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A ball of mass $m$ hits the floor with a speed $v$ making an angle of incidence $\theta$ with the normal. The coefficient of restitution is $e$.
The speed of reflected ball and the angle of reflection of the ball will be
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The speed of reflected ball and the angle of reflection of the ball will be
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Verified Answer
The correct answer is:
$v^{\prime}=v, \theta=\theta^{\prime}$
The parallel component of velocity of the ball remains unchanged. This gives
$v^{\prime} \sin \theta^{\prime}=v \sin \theta$ ...(i)
For the components normal to the floor, the velocity of separation $=v^{\prime} \cos \theta^{\prime}$
and the velocity of approach $=v \cos \theta$
Hence, $\quad v^{\prime} \cos \theta^{\prime}=e v \cos \theta$ ... (ii)
From Eqs. (i) and (ii)
$\begin{aligned} v^{\prime} & =v \sqrt{\sin ^2 \theta+e^2 \cos ^2 \theta} \\ \text { and } \quad \tan \theta^{\prime} & =\frac{\tan \theta}{e}\end{aligned}$
For elastic collision, $e=1$, so that $\theta^{\prime}=\theta$ and $v^{\prime}=v$
$v^{\prime} \sin \theta^{\prime}=v \sin \theta$ ...(i)
For the components normal to the floor, the velocity of separation $=v^{\prime} \cos \theta^{\prime}$
and the velocity of approach $=v \cos \theta$
Hence, $\quad v^{\prime} \cos \theta^{\prime}=e v \cos \theta$ ... (ii)
From Eqs. (i) and (ii)
$\begin{aligned} v^{\prime} & =v \sqrt{\sin ^2 \theta+e^2 \cos ^2 \theta} \\ \text { and } \quad \tan \theta^{\prime} & =\frac{\tan \theta}{e}\end{aligned}$
For elastic collision, $e=1$, so that $\theta^{\prime}=\theta$ and $v^{\prime}=v$
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