Classical-Mechanics · Unit 12 · Video 2 · Interactive Practice

Where You Draw the Line: Why Only External Forces Change a System's Momentum

IKey Formulas

Formula	Name	What you need
$\vec{F}_i = \vec{F}_i^{\,\mathrm{ext}} + \displaystyle\sum_{j \neq i} \vec{F}_{j,i}$	Force on the $i$ -th member, split by boundary	A chosen system boundary
$\vec{F}_{j,i} + \vec{F}_{i,j} = \vec{0} \;\Longrightarrow\; \vec{F}^{\,\mathrm{int}} = \vec{0}$	Third law kills the internal sum	Both partners inside the boundary
$\vec{F}^{\,\mathrm{ext}} = \dfrac{d\vec{p}_{\mathrm{sys}}}{dt}, \qquad \vec{p}_{\mathrm{sys}} = \displaystyle\sum_{i=1}^{N} \vec{p}_i$	Second Law for a system	Every force crossing the boundary
$\Delta\vec{p}_{\mathrm{sys}} = \displaystyle\int_{t_0}^{t_f} \vec{F}^{\,\mathrm{ext}}\,dt \equiv \vec{I}_{\mathrm{sys}}$	Impulse on the system	The external force history

Key Insight: "External" and "internal" are not properties of a force — they are properties of your boundary. The spring still pulls on the cart with exactly the same $\vec{F}_{\mathrm{spring,cart}}$ either way; swallow its third-law partner and that force simply stops counting toward $\Delta\vec{p}_{\mathrm{sys}}$ .

IIMoving the Boundary

Nothing physical changes when the boundary moves — but the list of forces that count does.

💡 Enclose everything and only the Earth and the floor are left reaching in — which is why "the momentum of an isolated system is constant" is a statement about the boundary, not about the forces.

IIIThe Double Sum Cancels in Pairs

For $N$ particles the internal forces fill an $N \times N$ grid off the diagonal, and every cell has a mirror twin.

Particles N 5 Pairs cancelled 0

IVOnly the External Impulse Moves $\vec{p}_{\mathrm{sys}}$

Two carts shove each other while one force reaches in from outside — which of them bends the total?

Time 0.00 s External force 0.0 N

VQuiz Questions

Problem 1 · Counting What Crosses

Given: the cart on the incline, pulled up-slope by a spring whose far end is bolted to a sensor. The system is the cart alone — find how many external forces act on it and what $\vec{F}^{\,\mathrm{ext}}$ is.

✅ Correct! Three partners sit outside — the spring, the Earth, the plane — so all three forces cross the boundary and

\vec{F}^{\,\mathrm{ext}}

is their vector sum.

❌ Only one of the three. Gravity and the plane's contact force also come from outside the boundary; "external" does not mean "applied by hand", it means the partner is a non-member.

❌ A pair only cancels when both its members are inside. Here every reaction force (

\vec{F}_{\mathrm{cart,spring}}

\vec{F}_{\mathrm{cart,earth}}

\vec{F}_{\mathrm{cart,plane}}

) acts on an outside object, so nothing cancels within the system.

❌ Half of those act on the surroundings.

\vec{F}^{\,\mathrm{ext}}

sums only the forces acting on system members; the reaction forces act on the spring, the Earth and the plane.

Show solution

Step 1 — list the interactions of the cart. The cart touches or is pulled by exactly three partners: the spring, the Earth, the inclined plane.

Step 2 — locate each partner. The boundary encloses the cart only, so all three partners lie in the surroundings. Every one of their forces on the cart therefore crosses the boundary:

\vec{F}^{\,\mathrm{ext}} = \vec{F}_{\mathrm{spring,cart}} + \vec{F}_{\mathrm{earth,cart}} + \vec{F}_{\mathrm{plane,cart}}

Step 3 — why the reactions are not in the list. Each of these has a third-law partner ( $\vec{F}_{\mathrm{cart,spring}}$ and so on), but those act on objects outside the system. A pair cancels inside $\vec{F}^{\,\mathrm{ext}}$ only when both of its forces act on members, so with a one-object system no cancellation is possible.

With $\vec{F}^{\,\mathrm{ext}} = d\vec{p}_{\mathrm{sys}}/dt$ and $\vec{p}_{\mathrm{sys}} = m_{\mathrm{cart}}\vec{v}_{\mathrm{cart}}$ , this is just the ordinary free-body statement — the system language has not changed the physics yet.

Problem 2 · The Boundary Expands

Given: the same setup, but now the boundary is redrawn around the cart and the spring, with the spring's upper end attached to the sensor bolted at the top of the incline — find the correct expression for $\vec{F}^{\,\mathrm{ext}}$ .

✅ Correct! The spring–cart pair is now internal and cancels, while the sensor — still outside — pulls on the spring, a member, so its force joins the external list.

❌ That list is from the old boundary. With the spring inside,

\vec{F}_{\mathrm{spring,cart}}

has its partner

\vec{F}_{\mathrm{cart,spring}}

inside too, so the two sum to

\vec{0}

and neither can appear in

\vec{F}^{\,\mathrm{ext}}

❌ You dropped the internal pair correctly, but the boundary also gained a frontier. Enclosing the spring brings its other interaction along: the sensor now pulls on a member of your system.

❌ Right interaction, wrong force of the pair.

\vec{F}_{\mathrm{spring,sensor}}

acts on the sensor, which is outside;

\vec{F}^{\,\mathrm{ext}}

collects only forces acting on members — here

\vec{F}_{\mathrm{sensor,spring}}

❌ Not quite. Sort every force by two questions: does it act on a member, and is its partner also a member? Keep it only if the answers are yes and no.

Show solution

Step 1 — the spring–cart interaction goes internal. Both of its forces now act on members, and by the third law

\vec{F}_{\mathrm{spring,cart}} + \vec{F}_{\mathrm{cart,spring}} = \vec{0}

so this interaction contributes nothing to $\vec{F}^{\,\mathrm{ext}}$ . It has not vanished — the cart still feels it, and it still redistributes momentum within the system — it simply cannot change the total.

Step 2 — the surviving external partners. The Earth and the plane are still outside and still act on the cart. Enclosing the spring exposes a new frontier: the sensor, outside, pulls the spring's upper end. Hence

\vec{F}^{\,\mathrm{ext}} = \vec{F}_{\mathrm{earth,cart}} + \vec{F}_{\mathrm{plane,cart}} + \vec{F}_{\mathrm{sensor,spring}}

Step 3 — the bookkeeping rule. A force belongs in $\vec{F}^{\,\mathrm{ext}}$ exactly when it acts on a member and its third-law partner acts on a non-member. Expanding a boundary always deletes the pairs it swallows and promotes the interactions at the new edge.

Problem 3 · Internal Traffic vs the Total

Given: blocks $A$ ( $2.0\ \mathrm{kg}$ ) and $B$ ( $3.0\ \mathrm{kg}$ ) on a frictionless horizontal track with a compressed spring between them. A constant external force of $4.0\ \mathrm{N}$ along $+x$ acts on $A$ for $0.50\ \mathrm{s}$ ; over that interval $A$ 's momentum increases by $3.2\ \mathrm{kg\,m\,s^{-1}}$ . Take the system to be $A$ , $B$ and the spring — find $\Delta \vec{p}_{\mathrm{sys}}$ and $\Delta \vec{p}_B$ .

What is the $x$ -component of $\Delta \vec{p}_{\mathrm{sys}}$ ?

What is the $x$ -component of $\Delta \vec{p}_B$ ?

✅ Correct! The external impulse fixes the total at

+2.0\ \mathrm{kg\,m\,s^{-1}}

; since

A

took

+3.2

, the spring must have handed

B

a net

-1.2

— internal forces only shuffle momentum between members.

❌ The spring is internal, but the $4.0\ \mathrm{N}$ push is not. Momentum is constant only when

\vec{F}^{\,\mathrm{ext}} = \vec{0}

; here the external impulse is

(4.0)(0.50) = 2.0\ \mathrm{N\,s}

❌ You divided by the time. Impulse accumulates:

\Delta p = F^{\mathrm{ext}}\Delta t

, not

F^{\mathrm{ext}}/\Delta t

. Check the units —

\mathrm{N\,s}

, not

\mathrm{N\,s^{-1}}

❌ That is $\Delta p_A$ , not $\Delta p_{\mathrm{sys}}$ .

A

can gain more than the external impulse precisely because the spring is pushing it — that extra momentum is taken from

B

❌ Check the external impulse. Only the

4.0\ \mathrm{N}

force crosses the boundary horizontally, so

\Delta p_{\mathrm{sys}} = (4.0\ \mathrm{N})(0.50\ \mathrm{s})

❌ Use the total as the constraint.

\Delta p_A + \Delta p_B = \Delta p_{\mathrm{sys}}

, so

\Delta p_B = 2.0 - 3.2

— a negative number, because

B

is pushed backwards by the spring.

Show solution

Step 1 — sort the forces. The spring's two forces both act on members, so they are internal and sum to $\vec{0}$ . Gravity and the track's normal force are external but cancel vertically. Horizontally only the applied $4.0\ \mathrm{N}$ crosses the boundary:

F^{\mathrm{ext}}_x = 4.0\ \mathrm{N}

Step 2 — integrate for the system.

\Delta p_{\mathrm{sys},x} = \int_{t_0}^{t_f} F^{\mathrm{ext}}_x\,dt = (4.0\ \mathrm{N})(0.50\ \mathrm{s}) = 2.0\ \mathrm{kg\,m\,s^{-1}}

Notice that neither mass appears — the external impulse alone sets the total.

Step 3 — split the total. By definition $\vec{p}_{\mathrm{sys}} = \vec{p}_A + \vec{p}_B$ , so

\Delta p_{B,x} = \Delta p_{\mathrm{sys},x} - \Delta p_{A,x} = 2.0 - 3.2 = -1.2\ \mathrm{kg\,m\,s^{-1}}

Step 4 — read it physically. The spring pushed $A$ forward with an impulse of $+1.2\ \mathrm{N\,s}$ and pushed $B$ backward with $-1.2\ \mathrm{N\,s}$ : equal and opposite, exactly cancelling in the total. Internal forces are free to move momentum around inside a system; they can never add any.

Problem 4 · Transfer — One Force, Two Boundaries

Given: a $0.50\ \mathrm{kg}$ ball falls freely for $\Delta t = 0.40\ \mathrm{s}$ near the Earth's surface ( $g = 9.8\ \mathrm{m\,s^{-2}}$ ; ignore air) — find $\Delta \vec{p}_{\mathrm{sys}}$ for two different choices of system.

System = the ball alone

System = the ball and the Earth

✅ Excellent! The same gravitational interaction is external to the ball alone and internal to ball-plus-Earth:

mg\Delta t

downward in one bookkeeping, exactly

\vec{0}

in the other.

❌ Those are newtons.

mg = 4.9\ \mathrm{N}

is the force; the impulse it delivers is

mg\Delta t

, which still needs the

0.40\ \mathrm{s}

❌ Not for this boundary. With only the ball inside, the Earth is a non-member, so

\vec{F}_{\mathrm{earth,ball}}

crosses the boundary and is external — its partner is nowhere in the system to cancel it.

❌ You divided instead of multiplying.

\Delta p = mg\,\Delta t = (0.50)(9.8)(0.40)

; dividing by

g

leaves units of

\mathrm{kg\,s}

, which is not a momentum.

❌ Check the external impulse. The only force on the ball is gravity, constant at

mg = 4.9\ \mathrm{N}

downward, acting for

0.40\ \mathrm{s}

❌ Both partners are now inside.

\vec{F}_{\mathrm{earth,ball}}

and

\vec{F}_{\mathrm{ball,earth}}

are a third-law pair acting on two members, so they cancel in

\vec{F}^{\,\mathrm{ext}}

and nothing is left to change the total.

Show solution

Boundary 1 — the ball alone. Gravity's partner, $\vec{F}_{\mathrm{ball,earth}}$ , acts on the Earth, which is outside. So gravity is external:

|\Delta \vec{p}_{\mathrm{sys}}| = \int_{t_0}^{t_f} mg\,dt = mg\,\Delta t = (0.50)(9.8)(0.40) = 1.96 \approx 2.0\ \mathrm{kg\,m\,s^{-1}}

directed downward — the ball speeds up, exactly as $v = g\Delta t = 3.9\ \mathrm{m\,s^{-1}}$ predicts.

Boundary 2 — ball plus Earth. Now both members of the gravitational pair are inside:

\vec{F}_{\mathrm{earth,ball}} + \vec{F}_{\mathrm{ball,earth}} = \vec{0} \quad\Longrightarrow\quad \vec{F}^{\,\mathrm{ext}} = \vec{0} \quad\Longrightarrow\quad \Delta\vec{p}_{\mathrm{sys}} = \vec{0}

Reconciling the two answers. They do not contradict each other, because $\vec{p}_{\mathrm{sys}}$ means different sums. In the second case the Earth also gains momentum — $1.96\ \mathrm{kg\,m\,s^{-1}}$ upward, i.e. a speed of about $3\times10^{-25}\ \mathrm{m\,s^{-1}}$ — precisely cancelling the ball's gain:

\Delta\vec{p}_{\mathrm{ball}} + \Delta\vec{p}_{\mathrm{Earth}} = (+1.96 \;\text{down}) + (1.96 \;\text{up}) = \vec{0}

Nothing physical changed between the two calculations. Only the line you drew changed — and with it, which forces you were required to count.

Solved: 0 / 4

Where You Draw the Line: Why Only External Forces Change a System's Momentum

IKey Formulas

IIMoving the Boundary

IIIThe Double Sum Cancels in Pairs

IVOnly the External Impulse Moves p⃗sys\vec{p}_{\mathrm{sys}}p​sys​

VQuiz Questions

IVOnly the External Impulse Moves $\vec{p}_{\mathrm{sys}}$