Special Relativity kinetic energy

[PsychonautQQ]

Homework Statement

In reference frame S', two protons, each moving at .500c, approach each other head-on. a) calculate the total kinetic energy of the two protons in frame S'. b) Calculate the total kinetic energy of the protons as seen in the reference frame S, which is moving with one of the protons.

Homework Equations

E = mc^2 + K
K = γ1/2mv^2
Proton rest mass = 938 MeV/c^2

The Attempt at a Solution

So for part A I just found the kinetic energy of one of the protons given by γ(1/2)mv^2
(1.1547)(1/2)(938)(.25) * 2 = 270 MeV.. and the back of the book says 290 MeV.. but this textbook is known for having a lot of wrong answers, can anyone verify whether I am correct or incorrect?

 

[Simon Bridge]

I get 290.22MeV Note:
$K=(\gamma-1)mc^2$ because $E=\gamma mc^2$

 

[PsychonautQQ]

what is incorrect about my method of using the equation K = γ1/2mv^2 to obtain the answer?

 

[Doc Al]

what is incorrect about my method of using the equation K = γ1/2mv^2 to obtain the answer?

It looks like you just took the non-relativistic expression for KE (1/2mv^2), which is only good for low speeds, and thought you could make it relativistically correct by slapping on a gamma. But that doesn't work. Simon Bridge gave the correct formula.

 

[Simon Bridge]

what is incorrect about my method of using the equation K = γ1/2mv^2 to obtain the answer?

DocAl is correct - $K\neq \gamma\frac{1}{2}mv^2$
What lead you to believe the equation would be correct?