Solving Momentum Equalization of Electron and Proton

[Decan]

Homework Statement

An electron has a speed of v = 0.14c. At what speed will a proton have a momentum equal to that of the electron?

Homework Equations

p=mv

The Attempt at a Solution

MeVe = MpVp where p=proton and e=electron
so: MeVe/Mp=Vp
Me = 9.109 X 10^-31 kg
Mp = 1.672 X 10^-27 kg

Plugging this information in gives me an answer of 22873.91 m/s but the correct answer is 23139.3 m/s. What am I doing wrong? Thanks in advance for any help!

 

[Chi Meson]

Does the word "relativistic" mean anything to you? Has the name Einstein come up in class?

And what's the deal with all those digits? If the speed is given as ".14c" the correct answer should be 23,000 m/s .

 

[Decan]

We briefly talked about einstein and E = mc^2 but the professor didnt really go over it...I also understand that .14c means .14 X 3e8...but I don't understand why the answer should be 23,000...any hints?

 

[Chi Meson]

"Breifly went over E=mc^2"?

You mean "Lorenzian Transformation" "time dialation" "length contraction"
"gamma factor" and "mass increase" were never discussed? If not, then there is no business assigning this problem. You do not need "hints" you need about two weeks worth of Special Relativity lessons.

 

[Decan]

Like I said, briefly. We never really talked about these...the prof just gave us equations on the board and told us "use these to solve the homework, you won't be tested on them"

 

[Decan]

Also, the homework is due tomorrow and this is the only problem I have left...

 

[Chi Meson]

What a waste of time. "Just do the problems without knowing what's going on" right? I'm sorry for you.

Relativistic momentum: find the equation that resembles the following:

p=(mv)/SQRT[1-(v^2/c^2)]. You are in for an algebraic workout since v appears in two places.

Edit: That looks horrible. Here's the latex version

p=\frac{mv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

well that's not looking great either.

 

[Decan]

got it...thanks!