Question 97 of 130

A rod of mass $M$ and length $L$ is lying on a horizontal frictionless surface. A particle of mass $m$ travelling along the surface hits at one end of the rod with a velocity $u$ in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $\left(\dfrac{m}{M}\right)$ is $\dfrac{1}{x}.$ The value of $x$ will be

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Answer: 4

From momentum conservation, ${P}_{i}^{0}={P}_{f}$
$mu=Mv\ldots \left(i\right)$
From angular momentum conservation about $O$,
$mu\cdot \dfrac{L}{2}=\dfrac{M{L}^{2}}{12}\omega$
$\Rightarrow \omega =\dfrac{6mu}{ML}\ldots (ii)$
From $e=\dfrac{R.V\cdot S}{R.V\cdot A}$
$1=\dfrac{V+\dfrac{\omega L}{2}}{u}$
$V+\dfrac{\omega L}{2}=u$
$V+\dfrac{3mu}{M}=u$
$\dfrac{mu}{M}+\dfrac{3mu}{M}=u$
$\dfrac{4mu}{M}=u$
$\dfrac{m}{M}=\dfrac{1}{4}$
$x=4$