AP Physics C: Mechanics · Unit 4: Linear Momentum · Lesson 4.3

Deep Dive: Conservation of Linear Momentum

🔬 Deep Dive

This is your textbook for this topic. Take your time. Read it more than once.

4.3.A.14.3.A.1.i4.3.A.1.iiMath

Center of Mass Velocity of a System

A collection of objects, no matter how individually complicated their motions, can always be described as a single system moving at one center-of-mass velocity:

v⃗(cm) = Σpᵢ / Σmᵢ = Σ(mᵢv⃗ᵢ) / Σmᵢ

🔑This velocity is constant as long as no net external force acts on the system — regardless of what internal interactions (collisions, explosions, springs) happen between the objects inside it. Internal forces can completely rearrange how individual objects move without ever budging the system's own overall motion.

Explore this directly below — adjust two carts' masses and velocities, then trigger a collision between them and watch v(cm) survive untouched.

Two carts, each with its own mass and velocity. Watch the center-of-mass velocity respond — then toggle a collision between them and see v(cm) stay exactly the same, even though each cart's own velocity changes.

m₁4 kg

v₁6 m/s

m₂6 kg

v₂-2 m/s

Simulate a collision between the carts

v(cm) before = 1.20 m/s

Toggle the collision above to see v(cm) survive completely unchanged.

4.3.A.24.3.A.34.3.A.3.i–iiiConceptMath

Total Momentum and How It Balances

A system's total momentum is simply the sum of the momenta of its constituent parts. In the absence of net external forces, any change to one object's momentum within the system must be balanced by an equal and opposite change somewhere else in the system.

🔑This isn't a coincidence — it's a direct consequence of Newton's third law. The impulse one object exerts on a second object during an interaction is exactly equal and opposite to the impulse the second exerts back on the first. Whatever momentum one object gains, the other loses in exactly equal measure.

You get to choose your system so that its total momentum is constant — and if that total ever does change, the change is exactly equal to whatever impulse crossed the system's boundary:

J⃗ = Δp⃗(system)

4.3.A.44.3.B.14.3.B.24.3.B.3ConceptMath

Choosing a System: Conservation of Momentum

Momentum is conserved in all interactions — full stop, no exceptions, exactly parallel to how energy is always conserved (Lesson 3.4). Whether a specific system's total momentum stays constant depends on how you draw that system's boundary:

🔑If the net external force on your chosen system is zero, its total momentum is constant — an isolated system. If the net external force is nonzero, momentum is being transferred between the system and its environment, and the total will change by exactly that transferred amount.

Putting it all together gives the signature equation of this entire unit:

Σp⃗ᵢ = Σp⃗f   (isolated system)

Correct application of this equation lets you find velocities immediately before or after a collision or explosion — without ever needing to know the messy interaction forces involved. Explore the "P(system) constant" gauge from this unit's header art directly below.

The "P(system) constant" gauge from this unit's header art. Set each object's momentum, then drag the redistribution slider to simulate any internal interaction between them — the needle on the total never moves.

p₁40 kg·m/s

p₂-15 kg·m/s

internal interaction

p₁ (current) = 40.0p₂ (current) = -15.0Total = 25.0

No matter how the interaction slider redistributes momentum between the two objects, the needle on P(system) stays exactly still — this is what "isolated system" means in practice.

ExampleGuided Example — Finding a Velocity After a Collision

A 3 kg cart moving at 5 m/s collides with a stationary 2 kg cart. After the collision, the 3 kg cart continues at 1 m/s in the same direction. Find the 2 kg cart's velocity after the collision.

Step 1 — Choose the system and check for isolation

System: both carts together. If the track is frictionless and no other external force acts, this system is isolated — total momentum is conserved.

ExampleWorked Example — Non-Isolated System

The same two carts as above collide, but this time an external braking force also acts on the system for the duration of the collision, delivering an impulse of −4 N·s (opposing the initial motion). Find the 2 kg cart's final velocity under these conditions.

← Back to Lesson 4.3Ready for 4.4? Elastic and Inelastic Collisions closes out Unit 4 by classifying exactly what conservation of momentum allows — and what conservation of energy doesn't always guarantee.