OCR A Level Physics A: Newton's laws, momentum, impulse and collisions (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover OCR A Level Physics A (H556), sections 3.5.1 and 3.5.2 (specification Version 3.0, March 2026). Use g = 9.81 m s⁻². Questions 10 and 11 are two-dimensional, which is assessed at A level only. Read the study guide and revision notes first. See the OCR A Level Physics hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. A laptop rests on a desk.

(a) The desk pushes up on the laptop. Identify the other force in the same third-law pair. [2] (b) The laptop’s weight and the desk’s push are equal and opposite. Explain why they are not a third-law pair. [2]

2. A 0.43 kg football is kicked from rest and leaves the boot at 24 m s⁻¹.

(a) Calculate its momentum. [2] (b) Show that the unit N s is equivalent to kg m s⁻¹. [2]

3. A 0.024 kg squash ball hits a wall at 40 m s⁻¹ and rebounds along the same line at 31 m s⁻¹. Contact lasts 6.0 ms.

(a) Calculate the magnitude of the change in momentum. [3] (b) Find the mean force on the ball during contact. [2]

4. Sand falls vertically onto a horizontal conveyor belt at 18 kg s⁻¹. The belt moves at 1.2 m s⁻¹ and the sand ends up moving with it. Calculate the horizontal force the belt must exert on the sand, and state the force the sand exerts on the belt. [4]

5. A 0.43 kg football, initially at rest, is kicked. The force measured during contact is:

t / ms	0	1.0	2.0	3.0	4.0	5.0	6.0	7.0	8.0
F / N	0	420	900	1300	1100	640	260	60	0

(a) Estimate the impulse using trapeziums. [3] (b) Calculate the speed of the ball after the kick. [2] (c) Explain why your answer to (a) is an estimate. [1]

6. Explain, using F = Δp/Δt, how an air bag reduces the force on a driver in a collision, and state one risk of air bags. [4]

7. A 1.6 kg trolley moving at 0.75 m s⁻¹ hits a stationary 0.90 kg trolley and they stick together.

(a) Calculate their common speed. [2] (b) Calculate the kinetic energy lost. [2] (c) Is the collision elastic? Give a reason. [1]

8. A 0.50 kg trolley moving at 3.0 m s⁻¹ runs into a stationary 0.30 kg trolley. Afterwards the first trolley moves in its original direction at 1.2 m s⁻¹.

(a) Calculate the velocity of the second trolley. [2] (b) Show that the collision is inelastic. [3]

9. Two trolleys, 0.80 kg and 1.20 kg, are at rest, held together by a compressed spring. When released, the 1.20 kg trolley moves off at 0.50 m s⁻¹.

(a) Calculate the speed of the 0.80 kg trolley. [3] (b) Calculate the total kinetic energy after release. [2] (c) State where this energy came from. [1]

10. A 0.20 kg puck moves along the x-axis at 4.0 m s⁻¹ and hits a stationary 0.20 kg puck. The first puck leaves at 2.4 m s⁻¹ at 53.1° above the x-axis.

(a) Calculate the x- and y-components of the first puck’s velocity after the collision. [2] (b) Calculate the components of the second puck’s velocity. [3] (c) Calculate the speed and direction of the second puck. [2] (d) Show that the collision is perfectly elastic. [3]

11. A 1200 kg car travelling east at 14 m s⁻¹ and a 1800 kg van travelling north at 9.0 m s⁻¹ collide at a junction and move off together.

(a) Calculate the east and north components of total momentum before. [2] (b) Calculate the speed and direction of the wreckage immediately after. [3] (c) Calculate the kinetic energy lost. [3]

12. A 0.050 kg ball is dropped from 1.8 m onto the floor and rebounds to 1.2 m. Contact lasts 8.0 ms. Ignore air resistance.

(a) Calculate the speed just before and just after impact. [2] (b) Calculate the change in momentum. [2] (c) Calculate the average force exerted by the floor on the ball. [2]

Answers

1. (a) The laptop pushes down on the desk [1], equal in size and of the same (contact) type [1]. [2] (b) Both forces act on the laptop, but a pair acts on two different bodies [1]. They are also different types of force: gravitational and contact [1]. [2] Examiner insight: Naming the body each force acts on is the clearest way to show you understand a pair.

2. (a) p = mv = 0.43 × 24 [1] = 10.32, so 10 kg m s⁻¹ [1]. [2] (b) N = kg m s⁻² (from F = ma) [1], so N s = kg m s⁻² × s = kg m s⁻¹ [1]. [2] Examiner insight: When asked to show a result, write each substitution so the reasoning can be followed.

3. (a) Take the initial direction as positive. Δp = 0.024 × (−31) − 0.024 × 40 [1] = −0.744 − 0.960 [1]. Magnitude = 1.7 kg m s⁻¹ [1]. [3] (b) F = Δp/Δt = 1.704 ÷ 0.0060 [1] = 2.8 × 10² N [1]. [2] Examiner insight: Using 40 − 31 gives a change of 9 m s⁻¹ and an answer far too small; show the signs so the reversal is visible.

4. Momentum gained per second = 18 × 1.2 [1] = 21.6 kg m s⁻¹ per second [1], so the belt exerts 22 N (2 s.f.) horizontally on the sand [1]. By the third law, the sand exerts 22 N on the belt in the opposite direction [1]. [4] Examiner insight: Mass is not constant for a body that “keeps gaining sand”, so use F = Δp/Δt and say why.

5. (a) Area = width × (sum of interior values + half the end values) [1]. Sum = 420 + 900 + 1300 + 1100 + 640 + 260 + 60 = 4680 [1]. Impulse = 0.0010 × 4680 = 4.7 N s [1]. [3] (b) v = impulse ÷ m = 4.68 ÷ 0.43 [1] = 11 m s⁻¹ [1]. [2] (c) The curve between readings is approximated by straight lines [1]. [1] Examiner insight: Convert ms to s before multiplying; an area of 4680 with no conversion is a factor of 1000 out.

6. The driver’s change of momentum from the collision is fixed by their mass and speed [1]. The air bag increases the time over which they stop [1], so F = Δp/Δt gives a smaller average force on the body [1]. One risk: an air bag deploying forcefully can injure a small or close passenger, or it may deploy when not needed [1]. [4] Examiner insight: “It is soft” earns little; link the longer time to a smaller force for the same change in momentum.

7. (a) Momentum before = 1.6 × 0.75 = 1.2 kg m s⁻¹ [1]. v = 1.2 ÷ 2.5 = 0.48 m s⁻¹ [1]. [2] (b) KE before = ½ × 1.6 × 0.75² = 0.45 J; KE after = ½ × 2.5 × 0.48² = 0.288 J [1]. Loss = 0.16 J [1]. [2] (c) No: kinetic energy decreased, so it is inelastic [1]. [1] Examiner insight: Quote both kinetic energies, not just the difference, so your comparison is explicit.

8. (a) 0.50 × 3.0 = 0.50 × 1.2 + 0.30v [1]; 1.5 = 0.60 + 0.30v, so v = 3.0 m s⁻¹ [1]. [2] (b) KE before = ½ × 0.50 × 3.0² = 2.25 J [1]. KE after = ½ × 0.50 × 1.2² + ½ × 0.30 × 3.0² = 0.36 + 1.35 = 1.71 J [1]. KE has decreased, so the collision is inelastic [1]. [3] Examiner insight: A conclusion needs a comparison of two stated values; “some energy is lost” without numbers is incomplete.

9. (a) Total momentum before = 0 [1]. 0.80v = 1.20 × 0.50 [1], so v = 0.75 m s⁻¹, in the opposite direction [1]. [3] (b) KE = ½ × 1.20 × 0.50² + ½ × 0.80 × 0.75² = 0.15 + 0.225 [1] = 0.38 J [1]. [2] (c) The elastic potential energy stored in the compressed spring [1]. [1] Examiner insight: Kinetic energy rising from zero is not a contradiction; say which store it came from.

10. (a) vₓ = 2.4 cos 53.1° = 1.44 m s⁻¹ [1]; v_y = 2.4 sin 53.1° = 1.92 m s⁻¹ [1]. [2] (b) x: 0.20 × 4.0 = 0.20 × 1.44 + 0.20vₓ [1], so vₓ = 2.56 m s⁻¹ [1]. y: 0 = 0.20 × 1.92 + 0.20v_y, so v_y = −1.92 m s⁻¹ [1]. [3] (c) Speed = √(2.56² + 1.92²) = 3.2 m s⁻¹ [1]; angle = tan⁻¹(1.92 ÷ 2.56) = 36.9° below the x-axis [1]. [2] (d) KE before = ½ × 0.20 × 4.0² = 1.6 J [1]. KE after = ½ × 0.20 × 2.4² + ½ × 0.20 × 3.2² = 0.576 + 1.024 [1] = 1.6 J, equal to the initial value, so the collision is perfectly elastic [1]. [3] Examiner insight: Resolve along the same axes throughout and give the direction with its reference line, such as “below the x-axis”.

11. (a) East = 1200 × 14 = 16 800 kg m s⁻¹ [1]; north = 1800 × 9.0 = 16 200 kg m s⁻¹ [1]. [2] (b) Total mass = 3000 kg, so vₑ = 5.6 and vₙ = 5.4 m s⁻¹ [1]. Speed = √(5.6² + 5.4²) = 7.8 m s⁻¹ [1]; direction = tan⁻¹(5.4 ÷ 5.6) = 44° north of east [1]. [3] (c) KE before = ½ × 1200 × 14² + ½ × 1800 × 9.0² = 117 600 + 72 900 = 190 500 J [1]. KE after = ½ × 3000 × 7.779² = 90 780 J [1]. Loss = 1.0 × 10⁵ J [1]. [3] Examiner insight: Momentum is conserved on each axis separately; adding the two initial speeds or momenta directly is a common route to a wrong answer.

12. (a) v = √(2gh): before = √(2 × 9.81 × 1.8) = 5.9 m s⁻¹ [1]; after = √(2 × 9.81 × 1.2) = 4.9 m s⁻¹ [1]. [2] (b) Δp = 0.050 × (4.852 + 5.943) [1] = 0.54 kg m s⁻¹ upwards [1]. [2] (c) Resultant force = 0.5397 ÷ 0.0080 = 67.5 N upwards [1]. The floor force also supports the weight: 67.5 + 0.050 × 9.81 = 68 N [1]. [2] Examiner insight: The impulse equals the change in momentum for the resultant force; remember the weight acts during contact.

Where marks are usually lost

• Subtracting speeds when velocity reverses, instead of adding the magnitudes.
• Leaving milliseconds unconverted in F = Δp/Δt or in graph areas.
• Treating weight and the table’s push as a third-law pair.
• Using F = ma when mass flows onto or off a moving surface.
• Claiming a collision is elastic or inelastic without comparing stated kinetic energies.
• Stating a direction with no reference line or axis.
• In 2D, conserving momentum in only one direction.
• Forgetting that the force from a surface during impact includes the weight term.

Next steps

• Revise with the revision notes and the study guide.
• Check the OCR A Level Physics hub and the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, A Level Physics A (H556) specification, Version 3.0, March 2026 (first assessment in 2017), sections 3.5.1 and 3.5.2, Module 3: Forces and motion.