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183_notes:examples:momentumfast [2014/07/10 13:59] – caballero | 183_notes:examples:momentumfast [2014/07/10 14:35] – caballero | ||
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+ | ~~NOTOC~~ | ||
====== Example: Calculating the momentum of a fast-moving object ====== | ====== 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 |
==== Setup ==== | ==== Setup ==== | ||
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* An electron is in motion | * An electron is in motion | ||
- | * It has a velocity | + | * It has a velocity, $\vec{v}_e=\langle |
- | * This velocity | + | * The speed of the electron |
=== Lacking === | === Lacking === | ||
- | * The mass of the electron is not given, but can be [[http:// | + | * The mass of the electron is not given, but can be [[http:// |
=== Approximations & Assumptions === | === Approximations & Assumptions === | ||
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==== Solution ==== | ==== 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, | ||
+ |