If a m=74.7 kg person were traveling at v=0.990c, 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? What is the ratio of the person's relativistic momentum to the person's classical momentum?

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DeniceSandidge DeniceSandidge · Answer 1 · 2019-08-28T13:24:46+02:00

Answer:

ratio of the person's relativistic kinetic energy is 2.486

the ratio of the person's relativistic momentum is 1.8983

Explanation:

Given data

m=74.7 kg

v=0.990 c

c = speed of light

to find out

ratio of the person's relativistic kinetic energy and ratio of the person's relativistic momentum

solution

we will apply here relativistic constant equation that is

relativistic constant = 1 / √(1-v²/c²)

put here all value

relativistic constant = 1 / √(1-(0.850c)²/c²)

relativistic constant = 1.8983

so

relativistic kinetic energy = ( relativistic constant -1 ) mc²

and we know person kinetic energy = 1/2 mv²

so ration are = ( relativistic constant -1 ) mc² / 1/2 mv²

ratio = 2 ( 1.8983 -1 ) c² / (0.850c)²

ratio = 2.486

so ratio of the person's relativistic kinetic energy is 2.486

and

relativistic momentum = relativistic constant mv

and momentum = mv

so ration are = relativistic constant mv / mv

ratio = relativistic constant

ratio = 1.8983

so the ratio of the person's relativistic momentum is 1.8983