Edexcel IGCSE Physics: Forces and Motion — 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 and Motion revision notes

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Section A

1. State what the gradient and the area under a velocity–time graph represent. [2]

2. State Newton’s first law. [2]

3. Give two factors that increase thinking distance and two that increase braking distance. [4]

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Section B

4. A skydiver of mass 75 kg jumps from a plane and eventually reaches terminal velocity.

(a) Calculate her weight. (g = 10 N kg⁻¹) [1]

(b) State the value of the air resistance at terminal velocity, with a reason. [2]

(c) Explain fully, in four steps, how terminal velocity is reached. [4]

(d) She opens her parachute. Describe and explain what happens to her velocity. [3]

5. A car of mass 900 kg accelerates uniformly from 8.0 m s⁻¹ to 20 m s⁻¹ in 6.0 s.

(a) Calculate the acceleration. [2]

(b) Calculate the resultant force. [2]

(c) Calculate the change in momentum. [2]

(d) Calculate the distance travelled. [2]

6. Explain, in terms of momentum, why cars are fitted with crumple zones. [3]

7. A uniform metre rule is balanced at its centre. A 2.0 N weight is placed 30 cm from the pivot on one side. Calculate where a 3.0 N weight must be placed on the other side to balance it. [3]

8. State Hooke’s law, and explain what happens to a spring’s behaviour beyond the limit of proportionality. [3]

9. A spring extends 4.0 cm under a force of 2.0 N. Calculate the spring constant, and predict the extension for a force of 5.0 N, assuming the limit of proportionality is not exceeded. [3]

10. Explain why the upward support forces at each end of a beam change as a load is moved along its length. [2]

11. Ball A, of mass 0.20 kg, moves at 3.0 m/s and hits a stationary ball B.

(a) Compare the force that A exerts on B with the force that B exerts on A during the collision. [2]

(b) Calculate the momentum of ball A before the collision. [1]

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Answers

1. Gradient = acceleration [1]; area under the graph = distance travelled [1].

2. An object remains at rest or moves at constant velocity [1] unless acted on by a resultant force [1].

3. Thinking distance: higher speed, tiredness, alcohol or drugs, distraction — any two [1] [1]. Braking distance: higher speed, worn tyres or brakes, wet or icy road — any two [1] [1].

4. (a) W = mg = 75 × 10 = 750 N [1].

(b) 750 N [1], because at terminal velocity the resultant force is zero, so air resistance equals weight [1].

(c) Weight acts downwards, so she accelerates [1]. As speed increases, air resistance increases [1]. When air resistance equals weight, the resultant force is zero [1]. Acceleration is zero, so she falls at constant velocity [1].

(d) Air resistance increases suddenly and becomes greater than weight [1], so there is a resultant upward force and she decelerates [1]. As she slows, air resistance falls until it again equals weight, giving a new, lower terminal velocity [1].

5. (a) a = (20 − 8.0) ÷ 6.0 [1] = 2.0 m s⁻² [1].

(b) F = ma = 900 × 2.0 [1] = 1800 N [1].

(c) Δp = mΔv = 900 × 12 [1] = 10 800 kg m s⁻¹ [1].

(d) s = ((u + v) ÷ 2) × t = ((8.0 + 20) ÷ 2) × 6.0 [1] = 84 m [1].

6. In a collision the momentum change is fixed [1]. The crumple zone increases the time over which that change occurs [1], and since force is the rate of change of momentum, the force on the occupants is reduced [1].

7. Moment one side = 2.0 × 30 = 60 N cm [1]. For balance, 3.0 × d = 60 [1] d = 20 cm from the pivot [1].

8. Over an initial linear region, extension is directly proportional to the applied force [1]. Beyond the limit of proportionality, this linear relationship no longer holds [1], and if stretched far enough the spring may not return to its original length once the force is removed [1].

9. k = F ÷ x = 2.0 ÷ 0.040 [1] = 50 N/m. For F = 5.0 N: x = F ÷ k = 5.0 ÷ 50 [1] = 0.10 m (10 cm) [1]. Note that x must be converted to metres to give k in the standard unit of N/m; using centimetres directly would give a numerically different, non-standard result.

10. As the load moves closer to one support, that support carries a greater share of the load’s weight, while the more distant support carries less [1]. This is because taking moments about a support gives the reaction force at the other support: as the load moves closer to support A, the moment about A (which determines support B’s reaction) decreases, so B’s upward force decreases, while A’s upward force correspondingly increases [1]. The two support forces must always sum to the total weight being supported, however the load is positioned.

11. (a) The forces are equal in size [1] and opposite in direction [1]. (b) p = mv = 0.20 × 3.0 = 0.60 kg m/s [1].

Mark-scheme insight (Pearson Edexcel 4PH1 June 2024 mark scheme, Paper 2PR, Question 5(a)): separate marks are given for “same size” and “opposite directions”; “equal and opposite”, or “every action has an equal and opposite reaction”, scores only one mark when no other mark is awarded. Try the real question next: Pearson Edexcel International GCSE Physics 4PH1, June 2024, Paper 2PR, Question 5.

Source for the mark-scheme insights on this page: Pearson Edexcel International GCSE Physics 4PH1 Paper 2PR mark scheme (Summer 2024), paraphrased.

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Where marks are usually lost

• Reading area as gradient on a velocity–time graph.
• Saying the skydiver stops, or moves at constant speed only, at terminal velocity.
• Forgetting that opening a parachute gives a new lower terminal velocity, not zero.
• Saying a crumple zone “absorbs the force”.
• Omitting “perpendicular distance” in moment calculations.
• Forgetting to convert extension to metres before calculating a spring constant.
• Saying a spring “always” returns to its original length, regardless of how far it is stretched.