Relativistic and non-relativisitic kinetic energy

[lola2000]

Homework Statement

at what speed does the expression for relativistic kinetic energy vary from the non-relativistic expression by 5%?

Homework Equations

Relativistic kinetic energy K=(gamma-1)mc^2
Non-relativistic kinetic energy K=0.5mv^2

The Attempt at a Solution

I'm not sure how to attempt this!
Should I be finding the difference? ie K(relativistic)-K(non-relativistic)=0.05??

 

[gabbagabbahey]

Hi lola2000, welcome to PF!

Should I be finding the difference? ie K(relativistic)-K(non-relativistic)=0.05??

Close, you should be finding the relative difference |K_rel-K_non-rel|/K_non-rel=0.05 ... make sense?

 

[lola2000]

I see, that makes sense.

But when I do this I get
0.05 = [(gamma-1)mc^2 - 0.5mc^2] / 0.5mv^2
which simplifies to
0.6mv^2 = (gamma-1)mc^2
0.6v^2/c^2 +1 = gamma

which is really nasty to solve! Is there a trick I have missed?

 

[gabbagabbahey]

I see, that makes sense.

But when I do this I get
0.05 = [(gamma-1)mc^2 - 0.5mc^2[/color]] / 0.5mv^2

I assume this is a typo?

which simplifies to
0.6mv^2 = (gamma-1)mc^2

Really?

 

[lola2000]

You are right!
That was a typo - it should have been -0.5mv^2

But I am still stuck with the algebra

I have 0.525v^2/c^2 = gamma - 1

How do I rearrange this??

It is not simplifying!

 

[nrqed]

You are right!
That was a typo - it should have been -0.5mv^2

But I am still stuck with the algebra

I have 0.525v^2/c^2 = gamma - 1

How do I rearrange this??

It is not simplifying!

I did not check the previous steps so I cannot guarantee this is the correct equation. But assuming it is, you just need to add 1 to both sides (to have gamma isolated). Then square both sides and rename $$v^2/c^2 = X$$ . Then you will have a quadratic equation for X. Solve, keep only the positive root. That will be $$v^2/c^2$$. Then the answer is the square root of X.