OCR GCSE Physics A: Newton's second law, inertia, momentum and circular motion (J249) – 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 statements P2.2i, P2.2j, P2.2k, P2.2l (Higher Tier only) and P2.2q of OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026). Questions 8 and 9 are Higher Tier only. Learn the content first in the study guide or the revision notes. The course hub, printable checklist and free 10-minute diagnostics are linked for planning.

Questions

1. A car of mass 1250 kg accelerates at 2.4 m/s². Calculate the resultant force on the car. [2]

2. A sledge of mass 22 kg is pulled forward by a force of 85 N. Friction of 30 N acts against the motion. Calculate the acceleration of the sledge. [3]

3. A trolley is pulled by different forces and its acceleration is measured with light gates. The mass of the system is kept constant.

Force (N)	0.5	1.0	1.5	2.0
Acceleration (m/s²)	0.62	1.25	1.88	2.50

(a) Describe the relationship between force and acceleration shown by the data. [1] (b) Calculate the mass of the system, using the 2.0 N result. [2] (c) Suggest how the mass of the system can be kept constant as the force is increased. [1]

4. (a) Define inertial mass. [1] (b) The same resultant force of 12 N is applied separately to two objects. Object P accelerates at 3.0 m/s² and object R at 0.50 m/s². Calculate the inertial mass of each. [2] (c) State which object has more inertia and explain how you know. [1]

5. (a) A lorry of mass 7500 kg travels at 8.0 m/s. Calculate its momentum. [2] (b) A cyclist and bicycle have a combined mass of 90 kg and a momentum of 540 kg m/s. Calculate their velocity. [2]

6. A trolley of mass 0.60 kg moves at 2.0 m/s towards a stationary trolley of mass 0.40 kg. They stick together. Calculate the velocity of the joined trolleys, stating its direction. [4]

7. An astronaut of mass 80 kg floats at rest in space. She throws a 4.0 kg tool away from her at 5.0 m/s. Calculate the velocity of the astronaut, including its direction relative to the tool. [4]

8. (Higher Tier only) A cricket ball of mass 0.16 kg moving at 30 m/s is brought to rest by a fielder’s hands in 0.040 s. Calculate the average force on the ball. [3]

9. (Higher Tier only) A driver of mass 65 kg travelling at 15 m/s is stopped in a crash.

(a) Calculate the average force on the driver if the stopping time is 0.020 s with no air bag. [2] (b) Calculate the average force if an air bag makes the stopping time 0.20 s. [1] (c) Explain why the air bag reduces the risk of injury. [2]

10. A car travels round a roundabout at a constant speed of 9 m/s. (a) Explain why the velocity of the car is changing. [2] (b) A rubber bung is whirled in a horizontal circle on a string. The string breaks. Describe the motion of the bung afterwards. [2]

11. A bumper car A of mass 200 kg moves at 3.0 m/s and hits a stationary bumper car B of mass 150 kg. The cars lock together. (a) Calculate the momentum of A before the collision. [1] (b) Calculate the speed of the locked cars immediately after the collision. [2] (c) State what happens to the momentum of A during the collision. [1] (d) The locked cars then slow down because of a constant resultant force of 140 N. Calculate their deceleration. [2] (e) Calculate the time taken for the locked cars to stop. [2]

Answers

1. F = ma [1]. F = 1250 × 2.4 = 3000 N [1] Examiner insight: Writing the equation before substituting shows your method, and the unit is part of the answer.

2. Resultant force = 85 - 30 = 55 N [1]. a = F / m = 55 / 22 [1] = 2.5 m/s² [1] Examiner insight: Using 85 N alone in F = ma is the usual error. State the resultant first.

3. (a) Acceleration is directly proportional to force (doubling the force doubles the acceleration) [1]. (b) m = F / a = 2.0 / 2.50 [1] = 0.80 kg [1] (c) Move weights from the trolley to the hanger (so the total mass stays the same) [1]. Examiner insight: Answers on a controlled variable gain credit when they say how the control is achieved, not just “keep the mass the same”.

4. (a) Inertial mass is the ratio of force to acceleration [1]. (b) P: 12 / 3.0 = 4.0 kg [1]. R: 12 / 0.50 = 24 kg [1] (c) R has more inertia: for the same force it has the smaller acceleration, so it is harder to change its velocity [1]. Examiner insight: A comparison needs both objects and the reason, in terms of acceleration for the same force.

5. (a) p = mv [1] = 7500 × 8.0 = 60 000 kg m/s [1] (b) v = p / m = 540 / 90 [1] = 6.0 m/s [1] Examiner insight: Momentum has the unit kg m/s. Give it in full, not “N” or “J”.

6. Momentum before = 0.60 × 2.0 + 0.40 × 0 = 1.2 kg m/s [1]. Mass after = 0.60 + 0.40 = 1.0 kg [1]. 1.2 = 1.0 × v [1]. v = 1.2 m/s in the original direction of the 0.60 kg trolley [1] Examiner insight: Use the combined mass after the trolleys stick, and give the direction for a vector answer.

7. Momentum of tool = 4.0 × 5.0 = 20 kg m/s [1]. Total momentum before = 0, so the astronaut has 20 kg m/s the opposite way [1]. v = 20 / 80 [1] = 0.25 m/s, in the opposite direction to the tool [1] Examiner insight: Starting from “total momentum before = 0” makes the equal and opposite momenta clear.

8. Change in momentum = 0.16 × 30 - 0 = 4.8 kg m/s [1]. F = change in momentum / time [1] = 4.8 / 0.040 = 120 N [1] Examiner insight: Find the change in momentum first, then divide by time.

9. (a) Change in momentum = 65 × 15 = 975 kg m/s [1]. F = 975 / 0.020 = 48 750 N [1] (b) F = 975 / 0.20 = 4875 N [1] (c) The change in momentum is the same, but the air bag increases the stopping time [1], so the force on the driver is smaller (ten times smaller here) [1]. Examiner insight: An explanation that mentions only “cushioning” loses the key idea: same change in momentum, longer time, smaller force.

10. (a) The direction of travel keeps changing [1]. Velocity is a vector, so a change in direction is a change in velocity even at constant speed [1]. (b) There is no longer a resultant force towards the centre [1], so the bung moves in a straight line at constant velocity along the direction it was travelling when the string broke [1]. Examiner insight: Say “velocity” for the quantity that changes and “speed” for the one that stays constant.

11. (a) p = 200 × 3.0 = 600 kg m/s [1] (b) Mass after = 350 kg, so v = 600 / 350 [1] = 1.7 m/s [1] (c) The momentum of A decreases (some is transferred to B) [1]. (d) a = F / m = 140 / 350 [1] = 0.40 m/s² [1] (e) t = (v - u) / a = 1.714 / 0.40 [1] = 4.3 s [1] Examiner insight: Carrying the unrounded speed of 1.714 m/s into later steps avoids small rounding drift in the final answer.

Where marks are usually lost

• Using one force, not the resultant, in F = ma.
• Leaving out the unit, or using the wrong one for momentum.
• Not converting grams to kilograms before substituting.
• Forgetting the stationary object’s zero momentum, or using one mass when objects join.
• Missing the direction for velocity or momentum.
• Describing circular motion as “changing speed”.
• Describing a safety feature without the time and change-in-momentum link (Higher Tier only).

Next steps

• Revise from the revision notes.
• Re-read the study guide.
• Go to the OCR GCSE Physics hub and the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), first assessment 2018, published by OCR.