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Relativistic Momentum and Kinetic Energy

  • Thread starter erok81
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  • #1
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Homework Statement



What is the speed of an electron (me=0.511 MeV/c2) with a momentum p=5 MeV/c.

Homework Equations



Starting with momentum:

[tex]p=\gamma mu[/tex]

Then solving for u:

[tex]u=\pm\frac{cp}{\sqrt{c^2m^2+p^2}}[/tex]

The Attempt at a Solution



I've never dealt with these units before (MeV/c2 and MeV/c) so I am not sure what to do with them. Can one use those values to solve for speed? I solved it (getting 0.993c) but I converted everything to kg and kg*m/s and the solved from there.

Is there an easy way to convert (if I have to)? I ended up using a calculator online to find values. But I'd rather understand them better.
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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hi erok81! :smile:
(have a gamma: γ and a square-root: √ and try using the X2 and X2 icons just above the Reply box :wink:)
I've never dealt with these units before (MeV/c^2 and MeV/c) so I am not sure what to do with them. Can one use those values to solve for speed?
yes

if in doubt, look at the units given to you …

the whole of that square-root bracket is in units of MeV/c2 squared times c2, and (MeV/c)2, which are the same, and the top line is the square-root of that …

everything is in the same units, so the whole thing is just a dimensionless number, just like v/c (you don't even need to convert from MeV to eV) :wink:
 
  • #3
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Oh hey, what do you know, it worked.

I tried it once before after I solved it using my know units and it didn't match. Now I see why...I messed up the m2 under the radical.

My answer is slightly different, so I'll go back and change it. (0.995 vs. 0.993)

Thanks for the help.

And thanks for pointing out the super/subscript buttons. I never saw those and didn't like using LaTeX in the middle of a sentence just for that.
 

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