OCR GCSE Physics A: Gravity, weight and pressure force in fluids (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 P2.3g to P2.3j and P2.3n to P2.3o of Topic P2 Forces in the OCR GCSE (9-1) Physics A (Gateway Science) specification, Version 5.0 (July 2026), first assessment 2018. Questions 1 to 5 are both tiers. Questions 6 to 12 are marked (Higher Tier only). Use g = 10 N/kg at the Earth’s surface unless told otherwise. Learn the content first from the study guide and revision notes. Course pages: OCR GCSE Physics hub, printable checklist and free diagnostics.

Questions

1. Describe what is meant by a gravitational field, and explain why you do not notice the gravitational attraction between two people standing next to each other. [3]

2. A pupil hangs a 0.40 kg mass from a newtonmeter.

(a) State the quantity a newtonmeter measures and its unit. [1] (b) Calculate the reading on the newtonmeter. [2]

3. An astronaut and equipment have a total mass of 80 kg. Gravitational field strengths (supplied): Moon 1.6 N/kg, Mars 3.7 N/kg, Jupiter 25 N/kg.

(a) Calculate the total weight on the Moon. [2] (b) Calculate the total weight in the gravitational field at the top of Jupiter’s atmosphere. [2] (c) State the total mass on Mars. [1]

4. On the Moon an astronaut releases a hammer and a feather together from the same height.

(a) Recall the acceleration in free fall near the Earth’s surface. [1] (b) Explain why the hammer and feather land at the same time. [2] (c) On Earth a stone falls from rest in a vacuum chamber for 2.5 s. Calculate its speed at the end. [1]

5. Two balls, one of mass 2.0 kg and one of mass 0.50 kg, are dropped together in a vacuum on Earth.

(a) Calculate the weight of each ball. [2] (b) Use acceleration = resultant force / mass to calculate the acceleration of each ball. [2]

6. (Higher Tier only) A probe has weight 54 N on a planet where g = 3.6 N/kg.

(a) Calculate the mass of the probe. [2] (b) Calculate the weight of the probe on Earth. [2] (c) Explain why the probe’s mass is the same on both planets. [1]

7. (Higher Tier only) A pupil measures the weight of masses on an unknown planet.

Mass (kg)	0.20	0.40	0.60	0.80
Weight (N)	0.74	1.48	2.22	2.96

(a) Show that weight is directly proportional to mass. [1] (b) Calculate the gradient of a graph of weight against mass. [2] (c) State what the gradient represents. [1] (d) Predict the weight of a 3.0 kg mass on this planet. [2]

8. (Higher Tier only) A sealed balloon contains air. Explain the direction of the net force of the air on the balloon skin. [3]

9. (Higher Tier only) A skier has mass 72 kg.

(a) Calculate the skier’s weight. [1] (b) The boots have total contact area 0.032 m² on snow. Calculate the pressure on the snow. [2] (c) With snowshoes the contact area is 0.24 m². Calculate the pressure. [1] (d) Explain how snowshoes help the skier stay on top of soft snow. [1]

10. (Higher Tier only) An aircraft door has area 1.5 m². Cabin air pressure is 8.0 × 10⁴ Pa and the outside pressure is 3.0 × 10⁴ Pa.

(a) Calculate the pressure difference. [1] (b) Calculate the net force on the door. [2] (c) State the direction of this force relative to the door. [1]

11. (Higher Tier only) A garage jack has a small piston of area 2.0 × 10⁻⁴ m² and a large piston of area 4.0 × 10⁻² m². A force of 150 N acts on the small piston.

(a) Calculate the pressure in the liquid. [2] (b) Calculate the force on the large piston. [2] (c) A car has mass 1200 kg. Decide whether the jack can lift it. [2] (d) The small piston moves 0.50 m. Calculate how far the large piston moves. [1]

12. (Higher Tier only) A lander of mass 900 kg stands on three equal pads, each of area 0.20 m². On Mars g = 3.7 N/kg.

(a) Calculate its weight on Mars. [2] (b) Calculate the pressure of the pads on the ground. [2] (c) Calculate the pressure if the same lander stood on the Earth. [2]

Answers

1. All matter has a gravitational field [1] which causes attraction between masses [1]. The strength is much greater for massive objects, and people have tiny masses so the force is far too small to notice [1]. [3] Examiner insight: Answers that say only “gravity pulls things down” miss the point that the field belongs to all matter, so link the size of the force to mass.

2. (a) Weight (a force), in newtons [1]. (b) W = m g = 0.40 × 10 [1] = 4.0 N [1]. Examiner insight: A bare number with no unit is easy to lose credit on, so always write N.

3. (a) W = 80 × 1.6 [1] = 128 N [1]. (b) W = 80 × 25 [1] = 2000 N [1]. (c) 80 kg, since mass does not change [1]. Examiner insight: Part (c) is a common trap: writing a different mass for Mars suggests weight and mass are being confused.

4. (a) 10 m/s² (about 9.8 m/s²) [1]. (b) There is no air resistance on the Moon [1], so all objects accelerate equally whatever their mass [1]. (c) v = 10 × 2.5 = 25 m/s [1]. Examiner insight: An explanation that only says “the Moon has less gravity” does not explain equal landing times, so name the absence of air resistance.

5. (a) 2.0 × 10 = 20 N [1]; 0.50 × 10 = 5.0 N [1]. (b) 20 / 2.0 = 10 m/s² [1]; 5.0 / 0.50 = 10 m/s² [1]. Examiner insight: Show both substitutions, because the point of the question is that the two results are equal.

6. (a) m = W / g [1] = 54 / 3.6 = 15 kg [1]. (b) W = 15 × 10 [1] = 150 N [1]. (c) Mass is the amount of matter, which does not depend on the field [1]. Examiner insight: Finding the mass first is a clear route, and working with a ratio of g values also gains credit when the steps are shown.

7. (a) Weight / mass is constant at 3.7 for every pair [1]. (b) Gradient = change in weight / change in mass = (2.96 − 0.74) / (0.80 − 0.20) [1] = 3.7 N/kg [1]. (c) The gravitational field strength, g [1]. (d) W = 3.0 × 3.7 [1] = 11 N (11.1 N) [1]. Examiner insight: Gradient answers should include the unit N/kg, and using widely spaced points reduces the effect of a small reading error.

8. Air particles collide with the skin [1]. Each collision pushes at right angles to the skin [1]. The net force is therefore outwards, at right angles to the skin [1]. [3] Examiner insight: “Pushes outwards” alone is incomplete; stating that the force is at right angles is the detail the statement asks you to recall.

9. (a) W = 72 × 10 = 720 N [1]. (b) p = F / A = 720 / 0.032 [1] = 22 500 Pa [1]. (c) p = 720 / 0.24 = 3000 Pa [1]. (d) The same weight is spread over a larger area, so the pressure is smaller [1]. Examiner insight: In part (b), use the weight, not the mass, as the force; using 72 gives the wrong unit as well as the wrong value.

10. (a) 8.0 × 10⁴ − 3.0 × 10⁴ = 5.0 × 10⁴ Pa [1]. (b) F = p A = 5.0 × 10⁴ × 1.5 [1] = 75 000 N [1]. (c) Outwards, at right angles to the door [1]. Examiner insight: Using the cabin pressure alone ignores the air outside, so show the subtraction as a separate step.

11. (a) p = F / A = 150 / (2.0 × 10⁻⁴) [1] = 7.5 × 10⁵ Pa [1]. (b) F = p A = 7.5 × 10⁵ × 4.0 × 10⁻² [1] = 30 000 N [1]. (c) Weight = 1200 × 10 = 12 000 N [1]. The jack gives 30 000 N, which is greater, so it can lift the car [1]. (d) The area ratio is 200, so the distance is 0.50 / 200 = 0.0025 m [1]. Examiner insight: A comparison in part (c) needs both forces stated with units before the conclusion.

12. (a) W = 900 × 3.7 [1] = 3330 N [1]. (b) Total area = 3 × 0.20 = 0.60 m² [1]; p = 3330 / 0.60 = 5550 Pa [1]. (c) W = 900 × 10 = 9000 N, so p = 9000 / 0.60 = 15 000 Pa [1] [1]. Examiner insight: Using the area of one pad instead of all three is a common slip; state the total area explicitly.

Where marks are usually lost

• Leaving out units, or writing N for mass and kg for weight.
• Using g = 10 on another planet when the question supplies a different field strength.
• Changing the mass when the object moves to another planet.
• Explaining equal fall on the Moon without mentioning air resistance.
• Using mass in kg as the force in p = F / A.
• Using the area of one contact instead of the total area.
• Leaving area in cm² instead of converting to m².
• Forgetting to subtract the outside pressure when fluid acts on both sides of a surface.
• Saying the force from fluid pressure is “along” the surface rather than at right angles to it.
• Treating the hydraulic force gain as free: the small piston moves further.

Next steps

• Read the revision notes and the study guide.
• Return to the OCR GCSE Physics hub and tick off the printable checklist.
• Try all free 10-minute diagnostics: /diagnostics/.
• Book a free trial class: /trial/.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), first assessment 2018, Topic P2 Forces, statements P2.3g to P2.3j, P2.3n and P2.3o, with the equation list in the appendix.