Question

Two small particles $A$ and $B$ of masses $M$ and $2M$ respectively, are joined rigidly to the ends of a light rod of length $l$ as shown in the figure. The system translates on a smooth horizontal surface with a velocity $u$ in a direction perpendicular to the rod. A particle $P$ of mass $M$ kept at rest on the surface sticks to the particle $A$ as the particle $P$ collides with it. Then choose the correct option(s).

• A. The speed of particle $A$ just after collision is $\frac{u}{2}$
• B. The speed of particle $B$ just after collision is $\frac{u}{2}$
• C. The speed of particle $B$ just after collision is $u$
• D. The angular velocity of the system $(A+B+P+\text{rod})$ just after collision is $\frac{u}{2l}$

Answer 1

Answer: (A), (C), (D)

The angular velocity of the system $(A+B+P+\text{rod})$ just after collision is $\frac{u}{2l}$

COM of the system will be at the centre of the rod (distance $L/2$ from point $A$)
Just after collision, the rod will rotate in clockwise direction about the COM of the system.


$I_{\text{CM}}=2M{\left(\frac{l}{2}\right)}^2+M{\left(\frac{l}{2}\right)}^2+M{\left(\frac{l}{2}\right)}^2$
$\therefore I_{\text{CM}}=M{l}^2$
From linear momentum conservation,
$P_i=P_f$
$2M\left(u\right)+M\left(u\right)+0=4M{v}_{\text{CM}}$
$\Rightarrow v_{\text{CM}}=\frac{3u}{4}$
Applying angular momentum conservation about $\text{COM}$, taking anticlockwise sense of rotation as $+ve$
$L_i=L_f$
$\Rightarrow -2M\left(u\right)\left(\frac{l}{2}\right)+M\left(u\right)\left(\frac{l}{2}\right)+0=(-I_{\text{CM}}.\omega )+0$
$\because L_f=L_{CM}+M{v}_{\text{CM}}r=I_{\text{CM}}\omega +0$
$\Rightarrow \frac{Mul}{2}=M{l}^2.\omega$
$\therefore \omega =\frac{u}{2l}$

Velocity of $B$ just after collision:
$v_B=v_{\text{CM}}+(\omega \times \frac{l}{2})$
$\Rightarrow v_B=\frac{3u}{4}+(\frac{u}{2l}\times \frac{l}{2})$
$\therefore v_B=u$

Velocity of $A$ just after collision:
($\because v_{\text{CM}}$ and tangential velocity due to rotational effect are opposing)
$v_A=v_{\text{CM}}-(\omega \times \frac{l}{2})$
$v_A=\frac{3u}{4}-(\frac{u}{2l}\times \frac{l}{2})$
$\therefore v_A=\frac{3u}{4}-\frac{u}{4}=\frac{u}{2}$
Hence, options A, C and D are correct.