Question

A 3 kg steel ball falls down vertically and strikes a floor with a speed of 12 m/s. It bounces off upward with the same speed. If the ball is in contact with the floor for 0.12 s, what is the magnitude of the average force exerted on the ball by the floor?

Answer 1

Answer:

600N

Explanation:

If the ball bounces off upward with the same speed, the velocities before and after are equal and in opposite directions. That means the floor must have caused a momentum impact to change its direction.

By the law of momentum conservation:

$mv + \Delta P_f = mV$

where m = 3 kg is the ball mass. v = -12 m/s is the velocity right before the impact. V = 12m/s is the velocity right after the impact. ΔP_f is the momentum caused by the floor.

$\Delta P_f = m(V-v) = 3(12 - (-12)) = 3*24 = 72kgm/s$

Then the average force exerted during the 0.12s impact is

$F = \frac{\Delta P_f}{\Delta t} = \frac{72}{0.12} = 600N$