The momentum of a relativistic particle is given by
[tex]p= \gamma m_0 v[/tex]
where
[tex]\gamma= \frac{1}{ \sqrt{1- \frac{v^2}{c^2} } } [/tex] is the relativistic factor
[tex]m_0[/tex] 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:
[tex]m_0 = 207 m_e = 207 \cdot 9.1 \cdot 10^{-31} kg=1.88 \cdot 10^{-28} kg[/tex]
The muon is moving at speed 0.995 c, therefore its velocity is
[tex]v=0.995 c=0.995 \cdot 2.998 \cdot 10^8 m/s =2.983 \cdot 10^8 m/s[/tex]
And the relativistic factor is
[tex]\gamma = \frac{1}{ \sqrt{1- (\frac{0.995 c}{c})^2 } } =10.01[/tex]
If we plug these numbers into the first equation, we find the muon momentum:
[tex]p= \gamma m_0 v=(10.01)(1.88 \cdot 10^{-28} kg)(2.983 \cdot 10^8 m/s)=[/tex]
[tex]=5.61 \cdot 10^{-19} kgm/s[/tex]
