Question

In the arrangement shown in the figure, $m_A=2kg$ and $m_B=1kg$. String is light and inextensible. Find the acceleration of centre of mass of both the blocks :

Neglect friction everywhere.

• A. $g/3$ downwards
• B. $g/9$ upwards
• C. $g/9$ downwards
• D. $g/3$ upwards

Answer 1

Answer: (C)

$g/9$ downwards

As $m_A>m_B$ so acceleration of A is downward and B is upward. For downward assume acceleration is positive and for upward acceleration negative. i.e, $a_A=a$ and $a_B=-a$
Here, $(m_A+m_B)a=(m_A-m_B)g$

$(2+1)a=(2-1)g\Rightarrow a=\frac{g}{3}$

thus, $a_{cm}=\frac{m_A{a}_A+m_B{a}_B}{m_a+m_B}=\frac{2\left(g/3\right)+1(-g/3)}{2+1}=g/9$ downward. (as it is positive)