AP Physics 1 · Topic 4.3

Topic 4.3: Conservation of Linear Momentum

Unit 4: Linear Momentum10-15% of the multiple-choice section

A system's total momentum stays constant whenever the net external force on it is zero. Internal forces come in Newton's third law pairs, so they cancel in the total. A system's momentum changes only when something outside it pushes across the boundary, and the change equals that external impulse.

AP Physics: Unit 4 (topics 4.3 Conservation of Linear Momentum). AP Physics 1 Unit 4, Topic 4.3. Two learning objectives: 4.3.A, describe the behavior of a system using conservation of linear momentum, and 4.3.B, describe how the selection of a system determines whether the momentum of that system changes. The load-bearing statements are 4.3.B.1 (momentum is conserved in all interactions), 4.3.B.2 (if the net external force on the selected system is zero, the total momentum of the system is constant), 4.3.B.3 (if it is nonzero, momentum is transferred between the system and the environment), and 4.3.A.3.i (the impulses two objects exert on each other are equal and opposite, a direct result of Newton's third law). The boundary statement limits AP Physics 1 to a quantitative and qualitative treatment in one dimension and a semiquantitative treatment in two dimensions, and excludes questions requiring the solution of simultaneous equations. Unit 4 carries 10 to 15 percent of the multiple-choice section, and the suggested skills for this topic are 1.A, 2.A, 2.D, and 3.C.

What Topic 4.3 requires

Topic 4.3 carries two learning objectives, and the second is the reason this topic is worth slowing down for.

4.3.A Describe the behavior of a system using conservation of linear momentum.
4.3.B Describe how the selection of a system determines whether the momentum of that system changes.

Four essential knowledge statements sit under 4.3.A, with five sub-statements beneath them, and three sit under 4.3.B. The CED's suggested skills here are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Unit 4 is weighted at 10 to 15 percent of the multiple-choice section, and the CED allots it about 10 to 15 class periods. Within the unit, 4.3 is where linear momentum and impulse stop being definitions and become a method for solving problems.

Momentum is conserved in all interactions (4.3.B.1)

Essential knowledge 4.3.B.1 is a single sentence: momentum is conserved in all interactions. Read on its own that would seem to end the discussion, so it is worth being exact about what it claims.

It is a statement about interactions, not about whichever system you happened to draw. Momentum is never created and never destroyed; it only moves from one object to another. The next two statements say what that means for your system.

4.3.B.2 If the net external force on the selected system is zero, the total momentum of the system is constant.
4.3.B.3 If the net external force on the selected system is nonzero, momentum is transferred between the system and the environment.

Notice the vocabulary. The CED says momentum is conserved in all interactions, and separately says a particular system's total momentum is constant under a stated condition. Keeping those two words apart clears up most of the confusion around this topic. Nothing leaks out of the universe; things routinely cross the boundary of the box you drew.

Everyone still says momentum is not conserved here as shorthand for this system's momentum is not constant, including the headings on this page, so read it that way. The useful question is never is momentum conserved here? It is does anything push across my boundary?

When momentum is not conserved in a collision

A system's total momentum changes exactly when the net external force on that system is nonzero (4.3.B.3), and the size of the change is set by the external impulse. Statement 4.3.A.3.iii puts that as an equation, the same one from Topic 4.2:

\vec{J} = \Delta \vec{p}

Four everyday cases where the momentum of the obvious system is not constant:

A falling ball, taking the ball alone as the system. Earth pulls on it from outside the boundary, so its downward momentum grows steadily.
A cart coasting to a stop on carpet. Friction from the floor is external, and it removes momentum.
A ball bouncing off a wall. The wall sits outside the system, so the ball's momentum reverses.
A cart on a track pulled by a hanging mass over a pulley. Take only the cart and the string tension is external.

In every case the momentum went somewhere real: into Earth, into the floor, into the wall, into the falling mass and Earth. Draw the boundary wide enough to enclose whatever is pushing, up to and including Earth, and it stops going anywhere.

For collisions specifically there are only two ways this bites you. Either you drew the system too small and left one of the colliding objects outside it, or you compared two instants far enough apart that an outside force had time to matter. The second is why 4.3.A.4 says conservation of momentum determines the velocity of a system immediately before and immediately after a collision or explosion. Compare a crashed car with where it finally stops fifty meters down the road and its momentum has plainly changed. Compare the instant before impact with the instant after and it has not.

That short window is also why gravity and friction can be ignored during a real collision even though they never switch off. Statement 4.1.A.3.i defines a collision as a model for an interaction where the forces exerted between the involved objects are much larger than the net external force exerted on those objects during the interaction. A contact force tens or hundreds of times the weight, acting for a few hundredths of a second, delivers an impulse that dwarfs what a weight of a few newtons delivers over the same few milliseconds. The external impulse is not zero, it is negligible, and the CED's collision model is your license to treat it that way.

Choosing the system is the method (4.3.A.3.ii)

Statement 4.3.A.3.ii reads that a system may be selected so that the total momentum of that system is constant. That may be selected is doing real work: it is an instruction about how to set the problem up, and learning objective 4.3.B exists to test whether you can follow it.

The procedure is four steps.

List every object that takes part in the interaction. Both carts. The gun and the bullet. The skater and the backpack she throws. The firework shell and the fragments it becomes.
Draw the boundary around all of them. Every force of one on another is now internal.
Ask what still crosses the boundary. Gravity, the normal force, friction, tension, a hand.
If nothing external contributes a meaningful impulse over your interval, the total momentum is constant. Write it before and after, then solve.

Step 3 is where the marks are. On a horizontal track gravity and the normal force cancel each other, so nothing survives to cross the boundary. That is why a frictionless horizontal surface is the standard exam setup: it is the arrangement in which step 3 comes out empty.

One fair warning about step 1. The boundary statement under 4.2.B says AP Physics 1 does not require students to quantitatively analyze systems in which the mass of the system changes with respect to time. So a rocket burning fuel continuously is off the table as a calculation, while a one-shot separation such as a bullet leaving a gun or a spring pushing two carts apart is exactly on it.

Why internal forces cancel (4.3.A.3 and 4.3.A.3.i)

Statement 4.3.A.3 says that in the absence of net external forces, any change to the momentum of an object within a system must be balanced by an equivalent and opposite change of momentum elsewhere within the system. Its second sentence adds the other half: any change to the momentum of a system is due to a transfer of momentum between the system and its surroundings.

The reason is 4.3.A.3.i: the impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first, a direct result of Newton's third law. Two objects start touching at the same instant and stop touching at the same instant, so the contact time $\Delta t$ is shared. Equal and opposite forces acting over an identical interval give equal and opposite impulses:

\vec{J}_{1 \text{ on } 2} = -\vec{J}_{2 \text{ on } 1} \quad \Rightarrow \quad \Delta \vec{p}_2 = -\Delta \vec{p}_1

Add the two changes and you get zero, so the total does not move. Every internal force in any system pairs off this way, which is why internal forces can never change a system's total momentum no matter how violent they are.

The CED's Unit 4 overview says the unit gives students an opportunity to revisit misconceptions surrounding Newton's third law, and the classic one lives right here. When a loaded truck hits a parked hatchback, the forces on the two are equal in magnitude and the contact times are identical, so the impulses are equal in magnitude and the momentum changes are equal in magnitude. What is not equal is the velocity change, because $\Delta v = \Delta p / m$ and the truck's mass is many times the hatchback's. The hatchback is wrecked not because it was pushed harder but because the same push produced a far larger change in its velocity.

Total momentum and the center of mass (4.3.A.1 and 4.3.A.2)

Statement 4.3.A.2 defines what you are conserving: the total momentum of a system is the sum of the momenta of the system's constituent parts. Momentum is a vector, so sum means a vector sum. In one dimension that reduces to adding signed numbers, and a dropped minus sign is an easy way to lose a momentum question.

Statement 4.3.A.1 offers a second description of the same quantity. A collection of objects with individual momenta can be described as one system with one center-of-mass velocity, and 4.3.A.1.i gives the equation, which is printed on the AP Physics 1 equation sheet:

\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\sum m_i \vec{v}_i}{\sum m_i}

Multiply through by the total mass and the numerator is simply the total momentum, so $\sum \vec{p}_i = M_{\text{total}} \vec{v}_{\text{cm}}$ . That makes 4.3.A.1.ii, which says the velocity of a system's center of mass is constant in the absence of a net external force, the same claim as 4.3.B.2 written in different symbols. For a system of fixed total mass, and AP Physics 1 works only with those, total momentum constant and center-of-mass velocity constant mean identical things.

That hands you a free check on any answer. Two carts blow apart from rest on a frictionless track and their center of mass stays exactly where it was. An astronaut and a wrench drifting in deep space have a center of mass that keeps gliding at the same velocity through the throw, the flight, and the catch. If your final velocities move the center of mass, you have made an arithmetic error.

Setting up the before-and-after equation (4.3.A.4)

Statement 4.3.A.4 is the payoff: correct application of conservation of momentum can be used to determine the velocity of a system immediately before and immediately after collisions or explosions. It names both. Collisions bring objects together, and explosions, defined in 4.1.A.3.iii as an interaction in which forces internal to the system move objects within that system apart, are the same physics run the other way.

For two objects in one dimension the equation is

m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}

and the discipline that makes it work is dull and non-negotiable. Choose a positive direction and write it down. Give every velocity a sign that matches that choice. Never enter a speed without deciding its sign first. If an answer comes out negative, that is information rather than an error: the object moves the other way.

Suggested skill 1.A asks for diagrams, tables, charts, or schematics, and a before-and-after table is the representation worth building the habit around. One row per object, one column for mass, one for initial velocity, one for final velocity, and every empty cell is a labeled unknown.

The full step-by-step procedure, including what to do when both final velocities are unknown, lives in the conservation of momentum guide. To check your own numbers against a solver, use the momentum collision calculator.

What the exam will and will not ask

The boundary statement for this topic is unusually specific, and it tells you where to stop practicing.

AP Physics 1 includes a quantitative and qualitative treatment of conservation of momentum in one dimension and a semiquantitative treatment of conservation of momentum in two dimensions. Exam questions involving solution of simultaneous equations are not included in AP Physics 1, but the exam may include questions that assess whether students can set up the equations properly and reason about how changing a given mass, speed, or angle would affect other quantities. AP Physics 2 is where the full two-dimensional treatment lives, for problems that include one unknown final velocity.

Two consequences worth acting on. In two dimensions your job is to know that momentum is conserved separately along each axis and to write those two equations, not to grind them out. And because simultaneous equations are excluded, a problem you are asked to finish numerically will hand you enough information to leave one unknown at a time. If you find yourself with two unknowns and no second relation, reread the problem.

Skills 2.D and 3.C point at the qualitative half. Expect questions of the form if the second cart's mass doubled, what would happen to the final speed? and questions that ask you to justify a claim about lab data using conservation of momentum as the principle. From here, Topic 4.4 keeps momentum conservation exactly as it is and adds the second question every collision problem asks: what happened to the kinetic energy?

A spring explosion: momentum constant, kinetic energy not

Two carts sit at rest and touching on a frictionless horizontal track with a compressed spring held between them. Cart A has mass 0.25 kg and cart B has mass 0.75 kg. The spring is released. Cart B leaves at 0.40 m/s to the right. (a) Find cart A's velocity. (b) Verify that the system's center of mass stays at rest. (c) Find the kinetic energy before and after, and say what that means for conservation of momentum.

Choose the system as both carts plus the spring, so the spring's push on each cart is internal. The track is frictionless and horizontal, so gravity and the normal force cancel and no net external force acts. By 4.3.B.2 the total momentum is constant. Take rightward as positive.
(a) The carts start at rest, so the total initial momentum is zero: $p_i = (0.25)(0) + (0.75)(0) = 0$ . Setting $p_f = p_i$ gives $(0.25\ \text{kg})v_A + (0.75\ \text{kg})(+0.40\ \text{m/s}) = 0$ .
Solve: $(0.25)v_A = -0.30\ \text{kg}\cdot\text{m/s}$ , so $v_A = -1.2\ \text{m/s}$ . The minus sign says cart A moves left, which is what the physics demands: a system starting at rest must end with zero total momentum, so the pieces go opposite ways.
Check the ratio. The speeds came out as $1.2$ and $0.40\ \text{m/s}$ , a ratio of 3, and the masses are $0.75$ and $0.25\ \text{kg}$ , also a ratio of 3. In any explosion from rest the speeds are in inverse proportion to the masses, which is suggested skill 2.D in one line.
(b) Apply the center-of-mass equation: $v_{\text{cm}} = \frac{(0.25)(-1.2) + (0.75)(+0.40)}{0.25 + 0.75} = \frac{-0.30 + 0.30}{1.00} = 0$ . The center of mass was at rest before and is at rest after, exactly as 4.3.A.1.ii requires.
(c) Kinetic energy before is zero, since nothing is moving. After: $K_A = \frac{1}{2}(0.25)(1.2)^2 = 0.18\ \text{J}$ and $K_B = \frac{1}{2}(0.75)(0.40)^2 = 0.060\ \text{J}$ , so $K_f = 0.24\ \text{J}$ .

Cart A moves at 1.2 m/s to the left, the center of mass stays at rest, and the kinetic energy rises from 0 to 0.24 J. Conservation of momentum says nothing at all about kinetic energy: here the elastic potential energy stored in the compressed spring became kinetic energy, while the total momentum sat at zero throughout because every force involved was internal.

One bounce, two different systems

A 0.15 kg ball is dropped and strikes the floor moving downward at 5.0 m/s. It rebounds upward at 4.0 m/s. The ball is in contact with the floor for 0.020 s. (a) Taking the ball alone as the system, find its change in momentum and the average net force on it. (b) Find the average force the floor exerts and compare it with the ball's weight. (c) Now take the ball and Earth together as the system and find Earth's change in velocity.

Take upward as positive throughout, so $v_i = -5.0\ \text{m/s}$ and $v_f = +4.0\ \text{m/s}$ . The signs are most of this problem: a ball that reverses direction has a momentum change larger than either individual momentum.
(a) $\Delta p = mv_f - mv_i = (0.15\ \text{kg})(+4.0\ \text{m/s}) - (0.15\ \text{kg})(-5.0\ \text{m/s}) = 0.60 + 0.75 = +1.35\ \text{kg}\cdot\text{m/s}$ , directed upward. The ball alone does not have constant momentum, because the floor lies outside this system. That is 4.3.B.3 in action.
The average net force follows from $\vec{J} = \Delta \vec{p}$ : $F_{\text{net,avg}} = \frac{1.35\ \text{kg}\cdot\text{m/s}}{0.020\ \text{s}} = 67.5\ \text{N}$ upward, or $68\ \text{N}$ to two significant figures.
(b) Two external forces act during contact: the floor pushing up and gravity pulling down. Gravity's impulse over the contact is $mg\Delta t = (0.15)(9.8)(0.020) = 0.029\ \text{N}\cdot\text{s}$ downward, so the floor's impulse must be $1.35 + 0.029 = 1.379\ \text{N}\cdot\text{s}$ upward and the average floor force is $\frac{1.379}{0.020} = 69\ \text{N}$ .
Compare that with the ball's weight, $mg = (0.15)(9.8) = 1.5\ \text{N}$ . The contact force is about 47 times the weight, which is precisely the condition 4.1.A.3.i uses to define a collision, and it is why the gravitational impulse gets dropped without comment in ordinary collision problems.
(c) Enlarge the system to ball plus Earth. The contact force is now internal, the total momentum is constant, and Earth must absorb $-1.35\ \text{kg}\cdot\text{m/s}$ . With Earth's mass about $6.0 \times 10^{24}\ \text{kg}$ , its velocity changes by $\Delta v = \frac{1.35}{6.0 \times 10^{24}} \approx 2.3 \times 10^{-25}\ \text{m/s}$ .

(a) $\Delta p = +1.35\ \text{kg}\cdot\text{m/s}$ upward, and the average net force is about 68 N upward. (b) The floor pushes with about 69 N, roughly 47 times the ball's 1.5 N weight. (c) Earth's velocity changes by about $2.3 \times 10^{-25}\ \text{m/s}$ , unmeasurable but not zero. The same event is a momentum change or a momentum transfer depending only on where you drew the boundary, which is what learning objective 4.3.B is asking you to see.

Frequently asked questions

When is momentum not conserved in a collision?

When the net external force on the system you chose is not zero over the interval you are looking at. That happens two ways: you left one of the colliding objects outside the system, so its push counts as external, or you compared instants far enough apart for friction or gravity to build up an impulse. During the collision itself the contact forces are far larger than any external force, so the CED's collision model treats the total momentum as constant from the instant before impact to the instant after.

Is momentum conserved in every collision?

Essential knowledge 4.3.B.1 says momentum is conserved in all interactions, and 4.3.B.2 gives the condition for your particular system: if the net external force on the selected system is zero, the total momentum of the system is constant. So yes for the standard exam setup of two objects on a frictionless horizontal surface, and yes for elastic, inelastic, and perfectly inelastic collisions alike. Collision type changes what happens to kinetic energy, not to momentum.

Why can internal forces never change a system's momentum?

Because they come in Newton's third law pairs that act for the same length of time. Statement 4.3.A.3.i says the impulse one object exerts on a second is equal and opposite to the impulse the second exerts on the first, so the two momentum changes cancel exactly when you add them. This holds no matter how large the internal forces are, which is why an explosion inside a system still leaves the total momentum untouched.

How do you choose the system in a momentum problem?

Draw the boundary around every object that takes part in the interaction, so all of their forces on each other become internal. Then check what still crosses the boundary. On a frictionless horizontal track, gravity and the normal force cancel and nothing does, so the total momentum is constant. Statement 4.3.A.3.ii puts it as a choice rather than a discovery: a system may be selected so that the total momentum of that system is constant.

Does conservation of momentum apply to explosions?

Yes. Statement 4.3.A.4 names collisions and explosions together, and 4.1.A.3.iii defines an explosion as an interaction in which forces internal to the system move objects within that system apart. Because those forces are internal, a system that starts at rest must end with zero total momentum, so the fragments move in opposite directions with speeds in inverse proportion to their masses.