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Question

(A) Let the velocity after collision be $v'$ at an angle $\phi$ with the normal. Along the wall (parallel component),the velocity remains unchanged be

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{	an 	heta}{e}ight), v\sqrt{\sin^2 	heta + e^2 \cos^2 	heta}$

Vedclass · Answer

(A) Let the velocity after collision be $v'$ at an angle $\phi$ with the normal. Along the wall (parallel component),the velocity remains unchanged because there is no impulsive force: $v' \sin \phi = v \sin 	heta$. Along the normal (perpendicular component),the coefficient of restitution $e$ is defined as the ratio of relative velocity of separation to the relative velocity of approach: $v' \cos \phi = e v \cos 	heta$. Dividing the two equations: $\frac{v' \sin \phi}{v' \cos \phi} = \frac{v \sin 	heta}{e v \cos 	heta} \implies 	an \phi = \frac{	an 	heta}{e} \implies \phi = 	an^{-1}\left(\frac{	an 	heta}{e}ight)$. The magnitude of the velocity $v'$ is given by: $v' = \sqrt{(v' \sin \phi)^2 + (v' \cos \phi)^2} = \sqrt{(v \sin 	heta)^2 + (ev \cos 	heta)^2} = v \sqrt{\sin^2 	heta + e^2 \cos^2 	heta}$.

$A$ ball of mass $m$ is thrown at a wall with speed $v$ at an angle $\theta$ with the normal to the wall. If the coefficient of restitution is $e$,what will be the magnitude and direction of the velocity of the ball after the collision with respect to the wall?

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