Question

Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s and just after is 3.5 m/s. If the ball is in contact with the floor for 0.050 s, what is the magnitude of the average force applied by the floor on the ball?

Answer 1

F = 30 N Directed up

Step-by-step explanation:

We can solve this problem using the momentum and momentum relationship.

I = ΔP

I = ∫ F dt

As we are asked for the average force, that leaves the integral, giving

I = F t

F t = m
`v_(f)` - m v₀

F = m (
`v_(f)`-v₀) / t

Note that, if you define the upward positive direction, the initial velocity is negative

Let's calculate

F = 0.15 (3.5 - (-6.5)) / 0.050

F = 30 N

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