AP Physics 1- 2.1 Systems and Center of Mass- Study Notes- New Syllabus
AP Physics 1-2.1 Systems and Center of Mass – Study Notes
AP Physics 1-2.1 Systems and Center of Mass – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Properties and Interactions of a System
- Center of Mass of a System
Properties and Interactions of a System
A system is a collection of objects or particles that can be studied together. The behavior of the system depends on its internal properties and its interactions with the surroundings.
Properties of a System:
- Mass: The total mass of the system is the sum of the masses of its components.
- Momentum: The total momentum is the vector sum of the momenta of the components.
- Energy: Includes kinetic energy, potential energy, internal energy, etc.
- Center of Mass: Represents the average position of the system’s mass.
Interactions of a System:
- Internal Forces: Forces that components of the system exert on each other. These forces occur in action–reaction pairs and cancel out when considering the whole system.
- External Forces: Forces exerted by the environment on the system. These change the momentum or energy of the system.
Important Principle:
- If the net external force on the system is zero, the system’s total momentum remains constant.
- If the net external torque on the system is zero, its angular momentum remains constant.
- Energy is conserved in all interactions, though it can change form within the system (e.g., kinetic ↔ potential).
Relevant Equations:
Momentum: \( \vec{p}_{\text{system}} = \sum m_i \vec{v}_i \)
Newton’s 2nd Law for a system: \( \dfrac{d\vec{p}_{\text{system}}}{dt} = \vec{F}_{\text{ext}} \)
Kinetic Energy: \( KE_{\text{system}} = \sum \dfrac{1}{2} m_i v_i^2 \)
Example :
A two-cart system of masses 3 kg and 2 kg connected by a rope is pulled with 10 N on a frictionless surface. Find the acceleration of the system.
▶️Answer/Explanation
Total mass = 3 + 2 = 5 kg
Net external force = 10 N
Acceleration = \( a = \dfrac{F}{M} = \dfrac{10}{5} = 2 \, \text{m/s}^2 \)
Answer: The system accelerates at 2 m/s².
Example :
Two ice skaters push off each other. Skater A (60 kg) moves at 3 m/s to the right. Find the velocity of Skater B (40 kg).
▶️Answer/Explanation
Momentum conservation: \( m_A v_A + m_B v_B = 0 \)
\( 60 \times 3 + 40 v_B = 0 \)
\( v_B = -\dfrac{180}{40} = -4.5 \, \text{m/s} \)
Answer: Skater B moves at 4.5 m/s in the opposite direction.
Center of Mass of a System
The center of mass (COM) is the point that represents the average position of all the mass in a system. The motion of the system can be described as if all its mass were concentrated at this point.
The center of mass of a system of particles is the weighted average of the positions of its constituent parts, where the weights are their masses.
Equation (Discrete Masses):
\( x_{\text{cm}} = \dfrac{\sum m_i x_i}{\sum m_i}, \quad y_{\text{cm}} = \dfrac{\sum m_i y_i}{\sum m_i}, \quad z_{\text{cm}} = \dfrac{\sum m_i z_i}{\sum m_i} \)
Equation (Continuous Mass Distribution):
\( \vec{r}_{\text{cm}} = \dfrac{1}{M} \int \vec{r} \, dm \)
- \( m_i \): mass of each particle
- \( x_i, y_i, z_i \): coordinates of each particle
- \( M = \sum m_i \): total mass of system
Important Notes:
- If masses are equal, COM is at the geometric center of the system.
- For a rigid body with uniform density, COM coincides with its geometric center.
- In external force analysis, the whole system behaves as though all mass is concentrated at the COM.
Example :
Two particles of masses 2 kg and 3 kg are located at x = 0 m and x = 4 m on a line. Find the x-coordinate of the center of mass.
▶️Answer/Explanation
\( x_{\text{cm}} = \dfrac{2(0) + 3(4)}{2+3} \)
\( x_{\text{cm}} = \dfrac{12}{5} = 2.4 \, \text{m} \)
Answer: The COM is at x = 2.4 m.
Example :
A uniform rod of length 2 m has its center at the midpoint. Where is the center of mass located?
▶️Answer/Explanation
Since the rod has uniform mass distribution, its COM lies at the geometric center.
Answer: The COM is at the midpoint (1 m from either end).