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

The initial moment of inertia of the system about the vertical axis of rotation is

$I_i = m\left(\dfrac{L}{2}\right)^2 + m\left(\dfrac{L}{2}\right)^2 = \dfrac{mL^2}{2}$

The initial angular momentum of the system is

$L_i = I_i \omega = \dfrac{mL^2}{2} \omega$

When the stationary particle of mass $m$ sticks to ball $A$, the mass at $A$ becomes $2m$. The final moment of inertia of the system is

$I_f = 2m\left(\dfrac{L}{2}\right)^2 + m\left(\dfrac{L}{2}\right)^2 = \dfrac{3mL^2}{4}$

Since no external torque acts on the system about the axis of rotation, the angular momentum is conserved. Equating the initial and final angular momentum:

$L_i = L_f$

$\dfrac{mL^2}{2} \omega = \dfrac{3mL^2}{4} \omega'$

$\omega' = \dfrac{2\omega}{3}$