# 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 centre of mass of the system is

$\begin {array} {1 1} (a)\;(n-1)^2g & \quad (b)\;\bigg(\frac{n+1}{n-1}\bigg)^2 g \\ (c)\;\bigg(\frac{n-1}{n+1}\bigg)^2 g & \quad (d)\;\bigg(\frac{n+1}{n-1}\bigg)g \end {array}$

Given $\large\frac{m_1}{m_2}$$= \large\frac{n}{1}$$ = n$
Each mass will have the acceleration $a=\large\frac{(m_1-m_2)g}{m_1+m_2}$
However $m_1$ which is heavier will have the will have acceleration $a_1$ vertically down while the lighter mass $m_2$ will have acceleration $a_2$ vertically up $\rightarrow a_2 = -a_1$
The acceleration of the center of mass of the system, $a_{cm}=\large\frac{m_1 \bar {a_1}+m_2 \bar {a_2}}{m_1+m_2}$
Given that $a_2 = -a_1 \rightarrow a_{cm} = \large\frac{(m_1 - m_2)a_1}{m_1 + m_2}$ $=\large\frac{m_1 - m_2}{m_1 + m_2}$$\times \large\frac{(m_1-m_2)g}{m_1+m_2}$$ = \large\frac{(m_1-m_2)^2g}{(m_1+m_2)^2}$
Since $\large\frac{m_1}{m_2}$ $= n$, diving by $m_2$ and simplifying $\Rightarrow a_{cm} = \bigg(\large \frac{n-1}{n+1}\bigg)^2$$g$
Hence c is the correct answer
edited Jun 17, 2014