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--- 183_notes:examples:momentumfast [2014/07/10 13:50] – caballero
+++ 183_notes:examples:momentumfast [2014/07/10 14:20] – caballero
@@ Line 1: / Line 1: @@
+~~NOTOC~~
 ====== Example: Calculating the momentum of a fast-moving object ======
-An electron is observed to be moving with a velocity of $\langle 2.05e7, 6.02e7, 0\rangle\:\dfrac{m}{s}$. 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 ====
+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 of $\langle -2.05\times10^7, 6.02\times10^7, 0\rangle\:\dfrac{m}{s}$.
+  * This velocity is near the speed of light ($c = 3.00\times10^8 \dfrac{m}{s}$).
+=== Lacking ===
+  * The mass of the electron is not given, but can be [[http://lmgtfy.com/?q=electron+mass|found online]] ($m_e = 9.11\times10^{-31} kg$).
+=== 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.
+$$|\vec{v}| = \sqrt{v_x^2+v_y^2+v_z^2} = \sqrt{(-2.05\times10^7 \dfrac{m}{s})^2+(6.02\times10^7 \dfrac{m}{s})^2+(0)^2} = 6.36 \times 10^7 \dfrac{m}{s}$$
+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.
+$$\vec{p} = \gamma m \vec{v} = (1.02) (9.11\times10^{-31} kg) \langle -2.05\times10^7, 6.02\times10^7, 0\rangle\:\dfrac{m}{s} = \langle \rangle\:\dfrac{kg\:m}{s}$$