Question 2.6.6: A ball of mass 0.250 kg moves on a frictionless horizontal f...

A ball of mass 0.250 kg moves on a frictionless horizontal floor and hits a vertical wall with speed 5.0 m s^{-1}. The ball rebounds with speed 4.0 m s^{-1}. If the ball was in contact with the wall for 0.150 s, find the average force that acted on the ball.

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The magnitude of the change in the ball’s momentum is (remember that momentum is a vector)

\Delta p=p_{t}-p_{i}

= 0.250 × 4.0 – (-0.250 × 5.0)

= 1.0 – (-1.25)

= 2.25 Ns

Hence

\bar{F}=\frac{\Delta p}{\Delta t}

 

=\frac{2.25 Ns }{0.150 s }

= 15 N

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