Answer
(a) $1.0\times 10^{15}Kg.m/s$
(b) $1.5\times 10^{15}Kg.m/s$
Work Step by Step
(a) We can find the magnitude of classical momentum as
$p=mv$
We plug in the known values to obtain:
$p=(4.5\times 10^6)(0.75\times 3.00\times 10^8)$
$p=1.0\times 10^{15}Kg.m/s$
(b) The relativistic momentum can be determined as
$p=\frac{mv}{\sqrt{1-v^2/c^2}}$
We plug in the known values to obtain:
$p=\frac{4.5\times 10^6(0.75)(3.00\times 10^8)}{\sqrt{1-(0.75)^2}}=1.5\times 10^{15}Kg.m/s$
