Unit 4: Linear Momentum

Linear Momentum

Definition and Vector Nature

• Linear momentum is defined as the product of an object’s mass and its velocity: \[ \vec{p} = m\vec{v} \] Momentum is a vector quantity — it has both magnitude and direction, and it points in the same direction as velocity. Because velocity can change in direction as well as magnitude, momentum is especially useful in analyzing systems where direction matters, such as collisions and explosions. In multi-object systems, total momentum is the vector sum of the momentum of each object, not just the sum of speeds.
• Momentum provides a powerful alternative to Newton's laws for analyzing interactions. Instead of tracking forces and accelerations over time, we can analyze the net change in momentum between before and after an interaction. This is particularly useful when the forces involved are unknown or highly variable over short time intervals. In such cases, focusing on initial and final momentum is often more practical than force-based models. Momentum is conserved in isolated systems regardless of the complexity of the forces during the interaction.
• The SI unit of momentum is \( \text{kg} \cdot \text{m/s} \), and it is not interchangeable with energy even though both involve mass and velocity. Energy depends on the square of velocity, while momentum depends linearly on velocity. This distinction becomes especially important in collisions, where momentum is conserved but kinetic energy may not be. Confusing these two leads to incorrect interpretations of motion and conservation principles.

Impulse and the Impulse–Momentum Theorem

Impulse as the Integral of Force Over Time

• Impulse is defined as the change in momentum of an object and is given by: \[ \vec{J} = \Delta \vec{p} = \vec{F}_{\text{avg}} \cdot \Delta t \] In cases where the force varies over time, the full impulse is found by integrating the force: \[ \vec{J} = \int_{t_i}^{t_f} \vec{F}(t)\,dt \] This calculus-based definition allows precise modeling of time-dependent forces such as during a collision, explosion, or rocket thrust. Students must understand that impulse is not a force — it’s the cumulative effect of force over time.
• The impulse–momentum theorem: \[ \int \vec{F}(t)\,dt = \Delta \vec{p} \] is a restatement of Newton’s Second Law in integral form. This is especially useful in analyzing real-world events such as car crashes, where the force isn't constant but varies rapidly. The area under a force–time graph gives the impulse, which directly translates to change in velocity when mass is constant. AP questions often ask students to estimate impulse graphically or derive it analytically using integrals.
• Impulse always causes a momentum change in the object receiving the force, regardless of whether kinetic energy is conserved. For example, catching a ball causes your hand to experience a backward impulse, even though the ball may lose kinetic energy. This is why airbags, padding, and long-duration collisions are effective — they spread the force over more time, reducing the peak force. This deepens students’ understanding of both safety design and mathematical modeling of force as a function of time.

Momentum of a System of Particles

Extending Newton’s Laws to Multi-Object Systems

• The total momentum of a system of particles is the vector sum of the individual momenta: \[ \vec{p}_{\text{sys}} = \sum_i m_i \vec{v}_i \] Internal forces (like tension or contact forces between parts) cancel in pairs due to Newton’s Third Law. As a result, only external forces can change the momentum of the system. This lets us treat complex, interacting systems as a single particle located at the center of mass, greatly simplifying analysis. This idea becomes essential in collisions, rocket motion, and systems involving internal constraints.
• Just like for a single object, Newton’s Second Law for a system is: \[ \vec{F}_{\text{net, external}} = \frac{d\vec{p}_{\text{sys}}}{dt} \] When net external force is zero, total momentum is conserved regardless of what happens internally. This allows us to analyze explosions, collisions, or internal rearrangements without knowing the exact details of forces involved. The AP exam often expects students to clearly state whether momentum is conserved and justify it based on external force considerations.
• In variable-mass systems like rockets, the system’s mass changes as fuel is ejected. Newton’s Second Law still applies, but the momentum changes due to both mass and velocity changes: \[ \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt} \] This expanded derivative reflects the complexities of systems where mass is not constant. It’s a preview of topics in advanced mechanics and engineering and is testable as a conceptual challenge in AP Physics C.

Types of Collisions

Classifying Interactions Based on Energy and Momentum

• All collisions conserve total momentum if the system is isolated and no net external force acts. That means: \[ \vec{p}_{\text{initial}} = \vec{p}_{\text{final}} \quad \text{or} \quad m_1 \vec{v}_{1i} + m_2 \vec{v}_{2i} = m_1 \vec{v}_{1f} + m_2 \vec{v}_{2f} \] This is true whether the collision is elastic or inelastic, and whether the motion is in one or two dimensions. Students must always start collision problems by checking whether momentum conservation applies — usually by confirming the absence of net external forces. Recognizing this allows simplified modeling of even violent or complex interactions.
• In an elastic collision, both momentum and total kinetic energy are conserved. This occurs in idealized particle collisions, like gas molecules or smooth balls on a frictionless surface. These collisions obey: \[ \begin{cases} \vec{p}_{\text{initial}} = \vec{p}_{\text{final}} \\ KE_{\text{initial}} = KE_{\text{final}} \end{cases} \] Solving elastic collisions often involves solving two simultaneous equations — one for momentum, one for energy — and yields symmetric velocity swapping behavior when masses are equal. Elastic collisions are rare in real life but common in test problems to reinforce conservation principles.
• In an inelastic collision, momentum is still conserved, but kinetic energy is not. Some of the initial mechanical energy is converted into thermal energy, sound, or deformation of the objects involved. A special case is the perfectly inelastic collision, where the objects stick together after impact: \[ m_1 \vec{v}_{1i} + m_2 \vec{v}_{2i} = (m_1 + m_2)\vec{v}_f \] These problems typically require calculating how much kinetic energy was “lost” by comparing initial and final values. They are useful for modeling real-world crashes, tackles, and globs sticking together — and for testing whether students understand the limits of mechanical energy conservation.

Solving Collision Problems

Strategies Using Conservation Laws

• To solve collision problems, always start with conservation of momentum. For one-dimensional collisions, treat direction carefully by assigning a coordinate axis and using signs. If the collision is elastic, also write down the kinetic energy conservation equation. When possible, use center-of-mass or relative velocity techniques to simplify. The AP exam often rewards clear, structured problem-solving over brute-force algebra.
• In two-dimensional collisions, conserve momentum in each direction independently: \[ \sum p_{x,\text{initial}} = \sum p_{x,\text{final}}, \quad \sum p_{y,\text{initial}} = \sum p_{y,\text{final}} \] You may also need to use trigonometry or vector components if angles are involved. Kinetic energy is only conserved if the problem is elastic; otherwise, compute how much energy is lost. These problems test students’ ability to break vectors into components, use scalar and vector conservation laws simultaneously, and justify whether a collision was elastic or not.
• For calculus-based problems, sometimes the force during collision is given as a function of time. In that case, use impulse: \[ \Delta \vec{p} = \int_{t_1}^{t_2} \vec{F}(t)\,dt \] This lets you analyze the velocity change caused by the collision, even if the exact collision duration or acceleration is unknown. These problems connect Unit 4 to earlier work with integrals in Unit 3, and they require strong fluency with interpreting force–time graphs and evaluating definite integrals. Expect these in AP Physics C free-response or as justification-based multiple choice.

Kinetic Energy Loss in Inelastic Collisions

Quantifying Energy Dissipation

• The amount of kinetic energy lost in an inelastic collision can be calculated by comparing the total kinetic energy before and after the event: \[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \] This value will always be negative or zero for inelastic collisions, indicating a loss of usable mechanical energy. The “lost” energy has been converted into thermal energy, sound, or internal deformation. On the AP exam, this often appears as a final part of a multi-part collision question.
• In perfectly inelastic collisions, the kinetic energy loss is maximized for a given system because the objects stick together and move with a common velocity: \[ \Delta KE = \frac{1}{2}(m_1 + m_2)v_f^2 - \left[\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2\right] \] These problems provide insight into energy conservation beyond just motion — students must account for how energy is transformed and where it “goes.” This encourages deeper thinking than simply plugging into equations. It also prepares students for analyzing explosions, thermal losses, and rotational analogs of collisions in Unit 5.
• It’s important to emphasize that momentum is conserved regardless of energy loss. Students often confuse momentum and energy conservation and incorrectly assume that if kinetic energy is not conserved, then momentum is not either. Clear labeling and side-by-side comparisons of what’s conserved in which collision type help prevent this. Understanding this distinction prepares students for impulse-based questions, conservation proofs, and mixed-dimension collision analysis.

Defining the Center of Mass

Locating the Weighted Balance Point of a System

• The center of mass (COM) of a system is the average position of all mass in the system, weighted by mass. For a discrete set of particles in one dimension, it is given by: \[ x_{\text{COM}} = \frac{\sum m_i x_i}{\sum m_i} \] This formula can be extended to two or three dimensions by applying it separately to \( x \), \( y \), and \( z \) coordinates. The center of mass does not need to be located at any actual object — it’s a mathematical point where the system’s total mass can be treated as if concentrated. This is especially useful in systems involving multiple objects, variable mass, or internal interactions.
• For a continuous mass distribution, such as a rod or ring, the summation becomes an integral: \[ x_{\text{COM}} = \frac{1}{M} \int x\,dm \] where \( dm \) is a small element of mass. This formulation is used in more advanced mechanics to analyze non-uniform bodies or variable mass systems. While AP Physics C rarely requires full integration, understanding the connection between summation and integration helps prepare students for center-of-mass calculations in college physics. It also allows estimation of COM location using symmetry and density functions.
• The center of mass is not affected by internal forces like tension or collisions between parts of the system. Only external forces can change the velocity or acceleration of the center of mass. This principle allows complex systems — like explosions, rocket motion, or two blocks colliding — to be modeled as a single point mass located at the COM. Tracking the COM simplifies both energy and momentum analysis and links directly to Newton’s Second Law for systems.

Motion of the Center of Mass

Center of Mass (COM) Obeys Newton’s Second Law for the Whole System

• Even though individual particles may accelerate in various directions, the center of mass of a system moves as if all external forces act on a single particle of total mass. Mathematically, this is: \[ \vec{F}_{\text{net, external}} = M \vec{a}_{\text{COM}} \] where \( M \) is the total mass of the system. This means that to predict the overall motion of the system, you don’t need to analyze every internal detail — just the total external force and total mass. This principle is especially useful in explosion problems, where parts scatter but the COM still follows a predictable trajectory.
• If the net external force is zero, the center of mass moves at constant velocity, even if the parts of the system move unpredictably relative to each other. For example, if a firework explodes in mid-air, each fragment accelerates differently, but the center of mass of the entire firework continues to move in a parabolic path under gravity. This allows you to use projectile motion equations to track the COM, even when the system internally breaks apart. It also reinforces Newton’s First Law applied to systems.
• The center of mass connects directly to momentum, since: \[ \vec{p}_{\text{total}} = M \vec{v}_{\text{COM}} \] This relationship allows you to relate linear momentum to the motion of the COM, especially when applying impulse or conservation principles. If momentum is conserved, then the velocity of the COM remains constant. Understanding this link helps students visualize how momentum flows through systems, particularly in inelastic collisions or internal explosions.

Choosing a Reference Frame Using the Center of Mass

Analyzing Symmetry and Simplifying Calculations

• In many problems, it is helpful to set the origin of your coordinate system at the center of mass. This simplifies equations and often makes symmetry easier to identify. For example, in a perfectly symmetric object like a uniform rod or disk, the center of mass lies at the geometric center, making calculations easier. When two objects of equal mass collide or explode, placing the origin at the center of mass allows velocities to be measured relative to a moving reference frame. This simplifies the math and helps visualize momentum conservation.
• Using the center-of-mass frame is especially powerful in elastic collisions. In this frame, the total momentum is zero, and objects simply reverse direction if the collision is symmetric. After solving the problem in the COM frame, results can be transformed back to the lab frame using vector addition. Though this isn’t required by the AP curriculum, understanding it reinforces deep reasoning about relative motion, symmetry, and invariance. It prepares students for advanced mechanics and special relativity concepts.
• Students should avoid assuming the center of mass is at the midpoint unless the mass distribution is uniform. For example, in a system with a 2 kg mass at \( x = 0 \) and a 4 kg mass at \( x = 3 \), the COM is not at \( x = 1.5 \), but rather: \[ x_{\text{COM}} = \frac{2(0) + 4(3)}{2 + 4} = \frac{12}{6} = 2\,\text{m} \] Mistakes like this often lead to incorrect predictions about motion, torque, or force balance. The correct approach is always to use the weighted average based on mass, not distance alone.

Example Problem 1: Velocity of Center of Mass After an Explosion

A 5.0 kg object is at rest on a frictionless horizontal surface when it suddenly explodes into two fragments. One fragment has a mass of 3.0 kg and moves at 4.0 m/s to the right. What is the velocity of the second fragment? What is the velocity of the center of mass before and after the explosion?

Solution:

• Before the explosion, the system is at rest. Therefore, the total momentum is 0, and by conservation of momentum: \[ 0 = (3.0\,\text{kg})(+4.0\,\text{m/s}) + (2.0\,\text{kg})(v) \Rightarrow v = -6.0\,\text{m/s} \] The second object moves left at 6.0 m/s to conserve momentum.
• The velocity of the center of mass is found using: \[ v_{\text{COM}} = \frac{\sum m_i v_i}{\sum m_i} = \frac{(3)(4) + (2)(-6)}{5} = \frac{12 - 12}{5} = 0\,\text{m/s} \] The COM velocity remains 0 both before and after the explosion, because no net external force acted on the system.
• This example reinforces that even when objects within a system move violently, the COM motion depends only on external forces. Internal forces (like an explosion) cannot alter the system’s overall momentum or center of mass velocity.

Example Problem 2: Calculating Center of Mass of a Rod and Particle

A thin uniform rod of length 4.0 m and mass 3.0 kg lies on the x-axis from \( x = 0 \) to \( x = 4.0 \, \text{m} \). A 2.0 kg particle is attached at \( x = 5.0 \, \text{m} \). What is the x-coordinate of the center of mass of the system?

Solution:

• The center of mass of the rod alone is at its midpoint: \[ x_{\text{rod}} = \frac{0 + 4}{2} = 2.0\,\text{m} \] Treat the rod as a point mass located at that point.
• Now apply the center of mass formula: \[ x_{\text{COM}} = \frac{(3.0)(2.0) + (2.0)(5.0)}{3.0 + 2.0} = \frac{6 + 10}{5} = \frac{16}{5} = 3.2\,\text{m} \] So the system’s COM is at 3.2 meters from the origin.
• This type of question often appears in rotation and torque problems where a distributed object and a discrete mass are combined. It builds spatial reasoning and prepares students for problems involving variable mass distributions.

Example Problem 3: Center of Mass of a Two-Block System in Motion

Two blocks, \( m_1 = 2.0 \, \text{kg} \) and \( m_2 = 3.0 \, \text{kg} \), are sliding on a frictionless surface. At a certain moment, block 1 is at position \( x = 1.0 \, \text{m} \) and moving at 2.0 m/s. Block 2 is at \( x = 5.0 \, \text{m} \) and moving at 1.0 m/s. Find (a) the center of mass position and (b) the center of mass velocity at this instant.

Solution:

• (a) Position of center of mass: \[ x_{\text{COM}} = \frac{(2.0)(1.0) + (3.0)(5.0)}{2.0 + 3.0} = \frac{2 + 15}{5} = 3.4\,\text{m} \] This tells us where the average mass of the system is located.
• (b) Velocity of center of mass: \[ v_{\text{COM}} = \frac{(2.0)(2.0) + (3.0)(1.0)}{5} = \frac{4 + 3}{5} = 1.4\,\text{m/s} \] So the center of mass moves rightward at 1.4 m/s, even though one block is faster than the other.
• Questions like this test whether students can apply COM formulas to moving systems, not just static ones. It also reinforces that the center of mass doesn’t necessarily lie between the objects, especially when their masses and speeds are unequal.

Common Misconceptions: Center of Mass

• “The center of mass is always located at a physical object.”
  Many students assume that the center of mass must lie on one of the objects in the system. In reality, the COM can exist in empty space, especially in systems with widely spaced or asymmetrically distributed masses. For example, if two skaters push off from each other, their COM remains between them in space, even as they move apart. The COM is not a physical part of the system but a mathematical point representing the system’s mass-weighted balance. This misunderstanding often causes errors in diagram analysis and motion prediction.
• “If the parts of a system accelerate, the COM must also accelerate.”
  Students sometimes believe that if pieces of a system are moving in different directions, the center of mass must be accelerating too. This is false unless there’s a net external force. In fact, in an explosion on a frictionless surface, all pieces may accelerate outward, but the COM can remain stationary if no external force is applied. Only external forces influence the COM's acceleration, not the internal interactions within the system. This misconception reflects a deeper confusion about Newton’s laws at the system level.
• “Momentum conservation always applies to each object individually.”
  While momentum is conserved for the entire isolated system, it may not be conserved for each individual object. Internal forces can drastically change individual momenta, like during a collision or explosion. The key is that the total momentum of the system stays constant if no net external force acts, not the momentum of each object. Confusing object-level and system-level momentum leads to incorrect assumptions in conservation problems.
• “If two objects move apart, their COM also moves apart.”
  Students may assume that if two blocks or particles separate, the center of mass splits or follows one of them. In fact, the COM follows a path determined only by external forces. For instance, in projectile motion, an object that explodes midair has pieces that scatter, but the COM continues along the original parabolic arc. The COM behaves as if the object were intact, which helps simplify calculations and visualize outcomes.
• “The center of mass is always located halfway between two objects.”
  Students often think the COM lies at the midpoint between two objects, regardless of their masses. This is only true if the masses are equal. If one object is heavier, the COM lies closer to it, since mass acts like a weight in the averaging formula. Misplacing the COM leads to incorrect torque calculations and errors in predicting rotational motion or balance points.