Two particles of different mass and uniform velocity are moving towards each other as shown in the figure. Find the displacement of the $COM$ after $5\text{s}$.

The correct option is A $50\text{m}$ towards left
As we know,
$V_{com}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$
Consider rightward movement as positive.
$V_{com}=\frac{2\times \left(10\right)+\left(4\right)\times (-20)}{2+4}$
$V_{com}=-10\text{m/s}$ {It means $V_{com}$ is towards left}.

Distance travelled by the $COM$ in $5\text{s}$
$S=ut+\frac{1}{2}a{t}^2$
Here, $a=0$
$\therefore S=u\times t=10\times 5=50\text{m}$
(towards left)