A 2 kg ball strikes a horizontal floor vertically downward with speed 8 m/s. The coefficient of restitution between the ball and floor is 1/2. Neglect the impulse due to gravity during the short collision. If the impulse delivered by the floor is N N·s, find N.
Physics · Center of maas , Momentum and Collision · impuse and question based on it
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Question

A 2 kg ball strikes a horizontal floor vertically downward with speed 8 m/s. The coefficient of restitution between the ball and floor is 1/2. Neglect the impulse due to gravity during the short collision. If the impulse delivered by the floor is N N·s, find N.

Physics · Center of maas , Momentum and Collision · impuse and question based on it

G1 Physics · Accepted Answer

The rebound speed is e times the approach speed: v = (1/2)×8 = 4 m/s upward. Take upward as positive. Then initial velocity u = −8 m/s and final velocity v = +4 m/s. Therefore change in momentum is Δp = m(v−u) = 2[4−(−8)] = 24 kg·m/s. Since gravity's impulse is neglected during the collision, the floor impulse equals this change in momentum. Hence N = 24.