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(B) Let the velocity of the ball before collision be $v$. The horizontal component is $v_x = v \sin(45^o)$ and the vertical component is $v_y = v \cos

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) Let the velocity of the ball before collision be $v$. The horizontal component is $v_x = v \sin(45^o)$ and the vertical component is $v_y = v \cos(45^o)$.After the collision,the horizontal component remains $v_x' = v_x = v \sin(45^o)$ because the ground is smooth.The vertical component changes to $v_y' = e v_y = e v \cos(45^o)$,where $e$ is the coefficient of restitution $(0 \leq e \leq 1)$.Let the angle after collision with the vertical be $	heta$. Then $	an(	heta) = \frac{v_x'}{v_y'} = \frac{v \sin(45^o)}{e v \cos(45^o)} = \frac{1}{e} 	an(45^o) = \frac{1}{e}$.Since $e \leq 1$,it follows that $\frac{1}{e} \geq 1$.Therefore,$	an(	heta) \geq 1$,which implies $	heta \geq 45^o$.Thus,the angle with the vertical cannot be less than $45^o$. Among the given options,$30^o$ is less than $45^o$,so it is not possible.

$A$ ball strikes a smooth horizontal ground at an angle of $45^o$ with the vertical. What cannot be the possible angle of its velocity with the vertical after the collision? (Assume $e \leq 1$ ).

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