Momentum, conservation of momentum and Newton's third law (P) – IGCSE Physics exam questions with mark scheme

1Grade 6–7Triple only5 marks

A trolley A of mass 0.80 kg moves at 1.5 m/s towards a stationary trolley B of mass 0.40 kg.

The trolleys collide and stick together.

(a) Calculate the momentum of trolley A before the collision. [2]

(b) Calculate the velocity of the two trolleys immediately after the collision. [3]

(a) $p = mv = 0.80 \times 1.5$(1)
(a) 1.2 kg m/s allow N s; condone missing unit; wrong unit loses this mark(1)
(b) (total) momentum before = (total) momentum after or conservation of momentum used(1)
(b) $1.2 = (0.80 + 0.40) \times v$ allow ecf from (a)(1)
(b) 1.0 m/s(1)
Answer: (a) 1.2 kg m/s (b) 1.0 m/s
Examiner tip: After sticking together the trolleys move as one object — use the combined mass 1.20 kg, not 0.40 kg.

2Grade 6–7Triple only5 marks

A footballer kicks a stationary ball of mass 0.45 kg. The ball leaves the boot at 24 m/s.

The boot is in contact with the ball for 0.012 s.

(a) Calculate the average force exerted on the ball. [3]

(b) A goalkeeper catches the ball. As she catches it, she moves her hands backwards.

Explain why this reduces the force on her hands. [2]

(a) change in momentum $= 0.45 \times 24$ (= 10.8 kg m/s)(1)
(a) $F = \dfrac{10.8}{0.012}$ or $F = \dfrac{(mv - mu)}{t}$ with values substituted(1)
(a) 900 N(1)
(b) (moving hands back) increases the time taken for the ball to stop / for the momentum to change(1)
(b) (same change in momentum) so the rate of change of momentum is smaller, so the force is smaller allow F = Δp ÷ t with Δp unchanged and t larger; ignore 'absorbs the force / impact'(1)
Answer: (a) 900 N (b) longer stopping time → smaller rate of change of momentum → smaller force
Examiner tip: Saying the hands 'absorb the impact' scores nothing — link a longer time to a smaller rate of change of momentum.

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