Answers · Momentum and collisions

Momentum and collisions — return questions key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. CR1 · Return

Model answer. (a) Momentum = mass × velocity, p = mv. (b) m = p ÷ v = 7200 kg m/s ÷ 2.4 m/s = 3000 kg. (c) Momentum is conserved: 7200 kg m/s = (3000 + 1800) kg × v, so v = 7200 ÷ 4800 = 1.5 m/s. (d) No resultant force from outside acts on the two trucks during the collision.

A full answer includes

(a) I wrote: momentum is mass × velocity.
(b) I wrote m = 7200 kg m/s ÷ 2.4 m/s.
(b) I wrote 3000 kg.
(c) I wrote 4800 kg × v = 7200 kg m/s.
(c) I wrote 1.5 m/s.
(c) I did not use 3000 kg as the mass after.
(d) I wrote: no resultant force from outside acts on the trucks.

Also accept

(a) mass × velocity (b) 3000 kg (c) 1.5 m/s (d) no resultant force from outside

Watch for

“v = 2.4 m/s, because the momentum is conserved.” — The momentum is conserved, not the velocity. A bigger mass shares it, so the trucks slow down.
“m = 7200 × 2.4 kg.” — p = mv, so m = p ÷ v.

2. CR2 · Return

Model answer. (a) Impulse is force × the time for which the force acts, F∆t. (b) Impulse = ∆(mv) = 0.17 kg × 2.0 m/s − 0 = 0.34 N s. (c) Resultant force is the change in momentum per unit time, F = ∆p ÷ ∆t. (d) 4.0 ms = 0.0040 s, so F = 0.34 N s ÷ 0.0040 s = 85 N.

A full answer includes

(a) I wrote: force × the time for which it acts.
(b) I wrote impulse = 0.17 kg × 2.0 m/s.
(b) I wrote 0.34 N s.
(c) I wrote: the change in momentum per unit time.
(d) I changed 4.0 ms to 0.0040 s.
(d) I wrote F = 0.34 ÷ 0.0040.
(d) I wrote 85 N.
(d) I did not divide by 4.0.

Also accept

(a) force × time (b) 0.34 N s (c) change in momentum per unit time (d) 85 N

Watch for

“Impulse is the force of the hit.” — Impulse is force × time, measured in N s.
“F = 0.34 ÷ 4.0 = 0.085 N.” — Change 4.0 ms to 0.0040 s before you divide.