Two blocks of equal mass are tied with a light string which passes over a massless pulley as shown in figure. The magnitude of acceleration of centre of mass of both the blocks is (neglect friction every where):

A
$\frac{\sqrt{3} - 1}{4 \sqrt{2}} g$
B
$(\sqrt{3} - 1) g$
C
$\frac{g}{2}$
D
$(\frac{\sqrt{3} - 1}{\sqrt{2}}) g$

Solution:

mg sin $60^{0} - T = ma, T - mg \sin 30^{0} = ma$

Solving them, we get $a = \frac{g (\sqrt{3} - 1)}{4}$

$\vec{a_{1}} = - acos 60^{0} \hat{i} - asin 60^{0} \hat{j} = - \frac{a}{2} \hat{i} - \frac{a \sqrt{3}}{2} \hat{j}$

$\vec{a_{2}} = - a \cos 30^{0} \hat{i} + a \cos 30^{0} \hat{j} = - \frac{a \sqrt{3}}{2} \hat{i} + \frac{a}{2} \hat{j}$

Now

$\vec{a_{cm}} = \frac{m_{1} \vec{a_{1}} + m_{2} \vec{a_{2}}}{m_{1} + m_{2}} = \frac{- \frac{ma}{2} (1 + \sqrt{3}) \hat{i} + \frac{ma}{2} (1 - \sqrt{3}) \hat{j}}{2 m}$

= $\frac{a}{4} [(1 + \sqrt{3}) \hat{i} + (1 - \sqrt{3}) \hat{j}]$

$| \vec{a_{cm}} | = \frac{a}{4} [\sqrt{{(1 + \sqrt{3})}^{2} + {(1 - \sqrt{3})}^{2}}] = \frac{a}{\sqrt{2}} = g \frac{(\sqrt{3} - 1)}{4 \sqrt{2}}$

Solution:

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