System Models — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Defining systems and system boundary notation, distinguishing internal vs external forces, classifying open vs closed systems, applying Newton’s second law to multi-body systems, and system selection strategy for solving connected-object dynamics problems.

You should already know: Newton's three laws of motion, vector addition of forces, kinematics for constant acceleration.

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 System Models?

A system model is simply the choice of which objects to group together and analyze as a single collection, rather than analyzing each object individually. We mark the boundary of the system with a dashed line: every object inside the dashed line is part of the system, and everything outside (called the surroundings) is external. This topic is a core component of Unit 2: Dynamics, which makes up 12-18% of the total AP Physics 1 exam score, and system model concepts appear in both multiple choice and free response questions, most often in multi-object connected problems. System modeling is a problem-solving strategy, not just a conceptual definition: it simplifies complex problems by eliminating internal forces that cancel out by Newton’s third law, so you only need to solve for external forces first, then solve for internal forces if required. Synonyms you may see in exam questions include “combined system” or “system of particles”, both meaning the same grouping of objects for analysis. The key skill AP exams test here is your ability to choose the optimal system boundary to reduce the amount of algebra and avoid errors.

2. Internal vs. External Forces

By definition, any force acting on an object inside the system is either internal or external, depending on where the force originates. An internal force is an interaction between two objects that are both inside the system boundary. By Newton’s third law, every internal action force has an equal and opposite reaction force, also internal to the system. When summing net force for the entire system, these pairs cancel out: $\sum {\vec{F}}_{\text{int}}=0$, so internal forces never contribute to the acceleration of the system as a whole. An external force is an interaction between an object inside the system and an object outside the system (in the surroundings). There is no matching reaction force inside the system, so external forces do not cancel, and they determine the net force on the entire system. For any system, Newton’s second law becomes: $$\sum {\vec{F}}_{\text{ext}}=M{\vec{a}}_{cm}$$ where $M$ is the total mass of all objects in the system, and ${\vec{a}}_{cm}$ is the acceleration of the system’s center of mass. For connected objects that all move with the same acceleration, ${\vec{a}}_{cm}$ equals the acceleration of every object in the system.

Worked Example

Two blocks of mass $2.0\text{ kg}$ and $3.0\text{ kg}$ are connected by a massless inextensible string on a frictionless horizontal table. A $10\text{ N}$ horizontal pulling force is applied to the $3.0\text{ kg}$ block to the right. If we define the system as both blocks, identify all internal and external forces, then find the acceleration of the system.

1. Draw a dashed boundary around both blocks to mark the system.
2. Classify all forces: Internal forces include tension from the string (the string pulls both blocks, both inside the system, so tension is internal). External forces include: weight of both blocks (from Earth, outside the system), normal force from the table (from the table, outside the system), and the 10 N pulling force (from the puller, outside the system).
3. Sum vertical external forces: total weight equals total normal force, so they cancel out, leaving only the 10 N pulling force as net external force.
4. Substitute into Newton's second law for systems: $10\text{ N}=(2.0+3.0)a⟹a=2.0{\text{ m/s}}^2$ to the right.

Exam tip: Always confirm if a force is internal by checking if both the action and reaction forces are between objects inside your selected system. If yes, you can completely ignore it in net force calculations.

3. System Selection Strategy for Internal Force Calculations

You can choose any system boundary you want for a problem — there is no “wrong” choice, but some choices are far simpler than others. The standard strategy for connected-object problems (where you need to find an internal force like tension or normal force between two connected objects) is:

1. First select the combined system of all connected objects (all moving with the same acceleration) to find the acceleration of the whole system, ignoring all internal forces.
2. Then select a smaller subsystem (usually one of the individual objects) to solve for the internal force of interest, because the internal force for the combined system becomes an external force for the smaller subsystem.

This strategy cuts down on the number of equations you need to solve simultaneously, reducing the chance of algebra and sign errors. It works for any connected objects moving with the same acceleration, including blocks connected by strings, stacked blocks moving together, Atwood machines, and multiple trailers pulled by a truck.

Worked Example

Using the same 2.0 kg and 3.0 kg blocks from the previous example (10 N pull on the 3.0 kg block, frictionless table), find the tension in the connecting string between the two blocks.

1. We already used the combined system to find the acceleration of both blocks: $a=2.0{\text{ m/s}}^2$.
2. Select the subsystem that only includes the 2.0 kg trailing block. The only horizontal external force acting on this subsystem is the tension $T$ pulling it to the right.
3. Apply Newton's second law to this subsystem: $\sum F_{\text{ext}}=T=m_{2}a=(2.0\text{ kg})(2.0{\text{ m/s}}^2)=4.0\text{ N}$.
4. To confirm, select the 3.0 kg block as the subsystem: net force is $10\text{ N}-T=(3.0\text{ kg})(2.0{\text{ m/s}}^2)⟹T=4.0\text{ N}$, which matches our result.

Exam tip: When asked for an internal force between connected objects, always analyze the trailing/front object that only has the internal force as its horizontal external force — this eliminates the need for subtraction and avoids sign errors.

4. Open vs. Closed Systems

A closed system is defined as a system where no mass crosses the system boundary during the entire process we are analyzing. All mass that is inside the boundary at the start stays inside, and no mass enters from outside. An open system is a system where mass crosses the boundary during the process: mass can enter (like sand being dropped into a cart) or leave (like fuel being burned and expelled from a rocket).

For AP Physics 1, almost all dynamics problems you will solve use closed systems, because standard Newton’s second law for systems applies directly to closed systems, with no extra terms for mass flow. When mass does enter or leave an object (the cart, the rocket), you can almost always expand your system boundary to include all the mass involved in the process (the cart plus the sand before it is dropped, for example) to make the system closed, simplifying your calculation. AP Physics 1 only tests conceptual understanding of open vs closed systems, not the full rocket equation for open systems with mass flow.

Worked Example

A 15 kg cart is rolling at constant speed across frictionless ground, when a 5 kg bag of sand is dropped vertically into the cart from above. (a) Is the cart alone an open or closed system during this process? (b) Is the system that includes the cart and the sand (from before the drop) open or closed?

1. Recall the definition: a system is open if mass crosses the boundary during the process.
2. For the cart alone: Before the drop, only 15 kg of cart is inside the boundary. After the drop, the 5 kg of sand is inside the cart, so 5 kg of mass entered the boundary. This means the cart alone is an open system.
3. For the system that includes both the 15 kg cart and 5 kg sand before the drop: All 20 kg of mass is inside the boundary before and after the drop. No mass enters or leaves, so this is a closed system.
4. Conceptual note: Using the closed system lets us ignore the internal impact forces between the sand and cart, making it much easier to find the final speed of the loaded cart.

Exam tip: When solving any dynamics or momentum problem, always expand your boundary to create a closed system if possible — it eliminates extra complexity from mass flow and lets you use standard formulas directly.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Counting tension between two connected blocks as an external force when both blocks are included in the system. Why: Students often remember tension as a pulling force, so they incorrectly add it to the net external force sum. Correct move: After drawing your system boundary, check if both ends of the string are connected to objects inside the boundary — if yes, tension is internal, ignore it.
• Wrong move: Using a single acceleration for the whole system when individual objects accelerate in different directions or at different magnitudes. Why: Students assume all objects in a system share the same acceleration, which is only true for rigidly connected moving objects. Correct move: If objects have different accelerations, use the full form $\sum F_{\text{ext}}=m_1{a}_1+m_2{a}_2$ instead of $M{a}_{cm}$ with a single $a$.
• Wrong move: Treating the cart alone as a closed system when mass is being added or removed during the process. Why: Students default to closing the system around the obvious object (the cart) without checking for mass flow. Correct move: If mass is entering/leaving the obvious object, expand the system boundary to include all mass involved in the process from start to finish to make it closed.
• Wrong move: Solving for tension by analyzing the pulled block first, leading to sign errors when subtracting tension from the applied force. Why: Students pick the pulled object first out of habit, leading to incorrect sign when rearranging equations. Correct move: When solving for tension between two connected objects, always analyze the trailing object (the one only pulled by tension) first, it has only one horizontal force so no subtraction is needed.
• Wrong move: Forgetting to include weight and normal force as external forces when friction is present. Why: Students get focused on horizontal forces and ignore vertical forces, which are needed to calculate friction. Correct move: Always list all external forces (vertical and horizontal) first before summing, even if you expect vertical forces to cancel.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

Two blocks of mass 1.0 kg and 4.0 kg are stacked on top of each other on a frictionless horizontal floor. The 1.0 kg block sits on top of the 4.0 kg block. A horizontal 10 N force is applied to the top 1.0 kg block, and the two blocks move together without slipping. If the system is defined as both blocks together, what is the net external force on the system? A) $0\text{ N}$ B) $2\text{ N}$ C) $10\text{ N}$ D) $50\text{ N}$

Worked Solution: First, draw the system boundary around both blocks. The 10 N applied force comes from an object outside the system, so it is external. The static friction force between the two blocks is an interaction between two objects inside the system, so it is internal and sums to zero. Vertical forces (weight from Earth, normal force from the floor) cancel each other out, so the total net external force is just the 10 N applied force. The correct answer is C.

Question 2 (Free Response)

A simple Atwood machine has two blocks hanging from a massless, frictionless pulley: block 1 has mass $m_1=1.2\text{ kg}$, block 2 has mass $m_2=1.8\text{ kg}$. (a) Identify all internal and external forces if we take the system as both blocks. (b) Use system models to calculate the magnitude of the acceleration of the blocks. (c) Calculate the tension in the string connecting the blocks.

Worked Solution: (a) Internal forces: Tension exerted by each end of the string on the blocks (both ends act on objects inside the system, so tension is internal). External forces: Gravitational weight on $m_1$ from Earth, gravitational weight on $m_2$ from Earth, and the force from the pulley axle (the pulley is outside the system, so the support force is external). (b) Total mass of the system: $M=m_1+m_2=3.0\text{ kg}$. Net external force is the difference between the two weights: $$\sum F_{\text{ext}}=m_{2}g-m_{1}g=(0.6\text{ kg})(9.8{\text{ m/s}}^2)=5.88\text{ N}$$ By Newton's second law for systems: $a=\frac{\sum F_{\text{ext}}}{M}=\frac{5.88}{3.0}=1.96{\text{ m/s}}^2\approx 2.0{\text{ m/s}}^2$. (c) Select the subsystem of only $m_1$, which accelerates upward. Net force on $m_1$ is: $$T-m_{1}g=m_{1}a⟹T=m_1(g+a)=1.2(9.8+1.96)=14.1\text{ N}$$ Confirmation with $m_2$ gives the same result, so tension is $14\text{ N}$ (or $14.1\text{ N}$).

Question 3 (Application / Real-World Style)

A moving truck pulls two trailers behind it on level ground. The truck has a mass of 3000 kg, the first trailer (connected directly to the truck) has a mass of 2000 kg, and the second (rearmost) trailer has a mass of 1000 kg. The truck’s engine provides a total forward force of 12,000 N from friction with the road. All rolling friction is negligible. What is the tension in the hitch between the first trailer and the second trailer? Interpret your result in context.

Worked Solution:

1. First, treat the entire system (truck + two trailers) as a closed system to find acceleration: Total mass $M=3000+2000+1000=6000\text{ kg}$, so $a=\frac{F_{\text{net}}}{M}=\frac{12000}{6000}=2{\text{ m/s}}^2$.
2. We need tension between the first and second trailer, so select the subsystem of only the second (rearmost) trailer. The only forward external force on this subsystem is the hitch tension $T$.
3. Apply Newton's second law: $T=m_{\text{second}}a=(1000\text{ kg})(2{\text{ m/s}}^2)=2000\text{ N}$. In context: Hitch tension decreases for hitches further behind the truck, because each hitch only needs to accelerate the mass that comes after it. The tension between the truck and first trailer would be 6000 N, much larger than the 2000 N tension for the last hitch.

7. Quick Reference Cheatsheet

8. What's Next

System models are the foundation for every multi-object problem in the rest of AP Physics 1, from circular motion to momentum conservation to rotational dynamics. Next, you will apply system models to analyze friction and circular motion in the remaining parts of Unit 2 Dynamics: you will use combined system analysis to find acceleration of cars on banked curves, and to solve for tension in strings spinning objects in vertical circles. Without mastering the ability to distinguish internal vs external forces and select the optimal system boundary, you will struggle to avoid algebra errors and incorrectly calculate forces like tension and static friction. This topic also feeds directly into the study of momentum conservation in Unit 3, where choosing a closed system is the first step to applying conservation of momentum.

• Newton's Second Law
• Free-Body Diagrams
• Conservation of Momentum
• Atwood Machine Analysis