A 1.0 kg ball drops vertically onto a floor, hitting with a speed of v (initial)

ID: 1285974 • Letter: A

Question

A 1.0 kg ball drops vertically onto a floor, hitting with a speed of v (initial) = 24 m/s

It rebounds with a speed of v (final) = -24 m/s. The ball is in contact with the floor for 0.12 s.

(Please provide equations and step by step instruction to help me understand concept)

1) The change in momentum of the ball (in kg m/s) was...

2) The impulse to the ball (in N) was...

3) The magnitude of the average force on the ball was...

4) The magnitude of the maximum force on the ball was approximately...

change in momentum = m*(v1-v2)= 1*(24-(-24)) = 48 kg-mt/sec

impulse = change in momentum = 48 kg-mt/sec

magnitude of the average force on the ball = m* v1 /time = 24/0.12= 200N

magnitude of the maximum force on the ball = impulse/time =48/0.12= 400N

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