Answer to Question #116810 in Mechanics | Relativity for nicolee heath

Question

The relativistic kinetic energy of a particle is three times its rest mass energy. Find
the:
a) Lorentz factor and speed parameter
b) velocity (in m/s) of the particle.

Answer 1

Given data

Relativistic Kinetic energy of particle is $K=3E_0$

Here $E_0$  is the rest energy of particle.

a)

The expression for relativistic kinetic energy of particle is given by

$K=(\gamma-1)E_0$

Here $\gamma =\frac{1}{\sqrt{\beta^2-1}}$ is called relativistic factor or Lorentz factor.

$3E_0=(\gamma-1)E_0 ​$

$\gamma=4$

Hence the Lorentz factor is $\gamma=4$ .

But the Lorentz factor is

$\gamma =\frac{1}{\sqrt{\beta^2-1}}$

$4=\frac{1}{\sqrt{\beta^2-1}} ​$

Solve for $\beta$ ,

$\beta =0.968$

So the velocity factor is$β=0.968.$

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b)

The velocity factor is

$β=0.968$

$\frac{v}{c}=0.968$

Therefore, the velocity of the particle is

$v=0.968c=0.968(3*10^8 m/s)=2.90*10^8 m/s$