Question

a crane drops a .3 kg steel ball onto a steel plate. the ball's speeds just before impact and after are 4.5 m/s and 4.2 m/s respectively. If the ball is in contact with the plate for .03 s, what is the magnitude of the average force that the ball exerts on the plate during impact?

Answers

Presumably the steel ball bounced and changed directions. The momentum change is therefore 0.3 kg*(4.5 + 4.2 kg.s) = 2.61 kg m/s

Set that equal to the impulse (Force * time) and divide by the time interval to get the average force.
87 N
87

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