speed of a proton

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mario99

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Calculate the speed of a proton \(\displaystyle (m = 1.67 \times 10^{-27} kg)\) whose kinetic energy is exactly half its total energy.

Any help would be appreciated!
 
[imath]\frac{mv^2}{2} = \frac{mc^2}{2} \longrightarrow v = \pm c ?[/imath] This does not make much sense to me, but I am not a physicist.
 
I'm guessing this is supposed to be a SR problem? Recall that [imath]T = ( \gamma - 1) mc^2[/imath] and [imath]E = \gamma mc^2[/imath]. (m is the rest mass.)

-Dan
 
Thank you very much blamocur and topsquark for helping me.

blamocur said:
[imath]\frac{mv^2}{2} = \frac{mc^2}{2} \longrightarrow v = \pm c ?[/imath] This does not make much sense to me, but I am not a physicist.
Click to expand...
Good luck. Maybe next time.


topsquark said:
I'm guessing this is supposed to be a SR problem? Recall that [imath]T = ( \gamma - 1) mc^2[/imath] and [imath]E = \gamma mc^2[/imath]. (m is the rest mass.)

-Dan
Click to expand...
I don't understand this. I know that the total energy of the proton is \(\displaystyle E = K + mc^2\).
 
mario99 said:
Thank you very much blamocur and topsquark for helping me.


Good luck. Maybe next time.



I don't understand this. I know that the total energy of the proton is \(\displaystyle E = K + mc^2\).
Click to expand...
Well, [imath]E = \gamma mc^2[/imath] so [imath]K = E - mc^2 = ( \gamma - 1) mc^2[/imath]. Since K = E/2 then
[imath]( \gamma - 1) mc^2 = \gamma mc^2 / 2 \implies 2 ( \gamma - 1) = \gamma[/imath]

Can you solve that for v?

-Dan
 
\(\displaystyle \gamma = 2\)

\(\displaystyle \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 2\)

\(\displaystyle \sqrt{c^2 - v^2} = \frac{c}{2}\)

\(\displaystyle v^2 = c^2 - \frac{c^2}{4}\)

\(\displaystyle v = \frac{c\sqrt{3}}{2}\)
 
mario99 said:
\(\displaystyle \gamma = 2\)

\(\displaystyle \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 2\)

\(\displaystyle \sqrt{c^2 - v^2} = \frac{c}{2}\)

\(\displaystyle v^2 = c^2 - \frac{c^2}{4}\)

\(\displaystyle v = \frac{c\sqrt{3}}{2}\)
Click to expand...
Looks good!

-Dan
 
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