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

Answer:

F = 30 N  Directed up

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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