A crane drops a 0.40-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 0.030 s,what is the magnitude of the average force that the ball exerts on the plate during impact?

Question

Quizplus · Accepted Answer

We can use the impulse-momentum theorem to solve this problem. The impulse on the ball is equal to its change in momentum:
Impulse = m(vf - vi)
where m is the mass of the ball, vi and vf are the initial and final velocities, respectively.

Impulse = 0.40 kg(4.2 m/s - 4.5 m/s) = -0.12 Ns

The impulse is also equal to the average force multiplied by the time:
Impulse = F(avg)Δt

Knowing the impulse and the time, we can solve for the average force:
F(avg) = Impulse / Δt = -0.12 Ns / 0.030 s = -4 N

The negative sign indicates that the force is in the opposite direction of the initial velocity. Taking the magnitude of the force, we get:
|F(avg)| = 4 N

Therefore, the magnitude of the average force that the ball exerts on the plate during impact is 4 N, which is closest to answer choice B (116 N).

A Crane Drops a 0