AP Physics C Mechanics- 4.3 Conservation of Linear Momentum- Study Notes- New Syllabus
AP Physics C Mechanics- 4.3 Conservation of Linear Momentum – Study Notes
AP Physics C Mechanics- 4.3 Conservation of Linear Momentum – Study Notes – per latest Syllabus.
Key Concepts:
- Velocity of the Center of Mass of a System
- Total Momentum of a System
- Internal and External Forces
- Applying Conservation of Momentum to Collisions and Explosions
- System Selection and Conservation of Momentum
Velocity of the Center of Mass of a System
A system consisting of multiple objects can be analyzed as if all its mass were concentrated at a single point called the center of mass (CM). The velocity of the center of mass represents the overall motion of the system as a whole, taking into account the individual momenta and masses of all its components.
Momentum and Center of Mass Velocity
The total momentum of a system is the vector sum of the momenta of all individual objects:
\( \mathrm{\vec{p}_{total} = \sum_i \vec{p}_i = \sum_i m_i \vec{v}_i} \)
The system can be treated as a single equivalent particle of total mass \( \mathrm{M} \) moving with the center-of-mass velocity \( \mathrm{\vec{v}_{cm}} \):
\( \mathrm{\vec{p}_{total} = M \vec{v}_{cm}} \)
Hence, the center-of-mass velocity is defined as:
\( \mathrm{\vec{v}_{cm} = \dfrac{\sum_i m_i \vec{v}_i}{\sum_i m_i}} \)
- \( \mathrm{\vec{v}_{cm}} \): velocity of the system’s center of mass
- \( \mathrm{m_i} \): mass of each object
- \( \mathrm{\vec{v}_i} \): velocity of each object
- \( \mathrm{M = \sum_i m_i} \): total mass of the system
Physical Interpretation:
- The center-of-mass velocity describes the overall motion of the entire system, regardless of how the individual objects move relative to one another.
- Even if the internal motion of the system changes (e.g., due to collisions), the center of mass moves according to external forces only.
Example:
Two objects, \( \mathrm{m_1 = 2.0\,kg} \) and \( \mathrm{m_2 = 4.0\,kg} \), move in a straight line. Their velocities are \( \mathrm{v_1 = 3.0\,m/s} \) and \( \mathrm{v_2 = 1.0\,m/s} \), respectively, in the same direction. Find the velocity of the center of mass of the system.
▶️ Answer / Explanation
Step 1: Use the center-of-mass velocity formula:
\( \mathrm{\vec{v}_{cm} = \dfrac{\sum_i m_i \vec{v}_i}{\sum_i m_i}} \)
Step 2: Substitute values:
\( \mathrm{v_{cm} = \dfrac{(2.0)(3.0) + (4.0)(1.0)}{2.0 + 4.0}} \)
\( \mathrm{v_{cm} = \dfrac{6 + 4}{6} = \dfrac{10}{6} = 1.67\,m/s} \)
Result: The center of mass of the system moves with a velocity of \( \mathrm{1.67\,m/s} \) in the same direction as both objects.
Conceptual Note: Even though the two masses move at different speeds, the center of mass moves at an intermediate velocity that reflects their relative masses and velocities.
Total Momentum of a System
The total momentum of a system is the vector sum of the momenta of all the individual objects that make up the system. It represents the overall motion of the system as a whole and determines how the system will respond to external forces.
Expression for Total Momentum:
\( \mathrm{\vec{p}_{total} = \sum_i \vec{p}_i = \sum_i m_i \vec{v}_i} \)
- \( \mathrm{\vec{p}_i} \): momentum of the \( \mathrm{i^{th}} \) object
- \( \mathrm{m_i} \): mass of the \( \mathrm{i^{th}} \) object
- \( \mathrm{\vec{v}_i} \): velocity of the \( \mathrm{i^{th}} \) object
- \( \mathrm{\vec{p}_{total}} \): total (net) momentum of the entire system
Nature of Total Momentum:
- Total momentum is a vector quantity — both magnitude and direction matter.
- It represents the motion of the system’s center of mass.
- Even if objects within the system move relative to each other, the system’s total momentum depends only on the overall mass distribution and motion.
Example:
Two skaters push off each other on frictionless ice. Skater A has mass \( \mathrm{50\,kg} \) and moves at \( \mathrm{2.0\,m/s} \) to the right, while skater B has mass \( \mathrm{70\,kg} \) and moves to the left. What is the total momentum of the system before and after they push off?
▶️ Answer / Explanation
Step 1: Before they push off, both are at rest:
\( \mathrm{\vec{p}_{total,\,i} = 0} \)
Step 2: After pushing off:
- \( \mathrm{m_A = 50\,kg, \; v_A = +2.0\,m/s} \)
- Let \( \mathrm{v_B} \) be skater B’s velocity (to the left ⇒ negative).
Since total momentum is conserved:
\( \mathrm{m_A v_A + m_B v_B = 0} \)
Substitute values:
\( \mathrm{(50)(2.0) + (70)(v_B) = 0} \Rightarrow v_B = -1.43\,m/s} \)
Step 3: Total momentum before and after:
- Before: \( \mathrm{\vec{p}_{total,\,i} = 0} \)
- After: \( \mathrm{\vec{p}_{total,\,f} = 0} \)
Result: Total momentum of the system is conserved (zero both before and after), even though individual momenta are nonzero and opposite.
Conceptual Note: The skaters’ internal forces cancel, keeping the system’s total momentum unchanged — illustrating conservation in an isolated system.
Internal and External Forces
When objects within a system interact, the forces they exert on each other are internal forces.
- According to Newton’s Third Law, these internal forces are equal in magnitude and opposite in direction:
\( \mathrm{\vec{F}_{12} = -\vec{F}_{21}} \)
- Because they cancel in pairs, internal forces cannot change the total momentum of the system.
- Only external forces can change a system’s total momentum.
Conservation of Momentum (Isolated System)
In the absence of a net external force, the total momentum of a system remains constant:
\( \mathrm{\vec{F}_{net,\,ext} = 0 \;\Rightarrow\; \vec{p}_{total,\,i} = \vec{p}_{total,\,f}} \)
- Any change in momentum of one object within the system must be balanced by an equal and opposite change in another part of the system.
- This means momentum is transferred internally but not lost or gained overall.
Impulse and Momentum Change
If a net external force acts on a system, the system’s total momentum changes by an amount equal to the net impulse exerted on it:
\( \mathrm{\Delta \vec{p}_{system} = \vec{J}_{net,\,ext} = \int_{t_1}^{t_2} \vec{F}_{net,\,ext}(t)\,dt} \)
- This is the generalized form of the Impulse–Momentum Theorem applied to a system as a whole.
- The direction of the impulse determines the direction of the change in system momentum.
Newton’s Third Law and Impulse
The impulse exerted by one object on another is equal in magnitude and opposite in direction to the impulse received by the second object.
\( \mathrm{\vec{J}_{12} = -\vec{J}_{21}} \)
- This guarantees that the total momentum of an isolated two-object system remains constant, even though individual momenta change.
System Selection:
- A system can be chosen such that no net external forces act on it (isolated system).
- For such a system, total momentum remains constant: \( \mathrm{\vec{p}_{total} = constant} \).
- If external forces are present, the change in momentum equals the total external impulse: \( \mathrm{\Delta \vec{p}_{total} = \vec{J}_{ext}} \).
Example:
A \( \mathrm{0.15\,kg} \) ball moving at \( \mathrm{8.0\,m/s} \) collides head-on elastically with a \( \mathrm{0.10\,kg} \) ball moving at \( \mathrm{-6.0\,m/s} \). No external forces act on the system. Verify that total momentum is conserved.
▶️ Answer / Explanation
Step 1: Calculate initial momentum.
\( \mathrm{p_{i} = (0.15)(8.0) + (0.10)(-6.0) = 1.2 – 0.6 = 0.6\,kg \cdot m/s} \)
Step 2: Suppose after collision, velocities are \( \mathrm{v’_1 = -2.0\,m/s} \) and \( \mathrm{v’_2 = +10.0\,m/s} \).
\( \mathrm{p_{f} = (0.15)(-2.0) + (0.10)(10.0) = -0.3 + 1.0 = 0.7\,kg \cdot m/s} \)
Step 3: Compare initial and final momenta.
\( \mathrm{p_i \approx p_f} \Rightarrow \text{Momentum is conserved (small rounding difference due to data).} \)
Conceptual Note: During the collision, each ball exerts an equal and opposite impulse on the other, so the total system momentum remains constant, illustrating Newton’s Third Law and conservation of momentum.
Applying Conservation of Momentum to Collisions and Explosions
The principle of conservation of momentum states that in the absence of a net external force, the total momentum of a system remains constant. This principle can be applied to determine the velocity of a system immediately before and after a collision or explosion.
Conservation of Momentum in Collisions and Explosions
For any isolated system (no external forces):
\( \mathrm{\vec{p}_{total,i} = \vec{p}_{total,f}} \)
\( \mathrm{\sum_i m_i \vec{v}_{i} = \sum_i m_i \vec{v’}_{i}} \)
- Subscript \( \mathrm{i} \) denotes initial values (before interaction).
- Prime ( \( \mathrm{‘} \) ) denotes final values (after interaction).
- This relationship applies to both linear and multidimensional motion.
Application:
- For a collision (objects interact briefly and may stick together or rebound): Total momentum before impact = Total momentum after impact.
- For an explosion (objects separate from rest due to internal forces): Total momentum before explosion = Total momentum after explosion.
Even though individual momenta change, the vector sum remains the same.
Example:
A \( \mathrm{0.20\,kg} \) ball moving at \( \mathrm{5.0\,m/s} \) collides head-on and sticks to a \( \mathrm{0.30\,kg} \) ball at rest. Find their common velocity immediately after the collision.
▶️ Answer / Explanation
Step 1: Apply momentum conservation (no external force):
\( \mathrm{m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f} \)
Step 2: Substitute known values:
\( \mathrm{(0.20)(5.0) + (0.30)(0) = (0.20 + 0.30)v_f} \)
Step 3: Solve for \( \mathrm{v_f} \):
\( \math
System Selection and Conservation of Momentum
The conservation or change of momentum in a physical situation depends entirely on how the system is defined. By selecting appropriate boundaries for the system, one can determine whether external forces act on it — and therefore whether its total momentum remains constant or changes.
Conservation of Momentum Principle:
- Momentum is conserved in all interactions when the net external force acting on the system is zero.
- Internal forces between parts of the system occur in equal and opposite pairs and therefore do not change the total momentum.
- In this case, the total momentum of the system remains constant in both magnitude and direction.
Role of System Selection:
- The choice of the system determines whether forces are considered internal or external.
- If all interacting objects are included within the system, then the forces between them are internal and cancel in pairs (Newton’s Third Law).
- If some interacting objects are outside the system boundary, their forces become external and can change the system’s total momentum.
Effect of External Forces:
- If the net external force on a system is nonzero, the system’s momentum changes according to:
\( \mathrm{\vec{F}_{net,\,ext} = \dfrac{d\vec{p}_{system}}{dt}} \)
\( \mathrm{\Delta \vec{p}_{system} = \vec{J}_{ext} = \int_{t_1}^{t_2} \vec{F}_{net,\,ext}(t)\,dt} \)
- This means momentum is transferred between the system and its surroundings via an external impulse.
Example:
A person standing on a skateboard throws a ball forward. The ball has a mass of \( \mathrm{0.5\,kg} \) and is thrown at \( \mathrm{10\,m/s} \) relative to the ground. The person and skateboard together have a mass of \( \mathrm{60\,kg} \) and are initially at rest. Analyze the momentum of the system in two cases — (a) treating the person and skateboard as one system, and (b) treating the person as one system and the ball as another.
▶️ Answer / Explanation
Step 1: Identify system boundaries.
- (a) System 1: Person + skateboard + ball (isolated system).
- (b) System 2: Person + skateboard (non-isolated, external force from ball’s push).
Step 2: Before the throw:
\( \mathrm{p_{initial} = 0} \) (system at rest)
After the throw:
For the isolated system (person + ball):
\( \mathrm{m_{person}v_{person} + m_{ball}v_{ball} = 0} \)
Substitute values:
\( \mathrm{(60)v_{person} + (0.5)(10) = 0} \Rightarrow v_{person} = -0.083\,m/s} \)
Interpretation: The person and skateboard move backward at \( \mathrm{0.083\,m/s} \), conserving total momentum of the isolated system.
Step 3: System with external interaction (person only):
- The person experiences an external impulse due to throwing the ball forward.
- As a result, the person’s momentum changes — momentum is transferred to the environment (the ball).
Compare both cases.
| System Chosen | External Forces? | Momentum Conserved? |
|---|---|---|
| Person + Ball (Isolated System) | No | Yes — total \( \mathrm{p = 0} \) |
| Person Only | Yes (force from ball) | No — momentum changes |