Momentum and Newton's Third Law
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You should be comfortable with Newton's Second Law (\( \vec{F} = m\vec{a} \)) and the concept of work and energy from the previous lecture. Familiarity with vector addition and basic calculus is assumed.
Why this matters
Momentum conservation complements energy conservation, and is often the more useful of the two when interaction forces are complex or unknown.
While energy conservation tells us about the state of a system, momentum conservation governs the dynamics of interactions. In collisions, explosions, or rocket propulsion, the forces involved are often internal, brief, and difficult to measure directly. However, if the net external force is zero, the total momentum of the system remains constant. This principle allows us to predict the outcomes of such events without needing to know the detailed time-dependence of the internal forces, making it a cornerstone of classical mechanics.
The Intuitive Picture
Imagine two ice skaters standing still on a frictionless surface. If they push off each other, they move in opposite directions. The heavier skater moves slower, while the lighter skater moves faster. Even though they are moving, the "amount of motion" in one direction is exactly balanced by the "amount of motion" in the other. This balance is momentum. Newton's Third Law states that the force Skater A exerts on Skater B is equal in magnitude and opposite in direction to the force Skater B exerts on Skater A. Because these forces act for the same amount of time, the change in momentum for one skater is equal and opposite to the change for the other. If the system started at rest, it must end at rest in terms of total momentum. This idea generalizes to any number of particles: internal forces always come in pairs that cancel out, leaving only external forces to change the total momentum.
Formalizing Momentum Conservation
We define the linear momentum of a particle as the product of its mass and velocity, \( \vec{p} = m\vec{v} \). For a system of \( N \) particles, the total momentum is the vector sum: \[ \vec{P} = \sum_{i=1}^{N} m_i \vec{v}_i \] To find how this total changes with time, we differentiate with respect to time. Since mass is constant, we get: \[ \frac{d\vec{P}}{dt} = \sum_{i=1}^{N} m_i \frac{d\vec{v}_i}{dt} = \sum_{i=1}^{N} m_i \vec{a}_i \] By Newton's Second Law, \( m_i \vec{a}_i \) is the net force acting on particle \( i \). This net force can be split into external forces \( \vec{F}_i^{\text{ext}} \) and internal forces \( \vec{F}_{ij} \) exerted by other particles in the system. \[ \frac{d\vec{P}}{dt} = \sum_{i=1}^{N} \vec{F}_i^{\text{ext}} + \sum_{i=1}^{N} \sum_{j \neq i} \vec{F}_{ij} \] Newton's Third Law states that \( \vec{F}_{ij} = -\vec{F}_{ji} \). When we sum over all pairs \( i \) and \( j \), every internal force term cancels with its counterpart. For example, the force of 1 on 2 cancels the force of 2 on 1. Thus, the double sum of internal forces vanishes: \[ \sum_{i=1}^{N} \sum_{j \neq i} \vec{F}_{ij} = \vec{0} \] This leaves us with the fundamental result: \[ \frac{d\vec{P}}{dt} = \sum_{i=1}^{N} \vec{F}_i^{\text{ext}} \equiv \vec{F}^{\text{ext}}_{\text{net}} \] If the system is isolated, meaning \( \vec{F}^{\text{ext}}_{\text{net}} = \vec{0} \), then \( \frac{d\vec{P}}{dt} = \vec{0} \), which implies \( \vec{P} \) is constant.
Impulse and Variable Forces
In many real-world scenarios, such as a bat hitting a ball, the force is not constant but varies rapidly with time. We cannot simply use \( \vec{F} = m\vec{a} \) with a single value of \( \vec{F} \). Instead, we integrate the equation of motion over the duration of the interaction. \[ \int_{t_1}^{t_2} \vec{F}_{\text{net}} \, dt = \int_{t_1}^{t_2} \frac{d\vec{p}}{dt} \, dt \] The right-hand side simplifies directly to the change in momentum: \[ \int_{t_1}^{t_2} \vec{F}_{\text{net}} \, dt = \vec{p}_2 - \vec{p}_1 = \Delta \vec{p} \] The quantity \( \vec{J} = \int_{t_1}^{t_2} \vec{F}_{\text{net}} \, dt \) is called the impulse. The Impulse-Momentum Theorem states that the impulse delivered to a particle equals its change in momentum. For a system of particles, if we consider the total external impulse, we have: \[ \vec{J}_{\text{ext}} = \Delta \vec{P}_{\text{system}} \] This is particularly useful when the interaction time \( \Delta t \) is very short. We can approximate the average force \( \vec{F}_{\text{avg}} \) such that \( \vec{F}_{\text{avg}} \Delta t = \Delta \vec{p} \). This allows us to analyze collisions without knowing the exact force-time profile, provided we know the initial and final velocities.
Collisions and Elasticity
Collisions are a primary application of momentum conservation. We classify collisions based on kinetic energy. In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, momentum is conserved (assuming no external forces), but kinetic energy is not; some is converted to heat, sound, or deformation. In a perfectly inelastic collision, the objects stick together after impact, resulting in the maximum possible loss of kinetic energy consistent with momentum conservation. Consider a one-dimensional collision between two masses \( m_1 \) and \( m_2 \) with initial velocities \( u_1 \) and \( u_2 \) and final velocities \( v_1 \) and \( v_2 \). Momentum conservation gives: \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \] For an elastic collision, we also have: \[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \] Solving these two equations simultaneously yields the final velocities. A key insight from this algebra is that for elastic collisions, the relative velocity of separation is equal in magnitude and opposite in direction to the relative velocity of approach: \[ v_1 - v_2 = -(u_1 - u_2) \] This "coefficient of restitution" equals 1 for elastic collisions and less than 1 for inelastic ones.
The derivation
Momentum Conservation from Newton's Third Law
For an isolated system of \(N\) point particles interacting only through pairwise internal forces that obey Newton's third law, the total linear momentum \(\vec{P}=\sum_{i} m_i\vec{v}_i\) is constant in time, \(\frac{d\vec{P}}{dt}=\vec{0}\). The internal forces produce no net change in the system's momentum because they cancel in equal-and-opposite pairs, so \(\frac{d\vec{P}}{dt}\) equals the net external force alone.
Newton's Third Law in its simple form (\( \vec{F}_{12} = -\vec{F}_{21} \)) strictly applies to contact forces and static fields. In electrodynamics, where forces are mediated by fields that carry their own momentum, the simple pairwise cancellation can fail if one only considers particle momenta. The total momentum (particles plus field) is conserved, but the particle-only momentum may not be. In standard Newtonian mechanics, however, we assume instantaneous action-at-a-distance or contact forces, so the simple law holds. C
Worked example
Problem. A 1.0 kg cart moving at 4.0 m/s collides with a stationary 3.0 kg cart on a frictionless track. The carts stick together after the collision. Calculate the final velocity of the combined system and the fraction of kinetic energy lost.
- Identify the system: the two carts. External horizontal forces are zero, so momentum is conserved.
- Calculate initial total momentum: \( P_i = m_1 u_1 + m_2 u_2 = (1.0)(4.0) + (3.0)(0) = 4.0 \ \mathrm{kg \cdot m/s} \).
- Calculate final total momentum: \( P_f = (m_1 + m_2) v_f = (1.0 + 3.0) v_f = 4.0 v_f \).
- Equate \( P_i = P_f \): \( 4.0 = 4.0 v_f \), so \( v_f = 1.0 \ \mathrm{m/s} \).
- Calculate initial kinetic energy: \( K_i = \frac{1}{2} m_1 u_1^2 = \frac{1}{2}(1.0)(4.0)^2 = 8.0 \ \mathrm{J} \).
- Calculate final kinetic energy: \( K_f = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2}(4.0)(1.0)^2 = 2.0 \ \mathrm{J} \).
- Calculate energy lost: \( \Delta K = K_i - K_f = 8.0 - 2.0 = 6.0 \ \mathrm{J} \).
- Calculate fraction lost: \( \frac{\Delta K}{K_i} = \frac{6.0}{8.0} = 0.75 \).
Units check. Momentum is \( \mathrm{kg \cdot m/s} \). Velocity is \( \mathrm{m/s} \). Energy is \( \mathrm{J} \) (which is \( \mathrm{kg \cdot m^2/s^2} \)). The fraction is dimensionless. All units are consistent.
Three quick tests
- Does momentum conservation require the forces to be conservative?
Answer
No. Momentum conservation depends on the net external force being zero, not on the nature of the internal forces (conservative or not). - If two objects collide and stick together, is kinetic energy conserved?
Answer
No. This is a perfectly inelastic collision, and kinetic energy is not conserved, though momentum is. - Can an isolated system have a non-zero total momentum?
Answer
Yes. An isolated system can have any constant total momentum, including non-zero values, depending on its initial conditions.
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.