Question

If a ????=87.5 kgm=87.5 kg person were traveling at ????=0.900????v=0.900c , where ????c is the speed of light, what would be the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy?

Answer 1

`(K.E_r)/(K.E)=2.875`

Step-by-step explanation:

Given:

mass, m = 87.5kg

Velocity, V = 0.900c

now,

the relativistic kinetic energy id given as:

`K.E_r=(\gamma-1)mc^2` ...........(1)

where,

`\gamma` = relativistic factor, given as;
`\gamma=\frac{1}{\sqrt{1-(v^2)/(c^2)}}`

Now, the classical kinetic energy is given as:

`K.E = (1)/(2)mv^2` ..........(2)

Dividing the equation (1) by (2) we get

`(K.E_r)/(K.E)=((\gamma-1)mc^2)/((1)/(2)mv^2)`

or

`(K.E_r)/(K.E)=((\gamma-1)c^2)/((1)/(2)v^2)`

substituting the values in the equation we get,

`(K.E_r)/(K.E)=\frac{(\frac{1}{\sqrt{1-((0.90c)^2)/(c^2)}}-1)c^2}{(1)/(2)*(0.90c)^2}`

or

`(K.E_r)/(K.E)=2.875`