Skip to main content

# 14.3: Totally Inelastic Collision

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$

In a totally inelastic collision, particles stick together. A possible example is the absorption of a photon by a massive particle, resulting in an increase in its mass, as well as possibly a change in its momentum. Let’s consider, as an example, a particle of mass $$m$$ that is initially at rest, and absorbs an incoming photon with energy $$E_{\gamma}$$. There are now three ways to calculate the energy and momentum of the particle after this collision.

## Method 1

We have conservation of both (total) energy and momentum. Before the collision, the massive particle has energy $$E_{\mathrm{i}}=m c^{2}$$ (as it is standing still), and the total energy of the system $$E_{\gamma}+m c^{2}$$, which must be conserved. The total energy of the particle after the collision is $$E_{\mathrm{f}}=\gamma(v) m_{\mathrm{f}} c^{2}$$, where both the velocity $$v$$ and the mass $$m_{\mathrm{f}}$$ are unknown. The total momentum before the collision is $$E_{\gamma} / c$$, as the particle is initially standing still (and thus has momentum zero), while after the collision it is $$\gamma(v) m_{\mathrm{f}}v$$. We thus have:

\begin{align} E_{\gamma}+m c^{2} &=\gamma(v) m_{\mathrm{f}} c^{2} \label{14.4} \\[4pt] E_{\gamma} &=\gamma(v) m_{\mathrm{f}} vc \label{14.5} \end{align}

We thus have two equations with two unknowns ($$v$$ and $$m_{\mathrm{f}}$$). If we divide Equation \ref{14.5} by \ref{14.4}, we get an expression for the final velocity $$v$$, which we can substitute back in either equation to solve for $$m_{\mathrm{f}}$$ (and potentially use to calculate the momentum after the collision). This is not pretty though, as we’ll have complicated factors due to the presence of $$\gamma(v)$$.

## Method 2

The four-momentum of the system is conserved during the collision. We have $$\overline{\boldsymbol{p}}_{\gamma}$$ for the photon, $$\overline{\boldsymbol{p}}_{1}$$ for the massive particle before the collision, and $$\overline{\boldsymbol{p}}_{\mathrm{f}}$$ for that particle after the collision, given by the following equations:

\begin{align} \overline{\boldsymbol{p}}_{\gamma} &=\left(\frac{E_{\gamma}}{c}, \frac{E_{\gamma}}{c}, 0,0\right) \\[4pt] \overline{\boldsymbol{p}}_{1} &=(m c, 0,0,0) \\[4pt] \overline{\boldsymbol{p}}_{\mathrm{f}} &=\left(\frac{E_{\mathrm{f}}}{c}, p_{\mathrm{f}}, 0,0\right) \end{align}

From $$\overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{1}=\overline{\boldsymbol{p}}_{\mathrm{f}}$$ we can read off two equations:

$E_{\gamma}+m c^{2}=E_{\mathrm{f}}$

$E_{\gamma} / c=p_{\mathrm{f}}$

which immediately give us the final energy and momentum in terms of the initial ones. We can now find the final mass through Einstein’s equation (13.16):

\begin{align} m_{\mathrm{f}}^{2} c^{4}=E_{\mathrm{f}}^{2}-p_{\mathrm{f}}^{2} c^{2}&=\left(E_{\gamma}+m c^{2}\right)^{2}-E_{\gamma}^{2} \\[4pt] &=\left(E_{\gamma}+m c^{2}\right) m c^{2} \end{align}

This approach circumvents the use of the $$\gamma(v)$$ factor because we only use energy and momentum, not (classical) velocity. If we now want the velocity, we could still calculate it from the combination of $$m_{\mathrm{f}}$$ and either $$E_{\mathrm{f}}$$ or $$p_{\mathrm{f}}$$, but since it was the mass and momentum we were after, there’s no need to do so.

## Method 3

Since the total energy-momentum four-vector is conserved in the collision, so must be its length (or the square of the length), which is trivial to calculate (remember that $$\overline{\boldsymbol{p}} \cdot \overline{\boldsymbol{p}}=m^{2} c^{2}$$). We can often exploit this fact to make the maths much simpler. To see how this works, let’s consider the full four-vector equation for our example: $$\overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{\mathrm{i}}=\overline{\boldsymbol{p}}_{\mathrm{f}}$$, so

\begin{align} \left(\overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{\mathrm{i}}\right) \cdot\left(\overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{\mathrm{i}}\right) &= \overline{\boldsymbol{p}}_{\mathrm{f}} \cdot \overline{\boldsymbol{p}}_{\mathrm{f}} \label{14.9} \\[4pt] \overline{\boldsymbol{p}}_{\gamma} \cdot \overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{\mathrm{i}} \cdot \overline{\boldsymbol{p}}_{\mathrm{i}}+2 \overline{\boldsymbol{p}}_{\gamma} \cdot \overline{\boldsymbol{p}}_{\mathrm{i}} &=\overline{\boldsymbol{p}}_{\mathrm{f}} \cdot \overline{\boldsymbol{p}}_{\mathrm{f}} \label{14.10} \end{align}

$0+m^{2} c^{2}+2 E_{\gamma} m=m_{\mathrm{f}}^{2} c^{2}$

which immediately gives us $$m_{\mathrm{f}}$$. If we also want $$E_{\mathrm{f}}$$ or $$p_{\mathrm{f}}$$, we can again use Equations \ref{14.9} and \ref{14.10} for the components, but if we only wanted the final mass, we’re done in one step.

Note that although method 3 usually is the easiest route to your answer, it is not always - and it is a good idea to at least be aware of the other options.

This page titled 14.3: Totally Inelastic Collision is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Timon Idema (TU Delft Open) via source content that was edited to the style and standards of the LibreTexts platform.

• Was this article helpful?