# Relativistic energy

1. Oct 4, 2009

### dancergirlie

1. The problem statement, all variables and given/known data

What is the speed of a proton after being accelerated to a kinetic energy of 750keV?

2. Relevant equations

E=K+E0
E0=mc^2
K=(gamma-1)mc^2
gamma=1/(sqroot(1-v^2/c^2))
3. The attempt at a solution

Alright, so to find the speed, I used the equation:

K=(gamma-1)mc^2
which i simplified to:
1/(gamma-1)=mc^2/K

I converted K into joules which is equal to 1.202E-13 J
and what I ended up getting was that
sqroot(1-v^2/c^2) - 1=1250.42
but if i continue on with that I will need to take the square root of a negative number because i will bring v^2/c^2 to the right side and then I would have to do 1-1250.42 and take the square root of that which is undefined. Any help would be great!!!

2. Oct 4, 2009

### kuruman

Do yourself a favor and use symbols throughout your derivation. Plug in at the very end. It would be easier to trace your error(s) if you did. As it is, I cannot figure out what that 1250.42 means or where it came from. If you have gotten to the stage of special relativity you should be able to handle expressions symbolically. Doing so enables you to find your mistakes much more easily through dimensional analysis.