Question

A muon is traveling at 0.995

Answer 1

The momentum of a relativistic particle is given by
$p= \gamma m_0 v$
where
$\gamma= \frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }$ is the relativistic factor
$m_0$ is the rest mass of the particle
v is the speed particle

The rest mass of the muon is 207 times the rest mass of the electron:
$m_0 = 207 m_e = 207 \cdot 9.1 \cdot 10^{-31} kg=1.88 \cdot 10^{-28} kg$
The muon is moving at speed 0.995 c, therefore its velocity is
$v=0.995 c=0.995 \cdot 2.998 \cdot 10^8 m/s =2.983 \cdot 10^8 m/s$
And the relativistic factor is
$\gamma = \frac{1}{ \sqrt{1- (\frac{0.995 c}{c})^2 } } =10.01$

If we plug these numbers into the first equation, we find the muon momentum:
$p= \gamma m_0 v=(10.01)(1.88 \cdot 10^{-28} kg)(2.983 \cdot 10^8 m/s)=$
$=5.61 \cdot 10^{-19} kgm/s$