A uniform bar of length 6a& mass 8m lies on a smooth horizontal table. Two point masses 'm& 2m' moving in the same horizontal plane with speeds 2v & 'v' respectively strike the bar as shown & stick to the bar after collision then

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Total kinetic energy just after collision is $\frac{3}{2} {mv}^{2}$

Velocity of the centre of mass is zero

Angular velocity about centre of mass $= \frac{v}{5 a}$

Total kinetic energy just after collision is $\frac{3}{5} {mv}^{2}$

answer is A, B, C.

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Detailed Solution

From conservation of linear momentum

$2 mV - 2 mV = (11 m) V \Rightarrow V_{cm} = 0$
About C, from conservation of angular momentum

$(2 mV) a + m (2 V) 2 a + 0$ $= [\frac{8 m [6 a]^{2}}{12} + 2 {ma}^{2} + m (2 a)^{2}] ω \Rightarrow ω = \frac{v}{5 a}$

$K \cdot E = \frac{1}{2} [I_{system}] ω^{2} = \frac{1}{2} (30 {ma}^{2}) \frac{v^{2}}{25 a^{2}} = \frac{3}{5} {mv}^{2}$

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