A ball approaches a smooth horizontal floor with velocity components vx = 6 m/s and vy = −8 m/s. The coefficient of restitution with the floor is e = 1/2. If its speed immediately after collision is 2√N m/s, find N.
Physics · Center of maas , Momentum and Collision · collision and types of collision
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Question

A ball approaches a smooth horizontal floor with velocity components vx = 6 m/s and vy = −8 m/s. The coefficient of restitution with the floor is e = 1/2. If its speed immediately after collision is 2√N m/s, find N.

Physics · Center of maas , Momentum and Collision · collision and types of collision

G1 Physics · Accepted Answer

The floor is smooth, so the horizontal component remains unchanged: vx(after) = 6 m/s. The vertical component reverses and is multiplied by e: vy(after) = −e·vy(before) = −(1/2)(−8) = +4 m/s. Therefore final speed: v = √(6² + 4²) = √52 = 2√13 m/s. Hence N = 13.