Question

A muon is traveling at 0.995

c. what is its momentum? (the mass of such a muon at rest in the laboratory is 207 times the electron mass.)

Answer 1

The momentum of a relativistic particle is given by

`p= \gamma m_0 v`
where

`\gamma= \frac{1}{ \sqrt{1- (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- ((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`