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In the lab, a relativistic proton has a momentum of 1.00 x 10^-19 kg • m/s and a rest energy of 0.150 nanojoules (nJ). What is the speed of the...

To derive our formula, we will start with the generic formula for relativistic momentum,


`p=gammamv`


where gamma is `gamma=1/sqrt(1-(v^2/c^2))` , p is momentum, m is the rest mass of the particle, and v is the velocity. I will substitute G for p/m.


Then, I will solve for v.


`gammav=G`


I square both sides to get rid of the nasty square root in the gamma. From there, most of the simplification is straightforward.


`gamma^2v^2=G^2`


`(1/(1-(v^2/c^2)))v^2=G^2`


`v^2=G^2(1-v^2/c^2)`


`v^2=G^2-v^2G^2/c^2`


`v^2+v^2G^2/c^2=G^2`


`v^2(1+G^2/c^2)=G^2`


Finally, 


`v^2=G^2/(1+G^2/c^2)`


or with the G expanded,


`v^2=(p/m)^2/(1+(p/m)^2/c^2)`


Don't forget the square on the v. Now, we can substitute our numbers into the formula above to calculate the relative speed of the proton.


`v^2=((1*10^-19)/(1.67*10^-27))^2/(1+((1*10^-19)/(1.67*10^-27))^2/(3*10^8)^2)`


This gives v=5.87209*10^7 m/s, or .1957c.


Particles at this velocity do not behave the same as normal particles, so the Newtonian p=mv formula does not hold here.

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