Question

A photon with energy of 3m hits a particle of mass m at rest. the photon "back-scatters" from this interaction (meaning moves it the opposite direction) while the particle moves forward to conserve momentum. Find the back-scattered photon's energy E and the particles speed v.  

Answer 1

using conservation of momentum

3m/c + m (0) = E/c + mv

3m/c = E/c + mv

v = (3m - E)/(mc)                                  eq-1

using conservation of kinetic energy

3m = E + (0.5) m v²

3m = E + (0.5) m ((3m - E)/(mc) )²

(3m - E) = (0.5) m ((3m - E)/(mc) )²

1 = (0.5) m ((3m - E)/(mc)² )

2 m² c² = m ((3m - E)

2 m c² = ((3m - E)

E = 3 m - 2 m c²

using eq-1

v = (3m - (3 m - 2 m c²))/(mc)  

v = 2 m c²/(mc)

v = 2 c