A body dropped freely from a height h on to a horizontal plane, bounces up and down and finally comes to rest.The coefficient of restitution is e. The ratio of velocities at the beginning and after two rebounds is

1 : e

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e : 1

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1 : e^{3}

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e^{2} : 1

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Solution

The correct option is D $e^{2} : 1$
Let initial velocity is v at time of collision. $v = \sqrt{2 g h}$
after first re bound velocity $v_{1} = e v$
after second rebound velocity $v_{2} = e v_{1} = e^{2} v$
ratio $= \frac{v_{2}}{v} = \frac{e^{2} v}{v}$
$= e^{2} : 1$

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A body dropped freely from a height h on to a horizontal plane, bounces up and down and finally comes to rest.The coefficient of restitution is e. The ratio of velocities at the beginning and after two rebounds is

The correct option is D e2:1Let initial velocity is v at time of collision. v=√2ghafter first re bound velocity v1=evafter second rebound velocity v2=ev1=e2vratio =v2v=e2vv=e2:1

The correct option is D $e^{2} : 1$
Let initial velocity is v at time of collision. $v = \sqrt{2 g h}$
after first re bound velocity $v_{1} = e v$
after second rebound velocity $v_{2} = e v_{1} = e^{2} v$
ratio $= \frac{v_{2}}{v} = \frac{e^{2} v}{v}$
$= e^{2} : 1$