Question

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. What is the total energy of particle A?

Answer 1

E = 389 MeV

Step-by-step explanation:

The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,

Total Energy of A = E = Rest Mass Energy + Relative Energy

Using Einstein's Equation: E = mc²

E = m₀c² + mc²

From Einstein's Special Theory of Relativity, we know that:

m = m₀/[√(1-v²/c²)]

Therefore,

E = m₀c² + m₀c²/[√(1-v²/c²)]

E = m₀c²[1 + 1/√(1-v²/c²)]

where,

m₀c² = rest mass energy = 140 MeV

v = relative speed = 0.827 c

Therefore,

E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]

E = (140 MeV)(2.78)

E = 389 MeV