What is the Mass and Speed of the Second Object in a Relativistic Collision?

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Homework Statement



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

Homework Equations


1 u = 931.5 MeV/c2

γm1c2+γm2c2=γm3c2+γm4c2

KE = -mc^2 = KEf - KEi

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
 
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One for the momentum (γumu)conserved and the other for energy (γumc^2) conserved
Right. Alternatively, you can consider 4-vectors, where momentum and energy conservation are just two components of the more general enery-momentum conservation.
 
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
 
How does your m3 carry momentum? It is at rest.
You can get (different) mass<->momentum-relations for both conserved quantities.
 
Okay, then it would be

γ2794.5Mev+γ3726MeV=γm4u.

When I divide these two equations would it be this

0=5589Mev+γm4u
 
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.
 
heartstar89 said:
γm1c2+γm2c2=γm3c2+γm4c2
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##

heartstar89 said:
γ2794.5Mev+γ3726MeV=γm4u.
This one is wrong still.

When I divide these two equations would it be this

0=5589Mev+γm4u
I don't see how you got this at all.
 
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