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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

(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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