Question

A uniform rod of length $l$ and mass $4m$ lies on a frictionless horizontal surface on which it is free to move. A ball of mass $m$ moving with speed $v$ as shown in figure, collides with the rod at one of its ends. If the ball comes to rest immediately after the collision, then find the angular velocity $\omega$ of the rod just after collision.

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

Answer 1

The correct option is C $\frac{3v}{2l}$

On the system of (rod + ball), there is no external force. Hence, linear momentum will be conserved.
After collision:


$P_i=P_f$
$\Rightarrow mv=\left(4m\right)v^{\prime }+m\left(0\right)$
$\therefore v^{\prime }=\frac{v}{4}...\left(1\right)$

Also for (rod+ball) system, ${\tau }_{\text{ext}}=0$. Applying angular momentum conservation about centre $\left(O\right)$ of rod :
$L_i=L_f$
Taking anticlockwise sense of rotation as $+ve$,
$+mv\left(\frac{l}{2}\right)=L_{\text{Trans}}+L_{\text{Rot}}$
Here $L_{\text{Trans}}=M(\overrightarrow{r}\times {\overrightarrow{v}}_0)=0$, because $v_{\text{com}}\text{i.e}{v}^{\prime }$ passes through centre $O$, hence $(\overrightarrow{r}\times {\overrightarrow{v}}_0)=0$ about point $O$
$\Rightarrow mv\left(\frac{l}{2}\right)=0+I_{\text{COM}}\omega ...\left(i\right)$
$I_{\text{COM}}$for the rod,
$I_{\text{COM}}=\frac{\left(4m\right)l^2}{12}=\frac{m{l}^2}{3}$
Substituting in Eq $\left(i\right)$,
$mv\left(\frac{l}{2}\right)=\frac{m{l}^2}{3}\times \omega$
$\therefore \omega =\frac{3v}{2l}$