Conservation of Momentum for Isolated Systems

AP Physics 1· AP Physics 1 CED — Momentum· 14 min read

1. Core Law of Conservation of Momentum★★☆☆☆⏱ 3 min

Conservation of momentum is one of three fundamental conservation laws tested in AP Physics 1 Unit 5, making up 12–18% of total exam score. It appears in both multiple-choice and free-response sections, often combined with energy conservation or kinematics for multi-step problems.

📘 Definition

Law of Conservation of Momentum

\sum p_{initial} = \sum p_{final}

The total linear momentum of an isolated system remains constant over time, regardless of interactions between objects inside the system.

📘 Definition

Isolated System

A system where the net external force acting on the system equals zero. Momentum is only conserved for isolated systems, and can be conserved in one direction even if not conserved in another.

p_{total, initial} = p_{total, final} = \sum p_i = \sum p_f

Since momentum is a vector quantity, the law holds component-wise. This vector nature is the most commonly tested feature on the AP Physics 1 exam.

2. Classifying Internal vs External Forces★★☆☆☆⏱ 4 min

To apply conservation of momentum, your first step in any problem is to correctly define your system and classify forces as internal or external. This is the most frequently tested skill in AP momentum problems.

📘 Definition

Internal Force

A force exerted by one object inside the system on another object also inside the system. By Newton's third law, all internal forces cancel out, so they contribute zero net impulse to total system momentum.

📘 Definition

External Force

A force exerted by an object outside the system on an object inside the system. Only non-zero net external force changes total system momentum.

From the impulse-momentum theorem, we derive the conservation law:

\Delta p_{total} = J_{net} = \sum F_{ext} \Delta t \implies \sum F_{ext} = 0 \rightarrow \Delta p_{total} = 0

📐 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 $\sum F_{e x t, v er t} = 0$ . Horizontal direction: the track is frictionless, so there are no external horizontal forces, so $\sum F_{e x t, h or z} = 0$ .
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. One-Dimensional Collisions★★★☆☆⏱ 3 min

Collisions are the most common context for momentum conservation on the AP exam. During a collision, interaction time between objects is very short, so even if a small external force (like friction) acts, the impulse from the external force is negligible compared to the impulse from large internal collision forces. We almost always treat the system of colliding objects as isolated during the collision.

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, $v_{1 i} = + 20$ m/s; $m_{2} = 1000$ kg, $v_{2 i} = - 30$ m/s. After collision, combined mass $M = 1500 + 1000 = 2500$ kg.
2
Apply conservation of momentum:
3
$m_1 v_{1i} + m_2 v_{2i} = M v_f$
4
Substitute values:
5
$(1500)(20) + (1000)(-30) = 2500 v_f \implies 30000 - 30000 = 2500 v_f$
6
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★★★☆☆⏱ 3 min

Explosions are the reverse of perfectly inelastic collisions: a single object splits into two or more fragments due to internal forces from released stored 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} \implies 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 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:
4
$v_2 = -\frac{m_1}{m_2} v_1 = -\frac{2.0}{4.0}(-12) = +6 \text{ m/s}$
5
Conclusion: 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★★★☆☆⏱ 3 min

A key conceptual result of momentum conservation is that the velocity of the center of mass ( $v_{C M}$ ) 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 $p_{t o t a l}$ is constant for an isolated system, $v_{C M}$ must also be constant. If the system is initially at rest, $v_{C M} = 0$ forever as long as the system stays isolated. This is a great check for your calculations: if your final $v_{C M}$ doesn't match the initial $v_{C M}$ , 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 $v_{C M} = 0$ 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; $M_{t o t a l} = 6.0$ kg.
3
Calculate final $v_{C M}$ :
4
$v_{CM} = \frac{(2.0)(-12) + (4.0)(6)}{6.0} = \frac{-24 + 24}{6.0} = 0 \text{ m/s}$
5
The center of mass velocity stays zero, confirming our earlier calculation is correct.

Exam tip:

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

6. Concept Check★★★☆☆⏱ 2 min

✓ Quick check

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
Reveal answer
C —
The system of two carts is isolated, so total momentum is conserved from an initial value of zero. The compressed spring's elastic potential energy converts to kinetic energy, so total kinetic energy is non-zero after release.

7. Common Pitfalls

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, e.g. a block-cart system with ground friction

Why:

Students assume momentum is always conserved, regardless of system choice

Correct move:

Always check that $\sum F_{e x t} = 0$ (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 chosen 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

8. Quick Reference Cheatsheet

Category	Formula	Notes
Isolated system condition	$\sum F_{e x t} = 0$	Momentum can be conserved in one direction even if not another
Law of Conservation of Momentum	$\sum p_{ini t ia l} = \sum p_{f ina l}$	Vector law, apply component-wise; holds during collision/explosion when external impulse is negligible
Two-object 1D collision	$m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f}$	Applies to all collisions (elastic and inelastic) for isolated systems
Perfectly inelastic collision (stick together)	$m_{1} v_{1 i} + m_{2} v_{2 i} = (m_{1} + m_{2}) v_{f}$	Final velocities are equal for both objects
Explosion from rest	$0 = m_{1} v_{1 f} + m_{2} v_{2 f}$	Fragments move in opposite directions; total kinetic energy increases
Center of mass velocity	$v_{C M} = \frac{p _{t o t a l}}{M _{t o t a l}} = \frac{\sum m _{i} v _{i}}{\sum m _{i}}$	Constant for all isolated systems
Kinetic energy for momentum-conserving systems	Not necessarily conserved	Only conserved for elastic collisions
Impulse-momentum relation	$Δ p_{t o t a l} = F_{n e t, e x t} Δ t$	For non-isolated systems, relates momentum change to external impulse

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Isolated system force classification
2022 · FRQ
Explosion momentum conservation problem

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. This core law is a frequent topic in multi-step free-response questions that combine multiple concepts, so solidifying your understanding of system definition and force classification will pay off across the entire unit. 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 with energy concepts for higher scoring points on the AP exam.