Topic 4.1: Linear Momentum

What Topic 4.1 requires

Topic 4.1 carries one learning objective, 4.1.A: describe the linear momentum of an object or system. Three essential knowledge statements sit under it, and the last one adds three sub-statements of its own.

• 4.1.A.1 gives the equation $\vec{p} = m\vec{v}$.
• 4.1.A.2 states that momentum is a vector quantity and has the same direction as the velocity.
• 4.1.A.3 states that momentum can be used to analyze collisions and explosions.
• 4.1.A.3.i defines a collision as a model for an interaction where the forces exerted between the involved objects in the system are much larger than the net external force exerted on those objects during the interaction.
• 4.1.A.3.ii says that because only the initial and final states of a collision are analyzed, the object model may be used to analyze collisions.
• 4.1.A.3.iii defines an explosion as a model for an interaction in which forces internal to the system move objects within that system apart.

The boundary statement is a single line, and it is the reason the rest of this page can drop an adjective: unless otherwise stated, the general term "momentum" refers specifically to linear momentum. Angular momentum arrives in Unit 6, and the boundary statement is what lets you read an unlabeled momentum as the linear kind.

Notice what 4.1.A does not ask for. There is no conservation law here, no impulse, and no elastic-versus-inelastic sorting. Topic 4.1 is the definition and the vector bookkeeping, and it is deliberately placed before Topic 4.2 so that every later equation in the unit has a well defined quantity to move around.

The CED lists four suggested skills for this topic: 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Unit 4 is weighted at 10 to 15 percent of the multiple-choice section, and the CED suggests roughly 10 to 15 class periods for the whole unit.

Reading p = mv term by term

Mass in kilograms times velocity in meters per second gives momentum in kilogram meters per second. That unit has no special name, unlike the joule or the newton, so you write it out: kg·m/s. It is worth knowing that a newton second is the same unit, since $1 \, \mathrm{N} = 1 \, \mathrm{kg \cdot m/s^2}$ and multiplying by a second cancels one power of time. Impulse and momentum are measured in the same unit for exactly that reason, which is the first hint of the theorem waiting in Topic 4.2.

The mass is a positive scalar, so multiplying the velocity by it changes the length of the arrow and never its direction. That is the whole content of 4.1.A.2 in geometric form. In one dimension the practical consequence is simple: the sign of $p$ is always the sign of $v$.

Both variables appear to the first power. Double the speed and the momentum doubles. Double the mass and the momentum doubles. Nothing gets squared, which is a difference from translational kinetic energy, where the speed carries an exponent of 2. Keep that asymmetry in mind: it is behind the comparison questions this unit asks.

The magnitude alone is $p = mv$ with $v$ the speed. Use that form when a question asks how much momentum an object has without asking which way it points, and keep the full vector form whenever a system has parts moving in different directions.

Momentum is a vector, so signs are not optional (4.1.A.2)

Every momentum problem starts with one decision that never appears in the answer: which direction is positive. Write it down before any arithmetic. After that, a velocity of 6.0 m/s to the left is $-6.0$ m/s if right is positive, and a 4.0 kg object carrying it has $p = -24$ kg·m/s. The minus sign is not a mistake to clean up later; it is the direction, stored as arithmetic.

Three habits follow from that.

• Attach a sign to every velocity as you write it, not after you finish the algebra.
• Keep the same positive direction for the whole problem, including after a collision, when an object may have reversed.
• Read the sign of your answer back as a direction. A final momentum of $-2.0$ kg·m/s means 2.0 kg·m/s to the left, not a negative amount of anything.

In two dimensions the same rule runs on components: $p_x = mv_x$ and $p_y = mv_y$, each with its own sign. The boundary statement for Topic 4.3 restricts AP Physics 1 to a quantitative and qualitative treatment of conservation of momentum in one dimension and a semiquantitative treatment of it in two dimensions, so two-dimensional questions ask you to set up components and reason about them rather than grind out simultaneous equations.

The cleanest way to keep the vector nature in view is to hold momentum next to a scalar you already know.

The last two rows are where exam questions live. Two identical carts rolling toward each other at the same speed have zero total momentum and a perfectly ordinary, nonzero total kinetic energy at the same instant. Neither total is wrong, and noticing that they disagree is usually the point of the question.

The momentum of a system, not just an object

Learning objective 4.1.A says "object or system," and the system half is what the rest of the unit runs on. The total momentum of a system is the sum of the momenta of its parts, essential knowledge 4.3.A.2. The CED states that in words and prints no equation for it, so write the vector sum out yourself:

Sum with signs, never with magnitudes. Adding $|{+}12|$ and $|{-}10|$ to get 22 kg·m/s produces a number that means nothing; the system's momentum is $+2.0$ kg·m/s.

There is a second way to write the same total, and it is on the AP Physics 1 equation sheet. The velocity of a system's center of mass is

Rearranged, that says the total momentum equals the system's total mass times the velocity of its center of mass. A collection of objects with individual momenta can therefore be described as one system with one center-of-mass velocity, which is essential knowledge 4.3.A.1 and the reason Topic 2.1 spent its time on choosing systems. Choosing a good system in Unit 4 means choosing one whose center of mass does something simple.

A useful check on any multi-object answer: compute the total momentum, divide by the total mass, and ask whether the resulting center-of-mass velocity is believable. If two carts are approaching each other with equal and opposite momenta, their center of mass is not moving at all, before the collision or after it.

Momentum depends on the reference frame

Momentum is built from velocity, and velocity is measured relative to a frame, so a single object has as many momentum values as there are observers. This is not a subtlety to file away; it decides whether your before-and-after numbers are comparable.

Be precise about where this appears in the CED. Unit 4 does not print a frame-dependence statement, but Unit 1 does, and it is general: essential knowledge 1.4.A.1 says the choice of reference frame will determine the direction and magnitude of quantities measured by an observer in that reference frame. Momentum is one of those quantities. Unit 3 says the same thing for one specific quantity, at essential knowledge 3.1.A.3: different observers may measure different values of the translational kinetic energy of an object, depending on the observer's frame of reference. Momentum inherits the same dependence through the same route, because both quantities are built from a velocity that the frame defines. The difference is that a frame change can flip the sign of a momentum, while kinetic energy, built from a squared speed, only changes size.

In this course the mass is the same for every observer, so all of the frame dependence sits in $\vec{v}$. Frame conversions here are one-dimensional velocity arithmetic, the restriction set by the Topic 1.4 boundary statement: convert the velocity into the frame you want, keeping its sign, then multiply by the mass.

Two consequences are worth stating on their own.

• Every object has a frame in which its momentum is zero. It is the frame moving with the object. Nothing physical changed; the observer did.
• Every system has a frame in which its total momentum is zero. It is the frame moving with the system's center of mass, at the $\vec{v}_{\text{cm}}$ above. Explosions and head-on collisions are often set up in exactly this frame, which is why so many textbook problems begin with a total momentum of zero.

What does not depend on the frame is whether momentum is conserved. Pick any inertial frame, and if the net external force on your system is zero the total momentum in that frame stays constant. The value differs from frame to frame; the constancy does not. The one rule that keeps this from causing errors: choose a frame before you write the first equation and stay in it, because a before-value from one frame and an after-value from another compare nothing.

Collisions and explosions as models (4.1.A.3)

Essential knowledge 4.1.A.3 introduces the two situations the rest of the unit analyzes, and the CED defines both as models rather than as events, which matters for how you are allowed to use them.

A collision is a model for an interaction in which the forces the objects exert on each other are much larger than the net external force on those objects during the interaction (4.1.A.3.i). That inequality is a license, not a description of what a crash looks like. Because the internal contact forces dominate for the brief time they act, external forces such as gravity and track friction change the system's momentum by a negligible amount over that interval, and you may ignore them. It is why a problem can tell you a bat strikes a ball and expect you to leave the ball's weight out of the impulse calculation entirely. Check the license when the contact time is long or the external force is large, because then the approximation is the thing being tested.

The object model applies because only the initial and final states of a collision are analyzed (4.1.A.3.ii). Whatever happens during contact, the deformation, the sound, the heating, sits between the two states you actually use. Two carts can be treated as two points carrying mass and velocity, which is why collision problems never ask about the shape of the bumper.

An explosion is a model for an interaction in which forces internal to the system move objects within that system apart (4.1.A.3.iii). Recoil from a fired projectile, a compressed spring released between two carts, and a skater pushing off a partner are all the same model. The structure is a collision run backwards: one object before, several after, and the same internal-forces-dominate reasoning.

Both models pay off in Topic 4.3, where the choice of system turns those internal forces into a guarantee that the total momentum does not change, and in Topic 4.4, where the kinetic energy tells you which kind of collision you have. The conservation of momentum guide walks the full before-and-after procedure for both.

Momentum and kinetic energy answer different questions

Both quantities are built from mass and velocity, so students often treat them as interchangeable measures of "how much motion." Combining the two definitions shows why they are not. Substituting $v = p/m$ into $K = \frac{1}{2}mv^2$ gives

Neither of these is printed on the AP Physics 1 formula sheet; both come out of two lines of algebra using $p = mv$ and $K = \frac{1}{2}mv^2$, which are printed. Suggested skill 2.C is exactly the habit of comparing two scenarios like this, so practice reading them as scaling rules. In the table, object 2 is the heavier of the pair.

The physical reading of the middle two rows: momentum measures how hard it is to stop something in a given time, and kinetic energy measures how hard it is to stop it in a given distance. A loaded truck and a bullet can carry the same kinetic energy while the truck carries far more momentum, so the truck needs a much larger impulse to stop even though both take the same amount of work. Topic 4.2 supplies the time side of that sentence and the work-energy theorem supplies the distance side.

How Topic 4.1 is tested, and where it leads

Topic 4.1 usually shows up as the first step of a longer problem. It shows up as the first step of a longer problem and as the source of the sign errors that spoil the rest of it. The patterns to rehearse:

1. Compute a system's total momentum from a picture or a table of masses and velocities, with at least one object moving the other way (skills 2.B and 2.C).
2. Rank or compare the momenta of several objects given their kinetic energies, or the reverse (skill 2.C).
3. Sketch momentum against time or momentum against velocity for a described motion (skill 1.C). Against velocity the graph is a straight line through the origin with slope equal to the mass, which is a standard way to measure a mass from data.
4. Decide whether an interaction may be modeled as a collision, and justify it by comparing the internal forces with the net external force (skill 3.B and essential knowledge 4.1.A.3.i).
5. Explain why the total momentum of a system can be zero while its total kinetic energy is not.

From here the unit builds in one direction. Topic 4.2 asks what changes a momentum, and answers with impulse. Topic 4.3 asks when a total momentum cannot change, and answers with the net external force. Topic 4.4 sorts collisions by what happens to the kinetic energy. The momentum and collision calculator will check any before-and-after set of numbers you produce along the way, and the Unit 4 overview shows how the four topics fit together. The whole vector discipline returns in Unit 6 as angular momentum, so the sign habits you build here get used twice.