Question

Determine the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy (1/2mv2) when a particle has a speed of (a) 1.24 × 10-3c. and (b) 0.854c.

Answer 1

a)

v = 1.24 x 10^-3 c = 0.00124 c

KE_rel = relativistic kinetic energy = (1/sqrt(1 - (v/c)²) - 1) (m c²) = (1/sqrt(1 - (0.00124c/c)²) - 1) (m c²)

KE_rel = 6.92 x 10¹⁰ m J

KE_non = (0.5) m v² = (0.5) m (0.0124 x 3 x 10⁸)² = 6.92 x 10¹² m J

Ratio is given as

Ratio = KE_rel /KE_non = (6.92 x 10¹⁰ m) /(6.92 x 10¹² m) = 0.01

b)

v = 0.854 c

KE_rel = relativistic kinetic energy = (1/sqrt(1 - (v/c)²) - 1) (m c²) = (1/sqrt(1 - (0.854c/c)²) - 1) (m c²)

KE_rel = 8.3 x 10¹⁶ m J

KE_non = (0.5) m v² = (0.5) m (0.854 x 3 x 10⁸)² = 3.3 x 10¹⁶ m J

Ratio is given as

Ratio = KE_rel /KE_non = (8.3 x 10¹⁶ m) /(3.3 x 10¹⁶ m) = 2.52