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Q.

A uniform rod of length 6a and mass 8 m lies on a smooth horizontal table. Two particles of mass m and 2m moving in the same horizontal plane but in opposite directions with speeds 2V and V respectively strike the rod normally at perpendicular distances 2a and a from either side of centre of rod and stick to the rod. The total energy of rod after collision is

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a

$\frac{1}{2} {mv}^{2}$

b

$\frac{3}{5} {mv}^{2}$

c

${mv}^{2}$

d

$\frac{7}{10} {mv}^{2}$

answer is B.

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Detailed Solution

$L_{i} = L_{f} \begin{array}{l} (2 mva + 4 mva) = [8 m \cdot \frac{(6 a)^{2}}{12} + 2 m \cdot a^{2} + m (2 a)^{2}] ω \\ \Rightarrow ω = \frac{V}{5 a} and \\ E = \frac{1}{2} {Iω}^{2} = \frac{1}{2} (30 {ma}^{2}) {(\frac{V}{5 a})}^{2} = \frac{3}{5} {mv}^{2} \end{array}$

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