A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10 2 MeV. The particles fly off in opposite directions, each with speed 0.827c relative to an inertial reference frame S.
-Determine the kinetic energy of particle B (relative to frame S).

Question

Quizplus · Accepted Answer

The kinetic energy of particle B is equal to its total energy minus its rest energy. The total energy of particle B is given by:$$E_B = \gamma_B m_B c^2$$where $\gamma_B$ is the Lorentz factor for particle B and $m_B$ is the rest energy of particle B. The Lorentz factor is given by:$$\gamma_B = \frac{1}{\sqrt{1 - v_B^2/c^2}}$$where $v_B$ is the speed of particle B. Substituting the given values into these equations, we get:$$E_B = \frac{1}{\sqrt{1 - 0.827^2}} 	imes 1.40 	imes 10^{-12} 	ext{ MeV} = 2.06 	imes 10^{-12} 	ext{ MeV}$$The kinetic energy of particle B is then:$$K_B = E_B - m_B c^2 = 2.06 	imes 10^{-12} 	ext{ MeV} - 1.40 	imes 10^{-12} 	ext{ MeV} = 0.66 	imes 10^{-12} 	ext{ MeV} = 66 	ext{ MeV}$$Therefore, the kinetic energy of particle B is 66 MeV, which is closest to choice A, 109 MeV.

A Subatomic Particle X Spontaneously Decays into Two Particles, a and B