Question

A thin uniform rod of mass $m$ and length $l$ is kept on a smooth horizontal surface such that it can move freely. At what distance $x$ from the centre of the rod should a particle of mass $m$ strike on the rod such that the point $P$ at a distance $\frac{l}{3}$ from the end of the rod is instantaneously at rest just after the elastic collision ?

• A. $\frac{l}{2}$
• B. $\frac{l}{3}$
• C. $\frac{l}{6}$
• D. $\frac{l}{4}$

Answer 1

The correct option is A $\frac{l}{2}$


Let the impulse imparted due to the collision be $\overrightarrow{J}$.
Applying Linear Impulse-Momentum theorem on rod: $\overrightarrow{J}=\Delta \overrightarrow{P}$
$J=m{V}_{cm}\Rightarrow V_{cm}=\frac{J}{m}\dots \dots \left(1\right)$
Applying Angular Impulse-Momentum theorem:
$Jx=I\omega -0$
$Jx=\frac{m{l}^2}{12}\omega \Rightarrow \omega =\frac{12Jx}{m{l}^2}\dots \dots \left(2\right)$


For point $P$ to be at instantaneous rest,
$({\overrightarrow{V}}_{\text{net}}{)}_P=0\Rightarrow V_{cm}-\frac{l\omega }{6}=0\dots \dots (3)$

Substituting (1) and (2) in (3), we get:
$\frac{\text{/}J}{\text{/}m}-\frac{\text{/}l}{\text{/}6}\frac{\text{/}12\text{/}Jx}{\text{/}m{l}^{\text{/}2}}=0$
$\Rightarrow 1=\frac{2x}{l}$
$\Rightarrow x=\frac{l}{2}$