Asked by sam

a 1.2 kg ball drops vertically onto the floor, hitting with a speed of 25m/s. it rebounds with an initial speed of 10m/s. a) what impulse acts on the ball during the contact? b) if the ball is in contact with the floor for 0.20 s, what is the magnitude of the average force on the floor from the ball?

Answers

Answered by Damon
momentum down = 1.2 *25
momentum up = 1.2 * 10
change of momentum = 1.2 (25 + 10)
impulse is change of momentum

Force = change in momentum/change in time
= answer to part a / .2
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