Physics Problem 2

Question

A space vehicle is traveling at 4220 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative speed between the motor and the command module is then 93 km/h. 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?

Answer 1

There are several ways to do this.  Here's one, as an algebra problem.

 

Since we don't actually know the masses of the motor and the module, let's use variables and see what happens during the calculations.  So let's call the mass of the module M, and the mass of the motor is 4M.  The total mass of the vehicle, then, is 5M.  Let's assume they're in kg, though it doesn't matter.

 

So the momentum of the vehicle before separation is 4220 * 5M = 21100 M kg-km/h.

 

Let's call the velocity of the motor v.  So the velocity of the module must be v + 93.

 

So after the separation, the momentum of the motor is 4Mv, and the momentum of the module is M(v + 93) = Mv + 93M.  The total momentum is 4MV + Mv + 93M = 5Mv + 93M.  That has to be the same as the initial momentum.  That gives us the following equation:

 

21100 M = 5Mv + 93M

 

Divide both sides by M:

21100 = 5v + 93

 

(Notice M is gone from the equation now, so in fact it doesn't matter that we didn't know the actual masses.)

 

21007 = 5v

4201.4 = v

 

But v is the velocity of the motor, so we have to add 93 to get the velocity of the module, for a total of 4294.4.

 

So the command module is traveling at 4294.4 km/h after the separation.