In the system of two masses $m_{1}$ and $m_{2}$ tied through a light string passing over a smooth light pulley, find the acceleration of COM (Centre of mass).

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

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

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

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

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Solution

The correct option is B $\frac{(m_{1} - m_{2})^{2} g}{(m_{1} + m_{2})^{2}}$
FBD is shown above.
Let the acceleration of the masses be $a$ and tension in the string be $T$ .
For mass $m_{1}$ :
$m_{1} a = m_{1} g - T$ ....(1)
For mass $m_{2}$ :
$m_{2} a = T - m_{2} g$ ...(2)
Adding (1) and (2), we get $(m_{1} + m_{2}) a = (m_{1} - m_{2}) g$
$⟹ a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}$
Acceleration of centre of mass $a_{c m} = \frac{m_{1} a - m_{2} a}{m_{1} + m_{2}}$
$⟹ a = \frac{(m_{1} - m_{2})}{m_{1} + m_{2}} a = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}})^{2} g$

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