Answer
a) $220\%$
b) $205\%$
Work Step by Step
$\frac{p_2-p_1}{p_1}$
$p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2} }}$
a) $p_1=\frac{m(0.45c)}{\sqrt{1-\frac{(0.45c)^2}{c^2} }}=0.504mc$
$p_2=\frac{m(0.85c)}{\sqrt{1-\frac{(0.85c)^2}{c^2} }}=1.614mc$
$\frac{1.614mc-0.504mc}{0.504mc}\times100\%=220\%$
b) $p_1=1.614mc$
$p_2=\frac{m(0.98c)}{\sqrt{1-\frac{(0.98c)^2}{c^2} }}=4.925mc$
$\frac{4.925mc-1.614mc}{1.614mc}\times100\%=205\%$
