Asked by Marissa

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!!
Answered by drwls
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.
Answered by Marissa
wait, aren't we lookng for the mass, too?
Answered by drwls
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.
Answered by pearlsawme
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.
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Answered by VJC
what would happen if it became 1/4 the total energy instead? how would the equation be set up?
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