Posted by **Marissa** on Sunday, April 19, 2009 at 11:31pm.

OK, so I'm having some serious problems with this one question. It's on relativity. Here it is:

Calculate the mass of a proton (m0 = 1.67E-27 kg) whose kinetic energy is one sixth its total energy. How fast is it traveling?

THANKS!!

- Physics -
**drwls**, Sunday, April 19, 2009 at 11:55pm
Total energy is then

Etotal = m0c^2 + KE = (6/5) m0 c^2

= m0 c^2/sqrt[1 - (v/c)^2]

6/5 = 1/sqrt[1 - (v/c)^2]

25/36 = 1 - (v/c)^2

(v/c)^2 = 11/36

v/c = 0.306

v = 9.17*10^7 m/s

You don't need to use the rest mass.

There is a nonrelativistic way to do this also, but you will get a different (and wrong) answer. That would be to calculate m0c^2, take 1/5 of it for the kinetic energy, and set that equal to

(1/2) m0 v^2, then solve for v.

- Physics -
**Marissa**, Monday, April 20, 2009 at 12:09am
wait, aren't we lookng for the mass, too?

- Physics -
**drwls**, Monday, April 20, 2009 at 2:23am
Whoops, you are right. They asked for mass, too. It is the rest mass times the factor 1/sqrt[1 - (v/c)^2], which is (6/5) m0.

- Physics -
**pearlsawme**, Tuesday, April 13, 2010 at 7:46am
m = m0 / √ [1 - v^2/c^2]

Kinetic energy = [m-m0] c^2.

Total energy = mc^2.

It is given that [m-m0] c^2 = mc^2/6

m = [6/5] *m0 = [6*1.67e-27/5] = 2.004e-27 kg.

--------------------------------------…

v^2 /C^2=[ m^2 -m0^2] / m^2 =[ 2.004e-27 ^2 -1.67e-27^2] /2.004e-27^2

v = 0.553 C where C = 3e8m/s.

=================================

- Physics -
**VJC**, Wednesday, November 17, 2010 at 1:57pm
what would happen if it became 1/4 the total energy instead? how would the equation be set up?

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