Answer
a) $p=3.51\times10^{-18}kg\frac{m}{s}$
b) $0$
c) $p_{rel}=4.98\times10^{-17}kg\frac{m}{s}$
Work Step by Step
a) $p=\gamma mv=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{(1.67\times10^{-27}kg)(0.990c)}{\sqrt{1-\frac{(0.990c)^2}{c^2}}}$
$=3.51\times10^{-18}kg\frac{m}{s}$
b) The two protons are moving in opposite direction, so the total momentum is $0$
c) $v_{rel}=\frac{0.990c+0.990c}{1+\frac{(0.990c)(0.990c)}{c^2}}=0.99995c$
$p_{rel}=\gamma mv=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{(1.67\times10^{-27}kg)(0.99995c)}{\sqrt{1-\frac{(0.99995c)^2}{c^2}}}$
$=4.98\times10^{-17}kg\frac{m}{s}$