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

Question #116810
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.
1
2020-05-18T10:22:07-0400

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 \u200b"

"\\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}} \u200b"

Solve for "\\beta" ,

"\\beta =0.968"

So the velocity factor is"\u03b2=0.968."

----------------------------------------------------------------------------------

b)

The velocity factor is

"\u03b2=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"

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