Motion of the Centre of Mass

Fire a shell in a high arc and let it explode at the top. Fragments spray everywhere — forward, backward, straight up — a picture of pure chaos. Now plot one special point through the smoke: the mass-weighted average position of all the fragments. It sails on, serene, along exactly the parabola the intact shell would have flown. The explosion, for all its violence, was an internal affair — and internal forces, cancelling in third-law pairs, cannot lay a finger on the centre of mass. This is the payoff the last article's locating skills were bought for. The centre of mass is not just a balance point; it is the system's representative particle: total momentum is $M\vec v_{\rm cm}$ , and Newton's second law, applied to any system whatsoever — colliding, exploding, wobbling, alive — reads $\vec F_{\rm ext} = M\vec a_{\rm cm}$ . The whole zoo of internal drama drops out. One point obeys one clean law. The exam harvest is a family of elegant tricks. No external force and starting from rest? Then the COM never moves — which fixes how far a boat drifts backward when a man walks along it, where the second fragment lands when the first drops dead, and how two spring-loaded blocks share their displacements. JEE calls these "zero-momentum problems"; they look like magic and grade like arithmetic. This article builds the two laws, drills the tricks, and marks the fine print — when the COM's immunity ends (the moment an external force shows up) and what "walks 3 m" must mean before you trust it.

Key takeaways

The COM moves with the mass-weighted average velocity, and total momentum is exactly $\vec P = M\vec v_{\rm cm}$ — a system's momentum is its COM motion.
Newton's second law for any system: $\vec F_{\rm ext} = M\vec a_{\rm cm}$ — the COM accelerates as if it were one particle of the total mass feeling only the external forces. Internal forces (explosions, collisions, muscles, springs) cannot touch it.
$\vec F_{\rm ext} = 0$ : the COM coasts at constant velocity. If it also started at rest, it stays put forever — the zero-momentum anchor.
The anchor's arithmetic: $m_1\Delta x_1 + m_2\Delta x_2 = 0$ — displacements swap in inverse mass ratio (man-on-boat, blocks-and-spring).
Man of mass $m$ walking a length $L$ relative to the boat ( $M$ ): boat shifts $\dfrac{mL}{M+m}$ backward, man $\dfrac{ML}{M+m}$ forward — relative to the water.
A projectile's COM keeps flying the original parabola through an explosion — legal until any fragment lands (a new external force enters the books).
Key idea: only the outside world can move the centre of mass — internal violence rearranges the parts around a headquarters that obeys $\vec F_{\rm ext} = M\vec a_{\rm cm}$ and, with no external force and no initial motion, never moves at all.

The core insight

Because total momentum equals total mass times centre-of-mass velocity, Newton's second law for any system collapses to $\vec F_{\rm ext} = M\vec a_{\rm cm}$ : the centre of mass moves like a single particle driven by the external forces alone, so with zero external force it coasts — and if it started at rest, internal motions must trade displacements in inverse mass ratio about a centre of mass that never moves.

Where this lives today

The wobble of a star about an invisible shared centre of mass is now a planet-hunting industry: the radial-velocity method that found 51 Pegasi b — and hundreds of worlds since — measures the star's few-metres-per-second orbit around the star-plus-planet COM, this chapter's Earth–Moon calculation running at interstellar range. Spaceflight lives by both laws daily: an astronaut adrift can reorient her body (internal moves) but relocate her COM only by throwing mass or firing a thruster — momentum out, motion back — while the ISS as a whole flies a trajectory computed for its centre of mass, docking vessels and treadmill-running crews notwithstanding. Sports scientists track athletes' COM trajectories frame by frame: a long-jumper's flight arc is sealed at takeoff — the flailing "hitch-kick" repositions the legs around the COM for a better landing, never the COM itself — and divers' scoring somersaults are pure rotation-about-a-parabola. And every multi-stage rocket's guidance computer integrates $F_{\rm ext} = Ma_{\rm cm}$ for a vehicle whose parts move, drain and detach — the variable-mass frontier that the rocket-propulsion module opens next door.

The physical picture

Take the last article's point and set it in motion. Every particle of a system carries momentum $m_i\vec v_i$ ; add them up and something remarkable happens to the average: the total is exactly what a single particle of the whole mass $M$ would carry if it moved with the COM's velocity. The COM is not approximately representative — it is the system's momentum, incarnated as a point. Now push the system. Internal forces — the shell's explosive, the spring between the blocks, the man's legs against the deck — come in third-law pairs and cancel out of the total momentum books (the conservation article's central fact). So only external forces can change $\vec P = M\vec v_{\rm cm}$ : gravity, friction from outside, a wall. Divide by $M$ and you have the cleanest sentence in mechanics: the centre of mass accelerates as if all the external force acted on all the mass gathered at one point. The system may be a swarm of fragments, two skaters, or a gymnast mid-somersault; the headquarters obeys single-particle physics.

Figure

Hero — the explosion that couldn't: fragments scatter, the COM keeps the appointment

The most exam-productive special case is the humblest: no external force, starting from rest. Then $\vec v_{\rm cm} = 0$ now and forever — the COM is a fixed anchor in space. Whatever moves inside the system must be paid for by counter-motion: the man walks left, the boat slides right, and their mass-weighted displacements cancel to keep the anchor fixed. A fish cannot swim in a vacuum, an astronaut adrift cannot crawl home through nothing (she can throw something — exporting momentum is the one legal move), and no amount of pulling on your own bootstraps relocates your headquarters. Walking on solid ground works only because the ground is outside you: friction from the earth is an external force, and it, not your muscles directly, accelerates your COM.

The math: two laws and an anchor

Differentiate the locating formula in spirit (rates of change of a weighted average are the weighted average of rates — pure algebra over any interval):

\vec v_{\rm cm} = \frac{m_1\vec v_1 + m_2\vec v_2 + \cdots}{M} \;\;\Rightarrow\;\; \vec P = M\,\vec v_{\rm cm} \qquad (1)

where: $M = \sum m_i$ — total mass; $\vec P = \sum m_i \vec v_i$ — total momentum. Read it physically: a system's momentum is fully summarized by one velocity — the COM's. This is why the "common velocity" of a perfectly inelastic collision was the COM velocity, and why the conservation law can be restated: no external force ⇒ $\vec v_{\rm cm}$ constant. Apply the system law from the conservation article ( $\Delta \vec P = \vec F_{\rm ext}\Delta t$ ) to (1) and divide by $M$ :

\vec F_{\rm ext} = M\,\vec a_{\rm cm} \qquad (2)

where: $\vec F_{\rm ext}$ — the net external force only; every internal force has already cancelled. This is Newton's second law promoted to systems (Newton proved it as a corollary; Euler enthroned it as the first law of mechanics). It cuts problems in half: to track a system's overall trajectory you may ignore its internal life completely — the fragments' spray pattern is irrelevant to where their COM comes down. And the anchor. Set $\vec F_{\rm ext} = 0$ with the system initially at rest: $\vec v_{\rm cm} = 0$ for all time, so over any interval the displacements obey

m_1\,\Delta x_1 + m_2\,\Delta x_2 = 0 \qquad (3)

where: $\Delta x_i$ — each body's displacement relative to the ground, signs included. Displacements trade in inverse mass ratio — the moving statement of the see-saw. Equation (3) plus one relative-motion fact (how far the man walks along the boat) solves the entire boat family (derivation B).

the phrase in the problem	the move
"find the velocity of the COM"	$\vec v_{\rm cm} = \sum m\vec v / M$ — vector sum, then divide
"smooth surface, starts at rest, things move internally"	anchor: $m_1\Delta x_1 = -m_2\Delta x_2$
"man walks $L$ relative to the boat"	boat shifts $\frac{mL}{M+m}$ back; man $\frac{ML}{M+m}$ forward
"shell explodes mid-flight"	COM continues the original parabola (until first landing)
"one fragment lands at …, where does the other land?"	COM lands at the intact range; lever the positions
"blocks fly apart from a spring"	anchor + inverse-ratio displacements & speeds

Derivation AThe two system laws — momentum incarnate and the external-only second law

We derive equations (1) and (2) from the definitions, using finite intervals and algebra only. The strategy: the COM position is a mass-weighted average, so its change over any interval is the mass-weighted average of the changes; divide by the interval and speeds appear. Then feed the result through the conservation article's system law. Watch internal forces never even appear — they were exorcised once, in the conservation article, and every consequence here inherits the immunity.

Step 1. Over any interval $\Delta t$ , each particle moves $\Delta x_i$ . The COM's displacement, straight from the weighted-average definition:

\Delta x_{\rm cm} = \frac{\sum m_i\,\Delta x_i}{M}

Step 2. Divide by $\Delta t$ : each $\Delta x_i/\Delta t$ is that particle's velocity (average over the interval; instantaneous as the interval shrinks):

v_{\rm cm} = \frac{\sum m_i v_i}{M} \;\;\Rightarrow\;\; M v_{\rm cm} = \sum m_i v_i = P

Why this identity is the bridge: every momentum statement in the chapter can now be read as a COM-motion statement — "momentum conserved" and "COM coasts" are the same sentence.

Step 3. The conservation article proved $\Delta P = F_{\rm ext}\,\Delta t$ (internal pairs cancel). Substitute $P = Mv_{\rm cm}$ :

M\,\Delta v_{\rm cm} = F_{\rm ext}\,\Delta t \;\;\Rightarrow\;\; F_{\rm ext} = M\,a_{\rm cm}

Step 4. The two readings for exams. $F_{\rm ext} = 0$ : $v_{\rm cm}$ constant — collisions and explosions inside leave the COM's cruise untouched. $F_{\rm ext} = Mg$ (projectile system): $a_{\rm cm} = g$ — the COM flies the standard parabola whatever the fragments do to each other. Verified: single particle: (2) reduces to plain Newton ✓. Perfectly inelastic collision: common velocity $= (m_1u_1+m_2u_2)/M = v_{\rm cm}$ ✓ — sticking is "everyone agrees to travel with headquarters". Vectors throughout: each component obeys its own copy.

Go deeper · Derivation B · 1 minThe man on the boat — displacements from a fixed anchor

Derivation BThe man on the boat — displacements from a fixed anchor

We derive the standard results for a man (mass $m$ ) walking a length $L$ relative to a boat (mass $M$ ) on frictionless water, starting at rest: how far the boat drifts and how far the man actually travels. The strategy: the anchor equation (3) gives one relation between the two ground displacements; the phrase "relative to the boat" gives the second; solve the linear pair. Watch the relative-displacement line — it is where most marks die, because $L$ is the man's motion over the deck, not over the water.

Step 1. Signs: man walks in the $+$ direction. Let the man's ground displacement be $x_m$ and the boat's be $x_b$ (expected negative). No horizontal external force (frictionless water), starting at rest — the anchor holds:

m\,x_m + M\,x_b = 0

Step 2. The relative statement: he advanced $L$ along the deck, and the deck itself moved $x_b$ :

x_m = x_b + L

Why: displacement relative to the ground = displacement relative to the boat + the boat's own displacement — the relative-motion bookkeeping from kinematics, reused.

Step 3. Substitute and solve:

m(x_b + L) + Mx_b = 0 \;\Rightarrow\; x_b = -\frac{mL}{M+m}, \qquad x_m = \frac{ML}{M+m}

Step 4. Audits built into the answer: the two displacements are opposite ✓; their mass-weighted sum is zero ✓; they add (in magnitude) to $L$ ✓ — the walk is shared between man-forward and boat-backward, in inverse mass ratio. Verified: limits — $M \to \infty$ (walking on a pier): boat shift $\to 0$ , man advances the full $L$ ✓; $M \to 0$ (walking on a raft of nothing): the "boat" flees backward with nearly all of $L$ and the man barely advances ✓; $m = M$ : each moves $L/2$ ✓. Numbers (worked example 3): $60$ kg man, $120$ kg boat, $L = 3$ m: boat $-1.0$ m, man $+2.0$ m.

Figure

The boat problem — one walk, two displacements, one fixed anchor

The projectile that exploded — the landing-point lever

The parabola immunity has a beautiful cash value. A shell fired for range $R$ explodes at the apex into two pieces. Both pieces, at the apex, have zero vertical velocity (one "drops dead", the other flies on horizontally) — so both hit the ground at the same instant, and at that instant the COM, still flying its untouched parabola, is arriving at $x = R$ . The landing positions must therefore lever about $R$ : $m_1x_1 + m_2x_2 = (m_1+m_2)R$ . One fragment's landing spot fixes the other's by see-saw arithmetic — no trajectory integration, no explosion details. (The fine print: the trick reads the COM's position at the moment of touchdown; if the fragments land at different times, the first landing introduces a new external force — the ground's — and the naive lever needs care. Apex explosions with horizontal fragments are the setter's way of keeping it legal.)

Limiting cases

case	result	meaning
single particle	(2) = plain Newton	the system law contains the old one
$F_{\rm ext} = 0$	$v_{\rm cm}$ constant	collisions/explosions leave the cruise untouched
$F_{\rm ext} = 0$ , starts at rest	COM fixed forever	the anchor: displacement see-saw (3)
walking man, $M \to \infty$	boat shift $\to 0$	walking on a pier — the everyday limit
projectile system	$a_{\rm cm} = g$ regardless of internal forces	the parabola immunity
one fragment touches down	ground force joins $F_{\rm ext}$	immunity expires — the trick's legality gate

Common trapHe walked forward, so the system moved forward

The seductive error of the whole topic. The man's muscles are internal to the man-plus-boat system: they can redistribute the system's parts but cannot move its centre of mass, which — frictionless water, started at rest — is nailed to its spot. The man advances only because the boat retreats, the two displacements trading in inverse mass ratio; the "forward motion" of the man is exactly paid for, kilogram-metre for kilogram-metre. Every costume of this error is examined: the astronaut who "swims" in vacuum (nothing to push that is outside her), the fan bolted to a sailboat blowing into its own sail (internal pair — no net drive... a subtlety: with the sail catching the airflow the momentum mostly recirculates; genuinely external exhaust is what a rocket uses), the bird flying inside a sealed, frictionless box-car (the air pushes the bird up, the bird pushes the air down, the ground feels the same total weight). The un-trapping question is always the same: is anything pushing on the system from outside? Walking on land: yes — the Earth's friction, which is why your COM genuinely accelerates. On the frictionless boat: no — so the headquarters stays, whatever the legs do.

Worked examples

1 · Board level — the spectator velocity (CBSE pattern, 3 marks)

A $3.0\ \mathrm{kg}$ cart moves at $4.0\ \mathrm{m/s}$ toward a stationary $1.0\ \mathrm{kg}$ cart on a smooth track. Find the velocity of the centre of mass (a) before they collide, (b) after they collide and couple, (c) after they collide elastically instead.

Step 1. (a) $v_{\rm cm} = \dfrac{(3.0)(4.0) + 0}{4.0} = \boxed{3.0\ \mathrm{m/s}}$ , toward the light cart.

Step 2. (b) The collision is internal; no external horizontal force acts — $v_{\rm cm}$ is a spectator: $\boxed{3.0\ \mathrm{m/s}}$ . (Check: coupled carts move at $12/4 = 3.0$ m/s — the common velocity IS the COM velocity.)

Step 3. (c) Same answer, $\boxed{3.0\ \mathrm{m/s}}$ — elasticity changes how the parts share the motion (elastic forms give $2.0$ and $6.0$ m/s; check: $\frac{3(2)+1(6)}{4} = 3.0$ ✓), never the headquarters' cruise.

Step 4. The sentence worth a mark: the COM velocity is unchanged by any collision between the carts, because the collision forces are internal.

2 · Board level — where the other half lands (CBSE/ICSE pattern)

A shell fired for a range of $2.0\ \mathrm{km}$ explodes at the top of its flight into two equal pieces. One piece falls vertically from rest (relative to the ground) and lands directly below, at $1.0\ \mathrm{km}$ . Where does the other land?

Step 1. Both pieces leave the apex with zero vertical velocity (one at rest, one horizontal) → both land at the same instant, and the lever is legal.

Step 2. The COM, immune to the internal explosion, lands where the intact shell would: $x_{\rm cm} = 2.0$ km.

Step 3. Lever with equal masses: $\dfrac{x_1 + x_2}{2} = 2.0 \Rightarrow x_2 = 4.0 - 1.0 = \boxed{3.0\ \mathrm{km}}$ from the gun.

Step 4. Cross-check by momentum: at the apex the shell moved horizontally at $v$ ; conservation gives the survivor $2v$ , so it covers double the remaining $1.0$ km of range: lands at $1.0 + 2.0 = 3.0$ km ✓ — two routes, one answer.

3 · Competitive level — the man on the boat (JEE-Main pattern, ~3 minutes)

A $60\ \mathrm{kg}$ man stands at one end of a $120\ \mathrm{kg}$ , $3.0\ \mathrm{m}$ boat at rest on still (frictionless) water. He walks to the other end. Find (a) the boat's displacement, (b) the man's displacement relative to the water.

Step 1. Anchor legality: no horizontal external force, starts at rest → COM fixed. He walks $L = 3.0$ m relative to the boat.

Step 2. (a) Boat: $|x_b| = \dfrac{mL}{M+m} = \dfrac{60 \times 3.0}{180} = \boxed{1.0\ \mathrm{m}}$ backward.

Step 3. (b) Man: $x_m = \dfrac{ML}{M+m} = \dfrac{120 \times 3.0}{180} = \boxed{2.0\ \mathrm{m}}$ forward. (Audit: $2.0 + 1.0 = 3.0 = L$ ✓ and $60(2.0) = 120(1.0)$ ✓.)

Step 4. The disciplined read-back: had the problem said he walks $3.0$ m relative to the water, the boat's shift would instead come straight from the see-saw: $x_b = -\frac{60}{120}(3.0) = -1.5$ m. The reference phrase decides the equations — read it twice. Timing note: ~3 min. Write the two lines (anchor + relative statement) before any algebra; the formulas $mL/(M+m)$ and $ML/(M+m)$ are worth memorizing, but the phrase "relative to the boat/water/ground" is worth more — the mark scheme's fork lives there.

4 · Competitive level — two blocks and a spring, anchored (JEE-Main/Adv flavour, ~3 minutes)

Blocks of $1.0\ \mathrm{kg}$ and $2.0\ \mathrm{kg}$ rest on a smooth floor with a compressed spring held between them. The system is released. (a) When the $1.0\ \mathrm{kg}$ block has moved $30\ \mathrm{cm}$ , how far has the other moved? (b) At the instant the light block moves at $2.0\ \mathrm{m/s}$ , how fast is the heavy one? (c) Where is the COM after both blocks have left the spring and slid far away?

Step 1. Legality: horizontal external force zero, starts at rest → the COM is an anchor, forever.

Step 2. (a) Displacement see-saw: $1.0 \times 0.30 = 2.0 \times |x_2| \Rightarrow |x_2| = \boxed{0.15\ \mathrm{m}}$ , opposite direction.

Step 3. (b) Velocity see-saw (same law, per unit time): $1.0 \times 2.0 = 2.0 \times v_2 \Rightarrow v_2 = \boxed{1.0\ \mathrm{m/s}}$ , opposite. (Equivalently: total momentum stays zero.)

Step 4. (c) Exactly where it started — $\boxed{\text{the COM never moves}}$ , however far the blocks separate. The blocks recede forever, in inverse-ratio lockstep, about a point of empty floor that neither occupies. (Energy postscript: the spring's stored energy splits inversely with mass — the light block carries twice the KE of the heavy one, the recoil theorem from the conservation article.) Timing note: ~3 min. Every part is the same one-line see-saw — displacement, velocity, position. JEE-Advanced dresses this skeleton with a wedge, a rope, or a pulley; the anchor line " $F_{\rm ext} = 0$ + starts at rest ⇒ $\sum m\Delta x = 0$ " survives every costume. Check part (a) against (b): displacements and speeds carry the SAME ratio at every instant, because both are integrals of the same zero-sum momenta.

Exam traps

Relative-to-what. "Walks 3 m" relative to the boat, the water, or the ground are three different problems. The relative-to-boat phrasing needs derivation B's second line; misreading it flips the answer by the mass ratio — and the options are built to catch exactly that.

Anchor without its certificate. The COM stays fixed only if (i) net external force along that direction is zero AND (ii) it started at rest. A system already drifting keeps drifting ( $v_{\rm cm}$ constant ≠ zero); a system with friction underneath has no anchor at all.

Fragment-landing lever used across unequal landing times. The lever reads the COM at one instant. If one fragment lands early, the ground's normal force joins the external books and the naive "COM lands at R" needs re-examination. Apex explosions with zero vertical velocities (both land together) are the safe, and standard, case.

Internal forces smuggled into (2). Tension between the blocks, the explosion, the man's push on the deck — none of them belong in $F_{\rm ext}$ . If your "external" list contains a force exerted by one system member on another, redraw the boundary.

Forgetting the vertical books. On the boat, gravity and buoyancy do act — vertically, and they cancel. The anchor argument runs per component: horizontal momentum has no external force, so the horizontal COM is pinned; nothing is claimed vertically.

COM velocity computed as a plain average. $v_{\rm cm}$ is mass-weighted: $(\sum mv)/M$ , signs included. Two equal-and-opposite momenta give $v_{\rm cm} = 0$ no matter how fast the parts fly.

Claiming rotation moves the COM. Spin redistributes the parts around the COM; equation (2) is deaf to it. A diver somersaulting mid-air changes shape and spin at will — her COM's parabola was fixed at takeoff and cannot be renegotiated before splashdown.

"The spring pushed the blocks, so something external acted." The spring is inside the system. Its two pushes are a third-law pair; the anchor holds. External means across the boundary — and you drew the boundary.

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Basel, St Petersburg & Berlin

Newton proved, as Corollary IV of his laws, that a system's centre of mass "either is at rest or moves uniformly in a right line" when left alone — the anchor of this article, already in the Principia. But it was Leonhard Euler who made the COM the load-bearing wall of mechanics. In papers from the 1740s and his 1765 Theoria motus corporum solidorum, Euler recast dynamics as two independent laws for any body or system whatsoever: the rate of change of momentum equals the external force (acting as if at the centre of mass — this article's equation 2, still called Euler's first law), and the rate of change of angular momentum equals the external torque (the rotation chapter's coming attraction). That clean split — translation of the COM, rotation about the COM — is how every mechanics problem from tumbling wrenches to tumbling satellites has been organized since. Blind for the last third of his life, Euler dictated roughly half of his 850 works, including much of the mechanics that every "COM moves as if all forces acted there" exam sentence quietly cites.

History

1687

— Newton's Corollary IV: a system's centre of mass coasts uniformly when external forces are absent.

1740

s–1765 — Euler's laws of mechanics:

F_{\rm ext} = Ma_{\rm cm}

enthroned as the first law of any body's motion, rigid or not.

1846

— Le Verrier predicts Neptune from Uranus's wobbles — celestial bookkeeping about shared centres of mass.

1968

— Apollo astronauts experience the anchor personally: no swimming in vacuum; attitude control by reaction wheels and thrusters.

1995

— 51 Pegasi b: the first exoplanet found by watching a star orbit the system's centre of mass.

Going further

Texts. NCERT Physics Class 11, Ch. 7 (motion of the centre of mass) · HC Verma, Concepts of Physics Vol. 1, Ch. 9 (the boat and explosion problems) · Kleppner & Kolenkow, Ch. 4 — and Newton's Corollary IV for the seventeenth-century original.

Provenance. All results are standard. A verifier should recompute: $P = Mv_{\rm cm}$ and $F_{\rm ext} = Ma_{\rm cm}$ from the weighted average; the boat pair $x_b = -mL/(M+m)$ , $x_m = ML/(M+m)$ with its three audits; Example 1's spectator $3.0$ m/s in all three parts; Example 2's $3.0$ km by both the lever and the momentum route; Example 3's $1.0$ m / $2.0$ m; Example 4's $0.15$ m and $1.0$ m/s; the walking-Earth recoil estimate ( $\sim10^{-22}$ m).

Prerequisites. Locating the Centre of Mass · Conservation of Linear Momentum. Leads to. Rocket Propulsion & Variable Mass · Man on Boat / Plank on Ice (archetype) · Explosions, Recoil & Fragments (archetype) · Systems of Particles & Centre of Mass (UG deep dive).

Why exactly can't internal forces move the centre of mass?

Because the COM's motion is the total momentum in disguise ( $P = Mv_{\rm cm}$ ), and internal forces are momentum-neutral: each one arrives with its third-law partner, equal and opposite at every instant, so their impulses cancel from the total no matter how violent, prolonged or clever they are. Muscles, springs, explosions and magnets can shuffle momentum between members — that's what they do — but the ledger's bottom line only responds to forces from outside the boundary. The COM, being the bottom line's avatar, is unbribable from within.

Then how does walking move ME — my muscles are internal to my body?

Your muscles are internal to you, but the force that accelerates your COM is external to you: friction from the ground. Your leg muscles' real job is to make your foot push backward on the Earth so that the Earth pushes forward on you — that reaction is the external force in $F_{\rm ext} = Ma_{\rm cm}$ . Put the system boundary around you-plus-Earth instead and the bookkeeping flips context: those friction forces become internal, and the you-plus-Earth COM stays put — the planet recoils by about $10^{-22}$ m per stride. Same physics, two boundaries; the skill is always naming whose outside is doing the pushing.

In the boat problem, does the boat drift back forever or stop when the man stops?

It stops when he stops — and this is a good self-test of the mechanism. While he walks, momentum balances: man forward, boat backward. The instant he halts relative to the boat, the two must share one velocity; but the total momentum has been zero throughout, so that shared velocity is zero — everything is at rest again, displaced but stationary. (His stopping push on the deck is the internal impulse that kills the boat's drift.) With real water resistance the final rest state is the same but the displacements differ slightly — resistance is an external force and the anchor certificate fails; exam problems say "neglect water resistance" precisely to keep the anchor legal.

How can the COM land at the full range if the explosion blasted the pieces everywhere?

Because gravity is the only external force in flight, and it steers the COM along the same parabola whether the shell is whole or in fragments — the explosion redistributed momentum among the pieces but could not alter their total. The one legality gate is time: the lever "COM lands at $R$ " reads the fragment positions at the instant they are all landing, which is guaranteed when all pieces leave the apex with zero vertical velocity (they share the same fall time). If one lands early, the ground starts pushing on it — a new external force on the system — and the remaining flight needs honest bookkeeping. Setters keep to the legal case; examiners test whether you know why it's legal.

Is v_cm constant during the collision itself, or only before and after?

Throughout — including the crunch. The system law $F_{\rm ext} = Ma_{\rm cm}$ holds at every instant, and the collision forces, however monstrous, are internal at every instant. On a smooth track the COM velocity of two colliding carts is the same before, during and after contact; plot both carts' positions against time and the mass-weighted average traces an unbroken straight line through the drama. This is also the deepest way to see why a perfectly inelastic collision ends at the COM velocity: sticking means "everyone agrees to move with headquarters", and headquarters never changed speed.

Does spinning or tumbling affect the COM's path?

No — and the immunity is the foundation of the entire rotation chapter. External forces steer the COM (Euler's first law); external torques steer the spin about the COM (his second); the two books are kept separately. A thrown wrench's COM flies a clean parabola while the wrench gyrates around it; a diver can trade tuck for layout, changing her spin rate dramatically, without moving her COM's arc by a millimetre. What internal rearrangement can do is change how the body is distributed about the COM — which changes the rotation (the ice-skater effect) but never the translation. When the rotation chapter opens, this factorization is the map.