# Momentum in spacecraft

1. Feb 13, 2017

### Leyzorek

1. The problem statement, all variables and given/known data
Spacecraft, with vacuum inside, in free-fall, contains a small mass captured against a fully-compressed compression spring against one wall of the craft. Release the spring and mass. Mass is accelerated by recovering spring (elastic potential energy turning to kinetic), spacecraft is, of course, accelerated in opposite direction. Spring once fully extended also lifts off behind mass at its average velocity relative to the spacecraft, and while travelling oscillates in length about its unstressed length.

The kinetic energies of spring and mass on the one hand and the spacecraft on the other, must be equal and opposite.

Mass and spring eventually hit the opposite wall of the spacecraft. IF the collision is fully elastic, (and assuming axis of motion passes through center of gravity of system, nothing starts to spin) spacecraft should stop and accelerate in opposite direction while spring and mass return to their original position, repeat cycle..Correct so far?

What IF the collision of spring and mass with opposite wall is substantially inelastic, for instance into a sand-bag which converts significant portion of kinetic energy into frictional heat? Does the whole system end up with kinetic energy in direction opposite to the original spring-driven travel of the small mass?

2. Relevant equations

None
3. The attempt at a solution
Described above

2. Feb 13, 2017

### CWatters

No. Momentum is always conserved.

If the space craft was initially at rest it will end up at rest after the inelastic collision. The energy that was originally in the spring ends up as heat in the sand bag/mass.

If the system boundary only includes the KE of the spacecraft and mass (and doesn't include heat in the sand bag) then energy is not conserved

3. Feb 13, 2017

### Staff: Mentor

A description is not a solution unless there's math to back it up. The lack of relevant equations is worrisome. What conservation laws apply? Which one applies no matter if the collision is elastic or inelastic?

4. Feb 13, 2017

### Orodruin

Staff Emeritus
Kinetic energy is not a vector, it does not have a direction.

5. Feb 14, 2017

### Leyzorek

Of course energy is conserved, and if the collisions at each end of the craft are equally and perfectly elastic, then the center of mass will be stationary and the craft's hull will oscillate opposite to the sprung mass inside. But if some of that energy is converted to heat at one end how can that heat (RANDOM MOTION) reconvert to kinetic energy? IE , Orodruin, does the whole assembly (CG) end up in motion?

Thanks!

6. Feb 14, 2017

### jbriggs444

What makes you think that it does reconvert to kinetic energy?

7. Feb 14, 2017

### Leyzorek

I think it does not. So, does that mean that the CG of the whole system ends up in motion?

8. Feb 14, 2017

### jbriggs444

Is momentum conserved?

[Also, review @Orodruin's post #4 above]

9. Feb 14, 2017

### Leyzorek

If some of the momentum (kinetic energy) that would normally cancel the spacecrafts momentum in the other direction was converted to heat (with the sandbag) it would seem that energy would be conserved even though momentum would not be conserved.
that was the original question, is momentum conserved?

10. Feb 14, 2017

### jbriggs444

Kinetic energy and momentum are two different things. Please review post #2 by @CWatters.

11. Feb 14, 2017

### Leyzorek

I guess unfamiliarity with terms is preventing me from stating my question clearly. Let me try again.

Begin, box (spacecraft) stationary, in free fall. Inside box, against one wall is a precompressed spring with a small mass against the end of the spring opposite the wall.

Release spring. Its elastic potential energy turns to kinetic energy (and momentum), as it accelerates the box and the small mass in opposite directions.

IF the small mass reached the opposite wall of the box and was captured by a similar spring-and latch assembly so force curve equal and opposite to the original spring-release event was applied to that opposite wall, and all the kinetic energy and momentum of the small mass was reconverted to elastic potential energy, then of course the small mass and box would decelerate together (though in opposite directions) and all parts would return to their original resting state.

BUT if instead of being caught by a lossless perfectly elastic spring-and-latch, the small mass comes to rest in a sandbag which converts some of the energy to heat, then energy is not lost, but it seems that momentum is, and the box-and-small-mass would continue moving in the direction in which the box alone was originally accelerated by the energy released by the spring.

This does not FEEL right to me, I am eager to be told I am wrong but need to know why/where. The heat energy in the sand has to come out of the kinetic energy of the small mass, and will ( it seems) be unavailable to decelerate the box.

12. Feb 14, 2017

### Staff: Mentor

That's where your intuition is leading you astray. KE and momentum are separate things. In an isolated system momentum is always conserved independently of energy (KE, PE, heat,...).