OCR GCSE Physics A: Moments, levers and gears (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 P2.3k, P2.3l and P2.3m of OCR GCSE (9-1) Gateway Science Physics A (J249, Version 5.0, July 2026). The specification prints all three statements in bold, so every question is Higher Tier only. Where a question needs g, it is given as 10 N/kg. Revise first with the study guide and the revision notes. Course pages: the OCR GCSE Physics hub and the checklist.

Questions

1. (Higher Tier only) Define the moment of a force, write down the equation used to calculate it, and give the unit. [3]

2. (Higher Tier only) A see-saw is pivoted at its middle. A child on the right pushes down on the seat.

(a) State the direction of the turning effect of this force about the pivot. [1] (b) Describe what happens to the left-hand end of the see-saw. [1] (c) A second child sits directly above the pivot. Explain why this child’s weight has no turning effect. [1]

3. (Higher Tier only) A cyclist pushes down on a pedal with a force of 220 N. The crank is horizontal and 0.17 m long, and the force is vertical.

(a) Calculate the moment of the force about the axle. [2] (b) The crank is now vertical, with the same vertical force. State the moment and explain your answer. [2]

4. (Higher Tier only) A nut needs a moment of 85 N m to loosen it. A spanner has an effective length of 0.34 m.

(a) Calculate the force needed at the end of the spanner, at right angles to it. [2] (b) A pipe is slid over the handle so that the effective length becomes 0.85 m. Calculate the new force needed. [2]

5. (Higher Tier only) A uniform plank is pivoted at its centre. A girl of weight 520 N sits 1.5 m from the pivot. A boy sits on the other side, 1.95 m from the pivot, and the plank balances. The plank weighs 180 N.

(a) Calculate the weight of the boy. [3] (b) Calculate the upward force of the pivot on the plank and explain how you know it. [2]

6. (Higher Tier only) A uniform metre rule of mass 0.100 kg is pivoted at its 50 cm mark. A mass of 0.150 kg hangs from the 20 cm mark. A 0.100 kg mass is hung on the right-hand side to balance the rule.

(a) Calculate the distance from the pivot at which the 0.100 kg mass must hang. [4] (b) Explain why the weight of the rule can be left out of your calculation. [1]

7. (Higher Tier only) A uniform plank is 3.0 m long and weighs 240 N. It rests on a pivot 1.0 m from its left end. A person of weight 300 N stands on the left end. A weight is placed on the right end so the plank balances.

(a) Calculate the weight placed on the right end. [4] (b) Calculate the upward force of the pivot on the plank. [2] (c) The right-hand weight is removed. State, with a reason, the direction in which the plank starts to turn. [2]

8. (Higher Tier only) A crowbar is used to lift the lid of a crate. The crowbar presses on the lid 0.10 m from the pivot with a force of 1650 N. The hand pushes at right angles to the bar 1.1 m from the pivot.

(a) Calculate the force applied by the hand. [3] (b) Explain why a crowbar is called a force multiplier, and state one disadvantage of using it. [2]

9. (Higher Tier only) A driver gear with 18 teeth applies a moment of 4.2 N m and turns at 150 revolutions per minute. It meshes with a driven gear with 54 teeth. Ignore friction.

(a) Calculate the gear ratio. [1] (b) Calculate the moment on the driven gear. [2] (c) Calculate the speed of the driven gear. [2] (d) State the direction in which the driven gear turns, compared with the driver. [1]

10. (Higher Tier only) In a gear train, gear A (10 teeth) is on a motor shaft and drives gear B (40 teeth). Gear C (12 teeth) is on the same shaft as B and drives gear D (36 teeth). The motor supplies a moment of 0.80 N m at 3600 revolutions per minute. Ignore friction.

(a) Calculate the gear ratio for each pair. [2] (b) Calculate the overall ratio. [1] (c) Calculate the moment and the speed of gear D. [3]

11. (Higher Tier only) A uniform horizontal pole of length 1.2 m and weight 80 N is hinged at a wall. A sign of weight 150 N hangs from its far end. A cable from the wall to the pole holds it level. The perpendicular distance from the hinge to the line of the cable is 0.76 m.

(a) Calculate the total clockwise moment of the pole and the sign about the hinge. [3] (b) Calculate the force in the cable. [2] (c) Explain why the distance 0.76 m is used, and not the length of the cable or of the pole. [2] (d) Suggest why the force in the cable is greater than the total weight of 230 N. [1]

Answers

1. A moment measures how strongly a force tends to turn an object about a pivot [1]. Moment = force × perpendicular distance between the pivot and the force’s line of action [1]. The newton metre (N m) is the unit [1]. [3] Examiner insight: A definition that says “force × distance” without “perpendicular” is usually incomplete, and a unit given as N alone is a different quantity.

2. (a) The push is to the right of the pivot, so the turning effect is clockwise [1]. (b) The left-hand end moves upwards [1]. (c) The weight acts through the pivot, so the perpendicular distance is zero and so is the moment [1]. [3] Examiner insight: Name the perpendicular distance and the line of action, rather than saying the child is “in the middle”.

3. (a) M = F × d = 220 × 0.17 [1] = 37.4 N m [1]. (b) The moment is zero [1]. The line of action of the vertical force passes through the axle, so the perpendicular distance is zero [1]. [4] Examiner insight: Show the substitution and the unit, so that an arithmetic slip still leaves working to credit.

4. (a) F = M ÷ d = 85 ÷ 0.34 [1] = 250 N [1]. (b) F = 85 ÷ 0.85 [1] = 100 N [1]. [4] Examiner insight: Show the rearrangement before substituting. In part (b) say that the force falls because the distance has increased, since the moment needed has not changed.

5. (a) Moment of girl = 520 × 1.5 = 780 N m [1]. Boy’s moment = W × 1.95, equal to 780 for balance [1]. W = 780 ÷ 1.95 = 400 N [1]. (b) The plank is not accelerating, so the upward force equals the total downward force [1]. 520 + 400 + 180 = 1100 N [1]. [5] Examiner insight: The plank’s weight adds to the downward force but not to the moments here, so leaving it out of the total is a common slip.

6. (a) Weight of left mass = 0.150 × 10 = 1.5 N [1]. Moment = 1.5 × (0.50 − 0.20) = 0.45 N m [1]. Weight of right mass = 0.100 × 10 = 1.0 N [1]. d = 0.45 ÷ 1.0 = 0.45 m, so it hangs at the 95 cm mark [1]. (b) The rule is uniform, so its weight acts at the 50 cm mark, which is the pivot, and has zero moment [1]. [5] Examiner insight: Convert masses into weights before taking moments, and say whether the answer is a distance from the pivot or a position on the rule.

7. (a) The weight of the plank acts at the midpoint, 1.5 m from the left end, which is 0.5 m to the right of the pivot [1]. Clockwise moment of the plank = 240 × 0.5 = 120 N m [1]. Anticlockwise moment of the person = 300 × 1.0 = 300 N m [1]. W × 2.0 + 120 = 300, so W = 180 ÷ 2.0 = 90 N [1]. (b) The upward force equals the total downward force [1]. 300 + 240 + 90 = 630 N [1]. (c) The plank turns anticlockwise, so the left end goes down [1], because the anticlockwise moment (300 N m) is bigger than the clockwise moment (120 N m) [1]. [8] Examiner insight: When the pivot is not at the centre, state where the plank’s weight acts and its distance from the pivot, because that is the step most often missed.

8. (a) Effort × 1.1 = 1650 × 0.10 [1] = 165 [1]. Effort = 165 ÷ 1.1 = 150 N [1]. (b) The effort is smaller than the load because it acts further from the pivot (a ratio of 11) [1]. The hand must move through a larger distance than the lid moves [1]. [5] Examiner insight: A disadvantage must be specific, such as the larger distance moved. “It is harder” does not describe the trade-off.

9. (a) Ratio = 54 ÷ 18 = 3 [1]. (b) Output moment = 4.2 × 3 [1] = 12.6 N m [1]. (c) Output speed = 150 ÷ 3 [1] = 50 revolutions per minute [1]. (d) The driven gear turns in the opposite direction [1]. [6] Examiner insight: Say that the driven gear turns more slowly while its moment is larger. One without the other leaves the idea half explained.

10. (a) A to B: 40 ÷ 10 = 4 [1]. C to D: 36 ÷ 12 = 3 [1]. (b) Overall ratio = 4 × 3 = 12 [1]. (c) Moment = 0.80 × 12 = 9.6 N m [1]. Speed = 3600 ÷ 12 [1] = 300 revolutions per minute [1]. [6] Examiner insight: Multiply the ratios of the pairs rather than adding them, and write each ratio separately.

11. (a) Moment of pole = 80 × 0.6 (weight acts at the midpoint) = 48 N m [1]. Moment of sign = 150 × 1.2 = 180 N m [1]. Total = 48 + 180 = 228 N m [1]. (b) For balance, T × 0.76 = 228 [1]. T = 228 ÷ 0.76 = 300 N [1]. (c) The distance in a moment is measured at 90 degrees to the force, from the pivot [1]. The cable is at an angle to the pole, so its length is not that distance [1]. (d) The cable pulls at an angle and acts closer to the hinge than the weights do, so it needs a larger force for the same moment [1]. [8] Examiner insight: A reason that mentions the perpendicular distance to the line of action scores better than “because that is the right distance”.

Where marks are usually lost

• Using a distance in centimetres without converting to metres.
• Measuring to the end of the bar instead of to the pivot.
• Forgetting that the weight of a uniform bar acts at its midpoint when the pivot is elsewhere.
• Mixing up the driver and driven teeth, so the ratio is inverted.
• Describing a force multiplier without the trade-off in distance or speed.
• Adding clockwise moments to anticlockwise moments.

Next steps

• Go back over the study guide or the revision notes for anything you missed.
• Visit the OCR GCSE Physics hub and tick off the checklist.
• Try the free 10-minute diagnostics next.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249). Specification Version 5.0, July 2026; first assessment 2018. Topic P2.3, statements P2.3k to P2.3m.