Question

A muon has a kinetic energy equal to 4 times its rest energy of 105 MeV. (a) What is its velocity, in units of c?

(b) What is its momentum in energy units (i.e., units of MeV/c)?

Answer 1

v = 0.9798*c

Step-by-step explanation:

E0 = 105 MeV

The mass of a muon is

m = 1.78 * 10^-30 kg

The kinetic energy is:

`Ek = \frac{E0}{\sqrt{1 - (v^2)/(c^2)}}-E0`

The kinetic energy is 4 times the rest energy.

`4*E0 = \frac{E0}{\sqrt{1 - (v^2)/(c^2)}}-E0`

`4 = \frac{1}{\sqrt{1 - (v^2)/(c^2)}}-1`

`5 = \frac{1}{\sqrt{1 - (v^2)/(c^2)}}`

`\sqrt{1 - (v^2)/(c^2)} = (1)/(5)`

`1 - (v^2)/(c^2) = (1)/(25)`

v^2 / c^2 = 1 - 1/25

v^2 / c^2 = 24/25

v^2 = 24/25 * c^2

v = 0.9798*c