Answer to Question #115918 in Mechanics | Relativity for Joshua

Question

An alpha particle moves with velocity of 2.0 × 108 ms−1 in the positive x-axis direction. At a particular instant, it is located at the point x = 4 cm on the x-axis.

Answer 1

Its moment is

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

$$p=\frac{(6.6\cdot10^{-27})(2\cdot10^8)}{\sqrt{1-\frac{2^2}{3^2}}}=1.8\cdot10^{-18}\frac{kg\cdot m}{s}$$