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(B) Let the acceleration of the blocks be $a$. The forces along the incline are $Mg \sin 60^{\circ}$ and $Mg \sin 30^{\circ}$.The equation of motion i

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$\frac{5(\sqrt{3}-1)}{2 \sqrt{2}}$

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(B) Let the acceleration of the blocks be $a$. The forces along the incline are $Mg \sin 60^{\circ}$ and $Mg \sin 30^{\circ}$.The equation of motion is $Mg \sin 60^{\circ} - Mg \sin 30^{\circ} = (M+M)a$.$a = \frac{g(\sin 60^{\circ} - \sin 30^{\circ})}{2} = \frac{10(\frac{\sqrt{3}}{2} - \frac{1}{2})}{2} = \frac{5(\sqrt{3}-1)}{2} 	ext{ ms}^{-2}$.The acceleration of the block on the $60^{\circ}$ incline is $\vec{a}_1 = a(\cos 60^{\circ} \hat{i} - \sin 60^{\circ} \hat{j})$ and the block on the $30^{\circ}$ incline is $\vec{a}_2 = a(-\cos 30^{\circ} \hat{i} - \sin 30^{\circ} \hat{j})$.However,using the coordinate system aligned with the inclines as shown in the figure,$\vec{a}_1 = a \hat{i}'$ and $\vec{a}_2 = -a \hat{j}'$ where $\hat{i}'$ and $\hat{j}'$ are unit vectors along the inclines.The acceleration of the centre of mass is $\vec{a}_{cm} = \frac{M\vec{a}_1 + M\vec{a}_2}{2M} = \frac{\vec{a}_1 + \vec{a}_2}{2}$.The angle between the two inclines is $90^{\circ}$. Thus,the magnitude is $a_{cm} = \frac{\sqrt{a^2 + a^2}}{2} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.Substituting $a = \frac{5(\sqrt{3}-1)}{2}$,we get $a_{cm} = \frac{5(\sqrt{3}-1)}{2\sqrt{2}}$.

Two blocks of equal masses $M$ are tied with a light string passing over a massless pulley on a wedge with angles $60^{\circ}$ and $30^{\circ}$ (assuming frictionless surfaces). The acceleration of the centre of mass of the two blocks is $\left(g=10 \text{ ms}^{-2}\right)$.

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