\subsection{Recoil, Explosions, and the Center-of-Mass Viewpoint}

This subsection treats recoil and explosions as momentum-redistribution processes. In AP mechanics, internal energy may be released during the interaction, but if the net external impulse on the chosen system is zero, the total momentum and the center-of-mass motion do not change.

\dfn{Recoil/explosion interactions and the center-of-mass viewpoint}{Consider a closed system of particles labeled by $i=1,2,\dots,N$. Let $m_i$ denote the mass of particle $i$, let $\vec{v}_i$ denote its velocity, let
\[
\vec{p}_i=m_i\vec{v}_i
\]
denote its momentum, and let
\[
\vec{P}=\sum_{i=1}^N \vec{p}_i=\sum_{i=1}^N m_i\vec{v}_i
\]
denote the total momentum. Let
\[
M=\sum_{i=1}^N m_i
\]
denote the total mass, and let $\vec{v}_{\mathrm{cm}}$ denote the center-of-mass velocity. Then
\[
\vec{P}=M\vec{v}_{\mathrm{cm}}.
\]

A \emph{recoil} or \emph{explosion} interaction is a short internal interaction in which parts of the system push apart or are driven apart by released internal energy. If the net external impulse on the system during the interaction is zero or negligible, then
\[
\vec{P}_f=\vec{P}_i,
\]
so equivalently
\[
\vec{v}_{\mathrm{cm},f}=\vec{v}_{\mathrm{cm},i}.
\]
If the system is initially at rest, then $\vec{P}_i=\vec{0}$ and $\vec{v}_{\mathrm{cm}}=\vec{0}$ both before and after the interaction.}

\nt{In recoil and explosion problems, the internal forces between the parts of the system can be very large, but they occur in equal-and-opposite pairs and therefore only redistribute momentum within the system. They can change the individual velocities and can increase the total kinetic energy if internal energy is released, but they do not change the total momentum of an isolated system. From the center-of-mass viewpoint, the whole event is just internal rearrangement: the center of mass continues to remain at rest or to move with constant velocity if the external impulse is zero.}

\ex{Illustrative example}{A student of mass $m_s=60.0\,\mathrm{kg}$ stands at rest on a frictionless skateboard and throws a backpack of mass $m_b=5.0\,\mathrm{kg}$ horizontally backward with velocity
\[
\vec{v}_b=(-8.0\hat{\imath})\,\mathrm{m/s}.
\]
Let $\vec{v}_s=v_s\hat{\imath}$ denote the student's recoil velocity after the throw.

Because the student-backpack system starts at rest and the external horizontal impulse is negligible,
\[
\vec{P}_f=\vec{0}.
\]
Thus,
\[
m_s\vec{v}_s+m_b\vec{v}_b=\vec{0}.
\]
Substitute the values:
\[
(60.0\,\mathrm{kg})v_s\hat{\imath}+(5.0\,\mathrm{kg})(-8.0\hat{\imath}\,\mathrm{m/s})=\vec{0}.
\]
So,
\[
60.0v_s-40.0=0,
\]
which gives
\[
v_s=0.667\,\mathrm{m/s}.
\]
Therefore,
\[
\vec{v}_s=(0.667\hat{\imath})\,\mathrm{m/s}.
\]
The student recoils forward while the backpack moves backward, and the total momentum remains zero.}

\mprop{Useful recoil and center-of-mass relations}{Let a system of total mass $M$ have velocity $\vec{V}_0$ just before a recoil or explosion event. Let the final pieces have masses $m_1,m_2,\dots$ and velocities $\vec{v}_1,\vec{v}_2,\dots$. If the net external impulse during the short interaction is zero or negligible, then:

\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item Total momentum is conserved:
\[
\sum \vec{p}_{f}=\sum \vec{p}_{i},
\qquad
\sum m_k\vec{v}_k=M\vec{V}_0.
\]

\item The center-of-mass velocity stays constant:
\[
\vec{v}_{\mathrm{cm}}=\frac{\vec{P}}{M}=\vec{V}_0.
\]
If the system starts from rest, then $\vec{v}_{\mathrm{cm}}=\vec{0}$ and
\[
\sum m_k\vec{v}_k=\vec{0}.
\]

\item For a two-piece explosion from rest,
\[
m_1\vec{v}_1+m_2\vec{v}_2=\vec{0},
\]
so the two final momenta are equal in magnitude and opposite in direction:
\[
m_1\vec{v}_1=-m_2\vec{v}_2.
\]

\item In two dimensions, conserve momentum component-by-component:
\[
\sum p_{x,f}=\sum p_{x,i},
\qquad
\sum p_{y,f}=\sum p_{y,i}.
\]
This is usually the most direct way to find unknown fragment velocities.
\end{enumerate}}

\qs{Worked example}{A firework shell of total mass $M=4.0\,\mathrm{kg}$ is moving with velocity
\[
\vec{V}_0=(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}
\]
just before it explodes into two fragments. Fragment 1 has mass $m_1=1.5\,\mathrm{kg}$ and velocity
\[
\vec{v}_1=(18.0\hat{\imath}+2.0\hat{\jmath})\,\mathrm{m/s}
\]
immediately after the explosion. Fragment 2 has mass $m_2=2.5\,\mathrm{kg}$ and velocity
\[
\vec{v}_2=v_{2,x}\hat{\imath}+v_{2,y}\hat{\jmath}.
\]
Assume the net external impulse during the explosion is negligible.

Find $\vec{v}_2$, find the center-of-mass velocity after the explosion, and determine whether the explosion could have increased the total kinetic energy.}

\sol Because the explosion is brief and the net external impulse is negligible, the shell-plus-fragments system conserves momentum:
\[
\vec{P}_i=\vec{P}_f.
\]

First compute the initial total momentum. Since the shell has total mass $M=4.0\,\mathrm{kg}$ and velocity $\vec{V}_0$ just before the explosion,
\[
\vec{P}_i=M\vec{V}_0.
\]
Substitute the given values:
\[
\vec{P}_i=(4.0\,\mathrm{kg})(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}.
\]
Therefore,
\[
\vec{P}_i=(40.0\hat{\imath}+24.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
\]

Next compute the momentum of fragment 1:
\[
\vec{p}_1=m_1\vec{v}_1=(1.5\,\mathrm{kg})(18.0\hat{\imath}+2.0\hat{\jmath})\,\mathrm{m/s}.
\]
So,
\[
\vec{p}_1=(27.0\hat{\imath}+3.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
\]

Let $\vec{p}_2=m_2\vec{v}_2$ denote the momentum of fragment 2. Since
\[
\vec{P}_f=\vec{p}_1+\vec{p}_2,
\]
momentum conservation gives
\[
\vec{p}_2=\vec{P}_i-\vec{p}_1.
\]
Thus,
\[
\vec{p}_2=(40.0\hat{\imath}+24.0\hat{\jmath})-(27.0\hat{\imath}+3.0\hat{\jmath}).
\]
Therefore,
\[
\vec{p}_2=(13.0\hat{\imath}+21.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
\]

Now divide by $m_2=2.5\,\mathrm{kg}$ to find the velocity of fragment 2:
\[
\vec{v}_2=\frac{\vec{p}_2}{m_2}=\frac{(13.0\hat{\imath}+21.0\hat{\jmath})\,\mathrm{kg\cdot m/s}}{2.5\,\mathrm{kg}}.
\]
Hence,
\[
\vec{v}_2=(5.2\hat{\imath}+8.4\hat{\jmath})\,\mathrm{m/s}.
\]

Now find the center-of-mass velocity after the explosion. The total momentum after the explosion is still
\[
\vec{P}_f=\vec{P}_i=(40.0\hat{\imath}+24.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
\]
Using
\[
\vec{v}_{\mathrm{cm}}=\frac{\vec{P}}{M},
\]
we get
\[
\vec{v}_{\mathrm{cm},f}=\frac{\vec{P}_f}{M}=\frac{(40.0\hat{\imath}+24.0\hat{\jmath})\,\mathrm{kg\cdot m/s}}{4.0\,\mathrm{kg}}.
\]
So,
\[
\vec{v}_{\mathrm{cm},f}=(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}.
\]
This is exactly the same as the pre-explosion velocity $\vec{V}_0$, which is the center-of-mass viewpoint: the fragments fly apart, but the center of mass continues with the same velocity because the external impulse is negligible.

Finally, check whether the total kinetic energy could have increased.

Before the explosion,
\[
K_i=\tfrac12 M|\vec{V}_0|^2=\tfrac12 (4.0)\left[(10.0)^2+(6.0)^2\right].
\]
Thus,
\[
K_i=2.0(136)=272\,\mathrm{J}.
\]

After the explosion,
\[
K_f=\tfrac12 m_1|\vec{v}_1|^2+\tfrac12 m_2|\vec{v}_2|^2.
\]
For fragment 1,
\[
|\vec{v}_1|^2=(18.0)^2+(2.0)^2=328,
\]
so
\[
K_1=\tfrac12 (1.5)(328)=246\,\mathrm{J}.
\]
For fragment 2,
\[
|\vec{v}_2|^2=(5.2)^2+(8.4)^2=27.04+70.56=97.60,
\]
so
\[
K_2=\tfrac12 (2.5)(97.60)=122\,\mathrm{J}.
\]
Therefore,
\[
K_f=246\,\mathrm{J}+122\,\mathrm{J}=368\,\mathrm{J}.
\]

Since
\[
K_f>K_i,
\]
the total kinetic energy increased by
\[
\Delta K=368\,\mathrm{J}-272\,\mathrm{J}=96\,\mathrm{J}.
\]
That is possible because the explosion released internal energy. Momentum stayed constant because the system was isolated during the brief explosion, but kinetic energy did not have to remain constant.

So the results are
\[
\vec{v}_2=(5.2\hat{\imath}+8.4\hat{\jmath})\,\mathrm{m/s},
\]
and
\[
\vec{v}_{\mathrm{cm},f}=(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}.
\]
The explosion changed the fragment velocities and increased the total kinetic energy, but it did not change the total momentum or the motion of the center of mass.
