Question

A space vehicle is traveling at 5425 km/h relative to the Earth when the exhausted rocket motor (mass 4m) is disengaged and sent backward with a speed of 81 km/h relative to the command module (mass m). What is the speed of the command module relative to Earth just after the separation

Answer 1

Answer:

$V_{cE}=1489m/s$

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

$p_{i}=p_{f}$

Since the motion in on direction we can drop the unit vector direction

$MV_{i}=4mV_{mE}+mV_{CE}$

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

$V_{mE}=V_{mc}+V_{cE}$

Where Vmc is the velocity of motor relative to command

This yields

$5mV_{i}=4m(V_{mc}+V_{cE})+mV_{cE}\\5V_{i}=4V_{mc}+5V_{cE}$

Substitute the given values

$V_{cE}=\frac{5V_{i}-4V_{mc}}{5}\\ V_{cE}=5425*\frac{1000}{60*60}(m/s)-\frac{4}{5}*81\frac{1000}{60*60}(m/s)\\ V_{cE}=1489m/s$