183_notes:examples:momentumfast [Projects & Practices in Physics]

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====== 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.

==== 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, $\vec{v}_e=\langle -2.05\times10^7, 6.02\times10^7, 0\rangle\dfrac{m}{s}$.
  * The speed of the electron is near the speed of light ($c = 3.00\times10^8 \dfrac{m}{s}$).

=== Lacking ===

  * The mass of the electron is not given, but can be [[https://en.wikipedia.org/wiki/Electron_mass|found online]] ($m_e = 9.11\times10^{-31} kg$).

=== Approximations & Assumptions ===

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

=== Representations ===

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

==== Solution ====

First, we compute the speed of the electron.

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

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.

$$\vec{p}_e = \gamma m_e \vec{v}_e = (1.02) (9.11\times10^{-31} kg) \langle -2.05\times10^7, 6.02\times10^7, 0\rangle\dfrac{m}{s} = \langle -1.91 \times 10^{-23}, 5.61\times10^{-23},0\rangle\dfrac{kg\:m}{s}$$