A uniform rod of length `lambda` lies on a smooth horizontal table A particle moving on the table has a mass m and a speed v before the collision and it sticks to the rod after the collision. The rod has a mass M then find out.
The velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision.

Explanation

Figure shows the situation of system just before and just after collision.
Initially the centre of mass of the rod is at point O. After collision when the particle sticks to the rod. Centre of mass is shifted from point O to C as shown in figure. Now the system is rotated about axis passing through C
Now from linear momentum conservation
` mv = (M + m)v' rArr v' = (mv)/(M + m) `
From Angular momentum conservation about A
`L_(i) = L_(f)`
` 0 + 0= I omega - (m + M) v' l_(1)`
`rArr I omega = (m + M) v' l_(1)`
Put the value of `v',& l_(1)` we get
`omega = ( 6mv)/((M + 4m)l)`