# Relativistic energy/momentum

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## Homework Statement

An electron has a speed of .75c. Find the speed of a proton that has:

a) The same kinetic energy as the electron

b) The same momentum as the electron

γmc2 = K + mc2

P = γmv

## The Attempt at a Solution

PART A:

me (electron mass) = .511MeV/c2
Ve (electron velocity) = .75c

Ke (KE of electron) = (γ-1)mec2
Ke = 1.286(.511) = .657MeV

mp (proton mass) = 938.27MeV/c2
Vp (proton velocity) = ?

So to find Vp:

.657MeV = (γ-1)(938.27MeV) and after simplification, vp = .0374c

Answer is not in the back, so I can't be sure if I did this correctly.

PART B: (I used the kg representation of mass for this part, because I am confused by the eV units.)

me (electron mass) = 9.1E-31 kg
ve = .75c

Pe (electron momentum) = γmeve
Pe = 4.68E-22 kg*m/s

4.68E-22 kg*m/s = γmpvp

I simplified it down to:

2.8E5 = Vp - Vp3/c2

but this looks like a cubic function. I get the feeling that I am doing something wrong.

EDIT: Ok I went back and tried it, this time without plugging any values in, and I arrived at

$V_p = \sqrt{P_{e}^{2} (m_{p}^{2} + \frac{P_{e}^{2}}{c^2})^{-1}}$ and I think that is right. I came up with V_p = 280,239.5 m/s

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