# System Models

> AP Physics 1 · Unit 2: Dynamics
> Source: https://www.owlsprep.com/study/ap-physics-1-u2-system-models/

This guide covers core concepts of system models for AP Physics 1: system boundary definition, internal vs external forces, open vs closed systems, and optimal system selection strategy for connected-object dynamics problems.

**Prerequisites:** Newton's three laws of motion; Vector addition of forces; Kinematics for constant acceleration

## Learning objectives

- Define systems and system boundaries for dynamics analysis
- Distinguish between internal and external forces for any defined system
- Classify systems as open or closed based on mass flow across boundaries
- Apply Newton's second law to combined multi-body systems
- Select optimal system boundaries to simplify connected-object dynamics problems

## What is a System Model?

A system model is a problem-solving strategy that involves grouping selected objects together for analysis, rather than analyzing each object individually. You mark a system boundary with a dashed line: any object inside the line is part of the system, everything outside is called the surroundings.

This method simplifies complex multi-object problems by eliminating internal forces that cancel out via Newton's third law, leaving only external forces to calculate. The AP Physics 1 exam tests your ability to choose the optimal system boundary to reduce algebra and avoid errors. System models make up a core part of Unit 2 Dynamics, which accounts for 12-18% of your total exam score.

**System Model** — A problem-solving framework that groups a set of objects together for analysis, separated from the surroundings by an arbitrary user-chosen boundary.

*Example:* Two connected blocks can be analyzed as a single combined system or two separate individual systems.

## Internal vs. External Forces

**Internal Force** — A force resulting from an interaction between two objects that are both inside the system boundary. By Newton's third law, all internal force pairs cancel out, so they do not contribute to the net acceleration of the entire system.

**External Force** — A force resulting from an interaction between an object inside the system boundary and an object outside the system (in the surroundings). External forces do not cancel out, and determine the net acceleration of the system.

For any system, Newton's second law can be written in terms of external forces only:

$$\sum \vec{F}_{\text{ext}} = M\vec{a}_{cm}$$

Where $M$ is the total mass of the system, and $\vec{a}_{cm}$ is the acceleration of the system's center of mass. For connected rigid objects moving together, $\vec{a}_{cm}$ 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 (both ends act on blocks 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:
5. $$10 \text{ N} = (2.0 + 3.0)a \implies a = 2.0 \text{ m/s}^2$$
6. The acceleration of the system is $2.0 \text{ m/s}^2$ to the right.

> **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.

## System Selection Strategy for Internal Force Calculations

You can choose any system boundary you want for a problem; there is no technically 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) 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 eliminates the need to solve simultaneous equations, 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, 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:
4. $$T = m_2 a = (2.0 \text{ kg})(2.0 \text{ m/s}^2) = 4.0 \text{ N}$$
5. To confirm, selecting the 3.0 kg block as the subsystem gives the same result: $10 \text{ N} - T = (3.0 \text{ kg})(2.0 \text{ m/s}^2) \implies T = 4.0 \text{ N}$.

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

## Open vs. Closed Systems

**Closed System** — A system where no mass crosses the system boundary during the process being analyzed. All mass inside the boundary at the start remains inside, and no mass enters from outside.

**Open System** — A system where mass crosses the system boundary during the process being analyzed. Mass can enter (e.g., sand dropped into a cart) or leave (e.g., fuel expelled from a rocket).

For AP Physics 1, almost all dynamics problems use closed systems, because the standard form of 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, you can almost always expand your system boundary to include all mass involved in the process to make the system closed, simplifying calculations. AP Physics 1 only tests conceptual understanding of open vs closed systems, not complex mass flow calculations.

**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. Using the closed system lets us ignore internal impact forces between the sand and cart, making it much easier to find the final speed of the loaded cart.

> **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.

## AP Style Concept Check

**Check your understanding**

Test your understanding of core system model concepts:

1. 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?

   - 0 N
   - 2 N
   - 10 N
   - 50 N

   *Why:* The 10 N applied force is from an object outside the system, so it is external. Static friction between the two blocks is an interaction between two objects inside the system, so it is internal and cancels out. Vertical forces (weight and normal force) cancel each other, so the total net external force is 10 N.

## Common pitfalls

- **Wrong:** Counting tension between two connected blocks as an external force when both blocks are included in the system.
  - Why it fails: Students often remember tension as a pulling force, so they incorrectly add it to the net external force sum.
  - Correct: 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:** Using a single acceleration for the whole system when individual objects accelerate in different directions or at different magnitudes.
  - Why it fails: Students assume all objects in a system share the same acceleration, which is only true for rigidly connected moving objects.
  - Correct: If objects have different accelerations, use the full form $\sum F_{\text{ext}} = m_1a_1 + m_2a_2$ instead of $Ma_{cm}$ with a single $a$.
- **Wrong:** Treating the cart alone as a closed system when mass is being added or removed during the process.
  - Why it fails: Students default to closing the system around the obvious object (the cart) without checking for mass flow.
  - Correct: 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:** Solving for tension by analyzing the pulled block first, leading to sign errors when subtracting tension from the applied force.
  - Why it fails: Students pick the pulled object first out of habit, leading to incorrect sign when rearranging equations.
  - Correct: 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:** Forgetting to include weight and normal force as external forces when friction is present.
  - Why it fails: Students get focused on horizontal forces and ignore vertical forces, which are needed to calculate friction.
  - Correct: Always list all external forces (vertical and horizontal) first before summing, even if you expect vertical forces to cancel.

## Cheatsheet

| Category | Formula/Rule | Notes |
| --- | --- | --- |
| General System Definition | N/A (conceptual) | Any collection of objects bounded by a dashed line; inside = system, outside = surroundings |
| Internal Force | $\sum \vec{F}_{\text{int}} = 0$ | Forces between objects inside the system; cancel by Newton's third law, do not contribute to net system force |
| External Force | N/A (conceptual) | Forces from outside the system acting on inside objects; all net force on the system comes from external forces |
| Newton's Second Law for Systems | $\sum \vec{F}_{\text{ext}} = M \vec{a}_{cm} = \sum m_i \vec{a}_i$ | $M$ = total system mass, $a_{cm}$ = center of mass acceleration, $a_i$ = individual object acceleration |
| Closed System | N/A | No mass crosses the system boundary during the process; use for all standard dynamics calculations |
| Open System | N/A | Mass crosses the system boundary during the process; expand system to include all mass to make it closed if possible |
| Internal Force Calculation | 1. Find $a$ from combined system; 2. Analyze subsystem to find internal force | Works for tension, normal force, hitch tension, and all internal forces between connected objects |
| Optimal System Selection | N/A | Choose the smallest possible closed system that includes all objects for the simplest calculation |

## 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. Mastering system selection will help you avoid unnecessary algebra errors and simplify complex problems on every exam topic. Next, you will apply system models to analyze friction and circular motion in the remaining parts of Unit 2 Dynamics, using combined system analysis to find acceleration of cars on banked curves and solve for tension in strings spinning objects in vertical circles. This topic also directly prepares you for momentum conservation in Unit 3, where choosing the correct closed system is the key to solving collision and explosion problems correctly.

- [Unit 2 Dynamics Overview](https://www.owlsprep.com/study/ap-physics-1-u2-overview/)
- [Gravitational Force and Weight](https://www.owlsprep.com/study/ap-physics-1-u2-gravitational-force-and-weight/)
- [Newton's First Law](https://www.owlsprep.com/study/ap-physics-1-u2-newton-s-first-law/)

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