Answer to Question #257952 in Physics for Anand

Question

Calculate the linear momentum, kinetic energy and total energy (in J and MeV) of a particle travelling with speed 0.90c given that its rest mass 1.68×10⁻²⁷ kg and 1MeV = 1.6×10⁻¹³ J

Answer 1

Given:

$v=0.90c$

$m=1.68*10^{-27}\:\rm kg$

The linear momentum

$$p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$p=\frac{1.68*10^{-27}\:\rm kg*0.90*3*10^8\: m/s}{\sqrt{1-0.90^2}}=1.04*10^{-18}\:\rm kg\cdot m/s$$

The total energy

$$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$E=\frac{1.68*10^{-27}\:\rm kg*(3*10^8\: m/s)^2}{\sqrt{1-0.90^2}}\\ =3.47*10^{-10}\:\rm J=2168\: MeV$$