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)`

...

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.