# A rod of mass M and length L is lying on a horizontal

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 $\left(\frac{\mathrm{m}}{\mathrm{M}}\right)$ is $\frac{1}{\mathrm{x}}$. The value

of ' $x$ ' will be_______.

Solution:

Just before collision

Just after collision

From momentum conservation, $\mathrm{P}_{\mathrm{i}}^{0}=\mathrm{P}_{\mathrm{f}}$

$\mathrm{mu}=\mathrm{Mv}$ $\ldots .$ (i)

From angular momentum conservation about $\mathrm{O}$,

$\mathrm{mu} \cdot \frac{\mathrm{L}}{2}=\frac{\mathrm{ML}^{2}}{12} \omega$

$\Rightarrow \omega=\frac{6 \mathrm{mu}}{\mathrm{ML}}$  $\ldots$ (ii)

From $\mathrm{e}=\frac{\mathrm{R} . \mathrm{V.S}}{\mathrm{R} . \mathrm{V} . \mathrm{A}}$

$1=\frac{\mathrm{V}+\frac{\omega \mathrm{L}}{2}}{\mathrm{u}}$

$\mathrm{v}+\frac{\omega \mathrm{L}}{2}=\mathrm{u}$

$\mathrm{v}+\frac{3 \mathrm{mu}}{\mathrm{M}}=\mathrm{u}$

$\frac{\mathrm{mu}}{\mathrm{M}}+\frac{3 \mathrm{mu}}{\mathrm{M}}=\mathrm{u}$

$\frac{4 \mathrm{mu}}{\mathrm{M}}=\mathrm{u}$

$\frac{m}{M}=\frac{1}{4}$

$X=4$

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