A 0.15kg steel ball is dropped onto a steel plate where its speedjust before imp

Question

A 0.15kg steel ball is dropped onto a steel plate where its speedjust before impact and after impact is 4.5 m/s and 4.2 m/s,respectively. If the ball is in contact with the plate for 0.03seconds, what is the magnitude of the average force (in N) appliedby the plate on the ball?

Explanation / Answer

The method above that the poster described isnt probably what youwant to use. Those kinematics equations assume constantacceleration, but that is probably not the case here. Use impulseand momentum. We can use momentum here. Impulse is defined as two ways => J = P J = MV2 - MV1 Take the direction of V1 (downward since it is dropped) to bepositive, thus V2 must be negative ( in the opposite direction) => J = (.15kg)(-4.2m/s) - (.15)(4.5m/s) = -1.305 kg * m/s Impulse is also defined as => J = Favg T -1.305 kg * m/s = Favg x .03 seconds Favg = -43,5 N, which means the force opposes the initial directionof velocity, so the average force is 43.5N upward

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