Two blocks are connected by a string passing over a pulley as shown. Wedge is fixed. M is sliding down with speed 12 m/s at an instant and suddenly the string breaks. The coefficient of friction between the blocks and wedge is $\frac{1}{\sqrt{2}}$ . Find the velocity (in m/s) of block M with respect to m after t = 2 seconds from the moment the string breaks. [Take $g = 10 m s^{- 2}$ and $\sqrt{\frac{3}{2}} = 1.2$ . Assume that the length of incline is sufficiently long.]

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Detailed Solution

$a_{M} = g \sin 30^{\circ} - \frac{g \cos 30^{\circ}}{\sqrt{2}} = 5 - 6 = - 1 m / s^{2}$

$a_{m} = - (g \sin 30^{\circ} + \frac{g \cos 30^{\circ}}{\sqrt{2}}) = - 11 m / s^{2}$

As limiting friction is greater than mgsin $θ$ , block ‘m’ will not move after $t = \frac{12}{11} \sec .$

$v_{M} = 12 - 1 \times 2 = 10 m / s$

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