AP Physics 1- 4.3 Conservation of Linear Momentum- Study Notes- New Syllabus
AP Physics 1-4.3 Conservation of Linear Momentum – Study Notes
AP Physics 1-4.3 Conservation of Linear Momentum – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Conservation of Linear Momentum
- Effect of System Selection on Momentum
Conservation of Linear Momentum
Total Momentum of a System
The total linear momentum of a system of objects remains constant if no external forces act on the system. The total momentum of a system is the sum of the momenta of its constituent parts:
\( \vec{P}_\text{total} = \sum_i \vec{p}_i \)
Center-of-Mass Velocity
A collection of objects with individual momenta can be described as one system with one center-of-mass velocity.
Velocity of the system’s center of mass:
\( \vec{v}_\text{cm} = \dfrac{\sum_i m_i \vec{v}_i}{\sum_i m_i} \)
- The center-of-mass velocity is constant if no net external force acts on the system.
Internal and External Forces
Internal forces redistribute momentum among objects but do not change the total system momentum. External forces acting on the system change the total momentum:
\( \vec{F}_\text{net, ext} = \dfrac{d\vec{P}_\text{system}}{dt} \)
Momentum Transfer Within a System
In the absence of external forces, any change in momentum of one object is balanced by an equal and opposite change of momentum elsewhere within the system. The impulse exerted by one object on a second object is equal and opposite to the impulse exerted back (Newton’s Third Law).
Impulse and System Momentum
If the total momentum of a system changes, that change is equivalent to the impulse exerted on the system:
\( \vec{J}_\text{ext} = \Delta \vec{P}_\text{system} \)
Applications
- Correct application of conservation of momentum can determine velocities immediately before and after collisions or explosions.
- Choosing a system properly allows analysis where total momentum is constant or changes only due to external impulses.
Example:
Two ice skaters, Skater A (50 kg) and Skater B (70 kg), are initially at rest on a frictionless ice surface. Skater A pushes Skater B. After the push, Skater A moves backward at 1.5 m/s. Find the velocity of Skater B and the center-of-mass velocity of the system.
▶️Answer/Explanation
Step 1: Apply conservation of momentum: \( m_A v_A + m_B v_B = 0 \) → \( 50(-1.5) + 70 v_B = 0 \)
Step 2: Solve for \( v_B \) → \( 70 v_B = 75 \) → \( v_B = 1.07 \, \text{m/s} \)
Step 3: Center-of-mass velocity: \( v_\text{cm} = \dfrac{m_A v_A + m_B v_B}{m_A + m_B} = \dfrac{50(-1.5) + 70(1.07)}{50+70} \approx 0 \, \text{m/s} \)
Final Answer: Skater B moves at 1.07 m/s forward; center-of-mass velocity ≈ 0 m/s.
Effect of System Selection on Momentum
Momentum Conservation
Momentum is conserved in all interactions. The total momentum of a system depends on how the system is defined.
System with Zero Net External Force
If the net external force on the selected system is zero, the total momentum of the system remains constant. Internal forces may redistribute momentum among the objects, but the total system momentum does not change.
System with Nonzero Net External Force
If the net external force on the selected system is nonzero, momentum is transferred between the system and its environment. The change in system momentum is equal to the impulse exerted by the external forces:
\( \Delta \vec{P}_\text{system} = \vec{J}_\text{ext} \)
Example:
Two blocks, Block A (3 kg) and Block B (2 kg), are initially at rest on a frictionless surface. A person pushes Block A with a constant horizontal force of 10 N for 3 seconds. Determine the change in momentum of:
- (a) Block A alone
- (b) The system consisting of both Block A and Block B
▶️Answer/Explanation
Step 1: Momentum change for Block A (considering Block A as the system)
- Impulse on Block A = \( F \Delta t = 10 \cdot 3 = 30 \, \text{N·s} \)
- Change in momentum of Block A = 30 kg·m/s in the direction of the force.
Step 2: Momentum change for system of Block A + Block B
- If the system is Block A + Block B and no external horizontal forces act on this combined system (assuming the person is outside the system), net external force = 10 N → system momentum changes.
- If the system includes the person applying the force, net external force = 0 → total momentum of the system remains constant.
Key Takeaway: Whether momentum changes depends on how you define the system and whether external forces act on it.