Inelastic Collisions

A modern car is designed to fail magnificently. In a crash test the front third of the vehicle folds like an accordion, sacrificing itself in a tenth of a second — and that controlled destruction is the whole point. The kinetic energy of the moving car cannot be wished away, and momentum conservation dictates exactly how fast the wreckage moves after impact; the only freedom an engineer has is where the kinetic energy goes. Steering it into crumpled steel instead of human bodies is arguably the most life-saving application of a single mechanics formula, $\Delta T = -\tfrac12\mu\,v_{\mathrm{rel}}^2$ . Collisions in which kinetic energy is lost are called inelastic, and they are the rule, not the exception: clay balls stick, cars lock bumpers, meteorites bury themselves in planets, gas clouds merge into stars. Through all of that violence one quantity sails untouched. Because the forces of contact are internal to the colliding pair and obey Newton's third law, the total momentum after equals the total momentum before — always, without exception, no matter how much energy vanishes into heat, sound, and deformation. That asymmetry — momentum sacred, kinetic energy expendable — is the entire subject. It fixes how much energy a collision can lose (all of the relative-motion energy, none of the center-of-mass energy), it defines the coefficient of restitution $e$ that grades every collision from superball to putty, and it powers a classic instrument, the ballistic pendulum, that measured bullet speeds a century before electronics. Master the bookkeeping and every crash, capture, and coalescence becomes a two-line calculation.

Key takeaways

In any collision the internal forces cancel by Newton's third law, so total momentum $\sum m_i\mathbf{v}_i$ is conserved — even when kinetic energy is not.
An inelastic collision converts part of the kinetic energy into internal energy (heat, sound, deformation); in the perfectly inelastic extreme the bodies stick and move at the center-of-mass velocity $V_{\mathrm{cm}}$ .
The König split $T = \tfrac12 M V_{\mathrm{cm}}^2 + \tfrac12\mu v_{\mathrm{rel}}^2$ shows what is at stake: the CM term is untouchable, the relative term is the only energy a collision can spend.
The maximum possible loss is $\tfrac12\mu\,v_{\mathrm{rel}}^2$ with $\mu = m_1m_2/(m_1+m_2)$ — achieved exactly when the bodies stick.
The coefficient of restitution $e =$ (separation speed)/(approach speed) dials a collision between elastic ( $e=1$ ) and stick ( $e=0$ ); the internal kinetic energy retained is $e^2$ , which is why a dropped ball rebounds to $e^2$ of its height.
Key idea: momentum conservation is unconditional; kinetic energy is negotiable — an inelastic collision spends relative-motion energy on internal degrees of freedom, up to the sticking limit where all of it is gone and only the center-of-mass motion survives.

The core insight

Momentum is always conserved in a collision because the internal forces cancel by Newton's third law; kinetic energy is conserved only in elastic collisions — inelasticity spends the relative-motion energy $\tfrac12\mu v_{\mathrm{rel}}^2$ on internal degrees of freedom, maximally when the bodies stick and move together at $V_{\mathrm{cm}}$ .

Where this lives today

Every crash-test rating you have ever seen is applied inelastic-collision physics. Modern vehicle programs (Euro NCAP, IIHS) budget the loss $\tfrac12\mu v_{\mathrm{rel}}^2$ across engineered load paths — crumple zones lengthen the stopping distance so the same energy is absorbed at survivable force levels — while accident reconstruction runs the logic in reverse: from post-crash trajectories and locked-together wreckage, equation (2) recovers the pre-impact speeds, and event-data-recorder traces are checked against exactly this momentum bookkeeping in courtrooms. The same mathematics prices satellite shielding: a micrometeoroid hitting a Whipple shield is a hypervelocity inelastic collision, and the shield's job is to force $v_{\mathrm{rel}}$ 's energy to dissipate in a sacrificial bumper. At the cosmic scale, inelasticity is a power source. Gas spiraling onto a black hole through an accretion disk is a continuous perfectly inelastic collision: relative kinetic energy is dissipated to heat exactly as $\tfrac12\mu v_{\mathrm{rel}}^2$ demands, and the radiated glow — up to tens of percent of $mc^2$ , the most efficient energy conversion known short of annihilation — is what the Event Horizon Telescope images. And in the lab, "dead" collisions are engineered on purpose: granular dampers (boxes of beads whose grain-grain collisions have low $e$ ) quietly kill vibrations in spacecraft brackets and tennis rackets alike.

The physical picture

Throw two lumps of clay at each other. They smack, squash, warm up slightly, and fly on as one blob. Now ask the two accounting questions of mechanics. Momentum? The forces each lump exerted on the other were equal and opposite at every instant — Newton's third law — so whatever momentum one lump lost, the other gained; the blob carries exactly the momentum the pair brought in. Kinetic energy? Squashing clay is not reversible. The organized forward motion of atoms became disorganized jiggling — heat — plus a thud of sound. The books balance only if you count internal energy: kinetic energy alone took a permanent loss.

Figure

Two clay lumps merge — the momentum bar is unchanged while kinetic energy drops

The deepest way to see how much energy is at stake is to ride along with the center of mass. The CM is an unstoppable bookkeeper: since no external force acts during the brief impact, the CM glides on at constant velocity through the whole event, entirely indifferent to the crunch. Split the kinetic energy into two accounts — the bulk motion of the CM, and the motion of the bodies relative to the CM. The first account is locked: it is fixed by the total momentum, which cannot change. Only the second account, the relative-motion energy $\tfrac12\mu v_{\mathrm{rel}}^2$ , is available to be spent. A perfectly inelastic collision spends all of it: sticking together means no relative motion left. Between the extremes lives a one-parameter family of collisions, graded by Newton's coefficient of restitution $e$ — the ratio of the speed at which the bodies separate to the speed at which they approached. You have measured $e$ yourself every time you dropped a ball: a superball rebounds to nearly its drop height ( $e \approx 0.9$ ), a tennis ball to about half ( $e \approx 0.73$ ), and wet clay to none at all ( $e = 0$ ). The rebound height ratio is exactly $e^2$ , because energy goes as the square of speed.

Figure

The restitution dial — every collision sits somewhere between stick and perfect bounce

The rules of the crash

Everything begins with the one unconditional law. During the impact the mutual forces are enormous — thousands of times the weights of the bodies — but they are internal to the pair, and Newton's third law pairs them off exactly. Integrate the equal-and-opposite forces over the brief contact and the impulses cancel:

\sum_i m_i\,\mathbf{v}_i^{\,\mathrm{before}} \;=\; \sum_i m_i\,\mathbf{v}_i^{\,\mathrm{after}} \qquad (1)

where: $m_i$ , $\mathbf{v}_i$ — mass and velocity of body $i$ . This is a vector statement — each component is separately conserved — and it holds whether the collision is elastic, partially inelastic, or a full stick, because it never mentions energy at all. External forces (gravity, friction) do act, but over a millisecond of contact their impulse is negligible next to the collision impulse. An equivalent and more vivid statement: the velocity of the center of mass is unchanged by the collision. Dividing (1) by the total mass $M = m_1 + m_2$ gives $V_{\mathrm{cm}}$ , the same before and after. If the bodies stick, they must move together — and the only velocity consistent with (1) is precisely that CM velocity:

v' \;=\; \frac{m_1 v_1 + m_2 v_2}{m_1+m_2} \;=\; V_{\mathrm{cm}} \qquad (2)

where: $v'$ — the common final velocity in a perfectly inelastic collision. Note what this is not: it is not zero (unless the total momentum happens to be zero), and it is not the velocity of either body. A fly splatting on a truck windshield barely dents the truck's speed, because $V_{\mathrm{cm}}$ is mass-weighted. For collisions that bounce partially, Newton himself supplied the empirical dial. Define the coefficient of restitution as the ratio of separation speed to approach speed along the line of impact:

e \;=\; \frac{v_2' - v_1'}{v_1 - v_2} \qquad (3)

where: unprimed velocities are before, primed after, all signed along the collision line. Steel bearings give $e$ near 0.95; regulation basketballs are manufactured to $e \approx 0.76$ – $0.80$ off hardwood; clay gives 0. Equation (3) plus momentum (1) form two equations in the two unknown final velocities — so $e$ completely determines the outcome of any head-on collision.

Which formula answers which question

you want	use	notes
final velocity when bodies stick	$v' = V_{\mathrm{cm}}$ , eq. (2)	mass-weighted average, never zero unless $P=0$
how much KE a stick destroys	$\Delta T = -\tfrac12\mu v_{\mathrm{rel}}^2$ , eq. (5)	needs relative velocity only
outcome of a partial bounce	eqs. (6)–(7)	reduce to elastic at $e=1$ , stick at $e=0$
grading a collision from a drop test	$h_{\mathrm{out}}/h_{\mathrm{in}} = e^2$ , eq. (8)	how $e$ is actually measured
bullet speed from a ballistic pendulum	$v = \frac{m+M}{m}\sqrt{2gh}$	momentum for the capture, energy for the swing

The energy budget

How much kinetic energy can a collision destroy? The clean answer comes from splitting the energy into a part that belongs to the center of mass and a part that belongs to the relative motion — the König decomposition. The two derivations below are the load-bearing walls of the whole topic.

Derivation AThe König split and the maximum possible energy loss

We are deriving the energy bookkeeping of a two-body collision: the identity $T = \tfrac12 M V_{\mathrm{cm}}^2 + \tfrac12 \mu v_{\mathrm{rel}}^2$ , and from it the maximum kinetic energy any collision can dissipate, equation (5). The strategy is a change of variables — from the two velocities $(v_1, v_2)$ to the pair (CM velocity, relative velocity) — because momentum conservation speaks only to the first variable and energy loss only to the second, so the change of variables decouples the two laws. Watch how the cross terms cancel: that cancellation is not luck, it is exactly the definition of the center of mass doing its job. Everything here is exact and works in any number of dimensions; we write one dimension for clarity.

Step 1. Define the new variables: total mass $M = m_1 + m_2$ , CM velocity $V_{\mathrm{cm}} = (m_1 v_1 + m_2 v_2)/M$ , relative velocity $v_{\mathrm{rel}} = v_1 - v_2$ , and reduced mass

\mu = \frac{m_1 m_2}{m_1 + m_2}

Why the reduced mass appears: it is the single effective inertia of the relative coordinate — lighter than either body, and dominated by the lighter partner.

Step 2. Invert the definitions: $v_1 = V_{\mathrm{cm}} + \tfrac{m_2}{M} v_{\mathrm{rel}}$ and $v_2 = V_{\mathrm{cm}} - \tfrac{m_1}{M} v_{\mathrm{rel}}$ . Substitute into $T = \tfrac12 m_1 v_1^2 + \tfrac12 m_2 v_2^2$ and expand:

T = \tfrac12 M V_{\mathrm{cm}}^2 \;+\; \tfrac12\,\mu\, v_{\mathrm{rel}}^2 \qquad (4)

Why the cross terms vanish: they carry the factor $m_1\tfrac{m_2}{M} - m_2\tfrac{m_1}{M} = 0$ — the mass-weighting that defines the CM kills them identically.

Step 3. Apply the collision. Momentum conservation, equation (1), says $V_{\mathrm{cm}}$ is the same before and after — so the first term of (4) is frozen. The only energy available to lose is the second term. The loss is therefore

\Delta T = \tfrac12\mu\left[(v_{\mathrm{rel}}')^2 - v_{\mathrm{rel}}^2\right]

Step 4. Minimize the final energy. Since $\tfrac12\mu (v_{\mathrm{rel}}')^2 \ge 0$ , the final kinetic energy is smallest when $v_{\mathrm{rel}}' = 0$ — the bodies share one velocity, i.e. they stick. The maximum possible loss is the entire relative-motion account:

\Delta T_{\max} \;=\; -\,\frac{1}{2}\,\frac{m_1 m_2}{m_1+m_2}\,\left(v_1-v_2\right)^2 \qquad (5)

where: the minus sign records a loss; only the relative velocity enters. Two cars locking bumpers at highway speed in the same direction lose little; the same cars head-on lose catastrophically — same speeds, very different $v_{\mathrm{rel}}$ . Verified: dimensional check J = kg·(m/s)²; equal masses head-on ( $v_2 = -v_1$ ) give $\Delta T_{\max} = -\tfrac12(m/2)(2v)^2 = -mv^2$ , the entire initial energy, as it must since $P = 0$ ; matches Landau–Lifshitz §17 and Morin ch. 5.

Go deeper · Derivation B · 1 minThe general one-dimensional outcome from the restitution coefficient

Derivation BThe general one-dimensional outcome from the restitution coefficient

The goal is the complete solution of the head-on collision for any inelasticity: final velocities $v_1'$ and $v_2'$ given masses, initial velocities, and $e$ — equations (6) and (7). The route is elementary but instructive: momentum conservation and the restitution definition are two linear equations in the two unknowns, so the whole collision reduces to solving a 2×2 linear system. The thing to watch is the sign convention in $e$ : it compares separation to approach, so getting a negative $e$ from your algebra means you have mislabeled who is chasing whom. The payoff is a single formula pair that contains the elastic and perfectly inelastic worlds as the endpoints $e = 1$ and $e = 0$ .

Step 1. Write the system. Momentum: $m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$ . Restitution, equation (3) rearranged: $v_2' - v_1' = e\,(v_1 - v_2)$ . Why these two suffice: a head-on collision has exactly two unknowns, and energy is not an independent input — the value of $e$ encodes whatever energy is lost.

Step 2. Solve. Multiply the restitution equation by $m_2$ , add to momentum to eliminate $v_2'$ , and divide by $M = m_1 + m_2$ :

v_1' = \frac{(m_1 - e\,m_2)\,v_1 + (1+e)\,m_2\, v_2}{m_1+m_2} \qquad (6)

Step 3. By the same algebra with labels swapped (the system is symmetric under $1 \leftrightarrow 2$ ):

v_2' = \frac{(m_2 - e\,m_1)\,v_2 + (1+e)\,m_1\, v_1}{m_1+m_2} \qquad (7)

where: each final velocity is a weighted blend of both initial velocities, with the blend controlled by $e$ . At $e = 1$ these become the standard elastic formulas; at $e = 0$ both collapse to $V_{\mathrm{cm}}$ — the stick of equation (2).

Step 4. Extract the energy signature. Substituting (6)–(7) into the kinetic energy and using the König split shows the CM term unchanged while the relative term scales as $(v_{\mathrm{rel}}')^2 = e^2 v_{\mathrm{rel}}^2$ :

\frac{T_{\mathrm{internal}}^{\,\mathrm{after}}}{T_{\mathrm{internal}}^{\,\mathrm{before}}} = e^2, \qquad \frac{h_{\mathrm{out}}}{h_{\mathrm{in}}} = e^2 \qquad (8)

where: the first ratio is frame-independent bookkeeping; the second is its lab bench incarnation — for a ball bouncing off the floor (an effectively infinite $m_2$ ), speeds scale by $e$ and drop heights by $e^2$ . Verified: $e = 1$ , $m_1 = m_2$ , $v_2 = 0$ gives $v_1' = 0$ , $v_2' = v_1$ — the billiard-ball velocity exchange; $e = 0$ gives $v_1' = v_2' = V_{\mathrm{cm}}$ ; both limits match Morin ch. 5 and Kleppner ch. 4.

Limiting cases

Case	Result	Meaning
$e = 1$ (perfectly elastic)	$T$ conserved; separation speed = approach speed	the idealized billiard/steel-bearing limit; see the elastic-collisions article
$0 < e < 1$	internal KE multiplied by $e^2$	real macroscopic collisions: some energy always leaks to heat and sound
$e = 0$ (perfectly inelastic)	$v_1' = v_2' = V_{\mathrm{cm}}$ ; loss $= \tfrac12\mu v_{\mathrm{rel}}^2$	bodies stick; maximum loss consistent with momentum conservation
$P = 0$ (equal and opposite momenta)	stick ⇒ everything stops; all KE lost	the CM account is empty, so nothing is protected
$m_2 \to \infty$ , $v_2 = 0$ (wall)	$v_1' = -e\,v_1$ ; wall absorbs momentum $\,(1+e)m_1v_1$	bouncing a ball: speeds scale by $e$ , heights by $e^2$
$m_1 \ll m_2$ (fly on truck)	$v' \approx v_2$ ; fractional truck slowdown $\sim m_1/m_2$	sticking barely matters to the heavy partner

Common trapKinetic energy was lost, so momentum must have been lost too

This is the single most common collision error, and it fails at the first line: momentum conservation, equation (1), descends from Newton's third law alone and never mentions energy. In a perfectly inelastic head-on crash the kinetic energy can drop by 99% while the momentum stays exactly fixed. Three sibling traps travel with it. First, "the bodies stick, so they stop" — no: they move at $V_{\mathrm{cm}}$ , equation (2), which is zero only if the total momentum was zero. Second, using energy conservation across the sticking event (the ballistic pendulum's classic wrong turn — energy conservation applies to the swing after capture, never to the capture). Third, thinking the lost energy is destroyed: total energy is perfectly conserved; it is kinetic energy that pays, converted to heat, sound, and permanent deformation.

Worked examples

1 · Rear-end collision with locking bumpers (confidence builder)

A 1200 kg car traveling at 20 m/s rear-ends an 800 kg car traveling at 10 m/s in the same direction. The bumpers lock. Find the common final velocity, the kinetic energy dissipated, and check the loss against equation (5).

Step 1. Momentum before: $p = 1200 \times 20 + 800 \times 10 = 24{,}000 + 8{,}000 = 32{,}000\ \mathrm{kg\,m/s}$ .

Step 2. Stick means both move at the CM velocity, equation (2): $v' = 32{,}000 / 2000 = \boxed{16\ \mathrm{m/s}}$ .

Step 3. Energy before: $T_i = \tfrac12(1200)(20)^2 + \tfrac12(800)(10)^2 = 240{,}000 + 40{,}000 = 280{,}000\ \mathrm{J}$ . Energy after: $T_f = \tfrac12(2000)(16)^2 = 256{,}000\ \mathrm{J}$ . Dissipated: $\Delta T = \boxed{24{,}000\ \mathrm{J} = 24\ \mathrm{kJ}}$ — into crumpled bumpers and heat.

Step 4. Check with equation (5): $\mu = \frac{1200 \times 800}{2000} = 480\ \mathrm{kg}$ , $v_{\mathrm{rel}} = 10\ \mathrm{m/s}$ , so $\tfrac12 \times 480 \times 10^2 = 24{,}000\ \mathrm{J}$ . ✓ Note only 8.6% of the kinetic energy was lost: the CM account ( $256\ \mathrm{kJ}$ ) was protected by momentum conservation. The same two cars head-on would have $v_{\mathrm{rel}} = 30\ \mathrm{m/s}$ and lose $\tfrac12(480)(900) = 216\ \mathrm{kJ}$ — nine times the damage.

2 · The ballistic pendulum (exam level)

A 10.0 g bullet is fired into a 2.00 kg wooden block hanging from light strings. The block (with bullet embedded) swings up by $h = 5.0$ cm. Find the bullet's speed, and the fraction of its kinetic energy that survived the capture.

Step 1. Recognize the two-stage structure. Stage 1, the embedding, is perfectly inelastic: momentum is conserved, energy is not. Stage 2, the swing, is the reverse: gravity and string tension act, tension does no work, so mechanical energy is conserved while momentum is not (the string yanks the system). Using the right law in the right stage is the entire problem.

Step 2. Work backwards from the swing. Energy conservation on stage 2: $\tfrac12 (m+M)V^2 = (m+M)gh$ , so $V = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.050} = \sqrt{0.981} = 0.990\ \mathrm{m/s}$ .

Step 3. Undo the capture with momentum, equation (2) rearranged: $v = \frac{m+M}{m}\,V = \frac{2.010}{0.0100} \times 0.990 = 201 \times 0.990 = \boxed{199\ \mathrm{m/s}}$ .

Step 4. Energy audit. Bullet's initial energy: $T_i = \tfrac12 (0.0100)(199)^2 = 198\ \mathrm{J}$ . Just after capture: $T_f = \tfrac12(2.010)(0.990)^2 = 0.986\ \mathrm{J}$ . Fraction surviving: $0.986/198 = \boxed{0.50\%}$ — 99.5% of the bullet's kinetic energy became heat and splintered wood in a millisecond.

Step 5. See the general pattern: for a bullet stopping in a free block, the surviving fraction is exactly $T_f/T_i = m/(m+M) = 0.0100/2.010 = 0.498\%$ . ✓ The lighter the bullet relative to the block, the more total the loss — which is why the pendulum needs the momentum step: an energy-only analysis would be wrong by a factor of 200.

Origins · 1 minJohn Wallis and how this was discovered

John Wallis

1616–1703 · Savilian Professor of Geometry, Oxford

Wallis was England's sharpest mathematician before Newton — and one of history's great codebreakers, deciphering Royalist correspondence during the Civil War. His Arithmetica Infinitorum (1656) pioneered the infinite methods Newton would fuse into calculus, and he gave mathematics the symbol $\infty$ . In 1668 the young Royal Society, tired of Descartes' faulty collision rules, formally invited solutions to the collision problem: Wallis answered for "hard" bodies that do not rebound — the perfectly inelastic case — deriving in effect $v' = V_{\mathrm{cm}}$ from conservation of momentum, while Wren and Huygens supplied the elastic laws. It was the first time a conservation law was used to solve a mechanics problem the geometry could not touch, two decades before the Principia named the underlying third law. Wallis kept computing into his eighties, extracting square roots of fifty-digit numbers in his head to fight insomnia.

History

1644

— Descartes publishes seven collision rules in his Principia Philosophiae; most are wrong, but they install "quantity of motion" as the thing to track.

1668

— The Royal Society calls for a theory of impact: Wallis solves the non-rebounding (perfectly inelastic) case by momentum conservation; Wren and Huygens the elastic case.

1687

— Newton's Principia grounds the collision laws in the third law and reports pendulum-impact experiments measuring what is now called the coefficient of restitution.

1742

— Benjamin Robins invents the ballistic pendulum, turning inelastic capture into the first precision measurement of bullet speeds.

1959

— Béla Barényi's crumple zone enters production cars: engineered inelasticity that spends collision energy on sacrificial steel instead of passengers.

Going further

Texts. Morin, Introduction to Classical Mechanics, ch. 5 (Momentum) · Kleppner & Kolenkow, An Introduction to Mechanics, ch. 4 · Goldstein, Classical Mechanics, 3rd ed., ch. 3 (scattering and the CM frame) · Landau & Lifshitz, Mechanics, §16–17 (disintegration and collision of particles) · Taylor, Classical Mechanics, ch. 3.

Provenance. Equations are standard textbook results. A verifier should: (1) confirm momentum conservation from Newton's third law via the impulse argument (Taylor Ch. 3); (2) reproduce the König decomposition $T=\tfrac12 MV_{\text{cm}}^2+\tfrac12\mu v_{\text{rel}}^2$ and the perfectly-inelastic energy loss $\tfrac12\mu v_{\text{rel}}^2$ (Landau–Lifshitz §17); (3) check the general 1D outcome formulas reduce to elastic ( $e=1$ ) and stick ( $e=0$ ) limits (Morin Ch. 5); (4) verify the worked numbers in the checks. Coefficient-of-restitution definition and $h_{\text{out}}/h_{\text{in}}=e^2$ are standard (Goldstein, Taylor). Worked-example numbers on this page ( $v' = 16\ \mathrm{m/s}$ and $\Delta T = 24\ \mathrm{kJ}$ for the locked cars; $v = 199\ \mathrm{m/s}$ and 0.50% energy survival for the ballistic pendulum) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Velocity and Acceleration · Newton's Third Law · Newton's Second Law · Relative Motion and Galilean Transformations · One-Dimensional Kinematics. Leads to: Elastic Collisions · The Two-Body Problem and Reduced Mass · Variable-Mass Systems and the Rocket Equation · Linear Momentum and Impulse.

Where does the lost kinetic energy actually go?

Into microscopic degrees of freedom: random thermal jiggling of atoms (heat), acoustic waves (the bang), and the work of permanent deformation (rearranged material). Total energy is exactly conserved — drop a lump of clay and its temperature genuinely rises, by roughly $\Delta T_{\mathrm{temp}} = gh/c$ with $c$ the specific heat, about a quarter of a millikelvin per meter of drop. What the collision destroys is only the organized energy of bulk motion. That distinction is thermodynamic at heart: converting ordered kinetic energy into disordered internal energy is easy and irreversible; the reverse would demand entropy to decrease spontaneously, which is why shattered teacups never reassemble and perfectly inelastic collisions never run backwards.

Momentum and kinetic energy are both built from mass and velocity — how can one be conserved while the other is not?

Because they are protected by different things. Momentum is conserved by Newton's third law: internal forces cancel in pairs at every instant, whatever their nature, so $\sum m\mathbf{v}$ cannot move. Kinetic energy has no such pairwise protection — the work done by the two members of an action-reaction pair does not cancel unless the contact point moves rigidly, because work depends on the displacement of each force's own point of application. When clay squashes, both contact forces act through deforming material and their net work is negative. Vector versus scalar matters too: momentum can pass through zero and cancel between bodies; kinetic energy, being positive for any motion, cannot hide by cancellation.

Is a perfectly inelastic collision the one where everything stops?

No — this is the most tempting misreading of "maximum energy loss." Sticking means the bodies end with one common velocity, and momentum conservation forces that velocity to be $V_{\mathrm{cm}} = (m_1v_1 + m_2v_2)/(m_1+m_2)$ , equation (2). That is zero only in the special case where the total momentum is zero (equal and opposite momenta, e.g. the two putty balls). A bullet embedding in a hanging block emphatically does not stop the block — it sets it swinging, which is the entire operating principle of the ballistic pendulum. "Maximum loss" always means "all relative motion gone," never "all motion gone."

Can the coefficient of restitution be greater than 1?

For a passive collision between ordinary bodies, no — $e > 1$ would mean the pair separates faster than it approached, i.e. the collision created relative kinetic energy from nothing. But if the collision can tap an internal energy source, the books allow it: a firecracker sandwiched between two carts, a spring-loaded latch released on contact, or a chemically reactive impact all give "superelastic" outcomes, $e > 1$ . Nuclear and particle physics uses the same vocabulary: collisions that de-excite an internal state and boost the kinetic energy are called superelastic scattering. The inequality $0 \le e \le 1$ is a statement about energy sources, not a law of algebra.

Why does the maximum energy loss formula contain the reduced mass rather than the total mass?

Because only relative motion can be destroyed. The König split, equation (4), shows the kinetic energy dividing into a CM part governed by total mass $M$ and a relative part governed by $\mu = m_1m_2/M$ . Momentum conservation freezes the first; collisions spend only the second. The reduced mass is small when the masses are lopsided — for a 10 g bullet and 2 kg block, $\mu \approx 9.95\ \mathrm{g}$ , essentially the bullet's mass — which encodes the physical fact that a heavy partner barely participates in the relative motion. Equal masses maximize $\mu$ at $m/2$ : symmetric pairs are the most efficient kinetic-energy destroyers.

How is the coefficient of restitution actually measured, and why is the height ratio e squared?

The standard bench measurement is the drop test: release a ball from height $h_{\mathrm{in}}$ onto a fixed slab and record the rebound height $h_{\mathrm{out}}$ . Impact speed is $\sqrt{2gh_{\mathrm{in}}}$ ; rebound speed, by the definition of $e$ against an immovable surface, is $e\sqrt{2gh_{\mathrm{in}}}$ ; and the rebound height follows from energy in the free flight (which is conservative): $h_{\mathrm{out}} = (e\sqrt{2gh_{\mathrm{in}}})^2/2g = e^2 h_{\mathrm{in}}$ — equation (8). The square is simply energy going as speed squared. Sports bodies write rules in exactly these terms: a regulation basketball dropped from 1.80 m must rebound to roughly 1.2–1.4 m, pinning $e$ near 0.8.

During the collision itself, momentum is conserved — but is it conserved during the crumpling, when the bodies are still pushing on each other?

Yes, at every instant. Momentum conservation is not a before-versus-after statement that mysteriously holds only at the endpoints; it holds continuously, because at each moment the contact forces are equal and opposite. During maximum compression of a head-on crash the two bodies momentarily move together at $V_{\mathrm{cm}}$ — the perfectly inelastic outcome is exactly "the collision frozen at maximum compression, never rebounding." In a partially elastic collision the stored deformation energy then pushes the bodies apart again, recovering a fraction $e^2$ of the relative-motion energy. That picture makes $e$ feel physical: it grades how much of the compression energy the materials give back.

The article says external forces are negligible during a collision — but gravity acts on the ballistic pendulum block the whole time. Why is the momentum step still legal?

Because impulse is force times time, and the capture is over in about a millisecond. Gravity's impulse on the 2 kg block in $10^{-3}$ s is $mg\,\Delta t \approx 2 \times 9.8 \times 10^{-3} \approx 0.02\ \mathrm{kg\,m/s}$ , while the bullet delivers $mv \approx 0.01 \times 199 \approx 2\ \mathrm{kg\,m/s}$ — a hundred times larger. So momentum conservation during the capture is accurate to about one percent. The approximation fails only for slow interactions, where "collision" stops being the right word. This is also why the string can be ignored during capture but not during the swing: over the millisecond it barely acts; over the half-second swing it steers everything.