• # question_answer 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 everywhere, inclined wedge is fixed at floor) - A) $\left( \frac{\sqrt{3}-1}{4\sqrt{2}} \right)g$          B) $(\sqrt{3}-1)g$ C) $\frac{g}{2}$                            D) $\left( \frac{\sqrt{3}-1}{\sqrt{2}} \right)g$

[A]  NLM on B         $mg\sin 30{}^\circ -T=m\times a$                      ?(1) $T-mg\,\,\sin 60{}^\circ =ma$                                                   ?(2) $a=\frac{mg\,\,\sin 60{}^\circ -mg\,\,\sin 30{}^\circ }{2m}$
 Where a is Acceleration of system Here, m=mass of each block  or $a=\left( \frac{\sqrt{3}-1}{4} \right)g$ Acceleration of center of mass $\to {{a}_{com}}$ Now $\to {{\overrightarrow{a}}_{\,com}}=\frac{m{{\overrightarrow{\,a}}_{1}}+m\,{{\overrightarrow{a}}_{2}}}{2m}$ Here, ${{\overrightarrow{a}}_{1}}$ and ${{\overrightarrow{a}}_{2}}$are at $\left( \frac{\sqrt{3}-1}{4} \right)g$right angles. Hence, $\left| \to {{\overrightarrow{a}}_{com}} \right|=\frac{\sqrt{2}}{2}a=\left( \frac{\sqrt{3}-1}{4\sqrt{2}} \right)g$