Question

A uniform rod of mass $M$ and length $2\text{m}$ lies on a smooth horizontal table. A particle of mass $m$ moving on the table with a velocity $u$ strikes the rod perpendicularly at an end and stops there. Find the distance travelled by the center of the rod (in metres) by the time it turns through a right angle. Given $M=4m$.

• A. $\pi /12$
• B. $\pi /6$
• C. $\pi /3$
• D. $\pi$

Answer 1

The correct option is B $\pi /6$

Let $S$ be the distance travelled by the $\text{COM}$ of the rod.


Since, $F_{\text{ext}}=0$ on system of (rod+particle), hence linear momentum will be conserved.
$P_i=P_f$
$\Rightarrow mu=m\left(0\right)+Mv$
$\Rightarrow v=\left(\frac{mu}{M}\right)...\left(1\right)$
Here, $v$ is the velocity of centre of mass of the rod after the impact.

Also ${\tau }_{\text{ext}}=0$, hence applying angular momentum conservation about reference point as $\text{COM}$.
$L_i=L_f...\left(i\right)$
where $L_f=L_{\text{trans}}+L_{\text{Rot}}$
$r=0$, as velocity vector passes through centre, hence $\therefore L_{\text{trans}}=M{v}_{0}r=0$
Putting in Eq. $\left(i\right)$,
$\Rightarrow mu\left(\frac{l}{2}\right)=I\omega$
$\Rightarrow mu\left(\frac{l}{2}\right)=\frac{M{l}^2}{12}\times \omega$
$\therefore \omega =\frac{6mu}{Ml}$

${\tau }_{\text{ext}}=0\Rightarrow \alpha =0$
Applying kinamatic equation for angular displacement of rod,
$\theta =\omega t...\left(ii\right)$,
Here, angle rotated by rod $\theta =\pi /2$
$\Rightarrow t=\frac{\pi /2}{\left(\frac{6mu}{Ml}\right)}$
$t=\frac{\pi Ml}{12mu}...\left(iii\right)$
In that time interval, linear distance travelled by the rod's centre of mass is:
$S=v.t$
Putting $v$ and $t$ from Eq. $\left(i\right)\&\left(iii\right)$,
$S=\frac{mu}{M}\times \frac{\pi Ml}{12mu}$
$S=\frac{\pi l}{12}$
$\therefore S=\frac{\pi \times 2}{12}=\frac{\pi }{6}\text{m}$