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--- 183_notes:examples:momentumfast [2014/07/10 14:12] – [Solution] caballero
+++ 183_notes:examples:momentumfast [2014/07/10 14:21] – [Solution] caballero
@@ Line 32: / Line 32: @@
 $$|\vec{v}| = \sqrt{v_x^2+v_y^2+v_z^2} = \sqrt{(-2.05\times10^7 \dfrac{m}{s})^2+(6.02\times10^7 \dfrac{m}{s})^2+(0)^2} = 6.36 \times 10^7 \dfrac{m}{s}$$
+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.
+$$\vec{p} = \gamma m \vec{v} = (1.02) (9.11\times10^{-31} kg) \langle -2.05\times10^7, 6.02\times10^7, 0\rangle\dfrac{m}{s} = \langle -1.91 \times 10^{-23}, 5.61\times10^{-23},0\rangle\:\dfrac{kg\:m}{s}$$