A tennis ball hits the floor with a speed v a | vedclass

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Question

(A) Let the velocity of the ball before collision be $v$. The velocity components are $v_x = v \sin 	heta$ (parallel to the floor) and $v_y = v \cos

Vedclass · Accepted Answer

$	an ^{-1}\left(\frac{	an 	heta}{\varepsilon}ight)$

Vedclass · Answer

(A) Let the velocity of the ball before collision be $v$. The velocity components are $v_x = v \sin 	heta$ (parallel to the floor) and $v_y = v \cos 	heta$ (perpendicular to the floor).Since the floor is smooth,there is no impulsive force parallel to the floor,so the tangential component of velocity remains unchanged: $v'_x = v \sin 	heta$.For the normal component,the coefficient of restitution $\varepsilon$ is defined as the ratio of the velocity of separation to the velocity of approach along the normal: $\varepsilon = \frac{v'_y}{v_y}$.Thus,$v'_y = \varepsilon v_y = \varepsilon v \cos 	heta$.Let $	heta'$ be the angle of reflection with the normal. Then,$	an 	heta' = \frac{v'_x}{v'_y} = \frac{v \sin 	heta}{\varepsilon v \cos 	heta} = \frac{	an 	heta}{\varepsilon}$.Therefore,$	heta' = 	an ^{-1}\left(\frac{	an 	heta}{\varepsilon}ight)$.

$A$ tennis ball hits the floor with a speed $v$ at an angle $\theta$ with the normal to the floor. If the collision is inelastic and the coefficient of restitution is $\varepsilon$,what will be the angle of reflection?

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