A2 Daily A Level Physics question
On a low-friction track, a 0.50 kg ball moving at +6.0 m s⁻¹ collides head-on with a 1.50 kg cart initially at rest. Two trials are run with identical setup: Trial 1 the ball sticks to the cart; Trial 2 the ball bounces back in a nearly perfectly elastic collision. A force sensor on the cart records the magnitude of the impulse it receives during impact. Which statement must be true about the sensor readings, and why?
Answer
The correct answer is D.
Correct: D — The elastic bounce gives the larger impulse on the cart, about twice that for sticking (roughly 4.5 N s vs 2.25 N s), because the ball reverses direction. In Trial 1 (stick), the common speed is (0.50/(0.50+1.50))×6.0 = 1.5 m s⁻¹, so the cart’s impulse is 1.50×1.5 = 2.25 N s. In Trial 2 (nearly elastic), the cart’s speed is (2×0.50/(0.50+1.50))×6.0 = 3.0 m s⁻¹, giving 1.50×3.0 = 4.5 N s, i.e. about double; the key is the ball’s reversal of direction, which increases its momentum change and thus the equal-and-opposite impulse on the cart. A Momentum is conserved for the pair, but that does not make the individual impulses identical; the ball reversing in Trial 2 gives a larger momentum change (and impulse magnitude) than in sticking. B Sticking does not give the larger impulse here; the quoted numbers and ordering are wrong—sticking gives about 2.25 N s, elastic about 4.5 N s. C ‘No kinetic energy lost’ in an elastic bounce does not imply a smaller impulse; in fact, reversal of the ball means a larger change in momentum and thus a larger impulse on the cart. D Matches both the direction-change reasoning and the calculated values; as a limiting check, for a very heavy cart the elastic case tends to about twice the sticking impulse.