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Q.

A hardened steel ball, moving with velocity u, strikes a smooth horizontal surface at an angle of $60^{o}$ with the surface. Just after the impact the ball starts moving making an angle of $45^{o}$ with the surface. Then coefficient of restitution is

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a

$\frac{1}{\sqrt{2}}$

b

$\frac{1}{3}$

c

$\frac{1}{2}$

d

$\frac{1}{\sqrt{3}}$

answer is B.

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Detailed Solution

Since the surface is smooth, $V \cos 45^{o} = u \cos 60^{o} \Rightarrow \frac{u}{v} = \sqrt{2}$

$Again e = \frac{V \sin 45^{o} - 0}{0 - (- u \sin 60^{o})} = \frac{v}{u} . \sqrt{\frac{2}{3}}$

$∴ e = \frac{1}{\sqrt{2}} . \sqrt{\frac{2}{3}} = \frac{1}{\sqrt{3}}$

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A hardened steel ball, moving with velocity u, strikes a smooth horizontal surface at an angle of 60o with the surface. Just after the impact the ball starts moving making an angle of 45o with the surface. Then coefficient of restitution is