Question

Two particles $A$ and $B$ of mass $m$ each and moving with velocity $v$, hit the ends of a rigid bar of the same mass $m$ and length $l$ simultaneously and stick to the bar as shown in the figure. The bar is kept on a smooth horizontal plane. The linear and angular speed of the system (bar$+$particle) after the collision are

• A. $v_{cm}=0,\omega =\frac{12}{7}\frac{v}{\ell }$
• B. $v_{cm}=0,\omega =\frac{4v}{\ell }$
• C. $v_{cm}=0,\omega =\frac{6v}{5l}$
• D. $v_{cm}=0,\omega =\frac{v}{5\ell }$

Answer 1

Answer: (A)

$v_{cm}=0,\omega =\frac{12}{7}\frac{v}{\ell }$

$v_{cm}=0,w=\frac{12v}{7l}$

By conservation of linear momentum, we have,

$mv-mv=(m+m+m)v_{cm}$

$\Rightarrow v_{cm}=0$

using conservation of angular momentum, we get,

$mvl=Iw$

$=[\frac{m{l}^2}{12}+2m{\left(\frac{l}{2}\right)}^2]w$

$\therefore w=\frac{12v}{7l}$