Three identical balls each of mass $m=0.5kg$ are connected with each other as shown in figure and rest over a smooth horizontal table. At moment  $t=0$, ball B is imparted a horizontal velocity  $v_o=9m{s}^{-1}$, Calculate velocity of A just before it collides with ball C

Initial state at  $t = 0$ and final state at time t are shown in Figure.
Applying conservation of momentum at time t, we get
 $$\begin{gathered} m{v}_o=mv\text{'}+2m{v}_y \\ \Rightarrow v_0=v\text{'}+2v_y \end{gathered}$$
At the time of collision $\theta =0$ and  $v\text{'}=v_y$
 $\Rightarrow v\text{'}=v_y=\frac{v_0}{3}$
Applying conservation of energy, we get
$$\begin{gathered} \frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2+2\left[\frac{1}{2}m(v_x^2+v_y^2)\right] \\ \Rightarrow v_0^2=\frac{v_0^2}{9}+2(v_x^2+\frac{v_0^2}{9}) \\ \Rightarrow v_0^2(1-\frac{1}{9}-\frac{2}{9})=2v_x^2 \\ \Rightarrow v_0^2\left(\frac{9-3}{9}\right)=2v_x^2 \\ \Rightarrow v_x=\frac{v_0}{\sqrt{3}} \end{gathered}$$

So, net velocity of A at the time of collision is 

$$\begin{gathered} v_A=\sqrt{v_x^2+v_y^2} \\ \Rightarrow v_A=\sqrt{\frac{v_0^2}{3}+\frac{v_0^2}{9}}=v_0\sqrt{\frac{4}{9}}=\frac{2v_0}{3} \\ \Rightarrow v_A=\frac{2v_0}{3}=\left(\frac{2}{3}\right)\left(9\right)=6m{s}^{-1} \end{gathered}$$