A 0.400 kg trolley moves right at 1.50 m s⁻¹ before striking a buffer. The measured horizontal force on it has coordinates (0 ms, 0 N), (10 ms, −60 N), (25 ms, −60 N), and (45 ms, 0 N), joined by straight lines. Determine the impulse, final velocity and kinetic-energy change. Explain whether momentum of the trolley alone is conserved, identify where the missing momentum goes, and evaluate one limitation of integrating sampled force data.

Question

Accepted Answer

Impulse = −1.80 N s, final velocity = −3.00 m s⁻¹, and ΔK = +1.35 J. Trolley momentum is not conserved; the buffer and Earth receive the opposite impulse. Impulse is signed area under F–t. The first triangle is −½(0.010)(60) = −0.300 N s, the rectangle is −0.900 N s, and the final triangle is −0.600 N s. Total impulse is −1.80 N s. Initial momentum is (0.400)(1.50) = +0.600 kg m s⁻¹, so final momentum is −1.20 kg m s⁻¹. Final velocity is p_f/m = −1.20/0.400 = −3.00 m s⁻¹. Kinetic energy changes from 0.450 J to 1.80 J, so ΔK = +1.35 J. The trolley alone experiences an external impulse from the buffer and does not conserve momentum. In a larger trolley–buffer–Earth system, the buffer and Earth gain +1.80 N s, so total momentum remains conserved. Straight-line integration assumes no narrow force peak occurs between samples. A higher sampling rate and calibrated force zero reduce missing-area and offset errors; repeating impacts quantifies random variation.

Time / ms	Force / N
0	0
10	-60
25	-60
45	0

Impulse and rebound from force data

What a complete response should do

Measured buffer force

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