Question

A thin uniform bar of length $L$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ are moving in the same horizontal plane from opposite sides of the bar with speeds $2v$ and $v$ respectively. The masses stick to the bar after collision at a distance $\frac{L}{3}$ and $\frac{L}{6}$ respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be :

• A. $\frac{v}{5L}$
• B. $\frac{6 v}{5L}$
• C. $\frac{3 v}{5L}$
• D. $\frac{ v}{6L}$

Answer 1

Answer: (B)

Centre of mass after the collision will be at $O$. Therefore, $2 m \frac{L}{6}=m \frac{L}{3}$
Now, using conservation of angular momentum, we get $\quad 2 m \frac{L}{6} \times v+m \frac{L}{3} \times 2 v=I \omega$
Therefore, $ 2 m \frac{L}{6} \times v+m \frac{L}{3} \times 2 v =\left[8 m \frac{L^{2}}{12}+2 m\left(\frac{L}{6}\right)^{2}+m\left(\frac{L}{3}\right)^{2}\right] \omega $
$ \Rightarrow \left(\frac{L}{6}+\frac{1}{3}\right) 2 m v =\left[8 m \frac{L^{2}}{12}+\frac{2 m L^{2}}{36}+\frac{m L^{2}}{9}\right] \omega$
$ \Rightarrow \frac{3 L}{6} \times 2 m V=\frac{5}{6} m L^{2} \omega $
$\Rightarrow L m v =\frac{5}{6} m L^{2} \omega $
$\Rightarrow \omega=\frac{6 L m v}{5 m L^{2}} $
$\Rightarrow \omega=\frac{6}{5} \frac{v}{L}$