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Momentum, Impulse, and Conservation Laws

8.3 Conservation of Momentum for Isolated Systems

The Push-Off

Two ice skaters stand facing each other on a frozen lake, motionless. They place their palms together and push. The light skater flies backward. The heavy skater drifts backward more slowly. Nobody measured the push. Nobody timed it. The force between their hands was brief, complicated, and unmeasured.

And yet you can predict exactly how fast each skater moves afterward --- without knowing anything about the push itself.

This is the power of conservation of momentum. It does not care about the internal details of the interaction. It does not need a force profile, a time history, or a model of how hands deform on contact. It needs only one fact: the system was isolated. No external horizontal forces acted. And from that single fact, a precise quantitative prediction follows.

In Sections 8.1 and 8.2, you built the machinery: momentum $\vec{p} = m\vec{v}$, impulse $\vec{J} = \int \vec{F}\,dt$, and the impulse-momentum theorem $\vec{J} = \Delta\vec{p}$. Now comes the payoff. When you stop analyzing individual objects and start analyzing the system, the internal forces vanish and a conservation law appears.

Prediction

Before you read on: Two ice skaters start from rest on a frictionless surface. Skater A has mass 90 kg; skater B has mass 45 kg. They push off each other, and skater B moves to the right at 4 m/s.

How fast does skater A move, and in which direction?

(a) 2 m/s to the left

(b) 2 m/s to the right

(c) 4 m/s to the left

(d) 8 m/s to the left

Commit to your answer before continuing.

[Interactive: Predict-Then-Reveal. The student selects one of the four options. After committing, the student sees: "The answer is (a): 2 m/s to the left. The total momentum started at zero (both at rest), so it must remain zero. If the 45 kg skater carries momentum $45 \times 4 = 180\;\text{kg}!\cdot!\text{m/s}$ to the right, the 90 kg skater must carry 180 kg$\cdot$m/s to the left: $v = 180/90 = 2$ m/s. The heavier skater moves more slowly --- and in the opposite direction --- so that the total momentum stays zero."]

Notice what happened. You predicted the heavy skater's velocity without knowing the force, the duration of the push, or anything about the contact. All you needed was (1) total momentum was zero before the push, and (2) the system was isolated, so total momentum is still zero after the push. The rest followed from algebra.

This is not a trick. It is a conservation law.

The Guiding Question

What remains predictable in interactions even when the internal details are complicated?

Collisions are messy. Explosions are violent. The forces during a push-off between skaters are brief, large, and impossible to measure in real time. If you tried to apply Newton's second law to each skater individually, you would need the exact force history $\vec{F}(t)$ --- which you do not have.

But when you shift your attention from individual objects to the system as a whole, something remarkable happens. The messy internal forces cancel, and a clean, exact law survives. The total momentum of the system does not change. This section shows you why.

Exploration: Spring-Loaded Carts

[Interactive: Momentum Conservation Explorer. Two carts sit on a frictionless track with a compressed spring between them. The carts are held in place by a latch. When the student clicks "Release," the spring pushes the carts apart and they slide away in opposite directions. The display shows: - A momentum bar for each cart (colored, showing magnitude and direction). - A total momentum bar (the vector sum of the two). - Numerical readouts of each cart's mass, velocity, and momentum.

Sliders allow the student to adjust $m_1$ (1 kg to 10 kg) and $m_2$ (1 kg to 10 kg) independently. The spring stiffness is fixed. After each release, the momentum bars update to show the before and after states.]

Prompt 1: Set both carts to equal mass, say $m_1 = m_2 = 3$ kg. Release the spring. What do you notice about their speeds? What is the total momentum before and after the release?

Prompt 2: Now set $m_1 = 6$ kg and $m_2 = 2$ kg. Release the spring. Which cart moves faster? By what factor? What is the total momentum before and after?

Prompt 3: Try $m_1 = 8$ kg and $m_2 = 1$ kg. Predict the speed ratio before you release. Were you right?

Prompt 4: Watch the momentum bars carefully during the release (not just before and after). Does the total momentum bar ever flicker or change during the interaction? What does this tell you about when momentum is conserved --- only at the start and end, or continuously throughout the process?

Prompt 5: The spring pushes equally on both carts (Newton's third law). If the force on cart 1 is $\vec{F}$ at every instant, the force on cart 2 is $-\vec{F}$ at every instant. The impulses delivered to the two carts are equal and opposite: $\vec{J}_1 = -\vec{J}_2$. So $\Delta\vec{p}_1 = -\Delta\vec{p}_2$. What does this imply about $\Delta\vec{p}_1 + \Delta\vec{p}_2$?

If you worked through those prompts, you found the central result: the total momentum is zero before the release and zero after, no matter what masses you choose. The individual momenta change --- one cart gains momentum to the right, the other gains equal momentum to the left --- but the total never changes. And it does not just hold at the start and end. It holds at every instant during the interaction. The total momentum is conserved continuously.

The reason is Newton's third law. The spring is an internal force --- it acts between objects inside the system. It pushes cart 1 to the left and cart 2 to the right with equal magnitude at every instant. These pushes deliver equal and opposite impulses, producing equal and opposite momentum changes. The total momentum change is always zero.

Concept Reveal: The Conservation of Momentum

Let us make this precise.

Consider a system of two objects. The total momentum is:

$$\vec{p}_{\text{total}} = \vec{p}_1 + \vec{p}_2 = m_1\vec{v}_1 + m_2\vec{v}_2$$

Now differentiate with respect to time:

$$\frac{d\vec{p}_{\text{total}}}{dt} = \frac{d\vec{p}_1}{dt} + \frac{d\vec{p}_2}{dt}$$

By Newton's second law for each object, $d\vec{p}/dt$ equals the net force on that object. Each object experiences forces from two sources: forces from outside the system (external forces) and forces from the other object (internal forces). So:

$$\frac{d\vec{p}{\text{total}}}{dt} = \bigl(\vec{F}}} + \vec{F{2 \to 1}\bigr) + \bigl(\vec{F}\bigr)$$}} + \vec{F}_{1 \to 2

By Newton's third law, $\vec{F}{2 \to 1} = -\vec{F}$. These internal forces cancel exactly:

$$\frac{d\vec{p}{\text{total}}}{dt} = \vec{F}}} + \vec{F{\text{ext on 2}} = \sum\vec{F}$$}

This gives us the fundamental equation for systems:

$$\boxed{\sum\vec{F}{\text{ext}} = \frac{d\vec{p}$$}}}{dt}

This equation says: the rate of change of the total momentum of a system depends only on the external forces. Internal forces --- no matter how large, how complicated, how brief --- do not affect the total momentum. They always cancel in pairs.

The conservation law follows immediately. If the net external force is zero --- that is, if the system is isolated --- then:

$$\frac{d\vec{p}_{\text{total}}}{dt} = \vec{0}$$

which means:

$$\boxed{\vec{p}_{\text{total}} = \text{constant} \quad \text{(isolated system)}}$$

In component form, this gives independent conservation equations for each direction:

$$p_{x,\text{total}} = \text{constant}, \qquad p_{y,\text{total}} = \text{constant}$$

This is the conservation of momentum: when the net external force on a system is zero, the total momentum of the system does not change.

Why This Matters: System Reasoning vs. Object Reasoning

In Section 5.5, you learned the difference between analyzing individual objects and analyzing a whole system. You saw that when you draw a system boundary around multiple objects, internal forces --- like the tension in a rope connecting two blocks --- disappear from the equations. Only external forces survive.

Momentum conservation is the grand payoff of that idea. When the system boundary is drawn so that no net external force acts, you get something extraordinary: a conserved quantity. The total momentum before an interaction equals the total momentum after, regardless of how violent or complicated the interaction was internally.

This is why conservation laws feel different from force-by-force analysis. Force analysis requires knowing every detail: what forces act, in what direction, for how long. Conservation of momentum says: skip all of that. If the system is isolated, the total momentum is the same before and after. Period.

But this power comes with a discipline. You must be careful about your system boundary. Momentum is conserved for the system, not for individual objects within it. Each individual object can (and usually does) change its momentum. It is only the total that remains constant.

When Is a System Isolated?

A system is isolated (for the purpose of momentum conservation) when the net external force on it is zero: $\sum\vec{F}_{\text{ext}} = \vec{0}$.

In practice, perfectly isolated systems are rare. But approximately isolated systems are everywhere:

  • Collisions on frictionless surfaces. Gravity and the normal force act vertically and cancel. No horizontal external force acts. Horizontal momentum is conserved.

  • Explosions (fireworks, push-offs, spring releases). The interaction forces are enormous but internal. External forces like gravity may act, but if the explosion is fast enough, the impulse from gravity during the brief explosion is negligible. Momentum is approximately conserved during the event.

  • Collisions on surfaces with friction. Friction is an external force (from the ground, outside the system). Momentum is not strictly conserved. But if the collision is fast (milliseconds), the friction impulse during the collision is tiny compared to the collision impulse, and momentum is approximately conserved during the impact.

The key judgment call is always: are the external forces negligible compared to the internal interaction forces during the time interval of interest? If yes, treat the system as isolated.

Worked Example: Recoil of a Cannon

A cannon of mass $M = 500$ kg fires a cannonball of mass $m = 5$ kg. The cannonball exits the barrel at $v_{\text{ball}} = 200$ m/s horizontally. What is the recoil velocity of the cannon?

System: Cannon + cannonball.

Is the system isolated? The firing happens over a very short time. Gravity and the normal force act vertically and cancel. Friction with the ground is small compared to the explosive force. We treat horizontal momentum as conserved.

Before firing: Both are at rest. $\vec{p}_{\text{total}} = \vec{0}$.

After firing: The cannonball moves to the right at 200 m/s. The cannon moves to the left at speed $V$.

Momentum conservation (horizontal):

$$0 = mv_{\text{ball}} + M V_{\text{cannon}}$$

$$V_{\text{cannon}} = -\frac{m}{M}v_{\text{ball}} = -\frac{5}{500}(200) = -2 \text{ m/s}$$

The cannon recoils at 2 m/s --- in the opposite direction from the cannonball. The negative sign confirms the direction: if the ball goes right, the cannon goes left.

Notice: the cannon is 100 times more massive than the ball, so it moves 100 times more slowly. The momenta are equal in magnitude and opposite in direction, as required by conservation.

Connection to Section 5.5

In Section 5.5, you saw that internal forces (like the tension between two connected blocks) disappear when you analyze the whole system. At the time, the payoff was a faster route to the acceleration: you could use $\sum\vec{F}_{\text{ext}} = (m_1 + m_2)a$ without needing to find the tension.

Now you see the deeper payoff. When $\sum\vec{F}{\text{ext}} = \vec{0}$, the whole-system equation becomes $d\vec{p}$ --- a conservation law. The internal forces did not just simplify the equation. They vanished entirely, leaving behind an exact, powerful, and universally applicable principle.}}/dt = \vec{0

This pattern --- internal forces canceling to reveal a conservation law --- is one of the deepest ideas in physics. You will see it again when you study angular momentum and, in more advanced courses, when conservation laws are derived from symmetries of nature.

Spaced Retrieval

Before moving to practice, test your recall of earlier material.

Recall prompt 1: What is the impulse-momentum theorem, and how does it relate force to momentum? (Section 8.1)

Recall prompt 2: If two forces deliver the same impulse but one acts over a shorter time, what is different about the peak force? (Section 8.2)

Recall prompt 3: What is the difference between an internal force and an external force? What determines which category a force falls into? (Section 5.5)

Practice

Layer 1: Concrete

Problem 1. A 60 kg astronaut floating at rest in space throws a 2 kg wrench to the right at 12 m/s. What is the astronaut's velocity after the throw?

Check your answer **System:** Astronaut + wrench. No external forces act (deep space). The system is isolated. **Before:** Total momentum is zero (both at rest). **After:** Momentum conservation gives: $$0 = m_{\text{wrench}} v_{\text{wrench}} + m_{\text{astro}} v_{\text{astro}}$$ $$v_{\text{astro}} = -\frac{m_{\text{wrench}}}{m_{\text{astro}}} v_{\text{wrench}} = -\frac{2}{60}(12) = -0.4 \text{ m/s}$$ The astronaut drifts to the left at 0.4 m/s. The negative sign indicates the direction opposite to the wrench. This is a recoil problem --- the same structure as the cannon example.

Problem 2. A 1200 kg car traveling east at 15 m/s collides with a 800 kg car traveling west at 10 m/s. They lock bumpers and move together after the collision. Find the velocity of the combined wreck.

Check your answer **System:** Both cars. During the brief collision, external forces (friction, gravity) are negligible compared to the collision forces. Treat horizontal momentum as conserved. **Before:** Taking east as positive: $$p_{\text{total}} = m_1 v_1 + m_2 v_2 = (1200)(15) + (800)(-10) = 18{,}000 - 8{,}000 = 10{,}000\;\text{kg}\!\cdot\!\text{m/s}$$ **After:** They move together with mass $m_1 + m_2 = 2000$ kg at velocity $v_f$: $$p_{\text{total}} = (m_1 + m_2) v_f$$ $$10{,}000 = 2000 \, v_f$$ $$v_f = 5 \text{ m/s (east)}$$ The wreck moves east at 5 m/s. The heavier, faster car "wins" the direction, but the combined speed is less than either initial speed. This makes sense: the two momenta partially cancel, and the total mass increases.

Layer 2: Pattern

Problem 3. For each scenario below, decide whether the total momentum of the stated system is conserved. Justify your answer by identifying the external forces.

(a) A ball falls freely under gravity. System: the ball.

(b) A ball falls freely under gravity. System: the ball + the Earth.

(c) Two hockey pucks collide on ice with negligible friction. System: both pucks.

(d) A car brakes to a stop on a road. System: the car.

(e) A rifle fires a bullet. System: the rifle + the bullet. (Ignore gravity during the brief firing.)

Check your answer **(a)** Not conserved. The system is the ball alone, and gravity is an external force (exerted by the Earth, which is outside the system). The ball's momentum changes as it accelerates downward. **(b)** Conserved. The system now includes the Earth. Gravity is an internal force --- the Earth pulls the ball down, and the ball pulls the Earth up (Newton's third law). No external forces act on the ball-Earth system (assuming we ignore all other celestial bodies). The ball gains downward momentum, and the Earth gains an equal amount of upward momentum. The total is constant. (The Earth's velocity change is immeasurably tiny because its mass is enormous.) **(c)** Conserved. Friction is negligible, so no horizontal external forces act. The collision forces are internal. Vertical forces (gravity, normal) cancel. **(d)** Not conserved. Friction from the road is an external force on the car. (If you expanded the system to include the road and the Earth, then momentum would be conserved --- the car's lost momentum is transferred to the Earth.) **(e)** Conserved. The explosive force is internal (rifle pushes bullet forward, bullet pushes rifle backward). Ignoring gravity during the brief firing, no external forces act. The rifle recoils. **The pattern:** Whether momentum is conserved depends entirely on your system boundary. Expanding the system to include more objects can turn external forces into internal ones. The question is always: are there net external forces on the system you have defined?

Layer 3: Structure

Problem 4. A student asks: "Why does Newton's third law guarantee that internal forces don't change the total momentum? Can you show me, step by step?"

Write a clear explanation that starts from Newton's third law and arrives at the conclusion that internal forces contribute zero net impulse to a system.

Check your answer Consider two objects, A and B, inside a system. They exert forces on each other. By Newton's third law: $$\vec{F}_{A \to B} = -\vec{F}_{B \to A}$$ These forces act for the same duration $\Delta t$ (the interaction is simultaneous). The impulse delivered to B by A is: $$\vec{J}_{A \to B} = \vec{F}_{A \to B}\,\Delta t$$ The impulse delivered to A by B is: $$\vec{J}_{B \to A} = \vec{F}_{B \to A}\,\Delta t = -\vec{F}_{A \to B}\,\Delta t = -\vec{J}_{A \to B}$$ The two impulses are equal and opposite. The momentum gained by B is exactly the momentum lost by A: $$\Delta\vec{p}_B = -\Delta\vec{p}_A$$ So the change in total momentum from this internal interaction is: $$\Delta\vec{p}_A + \Delta\vec{p}_B = \vec{0}$$ This holds for *every* internal force pair. No matter how many objects interact inside the system, every interaction contributes zero to the total momentum change. Only external forces --- forces from objects *outside* the system --- can change the total. This is why Newton's third law is not just a "reaction force" rule. It is the engine that drives conservation of momentum.

Layer 4: Debug

Problem 5. A student analyzes a ball bouncing off a wall. The ball approaches at 5 m/s and rebounds at 5 m/s. The student writes:

"The ball's momentum before is $p = mv$ and after is $p = -mv$. The change is $\Delta p = -2mv$. But momentum is conserved, so $\Delta p = 0$. Contradiction!"

What is the student's error?

Check your answer The student is applying momentum conservation to the **ball alone**. But the ball-alone system is not isolated. The wall exerts an external force on the ball during the bounce (a large normal force over a short time). Since there is a net external force on the system, momentum of the ball is *not* conserved. The impulse from the wall is exactly $\Delta p = -2mv$ --- it accounts for the ball's momentum change. If the student expanded the system to include the ball *and the wall* (and the Earth the wall is attached to), then the total momentum would be conserved. The ball's momentum change of $-2mv$ would be matched by the wall/Earth gaining momentum $+2mv$ (an imperceptibly tiny velocity change for the Earth). **The lesson:** Conservation of momentum does not apply to any single object in isolation. It applies to a *system* for which the net external force is zero. Before invoking conservation, always ask: "Is my system isolated?" If the answer is no, either expand the system boundary or do not use conservation --- use the impulse-momentum theorem instead.

Problem 6. Two ice skaters push off from rest. Skater A (70 kg) moves left at 3 m/s. A student calculates skater B's velocity:

"$p_{\text{total}} = 0$, so $m_A v_A + m_B v_B = 0$. Therefore $v_B = -m_A v_A / m_B = -(70)(3)/m_B$."

The student then writes: "$v_B = -210/m_B$, so skater B moves to the right."

Is this correct? What should the student check?

Check your answer The algebra is correct, but the sign reasoning needs care. Let us define rightward as positive. If skater A moves left at 3 m/s, then $v_A = -3$ m/s. Then: $$v_B = -\frac{m_A v_A}{m_B} = -\frac{(70)(-3)}{m_B} = \frac{210}{m_B}$$ This is positive --- so yes, skater B moves to the right. The student's final conclusion is correct. But the student wrote $v_A = 3$ (positive) while saying A moves *left*. If leftward is positive in the student's convention, then the answer $v_B = -210/m_B$ is negative in that convention, meaning B moves rightward (opposite to the positive direction). The direction is correct either way, but the student should be explicit about which direction is positive. **The check:** Always state your sign convention at the start. Then verify that the signs of your answers match the physical directions. Momentum is a vector --- signs carry directional meaning, and being sloppy with them leads to errors, especially in 2D problems.

Reflection

What is the relationship between Newton's third law and conservation of momentum?

Think about this carefully. Newton's third law says that forces between two objects are always equal and opposite. Conservation of momentum says that the total momentum of an isolated system does not change. These sound like different statements --- one about forces, the other about momentum.

But the derivation in this section showed they are intimately connected. Newton's third law is the reason internal forces cancel when you sum over the whole system. And when internal forces cancel, only external forces affect the total momentum. Remove the external forces (isolate the system), and total momentum cannot change.

Conservation of momentum is not an independent law. It is a consequence of Newton's second and third laws combined. But it is a consequence so powerful, so broadly applicable, and so easy to use that it deserves --- and receives --- the status of a standalone principle. It lets you solve problems that would be intractable by direct force analysis: collisions where you do not know the forces, explosions where the forces are too brief to measure, interactions where the internal details are hopelessly complicated.

As you move forward, remember: the conservation law is not magic. It is Newton's third law, summed over a system, with the external forces set to zero. The principle feels profound because the conclusion is so much simpler than the messy interactions it summarizes.

Looking Ahead

You now know when momentum is conserved (isolated systems) and why (Newton's third law makes internal forces cancel). You can apply it to explosions, recoil, and push-off problems.

But there is a question we have not yet addressed: what happens to energy during these interactions? In the skater push-off, the total momentum is conserved --- but kinetic energy increases (both skaters go from rest to moving). Where did that energy come from? In a head-on collision where two cars crumple together, momentum is conserved --- but kinetic energy decreases. Where did it go?

Momentum conservation alone cannot answer these questions. In Section 8.4, you will classify collisions by what happens to kinetic energy: elastic collisions (kinetic energy is conserved), inelastic collisions (kinetic energy decreases), and perfectly inelastic collisions (the objects stick together, and the maximum kinetic energy is lost). You will see that momentum conservation and energy conservation are complementary tools --- and that using them together unlocks a new class of problems.