Question #154176

A 500 g rubber ball is thrown to the right at a wall at a speed of 5.00 m / s. The ball bounces to the left at a speed of 4.50 m / s. If the collision between the ball and the wall lasted 0.250 s, what is the force exerted by the wall on the ball?


1
Expert's answer
2021-01-10T18:29:21-0500

F=ΔpΔt=p2p1Δt=mv2(mv1)Δt=m(v2+v1)Δt=0.59.50.25=19 N.F=\frac{\Delta p}{\Delta t}=\frac{p_2-p_1}{\Delta t}=\frac{mv_2-(-mv_1)}{\Delta t}=\frac{m(v_2+v_1)}{\Delta t}=\frac{0.5\cdot 9.5}{0.25}=19~N.


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