OCR GCSE Physics: Forces — 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 available free from your board.

Related: Forces study guide | Forces revision notes

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OCR’s real J249 papers open with a whole-paper Section A of 15 multiple-choice questions (15 marks) before a longer Section B of short-answer and extended questions. The set below now matches that real 15-mark weighting, but every question is drawn from this topic alone, rather than sampling across the whole specification as a genuine paper’s Section A would — so it is still not a reproduction of a full paper’s Section A.

Section A — Multiple choice (topic sample)

1. A stone of mass 2 kg is taken from the Earth’s surface (gravitational field strength 10 N/kg) to the Moon’s surface (gravitational field strength 1.6 N/kg). Which statement correctly describes what happens to the stone’s mass and weight? A) Both mass and weight stay the same B) Mass stays the same; weight decreases C) Mass decreases; weight stays the same D) Both mass and weight decrease [1]

2. A resultant force of 20 N acts on an object of mass 5 kg. What is the object’s acceleration? A) 0.25 m/s² B) 4 m/s² C) 15 m/s² D) 100 m/s² [1]

3. A distance-time graph for a moving object is a straight line with a constant, positive gradient. What does this tell us about the object’s motion? A) It is accelerating uniformly B) It is moving at a constant, non-zero speed C) It is stationary D) It is decelerating uniformly [1]

4. On a velocity-time graph, what physical quantity does the gradient of the line represent? A) Distance travelled B) Speed C) Acceleration D) Displacement [1]

5. A book rests on a table. Which pair of forces is a genuine Newton’s third law pair? A) The weight of the book and the normal contact force from the table on the book B) The weight of the book (Earth’s pull on the book) and the book’s gravitational pull on the Earth C) The normal contact force from the table on the book and the weight of the book D) The push of the book down on the table and the weight of the book [1]

6. A resultant force of 15 N acts on an object of mass 3 kg. What is the object’s acceleration? A) 5 m/s² B) 0.2 m/s² C) 12 m/s² D) 45 m/s² [1]

7. Which equation correctly defines momentum? A) momentum = force × time B) momentum = mass × velocity C) momentum = mass ÷ velocity D) momentum = force ÷ mass [1]

8. A spring obeys Hooke’s law. A force of 6 N produces an extension of 0.03 m. What is the spring constant? A) 0.18 N/m B) 2 N/m C) 200 N/m D) 18 N/m [1]

9. What physical quantity is represented by the area under a velocity-time graph? A) Acceleration B) Distance travelled C) Force D) Momentum [1]

10. An object has a resultant force of zero acting on it at all times. Which statement is correct? A) The object must be stationary B) The object is either stationary or moving at a constant velocity C) The object must be accelerating D) The object must be decelerating [1]

11. A skydiver has reached terminal velocity. Which statement correctly describes the forces acting on them? A) There are no forces acting on the skydiver B) Air resistance is greater than weight C) Weight and air resistance are equal in size and opposite in direction, so the resultant force is zero D) Weight is much greater than air resistance [1]

12. A force of 8 N is applied perpendicular to a spanner at a distance of 0.2 m from a bolt. What is the moment of the force? A) 1.6 N·m B) 40 N·m C) 0.025 N·m D) 8.2 N·m [1]

13. A vehicle’s total stopping distance is made up of which two components? A) Braking distance and reaction distance only, which are always equal B) Thinking distance and braking distance C) Acceleration distance and deceleration distance D) Reaction time and speed [1]

14. A spring is stretched beyond its limit of proportionality. Which statement is correct? A) Extension is still directly proportional to force B) The spring returns to its original length once the force is removed, regardless of the extension C) Extension is no longer directly proportional to force D) The spring constant increases beyond this point, but the relationship remains linear [1]

15. An object moves in a circle at constant speed. In which direction does the resultant (centripetal) force act? A) In the direction of motion, tangential to the circle B) Outward, away from the centre of the circle C) Toward the centre of the circle D) There is no resultant force, since the speed is constant [1]

Section B — Short answer and extended questions

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

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

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

19. 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]

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

21. 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]

22. 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]

23. 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]

24. 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]

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

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Answers

1. B) Mass is the amount of matter in an object and does not change with location, so it stays the same on the Moon; weight is the force of gravity on that mass (weight = mass × gravitational field strength), and since the Moon’s gravitational field strength is smaller, the weight decreases [1].

2. B) acceleration = force ÷ mass = 20 ÷ 5 = 4 m/s²; option A inverts the formula (mass ÷ force), option C subtracts the values, and option D multiplies them instead of dividing [1].

3. B) A straight line on a distance-time graph means the object covers equal distances in equal times, i.e. constant speed; a curved line, not a straight one, would indicate acceleration or deceleration, and a horizontal line (zero gradient) would indicate the object is stationary [1].

4. C) The gradient of a velocity-time graph is the change in velocity divided by the change in time, which is the definition of acceleration; the area under a velocity-time graph (not its gradient) gives distance travelled [1].

5. B) A genuine Newton’s third law pair acts on two different objects, is the same type of force, and consists of equal and opposite forces: the Earth pulling the book down and the book pulling the Earth up are such a pair. The weight of the book and the normal contact force from the table (options A and C) are a balanced pair acting on the same object (the book), not a Newton’s third law pair, even though they are also equal and opposite [1].

6. A) acceleration = force ÷ mass = 15 ÷ 3 = 5 m/s² [1].

7. B) Momentum = mass × velocity; it is not the same as force × time (which gives impulse, a change in momentum) [1].

8. C) Spring constant = force ÷ extension = 6 ÷ 0.03 = 200 N/m [1].

9. B) The area under a velocity-time graph gives distance travelled; the gradient of the same graph instead gives acceleration [1].

10. B) A zero resultant force means the forces on the object are balanced; by Newton’s first law, this means the object continues at whatever constant velocity it already has, including a constant velocity of zero (stationary) — it does not have to be stationary [1].

11. C) At terminal velocity the forces are balanced, not absent: air resistance (upward) has increased until it exactly equals the skydiver’s weight (downward), so the resultant force is zero and velocity stays constant [1].

12. A) Moment = force × perpendicular distance = 8 × 0.2 = 1.6 N·m [1].

13. B) Total stopping distance = thinking distance (during the driver’s reaction time) + braking distance (once the brakes are applied); the two are not generally equal [1].

14. C) Beyond the limit of proportionality, extension is no longer directly proportional to force, so Hooke’s law no longer applies, and the spring may not return to its original length once the force is removed [1].

15. C) An object moving in a circle at constant speed still has a continuously changing velocity, because its direction keeps changing; this requires a resultant (centripetal) force directed toward the centre of the circle [1].

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

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

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

19. 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].

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

21. 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].

22. 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].

23. (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].

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

25. 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.