Question

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

• A. $\frac{g}{4}$
• B. $\frac{g}{2}$
• C. $\frac{g}{6}$
• D. g

Answer 1

The correct option is B

$\frac{g}{2}$

Let us consider incline surfaces of wedge as x-axis and y-axis respectively. We can write accelerations of blocks A and B

${\overrightarrow{a}}_A=mgsin{30}^{\circ }\hat{i}and{\overrightarrow{a}}_B=mgcos{30}^{\circ }\hat{j}$
The acceleration of center of mass of two blocks can be written as
${\overrightarrow{a}}_{cm}=\frac{m_A{\overrightarrow{a}}_A+m_B{\overrightarrow{a}}_B}{m_A+m_B}=\frac{mgsin{30}^{\circ }\hat{i}+mgsin{60}^{\circ }\hat{j}}{m+m}$
${\overrightarrow{a}}_{cm}=\frac{gsin{30}^{\circ }\hat{i}+gsin{60}^{\circ }\hat{j}}{2}{\overrightarrow{a}}_{cm}=\frac{g}{4}\hat{i}+\frac{\sqrt{3}g}{4}\hat{j}\Rightarrow \left|{\overrightarrow{a}}_{cm}\right|=\frac{g}{2}$