A ball of mass m strikes a smooth ground at a | vedclass

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(A) Since the ground is smooth,there is no impulsive force acting parallel to the ground. Therefore,the component of velocity parallel to the ground r

Vedclass · Accepted Answer

$	an \alpha / 	an \beta$

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(A) Since the ground is smooth,there is no impulsive force acting parallel to the ground. Therefore,the component of velocity parallel to the ground remains constant.$v \sin \beta = u \sin \alpha$  --- $(1)$The coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach along the normal direction.$e = \frac{	ext{velocity of separation along normal}}{	ext{velocity of approach along normal}} = \frac{v \cos \beta}{u \cos \alpha}$  --- $(2)$From equation $(1)$,we have $\frac{v}{u} = \frac{\sin \alpha}{\sin \beta}$.Substituting this into equation $(2)$:$e = \left( \frac{\sin \alpha}{\sin \beta} ight) \left( \frac{\cos \beta}{\cos \alpha} ight)$$e = \frac{\sin \alpha}{\cos \alpha} \cdot \frac{\cos \beta}{\sin \beta}$$e = \frac{	an \alpha}{	an \beta}$

$A$ ball of mass $m$ strikes a smooth ground at an angle $\alpha$ as shown in the figure and is deflected at an angle $\beta$. The coefficient of restitution will be

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