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Question:

A thin uniform bar of length L and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to the bar after collision at a distance L/3 and L/6 respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be:

v/6L

v/5L

6v/5L

3v/5L

Solution:

M1=8m, L, m, 2v, L/3, m, v, L/6
Linear momentum is conserved
2mv - mv = (2m + m + M)Vcm ⇒ Vcm = 0
Angular momentum is conserved
2mv(L/6) + 2mv × (L/3) + 0 = Iw (1)
I = M.I. of rod + M.I. of mass 2m + M.I of mass m
= 8mL²/12 + 2m(L/6)² + m(L/3)² = mL²(2/3 + 1/18 + 1/9) = 56mL²/18 = 28mL²/9
eqn (1) becomes
(mvL/3 + 2/3mvL) = (28mL²/9)w
5mvL/3 = (28mL²/9)w
(mvL/3 + 2/3mvL) = (28/9)mL²w
6v/5L = w