A sphere strikes a wall and rebounds with a c | vedclass

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(D) Let the velocity of the sphere before impact be $u$ and after impact be $v = 0.1 \ m/s$. The angle of rebound with the normal is $	heta = 60^{\ci

Vedclass · Accepted Answer

$66 \frac{2}{3}\%$

Vedclass · Answer

(D) Let the velocity of the sphere before impact be $u$ and after impact be $v = 0.1 \ m/s$. The angle of rebound with the normal is $	heta = 60^{\circ}$.Since the wall is smooth,the component of velocity parallel to the wall remains unchanged: $u \sin 	heta = v \sin 60^{\circ}$.The component of velocity perpendicular to the wall changes according to the coefficient of restitution $e$: $v \cos 60^{\circ} = e (u \cos 	heta)$.Given $e = 1/3$,$v = 0.1$,and $	heta = 60^{\circ}$:$v \sin 60^{\circ} = u \sin 	heta \implies u \sin 	heta = 0.1 	imes \frac{\sqrt{3}}{2} = 0.05\sqrt{3}$.$v \cos 60^{\circ} = e (u \cos 	heta) \implies 0.1 	imes \frac{1}{2} = \frac{1}{3} (u \cos 	heta) \implies u \cos 	heta = 0.15$.Initial kinetic energy $K_i = \frac{1}{2} m u^2 = \frac{1}{2} m ((u \sin 	heta)^2 + (u \cos 	heta)^2) = \frac{1}{2} m ((0.05\sqrt{3})^2 + (0.15)^2) = \frac{1}{2} m (0.0075 + 0.0225) = \frac{1}{2} m (0.03)$.Final kinetic energy $K_f = \frac{1}{2} m v^2 = \frac{1}{2} m (0.1)^2 = \frac{1}{2} m (0.01)$.Loss in kinetic energy $\Delta K = K_i - K_f = \frac{1}{2} m (0.03 - 0.01) = \frac{1}{2} m (0.02)$.Percentage loss $= \frac{\Delta K}{K_i} 	imes 100 = \frac{0.02}{0.03} 	imes 100 = \frac{2}{3} 	imes 100 = 66 \frac{2}{3}\%$.

$A$ sphere strikes a wall and rebounds with a coefficient of restitution $e = 1/3$. If it rebounds with a velocity of $0.1 \ m/s$ at an angle of $60^{\circ}$ to the normal to the wall,the percentage loss of kinetic energy is:

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