# Relativistic Mechanics (momentum)

1. Sep 20, 2009

### Cmertin

I did the problem, though I don't think that it is right. I think that the number is to big so I think I might have screwed up somewhere though I don't know where.

1. The problem statement, all variables and given/known data
What is the momentum (in units of MeV/c) of an electron with a kinetic energy of 1.00 MeV?

2. Relevant equations
Energy$$_{kinetic}$$ = Energy
E$$^{2}$$ = m$$^{2}_{0}$$c$$^{4}$$ + p$$^{2}$$c$$^{2}$$
c = 3E8 m/s
m0 = rest mass = .511 MeV c$$^{-2}$$

3. The attempt at a solution
E$$^{2}$$ = m$$^{2}_{0}$$c$$^{4}$$ + p$$^{2}$$c$$^{2}$$
p$$^{2}$$c$$^{2}$$ = m$$^{2}_{0}$$c$$^{4}$$ - E$$^{2}$$
p$$^{2}$$c$$^{2}$$ = (.511 MeV c$$^{-2}$$)$$^{2}$$(3x10$$^{8}$$m/s)$$^{4}$$ - (1.00 MeV)$$^{2}$$
p$$^{2}$$c$$^{2}$$ = 2.35E16 MeV$$^{2}$$
pc = 1.53E8 MeV
p = 1.53E8 MeV c$$^{-1}$$

Last edited: Sep 20, 2009
2. Sep 20, 2009

### Fightfish

Firstly, $$p^{2} c^{2} = E^{2} - m_{0}^{2} c^{4}$$ - rearrange your terms carefully.

Next, $$( 0.511 MeV c^{-2} )^{2} ( c^{4} ) = 0.511 MeV^{2}$$
Note that the $$c^{-2}$$ is also squared!

3. Sep 20, 2009

### Cmertin

OK, thanks for your help, though the left and the right side of the equations are not matching up when I go to plug the answers back in.

I got that p = .8596 MeV c-1

Plugging it back in, I get
p2c2 = E - m02c4
(.8596 [STRIKE]c-1[/STRIKE])2[STRIKE]c2[/STRIKE] = 1 - (.26112 MeV2 [STRIKE]c-4[/STRIKE])[STRIKE]c4[/STRIKE]
.7389 = .2611

4. Sep 20, 2009

### Fightfish

Err..it should be 0.7389 = 1 - 0.2611 *points to your 2nd last equation* which is coherent.

5. Sep 20, 2009

### Cmertin

My bad, I'm a retard today... Thanks for your help (I forgot that the 1 was there...)