# 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))