JEE Main 2026 — Center of Mass Momentum and Collision Question with Solution

The acceleration of the blocks in the Atwood machine is given by:

$a=\frac{m_1-m_2}{m_1+m_2}g$

Substituting the given values ($m_1=2$ kg, $m_2=1$ kg):

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

Taking the downward direction as positive, the acceleration of the $2$ kg block is $a_1=\frac{g}{3}$ and the acceleration of the $1$ kg block is $a_2=-\frac{g}{3}$.

The acceleration of the centre of mass is:

$a_{cm}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}$

$a_{cm}=\frac{2\left(\frac{g}{3}\right)+1\left(-\frac{g}{3}\right)}{2+1}=\frac{\frac{g}{3}}{3}=\frac{g}{9}$

Since the system is released from rest, the initial velocity of the centre of mass is zero.

The distance traversed by the centre of mass in $t=2$ s is:

$S_{cm}=\frac{1}{2}{a}_{cm}{t}^2$

$S_{cm}=\frac{1}{2}\left(\frac{10}{9}\right)(2{)}^2=\frac{20}{9}\approx 2.22\text{ m}$

Answer: $2.22$