IGCSE Physics: Motion, Forces and Energy — 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: Motion, Forces and Energy revision notes

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Questions

1. A student runs 400 m in 80 s. Calculate their average speed. [2]

2. A ball is dropped and falls freely under gravity for 2 s. Using g = 9.8 m/s², calculate its speed after 2 s, assuming air resistance is negligible. [2]

3. An object has a mass of 5 kg.

(a) Calculate its weight on Earth, where g = 9.8 N/kg. [2] (b) State its mass and weight on the Moon, where g = 1.6 N/kg. [2]

4. A solid block has a volume of 0.002 m³ and a mass of 5.4 kg. Calculate its density in kg/m³. [2]

5. A spring has a spring constant of 40 N/m. Calculate the extension produced by a force of 8 N, and state the law this uses. [3]

6. A trolley of mass 3 kg moving at 4 m/s collides head-on with a stationary trolley of mass 1 kg, and they stick together.

(a) Calculate the total momentum before the collision. [2] (b) Calculate their common velocity after the collision. [2]

7. A crate is pushed 6 m across a floor by a horizontal force of 50 N.

(a) Calculate the work done. [2] (b) If this takes 12 s, calculate the power developed. [2]

8. A box exerts a force of 120 N on the ground through a base of area 0.4 m². Calculate the pressure the box exerts on the ground. [2]

9. Explain, in terms of forces, why a skydiver falling through the air eventually reaches terminal velocity. [3]

10. A ball of mass 0.50 kg falls from rest through a height of 5.0 m, with air resistance negligible. (g = 9.8 N/kg)

(a) Calculate the loss in gravitational potential energy as it falls. [2]

(b) Using conservation of energy, calculate the speed of the ball just before it hits the ground. [3]

11. A uniform beam is pivoted at its centre. A force of 20 N acts downward at a distance of 0.6 m from the pivot on one side.

(a) Calculate the moment of this force about the pivot. [2]

(b) A second force acts on the other side of the pivot, at a distance of 0.3 m from the pivot, holding the beam in equilibrium. Calculate the size of this second force. [2]

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Answers

1. speed = distance / time = 400 / 80 = 5 m/s [2].

2. Acceleration is the change in velocity per unit time: a = Δv ÷ t [1]. Rearranging, the change in velocity Δv = a × t = 9.8 × 2 = 19.6 m/s; since the ball starts from rest, this change in velocity is its final speed, 19.6 m/s [1].

3. (a) W = mg = 5 × 9.8 = 49 N [2]. (b) Mass is unchanged: 5 kg [1]. Weight = mg = 5 × 1.6 = 8 N [1].

4. ρ = m/V = 5.4 / 0.002 = 2700 kg/m³ [2].

5. F = kx → x = F/k = 8 / 40 = 0.2 m [2]. This uses Hooke’s law (extension x is directly proportional to force, within the limit of proportionality) [1].

6. (a) momentum = mv = (3 × 4) + (1 × 0) = 12 kg m/s [2]. (b) Momentum is conserved: 12 = (3 + 1) × v [1] → v = 12/4 = 3 m/s [1].

7. (a) W = Fd = 50 × 6 = 300 J [2]. (b) P = W/t = 300 / 12 = 25 W [2].

8. p = F/A = 120 / 0.4 = 300 Pa (N/m²) [2].

9. As the skydiver falls, their speed increases, so air resistance (which increases with speed) increases [1]. Eventually air resistance becomes equal in size to weight, so the resultant force on the skydiver becomes zero [1]. With no resultant force, the skydiver stops accelerating and falls at a constant (terminal) velocity [1].

10. (a) GPE lost = mgh = 0.50 × 9.8 × 5.0 [1] = 24.5 J [1]. (b) By conservation of energy, all the GPE lost converts to KE: ½mv² = 24.5 [1]. v² = (2 × 24.5) ÷ 0.50 = 98 [1]. v = √98 = 9.9 m/s [1].

11. (a) moment = force × perpendicular distance = 20 × 0.6 [1] = 12 N m [1]. (b) At equilibrium, clockwise moment = anticlockwise moment [1]: F × 0.3 = 12, so F = 12 ÷ 0.3 = 40 N [1].

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Where marks are usually lost

• Forgetting that GPE lost equals KE gained only when air resistance is negligible — stating this assumption explicitly is often worth its own mark.
• Taking a square root too early, or forgetting to square v correctly when rearranging ½mv² = GPE for v.
• Confusing “moment” (force × perpendicular distance, in N m) with “force” alone, or forgetting the perpendicular distance must be measured at right angles to the line of action of the force, not simply the distance along the beam.
• Setting clockwise and anticlockwise moments equal without first checking the beam is actually in equilibrium — the principle of moments only applies once that’s established.

Questions 10 and 11 draw on the energy and moments sections of the Motion, Forces and Energy revision notes — kinetic and gravitational potential energy, and the principle of moments, neither previously tested by the questions above.