A relativistic proton has a momentum of 1.0 * 10-19 kg · m/s and a rest energy of 0.15 nJ. What is the kinetic energy of this proton? (c = 3.00 * 108 m/s, mproton = 1.67 * 10-27 kg)

The kinetic energy is 0 J.

Explanation:

The kinetic energy of a relativistic proton is determined as follows:

KE = mc² - m₀c²

To determine m₀, we use the expression

E_rest = m₀c²

where

E_rest is the rest energy c is the speed of light m₀ is the rest mass

Thus, rearranging the above expression we get

m₀ = E_rest/c²

= (0.15*10⁻⁹) / (3*10⁸) ²

= 1.67*10⁻²⁷ kg

Therefore,

KE = (1.67*10⁻²⁷) (3*10⁸) ² - (1.67*10⁻²⁷) (3*10⁸) ²

= 0 J