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Question
A ball of mass m is dropped onto a floor from a certain height. The collision is perfectly elastic and the ball rebounds to the same height and again falls. Find the average force exerted by the ball on the floor during a long time interval.
Solution
It is given that the mass of the ball is m.
Let the ball be dropped from a height h.
The speed of ball before the collision is v1.
\[\therefore \text{v}_1 = \sqrt{2\text{gh}}\]
The speed of ball after the collision is v2.
\[\text{v}_2 = - \sqrt{2\text{gh}}\]
\[\text{ Rate of change of velocity = acceleration }\]
\[ \Rightarrow a = \frac{2\sqrt{2gh}}{t}\]
\[ \therefore \text{ Force}, F = \frac{m \times 2\sqrt{2gh}}{t} \ldots(1)\]
Using Newton's laws of motion, we can write:
\[v = \sqrt{2gh}, s = h, u = 0\]
\[ \Rightarrow \sqrt{2gh} = gt\]
\[\Rightarrow t = \sqrt{\frac{2h}{g}}\]
\[ \therefore \text{ Total time }= 2\sqrt{\frac{2h}{g}}\]
Substituting this value of time t in equation (1), we get:
F = mg
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