Relativistic Mechanics (momentum)

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Cmertin
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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[tex]_{kinetic}[/tex] = Energy
E[tex]^{2}[/tex] = m[tex]^{2}_{0}[/tex]c[tex]^{4}[/tex] + p[tex]^{2}[/tex]c[tex]^{2}[/tex]
c = 3E8 m/s
m0 = rest mass = .511 MeV c[tex]^{-2}[/tex]

The Attempt at a Solution


E[tex]^{2}[/tex] = m[tex]^{2}_{0}[/tex]c[tex]^{4}[/tex] + p[tex]^{2}[/tex]c[tex]^{2}[/tex]
p[tex]^{2}[/tex]c[tex]^{2}[/tex] = m[tex]^{2}_{0}[/tex]c[tex]^{4}[/tex] - E[tex]^{2}[/tex]
p[tex]^{2}[/tex]c[tex]^{2}[/tex] = (.511 MeV c[tex]^{-2}[/tex])[tex]^{2}[/tex](3x10[tex]^{8}[/tex]m/s)[tex]^{4}[/tex] - (1.00 MeV)[tex]^{2}[/tex]
p[tex]^{2}[/tex]c[tex]^{2}[/tex] = 2.35E16 MeV[tex]^{2}[/tex]
pc = 1.53E8 MeV
p = 1.53E8 MeV c[tex]^{-1}[/tex]
 
Last edited:
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Firstly, [tex]p^{2} c^{2} = E^{2} - m_{0}^{2} c^{4}[/tex] - rearrange your terms carefully.

Next, [tex]( 0.511 MeV c^{-2} )^{2} ( c^{4} ) = 0.511 MeV^{2}[/tex]
Note that the [tex]c^{-2}[/tex] is also squared!
 
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
 
Err..it should be 0.7389 = 1 - 0.2611 *points to your 2nd last equation* which is coherent.
 
Fightfish said:
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...)