Answer
(a) $p = 1.12\times 10^{-17}~kg~m/s$
(b) The proton's relativistic momentum exceeds its Newtonian momentum by the factor $22.4$
Work Step by Step
(a) We can find $\gamma$:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\gamma = \frac{1}{\sqrt{1-\frac{(0.999~c)^2}{c^2}}}$
$\gamma = \frac{1}{\sqrt{0.001999}}$
$\gamma = 22.4$
We can find the proton's relativistic momentum:
$p = \gamma ~m~v$
$p = (22.4)(1.67\times 10^{-27}~kg)(0.999)(3.0\times 10^8~m/s)$
$p = 1.12\times 10^{-17}~kg~m/s$
(b) The proton's momentum exceeds its Newtonian momentum by the factor $\gamma$
Therefore, the proton's relativistic momentum exceeds its Newtonian momentum by the factor $22.4$
