A 1.0 kg ball drops vertically onto a floor | vedclass

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(C) The velocity of the ball just before the collision is given by $u = \sqrt{2gh}$,where $h = 25 \ cm = 0.25 \ m$.$u = \sqrt{2 	imes 9.8 	imes 0.25

Vedclass · Accepted Answer

$0.40$

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(C) The velocity of the ball just before the collision is given by $u = \sqrt{2gh}$,where $h = 25 \ cm = 0.25 \ m$.$u = \sqrt{2 	imes 9.8 	imes 0.25} = \sqrt{4.9} \ m/s$.The velocity of the ball just after the collision is given by $v = \sqrt{2gh'}$,where $h' = 4 \ cm = 0.04 \ m$.$v = \sqrt{2 	imes 9.8 	imes 0.04} = \sqrt{0.784} \ m/s$.The coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach: $e = \frac{v}{u}$.$e = \sqrt{\frac{2gh'}{2gh}} = \sqrt{\frac{h'}{h}}$.Substituting the values: $e = \sqrt{\frac{4 \ cm}{25 \ cm}} = \sqrt{\frac{4}{25}} = \frac{2}{5} = 0.4$.

$A$ $1.0 \ kg$ ball drops vertically onto a floor from a height of $25 \ cm$. It rebounds to a height of $4 \ cm$. The coefficient of restitution for the collision is

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