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.
find:
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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