OCR GCSE Physics: Forces — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Forces study guide | Forces revision notes

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Section A

1. State whether each of the following is a vector or a scalar: (a) speed (b) velocity (c) momentum (d) kinetic energy. [4]

2. State Newton’s first law of motion. [2]

Section B

3. A cyclist travels 240 m in 30 seconds at a constant speed. Calculate the cyclist’s speed. [2]

4. A velocity-time graph shows an object accelerating uniformly from 0 to 15 m/s over 5 seconds, then travelling at a constant 15 m/s for a further 8 seconds. Calculate the total distance travelled. [5]

5. A resultant force of 24 N acts on a trolley of mass 8 kg. Calculate the trolley’s acceleration. [2]

6. A car of mass 1200 kg travelling at 15 m/s collides with a stationary car of mass 800 kg, and the two cars lock together. Calculate their combined velocity immediately after the collision. [4]

7. A skydiver reaches terminal velocity during free fall. Explain, in terms of the forces acting, why the skydiver’s velocity stops increasing at this point. [4]

8. A spring has a spring constant of 40 N/m. Calculate: (a) the extension produced by a force of 8 N [2] (b) the elastic potential energy stored at this extension [3]

9. A force of 12 N is applied perpendicular to a spanner at a distance of 0.25 m from a bolt. Calculate the moment of the force. [2]

10. Explain why an object moving in a circle at constant speed is still accelerating. [3]

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Answers

1. (a) scalar [1]. (b) vector [1]. (c) vector [1]. (d) scalar [1].

2. An object continues moving at a constant velocity (or remains at rest) unless a resultant force acts on it [1] [1].

3. Speed = distance ÷ time = 240 ÷ 30 = 8 m/s [1] [1].

4. Phase 1 (0–5 s): distance = area of triangle = ½ × 5 × 15 = 37.5 m [1] [1]. Phase 2 (5–13 s): distance = area of rectangle = 8 × 15 = 120 m [1] [1]. Total distance = 37.5 + 120 = 157.5 m [1].

5. acceleration = force ÷ mass = 24 ÷ 8 = 3 m/s² [1] [1].

6. Using conservation of momentum: total momentum before = total momentum after [1]. Momentum before = (1200 × 15) + (800 × 0) = 18,000 kg m/s [1]. Combined mass after = 1200 + 800 = 2000 kg [1]. Combined velocity = 18,000 ÷ 2000 = 9 m/s [1].

7. As the skydiver falls, air resistance increases as speed increases [1]. Terminal velocity is reached when the air resistance (upward) becomes equal in size to the skydiver’s weight (downward) [1] [1], so the resultant force becomes zero, meaning there is no further acceleration and velocity stays constant [1].

8. (a) extension = force ÷ spring constant = 8 ÷ 40 = 0.2 m [1] [1]. (b) elastic potential energy = ½ × spring constant × extension² = ½ × 40 × 0.2² = ½ × 40 × 0.04 = 0.8 J [1] [1] [1].

9. Moment = force × perpendicular distance = 12 × 0.25 = 3 N·m [1] [1].

10. Although the object’s speed stays constant, its direction is continuously changing as it moves around the circle [1] [1]. Since velocity is a vector that depends on both speed and direction, a change in direction alone means the velocity is changing, and a changing velocity means the object has acceleration (directed toward the centre of the circle) even though its speed never changes [1].

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Where marks are usually lost

• Confusing vector and scalar quantities, particularly momentum (vector) with kinetic energy (scalar).
• Estimating the area under a velocity-time graph in one step rather than splitting it into separate geometric shapes (triangle, rectangle) and calculating each area individually.
• Treating a stationary or terminal-velocity object as having no forces acting on it, rather than explaining that the forces are balanced.
• Using the straight-line distance to the pivot in a moments calculation instead of the perpendicular distance.
• Explaining circular motion by saying speed changes, when it is direction (and therefore velocity) that changes at constant speed.

Approaching forces questions

For any velocity-time graph question, split the graph into separate geometric shapes and calculate the area of each individually before summing them, rather than attempting to estimate the total area in one step — this reliable technique works whether the graph shows constant acceleration, constant velocity, or a mixture of both. For any question involving balanced or terminal-velocity forces, explicitly state that the forces are present and equal in size rather than absent, since this is the specific misconception the specification flags and examiners routinely test. For moments and Hooke’s law calculations, always confirm the distance used is the perpendicular distance from the pivot (not the straight-line distance to a point of application at an angle), and practise rearranging each formula for every variable rather than only the form given in the specification, since exam questions frequently ask for the “wrong” variable. For momentum-conservation questions, always write the total momentum before and the total momentum after as separate sums before equating them, including a zero term explicitly for any object that starts at rest, since omitting a zero-velocity object’s mass from the total is a common source of an otherwise fully correct method losing marks.