Answer:
[tex]V_{cE}=1489m/s[/tex]
Explanation:
Given data
Space vehicle speed=5425 km/h relative to earth
The rocket motor speed=81 km/h and mass 4m
The command has mass m
From the conservation of momentum as the system isolated
[tex]p_{i}=p_{f}\\[/tex]
Since the motion in on direction we can drop the unit vector direction
[tex]MV_{i}=4mV_{mE}+mV_{CE}[/tex]
Where M is the mass of space vehicle which equals to sum of the motors mass and command mass.
The velocity of the motor relative to the earth equals the velocity of the motor relative to command plus the velocity of the command relative to earth
[tex]V_{mE}=V_{mc}+V_{cE}[/tex]
Where Vmc is the velocity of motor relative to command
This yields
[tex]5mV_{i}=4m(V_{mc}+V_{cE})+mV_{cE}\\5V_{i}=4V_{mc}+5V_{cE}[/tex]
Substitute the given values
[tex]V_{cE}=\frac{5V_{i}-4V_{mc}}{5}\\ V_{cE}=5425*\frac{1000}{60*60}(m/s)-\frac{4}{5}*81\frac{1000}{60*60}(m/s)\\ V_{cE}=1489m/s[/tex]
