---
title: Multiparticle Work
module: Energy
moduleNumber: 3
lessonNumber: 3
order: 303
summary: >
  A single particle has one velocity and one kinetic energy; a system of many can
  spin, deform, explode, and warm up while its centre of mass glides along as if
  nothing happened. Splitting the motion into a centre-of-mass part and an
  internal part separates the energy that momentum already fixes from the energy
  left free for relative motion, $K=\tfrac12MV_{\rm cm}^2+K'$. We derive the
  centre-of-mass work theorem, see why an explosion or a released spring can raise
  total kinetic energy with no external work at all, and use the reduced-mass and
  centre-of-mass frames to make collisions and internal transfers clean.
topics: [Energy]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 6 — Work and Kinetic Energy; §6-5"
---

## System kinetic energy and external work

For particles of total mass $M$, write each velocity as a centre-of-mass part plus
a relative part,

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

The total kinetic energy separates as

$$
K=\sum_i\frac12m_iv_i^2
=\frac12MV_{\rm cm}^2+\sum_i\frac12m_i{v_i'}^2
=K_{\rm cm}+K'.
$$

The cross term vanishes because $\sum_i m_i\vec v_i'=0$. The first term is
translation of the system centre of mass. The second is kinetic energy relative to
the centre-of-mass frame: rotation, vibration, and other internal motion. A rigid
body translating without rotation has $K'=0$; a spinning body at rest at its centre
of mass has $K_{\rm cm}=0$ but can have large $K'$.

External force determines centre-of-mass motion:

$$
M\vec A_{\rm cm}=\sum\vec F_{\rm ext}.
$$

The work associated with the net external force and centre-of-mass displacement is

$$
W_{\rm cm}=\int\left(\sum\vec F_{\rm ext}\right)
\cdot\d\vec R_{\rm cm}=\Delta K_{\rm cm}.
$$

Centre-of-mass work measures translation only. Total external work uses the
application-point displacement of every force and can change internal kinetic energy
as well as centre-of-mass motion.

$$
% caption: External resultant force changes center-of-mass translation, while
% forces acting at different points can also create internal rotation or vibration.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[thick] (0,0) ellipse (1.3 and 0.75);
  \draw[fill=white, draw=black, thick] (0,0) circle (2pt);
  \node[anchor=north west, black] at (0.08,-0.06) {CM};
  \draw[->, acc, very thick] (0,0) -- (1.75,0) node[right] {$\mathbf V_{\rm cm}$};
  \draw[->, black, thick] (0.45,0.60) -- (1.55,1.45) node[above right] {$\mathbf F$};
  \draw[->, acc, thick] (-1.55,-0.5) to[bend left=45] (-1.30,0.55) node[above left] {rotation};
\end{tikzpicture}
$$

**Work of individual external forces.**

The total work of external forces is the sum of work at their individual points of
application,

$$
W_{\rm ext}=\sum_a\int \vec F_a^{\rm ext}\cdot\d\vec r_a.
$$

Application points generally do not share the centre-of-mass displacement. A force on a rotating
wheel can do work even if the centre of mass is fixed; a force through the centre
of mass can translate the body without changing its rotation. For a rigid body,
the displacement of a point can be separated into translational and rotational
parts. The external work then contains a centre-of-mass contribution and a torque
contribution,

$$
P_{\rm ext}=\left(\sum\vec F_{\rm ext}\right)\cdot\vec V_{\rm cm}
+\boldsymbol\tau_{\rm cm}\cdot\boldsymbol\omega.
$$

Net force and net torque govern distinct power terms. A pure couple has zero
resultant force but nonzero torque and can increase rotational kinetic energy. A
force whose line of action passes through the centre of mass can have zero torque
but nonzero translational power.

Force and velocity must be read at the same instant and material point. An off-centre
actuator, a rotating shaft, and a translating support can share a force magnitude yet
deliver different power, because their application-point velocities differ. Force
magnitude alone does not fix the energy transfer without the matching kinematics.

## Internal work, rolling, and center-of-mass scope

A pair of particles has equal-and-opposite internal forces, but their works need not
cancel because the particles generally have different displacements. A stretched
spring pulls both ends; its forces sum to zero, yet its potential energy can become
kinetic energy of the attached masses. Internal force cancellation governs the
momentum sum; work requires the individual application-point displacements.

For conservative internal interactions, potential energy accounts for the work:

$$
W_{\rm int}=-\Delta U.
$$

For dissipative interactions, internal work can raise thermal energy or permanent
deformation. A complete system-energy statement has the form

$$
\Delta K_{\rm cm}+\Delta K'+\Delta U+\Delta E_{\rm th}=W_{\rm ext}.
$$

The division into internal and external terms depends on the system boundary. For a
crate-only system, a person's push is external work. For a person-crate system, the
contact force is internal, while chemical energy in the person decreases.

$$
% caption: A spring transfers internal potential energy into relative kinetic
% energy. The center of mass remains fixed when no external force acts.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[thick] (0,0) rectangle (0.9,0.85);
  \draw[thick] (4.5,0) rectangle (5.4,0.85);
  \draw[black, thick] (0.9,0.42) -- (1.25,0.42)
    -- (1.45,0.66) -- (1.70,0.18) -- (1.95,0.66) -- (2.20,0.18)
    -- (2.45,0.66) -- (2.70,0.18) -- (2.95,0.66) -- (3.20,0.18)
    -- (3.45,0.66) -- (3.70,0.18) -- (3.90,0.42) -- (4.5,0.42);
  \draw[dashed, black] (2.7,-0.55) -- (2.7,1.20);
  \draw[fill=acc!12, draw=acc, thick] (2.7,0.42) circle (2.2pt);
  \node[anchor=north, black] at (2.7,-0.55) {CM held};
  \draw[->, black, thick] (0.45,1.20) -- (-0.30,1.20);
  \draw[->, black, thick] (4.95,1.20) -- (5.70,1.20);
\end{tikzpicture}
$$

> **Worked example (explosion at rest).** Two fragments of masses $m$ and $2m$
> separate from a package initially at rest. With negligible external impulse, their
> momenta are opposite:
> $$
> mv_1+2mv_2=0,
> \qquad v_1=-2v_2.
> $$
> The centre of mass stays at rest, so $K_{\rm cm}=0$ before and after. An internal
> chemical release $E$ becomes $K'$ after the explosion. The lighter fragment carries
> the greater speed and kinetic energy, since equal momentum is not equal kinetic
> energy. No external work drives the rise in total kinetic energy; internal chemical
> energy supplies it.

**Rolling and contact work.**

In ideal rolling on a fixed surface, static friction does no work on the rolling body:
the contact point is instantaneously at rest. It still supplies torque and changes
angular speed. Kinetic friction does work because the contact point slides, and a
moving conveyor or accelerating support shifts the contact-point displacement so that
even static friction can transfer energy.

**Derivation of centre-of-mass work.**

From $\vec R=M^{-1}\sum_i m_i\vec r_i$, two differentiations give

$$
M\vec A=\sum_i m_i\vec a_i.
$$

Newton's second law expresses each particle acceleration through external and
internal forces. Equal-and-opposite internal pairs cancel in the sum, leaving

$$
M\vec A=\sum\vec F_{\rm ext}.
$$

Dot multiplication by $\d\vec R=\vec V\d t$ gives

$$
\int\sum\vec F_{\rm ext}\cdot\d\vec R
=\int M\vec A\cdot\vec V\,\d t
=\Delta\left(\frac12MV^2\right).
$$

Hence $W_{\rm cm}=\Delta K_{\rm cm}$. The theorem uses center-of-mass
displacement, not the displacement of an arbitrary material point. That distinction
is what excludes rotational and deformational energy from $K_{\rm cm}$.

**Total particle work.**

Applying the work-energy theorem to every particle gives

$$
\Delta(K_{\rm cm}+K')=W_{\rm ext}+W_{\rm int}.
$$

The external work is evaluated at each force's point of application. For
conservative internal forces, $W_{\rm int}=-\Delta U$. For dissipative internal
forces, it can increase thermal energy and permanent deformation. The recoverable
system-energy form is

$$
\Delta K_{\rm cm}+\Delta K'+\Delta U+\Delta E_{\rm th}=W_{\rm ext}.
$$

Internal forces cancelling in the net-force equation does not require their work
to cancel. The two ends of a spring have different displacements, so spring forces
can convert elastic energy into relative kinetic energy while producing no
centre-of-mass acceleration.

## Deformation, rotation, and accelerating frames

Two carts connected by a compressed spring provide a direct example. The system is
initially at rest and has no external force. After release, the carts have opposite
momenta, so $K_{\rm cm}=0$ remains unchanged. The spring potential energy decreases
and relative kinetic energy increases:

$$
\Delta K'= -\Delta U_s.
$$

An inelastic collision reverses this energy path. The centre-of-mass velocity is
unchanged when external impulse is negligible, but relative kinetic energy becomes
internal deformation and thermal energy. Centre-of-mass work alone cannot predict
that heating because it contains no relative-motion term.

A rigid body has external power with translational and rotational pieces,

$$
P_{\rm ext}=\vec F_{\rm net}\cdot\vec V_{\rm cm}
+\boldsymbol\tau_{\rm cm}\cdot\boldsymbol\omega.
$$

The first term changes centre-of-mass translation and the second changes rotation.
A pure couple has zero net force but nonzero torque and can supply rotational power.
A force through the centre of mass can supply translational power with zero torque.

> **Worked example (off-centre force).** A force applied tangentially to the rim of a
> disk both translates and rotates it. Its application point moves differently from the
> centre of mass, so the work there equals the increase in translational plus
> rotational kinetic energy. Using only $F\Delta R_{\rm cm}$ omits the rotational
> contribution and holds only when the force has no torque about the centre of mass.

## Checks and boundaries

All energy terms have units of joules. An isolated system has $W_{\rm cm}=0$ and
constant centre-of-mass speed, but it can have changing total kinetic energy.
System boundaries determine whether a force is external or internal. A rope force
is external to a block, internal to a block-rope system, and changes category again
when a person pulling the rope is included. The force itself is unchanged; only the
energy accounting changes.

**Translational and rotational work for rigid bodies.**

The velocity of point $a$ on a rigid body is

$$
\vec v_a=\vec V_{\rm cm}+\boldsymbol\omega\times\vec r_a',
$$

where $\vec r_a'$ runs from the centre of mass to the point. Substitution into
the sum of force powers gives

$$
\sum_a\vec F_a\cdot\vec v_a
=\left(\sum_a\vec F_a\right)\cdot\vec V_{\rm cm}
+\left(\sum_a\vec r_a'\times\vec F_a\right)\cdot\boldsymbol\omega.
$$

The second parenthesis is torque about the centre of mass. The common formula $P=Fv$
holds only at the particular point where the force acts. When a body rotates, its
points have different velocities, so a force may do positive work at one point while
the centre of mass has no displacement along it. Power accounting begins with the
point of application.

| Description | Power expression | Required velocity |
| --- | --- | --- |
| One applied force | $P_a=\vec F_a\cdot\vec v_a$ | velocity at point $a$ |
| Net translation | $\vec F_{\rm net}\cdot\vec V_{\rm cm}$ | centre-of-mass velocity |
| Rotation about CM | $\boldsymbol\tau_{\rm cm}\cdot\boldsymbol\omega$ | angular velocity and moment arm |

An ideal hinge is a limiting case. A force at a fixed hinge point does
zero work because that point has zero displacement, yet other forces can rotate the
body and change its kinetic energy. The hinge force can be large without supplying
energy. A motor shaft is different: it exerts torque through an angle and supplies
work $W=\int\tau\,\d\theta$.

**Work in accelerating frames.**

The centre-of-mass theorem is most direct in an inertial frame. In a uniformly
accelerating frame, an inertial force $-M\vec a_{\rm frame}$ is included in the
effective external-force sum. The centre of mass then obeys the frame's component
equation. Apparent-work terms can be represented by an effective potential only
under restricted conditions. Mixing a centre-of-mass velocity measured in one frame
with force work measured in another produces an inconsistent energy calculation.

**Detailed deformable-body example.**

A bullet embeds in a suspended block. During the short collision, external impulse
is negligible in the horizontal direction, so horizontal centre-of-mass velocity is
approximately constant. The block and bullet deform, heat, and acquire internal
vibration. Their total kinetic energy decreases even though $K_{\rm cm}$ is
unchanged. After the collision, the combined body rises; during that separate stage,
the centre-of-mass kinetic energy converts to gravitational potential energy. The
collision equation and the subsequent energy equation cannot be merged because the
internal energy transfer occurs only during impact.

The same bookkeeping applies to a falling deformable object striking the ground.
An object-ground system classifies contact forces as internal. Gravitational
potential energy can become thermal energy of both surfaces, sound, and residual
deformation. An object-only system classifies the ground's contact force as external
work. Both system
boundaries are valid if their energy terms are classified consistently.

**Two failure modes.**

- **An isolated internal explosion accelerating the centre of mass.** This violates
  momentum conservation; a zero external resultant keeps $V_{\rm cm}$ fixed.
- **Internal force pairs always doing zero total work.** This confuses
  equal-and-opposite forces with equal displacements, since spring ends move by
  different amounts.

The translational theorem and the full system-energy balance answer different
questions and draw on different displacement data.

| Requested quantity | Relation | Required kinematic data |
| --- | --- | --- |
| Centre-of-mass translation | $W_{\rm ext,cm}=\Delta K_{\rm cm}$ | centre-of-mass displacement and net external force |
| Total power | $P=\sum_a\vec F_a\cdot\vec v_a$ | velocity of every force application point |
| Rigid-body power | $P=\vec F_{\rm net}\cdot\vec V_{\rm cm}+\boldsymbol\tau_{\rm cm}\cdot\boldsymbol\omega$ | resultant force, torque, and angular speed |
| Internal storage | change in relative, elastic, thermal, or deformation energy | declared system boundary |

State the approximation explicitly. If external impulse is appreciable, $V_{\rm cm}$
changes and the matching external work stays in the balance. Internal energy terms may
be dropped only once the model has shown that they are unchanged.

[^tipler65]: Tipler and Mosca, _Physics for Scientists and Engineers_, 6th ed., §6-5.

## External and internal work at distinct points

The external-work sum uses the displacement of the point where each external force
acts. Substituting the centre-of-mass displacement everywhere holds only for special
cases, such as a rigid body under a resultant force with no torque contribution. An
off-centre force can translate the system and change its rotational or vibrational
energy at the same time.

Internal forces need the same point-by-point care. Equal and opposite forces need not
do equal and opposite work, because their application points can move through
different displacements: a relaxing spring can do positive work on both attached masses
while its stored potential energy decreases. The internal-force sum is zero at each
instant, yet the internal work still converts energy among the system's relative
degrees of freedom. The system boundary sets the classification, not the physical
identity of the force. A hand's push is external to a cart-only system; for a hand-cart
system the same contact force is internal, and the chemical-energy change in the hand
joins the account.

**Boundary sets the bookkeeping.**

A person pushes a crate $2.0\ \mathrm m$ with a $40\ \mathrm N$ horizontal force,
supplying $+80\ \mathrm J$ of external work to the crate-only system. In the
person-crate system the push is internal, so an $80\ \mathrm J$ chemical-energy
decrease and any thermal change carry the account instead. Entering both the external
push work and the internal chemical-energy decrease in one equation double-counts the
same $80\ \mathrm J$.

**Center-of-mass energy versus relative energy.**

$K_{\rm cm}=\tfrac12MV_{\rm cm}^2$ carries translation of the centre of mass; the
remaining $K'$, measured in the centre-of-mass frame, holds the opposite motions of
two separating fragments, rotation about the centre, and vibration. The external
resultant controls only $K_{\rm cm}$: a zero resultant fixes $V_{\rm cm}$ yet leaves
$K'$ free to change, as in an explosion at rest ($K_{\rm cm}=0$, chemical energy
feeding $K'$) or an inelastic collision ($K_{\rm cm}$ fixed, $K'$ falling into heat).

Mass weighting shapes the split. The centre of mass sits nearer the heavier particle,
and for equal and opposite momenta the lighter particle moves faster and takes the
larger share of $K'$, because at fixed momentum magnitude the kinetic energy $p^2/2m$
grows as $m$ shrinks.

$$
% caption: Two particles with opposite centre-of-mass-frame momenta have a fixed
% center of mass but nonzero relative kinetic energy. The lighter particle moves
% faster, so the internal kinetic-energy partition is mass-weighted even though the
% system center of mass remains at rest.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[fill=black!12, draw=black, thick] (-2.30,0) circle (0.34);
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  \draw[fill=black!12, draw=black, thick] (1.90,0) circle (0.64);
  \node[black] at (1.90,0) {$m_2$};
  \draw[dashed, black] (0,-1.15) -- (0,1.15);
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  \draw[->, black, thick] (2.54,0) -- (3.80,0) node[right] {$\mathbf p$};
  \node[anchor=north, black] at (-2.30,-0.50) {small, fast};
  \node[anchor=north, black] at (1.90,-0.80) {heavy, slow};
\end{tikzpicture}
$$

> **Worked example (kinetic-energy split).** Particles of masses $1.0$ and $3.0\ \mathrm{kg}$ have equal and opposite momenta
> of magnitude $6.0\ \mathrm{kg\,m\,s^{-1}}$ in their centre-of-mass frame. Their
> speeds are $6.0$ and $2.0\ \mathrm{m\,s^{-1}}$, respectively. Their relative
> kinetic energies are $18\ \mathrm J$ and $6.0\ \mathrm J$, for
> $K'=24\ \mathrm J$. The centre-of-mass kinetic energy is zero in this frame despite
> the substantial total kinetic energy.

**Two-body spring release.**

A compressed spring between two free masses gives a clean centre-of-mass picture. On a
frictionless surface starting from rest, external force and impulse vanish, so the
centre of mass stays fixed. Release drops the elastic potential energy and gives the
masses opposite momenta and positive relative kinetic energy, drawn from internal
spring deformation rather than external work.

Momentum conservation sets the ratio of final speeds; the spring-energy decrease sets
their scale. With final momenta of equal magnitude $p$ and opposite direction,

$$
K'=\frac{p^2}{2m_1}+\frac{p^2}{2m_2},
$$

and equating $K'$ to the released spring energy fixes $p$. The lighter mass leaves
faster and with the larger share of kinetic energy, while the centre of mass holds its
original position.

> **Worked example (spring release).** Masses $m_1=1.0\ \mathrm{kg}$ and
> $m_2=3.0\ \mathrm{kg}$ are separated by a compressed spring storing $24\ \mathrm J$.
> With final momentum magnitude $p$, energy gives
> $$
> 24=\frac{p^2}{2(1.0)}+\frac{p^2}{2(3.0)}=\frac{2p^2}{3},
> $$
> so $p=6.0\ \mathrm{kg\,m\,s^{-1}}$. The speeds are $6.0$ and $2.0\ \mathrm{m\,s^{-1}}$
> in opposite directions. The centre-of-mass kinetic energy stays zero, and all
> $24\ \mathrm J$ appears as relative kinetic energy.

The centre-of-mass theorem and the full system-energy balance are complementary: the
first tracks translation, the second the internal changes translation cannot resolve.
Both require one consistent system boundary and inertial frame.

## Deformation and relative energy

External work on a deformable body is generally not the resultant force times
centre-of-mass displacement, since material points move by different amounts; the work
sum uses each external force with its own application-point displacement. Equal and
opposite external forces can have zero resultant and zero centre-of-mass acceleration
yet still do positive work by stretching the body, storing energy as elastic
deformation, vibration, or heat rather than translation.

A bar pulled at both ends illustrates the distinction. If the end forces pull
outward while the ends separate, each force does positive work on the bar. Their
resultant is zero, so $W_{\rm cm}=0$, but the bar's internal energy increases. A
centre-of-mass work calculation alone would report no translational kinetic-energy
change and would miss the energy stored in strain. Releasing the bar can reverse the
transfer and launch waves or masses attached to its ends.

An ideal elastic deformation stores external work reversibly as elastic potential
energy. Plastic deformation or internal damping converts part to thermal energy that
is not recovered on unloading. The same external-force pattern can
therefore produce different final energy partitions depending on the material model.
The system boundary and constitutive assumption must be stated before assigning a
single “internal energy” term.

$$
% caption: Equal and opposite external pulls can deform a bar without moving its
% center of mass. Each force acts through an outward endpoint displacement and does
% positive work; the zero external resultant leaves CM translation unchanged while
% the work appears as internal strain or thermal energy.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[acc, fill=acc!12, thick] (-1.65,0) rectangle (1.65,0.78);
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  \draw[->, black, very thick] (-1.65,0.39) -- (-3.05,0.39) node[left] {pull};
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  \draw[->, black, dashed] (1.65,1.20) -- (2.35,1.20) node[above] {end moves};
\end{tikzpicture}
$$

> **Worked example (deformation work).** Two $50\ \mathrm N$ outward pulls each move their point of application
> $0.040\ \mathrm m$ outward while stretching a bar. Each force does
> $+2.0\ \mathrm J$ of work, so external work totals $+4.0\ \mathrm J$. The centre
> of mass remains fixed because the forces balance, yet the bar acquires $4.0\ \mathrm
> J$ of internal energy in the ideal elastic model.

**Energy in the center-of-mass frame.**

The centre-of-mass frame removes the translational kinetic-energy term by definition:
$\vec V_{\rm cm}=\vec 0$, so the total kinetic energy measured in that frame
is $K'$. This makes the centre-of-mass frame well suited to collisions,
explosions, and two-body interactions. It isolates the energy available for relative
motion from the energy associated with the system moving as a whole past an external
observer.

For two particles, relative kinetic energy can be expressed with the reduced mass
$\mu=m_1m_2/(m_1+m_2)$ and relative speed $v_{\rm rel}$:

$$
K'=\frac12\mu v_{\rm rel}^2.
$$

The reduced-mass form replaces the two-body motion with one relative coordinate and gives the
same value as summing individual centre-of-mass-frame kinetic energies, but it makes
clear that $K'$ depends on how rapidly the particles approach or separate. The
reduced mass is smaller than either total mass and is weighted toward the lighter
particle, reflecting the unequal shares of motion in the centre-of-mass frame.

Frame choice changes the numerical total kinetic energy but not the relative speed
or the internal kinetic energy $K'$. Two particles observed from a train and from
the ground have different centre-of-mass velocities, yet their separation rate and
reduced-mass energy are the same under a constant-velocity frame transformation.
$K'$ is therefore the frame-independent energy quantity for internal interactions.

> **Worked example (reduced mass).** Masses $2.0\ \mathrm{kg}$ and $6.0\ \mathrm{kg}$
> approach with relative speed $5.0\ \mathrm{m\,s^{-1}}$. Their reduced mass is
> $\mu=(2.0)(6.0)/(8.0)=1.50\ \mathrm{kg}$, so
> $K'=\tfrac12(1.50)(5.0)^2=18.75\ \mathrm J$. This is the kinetic energy available in
> the centre-of-mass frame before any collision or internal interaction.

**Energy transfer through an internal force.**

Internal forces transfer energy between parts of a system even when their vector sum is
zero. For an ideal spring the two internal works sum to the negative change in spring
potential energy, and each end's work depends on that end's own motion. Momentum
conservation follows from the cancellation of internal force pairs, but energy transfer
needs the velocities of the application points. A damper is the contrasting case: its
forces also sum to zero, yet the work it removes from relative motion becomes heat
rather than recoverable elastic energy.

The total internal power is the sum of the two force-velocity dot products: negative
while spring potential energy rises, positive while it falls. The centre-of-mass
velocity can stay constant throughout, so this internal power changes $K'$ and $U$
without touching $K_{\rm cm}$.

**Internal-power interpretation.**

When two masses move apart and stretch their spring, each spring force opposes its
endpoint velocity, so the spring does negative work on both: relative kinetic energy
falls and spring potential energy rises. On release the velocities reverse relative to
the forces, the spring does positive work on both, and relative kinetic energy returns.
An isolated system holds the same centre-of-mass velocity throughout, since the internal
forces have no resultant.

## Constraints, collisions, and energy audits

An ideal constraint does not automatically imply zero work on every member of a
system. A fixed, smooth guide exerts a normal force perpendicular to its contact
point displacement and therefore does no work on the guided particle. A moving guide
can exert a normal force through a nonzero displacement and transfer energy. A taut
string of fixed length can transfer energy between its endpoints even when the net
work of tension on the complete ideal string-and-mass system is zero. The relevant
question is always which point moves under which force.

Constraint geometry often links velocities and therefore links power. In a simple
massless rope over an ideal pulley, tension acts along each rope segment. The power
on one attached mass is tension times that endpoint's velocity component along the
rope; the power on the other mass has the opposite sign when the endpoints move in
opposite directions with equal speeds. Tension can remove kinetic energy from one
object while delivering it to the other without changing the total energy of the
ideal constrained pair.

Nonideal constraints add internal energy channels. Pulley axle friction, rope
stretch, sliding contact, or deformation can convert part of the transferred
mechanical energy to thermal energy. The common-tension and zero-net-constraint-work
assumptions then need revision. A diagram that labels an ideal constraint is making
a quantitative energy statement about the constraint.

> **Worked example (constraint power).** In an ideal pulley system, a $30\ \mathrm N$ tension pulls one mass upward at
> $0.40\ \mathrm{m\,s^{-1}}$ and the other downward at the same speed. Tension power
> on the rising mass is $+12\ \mathrm W$ when the tension direction matches its
> velocity, while tension power on the descending mass is $-12\ \mathrm W$. The rope
> transfers energy between masses without creating or destroying it under the ideal
> model.

**Collision energy and center-of-mass motion.**

During a short collision with negligible external impulse, the centre-of-mass
velocity is unchanged. That does not mean total kinetic energy is unchanged. The
centre-of-mass kinetic term is fixed by the conserved total momentum, while the
relative kinetic term can decrease as the objects deform, heat, or vibrate. In a
perfectly elastic collision, relative kinetic energy is preserved. In an inelastic
collision, some or all of it becomes internal energy, even though the centre of mass
continues with the same velocity.

The centre-of-mass frame makes this contrast direct. Before a head-on collision,
the particles approach with opposite momenta. After a perfectly inelastic collision,
the combined object is at rest in the centre-of-mass frame and the relative kinetic
energy is zero. In a laboratory frame, that same combined object may continue to
move with substantial kinetic energy because the centre of mass is translating. The
lab-frame final kinetic energy is then precisely the centre-of-mass translational
term of the combined mass.

Energy lost from relative motion is not negative; it is added deformation, thermal, and
sometimes sound energy inside the larger system. Momentum conservation alone sets the
common post-collision centre-of-mass velocity; an energy statement is needed to find
how much relative kinetic energy remains and how much turns internal.

$$
% caption: In a short inelastic collision, the center of mass continues uniformly
% when external impulse is negligible, while relative kinetic energy collapses into
% deformation and thermal energy. The laboratory-frame motion after sticking can
% remain nonzero because it contains the unchanged center-of-mass translation term.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.2,0) node[right] {position};
  \draw[fill=black!12, draw=black, thick] (0.90,0.20) rectangle (1.90,1.00);
  \draw[fill=black!12, draw=black, thick] (4.40,0.20) rectangle (5.60,1.00);
  \draw[->, black, thick] (2.05,0.60) -- (3.05,0.60);
  \node[anchor=south, black] at (2.40,0.66) {$v_1$};
  \draw[->, black, thick] (4.25,0.60) -- (3.25,0.60);
  \node[anchor=south, black] at (3.90,0.66) {$v_2$};
  \draw[dashed, black] (3.15,-0.55) -- (3.15,1.35);
  \node[anchor=north, black] at (3.15,-0.55) {stick here};
  \draw[->, acc, very thick] (2.50,-1.15) -- (3.90,-1.15) node[right] {$\mathbf V_{\rm cm}$ uniform};
\end{tikzpicture}
$$

> **Worked example (inelastic collision).** A $1.0\ \mathrm{kg}$ cart at
> $6.0\ \mathrm{m\,s^{-1}}$ strikes and sticks to a $2.0\ \mathrm{kg}$ cart at rest.
> The common velocity is $(1.0)(6.0)/(3.0)=2.0\ \mathrm{m\,s^{-1}}$. Initial kinetic
> energy is $18\ \mathrm J$, and final kinetic energy is
> $\tfrac12(3.0)(2.0)^2=6.0\ \mathrm J$. The $12\ \mathrm J$ decrease is lost relative
> kinetic energy; the final $6.0\ \mathrm J$ is the unchanged translation of the system
> centre of mass in the original frame.

**Complete two-object system energy audit.**

For two carts joined by a spring and pulled by an external agent, the external work can
change centre-of-mass translation, spring deformation, and relative motion at once. The
internal spring's own work is not entered again as an external term; its transfer shows
up through the change in elastic potential and relative kinetic energy. Adding both the
spring work and the spring-potential change to the same total-system balance
double-counts it.

$$
% caption: A two-object energy audit separates center-of-mass translation from
% relative motion and internal storage. External work crosses the system boundary;
% spring deformation and thermal energy are internal terms whose changes explain
% energy transfer even when the external resultant and CM motion are zero.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,0) rectangle (5.40,2.90);
  \node[anchor=north west, black] at (0.20,2.75) {two-object system};
  \draw[fill=black!12, draw=black, thick] (0.70,1.10) rectangle (1.60,1.85);
  \draw[fill=black!12, draw=black, thick] (3.80,1.10) rectangle (4.70,1.85);
  \draw[black, thick] (1.60,1.475) -- (1.85,1.475)
    -- (2.05,1.68) -- (2.30,1.27) -- (2.55,1.68) -- (2.80,1.27)
    -- (3.05,1.68) -- (3.30,1.27) -- (3.55,1.475) -- (3.80,1.475);
  \node at (2.70,0.55) {$K_{\rm cm}+K_{\rm rel}+U+E_{\rm th}$};
  \node[anchor=west] at (5.55,2.30) {external work};
  \draw[->, acc, very thick] (6.30,1.475) -- (5.40,1.475);
\end{tikzpicture}
$$

> **Worked example (two-object audit).** Two carts and their compressed spring form a system initially at rest. An external
> agent does $10\ \mathrm J$ of work while further compressing the spring, and no
> centre-of-mass motion occurs because equal external pulls act at the two ends. The
> $10\ \mathrm J$ appears as increased spring potential energy, not $K_{\rm cm}$.
> After the pulls are removed, the spring releases internally: $K_{\rm cm}$ remains
> zero while relative kinetic energy rises by $10\ \mathrm J$ if losses are negligible.
> The external-work term belongs only to the compression stage.

## Variable mass and coupled bodies

An energy statement for a system whose mass changes requires special care because
matter can carry kinetic energy, internal energy, and momentum across the boundary.
A rocket-only system, for example, loses mass through its nozzle. The expelled fuel
crosses the boundary with velocity relative to the chosen frame and carries kinetic
energy with it. A simple fixed-mass equation for the rocket
alone omits this energy flux unless it is written explicitly.

One remedy is to choose a larger system that includes the rocket and all exhaust
under consideration. Then the chemical-energy decrease, kinetic energy of the rocket,
and kinetic energy of the exhaust are internal terms. Another is to retain a smaller
system and account for energy carried across its boundary by the entering or leaving
mass. Neither choice is automatically simpler; the correct choice is the one whose
boundary terms can be identified without mixing frames or velocities.

The same caution applies to a conveyor dropping material into a cart, leaking fluid,
or a sandbag landing on a moving wagon. The system mass changes and the incoming or
outgoing material has a velocity not generally equal to the centre-of-mass velocity
of the retained system. Centre-of-mass work and fixed-mass kinetic-energy formulas
must be applied only after the system definition has been made explicit.

$$
% caption: A variable-mass system boundary must account for energy carried by matter
% crossing it. Exhaust leaving a rocket has its own kinetic energy in the selected
% frame; enlarging the boundary to include rocket and exhaust converts that boundary
% flux into internal kinetic and chemical-energy changes.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[dashed, black, thick] (0,0.30) rectangle (4.0,2.55);
  \node[anchor=north, black] at (2.00,0.15) {rocket-only boundary};
  \draw[black, thick] (1.00,1.05) -- (2.40,1.05) -- (3.00,1.40) -- (2.40,1.75) -- (1.00,1.75) -- cycle;
  \draw[->, black, thick] (1.20,1.98) -- (2.65,1.98);
  \node[anchor=south, black] at (1.95,2.05) {rocket speed};
  \draw[->, acc, thick] (1.00,1.40) -- (-1.30,1.40) node[left] {exhaust KE};
\end{tikzpicture}
$$

**Exhaust energy is not negligible.**

If $0.20\ \mathrm{kg}$ of exhaust leaves a vehicle at high speed, its kinetic energy
matters even though the remaining vehicle is far heavier. A vehicle-only equation needs
an outflow-energy term; a vehicle-plus-exhaust equation makes that term internal but
must include the chemical-energy decrease that drove both motions. The two accounts
agree only for the same frame and the same material crossing.

**Coupled-body energy calculation.**

Two bodies linked by an ideal string offer a compact system-energy calculation when
their motions are constrained. The string enforces a relation between their speeds,
and tension is internal to the two-body-string system. Its individual works can be
nonzero on each mass, yet they cancel in the total ideal-system energy balance when
the string is massless and taut. Gravitational potential changes, kinetic energy of
both masses, and any friction on a contact surface remain as the relevant terms.

The system approach is efficient for finding the common speed after a displacement.
It does not immediately supply tension, because tension has been removed as an
internal interaction. To obtain tension, return to an individual-body force equation
after the energy calculation has established the speed or acceleration. This division
of labour avoids treating the internal tension as both cancelled and externally
available in the same line of reasoning.

Energy methods determine reachability. If the decrease in gravitational potential
of a descending mass is smaller than the increase required by the rising mass plus
friction work, the assumed displacement cannot occur from rest without an external
energy source. A negative result for speed squared indicates a physically inaccessible
configuration, not a negative speed.

$$
% caption: For two masses coupled by an ideal string, tension is internal to the
% enlarged two-body system while gravity and surface friction supply the net energy
% change. The linked displacements give a common speed magnitude, allowing one
% energy equation to determine motion without first solving for tension.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,0) -- (3.30,0);
  \draw[fill=acc!12, draw=acc, thick] (0.80,0) rectangle (1.80,0.80);
  \draw[black, thick] (1.80,0.40) -- (3.30,0.40);
  \draw[black, thick] (3.40,0.40) circle (0.10);
  \draw[black, thick] (3.50,0.40) -- (3.50,-1.20);
  \draw[fill=black!12, draw=black, thick] (3.05,-2.00) rectangle (3.95,-1.20);
  \draw[->, acc, thick] (1.95,1.05) -- (2.80,1.05) node[right] {$\mathbf v_1$};
  \draw[->, black] (1.15,0.80) -- (1.15,1.55) node[above] {normal};
  \draw[->, black, thick] (4.25,-1.20) -- (4.25,-2.10) node[below] {$\mathbf v_2$};
\end{tikzpicture}
$$

> **Worked example (coupled-body energy).** A $2.0\ \mathrm{kg}$ block on a smooth table is attached over an ideal pulley to a
> $1.0\ \mathrm{kg}$ hanging mass. Starting from rest, the hanging mass falls
> $0.60\ \mathrm m$. Its gravitational potential decreases by
> $(1.0)(9.8)(0.60)=5.88\ \mathrm J$. Both masses have the same speed magnitude, so
>
> $$
> 5.88=\frac12(2.0+1.0)v^2,
> \qquad v=1.98\ \mathrm{m\,s^{-1}}.
> $$
>
> The energy equation contains no tension term because the string is included in the
> system and is ideal. A separate force equation for the table block would then give
> the tension if it were requested.

## Linkages and pulley work

A pulley or rigid linkage can redirect force and motion while transferring mechanical
energy between components. For an ideal fixed pulley, the two rope endpoints move
equal distances, so equal tension magnitudes produce equal and opposite tension
works on the attached bodies. The pulley changes direction but does not create energy.
A movable pulley has endpoint distances differing by a geometric factor. The force
ratio and displacement ratio compensate so that ideal input and output work match.

The work balance avoids treating mechanical advantage as an energy
advantage. A linkage that lets a person lift a load with half the force generally
requires twice the input displacement. Friction, pulley rotation, and rope stretch
reduce output energy relative to input, but no ideal arrangement can supply more
mechanical energy than it receives. A system-energy diagram should show where any
losses are assigned: axle heating, rope deformation, or contact friction.

Power has the same conservation structure. In an ideal linkage, input force times
input speed equals output force times output speed at each instant, with signs set by
the motion directions. A slow, high-force load motion can therefore be driven by a
fast, lower-force input motion. This is an instantaneous constraint relation, not an
additional energy source.

$$
% caption: A movable pulley trades force for displacement while preserving ideal
% work. Two supporting rope segments lift the load, so the input end moves twice as
% far as the load; the doubled load-side tension force is balanced by the doubled
% input displacement in the ideal energy transfer.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0.40,3.00) -- (5.00,3.00);
  \draw[black, thick] (2.00,3.00) -- (2.00,1.20);
  \draw[black, thick] (2.80,1.20) -- (3.60,3.00);
  \draw[black, thick] (2.40,1.20) circle (0.42);
  \draw[black, thick] (3.60,3.00) circle (0.10);
  \draw[black, thick] (3.70,3.00) -- (3.70,1.00);
  \draw[fill=acc!12, draw=acc, thick] (1.85,0.20) rectangle (2.95,0.78);
  \draw[->, acc, thick] (1.40,0.50) -- (1.40,1.10) node[above] {load $d$};
  \draw[->, black, thick] (3.70,1.00) -- (3.70,-0.30) node[below] {input $2d$};
\end{tikzpicture}
$$

> **Worked example (pulley work).** An ideal movable pulley supports a load with two rope segments, each carrying
> $80\ \mathrm N$ tension. Raising the load $0.30\ \mathrm m$ increases its mechanical
> energy by $(160)(0.30)=48\ \mathrm J$. The free end moves $0.60\ \mathrm m$ and is
> pulled with $80\ \mathrm N$, so the input work is also $48\ \mathrm J$. The force
> advantage has been exactly offset by the larger input displacement.

**Boundary and losses.**

State whether the string and pulley are inside the system. When they are ideal and
included, tension is internal and cancels from the total energy balance; a hand pulling
the free end does work across the boundary; modelled axle friction or rope stretch adds
a thermal or elastic term. The displacement ratio must come from the actual string or
linkage geometry. Using a force advantage without its matching motion constraint invents
energy and signals an inconsistent model, and the same geometry must hold for power at
every instant.