A uniform rod of length 6a and mass 8m lies on a smooth horizontal table. Two point masses 'm' and '2m, moving in the same horizontal plane with speed '2u' and 'u' respectively strikes the bar as shown and stick to the rod. Its angular velocity about centre of mass is
Using angular momentum conservation:
Initial a.m = 2m(v)(2a)+m(2v)(a)
= 6mva
Final a.m = [6m(8a)²/12+2m(2a)²+ma²]w
Where w is the angular velocity
Equating above two equations we get
W=6v/41a
