Relativistic Momentum and Energy

[Peter G.]

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!

 

[NoPoke]

the book answer is correct.

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

can you do a dimension analysis on MeVc-1

 

[Peter G.]

Oh, got it! Thanks! I got the number!

 

[Peter G.]

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!

 

[Nickie]

Hi there,

I understand your frustration with not getting the exact answer that the book provides. However, it is important to remember that in science, there is often more than one way to arrive at a solution and small variations in calculations can lead to slightly different results.

In this case, both of your methods seem correct and the differences in your answers could be due to rounding errors or slight variations in the values used for the proton's rest mass and the speed of light. It is also possible that the book's answer is an approximation or rounded off for simplicity.

Regardless, your answers are very close to the book's answer and can be considered accurate within a reasonable margin of error. As long as you understand the concepts and methods used to arrive at your solution, you are on the right track. Keep up the good work!