- #1

ElijahRockers

Gold Member

- 270

- 10

## 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

## Homework Equations

γmc

^{2}= K + mc

^{2}

P = γmv

## The Attempt at a Solution

PART A:

m

_{e}(electron mass) = .511MeV/c

^{2}

V

_{e}(electron velocity) = .75c

K

_{e}(KE of electron) = (γ-1)m

_{e}c

^{2}

K

_{e}= 1.286(.511) = .657MeV

m

_{p}(proton mass) = 938.27MeV/c

^{2}

V

_{p}(proton velocity) = ?

So to find V

_{p}:

.657MeV = (γ-1)(938.27MeV) and after simplification, v

_{p}= .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.)

m

_{e}(electron mass) = 9.1E-31 kg

v

_{e}= .75c

P

_{e}(electron momentum) = γm

_{e}v

_{e}

P

_{e}= 4.68E-22 kg*m/s

4.68E-22 kg*m/s = γm

_{p}v

_{p}

I simplified it down to:

2.8E5 = V

_{p}- V

_{p}

^{3}/c

^{2}

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

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

Last edited: