Motion of centre of mass — external force determines a_CM: Question bank

Figure — Motion of centre of mass — external force determines a_CM

True or false — justify

Diagram for the shell, skaters, boat and jump traps below:

True or false: An exploding shell's centre of mass changes its trajectory the instant it explodes.

False — the explosion is internal, so it cancels in pairs; only gravity (external, pointing

- y

) acts, so

a_{C M} = (0, - g)

before and after, and the CM stays on the same parabola until a fragment lands.

True or false: If total momentum is conserved, the CM must be at rest.

False — conserved momentum means

V_{C M}

is constant, which could be any fixed non-zero value; it is at rest only if it started at rest.

True or false: The centre of mass must lie inside the material of the object.

False — for a ring the CM sits at the hole's centre (empty space), for a boomerang it lies off the arms; the CM is a computed average point, not a chunk of matter (see the ring figure below).

True or false: Two skaters pushing apart on frictionless ice move their common CM.

False — the push is an internal

+ f, - f

pair and

F_{e x t, x} = 0

, so

X_{C M}

never changes no matter how hard they shove; skater 1 goes

- x

, skater 2 goes

+ x

, but the mass-weighted centre stays fixed.

True or false: A stronger internal force can shift the CM more than a weak one.

False — magnitude is irrelevant; every internal force has an equal-and-opposite partner, so the vector sum is exactly zero regardless of size.

True or false: For a system of equal masses, the CM is the geometric midpoint.

True — with equal

m_{i}

the weighting

\frac{\sum m _{i} r _{i}}{\sum m _{i}}

reduces to a plain positional average

\frac{1}{N} \sum r_{i}

, which is the geometric centre.

True or false: A jumping person can lift their own CM using only muscles, with no floor.

False — muscles are internal; the upward (

+ y

) launch needs the floor's external normal reaction. Once airborne,

F_{e x t} = (0, - M g)

and the CM only obeys gravity.

True or false: In a totally inelastic collision the CM keeps moving even though kinetic energy is lost.

True — momentum (hence

V_{C M}

) is conserved with no external force, while internal deformation drains kinetic energy; the CM glides on unchanged.

Spot the error

Reference schematic for the skater / boat / rocket errors:

Find the flaw: "A rocket in deep space can't accelerate its CM because ejecting fuel is an internal force."

The fuel leaves the system — once ejected it is external, so the exhaust is a genuine external force on the remaining rocket; see Rocket Propulsion.

Find the flaw: "The CM of a feather-and-anvil pair sits at their midpoint."

Wrong — plug into

x_{C M} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}

: with the anvil's

m_{2} ≫ m_{1}

the fraction is dominated by

m_{2} x_{2}

, so the CM sits almost on top of the anvil; only equal masses give a midpoint.

Find the flaw: "Since

P = M V_{C M}

, doubling every particle's speed doubles the CM speed only if masses are equal."

The relation holds for any masses; doubling all

v_{i}

doubles every term of

\sum m_{i} v_{i}

, hence doubles

V_{C M}

regardless of the mass distribution.

Find the flaw: "Friction from the ice is internal to the two-skater system, so it can't affect the CM."

Friction is exerted by the ice, which is outside the two-skater system, so it is external and would accelerate the CM — we only get a fixed CM by assuming it negligible.

Find the flaw: "The man walking on a boat moves the CM forward because he moves forward."

With

F_{e x t, x} = 0

the CM cannot move; from

m (+ Δ x_{man}) + M (- Δ x_{b o a t}) = 0

the boat slides in

- x

exactly enough to keep

X_{C M}

fixed while the man advances in

+ x

Find the flaw: "

F_{e x t} = M a_{C M}

needs each particle to feel the same external force."

No — it needs only the net external force; individual particles can feel wildly different externals, and only their vector sum drives the CM.

Why questions

Why do internal forces vanish from the CM equation even though they clearly move individual particles?

For any pair, Newton's 3rd law gives

f_{ij} = - f_{j i}

, so the pair sum is

0

; summed over all pairs the net internal force is zero — they rearrange the pieces but never shift the mass-weighted average (Newton's Third Law).

Why is the CM mass-weighted instead of a plain positional average?

\frac{\sum m _{i} r _{i}}{\sum m _{i}}

each position is scaled by its mass, so a heavier particle contributes more and drags the balance point toward it; a plain average

\frac{1}{N} \sum r_{i}

ignores this and mislocates the CM toward light parts.

Why does analysing collisions in the CM frame simplify them?

In the CM frame total momentum is zero, so the two objects always carry equal-and-opposite momenta

+ p

and

- p

before and after, making the algebra symmetric; see Collisions (Elastic & Inelastic).

Why can we treat a wobbling, spinning extended body as a single point when applying

F = M a

Sum Newton's 2nd law over every mass element; all internal stresses cancel in

+ f, - f

pairs, leaving

F_{e x t} = M a_{C M}

— the CM moves as one particle of mass

M

(foundation of Rigid Body Dynamics).

Why does

F_{e x t} = 0

immediately give conservation of momentum?

F_{e x t} = 0 \Rightarrow M a_{C M} = 0 \Rightarrow V_{C M}

constant, and since

P = M V_{C M}

with

M

fixed,

P

is constant too; this is Conservation of Linear Momentum.

Edge cases

The ring/boomerang "CM in empty space" edge case:

Edge case: A uniform ring — where is its CM?

By symmetry every mass element at

+ r

is matched by one at

- r

, so the integral

\int r d m = 0

and the CM sits at the geometric centre — inside the hole, with no material there.

Edge case: What is the CM velocity right at the top of the exploding shell's flight, the instant before it bursts?

At the apex

v_{y} = 0

(gravity has just cancelled the upward motion) while

v_{x}

is unchanged, so

V_{C M} = (v_{x}, 0)

— purely horizontal; the explosion leaves this and its future evolution untouched.

Edge case: A body with varying density (denser at one end) — does the discrete formula still work?

Use the continuous form

R_{C M} = \frac{1}{M} \int r ρ d V

; the CM shifts toward the denser end, exactly as heavy discrete masses pull the balance point.

Edge case: A single free particle — where is its CM and what governs it?

The CM is the particle itself, and

F_{e x t} = M a_{C M}

collapses to ordinary

F = m a

; the framework is consistent at

N = 1

Edge case: Two equal masses at the same point — is the CM defined?

Yes — it sits at that shared point; the formula stays valid as long as total mass

M > 0

Edge case: What happens to the CM the moment the last fragment of the exploded shell hits the ground?

The ground now supplies an external normal force, so the CM equation still holds but

F_{e x t}

is no longer just

(0, - M g)

— the clean parabola argument ends there.

Edge case: A closed system where every particle has zero velocity but forces act internally — does the CM stay put?

Yes momentarily and forever if

F_{e x t} = 0

V_{C M} = 0

stays zero because internal forces never contribute to

a_{C M}

Edge case: Can

V_{C M}

be non-zero while every individual particle momentarily has zero velocity?

No —

P = \sum m_{i} v_{i} = M V_{C M}

, so if all

v_{i} = 0

then

V_{C M} = 0

exactly at that instant.

Connections

The map below shows how each trap traces back to one of two roots: the definition of the mass-weighted CM, or Newton's 3rd law killing internal forces.

Conservation of Linear Momentum — the $F_{e x t} = 0$ special case behind half these traps.
Newton's Third Law — why internal forces always cancel.
Collisions (Elastic & Inelastic) — the CM-frame simplification questioned above.
Rocket Propulsion — the "internal vs ejected" boundary case.
Rigid Body Dynamics — CM as the point where $F = M a$ holds for extended bodies.