Conservation of Momentum for Isolated Systems — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Isolated system definition, internal vs external force classification, the law of conservation of momentum, one-dimensional collision/explosion problem solving, and center of mass velocity for isolated systems.

You should already know:

• Definition of linear momentum as a vector quantity
• The impulse-momentum theorem
• Newton's third law of motion

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Conservation of Momentum for Isolated Systems?

Conservation of momentum for isolated systems is one of three fundamental conservation laws tested in AP Physics 1 Unit 5 (Momentum), which accounts for 12–18% of the total exam score per the official CED. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with energy conservation or kinematics for multi-step problems. The law states that the total linear momentum of an isolated system remains constant over time, regardless of interactions between objects inside the system. An isolated system is defined as any system where the net external force acting on the system is zero. Synonyms for this law include the law of momentum conservation. The core notation for the law is $p_{total,initial}=p_{total,final}$, or more generally $\sum p_i=\sum p_f$. Since momentum is a vector, the law holds component-wise: momentum can be conserved in one direction even if it is not conserved in another, depending on the direction of net external force. This vector nature is the most commonly tested feature of the law on the AP exam.

2. Classifying Internal vs External Forces to Identify Isolated Systems

To apply conservation of momentum, your first step in any problem must be to correctly define your system and classify forces as internal or external. An internal force is any force exerted by one object inside the system on another object also inside the system. By Newton's third law, every internal force has an equal and opposite internal reaction force, so the total impulse from all internal forces adds to zero. An external force is any force exerted by an object outside the system on an object inside the system. A system is isolated if the sum of all external forces is zero ($\sum F_{ext}=0$). From the impulse-momentum theorem, $\Delta p_{total}=J_{net}=\sum F_{ext}\Delta t$, so if $\sum F_{ext}=0$, $\Delta p_{total}=0$, which gives us the core law: $$\sum p_{initial}=\sum p_{final}$$ It is extremely common for momentum to be conserved in one direction but not another. For example, a student jumping off a cart on a frictionless track has no net horizontal external force, so horizontal momentum is conserved, but there is a net vertical external force from the track during the jump, so vertical momentum is not conserved.

Worked Example

A 60 kg student stands on a 15 kg cart at rest on a horizontal frictionless track. If we define the system as the student + cart, classify all forces acting on the system and confirm if the system is isolated in the horizontal direction.

1. List all forces acting on the system: gravity on the student, gravity on the cart, normal force from the track on the cart, the contact force the student exerts on the cart, and the contact force the cart exerts on the student.
2. Classify forces: Gravity and the normal force come from objects outside the system (Earth and the track), so they are external. The two contact forces act between objects inside the system, so they are internal.
3. Calculate net external force by direction: Vertical direction: total downward gravitational force equals the upward normal force, so $\sum F_{ext,vert}=0$. Horizontal direction: the track is frictionless, so there are no external horizontal forces, so $\sum F_{ext,horz}=0$.
4. Conclusion: The system is isolated in the horizontal direction, so momentum is conserved horizontally.

Exam tip: Never assume a system is fully isolated across all directions. Always check net external force direction by direction — AP exam questions intentionally design problems where momentum is only conserved in one direction to test this skill.

3. Conservation of Momentum in One-Dimensional Collisions

Collisions are the most common context for momentum conservation on the AP exam. During a collision, the interaction time between the two objects is very short, so even if a small external force (like friction) acts on the system, the impulse from the external force ($F_{ext}\Delta t$) is negligible compared to the impulse from the large internal collision force. This means we can almost always treat the system of colliding objects as isolated during the collision, so momentum is conserved. For two objects colliding along a straight line, the conservation equation becomes: $$m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}$$ All velocities are signed based on your chosen positive direction. Critically, this equation applies to all collisions (both elastic and inelastic) for isolated systems — only kinetic energy changes between collision types, momentum is always conserved.

Worked Example

A 1500 kg pickup truck traveling north at 20 m/s collides head-on with a 1000 kg sedan traveling south at 30 m/s. The two vehicles lock together on impact. What is their velocity immediately after the collision?

1. Choose north as the positive direction. Assign values: $m_1=1500$ kg, $v_{1i}=+20$ m/s; $m_2=1000$ kg, $v_{2i}=-30$ m/s. After collision, combined mass $M=1500+1000=2500$ kg.
2. Apply conservation of momentum: $m_1{v}_{1i}+m_2{v}_{2i}=M{v}_f$
3. Substitute values: $(1500)(20)+(1000)(-30)=2500v_f⟹30000-30000=2500v_f$
4. Solve: $v_f=0$ m/s. The two vehicles come to rest immediately after impact.

Exam tip: The answer zero is a perfectly valid result for momentum problems! Don't automatically assume you made a mistake if your final velocity is zero — it just means the initial momenta of the two objects canceled out exactly.

4. Conservation of Momentum in Explosions

Explosions are the reverse of perfectly inelastic collisions: a single object splits into two or more fragments due to internal forces from released energy (chemical, elastic potential, etc.). Like collisions, explosions happen over a very short time, so external impulse is negligible, and momentum is conserved for the system of fragments. A common exam scenario is an object that is momentarily at rest before exploding, so total initial momentum is zero. This gives the simplified equation: $$0=m_1{v}_{1f}+m_2{v}_{2f}⟹v_{2f}=-\frac{m_1}{m_2}{v}_{1f}$$ The negative sign confirms the two fragments move in opposite directions. Unlike collisions, explosions always have an increase in total kinetic energy, as stored potential energy is converted to motion of the fragments.

Worked Example

A 6.0 kg model rocket is at rest horizontally at the top of its flight when it suffers a structural failure and splits into two fragments: a 2.0 kg engine section and a 4.0 kg payload section. The engine section moves west at 12 m/s immediately after the split. What is the velocity of the payload section?

1. The system of the two fragments is isolated horizontally (gravity acts vertically, so no net horizontal external force). Initial total momentum is zero, since the rocket is at rest before splitting.
2. Choose east as the positive direction, so the engine velocity is $v_1=-12$ m/s.
3. Substitute into the explosion equation: $v_2=-\frac{m_1}{m_2}{v}_1=-\frac{2.0}{4.0}(-12)=+6$ m/s.
4. The payload section moves east at 6 m/s immediately after the split.

Exam tip: Always check if the original object is moving or at rest before the explosion. Don't default to zero initial momentum if the problem states the object is moving horizontally before exploding.

5. Center of Mass Velocity for Isolated Systems

A key conceptual result of momentum conservation is that the velocity of the center of mass ($v_{CM}$) of an isolated system never changes, even if objects inside the system move relative to each other. The formula for center of mass velocity is: $$v_{CM}=\frac{\sum m_i{v}_i}{M_{total}}=\frac{p_{total}}{M_{total}}$$ Since $p_{total}$ is constant for an isolated system, $v_{CM}$ must also be constant. If the system is initially at rest, $v_{CM}=0$ forever, as long as the system stays isolated. This is a great check for your calculations: if your final $v_{CM}$ doesn't match the initial $v_{CM}$, you made an algebra or sign error.

Worked Example

For the rocket split in the previous worked example, confirm that the center of mass velocity after the split matches the initial value for the isolated system.

1. Initial velocity of the rocket is 0 m/s, so initial $v_{CM}=0$ m/s.
2. List final values: $m_1=2.0$ kg, $v_1=-12$ m/s; $m_2=4.0$ kg, $v_2=+6$ m/s; $M_{total}=6.0$ kg.
3. Calculate final $v_{CM}$: $v_{CM}=\frac{(2.0)(-12)+(4.0)(6)}{6.0}=\frac{-24+24}{6.0}=0$ m/s.
4. The center of mass velocity stays zero, confirming our earlier calculation is correct.

Exam tip: For conceptual FRQ questions asking why $v_{CM}$ doesn't change, always structure your answer as: 1) The system is isolated, so $\sum F_{ext}=0$; 2) Therefore total momentum is conserved; 3) $v_{CM}=p_{total}/M_{total}$, so $v_{CM}$ is constant. This is the exact reasoning AP graders look for.

Common Pitfalls (and how to avoid them)

• Wrong move: Classifying gravity or normal force as internal when the system does not include Earth. Why: Students forget that Earth is outside the system unless explicitly included, so all forces from Earth are external. Correct move: Always list all objects inside your system at the start of the problem, and mark any force from an unlisted object as external.
• Wrong move: Applying momentum conservation to a system with non-zero net external force. For example, applying it to a block-cart system with friction between the cart and the ground. Why: Students assume momentum is always conserved, regardless of system choice. Correct move: Always check that $\sum F_{ext}=0$ (or external impulse is negligible) in the direction you are working before writing $p_i=p_f$.
• Wrong move: Dropping negative signs for velocities of objects moving opposite your positive direction. Why: Students treat momentum as a scalar instead of a vector, and add all speeds regardless of direction. Correct move: Explicitly write your chosen positive direction on the page, then assign negative signs before plugging values into the equation.
• Wrong move: Applying momentum conservation across the entire problem, including after the collision/explosion when external forces act. Why: Students confuse conservation during the collision event with conservation after, when external forces like friction change momentum. Correct move: Only apply conservation during the collision/explosion, when external impulse is negligible.
• Wrong move: Assuming kinetic energy is always conserved when momentum is conserved. Why: Students confuse the two conservation laws, and assume both hold for all isolated systems. Correct move: Only assume kinetic energy is conserved if the problem explicitly states the collision is elastic.

Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

Two stationary carts of equal mass sit on a frictionless horizontal track, with a compressed massless spring between them. When the spring is released, the carts push off each other. Which of the following correctly describes the total momentum and total kinetic energy of the system of two carts after the spring is released? A. Total momentum is zero, total kinetic energy is zero B. Total momentum is non-zero, total kinetic energy is zero C. Total momentum is zero, total kinetic energy is non-zero D. Total momentum is non-zero, total kinetic energy is non-zero

Worked Solution: The system of two carts is isolated on the frictionless track, so total momentum is conserved. The system starts at rest, so total initial momentum is zero, which means total final momentum must also be zero. The compressed spring stores elastic potential energy, which is converted to kinetic energy of the two carts when the spring expands, so total kinetic energy after release is non-zero. Only option C matches this description. Correct answer: C.

Question 2 (Free Response)

A 80 kg astronaut floats at rest next to a 240 kg space probe, both initially at rest relative to a nearby space station. The astronaut pushes off the probe, giving themselves a speed of 0.6 m/s away from the station. (a) Calculate the speed of the probe after the push. (b) Is the momentum of the astronaut alone conserved during the push? Explain your reasoning. (c) If the push lasts for 0.20 seconds, what is the magnitude of the average force the astronaut exerts on the probe?

Worked Solution: (a) The system of astronaut + probe is isolated, so total momentum is conserved. Initial momentum is zero, so $0=m_a{v}_a+m_p{v}_p$. Taking away from the station as positive, $v_a=+0.6$ m/s. Solving for $v_p$: $v_p=-\frac{m_a{v}_a}{m_p}=-\frac{(80)(0.6)}{240}=-0.2$ m/s. The speed of the probe is 0.2 m/s toward the station. (b) No, the momentum of the astronaut alone is not conserved. The probe exerts an external force on the astronaut during the push, so there is a non-zero net external force on the system of just the astronaut. The astronaut's momentum changes from zero to $m_a{v}_a$, so it is not conserved. (c) Use the impulse-momentum theorem on the probe: $\Delta p_p=F_{avg}\Delta t$. $\Delta p_p=m_p{v}_p-0=240(-0.2)=-48$ kg·m/s. The magnitude of the force is $|F_{avg}|=\frac{|\Delta p_p|}{\Delta t}=\frac{48}{0.20}=240$ N. By Newton's third law, this is the force the astronaut exerts on the probe.

Question 3 (Application / Real-World Style)

A 70 kg hunter holding a 4.0 kg rifle stands on frictionless ice on a frozen pond. The hunter fires a 12 g bullet horizontally with a muzzle velocity of 650 m/s. The hunter and rifle recoil together. What is the recoil speed of the hunter and rifle immediately after firing?

Worked Solution: Convert all units to SI: bullet mass $m_b=12$ g = 0.012 kg. Total mass of hunter + rifle $m_{hr}=70+4.0=74$ kg. The system is isolated horizontally on frictionless ice, initially at rest, so total initial momentum is zero. Conservation of momentum gives $0=m_b{v}_b+m_{hr}{v}_{hr}$. Solving for recoil speed: $|v_{hr}|=\frac{m_b|v_b|}{m_{hr}}=\frac{(0.012)(650)}{74}\approx 0.11$ m/s. A recoil speed of ~0.1 m/s matches real-world experience: it is slow enough that the hunter does not slide rapidly across the ice after firing.

Quick Reference Cheatsheet

What's Next

Mastering conservation of momentum for isolated systems is the foundation for all upcoming work on collisions, energy, and rotational motion in AP Physics 1. Next, you will apply this core law to classify collisions as elastic or inelastic, compare momentum and kinetic energy conservation in different collision types, and solve multi-step problems that combine momentum conservation with kinematics and energy conservation. Without a solid understanding of how to identify isolated systems and apply momentum conservation, you will struggle to solve collision problems, which are a major component of both MCQ and FRQ sections of the AP exam. This topic also generalizes to rotational motion, where conservation of angular momentum for isolated systems follows the same core logic.

• Impulse and Momentum Change
• Elastic and Inelastic Collisions
• Center of Mass Motion
• Conservation of Angular Momentum