Two small balls A and B , each having mass m , are rigidly...

Let the initial direction of motion of the rod be along the positive x-axis.

The system consists of three masses: ball $A$, ball $B$, and particle $P$.
Mass of ball $A$, $m_A = m$
Mass of ball $B$, $m_B = m$
Mass of particle $P$, $m_P = m$

Initial velocities of the components are:
$\vec{v}_A = v_0 \hat{i}$
$\vec{v}_B = v_0 \hat{i}$
$\vec{v}_P = 0$

The total initial linear momentum of the system $A + B + P$ is:
$\vec{P}_i = m_A \vec{v}_A + m_B \vec{v}_B + m_P \vec{v}_P$
$\vec{P}_i = m(v_0 \hat{i}) + m(v_0 \hat{i}) + m(0) = 2mv_0 \hat{i}$

Since there are no external horizontal forces acting on the system, the total linear momentum is conserved during the collision.
Let $\vec{v}_{cm}$ be the velocity of the centre of mass of the system after the collision. The total mass of the system is $M = m_A + m_B + m_P = 3m$.

The final linear momentum of the system is:
$\vec{P}_f = M \vec{v}_{cm} = 3m \vec{v}_{cm}$

By conservation of linear momentum:
$\vec{P}_i = \vec{P}_f$
$2mv_0 \hat{i} = 3m \vec{v}_{cm}$
$\vec{v}_{cm} = \dfrac{2}{3}v_0 \hat{i}$

Thus, the velocity of the centre of mass of the system is $\dfrac{2}{3}v_0$ along the initial motion of the rod.