Why negative masses don't break conservation of momentum

[robly18]

So when I first looked at negative masses' behavior the first thing that came through my mind was that they broke conservation of momentum. Because, y'know:

Gravity Fun at TestTubeGames.com: [ForceG: -2,Qual: 1,Zoom: 1,xSet: 1,ySet: -3], [x0: -10,y0: 0,vx: 0,vy: 0,t0: 0,who: 2,m: 100], [x0: 10,y0: 0,vx: 0,vy: 0,t0: 0,who: 2,m: -100]

But I did some research and I found the reason for why it actually doesn't break conservation of momentum!

So, the law goes as follows:

For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

(source: http://www.physicsclassroom.com/class/m ... /u4l2b.cfm )
So, momentum, or kinetic energy can be expressed as F = mv. What the sentence up there just said was basically that the overall kinetic energy of a closed system stays constant. So this means that m1 * v1 + m2 * v2 = vi

m1 and v1 being the mass and velocity of object 1, m2 and v2 being the same for object to, and vi being the initial velocity.

So let's try a little bit of a test with two positive masses. Take the following scenario:

Gravity Fun at TestTubeGames.com: [ForceG: -2,Qual: 1,Zoom: 1,xSet: 1,ySet: -3], [x0: -100,y0: 0,vx: 0,vy: 0,t0: 0,who: 2,m: 100], [x0: 100,y0: 0,vx: 0,vy: 0,t0: 0,who: 2,m: 100]

So we got 2 objects with mass 100, one at 100,0 and the other one at -100,0, both with 0 momentum.
At T=0 we got:

100 * 0 + 100 * 0 = 0 or 0 = 0. This holds true!

After one iteration(where G=1), let's say T=1 we get:

100 * 0.25 - 100 * 0.25 = 0

Yep, this checks out.
Now what happens if we add a negative mass into the mix?

Gravity Fun at TestTubeGames.com: [ForceG: -2,Qual: 1,Zoom: 1,xSet: 1,ySet: -3], [x0: -100,y0: 0,vx: 0,vy: 0,t0: 0,who: 2,m: 100], [x0: 100,y0: 0,vx: 0,vy: 0,t0: 0,who: 2,m: -100]

100 * 0 + (-100) * 0 = 0 or 0 = 0.

Yep, nothing wrong here. But, after one iteration...

100 * 0.25 + -100 * 0.25 = 0 or 25 + (-25) = 0 which means 0 = 0

Yep, this checks out too.

So what I mean is, these DON'T break conservation of momentum! Physics is happy again.

[A Random Player]

Or, y'know, just remember that increasing the speed of a negaparticle decreases overall momentum.

[robly18]

Or, y'know, just remember that increasing the speed of a negaparticle decreases overall momentum.

My way is more scientific

[testtubegames]

Or, y'know, just remember that increasing the speed of a negaparticle decreases overall momentum.

My way is more scientific

Yeah, things can get counter-intuitive really fast with negative masses. Put a +M mass next to a -M mass, and they can zoom anywhere - in any which way - without breaking conservation of momentum. (Since together they have zero mass... and never have momentum.)

...maybe that's how superman works?

[robly18]

(Since together they have zero mass... and never have momentum.)

I actually thought of that, but it doesn't work if you have a mass of 100 and another of -10 :\

EDIT: Actually, never mind that. Just found out that if the masses aren't the same, then they orbit around a point. That's interesting.

I also accidentally made a pretty pattern
Gravity Fun at TestTubeGames.com: [ForceG: -2,Qual: 1,Zoom: 1,xSet: 1,ySet: -3], [x0: -100,y0: 0,vx: 0,vy: 0,t0: 0,who: 2,m: 100], [x0: 100,y0: 0,vx: 0,vy: 0,t0: 0,who: 2,m: -100]