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---
title: "Elastic collision"
sort_title: "Elastic collision"
date: 2021-10-04
categories:
- Physics
- Classical mechanics
layout: "concept"
---
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:
$$\begin{aligned}
\begin{cases}
\quad\! m_1 v_1 +\:\:\: m_2 v_2
= \quad m_1 v_1' +\:\:\: m_2 v_2'
\\
\displaystyle\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2
= \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2
\end{cases}
\end{aligned}$$
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.
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