The two bodies of masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ respectively are tied to the ends of a string which passes over a light frictionless pulley. The masses are initially at rest and released. The acceleration of the centre of mass is

{(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} g

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(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g

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Solution

The correct option is A ${(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} g$
Let $a$ be acceleration of both the blocks,
the equations of motion of two blocks are
$m_{1} g - T = m_{1} a$
$T - m_{2} g = m_{2} a$
Eliminating $T$ , we get; $a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} g$ ........(1)
As there is no external force in horizontal direction, so the centre of mass of the system does not change along horizontal direction and it will only have a vertical acceleration.

$a_{C M} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}}$
Here $_{1} = a and_{2} = - a$ because if $m_{1}$ moves down, $m_{2}$ moves up.
$a_{C M} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} a$ .......(2)
However,
Putting equation (1) in equation (2)
We get
$a_{C M} = {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} g$

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The two bodies of masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ respectively are tied to the ends of a string which passes over a light frictionless pulley. The masses are initially at rest and released. The acceleration of the centre of mass is