AP Physics C: Mechanics - Night Before Review 🚀

Hey! Let's get you prepped and confident for tomorrow's exam. This guide is designed to be your quick, go-to resource. We'll break down the key concepts, highlight important points, and make sure you're ready to rock! Let's do this!

Properties and Interactions of Systems

System Properties from Interactions

• The interactions between objects within a system determine the system's properties. 🔍
• If the properties or interactions of individual parts aren't important, we can model the whole system as a single object.
• Systems can exchange energy or mass with their surroundings through interactions.
• Objects within a system can behave differently from each other and the system as a whole.
  • Think of a gas molecule in a container—it moves differently than the container itself.
• Analyzing a system requires considering its internal structure.
• External changes can alter a system's substructure.
  • For example, increasing temperature (external variable) can cause water (system) to change phase.

Systems as Single Objects

• Complex systems can be treated as a single object in certain scenarios.
  • A car's motion can be analyzed without considering its individual components.
• Simplifying a system to a single object allows for easier analysis.
• This works when the interactions and properties of individual components don't significantly impact the overall system behavior.
• This approach is best when focusing on macroscopic properties.

Energy and Mass Transfer

• Systems can exchange energy and mass with their surroundings. 🔄
• Open systems: Allow both energy and mass transfer.
  • Example: A cup of hot coffee cooling and evaporating.
• Closed systems: Allow energy transfer but not mass transfer.
  • Example: A sealed, insulated thermos.
• Isolated systems: Do not exchange energy or mass with the environment.
  • Example: An ideal, perfectly insulated container.

Individual vs. System Behavior

• Individual components can behave differently from each other and the system as a whole.
• Emergent properties arise from the collective interactions of individual components.
  • Think of individual neurons vs. the complex functioning of the brain.
• Analyzing individual components gives insights into the system's microscopic behavior.
• Macroscopic system behavior results from the cumulative effects of individual component interactions.

Internal Structure Effects

• A system's internal arrangement and composition influence its properties and behavior.
• Systems with identical macroscopic properties can have different internal structures.
  • Allotropes of carbon (diamond vs. graphite) have distinct properties due to their internal structures.
• Changing the internal structure can alter characteristics without changing composition.
  • Rearranging components of an electrical circuit changes its functionality.
• Understanding internal structure is key to predicting and explaining behavior.

External Variable Impacts

• External variables can change a system's substructure and properties.
• Modifying external conditions alters the interactions and arrangements of components.
  • Applying stress to a material changes its internal molecular structure.
• A system's response to external variables depends on its internal composition and structure.
• Controlling external variables allows for manipulation and optimization of system properties.
  • Adjusting temperature and pressure to fine-tune a chemical reaction. 🧪

Practice Question

Multiple Choice Questions

1. A closed system undergoes a process where its internal energy increases by 50 J. Which of the following is true about the energy transfer? (A) 50 J of work was done on the system. (B) 50 J of heat was transferred to the system. (C) 50 J of work was done by the system. (D) 50 J of heat was transferred from the system.

2. Which of the following best describes an isolated system? (A) A system that exchanges only energy with its surroundings. (B) A system that exchanges only mass with its surroundings. (C) A system that exchanges both energy and mass with its surroundings. (D) A system that exchanges neither energy nor mass with its surroundings.

Free Response Question

A 2.0 kg block is placed on a frictionless horizontal surface and is connected to a spring with a spring constant of 100 N/m. The block is initially at rest at the equilibrium position. A force of 20 N is applied to the block, causing it to move 0.5 m from the equilibrium position. Analyze the system and answer the following questions:

(a) Calculate the work done by the applied force. (b) Calculate the potential energy stored in the spring when the block is at the 0.5 m position. (c) Calculate the kinetic energy of the block at the 0.5 m position. (d) Calculate the total energy of the system.

Scoring Rubric

(a) Work done by the applied force: - Correctly use the formula: $W = Fd$ (1 point) - Correct substitution: $W = (20 N)(0.5 m)$ (1 point) - Correct answer: $W = 10 J$ (1 point) (b) Potential energy stored in the spring: - Correctly use the formula: $U = \frac{1}{2}kx^2$ (1 point) - Correct substitution: $U = \frac{1}{2}(100 N/m)(0.5 m)^2$ (1 point) - Correct answer: $U = 12.5 J$ (1 point) (c) Kinetic energy of the block: - Correctly use the work-energy theorem: $W = \Delta KE + \Delta U$ (1 point) - Correct substitution: 10 J = KE + 12.5 J (1 point) - Correct answer: $KE = -2.5 J$ (1 point) (d) Total energy of the system: - Correctly state the total energy: $E = KE + U$ (1 point) - Correct substitution: $E = -2.5 J + 12.5 J$ (1 point) - Correct answer: $E = 10 J$ (1 point)

Center of Mass Location

Symmetrical Mass Distributions

• For objects with symmetrical mass distributions, the center of mass is on the lines of symmetry.
• Symmetrical objects have their center of mass at the intersection of all symmetry lines.
  • A uniform rectangular sheet's center of mass is at its geometric center.
• The center of mass of a symmetrical object coincides with its centroid (geometric center).

Center of Mass Calculation

• The center of mass location along a given axis is calculated using:
  • $$\vec{x}_{\mathrm{cm}}=\frac{\sum m_{i} \vec{x}_{i}}{\sum m_{i}}$$
  • $\vec{x}_{\mathrm{cm}}$: position vector of the center of mass
  • $m_{i}$: mass of each individual object
  • $\vec{x}_{i}$: position vector of each object
• This equation is a mass-weighted average of the positions of all objects in the system.

Nonuniform Solid Calculations

• For nonuniform solids, the center of mass is calculated using:

  • $$\vec{r}_{\mathrm{cm}}=\frac{\int \vec{r} dm}{\int dm}$$
  • $\vec{r}_{\mathrm{cm}}$: position vector of the center of mass
  • $\vec{r}$: position vector of each differential mass element

• The linear mass density ($\lambda$) of a rod is:

  • $$\lambda=\frac{d}{d \ell} m(\ell)$$

• The total mass of a 3D object with variable density $\rho(r)$ is:

  • $$M_{\text {total }}=\int \rho(r) dV$$

  Center of mass calculations for both discrete objects and continuous mass distributions are crucial. Expect to see at least one problem that requires you to calculate the center of mass.

System Modeling at Center of Mass

• A system can be modeled as a single object located at its center of mass for simplicity.
• The motion of the center of mass represents the overall motion of the system.
  • The trajectory of a projectile can be analyzed by considering its center of mass.
• External forces acting on a system can be considered to act on the center of mass.
• Modeling a system at its center of mass simplifies calculations and analysis. 📊

Practice Question

Multiple Choice Questions

1. Two objects of masses $m_1$ and $m_2$ are located at positions $x_1$ and $x_2$ respectively on the x-axis. The center of mass of the system is located at: (A) $\frac{m_1x_1 + m_2x_2}{m_1 + m_2}$ (B) $\frac{m_1x_1 - m_2x_2}{m_1 + m_2}$ (C) $\frac{m_1 + m_2}{m_1x_1 + m_2x_2}$ (D) $\frac{m_1x_1 + m_2x_2}{m_1 - m_2}$

2. A uniform rod of length L has a mass M. Where is the center of mass of the rod located? (A) At the end of the rod (B) At L/4 from one end (C) At L/2 from one end (D) At 3L/4 from one end

Free Response Question

A system consists of three particles. Particle 1 has a mass of 2 kg and is located at (1, 2) m. Particle 2 has a mass of 3 kg and is located at (4, -1) m. Particle 3 has a mass of 5 kg and is located at (-2, 3) m. Calculate the center of mass of this system.

Scoring Rubric

• Correctly apply the center of mass formula for the x-coordinate: $\vec{x}_{cm} = \frac{\sum m_i \vec{x}_i}{\sum m_i}$ (1 point)
• Correct substitution for the x-coordinate: $\vec{x}_{cm} = \frac{(2 kg)(1 m) + (3 kg)(4 m) + (5 kg)(-2 m)}{2 kg + 3 kg + 5 kg}$ (2 points)
• Correct calculation for the x-coordinate: $\vec{x}_{cm} = \frac{4}{10} = 0.4 m$ (1 point)
• Correctly apply the center of mass formula for the y-coordinate: $\vec{y}_{cm} = \frac{\sum m_i \vec{y}_i}{\sum m_i}$ (1 point)
• Correct substitution for the y-coordinate: $\vec{y}_{cm} = \frac{(2 kg)(2 m) + (3 kg)(-1 m) + (5 kg)(3 m)}{2 kg + 3 kg + 5 kg}$ (2 points)
• Correct calculation for the y-coordinate: $\vec{y}_{cm} = \frac{16}{10} = 1.6 m$ (1 point)
• Correctly state the center of mass coordinates: (0.4 m, 1.6 m) (2 points)

Final Exam Focus 🎯

• High-Priority Topics:
  • Center of Mass: Calculating for both discrete and continuous systems. Expect to use integration for non-uniform objects.
  • System Interactions: Understanding how energy and mass are transferred in open, closed, and isolated systems.
  • Modeling Systems: Simplifying complex systems into single objects at their center of mass.
• Common Question Types:
  • Multiple Choice: Conceptual questions about system properties, energy transfer, and center of mass location.
  • Free Response: Problems requiring center of mass calculations, analysis of system behavior, and application of energy principles.
• Time Management Tips:
  • Quickly scan the exam and start with the questions you feel most comfortable with.