A 200-g ball is dropped from rest from a height of 1.4 m. It bounces and...

A 200-g ball is dropped from rest from a height of 1.4 m. It bounces and returns to the same position.

a) What is the change in momentum between the two points?

b) If the ball is in contact with the floor for 10-³ s, what impulse does the floor impart to the ball?

In these cases remember vectors.

Solutions

Expert Solution

Change in momentum equals the average net external force multiplied by the time this force acts. Δp = FnetΔt

The quantity FnetΔt is given the name impulse. Impulse is the same as the change in momentum.

Where Fnet is the force exerted by the ground during time Δt.

If you have any doubt in the solution, you can ask in the comment box.

genius_generous answered 1 year ago

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