Categories: Classical mechanics, Physics.

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}

1. M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.