Two particles $A$ and $B$, initially at rest, moves towards each other under a mutual force of attraction. At the instant when the speed of $A$ is $v$ and the speed $B$ is $2v$, the speed of centre of mass is :

The correct option is A $zero$
Force $F_A$ on particle $A$ is given by
$F_A=m_A{a}_A=\frac{m_{A}v}{t}...\left(1\right)$
Similarly $F_B=m_B{a}_B=\frac{m_B\times 2v}{t}...\left(2\right)$
Now $\frac{m_{A}v}{t}=\frac{m_B\times 2v}{t}(\because F_A=F_B)$
So $m_A=2m_B$
For the centre of mass of the system
$v=\frac{m_A{v}_A+m_B{v}_B}{m_A+m_B}$
Or $v=\frac{2m_{B}v-m_B\times 2v}{2m_B+m_B}=0$
Negative sign is used bacause the particles are travelling in opposite directions.
Alternatively, if
we consider the two masses in a system then no external force is acting
on the system. Mutual forces are internal forces. Since the centre of
mass is initially at rest, it well remain at rest.