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A uniform bar of length $6 a$ and mass $8 m$ lies on a smooth horizontal table. Two point masses $m$ and $2 m$ moving in the same horizontal plane with speeds $2 v$ , and $v$ , respectively, strike the bar (as shown in figure) and stick to the bar after collision. Denoting angular velocity, total energy and velocity of centre of mass by $ω, E$ and $V_{c}$ respectively, we have after collision

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$V_{c} = 0$

$ω = \frac{3 v}{5 a}$

$ω = \frac{v}{5 a}$

$E = \frac{3}{5} m v^{2}$

answer is A, C, D.

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Detailed Solution

Considering the two balls and bar as system

$P_{i} = 0 = P_{f} ∴ V_{c} = 0$

Taking $C$ for conservation of angular momentum, we have

$2 m v a + m \times 2 v \times 2 a = \{\frac{8 m {(6 a)}^{2}}{12} + 2 m a^{2} + m {(2 a)}^{2}\} ω ω = \frac{v}{5 a} E = \frac{1}{2} \times 30 m a^{2} \times ω^{2} = \frac{3}{5} m v^{2}$

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