A ball of mass m moving with speed u collides | vedclass

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Question

(D) The velocity of the ball before collision can be resolved into two components: horizontal component $u_x = u \cos 	heta$ and vertical component $

Vedclass · Accepted Answer

$m u(1+e) \sin 	heta$

Vedclass · Answer

(D) The velocity of the ball before collision can be resolved into two components: horizontal component $u_x = u \cos 	heta$ and vertical component $u_y = -u \sin 	heta$ (taking downward as negative).Since the surface is smooth,there is no impulse in the horizontal direction,so the horizontal velocity remains unchanged: $v_x = u \cos 	heta$.In the vertical direction,the coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach along the normal: $e = \frac{v_y}{u_y}$.Thus,the vertical velocity after collision is $v_y = e u \sin 	heta$ (upward).The impulse $I$ imparted to the ball by the surface is equal to the change in momentum of the ball: $\vec{I} = m(\vec{v} - \vec{u})$.Since there is no change in horizontal momentum,the impulse is only in the vertical direction: $I = m(v_y - u_y) = m(e u \sin 	heta - (-u \sin 	heta)) = m u(1+e) \sin 	heta$.The magnitude of the impulse imparted to the surface by the ball is equal to the magnitude of the impulse imparted to the ball by the surface,which is $m u(1+e) \sin 	heta$.

$A$ ball of mass $m$ moving with speed $u$ collides with a smooth horizontal surface at an angle $\theta$ with it as shown in the figure. The magnitude of the impulse imparted to the surface by the ball is [Coefficient of restitution of collision is $e$].

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