Question

In the arrangement shown in figure, $m_A=2\text{kg}$ and $m_B=1\text{kg}$. String is light and inextensible. The acceleration of centre of mass of both the blocks is
(Neglect friction)

• A. $\frac{g}{9}(\uparrow )$
• B. $\frac{g}{6}(\uparrow )$
• C. $\frac{g}{9}(\downarrow )$
• D. $g(\downarrow )$

Answer 1

The correct option is C $\frac{g}{9}(\downarrow )$

Given:
$m_A=2\text{kg};m_B=1\text{kg}$

As $m_A>m_B$,​ so acceleration of $\text{A}$ is downward and $\text{B}$ is upward.

The net pulling force on the system is,

$F_P=(m_A-m_B)g=(2-1)g=g$

The total mass being pulled is,

$m_T=m_A+m_B=3\text{kg}$

So,

$a=\frac{\text{Net pulling force}}{\text{Total mass}}=\frac{g}{3}$

Now,

$a_{\text{COM}}=\frac{m_A{a}_A+m_B{a}_B}{m_A+m_B}$

For downward motion, assume acceleration is negative and for upward motion, is acceleration positive. i.e, $a_A=-a=-g/3$ and $a_B=a=g/3$. So, accleration of centre of mass of system is

$a_{\text{COM}}=\frac{\left(2\right)(-a)+\left(1\right)\left(a\right)}{1+2}=-\frac{a}{3}$

$\therefore a_{\text{COM}}=\frac{g}{9}(\downarrow )$

Hence, option $\left(C\right)$ is the correct answer.