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(B) Let the acceleration of the system be $a$. Since the blocks are connected by a string,they move with the same magnitude of acceleration $a$.Applyi

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$\frac{1}{\sqrt{2}} \, m/s^2$

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(B) Let the acceleration of the system be $a$. Since the blocks are connected by a string,they move with the same magnitude of acceleration $a$.Applying Newton's second law to the system:$T - mg \sin 37^{\circ} = ma$$mg \sin 53^{\circ} - T = ma$Adding these two equations,we get:$mg(\sin 53^{\circ} - \sin 37^{\circ}) = 2ma$$a = \frac{g(\sin 53^{\circ} - \sin 37^{\circ})}{2} = \frac{10(0.8 - 0.6)}{2} = \frac{10(0.2)}{2} = 1 \, m/s^2$.The acceleration of the first block is $\vec{a}_1 = a \angle 53^{\circ}$ (up the incline) and the second block is $\vec{a}_2 = a \angle 37^{\circ}$ (down the incline).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 magnitude is $a_{cm} = \frac{1}{2} \sqrt{a^2 + a^2 + 2a^2 \cos(180^{\circ} - (53^{\circ} + 37^{\circ}))} = \frac{1}{2} \sqrt{2a^2 + 2a^2 \cos(90^{\circ})} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.Substituting $a = 1 \, m/s^2$,we get $a_{cm} = \frac{1}{\sqrt{2}} \, m/s^2$.

Two blocks,each of mass $m$,are connected to each other by a massless string and placed on two inclined planes as shown in the figure. After releasing the system from rest,find the magnitude of the acceleration of the centre of mass of the two blocks. (Take $g = 10 \, m/s^2$)

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