Answer
$ v_3=\dfrac{v_0}{\sqrt2} $
$45^\circ$
Work Step by Step
Let's write the velocities of the two pieces in component forms. We chose east and north to be our positive $x$ and $y$ directions respectively.
$$v_1=-v_0\;\hat i+0\;\hat j$$
$$v_2=0\;\hat i-v_0\;\hat j$$
After the explosion, there are no external forces exerted on the three pieces in the horizontal plane $x$-$y$ plane [East-North]. This means that during the explosion the momentum is conserved.
Thus, the momentum before the explosion is equal to the momentum after it.
$$p_i=p_f$$
where the momentum before the explosion was zero since the velocity of the three pieces [when they are still one unit] was zero.
Hence,
$$p_f=m_1v_1+m_2v_2+m_3v_3=0$$
We know that $m_1=m_2=m$ and that $m_3=m_1+m_2=2m$.
$$m v_1+m v_2+2m v_3=0$$
Solving for $v_3$;
$$v_3=\dfrac{-(m v_1+m v_2)}{2m}=\dfrac{-(v_1+v_2)}{2}$$
Plugging $v_1$ and $v_2$ from the first two formulas above;
$$v_3= \dfrac{-(-v_0\;\hat i+0\;\hat j+0\;\hat i-v_0\;\hat j)}{2}$$
$$v_3=0.5v_0\;\hat i+0.5v_0\;\hat j$$
Hence, the speed is given by applying the Pythagorean theorem.
$$v_3=\sqrt{v_x^2+v_y^2}=\sqrt{(0.5v_0)^2+(0.5v_0)^2}$$
$$\boxed{v_3=\dfrac{v_0}{\sqrt2}}$$
And its direction is given by
$$\tan\theta=\dfrac{v_y}{v_x}$$
$$\theta=\tan^{-1}\left(\dfrac{v_y}{v_x}\right)=\tan^{-1}\left(\dfrac{0.5v_0}{ 0.5v_0 }\right)=\color{red}{\bf45^\circ}$$
The velocity $v_3$ has a positive $x$-component which means that it has an east component and has a positive $y$-component which means it has a north component. So the velocity vector is on the first quadrant.
angle here of $45^\circ$ is north of the east or east of the north.
