Momentum of a System of Particles
When you watch a game of billiards, you see single balls colliding and scattering in every direction. But physicists often need to track not just one object, but an entire collection of objects at once — a system of particles. The momentum of a system is simply the vector sum of the momenta of every particle in that system. Because momentum is a vector quantity (it has both magnitude and direction), you must add the individual momenta as vectors, accounting for their directions, not just their magnitudes. This idea is more powerful than it might first appear. Once you know the total momentum of a system, you can apply Newton's second law to the system as a whole: the net external force on the system equals the rate of change of the system's total momentum. Crucially, internal forces — forces that particles in the system exert on each other — always come in equal-and-opposite pairs (Newton's third law) and therefore cancel out when you sum over the whole system. Only external forces can change the total momentum of a system. The concept of system momentum underpins the law of conservation of momentum: if the net external force on a system is zero, the total momentum is constant in time. This is why an astronaut floating in deep space who throws a wrench will recoil in the opposite direction — the system (astronaut + wrench) has no external horizontal force, so its total momentum, initially zero, must remain zero.
- Define the System
- Write the Momentum of Each Particle
- Sum as Vectors
- Connect to the Center of Mass
- Apply Newton's Second Law to the System
Define the System
Clearly identify which objects belong to your system and which are external. For example, in a rocket-exhaust problem, you might define the system as the rocket body plus the exhaust gases. Everything outside the system boundary is 'external' and can exert forces that change total momentum.
Write the Momentum of Each Particle
For each particle i with mass m_i and velocity vector v_i, its individual momentum is p_i = m_i * v_i. Remember that v_i is a vector, so it carries sign information (or x- and y-components in two dimensions).
Sum as Vectors
The total (system) momentum is P = p_1 + p_2 + ... + p_N = Σ m_i * v_i. In two or three dimensions, add x-components together and y-components together separately. A 3 kg object moving at +5 m/s and a 3 kg object moving at −5 m/s give a total system momentum of exactly zero, even though each piece has non-zero momentum.
Connect to the Center of Mass
There is an elegant shortcut: P = M * v_cm, where M = Σ m_i is the total mass of the system and v_cm is the velocity of the system's center of mass. This means the entire system behaves as if all its mass were concentrated at the center of mass moving with velocity v_cm. This is why the center of mass of a fireworks shell follows a smooth parabola even as fragments scatter in all directions — gravity (the only external force, ignoring air) acts on the total mass at the center of mass.
Apply Newton's Second Law to the System
The net external force on the system equals the time rate of change of the total momentum: F_net,ext = dP/dt. Internal forces (particle A pushing particle B and particle B pushing back on A) cancel in pairs and contribute nothing to dP/dt. If F_net,ext = 0, then P is constant — this is conservation of momentum.
The Key Equation
- \vec{P}_{\text{sys}} — System momentum (kg·m/s): The total linear momentum of all particles in the system, a vector quantity.
- m_i — Mass of particle i (kg): The mass of the i-th object in the system.
- \vec{v}_i — Velocity of particle i (m/s): The velocity vector of the i-th object; direction matters for the vector sum.
- M — Total system mass (kg): The scalar sum of all individual masses: M = Σ m_i.
- \vec{v}_{\text{cm}} — Center-of-mass velocity (m/s): The velocity of the system's center of mass; the system's total momentum equals M times this velocity.
Worked Example
Problem. Three hockey pucks slide on frictionless ice. Puck A has mass 0.20 kg and velocity (+6.0 î) m/s. Puck B has mass 0.30 kg and velocity (−4.0 î + 3.0 ĵ) m/s. Puck C has mass 0.50 kg and velocity (2.0 î − 5.0 ĵ) m/s. Find the total momentum of the system and the speed of the center of mass.
Answer. Total system momentum ≈ 1.89 kg·m/s at 58° south of east; center-of-mass speed ≈ 1.89 m/s.
Origins · 1 minRené Descartes and how this was discovered
Descartes was among the first to articulate a conservation principle for what he called 'quantity of motion' (the product of mass and speed) in his 1644 work Principia Philosophiae. Although his scalar version lacked the full vector character we now require, it planted the seeds for Newton's precise formulation of momentum and its conservation as a vector quantity — directly enabling the modern concept of system momentum. Descartes famously conducted his philosophical and scientific work while lying in bed until late morning, a habit he maintained for most of his adult life — reportedly because he believed his best thinking happened in a warm, undisturbed state.
Related Concepts
- Linear Momentum of a Single Particle — System momentum is built by summing single-particle momenta, so understanding p = mv for one object is the essential prerequisite.
- Conservation of Momentum — When net external force on a system is zero, total system momentum is conserved — the direct application of system momentum concepts.
- Impulse–Momentum Theorem — Impulse equals the change in momentum; applying this to a system connects external forces to changes in total system momentum over time.
- Center of Mass — Total system momentum equals total mass times center-of-mass velocity, making center of mass a key tool for analyzing system dynamics.