Linear momentum and relative velocity

[22steve]

Homework Statement

A space vehicle is traveling at 5300 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of 97 km/h relative to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation?

Homework Equations

p=mv

The Attempt at a Solution

I found the initial linear momentum of the space vehicle (5)(5300) = 26500
Then found the relative speed of the command modular by setting the momentum of the exhausted rocket and the module equal. (26.9 m/s)(4m) = (v)(1m) solving v=107.6 m/s or 387.6 km/h. I don't understand how to express the velocity relative to the Earth though.

 

[LowlyPion]

Homework Statement

A space vehicle is traveling at 5300 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of 97 km/h relative to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation?

Homework Equations

p=mv

The Attempt at a Solution

I found the initial linear momentum of the space vehicle (5)(5300) = 26500
Then found the relative speed of the command modular by setting the momentum of the exhausted rocket and the module equal. (26.9 m/s)(4m) = (v)(1m) solving v=107.6 m/s or 387.6 km/h. I don't understand how to express the velocity relative to the Earth though.

Welcome to PF.

The center of mass of the system has a velocity of 5300 km/h.

Focus then on the momentum of the two separating objects.

What does the conservation of momentum tell you about the momentum of each of the parts with respect to the other? Since you know the relative masses then shouldn't you be able to figure the relative proportion of the two velocities - in opposite directions - that together make for a speed of separation of 97 km/h?

 

[22steve]

Ah, yes, thank-you so much!