AP Physics C: Mechanics Unit 4: Systems of Particles and Linear Momentum Notes

Introduction

Momentum turns the motion of many particles into one vector ledger controlled by external impulse.

Individual momenta add vectorially, and their sum equals total mass times center-of-mass velocity. Internal interaction forces can rearrange that momentum among particles but cannot change the total system momentum.

The time integral of net external force changes total momentum. During a short collision, external impulse may be negligible even when external force is not exactly zero; that approximation must be stated and justified.

Mass and velocity

linear momentum

Force accumulated over time

impulse

Negligible external impulse

conserved system momentum

Momentum plus energy constraints

collision outcomes

Current exam guidance: Aligned to the current 2026–27 AP Physics C: Mechanics framework, with calculus-based modeling, derivations, experimental reasoning, and explicit assumptions.

4.1

Linear Momentum

How does particle momentum combine into the motion of an entire system?

Short answer

Linear momentum is $\vec p=m\vec v$ , and for constant total mass $\vec P=\sum_i m_i\vec v_i=M\vec v_{cm}$ .

Know firstVelocity vectors · Center of mass · Vector components

Momentum adds before its magnitude is taken

Momentum follows velocity direction and depends on the chosen frame. In a multiparticle system, add each Cartesian component. Opposing components may cancel even while every particle moves and the system retains positive kinetic energy.

The total momentum equals $M\vec v_{cm}$ for constant total mass. In a collision model, internal interaction forces dominate during a short interval, so compare states immediately before and after; an explosion model begins together and separates through internal forces. A center-of-momentum frame has $\vec P=0$ , but its particles need not be at rest relative to one another.

Physics model

Momentum follows velocity; kinetic energy follows speed squared

Reverse velocity and compare the signed momentum with the nonnegative kinetic energy.

Velocity4 m/s

Momentum, $m=2$ kg: 8 kg·m/s
Kinetic energy: 16 J

velocitymomentumkinetic energy

\vec p=m\vec v

Read this figure without the graphic

Reverse velocity and compare the signed momentum with the nonnegative kinetic energy.

Current quantity	Value
Momentum, $m=2$ kg	8 kg·m/s
Kinetic energy	16 J
Model relation	$\vec p=m\vec v$

Connect total momentum to center of mass

System: Particles with constant masses $m_i$ .
Frame: One inertial frame.
Assumptions: Total mass $M=\sum_i m_i$ is constant.
Givens: $\vec r_{cm}=M^{-1}\sum_i m_i\vec r_i$ .

Differentiate the center-of-mass definition with respect to time.
Constant masses pass through the derivative.
Multiply by $M$ to obtain $\sum_i m_i\vec v_i$ .

Result

\vec P=M\vec v_{cm}

Units:

\mathrm{kg\,m/s}

Boundary cases

$\vec P=0$ can coexist with $K>0$ .
An open system with changing membership requires explicit momentum-flux accounting.

The mass-weighted average velocity multiplied by total mass equals the vector sum of particle momenta.

Momentum and energy answer different questions

Quantity	Addition rule	Can opposite motion cancel?
$\vec P=\sum m_i\vec v_i$	vector components	yes
$K=\sum\tfrac12m_iv_i^2$	nonnegative scalar terms	no

Build a system momentum

Choose an inertial frame and coordinate axes.
List each included mass and velocity vector.
Add components with signs.
Use $\vec P/M$ for center-of-mass velocity.

Keep these three ideas

Momentum is vector-valued and frame dependent.
System momentum equals $M\vec v_{cm}$ .
Zero total momentum can coexist with motion.

Systems of Particles and Linear Momentum