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)

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

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
${\vec{a}}_{cm}=\frac{m_1{\vec{a}}_1+m_2{\vec{a}}_2}{m_1+m_2}=\frac{-\frac{ma}{2}(1+\sqrt{3})\hat{i}+\frac{ma}{2}(1-\sqrt{3})\hat{j}}{2m}$
$=\frac{a}{4}[(1+\sqrt{3})\hat{i}+(1-\sqrt{3})\hat{j}]$
$\left|{\vec{a}}_{cm}\right|=\frac{a}{4}\left[\sqrt{(1+\sqrt{3}{)}^2+(1-\sqrt{3}{)}^2}\right]$
$=\frac{a}{\sqrt{2}}=g\frac{(\sqrt{3}-1)}{4\sqrt{2}}$