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A boy is playing carom board. Striker is moving with velocity $u (= 10 m / s)$ and queen $Q$ is at rest. See angle $θ (= 37^{O})$ in the figure. Mass of S and Q are m and 2m. Friction is absent and coefficient of restitution is e. If after collision Q moves along SQ direction and S moves along perpendicular to SQ. Then the value of coefficient of restitution is

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Detailed Solution

Impulsive force N an Q acts along the direction of line of impact is towards SQ. So

$e v = (v_{2} - v_{1}) / u \cos θ$

$v_{2} - 0 = 8 e \Rightarrow v_{2} = 8 e$

Conservation of linear momentum is SQ direction gives: $m u \cos θ + 2 m (0) = m v_{1} + 2 m v_{2}$

$∴ 8 = 2 v_{2} \Rightarrow v_{2} = 4 m / s$

From Eqs. (i) and (ii), e=1/2

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