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(A) Let $u$ be the velocity of the hail stone before impact and $v$ be the velocity after impact.Since the surface is smooth,there is no impulsive for

Vedclass · Accepted Answer

$e=\frac{1}{3}$

Vedclass · Answer

(A) Let $u$ be the velocity of the hail stone before impact and $v$ be the velocity after impact.Since the surface is smooth,there is no impulsive force parallel to the surface. Thus,the component of velocity parallel to the surface remains unchanged:$u \sin 30^{\circ} = v \sin 60^{\circ}$  $(1)$For the component normal to the surface,the coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach:$v \cos 60^{\circ} = e (u \cos 30^{\circ})$  $(2)$Dividing equation $(2)$ by equation $(1)$:$\frac{v \cos 60^{\circ}}{v \sin 60^{\circ}} = \frac{e u \cos 30^{\circ}}{u \sin 30^{\circ}}$$\cot 60^{\circ} = e \cot 30^{\circ}$Since $\cot 60^{\circ} = \frac{1}{\sqrt{3}}$ and $\cot 30^{\circ} = \sqrt{3}$:$\frac{1}{\sqrt{3}} = e (\sqrt{3})$$e = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}$

Hail stones are observed to strike the surface of a frozen lake at $30^{\circ}$ with the vertical and rebound at $60^{\circ}$ with the vertical. Assuming the contact to be smooth,the coefficient of restitution is

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