A pulley mass system is shown in the figure.


The pulley and string are massless and all surfaces are frictionless. Find out the acceleration of $COM$ of the system. [Take $g=10{\text{m/s}}^2$ and consider upward direction as positive]

The correct option is A $-0.4{\text{m/s}}^2$
$FBD$ of the blocks are:


For Block $A$ and $B$:
$6g-T=6a$ and
$T-4g=4a$
Solving these equations,
$10a=6g-4g=2g$
$\Rightarrow a=\frac{2\times 10}{10}=2{\text{m/s}}^2$

Acceleration of block $A$ will be downward while acceleration of block $B$ will be upward.
Therefore, $a_{com}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}=\frac{6\times (-2)+4\times 2}{6+4}$
$a_{com}=\frac{-4}{10}=-0.4{\text{m/s}}^2$
It means $COM$ will be accelerating downwards.