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]
