Answer
The lower stage's velocity has a magnitude of $4500m/s$ and is directed in the same direction as the rocket at first.
Work Step by Step
We assume the rocket's direction of motion at the start is the positive direction.
At the start, 2 stages of the rocket acts as a system, with mass $m+M=1200+2400=3600kg$ and velocity $\vec{v_0}=+4900m/s$
After the explosion, the smaller stage $m=1200kg$ has velocity $\vec{v_f}=+5700m/s$. The larger stage $M=2400kg$ has velocity $\vec{V_f}$, which needs to be found.
The principle of conservation of linear momentum states that $$m\vec{v_f}+M\vec{V_f}=(m+M)\vec{v_0}$$ $$\vec{V_f}=\frac{(m+M)\vec{v_0}-m\vec{v_f}}{M}$$ $$\vec{V_f}=\frac{1.76\times10^7-6.84\times10^6}{2.4\times10^3}=+4500m/s$$
- Magnitude: $V=4500m/s$
- Direction: the positive sign means its direction is the same as the rocket's direction at first.
