Determine the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy (1/2mv2) when a particle has a speed of (a) 2.12 x 10-3c. and (b) 0.959c.

Answer 1

Relativistic kinetic energy is mₒc²{[1/√(1-β²)] – 1},
nonrelativistic kinetic energy is m v²/2 = 0.5mₒv²
(a) β=v/c=2.12•10^-3•c/c=2.12•10^-3
1/√(1-β²) =1.00000224
1/√(1-β²)] – 1 = 0.00000224=2.24 •10^-6
Relativistic kinetic energy is 2.24 •10^-6•mₒc²
The ratio is 2.24 •10^-6•mₒc²/[0.5•mₒ•(2.12•10^-3)² • c² ]=
=2.24 •10^-6/2.2472•10^-6 = 0.996796.

(b) β=v/c=0.959•c/c= 0.959
1/√(1-β²) = 3.5285,
1/√(1-β²)] – 1 =2.5285
Relativistic kinetic energy is 2.5285 •mₒc²
The ratio is 2.5285 •mₒc² /[0.5•mₒ•( 0.959)² • c² ]=5.499