From e28d2a982d0c65fcad9a2d2a4c20d06a9848fa8f Mon Sep 17 00:00:00 2001 From: Prefetch Date: Tue, 5 Oct 2021 19:31:12 +0200 Subject: Expand knowledge base --- content/know/concept/elastic-collision/index.pdc | 160 +++++++++++++++++++++++ 1 file changed, 160 insertions(+) create mode 100644 content/know/concept/elastic-collision/index.pdc (limited to 'content/know/concept/elastic-collision') diff --git a/content/know/concept/elastic-collision/index.pdc b/content/know/concept/elastic-collision/index.pdc new file mode 100644 index 0000000..9873144 --- /dev/null +++ b/content/know/concept/elastic-collision/index.pdc @@ -0,0 +1,160 @@ +--- +title: "Elastic collision" +firstLetter: "E" +publishDate: 2021-10-04 +categories: +- Physics +- Classical mechanics + +date: 2021-09-23T16:22:39+02:00 +draft: false +markup: pandoc +--- + +# Elastic collision + +In an **elastic collision**, +the sum of the colliding objects' kinetic energies +is the same before and after the collision. +In contrast, in an **inelastic collision**, +some of that energy is converted into another form, +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: + +$$\begin{aligned} + \begin{cases} + \quad\! m_1 v_1 +\quad m_2 v_2 + = \quad\, m_1 v_1' +\quad m_2 v_2' + \\ + \displaystyle\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2 + \end{cases} +\end{aligned}$$ + +After some rearranging, +these two equations can be written as follows: + +$$\begin{aligned} + \begin{cases} + m_1 (v_1 - v_1') + \qquad\quad\:\;\; = m_2 (v_2' - v_2) + \\ + m_1 (v_1 - v_1') (v_1 + v_1') + = m_2 (v_2' - v_2) (v_2 + v_2') + \end{cases} +\end{aligned}$$ + +Using the first equation to replace $m_1 (v_1 \!-\! v_1')$ +with $m_2 (v_2 \!-\! v_2')$ in the second: + +$$\begin{aligned} + m_2 (v_1 + v_1') (v_2' - v_2) + = m_2 (v_2 + v_2') (v_2' - v_2) +\end{aligned}$$ + +Dividing out the common factors +then leads us to a simplified system of equations: + +$$\begin{aligned} + \begin{cases} + \qquad\;\; v_1 + v_1' + = v_2 + v_2' + \\ + m_1 v_1 + m_2 v_2 + = m_1 v_1' + m_2 v_2' + \end{cases} +\end{aligned}$$ + +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: + +$$\begin{aligned} + m_1 v_1 + m_2 v_2 + &= m_1 (v_2 + v_2' - v_1) + m_2 v_2' + \\ + (m_1 + m_2) v_2' + &= 2 m_1 v_1 + (m_2 - m_1) v_2 +\end{aligned}$$ + +Dividing by $m_1 + m_2$, +and going through the same process for $v_1'$, +we arrive at: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + v_1' + &= \frac{(m_1 - m_2) v_1 + 2 m_2 v_2}{m_1 + m_2} + \\ + v_2' + &= \frac{2 m_1 v_1 + (m_2 - m_1) v_2}{m_1 + m_2} + \end{aligned} + } +\end{aligned}$$ + +To analyze this result, +for practicality, we simplify it by setting $v_2 = 0$. +In that case: + +$$\begin{aligned} + v_1' + = \frac{(m_1 - m_2) v_1}{m_1 + m_2} + \qquad \quad + v_2' + = \frac{2 m_1 v_1}{m_1 + m_2} +\end{aligned}$$ + +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: + +$$\begin{aligned} + \frac{m_2 v_2'}{m_1 v_1} + = \frac{2 m_2}{m_1 + m_2} + \qquad \quad + \frac{m_2 v_2'^2}{m_1 v_1^2} + = \frac{4 m_1 m_2}{(m_1 + m_2)^2} +\end{aligned}$$ + +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, +barely transferring any energy. +Object 2 ends up with twice object 1's momentum, +but $v_2'$ is very small and thus negligible: + +$$\begin{aligned} + \frac{m_2 v_2'}{m_1 v_1} + \approx 2 + \qquad \quad + \frac{m_2 v_2'^2}{m_1 v_1^2} + \approx \frac{4 m_1}{m_2} +\end{aligned}$$ + +If $m_1 \gg m_2$, object 1 barely notices the collision, +so not much is transferred to object 2: + +$$\begin{aligned} + \frac{m_2 v_2'}{m_1 v_1} + \approx \frac{2 m_2}{m_1} + \qquad \quad + \frac{m_2 v_2'^2}{m_1 v_1^2} + \approx \frac{4 m_2}{m_1} +\end{aligned}$$ + + + +## References +1. M. Salewski, A.H. Nielsen, + *Plasma physics: lecture notes*, + 2021, unpublished. -- cgit v1.2.3