Momentum: 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: Momentum revision notes

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Section A

1. Define momentum and state its unit. [2]

2. State the principle of conservation of momentum, including the condition under which it applies. [2]

Section B

3. A 0.16 kg ball travelling at 25 m s⁻¹ is struck and returns along the same line at 30 m s⁻¹.

(a) Calculate the change in momentum. [3] (b) The contact lasts 8.0 ms. Calculate the average force on the ball. [2] (c) Explain why the change in momentum is greater than 0.16 × 30. [2]

4. A 1500 kg van moving at 12 m s⁻¹ collides with a stationary 900 kg car. They move off together.

(a) Calculate their common velocity. [3] (b) Calculate the kinetic energy before and after, and state the type of collision. [4] (c) Explain where the “lost” energy has gone. [2]

5. A firework of mass 0.80 kg at rest explodes into two pieces. One piece of mass 0.30 kg moves left at 24 m s⁻¹. Calculate the velocity of the other piece. [4]

6. Explain, using momentum, how each of the following reduces injury in a crash: crumple zone, seat belt, air bag. [6]

Section C

7. A 2 kg trolley moving at 3 m/s collides with a stationary 1 kg trolley and they stick together. Calculate their common velocity. [3]

8. A 5 kg object at rest explodes into two parts. A 2 kg piece moves off at 6 m/s in one direction. Find the velocity of the remaining 3 kg piece, explaining the significance of the sign in your answer. [4]

9. State Newton’s second law in terms of momentum, and show that it is equivalent to F = ma for a constant mass. [3]

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Answers

1. Momentum = mass × velocity [1]; the unit is kg m s⁻¹ (or N s) [1]. It is a vector.

2. In a closed system with no external resultant force, the total momentum before an interaction equals the total momentum after it [1] [1].

3. (a) Taking the initial direction as positive: initial momentum = 0.16 × 25 = 4.0 kg m s⁻¹; final = 0.16 × (−30) = −4.8 kg m s⁻¹ [1] [1]; change = −4.8 − 4.0 = −8.8 kg m s⁻¹ (magnitude 8.8) [1]. (b) F = Δp ÷ t = 8.8 ÷ 0.0080 [1] = 1100 N [1]. (c) The ball’s direction reverses, so its velocity changes by 55 m s⁻¹, not 30 [1]; momentum is a vector, so the initial and final momenta have opposite signs and their magnitudes add [1].

4. (a) Momentum before = 1500 × 12 = 18 000 kg m s⁻¹ [1]; total mass after = 2400 kg [1]; v = 18 000 ÷ 2400 = 7.5 m s⁻¹ [1]. (b) KE before = ½ × 1500 × 12² = 108 000 J [1]. KE after = ½ × 2400 × 7.5² = 67 500 J [1]. Kinetic energy has fallen [1], so the collision is inelastic [1]. (c) It has been transferred to the internal (thermal) energy of the vehicles as they deform, and to sound [1] [1] — the deformation is permanent, so the energy is not recovered.

5. Total momentum before = 0, since the firework is at rest [1]. Momentum of first piece = 0.30 × (−24) = −7.2 kg m s⁻¹ [1]. So the second piece must have momentum +7.2 kg m s⁻¹ [1]; its mass is 0.50 kg, so v = 7.2 ÷ 0.50 = 14.4 m s⁻¹ in the opposite direction [1].

6. In each case the change in momentum is fixed — the occupant must be brought to rest [1]. Since force = change in momentum ÷ time, increasing the time taken reduces the force on the occupant [1]. A crumple zone deforms, so the car takes longer to stop [1]. A seat belt stretches slightly, extending the time over which the wearer decelerates and spreading the force over a larger area of the body [1] [1]. An air bag inflates and then deflates as the head presses into it, again extending the stopping time and spreading the force over a wider area [1].

7. momentum before = momentum after: (2 × 3) + (1 × 0) = (2 + 1) × v [1]; 6 = 3v [1]; v = 2 m/s [1].

8. momentum before = 0, since the object is at rest [1]. momentum after = (2 × 6) + (3 × v) = 0 [1], so 3v = −12, v = −4 m/s [1]. The negative sign shows the two pieces move in opposite directions, which is required for their total momentum to remain zero [1].

9. F = Δp ÷ Δt [1]. For constant mass, Δp = Δ(mv) = mΔv, so F = mΔv/Δt = ma [1] — the momentum form and F = ma describe the same law, but the momentum form also covers situations where mass itself changes [1].

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Where marks are usually lost

• Ignoring direction when a body rebounds.
• Assuming kinetic energy is conserved in a collision.
• Forgetting that total momentum before an explosion is zero.
• Explaining safety features by “absorbing the force” rather than extending the time.