Rotational MotionHard
Question
A nuniform bar of length 6a and mass 8 m lies on a smooth horizontal table. Two points masses m and 2m moving in the same horizontal plane with speed 2v and v respectively, strike the bar [as shoen in the figure] and stick to the bar after collision. Denoting angular velocity (about the centre of the mass), total energy and centre of mass velocity by ω, E and vc respectively, we have after collision.
Options
A.ω = 
B.Vc = 0
C.ω = 
D.E = 
mv2
mv2Solution
Pi = 0 ∴ Pf = 0 or vc = 0
Li = f or (2mv)a + (2mv)(2a) = Iω .....(i)
Here, I =
+ m(2a)2 + (2m)(a2) = 30ma2
Substituting in eq. (i), we get
ω =
Further, E =
Iω2 = × (30 ma2) 
∴ Correct option are (a), (c) and (d).
Li = f or (2mv)a + (2mv)(2a) = Iω .....(i)
Here, I =
+ m(2a)2 + (2m)(a2) = 30ma2Substituting in eq. (i), we get
ω =
Further, E =
Iω2 = × (30 ma2) ∴ Correct option are (a), (c) and (d).
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