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


If mass of block A is m and the mass of block B is nm.
nmg - T = nma
T - mg = ma
After solving
$a=\frac{(n-1)g}{n+1}$
Acceleration of the centre of mass of system.
$a_{COM}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}$
$=\frac{nma-ma}{(n+1)m}=\frac{na-a}{n+1}={\left(\frac{n-1}{n+1}\right)}^{2}g$