Two masses m and 2m are connected by a light | vedclass

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(B) Using the principle of conservation of mechanical energy, the loss in potential energy of the $2m$ mass equals the gain in kinetic energy of the s

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$2$

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(B) Using the principle of conservation of mechanical energy, the loss in potential energy of the $2m$ mass equals the gain in kinetic energy of the system.Loss in potential energy of $2m$ mass = $(2m)gh$.Gain in potential energy of $m$ mass = $mgh$.Net loss in potential energy = $(2m)gh - mgh = mgh$.This energy is converted into the kinetic energy of the two masses and the rotational kinetic energy of the pulley.Total kinetic energy = $K_{m} + K_{2m} + K_{pulley} = \frac{1}{2}mv^2 + \frac{1}{2}(2m)v^2 + \frac{1}{2}I\omega^2$.Since $I = \frac{1}{2}Mr^2 = \frac{1}{2}(30m)r^2 = 15mr^2$ and $\omega = \frac{v}{r}$, we have $K_{pulley} = \frac{1}{2}(15mr^2)(\frac{v^2}{r^2}) = 7.5mv^2$.Equating energy: $mgh = \frac{1}{2}mv^2 + mv^2 + 7.5mv^2 = 9mv^2$.$v^2 = \frac{gh}{9} = \frac{10 	imes 3.6}{9} = 4$.$v = 2 \ m/s$.

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