Question

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

• A. $\frac{g}{2}$ , downwards
• B. $g$ , downwards
• C. $\frac{g}{2}$ , upwards
• D. $\frac{3}{4}g\text{,}$ downwards

Answer 1

Answer: (A)

Acceleration of both the blocks will be $gsin45^{\circ }$ or $\frac{g}{\sqrt{2}}$ at right angles to each other. Now,

${\overrightarrow{a}}_{\text{com}}=\frac{{\text{m}}_{\text{A}}{\overrightarrow{a}}_{\text{A}}+{\text{m}}_{\text{B}}{\overrightarrow{a}}_{\text{B}}}{{\text{m}}_{\text{A}}+{\text{m}}_{\text{B}}}$
Here, $m_A=m_B$
$\therefore a_{com}=\frac{1}{2}\left(a_{\text{A}}+a_{\text{B}}\right)$
$=\frac{1}{2}g$ (downwards)