Edexcel IAL Physics: Mechanics — 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: Mechanics revision notes

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Section A

1. State Newton’s three laws of motion. [3]

2. Define momentum and impulse, and state the relationship between them. [3]

Section B

3. A 1200 kg car travelling at 20 m s⁻¹ is brought to rest in 4.0 s.

(a) Calculate the change in momentum. [2] (b) Calculate the average braking force. [2] (c) Explain how a crumple zone reduces the force on the occupants. [3]

4. A ball of mass 0.20 kg moving at 6.0 m s⁻¹ collides head-on with a stationary 0.30 kg ball; they move off together.

(a) Calculate their common velocity. [3] (b) Show whether the collision is elastic or inelastic. [4]

5. A projectile is launched at 25 m s⁻¹ at 30° to the horizontal. Take g = 9.81 m s⁻².

(a) Calculate the horizontal and vertical components of the initial velocity. [2] (b) Calculate the time of flight over level ground. [3] (c) Calculate the horizontal range. [2]

6. A pump raises 300 kg of water through 12 m in 40 s. Calculate the useful power output. [3]

7. A ball of mass 0.20 kg is thrown horizontally from a cliff at 15 m s⁻¹ and lands 2.0 s later. Take g = 9.81 m s⁻².

(a) Calculate the horizontal distance travelled. [2] (b) Calculate the vertical velocity on landing. [2]

8. A motor has an efficiency of 75% and a total power input of 800 W. Calculate the useful power output. [2]

9. Explain, using Newton’s first law, why a skydiver reaches a constant terminal velocity. [2]

10. Distinguish between mass and weight, including their units and whether each is a scalar or a vector. [2]

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Answers

1. An object stays at rest or moves at constant velocity unless acted on by a resultant force [1]. The resultant force is equal to the rate of change of momentum, giving F = ma for constant mass [1]. If A exerts a force on B, B exerts an equal and opposite force on A, of the same type [1].

2. Momentum = mass × velocity, a vector, in kg m s⁻¹ [1]. Impulse = force × time, in N s [1]. Impulse is equal to the change in momentum [1].

3. (a) Δp = mΔv = 1200 × (0 − 20) [1] = −24 000 kg m s⁻¹ (magnitude 24 000) [1]. (b) F = Δp ÷ t = 24 000 ÷ 4.0 [1] = 6000 N [1]. (c) The crumple zone deforms, increasing the time taken to stop [1]. Since force = change in momentum ÷ time and the change in momentum is fixed [1], a longer time gives a smaller force on the occupants [1].

4. (a) Momentum before = 0.20 × 6.0 = 1.2 kg m s⁻¹ [1]; total mass after = 0.50 kg [1]; v = 1.2 ÷ 0.50 = 2.4 m s⁻¹ [1]. (b) KE before = ½ × 0.20 × 6.0² = 3.6 J [1]. KE after = ½ × 0.50 × 2.4² = 1.44 J [1]. Kinetic energy has decreased [1], so the collision is inelastic — momentum is conserved but kinetic energy is not [1].

5. (a) Horizontal = 25 cos 30° = 21.7 m s⁻¹ [1]; vertical = 25 sin 30° = 12.5 m s⁻¹ [1]. (b) Time to highest point = 12.5 ÷ 9.81 = 1.27 s [1] [1]; total time of flight = 2.55 s [1]. (c) Range = 21.7 × 2.55 [1] = 55 m [1].

6. Work done = mgh = 300 × 9.81 × 12 [1] = 35 316 J [1]; power = 35 316 ÷ 40 = 880 W [1].

7. (a) Horizontal motion is at constant velocity (no horizontal force) [1]: distance = 15 × 2.0 = 30 m [1]. (b) Vertical motion starts from rest and accelerates at g [1]: v = u + at = 0 + 9.81 × 2.0 = 19.6 m s⁻¹ (downward) [1].

8. efficiency = useful power output ÷ total power input, so useful output = 0.75 × 800 [1] = 600 W [1].

9. Initially weight exceeds air resistance, so there is a resultant force and the diver accelerates [1]; as speed increases air resistance grows until it equals weight, giving zero resultant force, so by Newton’s first law the diver continues at constant velocity [1].

10. Mass is the amount of matter in an object, measured in kg, and is a scalar [1]. Weight is the force of gravity acting on that mass (W = mg), measured in N, and is a vector [1].

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Where marks are usually lost

• Forgetting momentum is a vector — take direction into account.
• Assuming kinetic energy is conserved in all collisions.
• Using the full launch speed for horizontal range instead of the horizontal component.
• Confusing work done with power.
• Using the launch/thrown speed for both horizontal and vertical motion in a projectile — the two components are independent.
• Applying total power input where useful (output) power is asked for, or vice versa, in an efficiency calculation.
• Confusing mass (a scalar, in kg) with weight (a vector force, in N).