Question

A uniform rod having mass $m$ and length $L$ lies on a horizontal surface. A particle of mass $\frac{m}{2}$ moving with speed $u$ on the horizontal surface strikes the rod perpendicularly at one end and sticks to the rod. Find the total angular momentum about the origin just after collision, if the angular velocity of the rod about its centre is $\omega$ just after collision.

• A. $[\frac{muL}{3}+\frac{m{L}^2\omega }{2}](-\hat{k})$
• B. $[\frac{muL}{3}-\frac{m{L}^2\omega }{2}](-\hat{k})$
• C. Zero
• D. $[\frac{-muL}{3}-m{L}^2\omega ](-\hat{k})$

Answer 1

The correct option is A $[\frac{muL}{3}+\frac{m{L}^2\omega }{2}](-\hat{k})$

COM of the (rod + particle) system w.r.t point $A$, just after collision is
$y_{\text{com}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}=\frac{m\left(\frac{L}{2}\right)+\frac{m}{2}\left(0\right)}{m+\frac{m}{2}}$
$\therefore y_{\text{com}}=\frac{L}{3}$
Angular momentum (just after collision) about point $B$,
${\overrightarrow{L}}_f={\overrightarrow{L}}_{Trans}+{\overrightarrow{L}}_{Rot}$
${\overrightarrow{L}}_f=M({\overrightarrow{r}}_0\times {\overrightarrow{v}}_0)+{\overrightarrow{L}}_{\text{com}}$
${\overrightarrow{L}}_f=(m+\frac{m}{2})[\frac{2L}{3}\hat{j}\times v^{\prime }\hat{i}]+I_{\text{com}}\omega (-\hat{k})...\left(1\right)$
[$\because$ from right hand rule we obtain, direction of $\overrightarrow{\omega }$ along $-\hat{k}$]


On the system of (rod+particle), ${\overrightarrow{F}}_{\text{ext}}=0$. Applying linear momentum conservation
i.e $P_i=P_f$
$\Rightarrow \frac{m}{2}u=(m+\frac{m}{2})v^{\prime }$
$\therefore v^{\prime }=\frac{u}{3}...\left(2\right)$
& MOI of system about its $\text{COM}$,
$I_{\text{com}}=I_{\text{rod}}+I_{\text{particle}}$
where from parallel axis theorem, $I_{\text{rod}}=[\frac{m{L}^2}{3}+m{\left(\frac{L}{3}\right)}^2]$
$\Rightarrow I_{\text{com}}=[\frac{m{L}^2}{3}+m{\left(\frac{L}{3}\right)}^2]+\frac{m}{2}{\left(\frac{L}{3}\right)}^2$
$\therefore I_{\text{com}}=\frac{m{L}^2}{2}...\left(3\right)$
From Eq. $\left(1\right),\left(2\right),\left(3\right)$,

$L_f=\frac{3m}{2}[-\frac{2Lu}{9}\hat{k}]+\frac{m{L}^2}{2}\times \omega (-\hat{k})$
$\therefore L_f=[\frac{muL}{3}+\frac{m{L}^2\omega }{2}](-\hat{k})$