AS Physics: Dynamics, Newton Laws and Momentum — 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: Dynamics revision notes

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Section A

1. State Newton’s second law in its general form, and say when F = ma is valid. [3]

2. Give the two conditions that must be satisfied by a Newton’s third law force pair. [2]

3. A book rests on a table. Explain why the book’s weight and the normal contact force are not a third law pair. [2]

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Section B

4. A car of mass 1200 kg travelling at 15 m s⁻¹ collides with a stationary van of mass 1800 kg. They lock together.

(a) Calculate their common velocity after the collision. [3]

(b) Calculate the kinetic energy before and after, and state what has happened to the difference. [4]

(c) State whether the collision is elastic or inelastic, with a reason. [1]

5. A tennis ball of mass 0.058 kg travelling at 25 m s⁻¹ is struck and returns along the same line at 30 m s⁻¹. The contact time is 6.0 ms.

(a) Calculate the change in momentum, taking the initial direction as positive. [3]

(b) Calculate the average force exerted on the ball. [2]

(c) The strings are loosened so the contact time increases to 9.0 ms for the same speeds. Calculate the new average force and comment on the result. [3]

6. Explain, in terms of momentum, how a car’s crumple zone reduces injury in a collision. [3]

7. A skydiver jumps from a stationary balloon and falls with air resistance acting.

(a) Explain why the skydiver’s acceleration decreases as their speed increases. [3]

(b) Explain what is meant by terminal velocity, and state the resultant force at that point. [2]

8. State the test used to determine whether a collision is elastic, in terms of relative speeds.

(a) A 2.0 kg trolley moving at 3.0 m s⁻¹ collides with a stationary 1.0 kg trolley and they stick together. Calculate their common velocity. [3]

(b) Calculate the kinetic energy before and after, and state whether the collision is elastic. [3]

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9. A ball of mass 0.050 kg hits the ground vertically at 6.0 m s⁻¹ and rebounds vertically at 4.0 m s⁻¹. Calculate the magnitude of the change in momentum of the ball. [2]

Answers

1. The rate of change of momentum is proportional to the resultant force and occurs in the direction of that force [1] [1]. F = ma is valid only when the mass is constant [1].

2. They act on two different bodies [1]; they are of the same type (both gravitational, both contact, etc.) [1].

3. They act on the same body — the book [1]. They are also not the same type: one is gravitational, the other a contact force [1]. The third-law partner of the book’s weight is the gravitational pull of the book on the Earth.

4. (a) Momentum before = 1200 × 15 = 18 000 kg m s⁻¹ [1]. Total mass after = 3000 kg [1]. v = 18 000 ÷ 3000 = 6.0 m s⁻¹ [1].

(b) Before: ½ × 1200 × 15² = 135 000 J [1]. After: ½ × 3000 × 6.0² = 54 000 J [1]. Difference = 81 000 J [1], transferred to thermal energy, sound and work done deforming the vehicles [1].

(c) Inelastic [1] — kinetic energy is not conserved.

5. (a) Initial momentum = 0.058 × 25 = +1.45 kg m s⁻¹. Final momentum = 0.058 × (−30) = −1.74 kg m s⁻¹ [1 for using a negative]. Δp = −1.74 − 1.45 [1] = −3.19 kg m s⁻¹ (magnitude 3.19) [1]. Treating both velocities as positive gives 0.29 — a very common error.

(b) F = Δp ÷ Δt = −3.19 ÷ 0.0060 [1] = −532 N [1] (magnitude 532 N, direction opposite to the ball’s initial motion, under the sign convention from part (a)).

(c) F = −3.19 ÷ 0.0090 [1] = −354 N [1] (magnitude 354 N, same direction as part (b)). The same momentum change spread over a longer time gives a smaller force [1].

6. The crumple zone increases the time over which the momentum change occurs [1]. Since force is the rate of change of momentum [1], a longer time for the same momentum change gives a smaller force on the occupants [1]. For the standard comparison (same collision speeds, with vs without the safety feature), the momentum change is unchanged and only the time varies — though real crumple zones can also affect the final speed itself, not just the time. Answers saying the crumple zone “absorbs the force” score nothing.

7. (a) As speed increases, air resistance (drag) increases [1], so the resultant force decreases [1], and since a = F/m (constant mass), the acceleration decreases [1].

(b) Terminal velocity is the constant speed reached when drag equals weight [1], so the resultant force is zero [1].

8. In an elastic collision, the relative speed of approach equals the relative speed of separation.

(a) m₁u₁ + m₂u₂ = (m₁ + m₂)v: (2.0 × 3.0) + (1.0 × 0) = (2.0 + 1.0)v [1]; 6.0 = 3.0v [1]; v = 2.0 m s⁻¹ [1].

(b) KE before = ½ × 2.0 × 3.0² = 9.0 J [1]. KE after = ½ × 3.0 × 2.0² = 6.0 J [1]. Kinetic energy is not conserved (9.0 J → 6.0 J), so the collision is inelastic [1].

9. Taking upwards as positive, the velocity changes from −6.0 m s⁻¹ to +4.0 m s⁻¹, a change of 10.0 m s⁻¹ [1]. Change in momentum = 0.050 × 10.0 = 0.50 kg m s⁻¹ (N s) [1].

Examiner insight (June 2025): many candidates ignored the change in direction and subtracted the speeds (giving 0.10 kg m s⁻¹) instead of adding them. Try the real question next: Cambridge International AS & A Level Physics 9702, June 2025, Paper 22, Question 4.

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Where marks are usually lost

• Quoting F = ma as Newton’s second law without the momentum form.
• Ignoring the vector nature of momentum in a rebound calculation.
• Saying kinetic energy is conserved in all collisions.
• Explaining safety features as “absorbing force” rather than extending time.
• Identifying weight and normal contact force as a third-law pair.