AP Physics 1 Linear Momentum: Conservation and Collisions

Conservation of Linear Momentum

What momentum is (and why it’s a “system” idea)

Linear momentum is a measure of “motion quantity” that depends on both how much mass is moving and how fast (and in what direction) it’s moving. For a single object, momentum is defined as

\vec{p} = m\vec{v}

Here \vec{p} is momentum (a vector), m is mass (a scalar), and \vec{v} is velocity (a vector). Because velocity has direction, momentum has direction too. That sounds simple, but it drives many of the most common mistakes: students often treat momentum like a scalar and accidentally use speed instead of velocity.

Momentum becomes especially powerful when you stop thinking about one object at a time and start thinking about a system—a chosen collection of objects you analyze together. In collisions and explosions, individual object momenta can change violently in a very short time, but the total momentum of the system may follow a simple rule.

For a system of multiple objects, total momentum is

\vec{p}_{\text{tot}} = \sum \vec{p}_i = \sum m_i\vec{v}_i

The conservation law (what it is)

Conservation of linear momentum says:

The total linear momentum of a system stays constant if the net external impulse on the system is zero.

In the most common AP Physics 1 collision situations (carts on a low-friction track, pucks on ice, objects interacting for a short time), the external forces are either negligible or act over such a short time that their effect is negligible during the interaction.

A standard mathematical statement for conservation in an isolated system is:

\vec{p}_{\text{tot, initial}} = \vec{p}_{\text{tot, final}}

This is a vector equation. In one dimension, that usually means you choose a positive direction and use signs. In two dimensions, it means momentum is conserved in each perpendicular direction separately (often x and y):

p_{\text{tot},x\,\text{initial}} = p_{\text{tot},x\,\text{final}}

p_{\text{tot},y\,\text{initial}} = p_{\text{tot},y\,\text{final}}

Why momentum is conserved (the mechanism)

It helps to see why this conservation law is true, because that tells you when you are allowed to use it.

When two objects in a system collide, they exert forces on each other. By Newton’s Third Law, these internal forces come in equal and opposite pairs. Internal forces can change each object’s momentum, but they cannot change the total momentum of the system, because the internal pushes and pulls “cancel out” when you add them over the whole system.

What can change total momentum is an external influence: friction from the ground, a wall pushing on the system, a hand applying a force, or any force from an object you did not include in your system. If external forces are negligible (or act symmetrically so the net effect is zero), then the total momentum stays the same.

A useful connection (often used to justify the conservation condition) is the impulse-momentum relationship for a system:

\Delta \vec{p}_{\text{system}} = \vec{J}_{\text{ext}}

If the net external impulse \vec{J}_{\text{ext}} is zero, then \Delta \vec{p}_{\text{system}} = \vec{0}, which means total momentum is constant.

How to use momentum conservation correctly (a reliable process)

When you see a collision, explosion, recoil, or “push apart” scenario, you should follow a consistent setup:

1. Choose the system. Most often, include all objects that interact (both colliding carts, both ice skaters, the gun and the bullet). If you leave one out, you usually introduce an external force that ruins simple conservation.
2. Check external forces during the interaction. A long interaction with significant friction means momentum might not be conserved. A brief collision on a low-friction track usually means it is.
3. Choose a coordinate direction and define positive. In 1D problems, this is essential for sign consistency.
4. Write total momentum before and after. Use velocities with signs.
5. Solve algebraically before plugging numbers (when possible). This helps you keep track of unknowns and signs.

Worked example 1: Recoil / “push apart” (explosion) in 1D

Two carts (masses m_1 and m_2) are initially at rest, coupled together on a nearly frictionless track. A spring releases, pushing them apart. Afterward cart 1 moves right at speed v_1. Find cart 2’s velocity v_2.

Concept first: This is not a collision but an internal interaction (a spring force) within the two-cart system. There is no significant external horizontal force, so total momentum stays zero because it started at zero.

Initial total momentum:

p_{\text{tot,i}} = 0

Final total momentum:

p_{\text{tot,f}} = m_1 v_1 + m_2 v_2

Conservation gives:

0 = m_1 v_1 + m_2 v_2

Solve for v_2:

v_2 = -\frac{m_1}{m_2}v_1

Interpretation: The negative sign means cart 2 moves left if cart 1 moves right. Also, the lighter cart ends up with the larger speed in magnitude.

Common pitfall to avoid: Students sometimes say “they push with equal force so they get equal speeds.” Equal force does not mean equal speed; the accelerations differ because masses differ.

Worked example 2: 2D momentum conservation (components)

A puck slides on frictionless ice. It explodes into two fragments A and B. Before the explosion, the puck moves east (positive x) with momentum magnitude p_0, and there is no north-south momentum. After the explosion, fragment A has momentum components p_{Ax} and p_{Ay}. Find fragment B’s momentum components.

Concept first: The explosion is internal to the system of fragments. With negligible external impulse, total momentum in each direction is conserved.

Before:

p_{\text{tot},x\,\text{i}} = p_0

p_{\text{tot},y\,\text{i}} = 0

After:

p_{\text{tot},x\,\text{f}} = p_{Ax} + p_{Bx}

p_{\text{tot},y\,\text{f}} = p_{Ay} + p_{By}

Conservation gives:

p_0 = p_{Ax} + p_{Bx}

0 = p_{Ay} + p_{By}

So:

p_{Bx} = p_0 - p_{Ax}

p_{By} = -p_{Ay}

Key takeaway: Even if the fragments fly off at angles, momentum bookkeeping stays clean if you break it into components.

Notation and sign conventions (small choices that prevent big errors)

Momentum problems often look different just because of notation. Here are equivalent ways you might see the same idea.

A simple but powerful habit: write velocities with signs from the beginning (for example, “left is negative”), and you will rarely need to “fix” anything later.

Exam Focus

• Typical question patterns:
  • “Two objects collide on a frictionless track; find a final velocity” using p_{\text{tot,i}} = p_{\text{tot,f}}.
  • “An object explodes into pieces; one piece’s motion is known; find the other’s velocity” (often starts with total momentum zero).
  • “A collision occurs at an angle; use conservation of x and y momentum separately.”
• Common mistakes:
  • Treating momentum as a scalar: using speeds instead of signed velocities (or forgetting components in 2D).
  • Applying conservation of momentum to one object instead of the whole interacting system.
  • Ignoring external forces when they matter (for example, long sliding with friction) or overthinking tiny external forces during a brief collision.

Elastic and Inelastic Collisions

What a “collision” means in physics

A collision is a short-time interaction in which objects exert large forces on each other, changing their velocities. The important AP Physics 1 idea is that collisions are analyzed using momentum conservation, but the type of collision determines what happens to kinetic energy.

In an idealized collision between two objects (1 and 2), momentum conservation is:

m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}

This is written for 1D motion. In 2D, you would write one conservation equation per direction.

Kinetic energy: conserved sometimes, not always

Kinetic energy is

K = \frac{1}{2}mv^2

During a collision, kinetic energy can transform into other forms: thermal energy (heating due to deformation and friction), sound, and internal energy associated with permanent deformation. That is why you must distinguish between collision types.

• In an elastic collision, total kinetic energy of the system is conserved.
• In an inelastic collision, total kinetic energy is not conserved (some becomes other forms).
• In a perfectly inelastic collision, the objects stick together after the collision—this is the maximum kinetic energy loss consistent with momentum conservation.

A crucial point: Momentum is conserved in all collisions only if the net external impulse is negligible. Collision type is about kinetic energy, not momentum.

Elastic collisions (what they are and how to solve them)

In a perfectly elastic collision, both momentum and kinetic energy are conserved:

m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}

\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2

How it works conceptually: In an elastic collision, the objects deform during contact but then “spring back” without lasting deformation, returning stored elastic potential energy back into kinetic energy. Think of ideal billiard balls: they briefly compress and then re-expand.

How to solve: In 1D, you typically have two unknown final velocities, and these two conservation equations let you solve for both. Algebra can be heavy, so staying organized matters.

Worked example: 1D elastic collision with a stationary target

A cart of mass m_1 moving right at speed v_{1i} collides elastically with a cart of mass m_2 initially at rest. Find expressions for v_{1f} and v_{2f}.

Concept first: Elastic means momentum and kinetic energy are conserved. Because cart 2 starts at rest, the equations simplify.

Momentum:

m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}

Kinetic energy:

\frac{1}{2}m_1 v_{1i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2

Solving this system (standard result for 1D elastic collisions) gives:

v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i}

v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i}

Interpretation checks (important on exams):

• If m_1 = m_2, then v_{1f} = 0 and v_{2f} = v_{1i}. The moving cart stops and the target cart takes off with the original speed—this matches the classic “Newton’s cradle” behavior.
• If m_2 is much larger than m_1, then v_{1f} is close to -v_{1i} (it bounces back) and v_{2f} is small (the heavy target barely moves).

Common pitfall to avoid: Students sometimes assume “elastic means speeds swap” in all cases. That only happens when masses are equal in 1D and one starts at rest.

Inelastic collisions (what they are and how to solve them)

In an inelastic collision, momentum is still conserved (assuming negligible external impulse), but kinetic energy is not:

m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}

What changes is that some kinetic energy becomes internal energy (heat, sound, deformation). This is why inelastic collisions often feel more realistic: real objects dent, compress, and generate sound.

Perfectly inelastic collisions (sticking together)

In a perfectly inelastic collision, the objects stick and move with the same final velocity v_f:

m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f

This is typically the easiest collision type algebraically because there is only one unknown final speed.

Why kinetic energy must decrease here: If the objects stick, they cannot “rebound” and restore the deformation energy back into motion. Momentum forces the combined mass to keep moving, but with less kinetic energy than before.

Worked example: perfectly inelastic collision (two carts stick)

Cart 1 (mass m_1 = 1.0\,\text{kg}) moves right at v_{1i} = 4.0\,\text{m/s}. Cart 2 (mass m_2 = 3.0\,\text{kg}) moves left at v_{2i} = -1.0\,\text{m/s}. They collide and stick. Find v_f.

Step 1: momentum conservation

m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f

Substitute:

1.0(4.0) + 3.0(-1.0) = (1.0 + 3.0)v_f

Compute:

4.0 - 3.0 = 4.0v_f

1.0 = 4.0v_f

v_f = 0.25\,\text{m/s}

Interpretation: Positive means the combined carts move to the right, but slowly. The heavier cart moving left had significant momentum, so it “nearly cancels” the rightward momentum.

Optional but insightful check: kinetic energy decreases

Initial kinetic energy:

K_i = \frac{1}{2}(1.0)(4.0)^2 + \frac{1}{2}(3.0)(1.0)^2

K_i = 8.0 + 1.5 = 9.5\,\text{J}

Final kinetic energy:

K_f = \frac{1}{2}(4.0)(0.25)^2

K_f = 0.125\,\text{J}

The kinetic energy loss is large, which is typical of sticking collisions.

Common pitfall to avoid: If you forget that cart 2’s velocity is negative, you would add momenta instead of subtracting and predict an incorrect (and too large) final speed.

Comparing elastic vs inelastic (what to look for in words and diagrams)

On AP problems, you are often told the collision type explicitly (“elastic,” “stick together”), but sometimes you infer it from the description.

A helpful memory aid: Momentum is (usually) conserved in collisions because external impulse is small; Kinetic energy is conserved only in the special Kind of collision: elastic. (“Only elastic keeps K.”)

How AP Physics 1 typically expects you to reason

AP Physics 1 emphasizes reasoning and representation as much as algebra. That means you should be comfortable with:

• Momentum bar charts / before-after reasoning: showing how total momentum stays constant while individual momenta change.
• System identification: explicitly stating what you include in the system and why external impulse is negligible.
• Qualitative predictions: for example, predicting whether an object reverses direction based on momentum signs and relative magnitudes.

A classic conceptual question: “If two objects collide and stick, what can you say about momentum and kinetic energy?” The correct reasoning is: momentum can be conserved (if isolated), kinetic energy decreases.

Mixed situations: when you need more than one idea

Sometimes a problem combines collision analysis with other topics (forces, energy, motion). Within the “conservation and collisions” scope, the key skill is knowing which conservation law applies to which time interval.

• During the short collision interval, you often use momentum conservation.
• Outside the collision (longer times), external forces like friction may matter, and you might use kinematics or energy with work done by friction.

A common mistake is trying to use kinetic energy conservation for a collision that is not elastic just because you want a second equation. If the collision is inelastic, you do not get to conserve kinetic energy; you must use other information (like “they stick,” or given one final speed, or an additional constraint).

Exam Focus

• Typical question patterns:
  • “Two carts collide; they stick; find final speed” (perfectly inelastic, momentum only).
  • “A collision is elastic; find both final speeds” (momentum and kinetic energy in 1D).
  • “Decide whether a collision is elastic or inelastic based on kinetic energy before/after or on a description.”
• Common mistakes:
  • Assuming kinetic energy is conserved in any collision where momentum is conserved.
  • Forgetting that momentum is a vector: using magnitudes only, missing negative signs, or failing to conserve separately in x and y.
  • Using the “stick together” equation without actually enforcing a shared final velocity (writing two different final velocities when the objects are attached).