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A rigid uniform rod of mass M and length ‘L’ is resting on a smooth horizontal table. Two marbles each of mass ‘m’ and travelling with uniform speed ‘V’ collide with two ends of the rod simultaneously and inelastically as shown. The marbles get stuck to the rod after the collision and continue to move with the rod. If m=M/6 and V=L m/sec, then the time taken by the rod to rotate through π/2 is

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1 s

2πs

πs

π/2 s

answer is D.

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Detailed Solution

P_i =P_f ; P_i =O hence P_f =0 there is no translatory motion for the rod
$\begin{array}{l} L_{i} = L_{f} \\ mV \frac{L}{2} + mV \frac{L}{2} = (m {(\frac{L}{2})}^{2} + m {(\frac{L}{2})}^{2} + \frac{{ML}^{2}}{12}) ω \end{array}$

$\Rightarrow \frac{MVL}{6} = (\frac{{ML}^{2}}{6}) ω$

$\begin{array}{l} \Rightarrow ω = 1 rad / s \\ t = \frac{θ}{ω} = \frac{π}{2 \times 1} = \frac{π}{2} S \end{array}$

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