Two blocks A and B of equal mass are released on two sides of a fixed wedge C as shown in the figure. Find…

Acceleration of both the blocks will be $g \sin 45°$ or $\frac{g}{\sqrt{2}}$ at right angles to each other. Now,



${\overrightarrow{\text{a}}}_{\text{com}}=\frac{{\text{m}}_{\text{A}}{\overrightarrow{\text{a}}}_{\text{A}}+{\text{m}}_{\text{B}}{\overrightarrow{\text{a}}}_{\text{B}}}{{\text{m}}_{\text{A}}+{\text{m}}_{\text{B}}}$

Here,    $m_{\mathrm{A}} = m_{\mathrm{B}}$

$\therefore a_{\mathrm{com}}=\frac{1}{2}\left(a_{\text{A}}+a_{\text{B}}\right)$

                     $=\frac{1}{2}g$   (downwards)