two small balls A and B,each of mass 'm' are connected by a light rod of length L. the system is lying on a frictionless horizontal surface. the particle of mass 'm' collides with rod horizontally with velocity 'u' perpendicular to the rod and gets stuck to it.
1)  the angular velocity of system after the collision
2)  the velocities of A  and B immediately after the collision
3)  the velocity of the centre of the rod when the rod rotates through 90 degree after the collision

Take the rod a distance d from the B, and therefore a distance L- d from A. Taking moments about C: 
(2 m)d = m(L-d) 
2d = L-d 
d = L/3 

If you were an observer situated at point C *before* the collision, the rod (with A and B) would have zero angular momentum (as the system is not rotating). But the particle approaching the rod would have an angular momentum about C of mvr = mv₀(L/3). 

So from your point of view at C, the total initial system angular momentum is: 
L = 0 + mv₀L/3 = mv₀L/3 

Angular momentum is conserved. So the final system angular momentum (as measured by you at C) is also L = mv₀L/3. 

The moment of inertia, I, of the final system is: 
I = (2m)d² + m(L-d)² 
. = m(2d² + L² - 2Ld + d²) 
. = m(3d² + L² - 2Ld) 

If the final angular velocity is ω, then since L = Iω 
ω = L/I 
. . = (mv₀L/3) / (m(3d² + L² - 2Ld)) 
. . = v₀L / (3(d² + L² - 2Ld)) 

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