Answers · Momentum and collisions

Momentum and collisions — final check key

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

1. CK1 · Final check

Model answer. (a) Momentum = mass × velocity. (b) p = 0.40 kg × 3.0 m/s = 1.2 kg m/s. (c) 1.2 kg m/s = (0.40 + 0.20) kg × v, so v = 2.0 m/s to the right. (d) Take right as positive: 1.2 = 0.40 × (−0.20) + 0.80 v = −0.080 + 0.80 v, so 0.80 v = 1.28 and v = +1.6 m/s, to the right. (e) The principle of conservation of momentum: when no resultant force from outside acts, the total momentum before a collision equals the total momentum after it.

A full answer includes

(a) I wrote: momentum is mass × velocity.
(b) I wrote p = 0.40 kg × 3.0 m/s.
(b) I wrote 1.2 kg m/s.
(c) I wrote 0.60 kg × v = 1.2 kg m/s.
(c) I wrote 2.0 m/s to the right.
(d) I wrote the bounce momentum as 0.40 × (−0.20) = −0.080 kg m/s.
(d) I wrote 1.2 = −0.080 + 0.80 v.
(d) I wrote 1.6 m/s to the right.
(d) I did not write 1.4 m/s.
(e) I named the conservation of momentum.
(e) I wrote that it holds when no resultant force from outside acts.

Also accept

(a) mass × velocity (b) 1.2 kg m/s (c) 2.0 m/s (d) 1.6 m/s to the right (e) conservation of momentum

Watch for

“In (d), 1.2 = 0.080 + 0.80 v, so v = 1.4 m/s.” — The bounce is to the left, so that momentum is −0.080 kg m/s.

2. CK2 · Final check

Model answer. (a) Impulse = force × the time for which it acts. (b) ∆p = 1400 kg × 15 m/s = 21 000 kg m/s. (c) F = ∆p ÷ ∆t = 21 000 ÷ 0.14 = 150 000 N. (d) F = 21 000 ÷ 0.050 = 420 000 N, 2.8 times bigger. (e) The resultant force is the change in momentum per unit time. The car’s change in momentum is fixed, so making the stop take longer makes the force smaller, and the people inside feel a smaller force too.

A full answer includes

(a) I wrote: force × the time for which it acts.
(b) I wrote ∆p = 1400 kg × 15 m/s.
(b) I wrote 21 000 kg m/s.
(c) I wrote F = 21 000 ÷ 0.14.
(c) I wrote 150 000 N.
(d) I wrote F = 21 000 ÷ 0.050.
(d) I wrote 420 000 N.
(e) I defined resultant force as the change in momentum per unit time.
(e) I wrote that the car’s change in momentum is the same with or without the crumple zone.
(e) I wrote that a longer stopping time gives a smaller force on the car and the people in it.
(e) I did not write that the crumple zone reduces the change in momentum.

Also accept

(a) force × time (b) 21 000 kg m/s (c) 150 000 N (d) 420 000 N (e) longer time, smaller rate of change of momentum, smaller force

Watch for

“The crumple zone reduces the car’s change in momentum.” — The car stops from 15 m/s either way; the crumple zone makes the stop take longer.