A rod of mass M and length L is lying on a horizontal frictionles… - Vidyadip

MCQ

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 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 ..... .

A
$5$
✓
$4$
C
$14$
D
$23$

✓

Answer

Correct option: B.

$4$

b
Just after collision

From momentum conservation, ${P}_{{i}}^{\circ}={p}_{{F}}$

$m u=M v \ldots \ldots(i)$

From angular momentum conservation about $O,$

$mu. \,l\frac{L}{2}=\frac{M L^{2}}{12} \omega$

$\Rightarrow \omega=\frac{6 {mu}}{{ML}} \quad \cdots . . (ii)$

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

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

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

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

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

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

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

${X}=4$

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