What Speed Makes Kinetic Energy Equal n Times Rest Energy?

[Pual Black]

Homework Statement

At what speed the kinetic energy of particle equal n times it's rest energy

Homework Equations

i use this equation
$T=(\gamma - 1)m_{0}c^{2}$

The Attempt at a Solution

$T=n(m_{0}c^{2})$

$n=\gamma -1$

$\gamma =n+1$

$\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=n+1$

$\frac{1}{1-\frac{v^{2}}{c^{2}}}=n^{2}+2n+1$

$\frac{1}{n^{2}+2n+1}=1-\frac{v^{2}}{c^{2}}$

$\frac{v^{2}}{c^{2}}=1-\frac{1}{n^{2}+2n+1}$

$\frac{v^{2}}{c^{2}}=\frac{n^{2}+2n+1-1}{n^{2}+2n+1}$

$v=\sqrt{\frac{n^{2}+2n}{n^{2}+2n+1}} ~~ c$

this is what i get.
Is that correct? isn't there
other solution?

 

[drvrm]

Homework Statement

At what speed the kinetic energy of particle equal n times it's rest energy

Homework Equations

i use this equation
$T=(\gamma - 1)m_{0}c^{2}$

The Attempt at a Solution

$T=n(m_{0}c^{2})$

$n=\gamma -1$

$\gamma =n+1$

$\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=n+1$

$\frac{1}{1-\frac{v^{2}}{c^{2}}}=n^{2}+2n+1$

$\frac{1}{n^{2}+2n+1}=1-\frac{v^{2}}{c^{2}}$

$\frac{v^{2}}{c^{2}}=1-\frac{1}{n^{2}+2n+1}$

$\frac{v^{2}}{c^{2}}=\frac{n^{2}+2n+1-1}{n^{2}+2n+1}$

$v=\sqrt{\frac{n^{2}+2n}{n^{2}+2n+1}} ~~ c$

this is what i get.
Is that correct? isn't there
other solution?

pl. check again the relations;
what is the the expression for relativistic kinetic energy?

 

[blue_leaf77]

pl. check again the relations;
what is the the expression for relativistic kinetic energy?

Is there something strange with $T=(\gamma - 1)m_{0}c^{2}$? At least it is consistent with the classical kinetic energy for small velocities.

 

[Pual Black]

pl. check again the relations;
what is the the expression for relativistic kinetic energy?

I checked it and it is correct
Or i can write it in this form
$E_{k} = \frac{m_{0}c^{2}}{\sqrt{1 - (v/c)^{2})}} - m_{0}c^{2}$

 

[ehild]

$v=\sqrt{\frac{n^{2}+2n}{n^{2}+2n+1}} ~~c$

this is what i get.
Is that correct?

It is correct.

 

[Pual Black]

Ok. Thank you all for your help