\subsection{Elastic, Inelastic, and Perfectly Inelastic Collisions}

This subsection classifies collisions by what happens to the system's kinetic energy. In AP mechanics, the first step is still to apply momentum conservation to an isolated system; only elastic collisions add a kinetic-energy conservation equation.

\dfn{Elastic, inelastic, and perfectly inelastic collisions}{Consider two objects of masses $m_1$ and $m_2$. Let $\vec{v}_{1,i}$ and $\vec{v}_{2,i}$ denote their velocities just before a collision, and let $\vec{v}_{1,f}$ and $\vec{v}_{2,f}$ denote their velocities just after it. Let the total kinetic energy before and after the collision be
\[
K_i=\tfrac12 m_1|\vec{v}_{1,i}|^2+\tfrac12 m_2|\vec{v}_{2,i}|^2
\]
and
\[
K_f=\tfrac12 m_1|\vec{v}_{1,f}|^2+\tfrac12 m_2|\vec{v}_{2,f}|^2.
\]

A collision is called \emph{elastic} if the total kinetic energy is unchanged, so
\[
K_f=K_i.
\]
It is called \emph{inelastic} if the total kinetic energy changes, so
\[
K_f\ne K_i.
\]
In the usual AP collision problems without stored energy being released during impact, an inelastic collision has $K_f<K_i$.

A collision is \emph{perfectly inelastic} if the objects stick together after the collision, so they share one final velocity:
\[
\vec{v}_{1,f}=\vec{v}_{2,f}=\vec{v}_f.
\]
Every perfectly inelastic collision is inelastic.}

\wc{Momentum is conserved in every collision type}{Total momentum of an isolated system is conserved in elastic, inelastic, \emph{and} perfectly inelastic collisions. The word ``elastic'' only tells you whether kinetic energy is also conserved. A common exam error is to drop the momentum equation for inelastic collisions --- \emph{never} do this.}

\wc{``Momentum is lost'' in a collision}{Momentum of the \emph{system} is never lost in any isolated collision. What is ``lost'' in an inelastic collision is \emph{kinetic energy}, which is converted to thermal energy, sound, deformation, etc. Momentum conservation holds regardless of how ``bouncy'' the collision is.}

Collisions connect momentum conservation (Unit 4) with energy conservation (Unit 3): elastic collisions conserve both, inelastic conserve only momentum. See also the coefficient of restitution in Section 4.5 for explosive separation.

\nt{Do not decide whether momentum is conserved by asking whether the collision is elastic. Those are different ideas. Momentum conservation depends on the net external impulse on the chosen system. If the system is isolated during the collision, then total momentum is conserved for elastic, inelastic, and perfectly inelastic collisions alike. Kinetic energy supplies an \emph{extra} condition only in the elastic case. In two-dimensional AP problems, conserve momentum separately in the $x$- and $y$-directions.}

\ex{Illustrative example}{On a frictionless track, cart 1 has mass $m_1=0.40\,\mathrm{kg}$ and initial velocity
\[
\vec{v}_{1,i}=(3.0\hat{\imath})\,\mathrm{m/s}.
\]
Cart 2 has mass $m_2=0.20\,\mathrm{kg}$ and initial velocity
\[
\vec{v}_{2,i}=\vec{0}.
\]
After the collision, the carts move together with common velocity
\[
\vec{v}_f=(2.0\hat{\imath})\,\mathrm{m/s}.
\]

Because the carts move together after impact, the collision is perfectly inelastic. The initial kinetic energy is
\[
K_i=\tfrac12(0.40)(3.0)^2=1.8\,\mathrm{J},
\]
while the final kinetic energy is
\[
K_f=\tfrac12(0.60)(2.0)^2=1.2\,\mathrm{J}.
\]
Since $K_f<K_i$, the collision is inelastic, as expected. Momentum is still conserved because
\[
\vec{P}_i=(0.40)(3.0\hat{\imath})=(1.2\hat{\imath})\,\mathrm{kg\cdot m/s}
\]
and
\[
\vec{P}_f=(0.60)(2.0\hat{\imath})=(1.2\hat{\imath})\,\mathrm{kg\cdot m/s}.
\]}

\mprop{Practical relations for isolated collisions}{Consider two objects that form an isolated system during a short collision. Let
\[
\vec{P}_i=m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}
\]
denote the total initial momentum and let
\[
\vec{P}_f=m_1\vec{v}_{1,f}+m_2\vec{v}_{2,f}
\]
denote the total final momentum.

\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item For any isolated collision,
\[
\vec{P}_i=\vec{P}_f,
\]
so
\[
m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}=m_1\vec{v}_{1,f}+m_2\vec{v}_{2,f}.
\]

\item In two dimensions, let $P_{x,i}$ and $P_{y,i}$ denote the initial momentum components, and let $P_{x,f}$ and $P_{y,f}$ denote the final momentum components. Then write two separate component equations:
\[
P_{x,i}=P_{x,f},
\qquad
P_{y,i}=P_{y,f}.
\]

\item If the collision is perfectly inelastic, then $\vec{v}_{1,f}=\vec{v}_{2,f}=\vec{v}_f$, so momentum conservation gives the shared final velocity directly:
\[
\vec{v}_f=\frac{m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}}{m_1+m_2}.
\]

\item If the collision is elastic, then in addition to momentum conservation,
\[
\tfrac12 m_1|\vec{v}_{1,i}|^2+\tfrac12 m_2|\vec{v}_{2,i}|^2
=
\tfrac12 m_1|\vec{v}_{1,f}|^2+\tfrac12 m_2|\vec{v}_{2,f}|^2.
\]
For a one-dimensional elastic collision, this is equivalent to the relative-speed relation
\[
v_{1,i}-v_{2,i}=-(v_{1,f}-v_{2,f}),
\]
where each $v$ is an $x$-component.
\end{enumerate}}

\qs{Worked example}{Choose the positive $x$-axis to the right. On a frictionless track, cart 1 has mass $m_1=1.0\,\mathrm{kg}$ and initial velocity
\[
\vec{v}_{1,i}=(4.0\hat{\imath})\,\mathrm{m/s}.
\]
Cart 2 has mass $m_2=3.0\,\mathrm{kg}$ and is initially at rest, so
\[
\vec{v}_{2,i}=\vec{0}.
\]
After the collision, the carts lock together and move with common final velocity $\vec{v}_f$.

Find $\vec{v}_f$, determine whether the collision is elastic, inelastic, or perfectly inelastic, and calculate the change in kinetic energy $\Delta K=K_f-K_i$.}

\sol Because the carts lock together, this is a perfectly inelastic collision by definition. During the short collision the track is frictionless, so the two-cart system is isolated horizontally. Therefore total momentum is conserved:
\[
\vec{P}_i=\vec{P}_f.
\]

The initial momentum is
\[
\vec{P}_i=m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}.
\]
Substitute the given values:
\[
\vec{P}_i=(1.0\,\mathrm{kg})(4.0\hat{\imath}\,\mathrm{m/s})+(3.0\,\mathrm{kg})(\vec{0})
=4.0\hat{\imath}\,\mathrm{kg\cdot m/s}.
\]

After the collision the carts move together, so their total mass is
\[
m_1+m_2=4.0\,\mathrm{kg}
\]
and their final momentum is
\[
\vec{P}_f=(m_1+m_2)\vec{v}_f.
\]
Thus,
\[
4.0\hat{\imath}\,\mathrm{kg\cdot m/s}=(4.0\,\mathrm{kg})\vec{v}_f,
\]
so
\[
\vec{v}_f=(1.0\hat{\imath})\,\mathrm{m/s}.
\]

Now compare the kinetic energies.

Initially,
\[
K_i=\tfrac12 m_1|\vec{v}_{1,i}|^2+\tfrac12 m_2|\vec{v}_{2,i}|^2
=\tfrac12(1.0)(4.0)^2+\tfrac12(3.0)(0)^2=8.0\,\mathrm{J}.
\]

Finally,
\[
K_f=\tfrac12 (m_1+m_2)|\vec{v}_f|^2
=\tfrac12(4.0)(1.0)^2=2.0\,\mathrm{J}.
\]

Therefore,
\[
\Delta K=K_f-K_i=2.0\,\mathrm{J}-8.0\,\mathrm{J}=-6.0\,\mathrm{J}.
\]

The negative sign means $6.0\,\mathrm{J}$ of kinetic energy was transformed into other forms of energy during the collision. Since the carts stick together and the kinetic energy decreases, the collision is inelastic, more specifically perfectly inelastic.

So the results are
\[
\vec{v}_f=(1.0\hat{\imath})\,\mathrm{m/s},
\qquad
\Delta K=-6.0\,\mathrm{J},
\]
and the collision is perfectly inelastic rather than elastic.
