Question

Two blocks $A$ and $B$ each of equal mass $m$ are released from the top of a smooth fixed wedge as shown in figure. Find the magnitude of acceleration of $COM$ of the two blocks.

• A. $2g$
• B. $g$
• C. $\frac{2g}{3}$
• D. $\frac{g}{2}$

Answer 1

The correct option is D $\frac{g}{2}$

Let us consider the inclined surfaces of the wedge as $x$-axis and $y$-axis respectively as shown in the figure below :-


Therefore, we can write acceleration of block $A$ and $B$ as :
${\overrightarrow{a}}_A=mg\sin {30}^{\circ }\hat{i}$ [$x$- axis direction]
${\overrightarrow{a}}_B=mg\sin {60}^{\circ }\hat{j}$ [$y$- axis direction]
Acceleration of COM
${\overrightarrow{a}}_{COM}=\frac{m_A{\overrightarrow{a}}_A+m_B{\overrightarrow{a}}_B}{m_A+m_B}$
${\overrightarrow{a}}_{COM}=\frac{mg\sin {30}^{\circ }\hat{i}+mg\sin {60}^{\circ }\hat{j}}{m+m}$
${\overrightarrow{a}}_{COM}=\frac{g}{4}i+\frac{\sqrt{3}g}{4}j$
$\Rightarrow |{\overrightarrow{a}}_{COM}|=g/2$