Question

A rigid rod of mass $m$ and length $L$ is shown in figure. A particle $P$ of mass $m$ moving with a speed $u$, normal to $AB$ strikes $A$ and sticks to it. Immediately after the impact, the velocity of $B$ w.r.t. $C$ is $n/10$, where $n$ is (in $\text{m/s}$)
[$C$ is the centre of mass of the system (rod+particle)]

Answer 1

Distance of COM of the system from point $A$
$y_{\text{COM}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}=\frac{m\left(0\right)+m\left(\frac{l}{2}\right)}{m+m}$
$y_{\text{COM}}=\frac{l}{4}$

Let $v_{\text{COM}}=$ velocity of COM of the system
$\omega =$ angular velocity of the system about COM.
Just after collision:


There is no external force on the system, hence linear momentum will be conserved. $mu+0=(m+m)v_{\text{COM}}$
or $v_{\text{COM}}=\frac{u}{2}$
There is no external torque, hence angular momentum will be conserved about COM.
$L_i=L_f$
$mu\left(\frac{l}{4}\right)=I_{\text{COM}}.\omega$ ... (1)
Here $I_{\text{COM}}={\left(I_{\text{rod}}\right)}_{\text{COM}}+{\left(I_{\text{particle}}\right)}_{\text{COM}}$
and $I_{\text{COM}}=[\frac{m{l}^2}{12}+m{\left(\frac{l}{4}\right)}^2]+\left[m{\left(\frac{l}{4}\right)}^2\right]$
$I_{\text{COM}}=\frac{7m{l}^2}{48}+\frac{m{l}^2}{16}=\frac{10m{l}^2}{48}$
So,
$mu\left(\frac{l}{4}\right)=\frac{5m{l}^2}{24}\times \omega$
$\Rightarrow \omega =\frac{6u}{5l}$

We know $\overrightarrow{V_B}=\overrightarrow{V_C}+\overrightarrow{\omega }\times \overrightarrow{r_{BC}}$
where $\overrightarrow{V_C}=v_{COM}\hat{i}=\frac{u}{2}\hat{i}$
Velocity of $B$ w.r.t $C$ will be:
$\overrightarrow{V_{BC}}=\overrightarrow{V_B}-\overrightarrow{V_C}=\overrightarrow{\omega }\times \overrightarrow{r}$
$=\frac{6u}{5l}(-\hat{k})\times \frac{3l}{4}(-\hat{j})$
$\overrightarrow{V_{BC}}=-\frac{9u}{10}\left(\hat{i}\right)$
Putting value of $u$,
$\left|V_{BC}\right|=\frac{9}{10}\text{m/s}$