Question

A thin uniform bar of length $L$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ 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. $\frac{6v}{5L}$
• C. $\frac{3v}{5L}$
• D. $\frac{v}{6L}$

Answer 1

The correct option is B $\frac{6v}{5L}$

$M_1=8m,L$

$m,2v,\frac{L}{3}$

$m,v,\frac{L}{6}$

Linear momentum is conserved

$2mv-2mv=(2m+m+M)V_{cm}$
$\Rightarrow V_{cm}=0$

Angular momentum is conserve

$2mv\frac{L}{6}+2mv\times \frac{L}{3}+0=Iw$ ....(1)

$I=\text{M.I. of rod}+\text{M.I. of mass 2m}+\text{M.I of mass m}$

$=\frac{8m{L}^2}{12}+2m{\left(\frac{L}{6}\right)}^2+m{\left(\frac{L}{3}\right)}^2$

$=m{L}^2(\frac{2}{3}+\frac{1}{18}+\frac{1}{9})=\frac{5}{6}m{L}^2$

eqn (1) becomes

$(\frac{mvL}{3}+\frac{2}{3}mvL)=\frac{5}{6}m{L}^{2}w$

$\frac{mL}{3}(V+2v)=\frac{5}{6}m{L}^{2}w$

$\frac{6V}{5L}=w$