Answer
(a) $v = 0.999998~c$
(b) $v = 0.99999997~c$
Work Step by Step
(a) We can find $\gamma$:
$E = \gamma~mc^2 = 500~GeV$
$\gamma~mc^2 = (500~GeV)(\frac{1.6\times 10^{-19}~J}{1~eV})$
$\gamma~mc^2 = (500\times 10^9~eV)(\frac{1.6\times 10^{-19}~J}{1~eV})$
$\gamma~mc^2 = 8.00\times 10^{-8}~J$
$\gamma = \frac{8.00\times 10^{-8}~J}{mc^2}$
$\gamma = \frac{8.00\times 10^{-8}~J}{(1.67\times 10^{-27}~kg)(3.0\times 10^8~m/s)^2}$
$\gamma = 532$
We can find the speed when $\gamma = 532$:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}$
$1-\frac{v^2}{c^2} = \frac{1}{\gamma^2}$
$\frac{v^2}{c^2} = 1-\frac{1}{\gamma^2}$
$v^2 = (1-\frac{1}{\gamma^2})~c^2$
$v = \sqrt{1-\frac{1}{\gamma^2}}~c$
$v = \sqrt{1-\frac{1}{(532)^2}}~c$
$v = \sqrt{0.9999964667}~c$
$v = 0.999998~c$
(b) We can find $\gamma$:
$E = \gamma~mc^2 = 2.0~GeV$
$\gamma~mc^2 = (2.0~GeV)(\frac{1.6\times 10^{-19}~J}{1~eV})$
$\gamma~mc^2 = (2.0\times 10^9~eV)(\frac{1.6\times 10^{-19}~J}{1~eV})$
$\gamma~mc^2 = 3.2\times 10^{-10}~J$
$\gamma = \frac{3.2\times 10^{-10}~J}{mc^2}$
$\gamma = \frac{3.2\times 10^{-10}~J}{(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2}$
$\gamma = 3903$
We can find the speed when $\gamma = 3903$:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}$
$1-\frac{v^2}{c^2} = \frac{1}{\gamma^2}$
$\frac{v^2}{c^2} = 1-\frac{1}{\gamma^2}$
$v^2 = (1-\frac{1}{\gamma^2})~c^2$
$v = \sqrt{1-\frac{1}{\gamma^2}}~c$
$v = \sqrt{1-\frac{1}{(3903)^2}}~c$
$v = \sqrt{0.999999934}~c$
$v = 0.99999997~c$
