the motor's power) is transferred as thermal energy (and sound)✓ 1
This happens because the wheat slides relative to the belt, with kinetic friction acting, until it reaches the belt's speed✓ 1

Common error: Assuming that all of the motor's power becomes kinetic energy of the load. When material is brought up to speed by sliding friction, exactly half of the work done is dissipated.

191IBP2S-003-01-13Drag & buoyancy Paper 2Medium7 marks

A quality-control laboratory checks the viscosity of glycerol by timing steel ball bearings falling at terminal speed through a tall cylinder of it. Density of steel = 7800 kg m−3. Density of glycerol = 1260 kg m−3. Stokes's law applies.

Radius r / mm: 1.0, 1.5, 2.0, 2.5
Terminal speed v / mm s−1: 10.2, 22.9, 40.7, 63.6

(a)

Outline why each ball bearing reaches a terminal speed.

(2)
(b)

Show that the data are consistent with v being proportional to r2.

(2)
(c)

Determine the viscosity of the glycerol.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
As the speed increases, the viscous drag force increases .✓ 1
Terminal speed is reached when drag + buoyant force (upthrust) = weight, so the resultant force, and hence the acceleration, is zero .✓ 1
Part (b)
Calculates v/r2 for at least three data points, e.g. 10.2, 10.2, 10.2, 10.2 mm−1 s−1 .✓ 1
States that the ratio is constant within the precision of the data, so v ∝ r2 .✓ 1
Part (c)
Uses 6πηrv = (4/3)πr3(ρs − ρf)g, leading to η = 2(ρs − ρf)g r2/(9v) .✓ 1
Uses v/r2 = 1.02×104 m−1 s−1, or any single data pair, correctly in SI units .✓ 1
η = 2 × 6540 × 9.8/(9 × 1.02×104) ≈ 1.4 Pa s .✓ 1
Allow error carried forward from the ratio found in the previous part.—

Exam tip: Convert v/r² to SI units before substituting. A value in mm s⁻¹ ÷ mm² must be multiplied by 10³ to give m⁻¹ s⁻¹.

192IBP2S-003-01-14Contact & field forces Paper 2Medium8 marks

A launch vehicle lifts off vertically from rest. At lift-off its mass is 5.4×105 kg and the engines provide a constant thrust of 7.6×106 N. Assume g = 9.8 m s−2 throughout the ascent.

(a)

Describe, in words, a free-body diagram for the vehicle at the instant of lift-off.

(2)
(b)

Calculate the acceleration of the vehicle at lift-off.

(2)
(c)

Later in the ascent, the mass of the vehicle has fallen to 3.0×105 kg and air resistance is 6.0×105 N. The thrust is unchanged. Determine the acceleration at this time.

(2)
(d)

Explain why the acceleration later in the ascent is greater than the acceleration at lift-off, even though air resistance now acts.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Thrust acting vertically upward and weight acting vertically downward, both drawn from the vehicle .✓ 1
Thrust arrow longer than weight arrow because the resultant force is upward; no drag arrow because v = 0 at lift-off .✓ 1
Part (b)
W = 5.4×105 × 9.8 = 5.29×106 N .✓ 1
a = (7.6×106 − 5.29×106)/(5.4×105) = 4.3 m s−2 upward .✓ 1
Part (c)
W = 3.0×105 × 9.8 = 2.94×106 N, so resultant = 7.6×106 − 2.94×106 − 6.0×105 = 4.06×106 N .✓ 1
a = 4.06×106/(3.0×105) = 13.5 ≈ 14 m s−2 upward .✓ 1
Part (d)
The weight has fallen by about 2.4×106 N, much more than the 6.0×105 N of drag added, so the resultant force is larger (4.06×106 N compared with 2.31×106 N) .✓ 1
The mass is also smaller, and a = F/m, so a smaller mass gives a larger acceleration for a given resultant force; both effects increase a .✓ 1

Common error: When the mass changes, the weight must be recalculated. Do not reuse the lift-off weight later in the flight.

193IBP2S-003-01-15Drag & buoyancy Paper 2Medium8 marks

A research balloon is filled with 12 m3 of helium of density 0.18 kg m−3. The envelope and instrument payload have a combined mass of 8.0 kg. The density of the surrounding air is 1.20 kg m−3. Assume the balloon volume and the air density stay constant from release until the balloon reaches terminal speed. Ignore the volume of the payload. The air resistance on the rising balloon is given by F = kv2, where k = 2.6 kg m−1.

(a)

Calculate the buoyant force on the balloon.

(2)
(b)

Show that the resultant force on the balloon immediately after release is about 42 N.

(2)
(c)

Calculate the initial acceleration of the balloon.

(2)
(d)

Determine the terminal speed of the balloon.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Fb = ρairVg .✓ 1
Fb = 1.20 × 12 × 9.8 = 141 N .✓ 1
Part (b)
Total weight = (0.18 × 12 + 8.0) × 9.8 = 10.16 × 9.8 = 99.6 N .✓ 1
Resultant = 141.1 − 99.6 = 41.5 N ≈ 42 N, upward .✓ 1
Part (c)
Total mass = 2.16 + 8.0 = 10.16 kg .✓ 1
a = 41.5 / 10.16 = 4.1 m s−2 .✓ 1
Part (d)
At terminal speed the drag equals the resultant of buoyancy and weight: kv2 = 41.5 N .✓ 1
v = √(41.5 / 2.6) = 4.0 m s−1 .✓ 1
Allow error carried forward from part 2.—

Common error: Include the mass of the helium in both the weight and the total mass. Even a light gas has a noticeable mass at a volume of 12 m³.

194A-2-07Connected bodies Paper 2Hard13 marks

A wooden block of mass 2.0 kg rests on a horizontal bench. A light, inextensible string attached to the block passes over a light, frictionless pulley at the edge of the bench and supports a hanging mass (Figure 1). The coefficient of static friction between the block and the bench is 0.40 and the coefficient of dynamic friction is 0.30.

A block on a horizontal bench joined by a string over a pulley at the edge of the bench to a hanging massblockhanging mass0.50 m0.90 m
Figure 1 — Diagram NOT accurately drawn
A dot representing the block, with the direction of motion to the rightblockdirection of motion
Figure 2 — for (b)(i)
(a)

A hanging mass of 0.60 kg is attached and the system is released from rest.

(i)

Determine whether the block moves.

(2)
(ii)

State the magnitude of the frictional force on the block.

(1)
(b)

The hanging mass is replaced by one of mass 1.2 kg, and the block slides.

(i)

Draw and label, on Figure 2, the forces acting on the block while it slides.

(2)
(ii)

Show that the acceleration of the system is about 1.8 m s−2.

(2)
(iii)

Calculate the tension in the string.

(2)
(iv)

Outline why the tension is less than the weight of the hanging mass.

(1)
(c)

The 1.2 kg mass is released from rest 0.50 m above the floor, with the block 0.90 m from the pulley. When the mass hits the floor it stops and the string becomes slack. Determine whether the block reaches the pulley.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
maximum static friction = 0.40 × 2.0 × 9.81 = 7.85 N✓ 1
weight of hanging mass = 5.89 N < 7.85 N so the block does not move✓ 1The decision must be supported by the comparison.
Part (a)(ii)
5.9 N «equal to the tension»✓ 1Do not accept 7.85 N.
Part (b)(i)
weight downwards and normal force upwards, of equal length, both labelled✓ 1
tension towards the pulley «in the direction of motion» AND friction opposite to the motion, shorter than the tension, both labelled✓ 1Award [1 max] if any extra force is drawn.
Part (b)(ii)
resultant force on the system = 1.2 × 9.81 − 0.30 × 2.0 × 9.81 = 5.89 N✓ 1ALTERNATIVE: two equations, m2g − T = m2a and T − μdm1g = m1a, solved together.
a = 5.89 / 3.2 «= 1.84 m s−2»✓ 1Must see full substitution OR answer to 3 s.f. Award [0] for 3.68 m s−2, which ignores friction.
Part (b)(iii)
T = 1.2(9.81 − 1.84) OR T = 2.0 × 1.84 + 5.89✓ 1
T = 9.56 N✓ 1Allow ECF from (b)(ii). Accept 9.5–9.6 N.
Part (b)(iv)
the hanging mass accelerates downwards, so the resultant force on it is downwards «its weight is greater than the tension»✓ 1
Part (c)
speed when the mass lands: v² = 2 × 1.84 × 0.50 «= 1.84 m² s−2»✓ 1
deceleration = 0.30 × 9.81 = 2.94 m s−2 so the further slide = 1.84 / (2 × 2.94) = 0.31 m✓ 1
total distance = 0.81 m < 0.90 m so the block stops «0.09 m» before the pulley✓ 1Allow ECF from (b)(ii). The decision must follow from the candidate’s numbers.

Answers: (a)(i) 7.85 N > 5.89 N; does not move  ·  (a)(ii) 5.9 N  ·  (b)(ii) 1.84 m s−2  ·  (b)(iii) 9.56 N  ·  (c) 0.81 m; does not reach the pulley (the remaining parts are explanations — see the table above)

195A-2-08Circular motion Paper 2Hard15 marks

A design for a space habitat is a ring of radius 110 m that rotates about its central axis. Astronauts stand on the inside surface of the outer wall of the ring (Figure 1). The ring makes one rotation every 30 s.

The habitat is in deep space, where the gravitational forces on the habitat and on everything inside it are negligible.

A rotating ring-shaped space habitat with an astronaut standing on the inside of the outer wall, head towards the axisaxis110 mastronautrotation
Figure 1 — Diagram NOT accurately drawn
(a)
(i)

Calculate the angular velocity of the ring.

(1)
(ii)

Show that the centripetal acceleration of an astronaut standing on the floor is about 4.8 m s−2.

(1)
(iii)

State and explain which force provides the centripetal force on the astronaut.

(2)
(iv)

Calculate the magnitude of this force for an astronaut of mass 72 kg.

(1)
(b)

The astronaut’s head is 1.8 m closer to the axis than her feet. Determine the percentage by which the centripetal acceleration at her head is smaller than at her feet.

(2)
(c)

The designers would like an acceleration of 9.81 m s−2 at the floor, but astronauts become unwell if the rotation rate is more than 2.0 revolutions per minute. Determine whether a ring of radius 110 m can meet both requirements.

(3)
(d)

An astronaut brings two timers from the Earth: a simple pendulum of length 0.50 m and a mass–spring oscillator. On the Earth’s surface each timer has a period of 1.42 s. In the habitat both timers are set up close to the floor of the ring.

(i)

Calculate the period of the pendulum in the habitat.

(2)
(ii)

State and explain the period of the mass–spring oscillator in the habitat.

(2)
(iii)

Outline why the period of the pendulum would be longer if it were set up on a platform closer to the axis.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
ω = 2π / 30 = 0.209 rad s−1✓ 1
Part (a)(ii)
a = ω²r = 0.209² × 110 «= 4.83 m s−2»✓ 1Must see full substitution OR answer to 3 s.f. Accept 4π²r/T².
Part (a)(iii)
the normal «contact» force from the floor on the astronaut✓ 1
directed towards the axis «centre of the ring»✓ 1Do not accept “centrifugal force”.
Part (a)(iv)
F = 72 × 4.83 = 347 N✓ 1Allow ECF from (a)(ii). Accept 346–349 N.
Part (b)
a ∝ r at constant ω «a = ω²r»✓ 1
1.8 / 110 × 100 = 1.6 %✓ 1Award [2] for CNA.
Part (c)
ω needed = √(9.81 / 110) = 0.299 rad s−1✓ 1
= 0.299 / 2π × 60 = 2.85 revolutions per minute✓ 1
2.85 > 2.0 so the ring cannot meet both requirements «it would need a radius of at least 224 m»✓ 1ALTERNATIVE: 2.0 rev min−1 is one rotation every 30 s, the present rate, which gives only 4.8 m s−2 [2]; so 9.81 m s−2 needs a faster rotation and the requirements cannot both be met [1].
Part (d)(i)
the acceleration of the floor replaces g: T = 2π√(l/a) with a = 4.83 m s−2✓ 1Allow ECF from (a)(ii).
T = 2π√(0.50/4.83) = 2.02 s✓ 1Accept 2.0–2.1 s. ALTERNATIVE: T = 1.42 × √(9.81/4.83). Award [2] for CNA. Award [0] for 1.42 s.
Part (d)(ii)
1.42 s «unchanged»✓ 1
T = 2π√(m/k) depends only on the mass and the spring constant, not on g «the acceleration of the floor changes only the equilibrium extension of the spring»✓ 1MP2 only scores if MP1 scores.
Part (d)(iii)
the acceleration a = ω²r is smaller at a smaller radius, and T = 2π√(l/a) so T increases as a decreases✓ 1OWTTE.

Answers: (a)(i) 0.209 rad s−1  ·  (a)(ii) 4.83 m s−2  ·  (a)(iv) 347 N  ·  (b) 1.6 %  ·  (c) 2.85 rev min−1 needed; not possible  ·  (d)(i) 2.02 s  ·  (d)(ii) 1.42 s (the remaining parts are explanations — see the table above)

196A-2-18Connected bodies Paper 2Hard15 marks

A car of mass 1200 kg tows a trailer of mass 600 kg along a straight, level road. They are joined by a rigid, horizontal tow-bar (Figure 1). The resistive forces on the car and on the trailer are 400 N and 200 N respectively and may be taken as constant. These resistive forces do not include braking forces.

A car towing a trailer with a rigid tow-bar; a dot marks the centre of the trailer; motion to the righttrailertow-barcardirection of motion
Figure 1 — Diagram NOT accurately drawn
(a)

The car and trailer accelerate at 1.5 m s−2.

(i)

Calculate the forward driving force on the car.

(2)
(ii)

Calculate the force exerted by the tow-bar on the trailer.

(2)
(b)

The driver brakes. Only the car’s brakes act, and the car and trailer decelerate at 3.0 m s−2.

(i)

Draw and label, on Figure 1, the horizontal forces acting on the trailer during braking.

(2)
(ii)

Determine the magnitude of the force in the tow-bar and state whether the tow-bar is in tension or in compression.

(3)
(iii)

Calculate the braking force on the car.

(2)
(c)

Heavy trailers must be fitted with their own brakes. Discuss, with reference to the forces on the car and on the trailer, why this is safer.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
F − (400 + 200) = (1200 + 600) × 1.5✓ 1
F = 3300 N✓ 1Award [2] for CNA.
Part (a)(ii)
T − 200 = 600 × 1.5✓ 1
T = 1100 N «forwards, tension»✓ 1Award [1 max] for 900 N, which ignores the resistive force on the trailer.
Part (b)(i)
resistive force «200 N» directed backwards, opposite to the direction of motion✓ 1
force from the tow-bar directed backwards AND longer than the resistive force, both labelled✓ 1Award [1 max] if a forward force is drawn.
Part (b)(ii)
resultant force on the trailer = 600 × 3.0 = 1800 N backwards✓ 1
tow-bar force = 1800 − 200 = 1600 N✓ 1
compression «the car pushes back on the trailer»✓ 1MP3 may be awarded for a backward push on the trailer even if the magnitude is wrong.
Part (b)(iii)
whole system: B + 600 = 1800 × 3.0✓ 1ALTERNATIVE: car alone: B + 400 − 1600 = 1200 × 3.0, the trailer pushing the car forwards.
B = 4800 N✓ 1Allow ECF from (b)(ii).
Part (c)
without trailer brakes the car’s brakes must decelerate the trailer as well as the car «larger braking force needed on the car»✓ 1
the trailer pushes forwards on the car through the tow-bar «compression», so the car needs a larger friction force from the road and is more likely to skid✓ 1
trailer brakes use the friction on the trailer’s own wheels, so a greater total braking force/deceleration is possible «shorter stopping distance»✓ 1
the force in the tow-bar is reduced, so the trailer is less likely to push the car sideways/off its path «jack-knife»✓ 1OWTTE. Award marks for any four distinct points.

Answers: (a)(i) 3300 N  ·  (a)(ii) 1100 N  ·  (b)(ii) 1600 N, compression  ·  (b)(iii) 4800 N (the remaining parts are explanations — see the table above)

197A-2-21Friction & inclined planes Paper 2Hard14 marks

A crate of mass 40 kg rests on the flat, horizontal floor of a delivery van, 2.0 m behind the front wall of the load space. The crate is not tied down. The coefficient of static friction between the crate and the floor is 0.35 and the coefficient of dynamic friction is 0.25.

Side view of a van. The crate, shown as a dot, rests on the flat floor of the load space, 2.0 m behind the front wall. An arrow shows the direction of the van’s acceleration.crate2.0 mfront wallacceleration of van
Diagram NOT accurately drawn
(a)

The van accelerates forwards from rest at 2.4 m s−2.

(i)

Draw and label, on the dot in the diagram, the forces acting on the crate.

(2)
(ii)

State the direction of the frictional force that the crate exerts on the floor of the van.

(1)
(iii)

Determine whether the crate slides on the floor.

(3)
(b)

The van then travels along a straight, level road at a constant velocity of 14 m s−1. State and explain the magnitude of the frictional force on the crate.

(2)
(c)

Later, the van is travelling at 14 m s−1 with the 40 kg crate 2.0 m behind the front wall. The van brakes with a constant deceleration of 5.0 m s−2 until it stops.

(i)

Explain why the crate slides along the floor.

(2)
(ii)

Determine whether the crate reaches the front wall before the van stops.

(3)
(iii)

Calculate the speed of the crate relative to the van when it reaches the front wall.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
weight «mg, W, Fg» vertically down AND normal force «FN, R» vertically up, of equal length✓ 1
frictional force «Ff» horizontal, pointing forwards «in the direction of the van’s acceleration»✓ 1Do not award MP2 for friction pointing backwards. Award [1] max if any additional force «eg a “forward force”» is drawn.
Part (a)(ii)
backwards «towards the rear of the van»✓ 1Newton’s third law partner of the friction on the crate.
Part (a)(iii)
friction needed = ma = 40 × 2.4 = 96 N✓ 1
maximum static friction = μsmg = 0.35 × 40 × 9.81 = 137 N✓ 1
96 N < 137 N, so the crate does not slide✓ 1ALTERNATIVE: greatest acceleration without sliding = μsg = 3.43 m s−2 ✓ > 2.4 m s−2 ✓ so it does not slide ✓. MP3 requires a comparison.
Part (b)
zero «no frictional force acts on the crate»✓ 1MP1 only scores if MP2 scores.
the crate moves with constant velocity, so the resultant force on it is zero «Newton’s first law»; friction is the only horizontal force on it, so it must be zero✓ 1
Part (c)(i)
to decelerate with the van the crate needs a frictional force of 40 × 5.0 = 200 N «or a deceleration of 5.0 m s−2»✓ 1
this exceeds the maximum static friction of 137 N «μsg = 3.4 m s−2 < 5.0 m s−2», so the crate slides «forwards relative to the van»✓ 1Allow ECF from (a)(iii).
Part (c)(ii)
deceleration of the sliding crate = μdg = 0.25 × 9.81 = 2.45 m s−2✓ 1
relative acceleration = 5.0 − 2.45 = 2.55 m s−2 and 2.0 = ½ × 2.55 × t² gives t = 1.25 s✓ 1ALTERNATIVE: in the 2.8 s the van takes to stop it travels 19.6 m and the crate 29.6 m; the difference 10 m > 2.0 m ✓✓
time for the van to stop = 14/5.0 = 2.8 s > 1.25 s, so the crate reaches the wall before the van stops✓ 1MP3 requires a comparison of times or distances.
Part (c)(iii)
vrel = 2.55 × 1.25 = 3.2 m s−1✓ 1Accept √(2 × 2.55 × 2.0). Allow ECF from (c)(ii).

Answers: (a)(iii) 96 N needed < 137 N; does not slide  ·  (c)(ii) 1.25 s < 2.8 s; reaches the wall  ·  (c)(iii) 3.2 m s−1 (the remaining parts are explanations — see the table above)

198A-2-31Circular motion Paper 2Hard13 marks

A space station moves in a circular orbit at a height of 420 km above the surface of the Earth. It completes one orbit every 92.8 minutes.

Radius of the Earth = 6.37 × 106 m; mass of the Earth = 5.97 × 1024 kg.

(a)
(i)

Calculate the orbital speed of the station.

(2)
(ii)

Show that the centripetal acceleration of the station is about 8.6 m s−2.

(1)
(b)

Determine whether the gravitational force of the Earth alone can provide the centripetal force on the station.

(2)
(c)

A communications satellite also moves in a circular orbit around the Earth. Its orbital period is 24.0 hours.

(i)

Explain why the speed of the space station is constant even though a resultant force acts on it.

(2)
(ii)

Determine the radius of the orbit of the satellite.

(2)
(d)

Equipment in the station generates thermal energy. The station gets rid of 70 kW by emitting radiation from radiator panels at a surface temperature of 280 K.

(i)

Outline why the station can lose energy to its surroundings only by radiation.

(1)
(ii)

Determine the total radiating area of the panels. Assume that they behave as black bodies and absorb no radiation.

(2)
(iii)

Suggest why the panels are turned so that they do not face the Sun.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
r = 6.37 × 106 + 4.2 × 105 = 6.79 × 106 m✓ 1
v = 2πr/T = 2π × 6.79 × 106/(92.8 × 60) = 7.66 × 103 m s−1✓ 1Award [1] max for 7.19 × 103 m s−1 «radius of the Earth used as r».
Part (a)(ii)
a = v²/r = (7.66 × 103)²/6.79 × 106 = 8.65 «m s−2»✓ 1Accept 4π²r/T². Must see full substitution OR answer to 3 s.f.
Part (b)
g = GM/r² = 6.67 × 10−11 × 5.97 × 1024/(6.79 × 106)² = 8.64 N kg−1✓ 1Award [0] for MP1 if R is used instead of r.
this equals the centripetal acceleration found in (a)(ii) «to within 0.1 %, the precision of the data», so gravity alone provides the centripetal force: yes✓ 1MP2 requires a comparison.
Part (c)(i)
the gravitational force on the station is always perpendicular to its velocity «directed towards the centre of the orbit»✓ 1Do not accept “the forces on the station are balanced”.
so it has no component along the direction of motion and does no work: it changes the direction of the velocity but not its magnitude «the kinetic energy stays constant»✓ 1
Part (c)(ii)
r³/T² is the same for both orbits «Kepler’s third law»: r³ = (6.79 × 106)³ × (24.0 × 60/92.8)²✓ 1
r = 4.2 × 107 m✓ 1Accept 4.22 × 107 m. Award [2] for CNA. Award [1] max for 3.6 × 107 m «height above the surface given». ALTERNATIVE for MP1: GMm/r² = 4π²mr/T² so r³ = GMT²/4π² with T = 8.64 × 104 s.
Part (d)(i)
space is a vacuum: there is no medium «no particles» for conduction or convection✓ 1
Part (d)(ii)
A = P/σT⁴ = 7.0 × 104/(5.67 × 10−8 × 280⁴)✓ 1
= 2.0 × 102 m²✓ 1Award [2] for CNA.
Part (d)(iii)
so that they do not absorb solar radiation, which would reduce the net rate of energy loss✓ 1OWTTE.

Answers: (a)(i) 7.66 × 103 m s−1  ·  (a)(ii) 8.65 m s−2  ·  (b) g = 8.64 N kg−1; yes  ·  (c)(ii) 4.2 × 107 m  ·  (d)(ii) 2.0 × 102 m² (the remaining parts are explanations — see the table above)

199A-2-34Collisions & energy Paper 2Hard13 marks

In a nuclear reactor, a fast neutron of mass 1.0 u moving at 1.80 × 107 m s−1 collides head-on with a stationary carbon-12 nucleus of mass 12 u in the graphite moderator. After the collision the neutron moves back along its original line with a speed of 1.52 × 107 m s−1.

1 u = 1.661 × 10−27 kg

Beforeneutron 1.0 u1.80×10⁷ m s⁻¹carbon-12, 12 uat restAfter1.52×10⁷ m s⁻¹v = ?
Diagram not to scale
(a)

Calculate the velocity of the carbon nucleus after the collision.

(2)
(b)

Show that the collision is elastic.

(2)
(c)

Show that the neutron transfers about 29 % of its kinetic energy to the carbon nucleus.

(1)
(d)

State and explain how the magnitude of the change in momentum of the carbon nucleus compares with the magnitude of the change in momentum of the neutron.

(2)
(e)

Some reactors use water as the moderator instead. The neutrons then collide mainly with hydrogen nuclei (protons), each of mass 1.0 u.

(i)

A student suggests that, in a head-on elastic collision with a stationary proton, the neutron stops and the proton moves off at 1.80 × 107 m s−1. Determine whether this outcome is consistent with the conservation of momentum and of kinetic energy.

(2)
(ii)

Explain why the neutrons in a reactor must be slowed down, and why a moderator made of light nuclei slows them down more effectively than one made of heavy nuclei.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
1.0 × 1.80 × 107 = 1.0 × (−1.52 × 107) + 12v✓ 1Masses may be in u throughout.
v = 2.77 × 106 m s−1 «in the original direction of the neutron»✓ 1Award [1] max for 2.3 × 105 m s−1 «rebound direction ignored».
Part (b)
Ek before = ½ × 1.661 × 10−27 × (1.80 × 107)² = 2.69 × 10−13 J✓ 1
Ek after = 1.92 × 10−13 + 7.63 × 10−14 = 2.68 × 10−13 J, equal to the value before «within the precision of the data», so the collision is elastic✓ 1Accept working in u (m s−1)². Allow ECF from (a).
Part (c)
1 − (1.52 × 107/1.80 × 107)² = 0.287 «≈ 29 %»✓ 1Accept 7.63 × 10−14/2.69 × 10−13. Accept 28–29 %. Must see full substitution OR answer to 3 s.f.
Part (d)
equal «in magnitude, opposite in direction»✓ 1MP1 only scores if MP2 scores.
the forces on the two bodies are equal and opposite «Newton’s third law» and act for the same time, so the impulses are equal and opposite✓ 1ALTERNATIVE: total momentum is conserved, so the two changes must cancel.
Part (e)(i)
momentum: 1.0 × 1.80 × 107 before AND 1.0 × 1.80 × 107 after, so momentum is conserved✓ 1
kinetic energy: ½mu² before AND ½mu² after «all transferred to the proton»; both are conserved, so the outcome is consistent✓ 1MP2 requires a conclusion.
Part (e)(ii)
slow «thermal» neutrons are much more likely than fast neutrons to cause fission of uranium-235✓ 1
so the chain reaction can be sustained✓ 1
a neutron loses a larger fraction of its kinetic energy in each collision when the target nucleus has a mass close to its own «up to 100 % for a proton, about 29 % for carbon»✓ 1
with a heavy nucleus the neutron rebounds with almost its original speed, so many more collisions «and more chances of being absorbed or escaping» would be needed✓ 1Accept “fewer collisions are needed with light nuclei”.

Answers: (a) 2.77 × 106 m s−1  ·  (b) 2.69 × 10−13 J before, 2.68 × 10−13 J after  ·  (c) 28.7 % (the remaining parts are explanations — see the table above)

200A-2-49Circular motion Paper 2Hard13 marks

A rally car of mass 1250 kg drives over the crest of a hill on a gravel road. Near the top, the road follows an arc of a vertical circle of radius 38 m (Figure 1). Treat the car as a point.

Side view of the road over a crest: an arc of a vertical circle of radius 38 m with the car, shown as a dot, at the top, moving to the rightcarto centre of circlemotion
Figure 1 — Diagram NOT accurately drawn
A box with a dot representing the car at the crestcar
Figure 2 — for (a)
(a)

The car passes over the crest at 15.0 m s−1. Draw and label, on the dot in Figure 2, the forces acting on the car at the crest. The relative lengths of your arrows should be correct.

(2)
(b)

Calculate the normal force on the car at the crest when its speed is 15.0 m s−1.

(2)
(c)

Show that the car loses contact with the road at the crest if its speed is greater than about 19 m s−1.

(1)
(d)

Explain, in terms of the forces on the car, why the car leaves the road at the crest if its speed is greater than this value.

(3)
(e)

The coefficient of friction between the tyres and the gravel is 0.65. The driver brakes as hard as possible at the crest while travelling at 15.0 m s−1. Determine the maximum deceleration that the brakes can give the car at the crest, and compare it with the maximum deceleration that the brakes could give the car on a level road.

(3)
(f)

A heavier car with identical tyres passes over the crest at 15.0 m s−1. State and explain whether its maximum braking deceleration at the crest is different from your answer to (e).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
downward arrow labelled weight and upward arrow labelled normal force✓ 1No other forces in the vertical plane; ignore a horizontal drag or driving force.
normal-force arrow clearly shorter than the weight arrow «about 0.4 of its length»✓ 1
Part (b)
mg − N = mv²/r so N = 1250 × (9.81 − 15.0²/38)✓ 1Award [0] for N = mg + mv²/r.
N = 4860 N✓ 1Accept 4.9 kN.
Part (c)
N = 0 when mg = mv²/r, v = √(gr) = √(9.81 × 38) = 19.3 «m s−1»✓ 1Must see full substitution OR answer to 3 s.f.
Part (d)
to follow the road the car needs a resultant downward «centripetal» force mv²/r✓ 1
the largest downward force available is its weight mg «when the normal force is zero»✓ 1
if mv²/r > mg the weight cannot provide enough centripetal force, so the path of the car is less curved than the road and the car leaves the surface✓ 1Do not award MP3 for “a centrifugal force lifts the car”.
Part (e)
maximum friction = μN = 0.65 × 4860 = 3160 N✓ 1Allow ECF from (b).
a = 3160/1250 = 2.53 m s−2✓ 1
on a level road a = μg = 0.65 × 9.81 = 6.38 m s−2, so the deceleration at the crest is only about 40 % of this✓ 1
Part (f)
no difference✓ 1No mark for the statement alone.
the normal force m(g − v²/r) and so the maximum friction are proportional to m, so the deceleration μ(g − v²/r) does not depend on mass✓ 1

Answers: (b) 4860 N  ·  (c) 19.3 m s−1  ·  (e) 2.53 m s−2 at the crest; 6.38 m s−2 on a level road (the remaining parts are explanations — see the table above)

201A-2-50Drag & buoyancy Paper 2Hard14 marks

Microplastic particles are small pieces of plastic that pollute the oceans. A particle can be modelled as a sphere of radius 5.0 × 10−5 m moving through still sea water of density 1025 kg m−3 and viscosity 1.2 × 10−3 Pa s. Two common plastics are PET, of density 1380 kg m−3, and polyethylene (PE), of density 920 kg m−3.

The drag force on a particle is given by Fd = 6πηrv. Particles reach their terminal speed almost immediately.

A box with a dot representing a PET particle sinking at its terminal speedPET particle
Figure 1 — for (a)
(a)

Draw and label, on the dot in Figure 1, the forces acting on a PET particle sinking at its terminal speed. The relative lengths of your arrows should be approximately correct.

(2)
(b)
(i)

Show that the buoyancy force on a particle is about 5.3 × 10−9 N.

(1)
(ii)

Determine the terminal speed of a PET particle.

(3)
(c)

Explain why the buoyancy force cannot be ignored for a PET particle in sea water, although it can be ignored for a water droplet falling through air.

(2)
(d)

Determine whether a PET particle released at the surface can reach the sea floor, 4.0 km below, within 30 days.

(2)
(e)

State and explain the motion of a PE particle of the same size released below the surface.

(2)
(f)

Over several weeks, algae grow on a PE particle. Its radius increases to 6.0 × 10−5 m and its average density becomes 1040 kg m−3. Deduce the terminal velocity of the particle.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
weight downwards; buoyancy and drag both upwards✓ 1Arrows must start on the dot.
weight arrow equal to buoyancy + drag, with the buoyancy arrow much longer «about three times» than the drag arrow✓ 1
Part (b)(i)
Fb = ρVg = 1025 × 4/3 π (5.0 × 10−5)³ × 9.81 = 5.26 × 10−9 «N»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
W = 1380 × 5.24 × 10−13 × 9.81 = 7.09 × 10−9 N✓ 1
W = Fb + 6πηrv so 6πηrv = 1.82 × 10−9 N✓ 1Award [0] for MP2 and MP3 if buoyancy is omitted.
v = 1.82 × 10−9/(6π × 1.2 × 10−3 × 5.0 × 10−5) = 1.6 × 10−3 m s−1✓ 1Award [3] for CNA.
Part (c)
the density of sea water is about 74 % of the density of PET, so Fb is a large fraction of the weight «Fb/W = ρfluid/ρparticle»✓ 1
the density of air is about 0.1 % of that of water, so for the droplet Fb is negligible compared with its weight✓ 1Accept a correct comparison of density ratios.
Part (d)
t = 4000/1.6 × 10−3 = 2.5 × 106 s = 29 days✓ 1Allow ECF from (b)(ii).
this is less than 30 days, so yes «just»✓ 1MP2 needs a comparison.
Part (e)
the buoyancy force is greater than the weight «PE is less dense than sea water», so the particle rises «to the surface»✓ 1
it rises at a terminal speed «≈ 4.8 × 10−4 m s−1» when weight + drag «now downwards» = buoyancy✓ 1Accept a correct statement that the drag acts downwards.
Part (f)
the particle is now denser than sea water, so it sinks✓ 1
v = (1040 − 1025) × 4/3 πr³g/(6πηr) = 9.8 × 10−5 m s−1 downwards✓ 1Accept the ratio method: v ∝ (ρ − ρw)r². Award [2] for CNA with direction.

Answers: (b)(i) 5.26 × 10−9 N  ·  (b)(ii) 1.6 × 10−3 m s−1  ·  (d) 29 days: yes  ·  (f) 9.8 × 10−5 m s−1 downwards (the remaining parts are explanations — see the table above)

202A-2-77Circular motion Paper 2Hard14 marks

The drum of a tumble dryer has a radius of 0.30 m and rotates about a horizontal axis at a constant 50 revolutions per minute. Ribs on the inside of the drum carry the wet clothes round with the drum wall, so that the clothes are lifted and then fall back across the drum. Figure 1 shows a cross-section of the drum; P is the lowest point and Q the highest point of the wall.

Treat each item of clothing as a particle that moves with the drum wall unless it loses contact with the wall. Air resistance is negligible.

Cross-section of the dryer drum, a circle of radius 0.30 m rotating anticlockwise about a horizontal axis through its centre O; P is the lowest point and Q the highest point of the drum wallOPQ0.30 mrotation
Figure 1 — Diagram NOT accurately drawn
A dot representing the towel at P, the lowest point of the drum
Figure 2 — for part (b)
(a)
(i)

Calculate the angular velocity of the drum.

(1)
(ii)

Calculate the speed of the drum wall.

(1)
(b)

Draw and label, on the dot in Figure 2, the forces acting on a towel at P. The relative lengths of your arrows should be approximately correct.

(2)
(c)

The mass of the towel is 0.80 kg. Calculate the normal force exerted on the towel by the drum at P.

(2)
(d)

Show that the towel cannot remain in contact with the drum wall at Q.

(2)
(e)

Describe and explain the motion of the towel after it loses contact with the wall.

(2)
(f)

Determine the greatest rotation rate, in revolutions per minute, at which the clothes still lose contact with the wall before they reach Q.

(2)
(g)

A manufacturer designs a dryer with a drum of larger radius. Deduce how the rotation rate in (f) changes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
ω = 50 × 2π/60 = 5.24 rad s−1✓ 1
Part (a)(ii)
v = ωr = 5.236 × 0.30 = 1.57 m s−1✓ 1Allow ECF from (a)(i).
Part (b)
Weight vertically downwards AND normal force vertically upwards, from the dot, labelled✓ 1Do not award MP1 if an outward «centrifugal» force or a force along the velocity is added.
Normal-force arrow clearly longer than the weight arrow «about 1.8 times as long»✓ 1Accept 1.5 to 2.2 times. Allow ECF from (c).
Part (c)
FN − mg = mω²r✓ 1Accept mv²/r.
FN = 0.80 × (9.81 + 8.22) = 14.4 N✓ 1Award [2] for CNA. Award [1] max for 6.6 N «centripetal force only» or 1.3 N «subtracted».
Part (d)
At Q the centripetal force needed is mω²r = 0.80 × 8.22 = 6.6 N, but the weight alone is 7.85 N «ω²r = 8.22 m s−2 < g»✓ 1Must see a comparison of ω²r with g OR of mω²r with mg.
The wall can only push towards the centre, so it would have to exert a negative «outward, pulling» normal force; so the towel loses contact before Q✓ 1OWTTE.
Part (e)
Only its weight acts on it, so it moves as a projectile «free fall» on a curved «parabolic» path✓ 1
It leaves the wall moving along the tangent, falls inside the drum and lands on the lower part of the wall, so the clothes tumble✓ 1OWTTE.
Part (f)
Contact just maintained at Q when ω²r = g: ω = √(9.81/0.30) = 5.72 rad s−1✓ 1
= 5.718 × 60/2π = 55 rev min−1✓ 1Accept 54–55 rev min−1. Award [2] for CNA.
Part (g)
ω = √(g/r) at the limit✓ 1
So a larger radius gives a smaller ω: the drum must rotate more slowly «rate ∝ 1/√r»✓ 1MP2 only scores if MP1 scores. OWTTE.

Answers: (a)(i) 5.24 rad s−1  ·  (a)(ii) 1.57 m s−1  ·  (c) 14.4 N  ·  (f) 55 rev min−1 (the remaining parts are explanations — see the table above)

203A-2-79Drag & buoyancy Paper 2Hard14 marks

In electrostatic crop spraying, pesticide droplets are given a negative charge as they leave the nozzle. The plants are connected to earth, so the droplets move in an electric field that pulls them towards the leaves. Near a leaf, a droplet moves through still air in a region where the electric field is uniform and horizontal, with field strength 3.0 × 104 N C−1, directed away from the leaf (Figure 1).

Model the droplet as a sphere of radius 1.0 × 10−5 m and density 1000 kg m−3, carrying a charge of −2.0 × 10−14 C. The viscous drag on the droplet is given by Fd = 6πηrv, where the viscosity of air η = 1.8 × 10−5 Pa s. The buoyancy force on the droplet is negligible.

A region of uniform horizontal electric field; the field lines point to the left; a dot in the middle represents the negatively charged droplet; an arrow to the right shows the direction of the leafelectric field linesdroplettowards the leaf
Figure 1 — Diagram NOT accurately drawn
(a)

Show that the weight of the droplet is about 4 × 10−11 N.

(1)
(b)

Calculate the magnitude of the electric force on the droplet and state its direction.

(2)
(c)

Draw and label, on the dot in Figure 1, the forces acting on the droplet when it moves at constant velocity. The relative lengths of your arrows should be approximately correct.

(2)
(d)

Explain why, when the droplet moves at constant velocity, the drag force on it is not horizontal.

(2)
(e)

Determine the speed of the droplet when it moves at constant velocity, and the angle that its velocity makes with the horizontal.

(3)
(f)

An uncharged droplet of the same size falls vertically at a constant speed of only 1.2 × 10−2 m s−1. Explain why charged droplets can be deposited on the undersides of leaves, whereas uncharged droplets land almost only on the upper surfaces.

(2)
(g)

As the droplet travels, water evaporates from it, so its radius decreases while its charge stays the same. Deduce how this affects the speed found in (e).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
W = 1000 × 4/3 π (1.0 × 10−5)³ × 9.81 = 4.11 × 10−11 «N»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)
F = qE = 2.0 × 10−14 × 3.0 × 104 = 6.0 × 10−10 N✓ 1
Horizontal, towards the leaf «opposite to the field, because the charge is negative»✓ 1Direction needed.
Part (c)
Weight vertically downwards AND electric force horizontal towards the leaf, labelled, with the electric-force arrow much longer «about 15 times» than the weight arrow✓ 1Allow ECF from (b).
Drag «viscous force» labelled, directed away from the leaf and slightly upwards, about as long as the electric-force arrow «opposite to the resultant of the other two»✓ 1Do not award MP2 for a horizontal drag arrow. Award [1] max if any other force is drawn.
Part (d)
At constant velocity the resultant force is zero «Newton's first law», so the drag must balance the vector sum of the weight and the electric force✓ 1
That sum points towards the leaf and slightly downwards, so the drag points away from the leaf and slightly upwards «the droplet's velocity is slightly below the horizontal»✓ 1OWTTE.
Part (e)
Resultant of weight and electric force = √((6.0 × 10−10)² + (4.1 × 10−11)²) = 6.0 × 10−10 N✓ 1Accept the weight neglected «6.0 × 10−10 N» if justified as negligible. Do not award MP1 for 6.4 × 10−10 N «forces added as scalars».
v = 6.0 × 10−10/(6π × 1.8 × 10−5 × 1.0 × 10−5) = 0.18 m s−1✓ 1Accept 0.17–0.18 m s−1. Allow ECF from MP1. Do not award MP2 for 0.089 m s−1 «diameter used in place of the radius».
Angle = tan−1(4.1 × 10−11/6.0 × 10−10) = 3.9° below the horizontal, towards the leaf✓ 1Accept 4°. Direction «below the horizontal» needed.
Part (f)
The electric force on a charged droplet is about 15 times its weight, so its motion is controlled by the electric field and it moves «almost» along the field lines; the field lines meet all surfaces of the earthed leaf, including the underside✓ 1
So a charged droplet can be pulled sideways and upwards onto the underside, whereas an uncharged droplet is acted on only by its weight and the drag, so it falls vertically onto upper surfaces✓ 1OWTTE.
Part (g)
At constant velocity 6πηrv equals «almost» the electric force, which is unchanged because q and E are unchanged✓ 1
v ∝ 1/r, so the speed increases as the radius decreases «and the weight also decreases»✓ 1MP2 only scores if MP1 scores. OWTTE.

Answers: (a) 4.11 × 10−11 N  ·  (b) 6.0 × 10−10 N, towards the leaf  ·  (e) 0.18 m s−1 at 3.9° below the horizontal (the remaining parts are explanations — see the table above)

204A-2-82Momentum & impulse Paper 2Hard13 marks

A rower races a single scull along a straight course. The total mass of the boat and the rower is 95 kg. Treat them as a single body, and ignore the motion of the rower relative to the boat.

Figure 1 shows how the horizontal propulsive force F exerted on the boat and rower by the water, through the oar blades, varies with time t during two stroke cycles. Each cycle lasts 2.0 s: a drive of 0.80 s, when the blades push on the water, followed by a recovery of 1.20 s, when F = 0. Over many cycles the average speed of the boat is constant at 4.5 m s−1.

Propulsive force F against time t for two stroke cycles: each drive is a triangle rising from 0 to 500 N at 0.30 s after the start of the cycle and falling to zero at 0.80 s; F is zero from 0.80 s to 2.0 s00.511.522.533.54t / s0100200300400500600F / N
Figure 1 — graph drawn to scale
Blank axes for the sketch: speed v from 3.5 to 5.5 m s−1 against time t from 0 to 4.0 s, with a dashed line at the average speed 4.5 m s−100.511.522.533.54t / s3.544.555.5v / m s⁻¹
Figure 2 — for part (e)
(a)

Determine the impulse of the propulsive force during one drive.

(2)
(b)

Explain why the impulse of the drag force on the boat during one complete cycle is equal in magnitude to your answer to (a).

(2)
(c)

Hence show that the average drag force on the boat is about 100 N.

(1)
(d)

Assume that the drag force is constant and equal to its average value. Show that the speed of the boat increases by about 1.3 m s−1 during each drive.

(2)
(e)

Sketch, on Figure 2, a graph to show how the speed of the boat varies with time during the two cycles shown in Figure 1. Use the assumption made in (d).

(3)
(f)

In reality the drag force is approximately proportional to the square of the speed. Explain why a boat whose speed varies during each cycle needs a greater average propulsive force than a boat moving at a constant 4.5 m s−1.

(2)
(g)

Estimate the average useful power developed by the rower.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Impulse = area under the F–t graph for one drive «triangle ½ × 0.80 × 500»✓ 1
= 200 N s✓ 1Award [2] for CNA.
Part (b)
The speed «momentum» of the boat is the same at the start of each cycle, so its change in momentum over a cycle is zero✓ 1
The total impulse equals the change in momentum, so the impulse of the drag is equal and opposite to the impulse of the propulsive force✓ 1OWTTE.
Part (c)
Average drag = 200/2.0 = 100 «N»✓ 1Must see the division by the cycle time of 2.0 s.
Part (d)
Net impulse during the drive = 200 − 100 × 0.80 = 120 N s✓ 1
Δv = 120/95 = 1.26 «m s−1»✓ 1Must see full substitution OR answer to 3 s.f. Award [1] max for 200/95 = 2.1 m s−1 «drag ignored».
Part (e)
Speed increases during each drive «0 to 0.80 s and 2.0 to 2.8 s», most steeply near 0.30 s and 2.3 s, on a curve✓ 1Accept a short initial decrease at the start of each drive, while F < 100 N.
Speed decreases linearly during each recovery, with the same gradient «≈ −1.05 m s−2» in both✓ 1
Rise and fall of about 1.3 m s−1, repeating each cycle, between about 3.8 and 5.1 m s−1 «centred on 4.5 m s−1»✓ 1Allow ECF from (d).
Part (f)
Drag ∝ v², so the extra drag when the boat moves faster than average is greater than the reduction when it moves slower «the average of v² is greater than the square of the average v»✓ 1
So the average drag, and hence «by (b)» the average propulsive force needed, is greater✓ 1MP2 only scores if MP1 scores.
Part (g)
P = Fv = 100 × 4.5 = 450 W✓ 1Allow ECF from (c). Accept 450–460 W.

Answers: (a) 200 N s  ·  (c) 100 N  ·  (d) 1.26 m s−1  ·  (g) 450 W (the remaining parts are explanations — see the table above)

205A-2-83Circular motion Paper 2Hard13 marks

A gymnast performs giant circles on a horizontal high bar: she swings round the bar with her body straight and her hands gripping the bar. In a simple model, her centre of mass moves in a vertical circle of radius 1.20 m centred on the bar (Figure 1). Treat the gymnast as a particle of mass 64 kg at her centre of mass. Air resistance and friction between her hands and the bar are negligible, and her body shape does not change during a circle, so her mechanical energy is conserved.

At the top of the circle, T, her speed is 1.80 m s−1. In parts (a) to (d), treat the bar as rigid.

Path of the centre of mass of the gymnast around the barbar O1.20 mTBmotionPath of the centre of mass (dashed)
Figure 1 — drawn to scale; T is the top and B the bottom of the circle
(a)

Consider the gymnast at T.

(i)

Calculate the centripetal acceleration of the gymnast at T.

(1)
(ii)

Determine the magnitude and the direction of the force exerted by the bar on the gymnast at T, and state whether the bar pushes or pulls on her.

(3)
(b)

Show that the speed of the gymnast at the bottom of the circle, B, is about 7.1 m s−1.

(2)
(c)

Draw and label, on the dot at B in Figure 1, the forces acting on the gymnast at B. The relative lengths of your arrows should be approximately correct.

(2)
(d)

Calculate the magnitude of the force exerted by the bar on the gymnast at B, and express it as a multiple of her weight.

(2)
(e)

In fact the bar bends slightly. It behaves like a spring that obeys Hooke’s law with a spring constant of 4.0 × 104 N m−1. Use the forces found in (a)(ii) and (d).

(i)

Determine the displacement of the middle of the bar from its unloaded position when the gymnast is at B.

(2)
(ii)

A student predicts that, when the gymnast is at T, the bar is displaced upwards. Deduce whether the student is correct.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
a = v2/r = 1.802/1.20 = 2.70 m s−2✓ 1Direction not required.
Part (a)(ii)
resultant force towards the centre «downwards»: mg − F = mv2/r OR F = 64 × (9.81 − 2.70)✓ 1
F = 455 N✓ 1Allow ECF from (a)(i). Award [1] max for 801 N «F = m(g + a)».
vertically upwards, so the bar pushes on her «her arms are compressed»✓ 1MP3 can be awarded for a correct direction consistent with a calculated F < mg.
Part (b)
½mvB2 = ½mvT2 + mg × 2.40 «height fallen is 2r» OR vB2 = 1.802 + 2 × 9.81 × 2.40✓ 1Do not accept a height of 1.20 m «gives 5.2 m s−1».
vB = 7.09 «m s−1»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)
weight «Fg, mg» vertically downwards AND force from the bar «F, tension, pull of the bar» vertically upwards, both labelled, from the dot✓ 1Do not award if a separate «centripetal force» arrow or any other force is drawn.
upward arrow clearly longer than the weight arrow «about 5 times as long»✓ 1Accept 4–6 times as long.
Part (d)
F − mg = mv2/r so F = 64 × (9.81 + 50.33/1.20) = 3.3 × 103 N✓ 1
= 5.3 × weight✓ 1Allow ECF from (b). Use of 7.1 m s−1 gives 3.3 × 103 N, 5.3 times. Award [1] max for 2.7 × 103 N «weight omitted».
Part (e)(i)
by Newton’s third law the gymnast exerts a force of 3.3 × 103 N vertically downwards on the bar✓ 1
x = F/k = 3.3 × 103/4.0 × 104 = 0.083 m downwards✓ 1Allow ECF from (d). Direction needed for MP2.
Part (e)(ii)
incorrect: at T the bar pushes up on the gymnast, so «by Newton’s third law» she pushes down on the bar and it is displaced downwards «by 1.1 cm»✓ 1Allow ECF from (a)(ii) for a consistent direction. A reason is needed.

Answers: (a)(i) 2.70 m s−2  ·  (a)(ii) 455 N upwards «a push»  ·  (b) 7.09 m s−1  ·  (d) 3.3 × 103 N, 5.3 × weight  ·  (e)(i) 0.083 m downwards (the remaining parts are explanations — see the table above)

206IBP2S-003-04-13Explosions & recoil Paper 2Hard7 marks

A satellite of mass 1200 kg is attached to its upper rocket stage of mass 300 kg. Springs between them are compressed and store 1500 J of elastic energy. Consider a reference frame in which the combination is initially at rest. The springs are released and push the satellite and the stage apart along a straight line. Assume all the stored energy becomes kinetic energy.

(a)

Explain why the total momentum of the satellite–stage system is unchanged by the separation.

(1)
(b)

Show that the speed of the stage is four times the speed of the satellite after separation.

(2)
(c)

Determine the speed of the satellite after separation.

(3)
(d)

Calculate the distance between the satellite and the stage 10.0 s after separation.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The spring forces are internal forces (equal and opposite by Newton's third law), and any external force has negligible effect during the brief separation, so the total momentum stays zero✓ 1
Part (b)
0 = 1200vsat − 300vstage✓ 1
vstage/vsat = 1200/300 = 4✓ 1
Part (c)
½(1200)v2 + ½(300)(4v)2 = 1500✓ 1
3000v2 = 1500✓ 1
v = 0.71 m s−1✓ 1
Part (d)
Relative speed = 0.707 + 2.83 = 3.54 m s−1, so distance = 3.54 × 10.0 = 35.4 m (accept 35–35.5 m from rounded speeds)✓ 1
ECF from (c)—

Exam tip: Use the momentum condition to link the two speeds before applying energy conservation. This turns a two-unknown problem into a one-unknown problem.

207IBP2S-003-04-15Momentum & impulse Paper 2Hard9 marks

On a construction site, a pile driver drops a hammer of mass 800 kg from rest through a height of 2.5 m onto the top of a vertical pile of mass 200 kg. The hammer and pile move together after impact. Together they drive the pile 0.12 m further into the ground before stopping. Ignore air resistance.

groundpile 200 kghammer 800 kg2.5 m
Diagram not to scale. The pile is driven 0.12 m into the ground.
(a)

Show that the speed of the hammer just before impact is about 7 m s−1.

(2)
(b)

Calculate the speed of the hammer and pile immediately after impact.

(2)
(c)

Determine the kinetic energy transferred to other forms during the impact.

(2)
(d)

Determine the mean resistive force exerted by the ground on the pile.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
½mv2 = mgh, so v = √(2 × 9.8 × 2.5)✓ 1
v = 7.0 m s−1✓ 1
Part (b)
Momentum conserved: 800 × 7.0 = (800 + 200)v✓ 1
v = 5.6 m s−1✓ 1
Part (c)
Ek before = ½ × 800 × 7.02 = 19 600 J and Ek after = ½ × 1000 × 5.62 = 15 680 J✓ 1
Energy transferred = 3.9 ×103 J (3920 J)✓ 1
Allow ECF from (b).—
Part (d)
Work done by the ground = loss of Ek + loss of gravitational potential energy, or deceleration a = v2/2s = 5.62/(2 × 0.12) = 131 m s−2✓ 1
F × 0.12 = 15 680 + 1000 × 9.8 × 0.12 = 16 856 J, or F − mg = ma, so F = 1000 × (131 + 9.8)✓ 1
F = 1.4 ×105 N✓ 1
Allow ECF from (b).—

Common error: While the pile sinks, gravity still does work on the hammer and pile. Leaving out the mgd term, or the weight in F − mg = ma, underestimates the resistive force.

208IBP2S-003-00-13Hooke's law & springs Paper 2Hard7 marks

A car of mass 1200 kg rests on four identical suspension springs, which share the load equally. Each spring is compressed by 4.5 cm. Assume throughout that the whole mass of the car, and of anything in it, is supported equally by the four springs.

(a)

Show that the spring constant of each spring is about 6.5×104 N m−1.

(2)
(b)

Passengers with a total mass of 300 kg get into the car. Using the spring constant from (a), calculate the additional compression of each spring.

(2)
(c)

The loaded car, of total mass 1500 kg, drives at a constant speed of 12 m s−1 over the top of a hump-backed bridge. At the top, the road curves in a vertical arc of radius 45 m. Determine the compression of each spring as the car passes over the top.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Force on each spring = 1200 × 9.81/4 = 2943 N .✓ 1
k = 2943/0.045 = 6.54×104 N m−1 .✓ 1
Part (b)
Extra force on each spring = 300 × 9.81/4 = 736 N .✓ 1
Δx = 736/6.54×104 ≈ 1.1×10−2 m (1.1 cm) . ECF from (a).✓ 1
Part (c)
At the top the resultant force is downward and provides the centripetal force: mg − Fsprings = mv2/r .✓ 1
Fsprings = 1500(9.81 − 122/45) = 1500 × 6.61 ≈ 9.9×103 N in total, i.e. ≈ 2.48×103 N per spring .✓ 1
x = 2480/6.54×104 ≈ 3.8×10−2 m (3.8 cm) . ECF from (a).✓ 1

Exam tip: Over a hump the supporting force is less than the weight, so the springs extend towards their natural length. The answer should therefore be less than the total compression of the loaded car at rest (4.5 cm + 1.1 cm ≈ 5.6 cm). This gives a useful check.

209IBP2S-003-05-15Friction & inclined planes Paper 2Hard9 marks

At a mine, a 50 kg sled of ore is hauled up a straight ramp inclined at 25° to the horizontal. A light, inextensible cable runs from the sled, parallel to the ramp, over a frictionless pulley at the top of the ramp, and down to a 45 kg counterweight that hangs vertically. The coefficient of kinetic friction between the sled and the ramp is 0.30. The sled moves up the ramp. Take g = 9.81 m s−2.

sled 50 kg45 kg25°
Diagram not to scale. The counterweight hangs vertically.
(a)

Outline the forces acting on the sled as it moves up the ramp.

(2)
(b)

Show that the frictional force on the sled is about 130 N.

(2)
(c)

Determine the acceleration of the sled.

(3)
(d)

Calculate the tension in the cable.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Weight acting vertically down, and a normal reaction perpendicular to the ramp✓ 1
Tension up the ramp, parallel to it, and kinetic friction down the ramp, opposing the motion✓ 1
Part (b)
N = mg cos 25° = 50 × 9.81 × cos 25° = 445 N✓ 1
F = μN = 0.30 × 445 = 133 N ≈ 130 N✓ 1
Part (c)
Component of the sled's weight along the ramp = 50 × 9.81 × sin 25° = 207 N✓ 1
Resultant force on the sled–counterweight system = 441 − 207 − 133 = 101 N, or equivalent pair of equations for each body✓ 1
a = 101/95 = 1.06 m s−2 ≈ 1.1 m s−2 . Allow error carried forward from the friction value.✓ 1
Part (d)
For the counterweight: mg − T = ma, so T = 45(9.81 − 1.06). Alternatively, for the sled: T = 50 × 1.06 + 207 + 133. Allow error carried forward from the acceleration✓ 1
T = 394 N ≈ 390 N✓ 1

Common error: The tension is not equal to the counterweight's weight (441 N) while the system accelerates. The counterweight accelerates downward, so T must be less than its weight.

210IBP2S-003-05-16Momentum & impulse Paper 2Hard8 marks

A sports scientist uses a force plate to analyse a vertical jump by an athlete of mass 62 kg. The plate measures the vertical force that the athlete exerts on it.

Phase — Plate reading / N — Duration / s
Standing still — 608 — (long)
Push-off — 1350 (average) — 0.25
Airborne — 0 — (until landing)

(a)

Explain, using Newton's laws, why the plate reads approximately 608 N while the athlete stands still.

(2)
(b)

Calculate the average resultant force on the athlete during push-off.

(2)
(c)

Determine the vertical speed of the athlete at take-off.

(2)
(d)

Estimate the maximum rise of the athlete's centre of mass after take-off. Air resistance is negligible.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The athlete is in equilibrium, so the upward normal force from the plate equals her weight, mg = 62 × 9.8 = 608 N (first law)✓ 1
By the third law, she pushes down on the plate with a force of the same magnitude, and this is the force the plate measures✓ 1
Part (b)
Fnet = 1350 − 62 × 9.8✓ 1
Fnet = 742 N ≈ 740 N, upward✓ 1
Part (c)
Impulse = FnetΔt = 742 × 0.25 = 186 N s (allow error carried forward from the resultant force calculated above)✓ 1
v = 186/62 = 3.0 m s−1✓ 1
Part (d)
h = v2/2g = 3.02/(2 × 9.8) (allow error carried forward from the take-off speed found above)✓ 1
h = 0.46 m✓ 1

Common error: Use the resultant force (plate reading minus weight) when calculating the impulse that changes momentum. Using the full 1350 N overestimates the take-off speed.

211IBP2S-003-02-16Collisions & energy Paper 2Hard9 marks

An ice-hockey goalkeeper of mass 60 kg (including equipment but not including the ball) stands at rest on ice holding a 4.0 kg training ball. The goalkeeper throws the ball. The coefficient of kinetic friction between the goalkeeper's skates and the ice is 0.020. Assume the throw is so brief that friction during the throw is negligible. All speeds are measured relative to the ice.

(a)

Explain, with reference to Newton's laws, why the goalkeeper moves backwards when the ball is thrown forwards.

(2)
(b)

The ball is thrown horizontally at 7.5 m s−1. Calculate the recoil speed of the goalkeeper.

(2)
(c)

Determine the distance the goalkeeper slides before stopping.

(2)
(d)

The ball is instead thrown at 7.5 m s−1 at 30° above the horizontal. Explain why only the horizontal momentum of the goalkeeper–ball system is conserved, and determine the new sliding distance.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The goalkeeper exerts a forward force on the ball, so by Newton's third law the ball exerts an equal and opposite (backward) force on the goalkeeper . This resultant backward force gives the goalkeeper a backward acceleration and so a backward velocity (Newton's second law); equivalently, the total momentum of the system remains zero .✓ 2
Part (b)
Conservation of momentum: 60v = 4.0 × 7.5 . v = 0.50 m s−1 .✓ 2
Part (c)
Deceleration a = μg = 0.020 × 9.8 = 0.196 m s−2 . s = v2/2a = 0.25/0.392 = 0.64 m . Allow ECF from (b).✓ 2
Part (d)
During the throw the ice exerts an extra upward normal force, which is external, so vertical momentum is not conserved; there is no external horizontal force, so horizontal momentum is conserved . Recoil speed = 4.0 × 7.5 cos 30°/60 = 0.43 m s−1 . s = 0.4332/(2 × 0.196) = 0.48 m . Allow ECF from (c).✓ 3

Common error: When an object is launched at an angle by someone standing on the ground, use only its horizontal velocity component to find the recoil. The ground takes up the vertical impulse.

212IBP2S-003-01-16Free-body diagrams Paper 2Hard8 marks

An autonomous underwater glider used for ocean monitoring has a mass of 52.0 kg and a volume of 5.00×10−2 m3. The density of sea water is 1025 kg m−3. While gliding, the water exerts a drag force opposite to the glider's velocity and a lift force perpendicular to its velocity.

30°constant velocity
(a)

Show that the net vertical force from the weight and the buoyant force is about 7.4 N downward.

(2)
(b)

The glider moves at constant velocity along a straight path directed 30° below the horizontal. Determine the magnitudes of the drag force, which acts along the path, and the lift force, which acts perpendicular to the path.

(3)
(c)

To ascend, the glider pumps oil into an external bladder. This increases its volume by 1.00×10−3 m3 while its mass is unchanged. Determine the new net vertical force from the weight and the buoyant force, and state its direction.

(2)
(d)

Explain why the glider does not immediately begin to move upward after its volume is increased.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
W = 52.0 × 9.8 = 509.6 N and Fb = 1025 × 5.00×10−2 × 9.8 = 502.3 N .✓ 1
Net force = 509.6 − 502.3 = 7.35 N ≈ 7.4 N downward .✓ 1
Part (b)
The resultant force is zero, so drag and lift together provide 7.35 N vertically upward .✓ 1
Drag = 7.35 sin 30° (= 7.35 cos 60°) = 3.7 N .✓ 1
Lift = 7.35 cos 30° = 6.4 N .✓ 1
Allow error carried forward from the net vertical force found in the first part.—
Part (c)
Fb = 1025 × 5.10×10−2 × 9.8 = 512.3 N .✓ 1
Net force = 512.3 − 509.6 = 2.7 N, upward .✓ 1
Part (d)
It already has a downward velocity component, and velocity cannot change instantly, so the upward resultant force must first decelerate its downward motion .✓ 1

Exam tip: In equilibrium problems, resolve along and perpendicular to the velocity. The drag then balances the component of the unbalanced force along the path, and the lift balances the component perpendicular to it.