Example: Calculating the momentum of a fast-moving object

An electron is observed to be moving with a velocity of $\langle -2.05\times10^7, 6.02\times10^7, 0\rangle\:\dfrac{m}{s}$. Determine the momentum of this electron.

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

• An electron is in motion
• It has a velocity of $\langle -2.05\times10^7, 6.02\times10^7, 0\rangle\:\dfrac{m}{s}$.
• This velocity is near the speed of light ($c = 3.00\times10^8 \dfrac{m}{s}$).

• The mass of the electron is not given, but can be found online ($m_e = 9.11\times10^{-31} kg$).

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

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

First, we compute the speed of the electron.

$$|\vec{v}| = \sqrt{v_x^2+v_y^2+v_z^2} = \sqrt{(-2.05\times10^7 \dfrac{m}{s})^2+(6.02\times10^7 \dfrac{m}{s})^2+(0)^2} = 6.36 \times 10^7 \dfrac{m}{s}$$

Next, we compute the gamma factor.

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

Finally, we compute the momentum vector.

$$\vec{p} = \gamma m \vec{v} = (1.02) (9.11\times10^{-31} kg) \langle -2.05\times10^7, 6.02\times10^7, 0\rangle\dfrac{m}{s} = \langle \rangle\:\dfrac{kg\:m}{s}$$