• # question_answer Two particles A and B, each of mass m and moving with velocity v, hit the ends of a rigid bar of mass M and length simultaneously and stick to the bar. The bar is kept on a smooth horizontal plane (as shown). Find the angular speed of the system (bar + particle) after the collision. $[M=6m]$            A) $\frac{2v}{\ell }$                              B)        $\frac{v}{\ell }$                  C) $\frac{3v}{\ell }$                      D)        zero

[b] Conservation of linear momentum ${{P}_{i}}=mv-mv=0$ $\Rightarrow \,\,\,\,\,\,\,\,\,{{V}_{cm}}=0$ Conservation of angular momentum $\Rightarrow \,\,\,\,\,\,{{L}_{initial}}={{L}_{final}}$ where ${{L}_{initial}}=\left| \frac{m\ell }{2}v+\frac{m\ell }{2}v \right|=mv\ell$ Let the system rotate about its cm 'O' with an angular speed $\omega$. $\Rightarrow \,\,\,\,\,\,\,\,{{L}_{final}}=\left( {{I}_{system}} \right)\omega$ where ${{I}_{system}}=\frac{M{{\ell }^{2}}}{12}+m{{\left( \frac{\ell }{2} \right)}^{2}}+m{{\left( \frac{\ell }{2} \right)}^{2}}=\left( \frac{M+6m}{12} \right){{\ell }^{2}}$ $\therefore \,\,\,\,\,\,\,\left( \frac{M+6m}{12} \right)\omega {{\ell }^{2}}=nv\ell$ $\Rightarrow \,\,\,\,\,\,\,\,\omega =\frac{12mv}{(M+6m)\ell }$.