Mass of the system vs. rest masses

1. Feb 2, 2012

danda

1. The problem statement, all variables and given/known data
Two particles of mass 4 and 5 kg (respectively) move towards each other. If the mass of the system is 20kg what can you say, qualitatively and quantitatively about the system in the center of mass frame?

2. Relevant equations

In the CM frame net momentum is zero:
gamma1*m1*v1 + gamma2*m2*v2 = 0

mass of the system:
gamma1*mass1 + gamma2*mass2 = 20

3. The attempt at a solution

I know that this problem with these two equations and two unknowns is solvable, but the algebra is extremely long and difficult. Is there a more elegant, short way to solve the problem? Perhaps using E^2 = (pc)^2 + (mc^2)^2 ?

2. Feb 3, 2012

ardie

in fact there's a very short way. use the relativistic mass formulae, which is basically
mass = gamma . rest mass
in CM of say the 5kg particle, the particle itself has no velocity, therefore no momentum, and just 5kg of mass. so since the net mass is 20kg, the other particle appears to have 15 kg of mass. so you know the rest mass of that particle is 4kg, and its relativistic mass is 15kg, you can workout the gamma and hence the relative velocity of the two particles.

3. Feb 3, 2012

vela

Staff Emeritus
It's a good idea to try avoid using velocities. It just makes the algebra harder, as you probably found out. Stick with energy E, momentum p, and mass m and use the equation you cited. In terms of those quantities, your equations become
\begin{align*}
\vec{p}_1 + \vec{p}_2 &= 0 \\
E_1 + E_2 &= Mc^2
\end{align*} where M is the mass of the system. Try squaring both equations.

4. Feb 6, 2012

danda

I did that and got reasonable answers (.96 and .95 c, respectively), but I'm not sure where to go from there.

5. Feb 6, 2012

vela

Staff Emeritus
I got different speeds. What did you find for the particles' energy and momentum?

6. Feb 6, 2012

danda

I got that in the frame where 1(4kg) is stationary 2 approaches at .9c, while in the frame where 2(5kg) is stationary 1 approaches at .92c.

I did this by writing: m(1) + (gamma2)(m(2)) = 20 and vice versa, solving for the u in the gamma.

So plugging in, in the frame where E(1) is stationary E(total) = m(1)c^2 + 3.16(m(2))c^2

and the same basic thing for the other velocity.

Momentum would be done the say way, I guess. But still, wouldn't I have to find the speed of the CM frame, and somehow write those equations over again with some dependence on the that value?

7. Feb 6, 2012

vela

Staff Emeritus
That can't be correct. If mass 1 moves with speed β relative to mass 2, mass 2 moves with the same speed relative to mass 1.

Did you try what I suggested in post 3?