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Diffstat (limited to 'source/know/concept/elastic-collision')
| -rw-r--r-- | source/know/concept/elastic-collision/index.md | 28 |
1 files changed, 14 insertions, 14 deletions
diff --git a/source/know/concept/elastic-collision/index.md b/source/know/concept/elastic-collision/index.md index 11c3115..ac15e06 100644 --- a/source/know/concept/elastic-collision/index.md +++ b/source/know/concept/elastic-collision/index.md @@ -19,8 +19,8 @@ for example heat. ## One dimension In 1D, not only the kinetic energy is conserved, but also the total momentum. -Let $v_1$ and $v_2$ be the initial velocities of objects 1 and 2, -and $v_1'$ and $v_2'$ their velocities afterwards: +Let $$v_1$$ and $$v_2$$ be the initial velocities of objects 1 and 2, +and $$v_1'$$ and $$v_2'$$ their velocities afterwards: $$\begin{aligned} \begin{cases} @@ -45,8 +45,8 @@ $$\begin{aligned} \end{cases} \end{aligned}$$ -Using the first equation to replace $m_1 (v_1 \!-\! v_1')$ -with $m_2 (v_2 \!-\! v_2')$ in the second: +Using the first equation to replace $$m_1 (v_1 \!-\! v_1')$$ +with $$m_2 (v_2 \!-\! v_2')$$ in the second: $$\begin{aligned} m_2 (v_1 + v_1') (v_2' - v_2) @@ -66,10 +66,10 @@ $$\begin{aligned} \end{cases} \end{aligned}$$ -Note that the first relation is equivalent to $v_1 - v_2 = v_2' - v_1'$, +Note that the first relation is equivalent to $$v_1 - v_2 = v_2' - v_1'$$, meaning that the objects' relative velocity is reversed by the collision. -Moving on, we replace $v_1'$ in the second equation: +Moving on, we replace $$v_1'$$ in the second equation: $$\begin{aligned} m_1 v_1 + m_2 v_2 @@ -79,8 +79,8 @@ $$\begin{aligned} &= 2 m_1 v_1 + (m_2 - m_1) v_2 \end{aligned}$$ -Dividing by $m_1 + m_2$, -and going through the same process for $v_1'$, +Dividing by $$m_1 + m_2$$, +and going through the same process for $$v_1'$$, we arrive at: $$\begin{aligned} @@ -96,7 +96,7 @@ $$\begin{aligned} \end{aligned}$$ To analyze this result, -for practicality, we simplify it by setting $v_2 = 0$. +for practicality, we simplify it by setting $$v_2 = 0$$. In that case: $$\begin{aligned} @@ -108,7 +108,7 @@ $$\begin{aligned} \end{aligned}$$ How much of its energy and momentum does object 1 transfer to object 2? -The following ratios compare $v_1$ and $v_2'$ to quantify the transfer: +The following ratios compare $$v_1$$ and $$v_2'$$ to quantify the transfer: $$\begin{aligned} \frac{m_2 v_2'}{m_1 v_1} @@ -118,15 +118,15 @@ $$\begin{aligned} = \frac{4 m_1 m_2}{(m_1 + m_2)^2} \end{aligned}$$ -If $m_1 = m_2$, both ratios reduce to $1$, +If $$m_1 = m_2$$, both ratios reduce to $$1$$, meaning that all energy and momentum is transferred, and object 1 is at rest after the collision. Newton's cradle is an example of this. -If $m_1 \ll m_2$, object 1 simply bounces off object 2, +If $$m_1 \ll m_2$$, object 1 simply bounces off object 2, barely transferring any energy. Object 2 ends up with twice object 1's momentum, -but $v_2'$ is very small and thus negligible: +but $$v_2'$$ is very small and thus negligible: $$\begin{aligned} \frac{m_2 v_2'}{m_1 v_1} @@ -136,7 +136,7 @@ $$\begin{aligned} \approx \frac{4 m_1}{m_2} \end{aligned}$$ -If $m_1 \gg m_2$, object 1 barely notices the collision, +If $$m_1 \gg m_2$$, object 1 barely notices the collision, so not much is transferred to object 2: $$\begin{aligned} |