Answer:
[tex]V_s = 1.8*10^5m/s[/tex]
Explanation:
There is no external force applied, therefore there is a moment's preservation throughout the trajectory.
Initial momentum = Final momentum.
The total mass is equal to
[tex]m_T= m_1 +m_2[/tex]
Where,
[tex]m_1 =[/tex] mass of ship
[tex]m_2 =[/tex] mass of fuell expeled.
As the moment is conserved we have,
[tex]0=V_fm_2+V_sm_1[/tex]
Where,
[tex]V_f =[/tex] Velocity of fuel
[tex]V_s =[/tex]Velocity of Space Ship
Solving and re-arrange to [tex]V_s[/tex]we have,
[tex]V_s = \frac{V_f m_2 }{m_1}[/tex]
[tex]V_s = \frac{3/5c}{10^6}[/tex]
[tex]V_s = 3.5*10^{-3}c[/tex]
Where c is the speed of light.
Therefore the ship be moving with speed
[tex]V_s = \frac{3}{5}*10^{-3}*3*10^8m/s[/tex]
[tex]V_s = 1.8*10^5m/s[/tex]
