A small ring of mass m attached at an end of a light string of length l = 0.6 m the other end of the string is tied to a small block B of mass 2 m. The ring is free to move on a fixed smooth horizontal rod. Find the velocity (In m/s) of the ring when the string becomes vertical. (g =10ms^-2 )

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answer is 4.

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Detailed Solution

Since there is no net external force in horizontal direction
$mu = 2 mv \Rightarrow u = 2 v$
From energy conservation
$\begin{array}{l} 2 mgl = \frac{1}{2} {mu}^{2} + \frac{1}{2} 2 {mv}^{2} = \frac{1}{2} {mu}^{2} + \frac{1}{2} 2 m \cdot \frac{u^{2}}{4} \\ \Rightarrow u = \sqrt{\frac{8 gl}{3}} = \sqrt{\frac{8 \times 10 \times 0 .6}{3}} = 4 m / s \end{array}$