A.2 Forces and Momentum Questions

1A-1A-05

Newton's laws & force pairs·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksIdentify

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?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingA.2 — Newton's three laws of motion; forces as interactions between bodies; identification of force pairs in various situations Command term: Identify

2A-1A-06

Contact & field forces·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksIdentify

Forces can be classified as contact forces or field forces.

Which row gives a contact force and a field force?

Contact forceField force

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingA.2 — forces as interactions between bodies; the nature and use of the contact forces (normal force, surface frictional force, tension, elastic restoring force, viscous drag, buoyancy) and of the field forces (gravitational force F_g, electric force F_e, magnetic force F_m) Command term: Identify

3A-1A-07

Circular motion·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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 point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingA.2 — that circular motion is caused by a centripetal force acting perpendicular to the velocity; that a centripetal force causes the body to change direction even if its magnitude of velocity may remain constant Command term: Deduce

4A-1A-08

Friction & inclined planes·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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 μ_d.

What is the magnitude of the acceleration of the block?

Show mark scheme

Marking point	Mark	Notes
Step 1Perpendicular to the plane there is no acceleration, so F_N = mg cos θ and the friction up the plane is μ_dmg cos θ.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Along the plane: ma = mg sin θ − μ_dmg cos θ, so a = g(sin θ − μ_d cos θ).	✓ 1	Answer 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: F_N = mg cos θ, friction = μ_dmg cos θ up the plane, and a = g(sin θ − μ_d cos θ).

Syllabus understandingA.2 — surface frictional force F_f on a body in motion as given by F_f = μ_dF_N; that free-body diagrams can be analysed to find the resultant force on a system Command term: Deduce

5A-1B-03

Resultant forces·A.2 Forces and momentum

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

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 / g	10	20	30	40	50	60
a / m s⁻²	0.164	0.342	0.555	0.741	0.950	1.130

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 point	Mark	Notes
Part (a)
so the total mass being accelerated stays constant and a depends only on the hanging weight «the accelerating force»	✓ 1	OWTTE. 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	✓ 1	Award [1] max for 0.0195 without conversion to kg.
19.5 m s⁻² kg⁻¹	✓ 1	Accept 19.0–20.1.
Part (d)
gradient = g/M so M = 9.81/gradient	✓ 1
M = 0.502 kg	✓ 1	Accept 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	✓ 1	Allow ECF.
Part (f)(i)
a constant frictional force «at the pulley or on the track» must be overcome before the glider accelerates	✓ 1	Accept 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	✓ 1	Accept 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)

Syllabus understandingA.2 — Newton's three laws of motion; that free-body diagrams can be analysed to find the resultant force on a system Command term: Determine

6A-1B-04

Circular motion·A.2 Forces and momentum

Paper 1BHard12 marks

Data-based question12 steps to full marksDetermine

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⁻¹	a / m s⁻²	lg(ω / rad s⁻¹)	lg(a / m s⁻²)
2.00	0.430	0.301	−0.367
3.00	0.926	0.477	−0.033
4.00	1.71	0.602	0.233
5.50	3.14
7.00	5.20	0.845	0.716
9.00	8.45	0.954	0.927

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 point	Mark	Notes
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	✓ 1	Both needed. Accept 2 or 3 d.p.
Part (c)
single straight line with points scattered evenly, extended to lg ω = 0	✓ 1	The line must reach the vertical axis for (f).
Part (d)
gradient from a large triangle on the line	✓ 1
n = 1.99	✓ 1	Accept 1.9–2.1.
Part (e)
yes: a = ω²r predicts n = 2, and the value found is 2 within experimental error	✓ 1	Allow ECF from (d).
Part (f)
intercept = lg r ≈ −0.97	✓ 1	Accept −1.00 to −0.94.
r = 10^−0.97 = 0.107 m	✓ 1	Accept 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	✓ 1	OWTTE.
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)

7A-2-04

Friction & inclined planes·A.2 Forces and momentum

Paper 2Easy11 marks

Short answer & extended response11 steps to full marksDraw

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.

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 point	Mark	Notes
Part (a)
weight «mg, F_g» vertically downwards from the dot	✓ 1
normal force «F_N» perpendicular to the ramp surface, away from it	✓ 1
friction «F_f» parallel to the ramp surface, directed up the slope	✓ 1	Each 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)
μ_s = F_f / F_N = mg sin 24° / mg cos 24° = tan 24° «= 0.445»	✓ 1	Must 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: F_f = 18 × 9.81 × sin 15°	✓ 1
F_f = 45.7 N	✓ 1	Do not accept μ_sF_N = 75.9 N: that is the maximum possible friction, not the actual friction.
Part (d)
F_f = 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	✓ 1	Accept a = g(sin θ − μ_d cos θ). Award [3] for CNA.
Part (e)
constant velocity so the resultant force is zero: mg sin θ = μ_d mg cos θ «tan θ = μ_d»	✓ 1
θ = tan⁻¹ 0.35 = 19°	✓ 1	Accept 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)

Syllabus understandingA.2 — that forces acting on a body can be represented in a free-body diagram; the nature and use of the following contact forces: normal force F_N; surface frictional force F_f ≤ μ_sF_N on a stationary body or F_f = μ_dF_N on a body in motion Command term: Draw

8A-2-05

Explosions & recoil·A.2 Forces and momentum

Paper 2Medium10 marks

Short answer & extended response10 steps to full marksShow that

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 E_α / E_Pb = m_Pb / m_α, 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 point	Mark	Notes
Part (a)
²⁰⁶₈₂Pb AND ⁴₂α	✓ 1	Both 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	✓ 1	OWTTE.
Part (c)
E = p² / 2m for each particle	✓ 1	Accept ½mv² with m_αv_α = m_Pbv_Pb.
p is the same for both, so E_α/E_Pb = (p²/2m_α) / (p²/2m_Pb) = m_Pb/m_α	✓ 1	Must see the momenta cancel.
Part (d)
E_α = 5.41 × 206 / (206 + 4.00)	✓ 1	Accept a ratio route: E_α = 51.5 E_Pb with E_α + E_Pb = 5.41 MeV.
E_α = 5.31 MeV	✓ 1	Award [2] for CNA. Award [0] for 5.41 MeV.
Part (e)(i)
E_α = 5.31 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.49 × 10⁻¹³ J AND m_α = 4.00 × 1.661 × 10⁻²⁷ kg	✓ 1	Allow ECF from (d).
v = √(2E/m) = 1.6 × 10⁷ m s⁻¹	✓ 1
Part (e)(ii)
v_Pb = 1.60 × 10⁷ × 4.00 / 206 = 3.1 × 10⁵ m s⁻¹	✓ 1	Allow ECF from (e)(i). Accept use of E_Pb = 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)

Syllabus understandingA.2 — that linear momentum as given by p = mv remains constant unless the system is acted upon by a resultant external force; explosions; energy considerations in elastic collisions, inelastic collisions, and explosions; A.3 — the kinetic energy of translational motion as given by E_k = ½mv² = p²/2m; E.3 — the radioactive decay equations involving α, β⁻, β⁺, γ Command term: Show that

9A-2-06

Drag & buoyancy·A.2 Forces and momentum

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksDetermine

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.

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 point	Mark	Notes
Part (a)(i)
m = 1000 × 4/3 × π × (12 × 10⁻⁶)³ = 7.2 × 10⁻¹² kg	✓ 1	Accept 7.24 × 10⁻¹² kg.
Part (a)(ii)
the droplet displaces its own volume of air, so F_b/W = ρ_airVg/ρ_waterVg = 1.2/1000 = 1.2 × 10⁻³ «0.12 %»	✓ 1	ALTERNATIVE: W = 7.1 × 10⁻¹¹ N and F_b = 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	✓ 1	Do 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»	✓ 1	Award MP2 only for a diagram with zero resultant force.
Part (c)
mg = 6πηrv «buoyancy negligible»	✓ 1	Allow inclusion of the buoyancy force.
v = 7.1 × 10⁻¹¹/(6π × 1.8 × 10⁻⁵ × 12 × 10⁻⁶) = 0.017 m s⁻¹	✓ 1	Accept 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	✓ 1	Accept 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	✓ 1	Accept 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»	✓ 1	OWTTE.
so the drag becomes equal to the weight at a much lower speed than the model predicts	✓ 1	Accept: 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)

Syllabus understandingA.2 — viscous drag force F_d acting on a small sphere opposing its motion through a fluid as given by F_d = 6πηrv; buoyancy F_b acting on a body due to the displacement of the fluid as given by F_b = ρVg; that forces acting on a body can be represented in a free-body diagram; A.1 — the qualitative effect of fluid resistance on projectiles, including terminal speed Command term: Determine

10A-2-07

Connected bodies·A.2 Forces and momentum

Paper 2Hard13 marks

Short answer & extended response13 steps to full marksDetermine

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.

Figure 1 — Diagram NOT accurately drawn

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 point	Mark	Notes
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	✓ 1	The decision must be supported by the comparison.
Part (a)(ii)
5.9 N «equal to the tension»	✓ 1	Do 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	✓ 1	Award [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	✓ 1	ALTERNATIVE: two equations, m₂g − T = m₂a and T − μ_dm₁g = m₁a, solved together.
a = 5.89 / 3.2 «= 1.84 m s⁻²»	✓ 1	Must 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	✓ 1	Allow 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	✓ 1	Allow 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)

Syllabus understandingA.2 — Newton’s three laws of motion; that free-body diagrams can be analysed to find the resultant force on a system; the nature and use of tension and of surface frictional force F_f ≤ μ_sF_N or F_f = μ_dF_N; A.1 — the equations of motion for solving problems with uniformly accelerated motion Command term: Determine

11A-2-08

Circular motion·A.2 Forces and momentum

Paper 2Hard15 marks

Short answer & extended response15 steps to full marksDetermine

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.

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 point	Mark	Notes
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»	✓ 1	Must 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»	✓ 1	Do not accept “centrifugal force”.
Part (a)(iv)
F = 72 × 4.83 = 347 N	✓ 1	Allow ECF from (a)(ii). Accept 346–349 N.
Part (b)
a ∝ r at constant ω «a = ω²r»	✓ 1
1.8 / 110 × 100 = 1.6 %	✓ 1	Award [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»	✓ 1	ALTERNATIVE: 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	✓ 1	Allow ECF from (a)(ii).
T = 2π√(0.50/4.83) = 2.02 s	✓ 1	Accept 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»	✓ 1	MP2 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	✓ 1	OWTTE.

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)

Syllabus understandingA.2 — that the motion along a circular trajectory can be described in terms of the angular velocity ω which is related to the linear speed v by the equation as given by v = 2πr/T = ωr; that circular motion is caused by a centripetal force acting perpendicular to the velocity; C.1 — the time period of a mass–spring system as given by T = 2π√(m/k) and of a simple pendulum as given by T = 2π√(l/g) Command term: Determine

12A-1A-22

Newton's laws & force pairs·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksIdentify

A parachutist is descending vertically at a constant speed.

Which statement about the forces on her is correct?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingA.2 — Newton's three laws of motion Command term: Identify

13A-1A-23

Resultant forces·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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?

Diagram NOT accurately drawn

Show mark scheme

Marking point	Mark	Notes
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).	✓ 1	Answer 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.

Syllabus understandingA.2 — that free-body diagrams can be analysed to find the resultant force on a system; the nature and use of tension; Newton's first law applied to translational equilibrium Command term: Deduce

14A-1A-24

Friction & inclined planes·A.2 Forces and momentum

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksIdentify

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?

Sketch graphs, not to scale

Show mark scheme

Marking point	Mark	Notes
Step 1While the crate is at rest the friction is static and balances the applied force exactly, F = P, up to the limit μ_sF_N.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Once P exceeds μ_sF_N the crate slides and the friction drops to the constant dynamic value μ_dF_N, which is smaller.	✓ 1	Answer 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 μ_dF_N, which is less than μ_sF_N.
BThis assumes the friction always equals the applied force, so the crate would never move. Static friction has a maximum value, μ_sF_N.
CCorrect: F = P while the crate is at rest (F_f ≤ μ_sF_N), then a drop to the constant sliding value F_f = μ_dF_N.
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.

Syllabus understandingA.2 — surface frictional force F_f acting in a direction parallel to the plane of contact between a body and a surface, on a stationary body as given by F_f ≤ μ_sF_N or a body in motion as given by F_f = μ_dF_N Command term: Identify

15A-1A-25

Drag & buoyancy·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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

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

Show mark scheme

Marking point	Mark	Notes
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 2F_b = ρVg with ρ the density of the water and V the volume displaced, which is the weight of the water displaced.	✓ 1	Answer 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 F_b = weight; and F_b = ρVg is the weight of the water displaced.
CV in F_b = ρVg is the volume of fluid displaced, which is only the submerged part of the block.
DThe density in F_b = ρVg is that of the fluid. The density of the wood only decides how much of the block is submerged.

Syllabus understandingA.2 — buoyancy F_b acting on a body due to the displacement of the fluid as given by F_b = ρVg where V is the volume of fluid displaced; Newton's three laws of motion Command term: Deduce

16A-1A-26

Drag & buoyancy·A.2 Forces and momentum

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

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 F_d = 6πηrv.

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

Show mark scheme

Marking point	Mark	Notes
Step 1At terminal speed the forces balance: F_d + F_b = weight, so 6πηrv = (ρ_steel − ρ_oil) × (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.	✓ 1	Answer 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 F_d ∝ 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.

Syllabus understandingA.2 — viscous drag force F_d acting on a small sphere opposing its motion through a fluid as given by F_d = 6πηrv; buoyancy F_b = ρVg Command term: Determine

17A-1A-27

Hooke's law & springs·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksCalculate

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?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingA.2 — elastic restoring force F_H following Hooke's law as given by F_H = –kx where k is the spring constant Command term: Calculate

18A-1A-28

Momentum & impulse·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Graph drawn to scale

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingA.2 — that the applied external impulse equals the change in momentum of the system Command term: Determine

19A-1A-29

Explosions & recoil·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksIdentify

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

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

20A-1A-30

Angular velocity & period·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

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 point	Mark	Notes
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).	✓ 1	Answer 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.

Syllabus understandingA.2 — centripetal acceleration as given by a = v²/r = ω²r; that the motion along a circular trajectory can be described in terms of the angular velocity ω which is related to the linear speed v by the equation as given by v = 2πr/T = ωr Command term: Determine

21A-1A-31

Variable mass & F = Δp/Δt·A.2 Forces and momentum

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingA.2 — that Newton's second law in the form F = ma assumes mass is constant whereas F = Δp/Δt allows for situations where mass is changing; A.3 — that power developed P is the rate of work done, as given by P = ΔW/Δt = Fv Command term: Deduce

22A-1B-10

Friction & inclined planes·A.2 Forces and momentum

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

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 / kg	F_N / N	F / N
0.250	2.45	0.97
0.450		1.51
0.650	6.38	2.22
0.850	8.34	2.78
1.050		3.50
1.250	12.3	4.05

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 F_N for the two missing rows.

(1)

(c)

Draw the line of best fit on the graph.

(1)

(d)

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

(3)

(e)

A data handbook gives μ_d = 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 μ_d.

(1)

(g)

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

(1)

Show mark scheme

Marking point	Mark	Notes
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)
F_N = mg: 4.41 N and 10.3 N	✓ 1	Both needed.
Part (c)
single straight line through all error bars, not forced through the origin	✓ 1
Part (d)
μ_d = gradient = 0.32	✓ 1	Accept 0.31–0.33.
maximum and minimum gradient lines drawn through all the error bars	✓ 1
Δμ_d = ½(max − min) ≈ 0.01, so μ_d = 0.32 ± 0.01	✓ 1	No 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»	✓ 1	Allow ECF from (d).
Part (f)(i)
a zero error in the force sensor «it reads about 0.16 N with no force applied»	✓ 1	Accept 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	✓ 1	Any 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)

Syllabus understandingA.2 — surface frictional force F_f acting in a direction parallel to the plane of contact between a body and a surface, on a body in motion as given by F_f = μ_dF_N; Tool 3 — lines of maximum and minimum gradient Command term: Determine

23A-1B-11

Explosions & recoil·A.2 Forces and momentum

Paper 1BEasy9 marks

Data-based question9 steps to full marksDetermine

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.

Trolley	Mass / kg	Time for card to pass gate / ms
A	0.512 ± 0.001	76.7
B	1.030 ± 0.001	156.6

(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 point	Mark	Notes
Part (a)
zero «both trolleys at rest»	✓ 1
Part (b)
v = 0.0500 / 0.0767 = 0.652 m s⁻¹	✓ 1	Must 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 %»	✓ 1	Accept omission of the timing term (0.13 %). Award [0] for adding absolute uncertainties.
p = 0.334 ± 0.004 kg m s⁻¹	✓ 1	MP3 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	✓ 1	Allow ECF from (c).
Part (e)
E_k = ½ × 0.512 × 0.652² + ½ × 1.030 × 0.319² = 0.161 J	✓ 1	Accept 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)

24A-1B-12

Momentum & impulse·A.2 Forces and momentum

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

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.

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 point	Mark	Notes
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	✓ 1	Accept 1.38–1.54 N s.
Part (b)
velocities are in opposite directions, so Δv = 5.40 + 3.62 = 9.02 m s⁻¹	✓ 1	Award [0] for 0.28 N s (speeds subtracted).
Δp = 0.160 × 9.02 = 1.44 N s	✓ 1	Accept kg m s⁻¹.
Part (c)
fractional uncertainty = 0.20/9.02 + 0.001/0.160 = 0.028 «2.8 %»	✓ 1	Absolute uncertainties of the two speeds add because they are summed.
Δ(Δp) = 0.04 N s	✓ 1	Accept 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»	✓ 1	Allow 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	✓ 1	Must see the contact time from the graph (8 ms).
Part (f)
F = J/Δt = 1.46/0.0080 = 182 ≈ 180 N	✓ 1	Accept 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	✓ 1	OWTTE.

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)

Syllabus understandingA.2 — that a resultant external force applied to a system constitutes an impulse J as given by J = FΔt; that the applied external impulse equals the change in momentum of the system Command term: Determine

25A-1B-13

Drag & buoyancy·A.2 Forces and momentum

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

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 / cm	0	1	2	3	4	5	6	7	8
F / N	2.70	2.59	2.45	2.35	2.22	2.09	1.99	1.85	1.74

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 point	Mark	Notes
Part (a)
the upthrust equals the weight of liquid displaced: F_b = ρVg with V = Ah	✓ 1
the cylinder is in equilibrium, so F + F_b = W, giving F = W − ρgAh	✓ 1
Part (b)
uses a large triangle on the line and converts cm to m	✓ 1	Award [1] max for −0.121 N cm⁻¹.
gradient = −12.1 N m⁻¹	✓ 1	Accept −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	✓ 1	Accept 1050–1120 kg m⁻³. Allow ECF from (b).
Part (d)
fractional uncertainty = 0.03 + 2 × (0.5/38.0) = 0.056	✓ 1	The 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	✓ 1	Accept 60–70 kg m⁻³.
Part (e)
range 1027–1149 kg m⁻³ includes 1100 kg m⁻³, so the result is consistent	✓ 1	Allow 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)

Syllabus understandingA.2 — buoyancy F_b acting on a body due to the displacement of the fluid as given by F_b = ρVg; that free-body diagrams can be analysed to find the resultant force on a system Command term: Determine

26A-2-16

Free-body diagrams·A.2 Forces and momentum

Paper 2Easy9 marks

Short answer & extended response9 steps to full marksDraw

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.

Figure 1 — graph drawn to scale

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 point	Mark	Notes
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	✓ 1	Do not award MP2 if any third force is drawn.
Part (c)
a = 2.4 / 2.0 = 1.2 m s⁻² AND F_N − mg = ma	✓ 1
F_N = 58(9.81 + 1.2) = 639 N	✓ 1	Award [2] for CNA.
Part (d)
569 N «equal to the weight: constant velocity so zero resultant force»	✓ 1	Allow ECF from (a).
Part (e)
a = −2.4 / 3.0 = −0.80 m s⁻²	✓ 1	Accept a deceleration of 0.80 m s⁻².
F_N = 58(9.81 − 0.80) = 523 N	✓ 1
Part (f)
the «downward» force exerted by the student on the scales	✓ 1	Do 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)

Syllabus understandingA.2 — Newton’s three laws of motion; that forces acting on a body can be represented in a free-body diagram; that free-body diagrams can be analysed to find the resultant force on a system Command term: Draw

27A-2-17

Newton's laws & force pairs·A.2 Forces and momentum

Paper 2Medium11 marks

Short answer & extended response11 steps to full marksExplain

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.

Figure 1 — Diagram NOT accurately drawn

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 point	Mark	Notes
Part (a)(i)
0.540 × 9.81 = 5.30 N	✓ 1
Part (a)(ii)
F_b = 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	✓ 1	Award [1 max] if any other force is drawn.
Part (b)(ii)
T = 5.30 − 1.96 «= 3.34 N»	✓ 1	Must 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	✓ 1	Allow 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	✓ 1	Allow 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)

Syllabus understandingA.2 — Newton’s three laws of motion; forces as interactions between bodies; buoyancy F_b acting on a body due to the displacement of the fluid as given by F_b = ρVg where V is the volume of fluid displaced; that forces acting on a body can be represented in a free-body diagram Command term: Explain

28A-2-18

Connected bodies·A.2 Forces and momentum

Paper 2Hard15 marks

Short answer & extended response15 steps to full marksDetermine

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.

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 point	Mark	Notes
Part (a)(i)
F − (400 + 200) = (1200 + 600) × 1.5	✓ 1
F = 3300 N	✓ 1	Award [2] for CNA.
Part (a)(ii)
T − 200 = 600 × 1.5	✓ 1
T = 1100 N «forwards, tension»	✓ 1	Award [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	✓ 1	Award [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»	✓ 1	MP3 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	✓ 1	ALTERNATIVE: car alone: B + 400 − 1600 = 1200 × 3.0, the trailer pushing the car forwards.
B = 4800 N	✓ 1	Allow 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»	✓ 1	OWTTE. 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)

Syllabus understandingA.2 — Newton’s three laws of motion; forces as interactions between bodies; that free-body diagrams can be analysed to find the resultant force on a system; the nature and use of tension Command term: Determine

29A-2-19

Variable mass & F = Δp/Δt·A.2 Forces and momentum

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksExplain

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 point	Mark	Notes
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»	✓ 1	Must see full substitution OR answer to 3 s.f.
Part (c)(i)
½ × 1.74 × 9.0² = 71 W	✓ 1	Accept 69–71 W.
Part (c)(ii)
electrical power = 14.8 × 7.5 = 111 W	✓ 1
efficiency = 70.6 / 111 = 0.64 «64 %»	✓ 1	Allow ECF from (c)(i).
Part (d)
2.7 × 10⁵ / 111 = 2.4 × 10³ s «≈ 41 min»	✓ 1	Allow 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»	✓ 1	MP3 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)

Syllabus understandingA.2 — that Newton’s second law in the form F = ma assumes mass is constant whereas F = Δp/Δt allows for situations where mass is changing; Newton’s three laws of motion; A.3 — that power developed P is the rate of work done, or the rate of energy transfer; efficiency η; B.5 — electrical power P = IV Command term: Explain

30A-2-20

Explosions & recoil·A.2 Forces and momentum

Paper 2Medium15 marks

Short answer & extended response15 steps to full marksDeduce

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.

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 point	Mark	Notes
Part (a)(i)
the total momentum was zero before the throw AND no «resultant» external horizontal force acts on the system during the throw	✓ 1	Do 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	✓ 1	Award [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»	✓ 1	The 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	✓ 1	Allow ± 3 kg m s⁻¹.
Part (c)(i)
½ × 12 × 4.5² + ½ × 180 × 0.30² = 121.5 + 8.1 = 129.6 «J»	✓ 1	Must 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»	✓ 1	Allow ECF from (a)(ii). An argument using only E_k = p²/2m with equal momenta scores [1] max.
Part (d)(i)
impulse = change in momentum = 54 N s	✓ 1	Allow ECF from (a)(ii).
F = 54/6.0 = 9.0 N	✓ 1	Award [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	✓ 1	Accept 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 %»	✓ 1	MP3 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)

Syllabus understandingA.2 — explosions; energy considerations in elastic collisions, inelastic collisions, and explosions; that linear momentum as given by p = mv remains constant unless the system is acted upon by a resultant external force; A.3 — that work done by a force is equivalent to a transfer of energy; the kinetic energy of translational motion as given by E_k = ½mv² Command term: Deduce

31A-2-21

Friction & inclined planes·A.2 Forces and momentum

Paper 2Hard14 marks

Short answer & extended response14 steps to full marksDetermine

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.

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 point	Mark	Notes
Part (a)(i)
weight «mg, W, F_g» vertically down AND normal force «F_N, R» vertically up, of equal length	✓ 1
frictional force «F_f» horizontal, pointing forwards «in the direction of the van’s acceleration»	✓ 1	Do 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»	✓ 1	Newton’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 = μ_smg = 0.35 × 40 × 9.81 = 137 N	✓ 1
96 N < 137 N, so the crate does not slide	✓ 1	ALTERNATIVE: greatest acceleration without sliding = μ_sg = 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»	✓ 1	MP1 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 «μ_sg = 3.4 m s⁻² < 5.0 m s⁻²», so the crate slides «forwards relative to the van»	✓ 1	Allow ECF from (a)(iii).
Part (c)(ii)
deceleration of the sliding crate = μ_dg = 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	✓ 1	ALTERNATIVE: 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	✓ 1	MP3 requires a comparison of times or distances.
Part (c)(iii)
v_rel = 2.55 × 1.25 = 3.2 m s⁻¹	✓ 1	Accept √(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)

Syllabus understandingA.2 — surface frictional force acting in a direction parallel to the plane of contact between a body and a surface, on a stationary body as given by F_f ≤ μ_sF_N or a body in motion as given by F_f = μ_dF_N; that forces acting on a body can be represented in a free-body diagram; Newton’s three laws of motion Command term: Determine

32A-2-22

Circular motion·A.2 Forces and momentum

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksDetermine

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.

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 point	Mark	Notes
Part (a)(i)
½mv² = mgr so v = √(2 × 9.81 × 3.2) = 7.92 «m s⁻¹»	✓ 1	Must 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)
F_N − mg = mv²/r	✓ 1	Accept a correct equation in words.
mv²/r = 62 × 7.92²/3.2 = 1.22 × 10³ N	✓ 1
F_N = 1.82 × 10³ N	✓ 1	Accept 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	✓ 1	Accept 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 F_N = 3mg for any radius: the claim is incorrect	✓ 1	MP2 only scores if MP1 scores. OWTTE.
Part (c)(i)
mgr − ½mv² = 62 × 9.81 × 3.2 − ½ × 62 × 7.2²	✓ 1
= 340 J	✓ 1	Accept 339 J. Award [2] for CNA.
Part (c)(ii)
smaller speed so a smaller centripetal force mv²/r is needed	✓ 1
F_N = mg + mv²/r is smaller «1.61 × 10³ N»	✓ 1	Award [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)

Syllabus understandingA.2 — that circular motion is caused by a centripetal force acting perpendicular to the velocity; normal force F_N is the component of the contact force acting perpendicular to the surface; that forces acting on a body can be represented in a free-body diagram; A.3 — that in the absence of frictional, resistive forces, the total mechanical energy of a system is conserved Command term: Determine

33A-2-23

Drag & buoyancy·A.2 Forces and momentum

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksDetermine

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.

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 point	Mark	Notes
Part (a)
buoyancy «upthrust, F_b» upwards; weight «mg» AND drag «F_d» downwards; all labelled	✓ 1	Do not award MP1 if drag is drawn upwards.
buoyancy arrow equal in length to the weight and drag arrows combined	✓ 1	By eye.
Part (b)(i)
F_b = ρVg = 1.2 × 4.0 × 9.81 = 47.1 «N»	✓ 1	Must 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	✓ 1	Award [3] for CNA. Award [2] max for 9.0 m s⁻² «mass of helium omitted».
Part (b)(iii)
at terminal speed kv² = F_b − mg = 15.9 N	✓ 1
v = √(15.9/0.65) = 4.9 m s⁻¹	✓ 1	Allow ECF from (b)(ii).
Part (c)(i)
V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 4.0 × (101/54) × (255/288)	✓ 1	Accept use of PV/T = constant.
= 6.6 m³	✓ 1	Award [1] max for 7.5 m³ «temperature change ignored».
Part (c)(ii)
F_b = 0.72 × 6.6 × 9.81 = 47 N	✓ 1	Allow ECF from (c)(i).
«about» the same as the 47 N at ground level: the lower air density is compensated by the larger volume	✓ 1	MP2 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»	✓ 1	Do 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)

Syllabus understandingA.2 — buoyancy F_b acting on a body due to the displacement of the fluid as given by F_b = ρVg where V is the volume of fluid displaced; that forces acting on a body can be represented in a free-body diagram; B.3 — that the ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature as given by PV/T = constant Command term: Determine

34A-1A-39

Collisions & energy·A.2 Forces and momentum

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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 point	Mark	Notes
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 E_k = p²/2m: E_after/E_before = (p²/2(m + M))/(p²/2m) = m/(m + M).	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: at equal momentum E_k ∝ 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.

Syllabus understandingA.2 — the elastic and inelastic collisions of two bodies; energy considerations in elastic collisions, inelastic collisions, and explosions; A.3 — E_k = ½mv² = p²/2m Command term: Deduce

35A-1A-40

Momentum & impulse·A.2 Forces and momentum

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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 point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingA.2 — that a resultant external force applied to a system constitutes an impulse J as given by J = FΔt where F is the average resultant force and Δt is the time of contact; that the applied external impulse equals the change in momentum of the system Command term: Deduce

36A-1A-41

Circular motion·A.2 Forces and momentum

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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?

Diagram NOT accurately drawn

Show mark scheme

Marking point	Mark	Notes
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).	✓ 1	Answer 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.

Syllabus understandingA.2 — that circular motion is caused by a centripetal force acting perpendicular to the velocity; that free-body diagrams can be analysed to find the resultant force on a system Command term: Deduce

37A-1A-42

Friction & inclined planes·A.2 Forces and momentum

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksIdentify

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?

Diagram NOT accurately drawn

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingA.2 — that forces acting on a body can be represented in a free-body diagram; normal force F_N is the component of the contact force acting perpendicular to the surface that counteracts the body; surface frictional force F_f acting in a direction parallel to the plane of contact between a body and a surface Command term: Identify

38A-1A-44

Collisions & energy·A.2 Forces and momentum

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

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?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingA.2 — the elastic and inelastic collisions of two bodies; energy considerations in elastic collisions, inelastic collisions, and explosions; that linear momentum as given by p = mv remains constant unless the system is acted upon by a resultant external force Command term: Deduce

39A-1B-19

Variable mass & F = Δp/Δt·A.2 Forces and momentum

Paper 1BHard12 marks

Data-based question12 steps to full marksDetermine

A horizontal jet of water from a nozzle of diameter 4.0 mm strikes a vertical plate attached to a force sensor, and the water then runs down the plate. The mass flow rate Q is varied with a tap and is found by collecting the water that leaves the nozzle in 20.0 s and weighing it. The force F on the plate is recorded for each flow rate.

The speed of the water leaving the nozzle is v = Q/(ρA), where A is the area of the nozzle and ρ = 1000 kg m⁻³. The student tests the hypothesis F = kQⁿ.

Mass collected / g	Q / kg s⁻¹	F / N	lg(Q / kg s⁻¹)	lg(F / N)
402	0.0201	0.0328	−1.697	−1.484
598	0.0299	0.0701	−1.524	−1.154
803		0.130		−0.886
1004	0.0502	0.199	−1.299	−0.701
1197	0.0599	0.287	−1.223	−0.542
1401	0.0701	0.388	−1.155
1598	0.0799	0.511	−1.097	−0.292

Graph drawn to scale

(a)

Complete the table.

(2)

(b)

Show that the speed of the water leaving the nozzle is about 5 m s⁻¹ when Q = 0.0599 kg s⁻¹.

(1)

(c)

The water is brought to rest horizontally by the plate. Using Newton's second law in the form F = Δp/Δt, show that F = Q²/(ρA).

(2)

(d)

Draw the line of best fit on the graph.

(1)

(e)

Determine n.

(2)

(f)

State whether the data support the relationship in (c).

(1)

(g)

Assuming n = 2, use a point on your line to determine the diameter of the nozzle implied by the data, and comment on the result.

(2)

(h)

In a later run some of the water splashes back from the plate towards the nozzle. State and explain the effect on the force on the plate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Q = 0.803 / 20.0 = 0.0402 kg s⁻¹ and lg Q = −1.396	✓ 1
lg F = lg 0.388 = −0.411	✓ 1	Accept 2 or 3 d.p. Award [1] for any two correct values.
Part (b)
v = 0.0599 / (1000 × π × (2.0 × 10⁻³)²) = 4.76 m s⁻¹	✓ 1	Must see full substitution OR answer to 3 s.f.
Part (c)
in time Δt a mass QΔt moving at v is stopped, so Δp = QΔt v and F = Δp/Δt = Qv	✓ 1
substituting v = Q/(ρA) gives F = Q²/(ρA)	✓ 1
Part (d)
single straight line with points evenly scattered on either side	✓ 1
Part (e)
gradient from a large triangle on the line	✓ 1
n = 2.0	✓ 1	Accept 1.9–2.1.
Part (f)
yes: the relationship predicts n = 2 and the value found is 2 within experimental error	✓ 1	Allow ECF from (e).
Part (g)
eg point (−1.34, −0.78): A = Q²/(ρF) = 10^−2.68/(1000 × 10^−0.78) = 1.25 × 10⁻⁵ m²	✓ 1	Any point on their line.
D = √(4A/π) = 4.0 mm, which agrees with the measured 4.0 mm	✓ 1	Accept 3.8–4.2 mm. Allow ECF.
Part (h)
the force is larger, because water that rebounds has a greater change in momentum than water that is only brought to rest	✓ 1	OWTTE.

Answers: (a) 0.0402 kg s⁻¹, −1.396, −0.411 · (b) 4.76 m s⁻¹ · (e) n = 2.0 · (g) 4.0 mm (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — that Newton's second law in the form F = ma assumes mass is constant whereas F = Δp/Δt allows for situations where mass is changing; Tool 3 — construct and interpret graphs using logarithmic scales Command term: Determine

40A-2-30

Variable mass & F = Δp/Δt·A.2 Forces and momentum

Paper 2Medium14 marks

Short answer & extended response14 steps to full marksExplain

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 point	Mark	Notes
Part (a)(i)
W = qV = 1.60 × 10⁻¹⁹ × 1200 = 1.92 × 10⁻¹⁶ «J»	✓ 1	Must 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	✓ 1	Award [2] for CNA.
Part (b)(i)
N = I/e = 1.8/1.60 × 10⁻¹⁹ = 1.1 × 10¹⁹ «s⁻¹»	✓ 1	Accept 1.13 × 10¹⁹.
Part (b)(ii)
mass ejected per second = 1.13 × 10¹⁹ × 2.18 × 10⁻²⁵ = 2.45 × 10⁻⁶ kg s⁻¹	✓ 1	Allow ECF from (b)(i).
F = Δp/Δt = (Δm/Δt) × v	✓ 1	Accept momentum given to the ions each second.
F = 2.45 × 10⁻⁶ × 4.20 × 10⁴ = 0.10 N	✓ 1	Allow 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	✓ 1	Accept 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	✓ 1	Accept 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	✓ 1	Accept 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)

Syllabus understandingA.2 — that Newton’s second law in the form F = ma assumes mass is constant whereas F = Δp/Δt allows for situations where mass is changing; Newton’s three laws of motion; B.5 — direct current as a flow of charge carriers as given by I = Δq/Δt; that the electric potential difference V is the work done per unit charge as given by V = W/q; D.2 — the direction of forces between the two types of electric charge; the conservation of electric charge Command term: Explain

41A-2-31

Circular motion·A.2 Forces and momentum

Paper 2Hard13 marks

Short answer & extended response13 steps to full marksExplain

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 point	Mark	Notes
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	✓ 1	Award [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»	✓ 1	Accept 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	✓ 1	Award [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	✓ 1	MP2 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»	✓ 1	Do 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	✓ 1	Accept 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	✓ 1	Award [2] for CNA.
Part (d)(iii)
so that they do not absorb solar radiation, which would reduce the net rate of energy loss	✓ 1	OWTTE.

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)

Syllabus understandingA.2 — that bodies moving along a circular trajectory at a constant speed experience an acceleration that is directed radially towards the centre of the circle as given by a = v²/r = 4π²r/T²; that circular motion is caused by a centripetal force acting perpendicular to the velocity; D.1 — that gravitational field strength g at a point is the force per unit mass as given by g = F/m = GM/r²; D.1 — Kepler’s three laws of orbital motion; B.1 — that conduction, convection and thermal radiation are the primary mechanisms for thermal energy transfer; the Stefan–Boltzmann law L = σAT⁴ Command term: Explain

42A-2-34

Collisions & energy·A.2 Forces and momentum

Paper 2Hard13 marks

Short answer & extended response13 steps to full marksExplain

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

(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 point	Mark	Notes
Part (a)
1.0 × 1.80 × 10⁷ = 1.0 × (−1.52 × 10⁷) + 12v	✓ 1	Masses may be in u throughout.
v = 2.77 × 10⁶ m s⁻¹ «in the original direction of the neutron»	✓ 1	Award [1] max for 2.3 × 10⁵ m s⁻¹ «rebound direction ignored».
Part (b)
E_k before = ½ × 1.661 × 10⁻²⁷ × (1.80 × 10⁷)² = 2.69 × 10⁻¹³ J	✓ 1
E_k 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	✓ 1	Accept working in u (m s⁻¹)². Allow ECF from (a).
Part (c)
1 − (1.52 × 10⁷/1.80 × 10⁷)² = 0.287 «≈ 29 %»	✓ 1	Accept 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»	✓ 1	MP1 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	✓ 1	ALTERNATIVE: 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	✓ 1	MP2 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	✓ 1	Accept “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)

Syllabus understandingA.2 — the elastic and inelastic collisions of two bodies; energy considerations in elastic collisions; that the applied external impulse equals the change in momentum of the system; Newton’s three laws of motion; E.4 — the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant Command term: Explain

43A-2-35

Hooke's law & springs·A.2 Forces and momentum

Paper 2Medium14 marks

Short answer & extended response14 steps to full marksDetermine

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 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 F_H = −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 point	Mark	Notes
Part (a)(i)
the force «exerted by a spring» is proportional to its extension or compression «within the elastic limit»	✓ 1	Accept F_H = −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»	✓ 1	Must 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	✓ 1	Accept 2π√(m/4k).
f = 1/T = 1.2 Hz	✓ 1	Award [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	✓ 1	Accept 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	✓ 1	OWTTE.
Part (d)(i)
L: starts at x₀, oscillates about zero with decreasing amplitude and «roughly» constant period	✓ 1	Period 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)

Syllabus understandingA.2 — elastic restoring force F_H following Hooke’s law as given by F_H = −kx where k is the spring constant; C.1 — the time period of a mass–spring system as given by T = 2π√(m/k); C.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; the effects of light, critical and heavy damping on the system Command term: Determine

44A-2-37

Circular motion·A.2 Forces and momentum

Paper 2Easy10 marks

Short answer & extended response10 steps to full marksExplain

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.

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 point	Mark	Notes
Part (a)(i)
T = 1/9.0 = 0.11 s	✓ 1	Accept 0.111 s.
Part (a)(ii)
ω = 2π × 9.0 = 56.5 «rad s⁻¹»	✓ 1	Accept 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	✓ 1	Accept v²/r.
≈ 36 g	✓ 1	Accept 35–36 g. Allow ECF.
Part (b)(ii)
F = ma = 5.0 × 10⁻⁵ × 352 = 0.018 N «towards the centre»	✓ 1	Allow ECF from (b)(i).
Part (c)
arrow at P tangent to the circle, pointing up the page «in the sense of rotation»	✓ 1	Do not award MP1 for a radial arrow.
path after leaving: a straight line continuing along the tangent at P	✓ 1	Do 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	✓ 1	Do 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)

Syllabus understandingA.2 — that bodies moving along a circular trajectory at a constant speed experience an acceleration that is directed radially towards the centre of the circle as given by a = v²/r = ω²r; that the motion along a circular trajectory can be described in terms of the angular velocity ω which is related to the linear speed v by the equation as given by v = 2πr/T = ωr; Newton’s three laws of motion Command term: Explain

A.2 Forces and momentum: IB Physics SL exam-style questions

44 practice questions on A.2 Forces and momentum

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

Every option, and why

No questions match that combination yet

Marks are lost on method, not on knowledge