Linear Momentum

What this unit covers

The topics below follow the published Physics C: Mech course framework for Unit 4. This unit is worth 10–20% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Lessons in this unit

• Impulse & Momentum13 min · 3 objectivesDefine linear momentum p = mv and express Newton's second law as ΣF = dp/dt · Define impulse as the time integral J = ∫F dt and relate it to the change in momentum · Apply the impulse-momentum theorem, including reading impulse as area under an F-t graph
• Conservation of Momentum & Collisions14 min · 3 objectivesState conservation of momentum for an isolated system and justify it from Newton's third law · Distinguish elastic from inelastic collisions by whether kinetic energy is conserved · Solve one-dimensional collision problems, including the perfectly inelastic case
• Center of Mass14 min · 3 objectivesLocate the center of mass of a system of discrete particles · Compute the center of mass of a continuous body using x_cm = (1/M)∫x dm · Relate the motion of the center of mass to the net external force
• Variable Mass & the Rocket Equation14 min · 3 objectivesAnalyze systems whose mass changes using conservation of momentum · Relate the thrust on a rocket to its exhaust speed and burn rate · Apply the ideal rocket equation Δv = v_ex ln(m_i/m_f)
• Impulse from Force–Time Graphs & Variable Forces14 min · 3 objectivesCompute impulse as the integral of force over time · Read impulse as the area under a force–time graph · Explain why extending collision time reduces peak force
• Two-Dimensional Collisions14 min · 3 objectivesApply conservation of momentum independently to perpendicular components · Solve perfectly inelastic collisions in two dimensions · Use kinetic energy to classify a collision and check a solution for consistency
• The Center-of-Mass Frame14 min · 3 objectivesCompute the velocity of the center of mass of a system · Explain why the center of mass moves at constant velocity absent external forces · Use the center-of-mass frame to simplify collision analysis

Formulas in Unit 4

Every term in Unit 4

All 42 terms we publish for Linear Momentum, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Center of mass by integration

x_cm = (1/M)∫x dm. Express dm using linear density λ dx for a rod, then integrate over its length.

Momentum and impulse as integrals

p = mv and J = ∫F dt = Δp. Impulse is the area under a force-time curve when force varies.

Conservation of momentum

Follows from Newton's third law: internal forces cancel in pairs, so total momentum changes only through external forces.

Center of mass

x_cm = Σmx/Σm for discrete masses, or (1/M)∫x dm for a continuous body.

Motion of the center of mass

Accelerates only under net external force, so it continues undisturbed through any explosion or collision.

Elastic collision in one dimension

Conserving both momentum and kinetic energy gives relative approach speed equal to relative separation speed — a shortcut avoiding the quadratic.

Perfectly inelastic collision

Objects move together at v = (m₁v₁ + m₂v₂)/(m₁ + m₂), and kinetic energy loss is maximized.

Variable-mass systems

Rocket propulsion requires ΣF = dp/dt including the dm/dt term, which is why F = ma is insufficient.

Two-dimensional collisions

Momentum conserves independently in each direction, giving two equations to solve simultaneously.

Deciding whether momentum is conserved

Ask whether an external force acts over the interval. During a brief collision, gravity and friction contribute negligible impulse.

Momentum conservation with a pivot

A pivot exerts an external force, so linear momentum is not conserved in a collision with a hinged rod — but angular momentum about the pivot is.

Ballistic pendulum sequence

Momentum conservation for the embedding collision, then energy conservation for the swing. Using energy for the collision loses the point.

Kinetic energy lost in a collision

ΔKE = KE_i − KE_f, maximized in a perfectly inelastic collision and zero in an elastic one.

Elastic collision special cases

Equal masses exchange velocities; a very heavy object striking a light one gives the light one nearly twice the heavy one's speed.

Rocket equation qualitatively

Thrust comes from expelling mass, so ΣF = dp/dt must include v(dm/dt). Final speed depends on exhaust speed and the mass ratio.

Impulse-momentum in two dimensions

Apply the theorem separately in each direction; the impulse vector points along the change in momentum, not along the velocity.

Momentum is a vector

p = mv points along the velocity. Two objects moving oppositely have momenta that partially or wholly cancel, unlike their kinetic energies, which always add.

Newton's second law in momentum form

ΣF = dp/dt is the general statement; F = ma is the special case for constant mass. Variable-mass problems such as rockets require the momentum form.

The impulse–momentum theorem

J = ∫F dt = Δp. It is the time-integrated counterpart of the work–energy theorem, which integrates over distance and gives ΔK.

Impulse as area under a force–time graph

The area gives Δp; the peak height gives the largest instantaneous force. Two collisions with the same area can have very different peaks, which is what safety design targets.

Average force in a collision

F_avg = J/Δt = Δp/Δt. It is defined so that a constant force acting for the same duration would deliver the same impulse — a derived quantity, not something measured directly.

Why extending collision time helps

Δp is fixed by the situation, so F_avg = Δp/Δt falls as Δt rises. Airbags, crumple zones, helmet padding and bending your knees on landing all lengthen Δt.

Internal versus external forces

Internal forces come in third-law pairs and cancel in the system total, so they cannot change the system's momentum. Only external forces can.

When momentum is conserved

Whenever the net external force is zero, or when the collision is brief enough that external impulses are negligible compared with the internal ones. Gravity acts during a collision but contributes almost no impulse in a few milliseconds.

Momentum conserves componentwise

Each perpendicular component is conserved independently, giving two equations in two dimensions. This is the entire content of two-dimensional collision analysis.

Perfectly elastic collision

Both momentum and kinetic energy are conserved. Genuinely elastic collisions are rare at macroscopic scale — billiard balls approximate it, and gas molecules realize it.

Perfectly inelastic collision and maximum energy loss

The objects stick together and move as one, losing the maximum kinetic energy consistent with momentum conservation. The final velocity is simply the total momentum divided by the total mass.

Coefficient of restitution

e = relative speed of separation divided by relative speed of approach. e = 1 is elastic, e = 0 perfectly inelastic, and intermediate values describe real collisions.

Equal-mass elastic collision

The moving object stops and the target moves off with the original velocity — the velocities are exchanged. This is why a Newton's cradle behaves as it does.

Light object striking a heavy one

The light object rebounds with nearly its original speed while the heavy one barely moves — a ball bouncing off a wall. Momentum is still transferred; the wall's velocity change is simply tiny.

Heavy object striking a light one

The heavy object continues nearly unchanged and the light one is launched at up to twice the heavy object's speed. This is the mechanism behind a golf drive.

Relative velocity reversal

In a one-dimensional elastic collision the relative velocity of approach equals the relative velocity of separation. Using this with momentum conservation avoids solving a quadratic.

Ballistic pendulum

Two stages that must not be mixed: momentum is conserved during the embedding, and mechanical energy is conserved during the swing afterward. Applying energy conservation to the embedding is the classic error.

Explosions

The reverse of a perfectly inelastic collision — momentum is conserved and kinetic energy INCREASES, supplied by stored chemical or elastic energy. Starting from rest, fragments carry equal and opposite momenta.

Center of mass position

x_cm = Σm_i x_i/M, the mass-weighted average position. It need not lie inside the object, as with a ring or a boomerang.

Center of mass of a continuous body

x_cm = (1/M)∫x dm, with dm expressed through the density. For a uniform symmetric body it coincides with the geometric center.

Center of mass velocity

v_cm = p_total/M. A system with zero total momentum has a stationary center of mass, whatever its parts are doing.

The center of mass obeys Newton's second law

ΣF_ext = Ma_cm. A tumbling wrench rotates chaotically while its center of mass traces a clean parabola, because internal forces cannot accelerate it.

The center-of-mass frame

The frame moving at v_cm, in which total momentum is zero. Colliding objects approach and separate with equal and opposite momenta there, which makes collision algebra far simpler.

Kinetic energy in the center-of-mass frame

Total K splits into ½Mv_cm² plus the kinetic energy relative to the center of mass. Only the second part is available to be lost in a collision, which is why a perfectly inelastic collision cannot lose everything.

Recoil

Firing a projectile from a free object gives the object backward momentum equal in magnitude to the projectile's forward momentum, so the heavier the launcher the smaller its recoil speed.

Thrust

A rocket is pushed by the momentum it throws backward: thrust = v_ex |dm/dt|, the exhaust speed times the mass flow rate. It works in vacuum precisely because it needs nothing to push against.

What examiners penalize here

• For a continuous body, the recipe is always the same: write dm using the density, integrate x dm for the numerator and dm for the total mass M, then divide. Never average the endpoints — that only works for a uniform object.
• The rocket equation comes from momentum conservation with changing mass, not from F = ma with constant mass. Watch the logarithm: the payoff for carrying more fuel diminishes, since Δv grows only as ln(m_i/m_f).
• Set up a small table with columns for x and y and rows for before and after. Filling it in before doing any algebra prevents the dominant error on these problems, which is not conceptual but bookkeeping — a dropped component or a sign flipped on one object.

Practice Physics C: Mech

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

Questions about this unit

All 7 units of AP Physics C: Mechanics

1. Unit 1 · Kinematics
2. Unit 2 · Force and Translational Dynamics
3. Unit 3 · Work, Energy, and Power
4. Unit 4 · Linear Momentum
5. Unit 5 · Torque and Rotational Dynamics
6. Unit 6 · Energy and Momentum of Rotating Systems
7. Unit 7 · Oscillations