Categories: Classical mechanics, Physics.

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.

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:

\[\begin{aligned} \frac{m_2 v_2'}{m_1 v_1} \approx \frac{2 m_2}{m_1} \qquad \quad \frac{m_2 v_2'^2}{m_1 v_1^2} \approx \frac{4 m_2}{m_1} \end{aligned}\]

- 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.