# Momentum & relativistic collision

1. Feb 5, 2013

### heartstar89

1. The problem statement, all variables and given/known data

A 3.000 u (1 u = 931.5 MeV/c2) object moving to the right through a laboratory at 0.8c collides with a 4.000 u object moving to the left through the laboratory at 0.6c. Afterward there are two objects, one of which is a 6.000 u object at rest.

A) Determine the mass and speed of the other object
B) Determine the change in kinetic energy of this collision

2. Relevant equations
1 u = 931.5 MeV/c2

γm1c2+γm2c2=γm3c2+γm4c2

KE = -mc^2 = KEf - KEi

3. The attempt at a solution

A) γ2794.5Mev+γ3726MeV=5589Mev+γm4c2

In this case, wouldn't there be two equations? One for the momentum (γumu)conserved and the other for energy (γumc^2) conserved

B) KE = KEf - KEi

Thank you for any help

2. Feb 5, 2013

### Staff: Mentor

Right. Alternatively, you can consider 4-vectors, where momentum and energy conservation are just two components of the more general enery-momentum conservation.

3. Feb 5, 2013

### heartstar89

γm1u1 +γm2u2=γm3u3+γm4u4

γ2794.5Mev+γ3726MeV=5589Mev+γm4u

Then, this would be the equation for momentum conserved. But how do we solve for m4?
I know if we have the equations for momentum and energy then we divide momentum over energy

4. Feb 5, 2013

### Staff: Mentor

How does your m3 carry momentum? It is at rest.
You can get (different) mass<->momentum-relations for both conserved quantities.

5. Feb 5, 2013

### heartstar89

Okay, then it would be

γ2794.5Mev+γ3726MeV=γm4u.

When I divide these two equations would it be this

0=5589Mev+γm4u

6. Feb 6, 2013

### Staff: Mentor

How (and why) do you divide the equations?
You can calculate those gamma factors by the way.

You should get two equations with two unknowns.

7. Feb 7, 2013

### vela

Staff Emeritus
You should have subscripts on the gamma factors since they're all different.
$\gamma_1 m_1 c^2 + \gamma_2 m_2 c^2 = \gamma_3 m_3 c^2 +\gamma_4 m_4 c^2$

This one is wrong still.

I don't see how you got this at all.