What is the speed of the proton? According to my professor, the answer is .98c.
Thus far I have... mc^2-moc^2 = KE x mc^2. Ahh I do not know where to go from here or if I have the right starting point. Explanation and work shown would be so helpful! Thank you in advance.
Thanks for the answer. But how did you get from the first step to the second?
Chris, how does that factor into the work for the problem above?
2 Answers
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Very essential. Whenever you have Anything above a speed of .2c, you must use the lorenz factor to account for special relativity.
L= gamma = 1/ (1-V^2/c^2)^(1/2)
The Lorenz factor affects the mass of the object. I think the answer above has the right equation and math, so I thought I should offer what he did with SR(number)
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You're close;
KE = mc^2 - m(o)c^2 = .8mc^2
.2mc^2 = m(o)c^2
.2m(o)/SR[1-v^2/c^2] = m(o)
.2 = SR[1-v^2/c^2]
.04 = 1 - v^2/c^2
v^2 = .96c^2
v = .98c
EDIT___________________________________
you know KE = mc^2 - m(o)c^2
and you're given KE = .8mc^2
So equate;
mc^2 - m(o)c^2 = .8mc^2
combine like terms
.2mc^2 = m(o)c^2