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--- 183_notes:examples:momentumfast [2014/07/10 14:17] – [Solution] caballero
+++ 183_notes:examples:momentumfast [2024/01/30 14:18] (current) – [Setup] hallstein
@@ Line 2: / Line 2: @@
 ====== 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.
+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 ====
@@ Line 11: / Line 11: @@
   * An electron is in motion
-  * It has a velocity of $\langle -2.05\times10^7, 6.02\times10^7, 0\rangle\:\dfrac{m}{s}$.
+  * It has a velocity, $\vec{v}_e=\langle -2.05\times10^7, 6.02\times10^7, 0\rangle\dfrac{m}{s}$.
-  * This velocity is near the speed of light ($c = 3.00\times10^8 \dfrac{m}{s}$).
+  * The 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 [[http://lmgtfy.com/?q=electron+mass|found online]] ($m_e = 9.11\times10^{-31} kg$).
+  * 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 ===
@@ Line 30: / Line 30: @@
 First, we compute the speed of the electron.
-$$|\vec{v}| = \sqrt{v_x^2+v_y^2+v_z^2} = \sqrt{(-2.05\times10^7 \dfrac{m}{s})^2+(6.02\times10^7 \dfrac{m}{s})^2+(0)^2} = 6.36 \times 10^7 \dfrac{m}{s}$$
+$$|\vec{v}_e| = \sqrt{v_x^2+v_y^2+v_z^2} = \sqrt{(-2.05\times10^7 \dfrac{m}{s})^2+(6.02\times10^7 \dfrac{m}{s})^2+(0)^2} = 6.36 \times 10^7 \dfrac{m}{s}$$
 Next, we compute the gamma factor.
-$$\gamma = \dfrac{1}{\sqrt{1-\left(\dfrac{|\vec{v}|}{c}\right)^2}} = \dfrac{1}{\sqrt{1-\left(\dfrac{6.36 \times 10^7 \dfrac{m}{s}}{3.00 \times 10^8 \dfrac{m}{s}}\right)^2}} = \dfrac{1}{\sqrt{1-(0.212)^2}}$$
+$$\gamma = \dfrac{1}{\sqrt{1-\left(\dfrac{|\vec{v}|}{c}\right)^2}} = \dfrac{1}{\sqrt{1-\left(\dfrac{6.36 \times 10^7 \dfrac{m}{s}}{3.00 \times 10^8 \dfrac{m}{s}}\right)^2}} = \dfrac{1}{\sqrt{1-(0.212)^2}}=1.02$$
+Finally, we compute the momentum vector.
+$$\vec{p}_e = \gamma m_e \vec{v}_e = (1.02) (9.11\times10^{-31} kg) \langle -2.05\times10^7, 6.02\times10^7, 0\rangle\dfrac{m}{s} = \langle -1.91 \times 10^{-23}, 5.61\times10^{-23},0\rangle\dfrac{kg\:m}{s}$$