Let third mass particle $(2 \mathrm{~m}$ ) moves making angle $\theta$ with $X$-axis.

The horizontal component of velocity of $2 \mathrm{~m}$ mass particle $=u \cos \theta$ and vertical component $=u \sin \theta$

From conservation of linear momentum in $X$-direction
$\begin{array}{rlrl}\text { or } & m_1 u_1+m_2 u_2 =m_1 v_1+m_2 v_2 \\ \text { or } & 0 =m \times 4+2 m(u \cos \theta) \\ \text { or } & -4 =2 u \cos \theta\end{array}$

Again, applying law of conservation of linear momentum in $Y$-direction.
$\begin{aligned}
0 & =m \times 6+2 m(u \sin \theta) \\
\Rightarrow \quad-\frac{6}{2} & =u \sin \theta
\end{aligned}$
or $\quad-3=u \sin \theta \quad \ldots$ (ii)
Squaring Eqs. (i) and(ii) and adding, we get
$\begin{aligned}
\text { (4) } +(9) & =u^2 \cos ^2 \theta+u^2 \sin ^2 \theta \\
& =u^2\left(\cos ^2 \theta+\sin ^2 \theta\right)
\end{aligned}$
$\begin{aligned} \text { or } 13=u^2 \\ \text { or } u=\sqrt{13} \mathrm{~ms}^{-1}\end{aligned}$