$X$ component of acceleration of $C O M$ is

(a_{c m})_{x} = \frac{m_{1} m_{2} g}{m_{1} + m_{2}}

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(a_{c m})_{x} = \frac{m_{1} m_{2} g}{(m_{1} + m_{2})^{2}}

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(a_{c m})_{x} = {(\frac{m_{2}}{m_{1} + m_{2}})}^{2} g

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(a_{c m})_{x} = {(\frac{m_{2}}{m_{1} + m_{2}})}^{2}

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Solution

The correct option is A $(a_{c m})_{x} = \frac{m_{1} m_{2} g}{m_{1} + m_{2}}$
For finding the acceleration of COM ,we have to first find the acceleration of blocks individually,

Let the tension in rope be T and acceleration of the block m $_{2}$ be a downwards which also implies that acceleration of bock m $_{1}$ will be a towards right

m $_{2}$ g-T=m $_{2}$ a
T=m $_{1}$ a

From where we get a = $\frac{m_{2}}{m_{1} + m_{2}} g$

(a $_{c m})_{x} = \frac{m_{1} m_{2} g}{(m_{2} + m_{1})}$

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