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

Answer:

$\frac{K.E_r}{K.E}=2.875$

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; $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Now, the classical kinetic energy is given as:

$K.E = \frac{1}{2}mv^2$    ..........(2)

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

$\frac{K.E_r}{K.E}=\frac{(\gamma-1)mc^2}{\frac{1}{2}mv^2}$

or

$\frac{K.E_r}{K.E}=\frac{(\gamma-1)c^2}{\frac{1}{2}v^2}$

substituting the values in the equation we get,

$\frac{K.E_r}{K.E}=\frac{(\frac{1}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}-1)c^2}{\frac{1}{2}\times(0.90c)^2}$

or

$\frac{K.E_r}{K.E}=2.875$