# Special Relativity(momentum)

1. Jan 29, 2008

### Winzer

1. The problem statement, all variables and given/known data
A particle of rest mass 1.30 MeV/c2 and kinetic energy 2.70 MeV collides with a stationary particle of rest mass 2.50 MeV/c2. After the collision, the particles stick together. Find the speed of the first particle before the collision

2. Relevant equations

$$KE=(1-\gamma)mc^2$$

3. The attempt at a solution
I am positive this is the right equation to use but I keep getting the wrong answer.
I don't know why!

2. Jan 29, 2008

### joelperr

Show us what you have so far, and we can go from there. That might be one of the equations that will be used, but you need to show some sort of attempt at a solution instead of simply stating that you're getting an incorrect answer.

3. Jan 29, 2008

### Winzer

fine.
That equation goes to:$$c*\sqrt(1-(\frac{E_rest}{KE-E_rest})^2)$$
I converted 1.30 MeV/c^2=2.314E-30 kg, included in E rest
and 2.70 MeV=4.325E-13 J, for KE

I get: 1.11E8 m/s

oh $$E_rest=mc^2$$

4. Jan 29, 2008

### joelperr

The best way to proceed is to write down the relevant equations for the conservation of momentum and conservation of energy:

$$E_{1} + E_{2} = E_{final}$$
$$\textbf{p}_{1} + 0 = \textbf{p}_{final}$$

Initially, the first particle has kinetic and rest mass energy, and the second particle has only rest mass energy (note that we are doing everything in the lab frame). The end product only has one 'particle', with a modified kinetic energy and rest mass energy. For momentum, the only change from initial to final is that the final combined 'particle' has more mass.

Use $$E^2 - \textbf{p}^2c^2 = m^2c^4$$ and then solve for the momentum of the final particle.

This problem can be solved in a few lines if you use 4-vector notation, and in a few more lines if you do everything component-wise.

Last edited: Jan 29, 2008