Question

Determine the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy (1/2mv²) when a particle has a speed of (a)2.61 � 10^-3c. and (b) 0.973c.

Please explain it step by step.

Answer 1

relativistic kinetic energy, K_rel = (m - m₀) c²                           { eq. 1}

where, m = rest mass relativistic mass of a particle

m₀ = rest mass

and m is given as :

m = m₀ / 1 - (v / c)²                                 { eq. 2 }

inserting the values of m in eq. 1,

K_rel = m₀ c² [ ( 1 / 1 - (v / c)² ) - 1 ]                                 { eq. 3 }

(a) relativistic kinetic energy to the nonrelativistic kinetic energy is given as :;

K_rel / K_{non rel} = ( c / v )² [ ( 1 / sqrt (1 - (v / c)² ) - 1 ]     { eq. 4 }

where, v = speed of particle = 2.61 x 10^-3 c = 0.00261 c

speed of light, c = 3 x 10⁸ m/s

and v/c = 2.61 x 10^-3 c / 3 x 10⁸ m/s = 8.7 x 10^-12

inserting the values in eq.4,

K_rel / K_{non rel} =  ( c / v )² [ ( 1 / sqrt (1 - (v / c)² ) - 1 ]    

K_rel / K_{non rel} = (1/0.00261)^2 [(1/sqrt (1-(0.00261)^2)) - 1]

K_rel / K_{non rel} = 0.5

(b) relativistic kinetic energy to the nonrelativistic kinetic energy is given as ::

given, v = 0.973 c

again, using eq. 4

K_rel / K_{non rel} = ( c / v )² [ ( 1 / sqrt (1 - (v / c)² ) - 1 ]    

K_rel / K_{non rel} = (1/0.973)^2 [(1/sqrt (1-(0.973)^2)) - 1 ]

K_rel / K_{non rel} = 3.5