Answer
(a) $1.00c$
(b) $3.6\times 10^{9}eV\lt \lt 1.0\times 10^{20}eV$
Work Step by Step
(a) We know that
$\frac{v}{c}=\sqrt{1-\frac{1}{(\frac{K}{m_{\circ}c^2}+1)^2}}$
We plug in the known values to obtain:
$\frac{v}{c}=\sqrt{1-\frac{1}{(\frac{1.0\times 10^{20}eV}{938.28\times 10^6eV})^2+1}}$
This simplifies to:
$\frac{v}{c}\approx1.00$
$v=1.00c$
(b) As $K=\frac{1}{2}mv^2$
We plug in the known values to obtain:
$K=\frac{1}{2}(15\times 10^{-6}Kg)(8.8\times 10^{-3}m/s)^2$
$K=3.6\times 10^{9}eV$
The kinetic energy of protons $=1.0\times 10^{20}eV$
$\implies 3.6\times 10^{9}eV\lt \lt 1.0\times 10^{20}eV$
