Unit 4.2 | Conservation of Momentum

Conservation of Momentum

When two objects interact, Newton’s 3rd law says they push on each other with equal and opposite forces, for exactly the same time. Equal and opposite forces over the same time give equal and opposite impulses. So whatever momentum one object gains, the other loses.

If no outside force acts on a system, its total momentum never changes.

[katex] \sum \vec{p}_i = \sum \vec{p}_f [/katex]

Here is that idea measured during a real-style collision between two carts on a track:

Momentum during a collision

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Choosing the System

Just like with energy, “conserved” depends on the system you choose.

• Include both objects that push on each other, so their forces are internal and cancel.
• Momentum is conserved when the net external force on the system is zero (or so brief it does not matter, like during a quick collision).

Two carts collide on a frictionless track. If the system is cart 1 only, cart 2 pushes on it from outside and its momentum changes. If the system is both carts, the push is internal and the total momentum is conserved.

▶ Watch: how to pick the system so momentum is conserved (starts at 4:20).

The Momentum Framework

Every conservation problem in this unit, from here to the hardest collision, uses the same three steps:

1. Pick the system (every object that pushes on another) and a positive direction.
2. Sketch before and after, with a velocity for each object. Unknowns get a letter.
3. Apply [katex] \sum p_i = \sum p_f [/katex] with signs, then solve.

Problem Type 1: Is momentum conserved?

How to spot it: “Which system”, “is momentum conserved”, “justify your answer”. Often there are no numbers at all.

Setup: List the forces from outside the system. If they add to zero, or the event is too quick for them to matter, momentum is conserved.

An astronaut floating in space throws a wrench. Is momentum conserved for the astronaut alone? For the astronaut and the wrench?

Astronaut alone: no. The wrench pushes back on the astronaut from outside the system. Astronaut and wrench: yes. That push is now internal, and nothing outside acts on the pair.

AP trap: Momentum being conserved says nothing about kinetic energy. A system can keep all its momentum and still lose kinetic energy (Lesson 4.3).

Problem Type 2: Something lands on or drops off

How to spot it: Sand, rain or a package dropped onto a moving cart, boat or sled. A person jumping onto a moving skateboard.

Setup: The new mass had no sideways momentum, so [katex] m_1v_1 = (m_1 + m_2)v’ [/katex].

A [katex] 20\,\text{kg} [/katex] cart rolls at [katex] 3\,\text{m/s} [/katex]. A [katex] 10\,\text{kg} [/katex] sack is dropped straight down into it. How fast does the cart move now?

[katex] (20)(3) = (20 + 10)v’ \rightarrow v’ = 2 \, m/s [/katex]. The same momentum is now shared by more mass, so the cart slows down.

AP trap: Thinking the speed stays the same because nothing pushed the cart backward. The sack has to be sped up sideways, and that momentum comes from the cart.

Problem Type 3: The center of mass keeps going

How to spot it: An explosion in mid-air, a collision, or “what happens to the center of mass?”.

Setup: A system’s total momentum is its total mass times the velocity of its center of mass: [katex] \vec{p}_{total} = M\vec{v}_{cm} [/katex]. If the total momentum is conserved, the center of mass keeps moving exactly as before.

AP trap: Saying the explosion “adds energy, so the center of mass speeds up”. Internal forces can add kinetic energy, but they can never change the motion of the center of mass.

▶ Watch: worked center of mass problems (starts at 37:45).

Lesson 4.3 Preview

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