A ball dropped from 1.8 m onto a floor bounces back up to 1.15 m…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A ball dropped from $1.8\ \text{m}$ onto a floor bounces back up to $1.15\ \text{m}$ . Taking $g = 9.8\ \text{m/s}^2$ , find

its speed just before it hits the floor;
its speed just after it leaves the floor;
the percentage of its mechanical energy lost in the bounce.

Need a reminder? The key ideaShow

When only conservative forces (gravity, springs) do work, mechanical energy is conserved: $K_1 + U_1 = K_2 + U_2$ . Forces that do no work, like a smooth surface's push or a string at right angles to the motion, do not spoil it.

The coefficient of restitution is $e = \dfrac{\text{speed of separation}}{\text{speed of approach}}$ . A ball bouncing on a floor keeps the fraction $e$ of its speed and $e^2$ of its height each bounce, so $e = \sqrt{\dfrac{h_1}{h_0}}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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