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 $u$ strikes the rod perpendicularly at one end and sticks to the rod as shown in the figure. Find the total angular momentum of the system about the origin just after the collision, if the angular velocity of the rod just after the collision is $\omega$ about its centre.

• 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. $\left(1\right),\left(2\right),\left(3\right)$,

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