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You throw a 0.350 kg tennis ball horizontally toward a wall. The speed of the ball...

You throw a 0.350 kg tennis ball horizontally toward a wall. The speed of the ball is 12 m/s before it hits the wall, and it bounces back with the same speed of 12 m/s. If the ball remains in contact with the wall during this collision for 0.025 s, what is the magnitude of the average force exerted on the ball?
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Answer #1

U = -12m/s

V = 12m/s

m = 0.35kg

∆t = 0.025

Favg = ∆P/∆t = m(V-U)/∆t = 0.35(12+12)/0.025 = 336N

=) Favg = 336N

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