Relativistic Mechanics (momentum)

[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.

Homework Statement

What is the momentum (in units of MeV/c) of an electron with a kinetic energy of 1.00 MeV?

Homework 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}

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}

 

[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!

 

[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

 

[Fightfish]

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

 

[Cmertin]

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

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