Two particles `A` and `B` of mass `m` each hit the ends of a rigid bar of mass `M` and length `l` inelastically from opposite sides the speeds `v` and perpendicular to the rod. The bar is kept on a smooth horizontal plane (as shown). Find the linear and angular speed of the system (bar`+`particle).

Explanation

Conservation of linear momentum. There is not external force acting on the system along `x`-axis. There fore the linear momentum of the system before and after the system remains constant.
The momentum of the system before impact.
`P_(i)=mv-mv=0`
The linear momentum of the system after the impact must be zero. therefore, the centre of mass of the system does not translate.
`implies v_(cm)=0`
Conservation of angular momentum
Since the impact forces are internal force, the net torque about a pperpendicular axis passing through `CM` of the system is zero. Thefore the angular momentum of the system before and after the impact will be equal.
`implies L_("initial")=L_("final") where L_("initial")|(ml)/2v+(ml)/2v|=mvl`
Let the system rotate about is `CMO` with an angular speed `omega`
`implies L_("final")=(I_("system"))omega`
where `I_("system")=MI` of the system about `0=(I_("rod"))_(0)+(I_(A))+(I_(B))_(0)`
`implies I_("system")=(Ml^(2))/12+m(l/2)^(2)+m(l/2)^(2)=((M+6m)/12)l^(2)`
Using the above equations we get
`F_(1)=1/2mralpha=1/3masintheta=mvlimpliesomega=(12mv)/((M+6m)l)`