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(A) Let the mass of each block be $m$. The forces acting along the incline are $mg \sin 60^{\circ}$ and $mg \sin 30^{\circ}$.Since $mg \sin 60^{\circ}

Vedclass · Accepted Answer

$\left(\frac{\sqrt{3}-1}{4 \sqrt{2}}\right) g$

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(A) Let the mass of each block be $m$. The forces acting along the incline are $mg \sin 60^{\circ}$ and $mg \sin 30^{\circ}$.Since $mg \sin 60^{\circ} > mg \sin 30^{\circ}$,the system accelerates such that the block on the $60^{\circ}$ incline moves down.For the block on the $60^{\circ}$ incline: $mg \sin 60^{\circ} - T = ma$  --- $(i)$For the block on the $30^{\circ}$ incline: $T - mg \sin 30^{\circ} = ma$  --- (ii)Adding $(i)$ and (ii): $mg(\sin 60^{\circ} - \sin 30^{\circ}) = 2ma$$a = \frac{g}{2} \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) = \frac{(\sqrt{3}-1)g}{4}$.The acceleration vectors for the blocks are $\vec{a}_1 = a(\cos 60^{\circ} \hat{i} - \sin 60^{\circ} \hat{j})$ and $\vec{a}_2 = a(-\cos 30^{\circ} \hat{i} - \sin 30^{\circ} \hat{j})$.The acceleration of the centre of mass is $\vec{a}_{cm} = \frac{m\vec{a}_1 + m\vec{a}_2}{2m} = \frac{\vec{a}_1 + \vec{a}_2}{2}$.$\vec{a}_{cm} = \frac{a}{2} [(\cos 60^{\circ} - \cos 30^{\circ}) \hat{i} - (\sin 60^{\circ} + \sin 30^{\circ}) \hat{j}]$.Using $\cos 60^{\circ} = 1/2, \cos 30^{\circ} = \sqrt{3}/2, \sin 60^{\circ} = \sqrt{3}/2, \sin 30^{\circ} = 1/2$:$|\vec{a}_{cm}| = \frac{a}{2} \sqrt{(\frac{1-\sqrt{3}}{2})^2 + (\frac{\sqrt{3}+1}{2})^2} = \frac{a}{2} \sqrt{\frac{1+3-2\sqrt{3} + 3+1+2\sqrt{3}}{4}} = \frac{a}{2} \sqrt{\frac{8}{4}} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.Substituting $a$: $|\vec{a}_{cm}| = \frac{(\sqrt{3}-1)g}{4\sqrt{2}}$.

Two blocks of equal masses are tied to the ends of a light string. The string passes over a massless pulley fixed on a frictionless surface as shown in the figure. The acceleration of the centre of mass of the blocks is ($g$ = acceleration due to gravity).

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