Question

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

Answer 1

Answer: (A)

$m g \sin 60^{\circ}-T=m a, T-m g \sin 30^{\circ}=m a$

Solving them, we get $a=\frac{g(\sqrt{3}-1)}{4}$
$\vec{a}_{1}=-a \cos 60^{0} \hat{i}-a \sin 60^{\circ} \hat{j}=-\frac{a}{2} \hat{i}-\frac{a \sqrt{3}}{2} \hat{j}$
$\vec{a}_{2}=-a \cos 30^{0} \hat{i}+a \cos 30^{\circ} \hat{j}=-\frac{a \sqrt{3}}{2} \hat{i}+\frac{a}{2} \hat{j}$
Now
$\vec{a}_{ cm }=\frac{m_{1} \vec{a}_{1}+m_{2} \vec{a}_{2}}{m_{1}+m_{2}}=\frac{-\frac{m a}{2}(1+\sqrt{3}) \hat{i}+\frac{m a}{2}(1-\sqrt{3}) \hat{j}}{2 m}$
$=\frac{a}{4}[(1+\sqrt{3}) \hat{i}+(1-\sqrt{3}) \hat{j}] $
$\left|\vec{a}_{ cm }\right| =\frac{a}{4}\left[\sqrt{(1+\sqrt{3})^{2}+(1-\sqrt{3})^{2}}\right]$
$=\frac{a}{\sqrt{2}}=g \frac{(\sqrt{3}-1)}{4 \sqrt{2}} $