Arelativistic 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

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

the correct answer to this question is 2