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Relativistic Momentum and Energy

  • Thread starter Peter G.
  • Start date
  • #1
442
0
Hello guys,

The question is: What is the momentum, in conventional SI units, of a proton of momentum 685 MeVc-1?

So, I tried two methods which yielded slightly different answers, both proximate to the actual answer (3.66 x 10-19)


What I first tried to do was to rearrange: ρ=γm0v

To find the particles velocity and then multiply it by the proton's rest mass in kg. I got 2.995 x 10-19 kgms-1

The second method, which gave me the result 3.45 x 10-19 kgms-1 was the following:

From: E2 = (m0)2c4 + p2c2 I got a value for total energy. I then subtracted the rest energy from that and I got a value for KE.

I equalled that to p2/2m and solved for p (using m in kg and the energy converted from MeV to J)

Are my answers wrong? The book says 3.66 x 10-19 and I simply can't get that!

Thanks in advance!
 

Answers and Replies

  • #2
125
1
the book answer is correct.

a question for you : What does MeVc-1 mean?

can you do a dimension analysis on MeVc-1
 
  • #3
442
0
Oh, got it! Thanks! I got the number!
 
Last edited:
  • #4
442
0
Oh, and, if you don't mind, could you (or anyone else) help me with the following problem please?

A proton initially at rest finds itself in a region of uniform electric field of magnitude 5.0 x 106 Vm-1. The electric field accelerates the proton for a distance of 1 km.

Find the kinetic energy of the proton.

So, what I did was the following:

KE = q * E * s

I then converted the result from J to MeV. I, however, get 5000 MeV and the book gets 500 MeV. Is my line of thought incorrect?

Thanks once again!
 

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