Answer:
Approximately [tex]\rm 2.5\; m \cdot s^{-1}[/tex].
Explanation:
Let the increase in the rocket's velocity be [tex]\Delta v[/tex]. Let [tex]v_0[/tex] represent the initial velocity of the rocket. Note that for this question, the exact value of [tex]v_0[/tex] doesn't really matter.
The momentum of an object is equal to its mass times its velocity.
Momentum is conserved in an isolated system like the rocket and its fuel. That is:
Sum of initial momentum = Sum of final momentum.
[tex]1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500)[/tex].
Note that [tex]1000\, v_0[/tex] appears on both sides of the equation. These two terms could hence be eliminated.
[tex]0 = 995\, \Delta v - 5\times 500[/tex].
[tex]\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}[/tex].
Hence, the velocity of the rocket increased by around 2.5 m/s.