Classical-Mechanics · Unit 12 · Video 4 · Interactive Practice

One Point Obeys Newton's Law: Center-of-Mass Motion and the Conservation of Momentum

IKey Formulas

Formula	Name	What you need
$\vec{p}_{\mathrm{sys}} = m_{\mathrm{sys}}\vec{V}_{\mathrm{cm}}$	Momentum of a system	Total mass and the center-of-mass velocity
$\vec{F}^{\,\mathrm{ext}} = m_{\mathrm{sys}}\vec{A}_{\mathrm{cm}}$	Newton's second law for a system	The external forces only
$\vec{F}^{\,\mathrm{ext}} = \vec{0} \;\Longrightarrow\; \Delta\vec{p}_{\mathrm{sys}} = \vec{0}$	Isolated system	A boundary nothing crosses
$\Delta\vec{p}_{\mathrm{sys}} + \Delta\vec{p}_{\mathrm{sur}} = \vec{0}$	Conservation of Momentum	System and surroundings counted together

Key Insight: $\vec{F}^{\,\mathrm{ext}} = m_{\mathrm{sys}}\vec{A}_{\mathrm{cm}}$ never mentions where a force is applied, so every external force may be slid onto the center of mass and added there as a vector. It is a promise about one point, not about the particles.

IISame Force, Three Places, One Acceleration

One identical push delivered at three points along the bat — what does the marked point do in each case?

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💡 The equation pins down $\vec{A}_{\mathrm{cm}}$ and nothing else. The rotation is governed by the torque $\vec{\tau} = \vec{r} \times \vec{F}$ , which does depend on where you push — that is the subject of a later chapter.

IIIWhat an Internal Force Cannot Do

A spring between two pucks trades velocity between them — can it bend the center of mass off its line?

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💡 Ride along with the center of mass and an isolated system always shows $\vec{p}_{\mathrm{sys}} = \vec{0}$ , since $\vec{V}_{\mathrm{cm}} = \vec{0}$ in that frame — which is why collisions are so often analyzed there.

IVWhere the Lost Momentum Goes

This system is not isolated: whatever momentum it loses, the surroundings must gain.

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VQuiz Questions

Problem 1 · Acceleration of the Center of Mass

Given: A bat of mass $m_{\mathrm{sys}} = 0.94\ \mathrm{kg}$ lies on a frictionless table. A single constant force of magnitude $4.7\ \mathrm{N}$ , perpendicular to the bat, is applied at the handle end — far from the center of mass. Find the magnitude of $\vec{A}_{\mathrm{cm}}$ .

✅ Correct! Only

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

and

m_{\mathrm{sys}}

appear in the law, and neither one knows where your hand was.

❌ Close, but… an off-center push is not "split" between translating and rotating. The center of mass receives the whole force:

\vec{A}_{\mathrm{cm}} = \vec{F}^{\,\mathrm{ext}}/m_{\mathrm{sys}}

, with no fraction in front of it.

❌ Not quite. Use

\vec{A}_{\mathrm{cm}} = \vec{F}^{\,\mathrm{ext}}/m_{\mathrm{sys}}

— divide the force by the mass, and check that your answer carries units of

\mathrm{m\,s^{-2}}

Show solution

The bat is one system, so apply the system form of the second law:

\vec{A}_{\mathrm{cm}} = \frac{\vec{F}^{\,\mathrm{ext}}}{m_{\mathrm{sys}}} = \frac{4.7\ \mathrm{N}}{0.94\ \mathrm{kg}} = 5.0\ \mathrm{m\,s^{-2}}

Two quantities appear on the right — the applied force and the total mass — and sliding your hand along the bat changes neither. Pushing at the handle, at the cross, or at the barrel therefore gives the same $A_{\mathrm{cm}} = 5.0\ \mathrm{m\,s^{-2}}$ .

The handle push does make the bat spin, and the motion of every individual particle is different from the pure-translation case. That is not a contradiction: the equation was never a statement about the particles.

Problem 2 · Momenta Add as Vectors

Given: An isolated system of two pucks. Puck A has $m_A = 3.0\ \mathrm{kg}$ moving east at $4.0\ \mathrm{m\,s^{-1}}$ ; puck B has $m_B = 2.0\ \mathrm{kg}$ moving north at $6.0\ \mathrm{m\,s^{-1}}$ . Find the magnitude of $\vec{p}_{\mathrm{sys}}$ .

✅ Correct! The two momenta are perpendicular, so the total is the hypotenuse:

\sqrt{12^2 + 12^2} = 12\sqrt{2} \approx 17\ \mathrm{kg\,m\,s^{-1}}

❌ Close, but… you added magnitudes.

\vec{p}_{\mathrm{sys}}

is a sum of vectors, added head to tail;

12 + 12 = 24

would only be right if both momenta pointed the same way.

❌ Not quite. Build each momentum in components first:

\vec{p}_A = (12, 0)

and

\vec{p}_B = (0, 12)

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

, then add componentwise.

Show solution

Take $\hat{x}$ east and $\hat{y}$ north.

\vec{p}_A = (3.0)(4.0)\,\hat{x} = 12\,\hat{x}, \qquad \vec{p}_B = (2.0)(6.0)\,\hat{y} = 12\,\hat{y}

\vec{p}_{\mathrm{sys}} = \vec{p}_A + \vec{p}_B = (12,\ 12)\ \mathrm{kg\,m\,s^{-1}}

|\vec{p}_{\mathrm{sys}}| = \sqrt{12^2 + 12^2} = 12\sqrt{2} \approx 17\ \mathrm{kg\,m\,s^{-1}}

Two remarks. First, "isolated" says the total cannot change, not that it is zero — option d confuses constancy with vanishing. Second, the vector sum genuinely can vanish: two fragments of equal and opposite momentum give $\vec{p}_{\mathrm{sys}} = \vec{0}$ while both are moving fast.

As a check, $\vec{V}_{\mathrm{cm}} = \vec{p}_{\mathrm{sys}}/m_{\mathrm{sys}} = (12, 12)/5.0 = (2.4,\ 2.4)\ \mathrm{m\,s^{-1}}$ , pointing northeast.

Problem 3 · Push-Off on Frictionless Ice

Given: Two skaters, $m_1 = 40\ \mathrm{kg}$ and $m_2 = 60\ \mathrm{kg}$ , stand at rest on frictionless ice and push off each other. Skater 1 then glides west at $3.0\ \mathrm{m\,s^{-1}}$ . Take the two skaters together as the system.

How fast does skater 2 glide?

How far does the center of mass travel in the first $5.0\ \mathrm{s}$ after the push-off?

✅ Correct! The push is internal, so

\vec{p}_{\mathrm{sys}}

stays at

\vec{0}

— the skaters separate about a center of mass that never budges.

❌ Check the mass ratio. Momentum conservation gives

m_1 v_1 = m_2 v_2

, so the heavier skater ends up slower:

v_2 = (m_1/m_2)v_1

❌ Check $\vec{V}_{\mathrm{cm}}$ first. The center of mass moves at

\vec{V}_{\mathrm{cm}} = \vec{p}_{\mathrm{sys}}/m_{\mathrm{sys}}

, and

\vec{p}_{\mathrm{sys}}

was zero before the push and cannot be changed by an internal force.

Show solution

Step 1 — Identify the system and the forces. The skaters push on each other: that force is internal. Gravity and the normal force cancel vertically, and the ice is frictionless horizontally, so $\vec{F}^{\,\mathrm{ext}} = \vec{0}$ along the direction of motion. The system is isolated and $\Delta\vec{p}_{\mathrm{sys}} = \vec{0}$ .

Step 2 — Momentum before and after. Take $\hat{x}$ east, so skater 1 moves at $-3.0\,\hat{x}$ :

\vec{p}_{\mathrm{sys},i} = \vec{0} = \vec{p}_{\mathrm{sys},f} = m_1 \vec{v}_{1,f} + m_2 \vec{v}_{2,f}

0 = (40)(-3.0) + (60)v_{2,f} \;\Longrightarrow\; v_{2,f} = \frac{120}{60} = +2.0\ \mathrm{m\,s^{-1}}

Skater 2 glides east at $2.0\ \mathrm{m\,s^{-1}}$ — heavier means slower, by exactly the mass ratio $40/60$ .

Step 3 — The center of mass.

\vec{V}_{\mathrm{cm}} = \frac{\vec{p}_{\mathrm{sys}}}{m_{\mathrm{sys}}} = \frac{\vec{0}}{100\ \mathrm{kg}} = \vec{0}

It was zero before the push, and an internal force cannot change it, so it is zero after: in $5.0\ \mathrm{s}$ the center of mass travels 0 m. The skaters drift apart symmetrically about a point fixed on the ice — skater 1 covers $15\ \mathrm{m}$ west while skater 2 covers $10\ \mathrm{m}$ east, and the mass-weighted average of those displacements is zero.

Problem 4 · Transfer — The Momentum Has to Go Somewhere

Given: A $0.15\ \mathrm{kg}$ ball travelling at $20\ \mathrm{m\,s^{-1}}$ strikes a wall bolted to the Earth and rebounds straight back along the same line at $16\ \mathrm{m\,s^{-1}}$ . Take the ball alone as the system, everything else as the surroundings, and $\hat{x}$ along the incoming direction. Find $\Delta\vec{p}_{\mathrm{sur}}$ .

✅ Correct! The ball loses

5.4\ \mathrm{kg\,m\,s^{-1}}

and the wall-plus-Earth gains exactly that much:

\Delta\vec{p}_{\mathrm{sys}} + \Delta\vec{p}_{\mathrm{sur}} = \vec{0}

❌ Close, but the ball reversed. Its final velocity is

-16\ \mathrm{m\,s^{-1}}

, not

+16

, so the change in speed

20 - 16 = 4

is not the change in momentum. Subtract the signed momenta.

❌ Not quite. An enormous mass makes the velocity change unmeasurably small, not the momentum change:

\Delta p = m\,\Delta v

stays finite as

m \to \infty

and

\Delta v \to 0

❌ Check the sign. That is

\Delta\vec{p}_{\mathrm{sys}}

, the change for the ball. The surroundings take the opposite change:

\Delta\vec{p}_{\mathrm{sur}} = -\Delta\vec{p}_{\mathrm{sys}}

Show solution

Step 1 — The system's momentum change. With $\hat{x}$ along the incoming direction, the rebound velocity is negative:

\vec{p}_{\mathrm{sys},i} = (0.15)(+20) = +3.0\ \mathrm{kg\,m\,s^{-1}}, \qquad \vec{p}_{\mathrm{sys},f} = (0.15)(-16) = -2.4\ \mathrm{kg\,m\,s^{-1}}

\Delta\vec{p}_{\mathrm{sys}} = -2.4 - (+3.0) = -5.4\ \mathrm{kg\,m\,s^{-1}}

Step 2 — Hand it to the surroundings. The wall's force on the ball is external, and its third-law partner acts on the surroundings, so the two changes must cancel:

\Delta\vec{p}_{\mathrm{sys}} + \Delta\vec{p}_{\mathrm{sur}} = \vec{0} \;\Longrightarrow\; \Delta\vec{p}_{\mathrm{sur}} = +5.4\ \mathrm{kg\,m\,s^{-1}}

Why the Earth seems unaffected. Its mass is about $6 \times 10^{24}\ \mathrm{kg}$ , so

\Delta V \approx \frac{5.4}{6 \times 10^{24}} \approx 9 \times 10^{-25}\ \mathrm{m\,s^{-1}}

— utterly unmeasurable, yet the momentum transferred is a perfectly ordinary $5.4\ \mathrm{kg\,m\,s^{-1}}$ . Enlarge the system to ball + wall + Earth and the wall force becomes internal: that system is isolated and its momentum does not change at all.

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