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Question

A 140g ball with speed 7.8m/s strikes a wall perpendicularly and rebounds in the opposite direction with the same speed. The collision lasts 3.80ms. What are the magnitudes of the (a) impulse and (b) average force on the wall from the ball during the elastic collision?

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Solution

(a) The magnitude of the impulse is equal to the change in momentum .
J=mνm(ν)=2mν=2(0.140kg)(7.80m/s)=2.18kgm/s
(b) Since in the calculus sense the average of a function is the integral of it divided by the corresponding interval, then the average force is the impulse divided by the time Δt .Thus, our result for the magnitude of the average force is 2mν/Δt With the given values, we obtain
Favg=2(0.140kg)(7.80m/s)0.00380s=575N

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