Question

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 $\left(\frac{m}{M}\right)$ is $\frac{1}{x}$. The value of ' $x$ ' will be ______.

Answer 1

Just before collision

Just after collision
From momentum conservation,
$P_{i}^{0} =P_{f} $
$m u =M v \ldots . . .$ (i)
From angular momentum conservation about $O$,
$m u \cdot \frac{L}{2}=\frac{M L^{2}}{12} \omega$
$\Rightarrow \omega=\frac{6 m u}{M L} \ldots \ldots$ (ii)
From e $=\frac{\text { R.V.S }}{\text { R.V.A }}$
$1=\frac{V+\frac{\omega L}{2}}{u}$
$v+\frac{\omega L}{2}=u$
$v+\frac{3 m u}{M}=u$
$\frac{m u}{M}+\frac{3 m u}{M}=u$
$\frac{4 m u}{M}=u$
$\frac{m}{M}=\frac{1}{4}$
$X=4$