Question

Two particles having mass ratio $n:1$ are interconnected by a light inextensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is :

• A. $(n-1{)}^{2}g$
• B. ${\left(\frac{n+1}{n-1}\right)}^{2}g$
• C. ${\left(\frac{n-1}{n+1}\right)}^{2}g$
• D. $\left(\frac{n+1}{n-1}\right)g$

Answer 1

The correct option is C ${\left(\frac{n-1}{n+1}\right)}^{2}g$


Let mass of block A is $m$ and the mass of block B is $nm$
$nmg-T=nma$
$T-mg=ma$
Adding the above equations, we get
$a=\frac{(n-1)g}{n+1}$
Acceleration of the centre of mass of system.
$a_{COM}=\frac{m_1\overrightarrow{a_1}+m_2\overrightarrow{a_2}}{m_1+m_2}$
$a_{COM}=\frac{ma-nma}{(n+1)m}=\frac{a-na}{n+1}=-{\left(\frac{n-1}{n+1}\right)}^{2}g$
Here $-ve$ sign indicated that COM is moving downwards.

$|a_{COM}|={\left(\frac{n-1}{n+1}\right)}^{2}g$