QJEEM18Ph83fe: JEE Main 2018 Physics - Class 11 JEEM18Ph83fe Solution: A thin uniform bar of length and mass m lies on a smooth…

Centre of mass after the collision will be at O. Therefore,

$2m{L \over 6} = m{L \over 3}$

Now, using conservation of angular momentum, we get

$2m{L \over 6} \times v + m{L \over 3} \times 2v = I\omega$

Therefore,

$2m{L \over 6} \times v + m{L \over 3} \times 2v = \left[ {8m{{{L^2}} \over {12}} + 2m{{\left( {{L \over 6}} \right)}^2} + m{{\left( {{L \over 3}} \right)}^2}} \right]\omega$

$\Rightarrow \left( {{L \over 6} + {L \over 3}} \right)2mv = \left[ {8m{{{L^2}} \over {12}} + {{2m{L^2}} \over {36}} + {{m{L^2}} \over 9}} \right]\omega$

$\Rightarrow {{3L} \over 6} \times 2mV = {5 \over 6}m{L^2}\omega$

$\Rightarrow Lmv = {5 \over 6}m{L^2}\omega \Rightarrow \omega = {{6Lmv} \over {5m{L^2}}} \Rightarrow \omega = {6 \over 5}{v \over L}$