Example: Calculating the momentum of a fast-moving object

An electron is observed to be moving with a velocity of $\langle - 2.05 \times 10^{7},6.02 \times 10^{7},0\rangle\frac{m}{s}$. Determine the momentum of this electron.

Setup

You need to compute the momentum of this electron using the information provided and any information that you can collect or assume.

Facts

An electron is in motion
It has a velocity, ${\overset{\rightarrow}{v}}_{e} = \langle - 2.05 \times 10^{7},6.02 \times 10^{7},0\rangle\frac{m}{s}$.
The speed of the electron is near the speed of light ($c = 3.00 \times 10^{8}\frac{m}{s}$).

Lacking

The mass of the electron is not given, but can be found online ($m_{e} = 9.11 \times 10^{- 31}kg$).

Approximations & Assumptions

The electron does not experience any interactions, so its velocity will remain unchanged.

Representations

The momentum of the electron is given by $\overset{\rightarrow}{p} = \gamma m\overset{\rightarrow}{v}$ where $\gamma = \frac{1}{\sqrt{1 - \left( \frac{|\overset{\rightarrow}{v}|}{c} \right)^{2}}}$.

Solution

First, we compute the speed of the electron.

$$ |{\overset{\rightarrow}{v}}_{e}| = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}} = \sqrt{( - 2.05 \times 10^{7}\frac{m}{s})^{2} + (6.02 \times 10^{7}\frac{m}{s})^{2} + (0)^{2}} = 6.36 \times 10^{7}\frac{m}{s} $$

Next, we compute the gamma factor.

$$ \gamma = \frac{1}{\sqrt{1 - \left( \frac{|\overset{\rightarrow}{v}|}{c} \right)^{2}}} = \frac{1}{\sqrt{1 - \left( \frac{6.36 \times 10^{7}\frac{m}{s}}{3.00 \times 10^{8}\frac{m}{s}} \right)^{2}}} = \frac{1}{\sqrt{1 - (0.212)^{2}}} = 1.02 $$

Finally, we compute the momentum vector.

$$ {\overset{\rightarrow}{p}}_{e} = \gamma m_{e}{\overset{\rightarrow}{v}}_{e} = (1.02)(9.11 \times 10^{- 31}kg)\langle - 2.05 \times 10^{7},6.02 \times 10^{7},0\rangle\frac{m}{s} = \langle - 1.91 \times 10^{- 23},5.61 \times 10^{- 23},0\rangle\frac{kg\ m}{s} $$