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.
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!!
5 answers
wait, aren't we lookng for the mass, too?
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.
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.
=================================
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.
=================================
what would happen if it became 1/4 the total energy instead? how would the equation be set up?
2 answers