Answer
$1.20\times 10^{-18}Kg.m/s$
Work Step by Step
We can find the momentum of the proton as follows:
$K.E=m_{\circ}c^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)$
This simplifies to:
$v=c\sqrt{1-(1+\frac{K.E}{m_{\circ}c^2})^{-2}}$
We plug in the known values to obtain:
$v=c\sqrt{1-(1+\frac{1.50\times 10^3 MeV}{938MeV})^{-2}}$
Now, $p=\frac{m_{\circ}v}{\sqrt{1-v^2/c^2}}$
We plug in the known values to obtain:
$p=(\frac{1.673\times 10^{-27})(2.77\times 10^8)}{\sqrt{1-(0.923)^2}}$
$p=1.20\times 10^{-18}Kg.m/s$
