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| 183_notes:examples:momentumfast [2014/07/10 14:11] – [Solution] caballero | 183_notes:examples:momentumfast [2024/01/30 14:18] (current) – [Setup] hallstein | ||
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| ====== 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 2.05\times10^7, | + | An electron is observed to be moving with a velocity of $\langle |
| ==== Setup ==== | ==== Setup ==== | ||
| Line 11: | Line 11: | ||
| * 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://lmgtfy.com/? | + | * The mass of the electron is not given, but can be [[https://en.wikipedia.org/ |
| === Approximations & Assumptions === | === Approximations & Assumptions === | ||
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| First, we compute the speed of the electron. | First, we compute the speed of the electron. | ||
| - | |→v|=√v2x+v2y+v2z=√(2.05×107ms)2+(6.02×107ms)2+(0)2=6.36×107ms | + | $$|\vec{v}_e| = \sqrt{v_x^2+v_y^2+v_z^2} = \sqrt{(-2.05\times10^7 \dfrac{m}{s})^2+(6.02\times10^7 \dfrac{m}{s})^2+(0)^2} = 6.36 \times 10^7 \dfrac{m}{s}$$ |
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| + | Next, we compute the gamma factor. | ||
| + | |||
| + | γ=1√1−(|→v|c)2=1√1−(6.36×107ms3.00×108ms)2=1√1−(0.212)2=1.02 | ||
| + | |||
| + | Finally, we compute the momentum vector. | ||
| + | |||
| + | →pe=γme→ve=(1.02)(9.11×10−31kg)⟨−2.05×107,6.02×107,0⟩ms=⟨−1.91×10−23,5.61×10−23,0⟩kgms | ||