AP Physics 1- 4.4 Elastic and Inelastic Collisions- Study Notes- New Syllabus
AP Physics 1-4.4 Elastic and Inelastic Collisions – Study Notes
AP Physics 1-4.4 Elastic and Inelastic Collisions – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Elastic Collisions
- Inelastic Collisions
Elastic Collisions
An elastic collision is a collision in which both momentum and of the system are conserved. There is no loss of kinetic energy to deformation, heat, or sound.
Conservation Principles
- Momentum Conservation: \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \)
- Kinetic Energy Conservation: \( \dfrac{1}{2} m_1 v_{1i}^2 + \dfrac{1}{2} m_2 v_{2i}^2 = \dfrac{1}{2} m_1 v_{1f}^2 + \dfrac{1}{2} m_2 v_{2f}^2 \)
Characteristics
- Velocities of the objects change according to both momentum and kinetic energy conservation.
- In one dimension, if the masses are equal, objects simply exchange velocities.
- Elastic collisions can be head-on (1D) or oblique (2D).
Example:
Two billiard balls collide elastically on a frictionless table. Ball A (0.5 kg) moves at 2 m/s toward stationary Ball B (0.5 kg). Find the final velocities of both balls.
▶️Answer/Explanation
Step 1: Apply momentum conservation: \( m_A v_{Ai} + m_B v_{Bi} = m_A v_{Af} + m_B v_{Bf} \) \( 0.5 \cdot 2 + 0.5 \cdot 0 = 0.5 v_{Af} + 0.5 v_{Bf} \) → \( 1 = 0.5 (v_{Af} + v_{Bf}) \) → \( v_{Af} + v_{Bf} = 2 \)
Step 2: Apply kinetic energy conservation: \( 0.5 \cdot 0.5 \cdot 2^2 = 0.5 \cdot 0.5 \cdot v_{Af}^2 + 0.5 \cdot 0.5 \cdot v_{Bf}^2 \) → \( 1 = 0.25 (v_{Af}^2 + v_{Bf}^2) \) → \( v_{Af}^2 + v_{Bf}^2 = 4 \)
Step 3: Solve the two equations:
- \( v_{Af} + v_{Bf} = 2 \)
- \( v_{Af}^2 + v_{Bf}^2 = 4 \)
Solution: \( v_{Af} = 0 \, \text{m/s}, \, v_{Bf} = 2 \, \text{m/s} \)
Final Answer: Ball A stops, Ball B moves at 2 m/s in the original direction of Ball A.
Inelastic Collisions
An inelastic collision is a collision in which the momentum of the system is conserved, but kinetic energy is not conserved. Some kinetic energy is transformed into other forms of energy such as heat, sound, or deformation.
In a perfectly inelastic collision, the colliding objects stick together after the collision.
Conservation Principle
- Momentum Conservation: \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \)
- Kinetic energy is not conserved in general: \( \dfrac{1}{2} m_1 v_{1i}^2 + \dfrac{1}{2} m_2 v_{2i}^2 \neq \dfrac{1}{2} m_1 v_{1f}^2 + \dfrac{1}{2} m_2 v_{2f}^2 \)
Characteristics
- Objects may stick together (perfectly inelastic) or move separately after collision (partially inelastic).
- Momentum is always conserved, but kinetic energy decreases.
- Used to model collisions in vehicles, balls with deformation, and other real-world interactions.
Example:
Two carts on a frictionless track collide and stick together. Cart A (3 kg) moves at 4 m/s, Cart B (2 kg) is at rest. Find the final velocity of the combined carts.
▶️Answer/Explanation
Step 1: Apply momentum conservation: \( m_A v_{Ai} + m_B v_{Bi} = (m_A + m_B) v_f \) \( 3 \cdot 4 + 2 \cdot 0 = (3+2) v_f \) → \( 12 = 5 v_f \)
Step 2: Solve for \( v_f \) → \( v_f = 2.4 \, \text{m/s} \)
Step 3: Kinetic energy check (not conserved): Initial KE = \( \frac{1}{2} \cdot 3 \cdot 4^2 = 24 \, \text{J} \) Final KE = \( \frac{1}{2} \cdot 5 \cdot 2.4^2 = 14.4 \, \text{J} \) → KE decreased by 9.6 J.
Final Answer: Combined carts move at 2.4 m/s after collision; kinetic energy is reduced.