(Sp. Relativity) Krel / Kcl in powers of (v/c)^2

[Kunhee]

Homework Statement

[/B]
By expanding Krel / Kcl in powers of (v/c)^2, estimate the value of v/c for which Krel differs from Kcl by 10%.

Homework Equations

Kcl = classical Kinetic Energy = 1/2 m0 v^2
Krel = relative Kinetic Energy = (y-1) (m0 c)^2

The Attempt at a Solution

I did a binomial expansion wherein x = (v/c)^2 and n = -1/2
The result is...
[1 - (v/c)^2]^(-1/2) = 1 + 1/2(v/c)^2
so If I plug this value into the Lorentz factor of Krel,
I can equate K rel to the Kcl equation.

But at which step of this expansion can I apply the 10% difference?
Do I need to set up an equation wherein Krel = 11/10 (Kcl)?

Thanks!

 

[Simon Bridge]

But at which step of this expansion can I apply the 10% difference?

You use, in addition to what you have done, the definition of "percentage difference".
What was your equation for Krel/Kcl ? (What were you exanding?)

Do I need to set up an equation wherein Krel = 11/10 (Kcl)?

Does that satisfy the definition?
Why not try it out and see what you get?

 

[Kunhee]

Hi Simon
Thanks for the help.

I think I am stuck at the expansion of K rel / K cl = (y-1) (m0 c^2) / (1/2)(m0 c^2)

Is there a website link or formula that can assist me with the expansion of the above equation in the power of (v/c)^2?

*and would it be okay for me to assume Kcl is (1/2)( m0 (v/c)^2 ) rather than with just c^2?

 

[Kunhee]

It's been solved. Thanks again.

 

[Simon Bridge]

Well done.
For the benifit of others stuck in the same way, how did you solve it?

Presumably younfigured out that Kcl = (1/2)mv^2 = (1/2)mc^2 (v/c)^2