---
title: Center-of-Mass Collisions
module: Momentum
moduleNumber: 4
lessonNumber: 2
order: 402
summary: >
  A two-body collision that looks asymmetric in the laboratory becomes almost
  trivial in the frame that rides along with the centre of mass, where the total
  momentum is zero and the two momenta stay equal and opposite. We build that frame,
  reduce the pair to a single relative coordinate carrying the reduced mass $\mu$,
  and show that an elastic collision there only rotates one momentum vector while its
  length holds fixed. Transforming back to the laboratory then handles elastic and
  inelastic collisions, scattering angles, and reaction thresholds with the same
  construction — and shows why relative speed, not laboratory kinetic energy,
  measures what a collision can convert.
topics: [Momentum]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 8 — Conservation of Linear Momentum; §8-4"
---

## The center-of-mass frame

The centre-of-mass frame moves at

$$
\vec V_{\rm cm}=\frac{m_1\vec v_1+m_2\vec v_2}{m_1+m_2}.
$$

Velocities in that frame are obtained by a Galilean transformation,

$$
\vec v_1'=\vec v_1-\vec V_{\rm cm},
\qquad
\vec v_2'=\vec v_2-\vec V_{\rm cm}.
$$

Their total momentum vanishes:

$$
m_1\vec v_1'+m_2\vec v_2'=0.
$$

Thus the two momenta are equal in magnitude and opposite in direction. A collision
that appears asymmetric in the laboratory can be geometrically simple in this frame.
After solving there, add $\vec V_{\rm cm}$ to every final velocity to return to
the laboratory frame.

$$
% caption: In the center-of-mass frame the two momenta $\vec p_1'$ and $\vec p_2'$
% are equal in magnitude and opposite in direction, before and after any collision
% with negligible external impulse; their vector sum is zero.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[fill=black!14, draw=black, thick] (2.7,1.3) circle (0.34);
  \draw[fill=acc!12, draw=acc, thick] (3.9,1.3) circle (0.56);
  \draw[->, black, very thick] (2.36,1.3) -- (0.8,1.3) node[above] {$p_1$};
  \draw[->, acc, very thick] (4.46,1.3) -- (6.0,1.3) node[above] {$p_2$};
  \draw[dashed, black] (0.4,0.4) -- (6.2,0.4);
  \node[anchor=north, black] at (3.3,0.32) {CM frame};
\end{tikzpicture}
$$

**Kinetic energy in the CM frame.**

The laboratory kinetic energy separates into translation of the centre of mass and
kinetic energy relative to it:

$$
K=\frac12MV_{\rm cm}^2+K',
\qquad
K'=\frac12m_1{v_1'}^2+\frac12m_2{v_2'}^2.
$$

Only $K'$ is available for conversion into internal energy in an isolated collision.
The centre-of-mass translational term cannot be removed by internal forces. For an
elastic collision, $K'$ is unchanged. For an inelastic collision, it decreases;
the difference becomes deformation, heating, sound, or internal excitation.

Using reduced mass

$$
\mu=\frac{m_1m_2}{m_1+m_2},
$$

and relative speed $u=|\vec v_1-\vec v_2|$, the relative kinetic energy is

$$
K'=\frac12\mu u^2.
$$

Relative speed is invariant under every Galilean transformation.

## Elastic and inelastic collisions

In an elastic collision, momentum conservation makes the final CM momenta opposite
and kinetic-energy conservation preserves their magnitudes. The collision therefore
rotates the two opposite momentum vectors without changing their length. In the CM
frame, an elastic two-body collision is a change of direction rather than a change
of speed.

$$
% caption: Elastic scattering in the center-of-mass frame rotates the equal and
% opposite momentum pair through the scattering angle $\theta$ while its magnitude is
% preserved; only the direction changes.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \fill[black] (3.0,1.5) circle (2pt);
  \draw[->, black, dashed] (3.0,1.5) -- (1.05,1.5) node[left] {initial};
  \draw[->, black, dashed] (3.0,1.5) -- (4.95,1.5);
  \draw[->, acc, very thick] (3.0,1.5) -- (1.55,2.65) node[above] {$p_1$};
  \draw[->, acc, very thick] (3.0,1.5) -- (4.45,0.35) node[below] {$p_2$};
  \draw[acc, thick] (1.9,1.5) arc (180:142:1.1);
\end{tikzpicture}
$$

For equal masses with one target initially at rest in the laboratory, the centre-of-
mass frame moves at half the incident speed. If the laboratory final velocities are
perpendicular, their vector sum equals the initial incident velocity. Momentum and
energy equations then reduce to a right triangle. This right-angle result is special
to equal masses, an elastic collision, and an initially stationary target.

> **Worked example.** An object of mass $m$ moving at $6.0\ \mathrm{m\,s^{-1}}$
> along $x$ strikes an identical object at rest elastically, so the CM speed is
> $3.0\ \mathrm{m\,s^{-1}}$. If the incident object scatters at $30^\circ$ in the
> laboratory, the perpendicular-final-velocity result sends the target off at
> $60^\circ$ below the original direction, and the components are
>
> $$
> v_{1f}=6.0\cos30^\circ=5.20\ \mathrm{m\,s^{-1}},
> \qquad
> v_{2f}=6.0\sin30^\circ=3.00\ \mathrm{m\,s^{-1}}.
> $$
>
> The squared speeds sum to $6.0^2$, as energy conservation for equal masses
> requires.

**Inelastic collisions.**

In a perfectly inelastic collision, both objects share a final laboratory velocity
equal to $\vec V_{\rm cm}$. In the CM frame both objects are at rest afterward,
so all relative kinetic energy is lost:

$$
\Delta K'= -\frac12\mu u^2.
$$

The maximum kinetic-energy loss compatible with momentum conservation for
the two bodies. Ordinary inelastic collisions retain some relative motion. The
coefficient of restitution describes the normal component of separation speed but
does not replace the vector momentum equation in an oblique collision.

**Laboratory transformation and checks.**

After determining final CM velocities, transform back through

$$
\vec v_{if}=\vec v_{if}'+\vec V_{\rm cm}.
$$

The same $\vec V_{\rm cm}$ is used before and after an isolated collision.
Momentum must match in each laboratory component. For an elastic event, total
kinetic energy must also match in one selected inertial frame. A calculation that
conserves CM kinetic energy but uses untransformed final velocities in the laboratory
has mixed frames and is invalid.

## Reduced mass and scattering geometry

The centre-of-mass condition is

$$
m_1\vec v_1'+m_2\vec v_2'=0.
$$

Solving for the individual CM velocities in terms of the relative velocity
$\vec u=\vec v_1-\vec v_2=\vec v_1'-\vec v_2'$ gives

$$
\vec v_1'=\frac{m_2}{m_1+m_2}\vec u,
\qquad
\vec v_2'=-\frac{m_1}{m_1+m_2}\vec u.
$$

Substitution into $K'=\tfrac12m_1{v_1'}^2+\tfrac12m_2{v_2'}^2$ yields

$$
K'=\frac12\left(\frac{m_1m_2}{m_1+m_2}\right)u^2
=\frac12\mu u^2.
$$

The reduced mass behaves as the effective inertia of the relative coordinate. When
$m_2\gg m_1$, $\mu\approx m_1$ and the heavier target is nearly stationary in the
CM frame. When masses are equal, $\mu=m/2$ and both objects have equal CM speeds.

**Scattering angles in two dimensions.**

In two dimensions, momentum conservation yields an $x$ and a $y$ equation.
An elastic collision of incident mass $m_1$ with stationary mass $m_2$ gives

$$
m_1v_i=m_1v_{1f}\cos\theta_1+m_2v_{2f}\cos\theta_2,
$$

$$
0=m_1v_{1f}\sin\theta_1+m_2v_{2f}\sin\theta_2.
$$

Kinetic-energy conservation adds a third scalar equation. A specified scattering
angle or a geometric relation adds the remaining condition. The
angles in these equations must be measured in one frame. Laboratory angles differ
from CM angles except in special symmetric cases.


For equal masses with the target initially at rest, elastic energy conservation and
momentum conservation imply

$$
\vec v_{1f}\cdot\vec v_{2f}=0.
$$

The final velocity vectors are perpendicular. This follows by squaring
$\vec v_i=\vec v_{1f}+\vec v_{2f}$ and comparing with
$v_i^2=v_{1f}^2+v_{2f}^2$. It is not true for unequal masses or for inelastic
collisions.

> **Worked example.** A $1.0\ \mathrm{kg}$ cart at $4.0\ \mathrm{m\,s^{-1}}$
> strikes a stationary $3.0\ \mathrm{kg}$ cart elastically. The one-dimensional
> elastic formulas give
>
> $$
> v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_i=-2.0\ \mathrm{m\,s^{-1}},
> \qquad
> v_{2f}=\frac{2m_1}{m_1+m_2}v_i=2.0\ \mathrm{m\,s^{-1}}.
> $$
>
> The same answer comes faster through the CM frame, whose speed is
> $1.0\ \mathrm{m\,s^{-1}}$. There the initial velocities are $+3.0$ and
> $-1.0\ \mathrm{m\,s^{-1}}$; an elastic collision reverses both, giving CM finals
> $-3.0$ and $+1.0\ \mathrm{m\,s^{-1}}$. Adding the CM speed back returns the
> laboratory velocities.

## Energy loss, models, and diagnostics

When objects stick, the final velocity is $\vec V_{\rm cm}$. The maximum kinetic
energy converted to internal forms is the initial relative kinetic energy,

$$
K_{\rm lost}=\frac12\mu u_i^2.
$$

With a stationary target, this becomes

$$
K_{\rm lost}=\frac12\frac{m_1m_2}{m_1+m_2}v_i^2.
$$

The loss is positive for nonzero relative speed and independent of the chosen
inertial frame because relative velocity is invariant under Galilean transformation.
Total kinetic energy itself is frame-dependent, so statements about a “fraction of
lab kinetic energy lost” require a specified frame.

**Experimental and modeling limits.**

Real scattering can involve rotation, deformation, multiple particles, and energy
transfer to internal states. Momentum remains conserved for an isolated complete
system, but a two-particle elastic model may fail. A detector commonly measures
laboratory angles and energies; CM angles are reconstructed by transformation. A
collision with a fixed wall is not a two-body isolated collision unless the wall and
Earth are included in the system. The wall's large mass then explains its nearly
unchanged velocity while it receives momentum.

**Diagnostic checks.**

Use one inertial frame per equation. Check momentum in each component. For an
elastic collision, check total kinetic energy in that same frame and check that the
CM momentum is zero after transformation. For a perfectly inelastic collision, the
shared final velocity must equal the initial centre-of-mass velocity. A result that
changes $\vec V_{\rm cm}$ without external impulse has violated momentum
conservation. A result that asserts CM kinetic energy is zero for every collision
has confused the CM frame's zero total momentum with zero relative motion.

## Laboratory and CM transformations

A particle emerging at CM-frame velocity $\vec v_1'$ has laboratory velocity
$\vec v_1=\vec V_{\rm cm}+\vec v_1'$. The
observed laboratory angle depends on both the CM scattering angle and the ratio of
$V_{\rm cm}$ to the CM speed. A forward-moving centre-of-mass frame compresses
laboratory scattering angles toward the incident direction. The transformation is
geometric rather than an additional collision law.

With a stationary target, $\vec V_{\rm cm}$ points along the incoming beam. A
CM particle scattered backward can still travel forward in the laboratory if the
centre-of-mass velocity exceeds the magnitude of its backward CM velocity. This is
common when the target is much heavier than the projectile. Conversely, a light
target can recoil through a large laboratory angle even when its CM deflection is
modest.

**Coefficient of restitution in the CM description.**

A direct one-dimensional collision gives the restitution relation in terms of relative speeds,

$$
e=\frac{|v_{2f}-v_{1f}|}{|v_{1i}-v_{2i}|}.
$$

Relative speed has the same value in every Galilean inertial frame, so $e$ may be
computed in the laboratory or CM frame. In the CM frame, the outgoing relative
speed is $e$ times the incoming relative speed. The kinetic energy in relative
motion therefore changes by

$$
K_f'=e^2K_i'.
$$

The missing fraction $(1-e^2)K_i'$ becomes internal energy for an isolated two-body
system under the one-dimensional collision model. The limits are $e=1$ for elastic
and $e=0$ for perfectly inelastic impact. Restitution does not describe tangential
friction, spin, or a collision that creates additional particles.

> **Worked example.** A $0.50\ \mathrm{kg}$ cart moving at $6.0\ \mathrm{m\,s^{-1}}$
> sticks to a $1.50\ \mathrm{kg}$ cart at rest. The common final speed is the CM
> speed,
>
> $$
> V_{\rm cm}=\frac{(0.50)(6.0)}{2.00}=1.50\ \mathrm{m\,s^{-1}},
> $$
>
> and the initial relative kinetic energy — the most a perfectly inelastic impact
> can convert — is
>
> $$
> K_i'=\frac12\left(\frac{(0.50)(1.50)}{2.00}\right)(6.0)^2=6.75\ \mathrm J.
> $$
>
> Laboratory kinetic energy falls from $9.0\ \mathrm J$ to $2.25\ \mathrm J$, a
> $6.75\ \mathrm J$ loss to deformation and heat, matching $K_i'$. The remaining
> $2.25\ \mathrm J$ is centre-of-mass translation, which the carts' internal
> sticking cannot remove.

**Scope of the two-body approximation.**

The CM construction applies to any number of particles, but the simple equal-and-
opposite momentum geometry is specific to two bodies. Three-body breakup has CM
momenta that form a closed triangle rather than one opposite pair. In particle and
nuclear reactions, newly created or excited products must be included in the final
state before applying conservation. An omitted photon or recoil particle can make a
correct event appear to violate energy or momentum conservation.

## Constructing and interpreting the CM frame

The centre-of-mass frame is an inertial description obtained by subtracting one
constant velocity from every laboratory velocity in the event. The subtraction
changes the coordinate rate while preserving relative position, relative speed,
contact time, and momentum transfer. A laboratory observer may see both particles
moving to the right before a collision, while a CM observer sees them approaching
from opposite directions.

In a one-dimensional collision with a stationary target, the sign of each CM
velocity is especially informative. With the positive axis along the projectile,
the projectile has positive CM velocity and the target has negative CM velocity.
Their speeds need not be equal. The lighter particle has the larger CM speed because
equal and opposite momenta require $m_1|v_1'|=m_2|v_2'|$. Confusing equal momentum
with equal speed is a common source of incorrect scattering sketches.

$$
% caption: Subtracting the common center-of-mass velocity turns two laboratory
% velocities (both rightward here) into oppositely directed center-of-mass
% velocities. The relative velocity — the gap between the two — is unchanged, so the
% collision speed is the same in both frames.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[black] at (0.15,1.25) {lab};
  \draw[->, black, very thick] (0.85,1.5) -- (4.25,1.5) node[right, black] {$v_1$};
  \draw[->, acc, very thick] (0.85,0.95) -- (1.85,0.95) node[right, acc] {$v_2$};
  \draw[dashed, black] (3.05,1.72) -- (3.05,0.68);
  \node[black, above] at (3.05,1.7) {$V_{cm}$};
  \node[black] at (0.15,-0.65) {CM};
  \draw[->, black, very thick] (2.6,-0.4) -- (3.8,-0.4) node[right, black] {$v_1$};
  \draw[->, acc, very thick] (2.6,-0.9) -- (1.4,-0.9) node[left, acc] {$v_2$};
\end{tikzpicture}
$$

The arrows in the CM rows are measured from their own tails, not from the
laboratory origin. Their head-to-head difference is the relative velocity, giving
the Galilean invariance
$\vec v_1-\vec v_2=\vec v_1'-\vec v_2'$. It also gives a direct
calculation check: after subtracting the computed $\vec V_{\rm cm}$, multiply
each CM velocity by its mass and verify that the vector sum is zero. A nonzero sum
usually indicates that the same frame was not used for both velocities or that a
component was omitted.

Compute the centre-of-mass velocity from simultaneous pre-collision velocities and
the masses selected for the event boundary. If a camera or timing system measures the
two objects at different instants, extrapolate them to a common time using the stated
pre-contact model. A transformation based on asynchronous velocities can create false
nonzero CM momentum even when the subsequent collision data are accurate.

**Unequal masses and the CM momentum pair.**

Mass asymmetry changes the velocity geometry without changing the momentum rule.
In an isolated two-body system's CM frame, the momentum vectors have the same
length and point in opposite directions at every instant. The velocities point in
opposite directions as well, but their lengths are inversely proportional to the
masses. A heavy target can appear almost stationary in the CM frame, whereas a light
projectile carries most of the relative motion. The limit approaches the familiar
approximation of a light object bouncing from a massive wall.

The ratio follows directly from the zero-momentum condition:

$$
\frac{|\vec v_1'|}{|\vec v_2'|}=\frac{m_2}{m_1}.
$$

When $m_2=4m_1$, particle one moves four times as fast as particle two in the CM
frame. The momentum magnitudes are nevertheless equal. During an elastic collision,
both speeds remain fixed and only directions rotate. During an inelastic collision,
both CM speeds shrink in the same ratio if the outgoing motion remains collinear;
for perfectly inelastic capture they both become zero.

$$
% caption: In the center-of-mass frame two unequal masses carry equal and opposite
% momenta but unequal speeds: the lighter mass moves faster, with speed ratio
% inversely proportional to mass. These are velocity arrows, so their lengths differ.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0.25,0) -- (6.4,0) node[right] {CM velocity};
  \draw[fill=black!14, draw=black, thick] (3.0,0.4) circle (0.26);
  \draw[fill=acc!12, draw=acc, thick] (4.15,0.4) circle (0.54);
  \draw[->, black, very thick] (2.74,0.4) -- (0.85,0.4);
  \draw[->, acc, very thick] (4.69,0.4) -- (5.35,0.4);
  \node[above, black] at (3.0,0.68) {$m_1$};
  \node[above, acc] at (4.15,0.96) {$m_2$};
\end{tikzpicture}
$$

The figure uses velocity arrows rather than momentum arrows, so unequal lengths are
expected. Replacing the arrows by $m\vec v'$ would make their lengths equal.
Keeping that distinction visible is valuable in reaction problems: a detector may
measure an outgoing speed, while conservation constrains momentum. Conversion
through the corresponding mass is required before vectors are compared. For several
final products, the special opposite-pair construction disappears, but the
mass-weighted vector sum remains zero in the CM frame.

**Laboratory angles as translated CM velocities.**

A laboratory scattering angle is produced by vector addition, not by simply copying
the CM scattering angle. For a given outgoing particle,
$\vec v_{\rm lab}=\vec V_{\rm cm}+\vec v_{\rm CM}$. The CM velocity rotates
during the collision; the translational velocity is the same fixed vector before and
after an isolated event. Graphically, the CM velocity is placed at the head of
$\vec V_{\rm cm}$, and the diagonal from the origin is the observed laboratory
velocity. The resulting direction can be smaller, larger, or even opposite the CM
angle depending on the mass ratio and scattering direction.

A forward CM velocity shifts every laboratory velocity toward the beam direction. A
backward CM trajectory can therefore appear in the forward laboratory hemisphere.
The limiting condition is simple in one dimension: a particle directed
backward in the CM frame still moves forward in the laboratory when
$V_{\rm cm}>|\vec v'|$. A laboratory claim that an object did not reverse
direction therefore does not establish that it did not backscatter in the CM frame.


The transformation must be applied to complete vectors. Adding $V_{\rm cm}$ to a
speed while leaving the direction unchanged is valid only for collinear motion.
In an oblique event, transform the horizontal and vertical components first, then
recover the speed and angle from the transformed pair. This procedure also avoids
mixing an angle measured by a laboratory detector with an energy equation written
for CM velocities.

**Restitution as radial contraction in the CM frame.**

The CM description makes restitution particularly compact in a direct collision.
Incoming CM velocities point along one line and outgoing CM velocities point along
the same line with reversed directions. The outgoing relative speed is $e$ times
the incoming relative speed, so each object's CM speed is multiplied by $e$ when
the masses are unchanged. The momentum pair remains equal and opposite throughout;
only its common magnitude contracts. Elastic impact corresponds to a rotation by
$180^\circ$ with no contraction, while perfectly inelastic capture contracts the
pair to the origin.

Because relative kinetic energy is proportional to the square of the relative
speed, the retained CM kinetic energy is $e^2K_i'$. This result applies to the
normal relative motion represented by the one-dimensional model. A real oblique
impact may retain tangential relative speed, add spin through friction, or have a
coefficient of restitution that depends on speed and material deformation. Applying
one scalar $e$ to an entire two-dimensional velocity vector without a contact-normal
model is generally unjustified.

$$
% caption: In a direct inelastic collision the equal-and-opposite CM momentum pair
% reverses and contracts by the restitution factor $e$. The dashed radius is the
% incoming magnitude; the solid radius is the smaller outgoing magnitude for $0<e<1$.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[dashed, black] (3.1,1.55) circle (2.0);
  \draw[acc!40] (3.1,1.55) circle (1.3);
  \draw[dashed, black, very thick] (3.1,1.55) -- (5.1,1.55);
  \draw[dashed, black, very thick] (3.1,1.55) -- (1.1,1.55);
  \draw[->, acc, very thick] (3.1,1.55) -- (4.4,1.55);
  \draw[->, acc, very thick] (3.1,1.55) -- (1.8,1.55);
  \fill[black] (3.1,1.55) circle (2pt);
  \node[black, above] at (4.85,1.6) {incoming};
  \node[acc, below] at (4.4,1.5) {outgoing};
\end{tikzpicture}
$$

The circles in the figure are momentum-space guides rather than physical paths. A
smaller final radius represents lower relative kinetic energy, not a smaller system
momentum; total CM momentum remains zero before and after. The laboratory final
velocities still include the unchanged translation $\vec V_{\rm cm}$. Thus a
strongly inelastic collision can leave both bodies moving rapidly in the laboratory
while their relative motion, and therefore their CM kinetic energy, has been nearly
eliminated.


Mass limits provide quantitative checks on any transformed result. When the target mass
greatly exceeds the projectile mass, the centre-of-mass velocity lies close to the
target's laboratory velocity. The projectile then has nearly the same speed in the
CM and laboratory descriptions, while the heavy target's CM motion is small. When
the projectile is much heavier, the centre of mass instead follows the projectile,
and a light target can receive a large laboratory deflection. Neither limit changes
the conservation laws; it changes how their vector geometry appears to a laboratory
observer.

A frame transformation should be reversed as a final arithmetic check. Starting
from reported CM velocities, add the same $\vec V_{\rm cm}$ to every product,
reconstruct laboratory momentum and energy, then subtract it again. The original
CM velocities and zero total CM momentum must return. This round trip detects a
misapplied sign, a velocity translated twice, or an angle measured in the wrong
frame before those errors propagate into a scattering interpretation.

## Relative motion and laboratory solutions

The reduced mass is the inertia associated with the separation coordinate between
two particles. With total mass
$M=m_1+m_2$, centre-of-mass position $\vec R$, and separation
$\vec r=\vec r_1-\vec r_2$, the kinetic energy separates exactly into
translation of the whole pair and motion of the separation:

$$
K=\frac12M V_{\rm cm}^2+\frac12\mu u^2,
\qquad
\mu=\frac{m_1m_2}{M}.
$$

The relative speed $u$ is the rate at which the particles approach or separate.
The corresponding CM momentum magnitude is $p'=\mu u$. Each particle has this same
momentum magnitude but generally a different CM speed. A light particle therefore
moves farther from the CM velocity origin than a heavy particle, even though their
momenta are equal and opposite.

The mass limits provide immediate checks. If $m_2\gg m_1$, then
$\mu\approx m_1$ and the light particle carries almost all the relative motion.
If the masses are equal, $\mu=m/2$. A calculation that uses the total mass in
$\tfrac12\mu u^2$ overestimates the energy available for deformation or a reaction.
The total mass governs translation; reduced mass governs relative motion.


The geometric centre is not usually the centre of mass. Its location is fixed by
$m_1\vec r_1'+m_2\vec r_2'=\vec 0$, so placing it halfway between unequal
masses produces incorrect CM velocities. This matters in molecular, nuclear, and
ordinary collision calculations alike. A detector may report speeds, but
conservation laws compare mass-weighted velocity vectors.

**Elastic collision solutions from CM reversal.**

A one-dimensional elastic collision has a particularly short CM solution. Zero total
CM momentum fixes the two momenta as an opposite pair, and kinetic-energy
conservation fixes their magnitudes. The only collinear elastic outcome is reversal:

$$
v_{1f}'=-v_{1i}',\qquad v_{2f}'=-v_{2i}'.
$$

Adding the unchanged $\vec V_{\rm cm}$ to both final CM velocities produces

$$
v_{1f}=\frac{(m_1-m_2)v_{1i}+2m_2v_{2i}}{m_1+m_2},
\qquad
v_{2f}=\frac{2m_1v_{1i}+(m_2-m_1)v_{2i}}{m_1+m_2}.
$$

These expressions allow either object to be initially moving. They are not general
two-dimensional formulas: an oblique collision needs vector components and a
contact-direction model. Their derivation also shows why the CM frame is efficient;
there is no quadratic root to select after energy and momentum equations are
combined.

$$
% caption: In a direct elastic collision the two center-of-mass velocities reverse
% direction while keeping their magnitudes (the longer arrow is the lighter, faster
% particle). Restoring the center-of-mass translation gives the laboratory velocities.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0.25,0) -- (6.6,0) node[right] {CM velocity};
  \node[left, black] at (0.7,1.35) {before};
  \draw[->, black, very thick] (3.2,1.35) -- (1.1,1.35);
  \draw[->, acc, very thick] (3.2,1.35) -- (3.9,1.35);
  \node[left, black] at (0.7,0.45) {after};
  \draw[->, black, very thick] (3.2,0.45) -- (5.3,0.45);
  \draw[->, acc, very thick] (3.2,0.45) -- (2.5,0.45);
  \fill[black] (3.2,1.35) circle (1.3pt);
  \fill[black] (3.2,0.45) circle (1.3pt);
\end{tikzpicture}
$$

Against a stationary target, a light incident object usually rebounds from a much
heavier target because the factor $(m_1-m_2)/(m_1+m_2)$ is negative. A heavy
incident object normally continues forward and transfers a large speed to a light
target. Equal masses exchange velocities in the ideal direct elastic limit. These
follow from one CM reversal rule and its mass-ratio limits.

**Recovering laboratory directions from CM scattering.**

Let the incoming beam define positive $x$, so
$\vec V_{\rm cm}=V_{\rm cm}\hat\imath$. A product with CM speed $v'$ at
CM angle $\theta'$ has laboratory components

$$
v_x=V_{\rm cm}+v'\cos\theta',
\qquad
v_y=v'\sin\theta'.
$$

Therefore

$$
\tan\theta_{\rm lab}=
\frac{v'\sin\theta'}{V_{\rm cm}+v'\cos\theta'}.
$$

The denominator determines the laboratory quadrant. A principal-value inverse
tangent without a sign check can assign a backward trajectory to the wrong side of
the beam. Component reconstruction, followed by a quadrant-aware angle calculation,
avoids that error.

$$
% caption: Final laboratory velocities are the fixed center-of-mass translation plus
% the rotated CM velocity, so their endpoints lie on a circle centered at the tip of
% $\vec V_{\rm cm}$. Equal angular steps in the CM frame map to unequal laboratory
% angles — the forward-focusing effect.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.2,0) -- (6.4,0) node[right] {$x$};
  \draw[->, black] (0,-0.35) -- (0,3.5) node[above] {$y$};
  \draw[acc!45, thick] (2.35,0) circle (1.42);
  \draw[->, black, very thick] (0,0) -- (2.35,0);
  \node[black] at (1.15,-0.35) {$V_{cm}$};
  \foreach \a in {40,90,140} {
    \draw[->, acc, very thick] (2.35,0) -- ($(2.35,0)+(\a:1.42)$);
  }
  \foreach \a in {40,90,140} {
    \draw[dashed, black] (0,0) -- ($(2.35,0)+(\a:1.42)$);
  }
\end{tikzpicture}
$$

The endpoints lie on a circle centred at the tip of $\vec V_{\rm cm}$, not on a
circle centred at the laboratory origin. This shifted circle explains forward
focusing: when the translational velocity is appreciable, many CM directions can
map to a narrow forward laboratory range. Adding $V_{\rm cm}$ to a speed while
leaving its direction unchanged is valid only for collinear motion; oblique
transformations require the complete vector.

**Energy thresholds in the CM description.**

The CM frame isolates energy available for a reaction or excitation from unavoidable
translation. If an internal channel requires energy $Q>0$, its nonrelativistic
threshold is

$$
K_i'=\frac12\mu u_i^2\ge Q.
$$

At threshold, final products have zero relative kinetic energy in the CM frame.
They may still all move together in the laboratory at $\vec V_{\rm cm}$, carrying
kinetic energy that cannot be converted into the internal requirement. A projectile
incident on a stationary target consequently needs more laboratory kinetic energy
than $Q$ alone.

For projectile mass $m_1$ and stationary target mass $m_2$,

$$
K_{{\rm lab},\,{\rm th}}=Q\frac{m_1+m_2}{m_2}.
$$

The result follows by equating the initial relative energy
$\tfrac12\mu v_i^2$ to $Q$ and comparing it with
$K_{\rm lab}=\tfrac12m_1v_i^2$. The multiplier exceeds one for every finite target
mass and approaches one only for an extremely massive target. It assumes
nonrelativistic kinematics and masses that are sufficiently unchanged for the
model's precision.

$$
% caption: At threshold, laboratory kinetic energy splits into center-of-mass
% translation, which total momentum forces the products to keep, and the relative
% energy that meets the internal requirement $Q$. The translation cannot be converted.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.5,0) node[right] {energy};
  \draw[fill=black!16, draw=black] (0.55,0.45) rectangle (2.25,1.18);
  \draw[fill=acc!12, draw=acc] (2.25,0.45) rectangle (5.6,1.18);
  \draw[fill=black!16, draw=black] (0.55,2.0) rectangle (2.25,2.73);
  \draw[fill=acc!12, draw=acc] (2.25,2.0) rectangle (5.6,2.73);
  \node[left, black] at (0.45,0.82) {lab input};
  \node[left, black] at (0.45,2.37) {threshold};
  \node[align=center] at (1.4,0.82) {c.m.};
  \node[align=center, acc] at (3.9,0.82) {relative energy};
  \node[align=center] at (1.4,2.37) {c.m.};
  \node[align=center, acc] at (3.9,2.37) {internal need};
\end{tikzpicture}
$$

Above threshold, excess CM energy can become final relative kinetic energy as well
as internal excitation. At threshold itself, zero final relative speed is the
least-energy configuration compatible with total momentum. A laboratory calculation
that transfers all incident kinetic energy to internal excitation violates this
constraint. The CM construction separates usable relative energy from the
translation that must accompany the products in every inertial laboratory frame.


> **Worked example.** A $1.0\ \mathrm{kg}$ cart moving at $4.0\ \mathrm{m\,s^{-1}}$
> strikes a $3.0\ \mathrm{kg}$ cart at rest in a direct elastic collision. The
> centre-of-mass speed is $V_{\rm cm}=1.0\ \mathrm{m\,s^{-1}}$, so the initial CM
> velocities are $+3.0\ \mathrm{m\,s^{-1}}$ for the light cart and
> $-1.0\ \mathrm{m\,s^{-1}}$ for the heavy one, with momenta $+3.0$ and
> $-3.0\ \mathrm{kg\,m\,s^{-1}}$.
>
> Elastic reversal sends the CM velocities to $-3.0$ and $+1.0\ \mathrm{m\,s^{-1}}$.
> Adding the $1.0\ \mathrm{m\,s^{-1}}$ translation returns laboratory velocities
> $-2.0$ and $+2.0\ \mathrm{m\,s^{-1}}$: the light cart rebounds, the heavy cart
> moves forward, and laboratory momentum stays $4.0\ \mathrm{kg\,m\,s^{-1}}$.
> Laboratory kinetic energy is $8.0\ \mathrm J$ before and
> $\tfrac12(1.0)(2.0)^2+\tfrac12(3.0)(2.0)^2=8.0\ \mathrm J$ after.

Both CM velocities reverse because the collision is elastic and collinear; only the
lighter cart reverses in the laboratory, because the positive translation is added
to both. The two frames describe one event through one invariant centre-of-mass
velocity, not two separate momentum calculations.

> **Worked example.** A $1.0\ \mathrm{kg}$ projectile strikes a stationary
> $4.0\ \mathrm{kg}$ target and opens an internal channel requiring
> $Q=3.0\ \mathrm J$. The nonrelativistic threshold laboratory kinetic energy is
>
> $$
> K_{{\rm lab},\,{\rm th}}
> =Q\,\frac{m_1+m_2}{m_2}
> =(3.0)\frac{1.0+4.0}{4.0}
> =3.75\ \mathrm J.
> $$
>
> Only $3.0\ \mathrm J$ reaches the channel as CM relative kinetic energy; the
> other $0.75\ \mathrm J$ is centre-of-mass translation the products must carry.
> Quoting $3.0\ \mathrm J$ as the laboratory threshold balances energy only by
> discarding that required momentum.

The calculation is deliberately nonrelativistic. For nuclear or particle reactions,
rest-energy changes and relativistic momentum-energy relations are required once
speeds or mass changes make the approximation inaccurate. The laboratory must still
supply both the internal threshold energy and the translational motion required by
total momentum. The CM frame identifies the part of the incoming energy available to
the new channel.

Three independent questions remain in any collision report. Momentum conservation
determines the unchanging centre-of-mass velocity when
external impulse is negligible. The scattering law determines how the relative
velocity changes: reversal for a direct elastic collision, contraction for a
partially inelastic collision, or a specified angular redistribution for elastic
scattering. Energy accounting then determines whether internal excitation or a new
reaction channel is possible. Keeping these roles separate prevents an energy
threshold from being mistaken for a momentum equation, or a measured laboratory
angle from being treated as a CM scattering angle without transformation.

## Advanced collision reconstruction

The centre-of-mass construction still organizes collisions beyond the symmetric
equal-mass example. Unequal masses change the speeds assigned to the two
CM momentum vectors, alter the laboratory angular map, and restrict which
laboratory outcomes are physically accessible. The invariant starting point remains
the same: for negligible external impulse, the total CM momentum is zero. Any
accepted solution must retain that condition after the final velocities have been
transformed, regardless of whether the interaction is elastic, inelastic, or opens
an internal channel.

A complete collision statement therefore needs more than a pair of laboratory
angles. It needs the masses, an initial velocity pair in one inertial frame, a
description of the interaction, and enough measured final information to close the
unknowns. In an elastic two-body event, energy conservation adds one scalar
condition in addition to vector momentum conservation. In a dissipative event, a
normal restitution measurement, an internal-energy measurement, or a specified
final connection such as sticking gives the additional information. The CM frame
organizes these conditions by separating the fixed translation from the part of
motion the interaction can change.

**Unequal-mass elastic scattering.**

With a stationary target, the incident particle and target have different CM speeds
unless their masses are equal. The initial CM momentum magnitude is
$p'=\mu v_i$. An elastic collision preserves $p'$ but can rotate the momentum pair.
The projectile CM speed is $p'/m_1$ and the target CM speed is $p'/m_2$. A heavier
body therefore traces a smaller velocity circle in the CM frame. After the common
translation $\vec V_{\rm cm}$ is added, those CM circles become laboratory
velocity circles whose centres are displaced along the beam direction.

The velocity-circle geometry determines several mass-ratio limits. A very heavy target has a small CM
velocity and a small CM speed, so the incident light particle can scatter through
large laboratory angles or rebound. A very light target has a large CM speed and
the centre of mass moves close to the projectile speed. The projectile's laboratory
direction is then strongly forward focused, while the light target can emerge at a
large angle. These qualitative statements follow from momentum and energy
conservation; they are not assumptions about which body “wins” the collision.

The CM velocity circles are velocity loci, not momentum loci. Their radii
differ inversely with mass, whereas the corresponding momentum arrows would have
equal lengths. For an ideal elastic event, every permitted CM scattering direction
lies somewhere on the appropriate circle. A laboratory detector samples the shifted
endpoint after addition of $\vec V_{\rm cm}$. Equal CM momenta therefore do not
justify an equal-radius velocity construction for unequal masses.

A numerical mass-ratio check is often enough to expose an impossible result. If a
$1.0\ \mathrm{kg}$ projectile meets a stationary $9.0\ \mathrm{kg}$ target, then
$V_{\rm cm}=0.10v_i$. The projectile's initial CM speed is $0.90v_i$ and the
target's is $0.10v_i$. Their momenta nevertheless have equal magnitude
$0.90\ \mathrm{kg}\,v_i$. Any proposed final elastic solution with unequal CM
momentum magnitudes, or with a changed CM speed, has violated a conservation law
before its laboratory angles are considered.

**Scattering angles and solution branches.**

Momentum components and energy conservation can permit more than one geometric
branch. A measured projectile laboratory angle alone may not identify a unique
target angle or speed when masses are unequal. The ambiguity is physical: distinct
CM scattering directions can project to related laboratory directions after the
velocity translation. A detector's energy measurement, time of flight, coincidence
measurement of the second particle, or a known interaction geometry selects the
branch.

An incident beam along $x$ obeys the laboratory-frame conservation equations

$$
m_1v_i=m_1v_{1f}\cos\theta_1+m_2v_{2f}\cos\theta_2,
\qquad
0=m_1v_{1f}\sin\theta_1+m_2v_{2f}\sin\theta_2.
$$

The second equation is not optional. The target's transverse momentum cancels the
projectile's transverse momentum, even when the detector lies in a single
horizontal plane. A nonzero out-of-plane component introduces a third momentum
component and cannot be hidden inside a two-dimensional sketch. Experimental
collisions are therefore reconstructed in three dimensions whenever the apparatus
cannot physically constrain all tracks to one plane.

A head-to-tail momentum construction does not trace particle paths. Drawing the
first final momentum from the origin and the second from its head, their resultant
reaches the initial momentum endpoint. Reversing the second vector and drawing both
from the origin gives the more familiar two-ray scattering picture, but the
head-to-tail form makes component cancellation easier to audit. A valid construction
must use momenta, not raw speeds, because unequal masses scale the vectors
differently.

At a limiting laboratory angle, the transformed velocity construction can become
tangent to a permitted locus. Beyond that angle no elastic two-body solution exists
for the stated masses and incident speed. This is a physical kinematic restriction,
not a detector failure. Reported angles should therefore be checked against momentum
and energy before a fit is interpreted as a new interaction effect.

**Restitution limits and tangential motion.**

The coefficient of restitution is a statement about the relative velocity along a
contact normal. For bodies touching at a well-defined contact point, let
$\hat n$ point from body one toward body two at impact. The normal relative
components obey

$$
(\vec v_{2f}-\vec v_{1f})\cdot\hat n
=-e(\vec v_{2i}-\vec v_{1i})\cdot\hat n.
$$

For passive direct impacts, $0\le e\le1$. The endpoints represent perfectly
inelastic capture and ideal elastic reversal along the normal. Values larger than
one are possible only when stored energy, a spring release, chemical energy, or an
explosion increases the outgoing relative kinetic energy; they are then called
superelastic. Restitution is not an energy-conservation law and it does not specify
the tangential outcome.

For smooth spheres, tangential contact impulse can be negligible, so tangential
relative velocity is approximately unchanged while the normal component contracts
and reverses. Rough surfaces transfer tangential impulse, produce spin, and couple
translation to rotation. A scalar restitution coefficient then leaves additional
unknowns. The CM frame still carries zero total momentum, but the translational
relative kinetic energy need not account for all energy stored in the final motion.

$$
% caption: The coefficient of restitution acts only on the normal (approach)
% component of relative velocity, which reverses and shortens for a passive inelastic
% impact. Tangential motion needs a separate contact model, so no tangential arrow is
% drawn.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[dashed, black] (3.3,-1.5) -- (3.3,2.0) node[above, black] {contact normal};
  \draw[fill=black!14, draw=black, thick] (2.25,0.55) circle (0.45);
  \draw[fill=acc!12, draw=acc, thick] (4.35,0.55) circle (0.45);
  \draw[->, black, thick] (1.9,1.35) -- (2.9,1.35);
  \draw[->, acc, thick] (4.7,1.35) -- (3.7,1.35);
  \node[black, above] at (2.9,1.4) {approach};
  \draw[->, black, thick] (2.25,-0.5) -- (1.55,-0.5);
  \draw[->, acc, thick] (4.35,-0.5) -- (5.05,-0.5);
  \node[black, below] at (1.55,-0.55) {separation};
\end{tikzpicture}
$$

The two upper arrows represent normal approach and the two lower arrows represent
normal separation. Their unequal lengths illustrate a relative speed reduction.
No tangential arrow is drawn because a single coefficient does not determine it.
When a problem states a smooth contact, zero tangential impulse is an added model
assumption; when it states roughness or rolling, angular momentum and rotational
kinetic energy enter the analysis.

**Endothermic channels and experimental reconstruction.**

An endothermic collision uses part of the initial CM relative kinetic energy to
create internal excitation, break a bound state, or form products with greater rest
energy. In nonrelativistic notation, the energy balance is

$$
K_i'+E_{{\rm int},i}=K_f'+E_{{\rm int},f}.
$$

The threshold discussed earlier is the smallest initial $K_i'$ for which the final
internal energy can be reached with $K_f'=0$. Above threshold, the excess may be
divided between product relative motion and additional internal excitation. An
exothermic event has the opposite sign: released internal energy increases the
final relative kinetic energy. Momentum conservation still fixes the CM velocity;
the energy release changes the radii of the final CM velocity or momentum loci.

Vector reconstruction retains more information than rounded speeds and angles. For
every detected product, form
$\vec p_i=m_i\vec v_i$ in one frame. Sum the final momenta and compare them
with the initial total. Transform the same products to the computed CM frame and
check that the mass-weighted velocity sum is zero. For an elastic hypothesis,
compare total kinetic energies in one frame; for an endothermic hypothesis, compare
the energy difference with the known internal requirement and its uncertainty.

$$
% caption: Experimental reconstruction tests whether the measured product momenta
% close the momentum polygon. The dashed residual is the gap to the initial momentum;
% a complete isolated reconstruction closes it to within measurement uncertainty.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.25,0) -- (6.3,0) node[right] {$x$};
  \draw[->, black] (0,-1.4) -- (0,3.1) node[above] {$y$};
  \draw[->, black, very thick] (0,0) -- (4.9,0.55);
  \node[black] at (2.05,-0.35) {initial};
  \draw[->, acc, very thick] (0,0) -- (2.35,2.1) node[above left] {product one};
  \draw[->, acc!70!black, very thick] (2.35,2.1) -- (4.55,0.35) node[midway, right] {product two};
  \draw[->, dashed, black, thick] (4.55,0.35) -- (4.9,0.55);
  \node[black, right] at (4.9,0.55) {residual};
  \fill[white] (4.9,0.55) circle (1.8pt);
  \draw[black, thick] (4.9,0.55) circle (1.8pt);
\end{tikzpicture}
$$

The dashed gap in the figure is a diagnostic, not a new force. If it is comparable
with the propagated measurement uncertainty, the event is consistent with closure.
If it is much larger, the reconstruction is incomplete. A neutral product, a
photon, undetected recoil, background impulse, or a mistaken particle assignment can
supply the missing momentum. The direction of the residual helps identify which
component of the apparatus or model needs examination.

The residual should be retained as a vector with a stated uncertainty rather than
compressed into one scalar closure number. Its direction distinguishes a missing
longitudinal carrier from an angular, timing, or calibration error.

| Residual pattern | Candidate source | Follow-up measurement |
| --- | --- | --- |
| Along the beam | unobserved recoil or energy-loss channel | downstream momentum acceptance |
| Transverse to the beam | angle or detector-position error | alignment and angular calibration |
| Common scale across products | velocity or timing calibration | reference event or clock comparison |

Experimental resolution also limits conclusions about restitution and thresholds.
A small measured kinetic-energy decrease can be indistinguishable from calibration
error, while a large momentum residual can dominate a fitted scattering angle.
Masses, timing, detector position, and energy calibration belong in the final
residual as stated uncertainty sources. A transparent report
states the frame, measured quantities, reconstructed CM velocity, momentum residual,
energy comparison, and the interaction model used to interpret the event.

The CM frame isolates the relative motion that internal forces can alter and gives
sharp consistency checks. Unequal masses, nonzero scattering angles, partial
restitution, and threshold channels use the same structure once the system boundary
and frame are fixed. Measured constraints make the collision solution testable.


A reconstruction should also distinguish a genuine threshold from an instrumental
cutoff. Near a reaction threshold, a finite detector threshold can remove slow
products from the record and imitate missing momentum or missing energy. Simulated
acceptance, calibration sources, and coincidence triggers establish which regions
of final-state momentum space the apparatus can observe. Without that information,
the absence of a reconstructed branch is not by itself evidence that the underlying
collision channel is forbidden.

Frame choice must remain consistent in the uncertainty analysis. An angular error in
the laboratory becomes a correlated error in CM speed and angle after the velocity
translation. A small laboratory momentum residual may grow after a transformation
if the centre-of-mass velocity was inferred from a poorly measured particle.
Conversely, a large laboratory energy change can be mostly unchanged centre-of-mass
translation and carry little information about the interaction. Reporting both
laboratory and CM residuals makes these distinctions visible.

Propagate timing, position, mass, and calibration uncertainties through the same
transformation used for the central velocities. A shared timing offset can correlate
the two measured laboratory velocities, while a common velocity-scale error can move
the inferred centre-of-mass velocity for every product. Preserve those correlations in
the residual test; treating all fitted angles and speeds as independent can make a
misreconstructed event appear more precise than the measurements support.

A closed two-body event has a definite audit order. Sum initial and
final momentum vectors in the laboratory frame; compute the centre-of-mass velocity
from the initial system; transform every measured product; test the zero CM momentum
condition; then apply the elastic, restitution, or threshold relation appropriate to
the stated model. A failed check identifies an assumption for revision: system
boundary, external impulse, event assignment, dimensionality, or collision law.

Reporting the laboratory and CM descriptions as two columns keeps the frame
transformation visible as a calculation step for the same physical system.

| Quantity | Laboratory representation | CM check |
| --- | --- | --- |
| Total momentum | $\sum_i m_i\vec v_i$ | determines $\vec V_{\rm cm}$ |
| Product velocity | measured $\vec v_i$ | $\vec v_i'=\vec v_i-\vec V_{\rm cm}$ |
| Momentum closure | initial versus final vector sum | $\sum_i m_i\vec v_i'=\vec 0$ |
| Energy claim | laboratory kinetic energies | relative energy, restitution, or threshold condition |

[^tipler84]: Tipler and Mosca, _Physics for Scientists and Engineers_, 6th ed., §8-4.