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(D) The velocity of the ball before collision is $\vec{v}_{initial} = a \hat{i} - b \hat{j}$.Since the surface is smooth,there is no friction,so the h

Vedclass · Accepted Answer

$a \hat{i} + e b \hat{j}$

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(D) The velocity of the ball before collision is $\vec{v}_{initial} = a \hat{i} - b \hat{j}$.Since the surface is smooth,there is no friction,so the horizontal component of velocity remains unchanged: $v_{x, final} = v_{x, initial} = a \hat{i}$.The vertical component of velocity changes due to the collision. According to the definition of the coefficient of restitution $e$,the velocity of separation along the normal is $e$ times the velocity of approach along the normal.$v_{y, final} = e 	imes |v_{y, initial}| = e b$.Since the ball bounces upward,the final vertical velocity is in the positive $y$-direction: $\vec{v}_{y, final} = e b \hat{j}$.Therefore,the velocity of the ball just after the collision is $\vec{v}_{final} = a \hat{i} + e b \hat{j}$.

$A$ glass ball collides with a smooth horizontal surface with a velocity $\vec{v} = a \hat{i} - b \hat{j}$. If the coefficient of restitution of the collision is $e$,find the velocity of the ball just after the collision. ($x$-axis is along the ground and $y$-axis is along the vertical).

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