Question

I'll call this classical vs actual

Answer 1

If a 87.5 kg person were travelling at 0.750c (where c is the speed of light), what would be the ratio of the person's actual kinetic energy to the person's classical kinetic energy?

Answer 2

So the same formula from before = 1/2mv^2? @theEric

Answer 3

\[K = \frac{ mc^2 }{ \sqrt{1-v^2/c^2} }-mc^2\]

Answer 4

I was never really good at these ratio questions, so would I just do \[\frac{ K }{ k } = \frac{ \frac{ mc^2 }{ \sqrt{1-v^2/c^2} }-mc^2 }{ 1/2mv^2 }\]

Answer 5

Looks good to me! And it looks like mass cancels out...

Answer 6

The velocity is expressed in terms of c, so that will simplify things a little, too.

Answer 7

Lol, Just to make it simpler and not go through the algebra I'll just do them separately, unless algebra does make it simpler...I'll try both ways XD.

Answer 8

Alright I got about 3, now I need the ratio for the person's actual momentum and classical momentum, @theEric just to make sure, actual momentum \[p = \gamma mv\] and classical is just p = mv right :P

Answer 9

Wikipedia agrees with your momentum! Yup!

Answer 10

Cool, thanks a ton Eric! ^.^

Answer 11

You're welcome! Did you get the first part correct? I didn't check it.

Answer 12

Wow so \[K_relativistic = (\gamma - 1)mc^2 \] I went overboard haha.

Answer 13

Second one is right, first one wasn't I'll fix it up and let you know

Answer 14

Yup, that's K! :)

Answer 15

Thank you for making me take a look at this stuff! I've wanted a refresher of this recently.

Answer 16

Yup! And I fixed it up and got it right :)!

Answer 17

Np, thank you for helping!

Answer 18

Bleh, openstudy is acting up again haha...

Answer 19

\(\dfrac{ K }{ k } = \dfrac{ (\gamma-1)mc^2 }{ \frac12mv^2 }=\dfrac{ 2(\gamma-1)c^2 }{ v^2 }=\dfrac{ 2(\gamma-1)c^2 }{ 0.75c^2 }=\\
\dfrac{ 2(\gamma-1) }{\frac34 }=\dfrac{4\dot\ 2(\gamma-1) }3=\dfrac83(\gamma-1)\)

Answer 20

I did the work for fun! I hope I did it well...

OS does that... Ah well! :) Congrats on getting them both!