From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 source/know/concept/elastic-collision/index.md | 28 +++++++++++++-------------
 1 file changed, 14 insertions(+), 14 deletions(-)

(limited to 'source/know/concept/elastic-collision')

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}
-- 
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