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(B) Let the acceleration of the system be $a$. Since the string is inextensible,both blocks move with the same magnitude of acceleration $a$.For the b

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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 string is inextensible,both blocks move with the same magnitude of acceleration $a$.For the block on the $53^{\circ}$ incline,the equation of motion is: $mg \sin 53^{\circ} - T = ma$.For the block on the $37^{\circ}$ incline,the equation of motion is: $T - mg \sin 37^{\circ} = ma$.Adding these two equations: $mg(\sin 53^{\circ} - \sin 37^{\circ}) = 2ma$.$a = \frac{g(\sin 53^{\circ} - \sin 37^{\circ})}{2} = \frac{10(0.8 - 0.6)}{2} = \frac{10(0.2)}{2} = 1 \, m/s^2$.The acceleration of the centre of mass is given by $\vec{a}_{cm} = \frac{m\vec{a}_1 + m\vec{a}_2}{2m} = \frac{\vec{a}_1 + \vec{a}_2}{2}$.The acceleration vectors are $\vec{a}_1 = a(\cos 53^{\circ} \hat{i} + \sin 53^{\circ} \hat{j})$ and $\vec{a}_2 = a(-\cos 37^{\circ} \hat{i} + \sin 37^{\circ} \hat{j})$.Since $\cos 53^{\circ} = \sin 37^{\circ} = 0.6$ and $\sin 53^{\circ} = \cos 37^{\circ} = 0.8$,we have $\vec{a}_1 = a(0.6 \hat{i} + 0.8 \hat{j})$ and $\vec{a}_2 = a(-0.8 \hat{i} + 0.6 \hat{j})$.$\vec{a}_{cm} = \frac{a}{2} [(0.6 - 0.8) \hat{i} + (0.8 + 0.6) \hat{j}] = \frac{a}{2} [-0.2 \hat{i} + 1.4 \hat{j}] = a(-0.1 \hat{i} + 0.7 \hat{j})$.The magnitude is $|\vec{a}_{cm}| = a \sqrt{(-0.1)^2 + (0.7)^2} = 1 \sqrt{0.01 + 0.49} = \sqrt{0.5} = \frac{1}{\sqrt{2}} \, m/s^2$.

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

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