AP Physics 1- 4.1 Linear Momentum- Study Notes- New Syllabus
AP Physics 1-Links – Study Notes
AP Physics 1-Links – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Linear Momentum
- Applications of Linear Momentum
Linear Momentum
Linear momentum is a measure of the motion of an object due to its mass and velocity. It is a vector quantity, having both magnitude and direction.
Mathematically, the linear momentum of an object is:
\( \vec{p} = m \vec{v} \), where \( m \) = mass of the object and \( \vec{v} \) = velocity.
- The SI unit of momentum is kg·m/s.
For a system of objects, the total momentum is the vector sum of individual momenta:
\( \vec{P}_\text{total} = \sum_i m_i \vec{v}_i \).
- Linear momentum is conserved in a system if no external forces act on it (Law of Conservation of Momentum).
Applications of Linear Momentum:
- Collisions and Explosions: Momentum conservation is used to analyze elastic and inelastic collisions.
- Rocket Propulsion: Rockets accelerate by expelling gas backward; the momentum of the rocket-gas system is conserved.
- Vehicle Safety: Seat belts and airbags work by changing momentum gradually to reduce force on passengers.
- Sports: Understanding momentum helps in catching, hitting, or kicking balls safely and effectively.
- Recoil of Guns: When a bullet is fired, the gun moves backward due to conservation of momentum.
Example:
Two carts on a frictionless track collide. Cart A: mass = 2 kg, velocity = 3 m/s to the right; Cart B: mass = 3 kg, velocity = -2 m/s (to the left). Find the velocity of the combined carts after an inelastic collision.
▶️Answer/Explanation
Step 1: Use conservation of momentum: \( m_A v_A + m_B v_B = (m_A + m_B) v_f \)
Step 2: Substitute values → \( 2 \cdot 3 + 3 \cdot (-2) = (2 + 3) v_f \)
Step 3: Calculate → \( 6 – 6 = 5 v_f \) → \( 0 = 5 v_f \)
Step 4: Solve for final velocity → \( v_f = 0 \, \text{m/s} \)
Final Answer: The combined carts come to rest after the collision.
Example:
Two ice skaters, Skater A (mass = 50 kg) and Skater B (mass = 70 kg), are initially at rest on a frictionless ice surface. Skater A pushes Skater B. After the push, Skater A moves backward with a velocity of 1.5 m/s. Find the velocity of Skater B.
▶️Answer/Explanation
Step 1: Define the system: The system consists of both Skater A and Skater B. No external horizontal forces act on the system.
Step 2: Apply conservation of momentum:
\( \text{Total momentum before push} = \text{Total momentum after push} \)
\( 0 = m_A v_A + m_B v_B \)
Step 3: Substitute known values:
\( 0 = 50 \cdot (-1.5) + 70 \cdot v_B \) (Skater A moves backward, take as negative)
Step 4: Solve for \(v_B\):
\( -75 + 70 v_B = 0 \) → \( 70 v_B = 75 \) → \( v_B \approx 1.07 \, \text{m/s} \)
Final Answer: Skater B moves forward with a velocity of 1.07 m/s.
Note: By choosing the system as both skaters, the internal push does not change the total momentum of the system.