Momentum & Impulse in AP Physics C: Mechanics (Unit 4 Core Ideas)

Linear Momentum

What momentum is (and why it’s a big deal)

Linear momentum is a measure of how hard it is to stop (or redirect) a moving object. Intuitively, a fast-moving truck is harder to stop than a slow-moving skateboard—not just because it’s faster, but because it also has much more mass. Momentum captures both effects at once.

Formally, the linear momentum \vec{p} of a particle of mass m moving with velocity \vec{v} is

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

Momentum matters because it’s the quantity that ties forces to motion in the most general way. In AP Physics C, you’ve already used Newton’s 2nd law in the form \vec{F}_{net} = m\vec{a}, but the deeper (and more general) statement is that net external force changes momentum. That idea becomes especially powerful when you analyze collisions, explosions, recoil, and any situation where forces are large but act over short times.

Momentum is a vector (direction and sign matter)

Because velocity is a vector, momentum is a vector. In one dimension, you often represent direction by a sign: choosing “right” as positive makes leftward momentum negative.

A common mistake is to treat momentum like speed (a scalar). On free-response questions, sign errors usually come from forgetting that \vec{p} points in the same direction as \vec{v}.

Units and interpretation

From \vec{p} = m\vec{v}, the SI units are

• m in \text{kg}
• \vec{v} in \text{m/s}
• \vec{p} in \text{kg}\cdot\text{m/s}

Momentum is also closely connected to impulse (covered in the next section), whose unit \text{N}\cdot\text{s} is equivalent to \text{kg}\cdot\text{m/s}.

Momentum of a system of particles

In Unit 4, you start thinking about systems—collections of objects. The total momentum of a system is the vector sum of individual momenta:

\vec{P} = \sum_i \vec{p}_i = \sum_i m_i\vec{v}_i

This is crucial because many interesting interactions (collisions, explosions) involve forces between objects inside the system. Those internal forces often complicate a single-object view, but they become manageable when you treat the objects as a system.

How momentum connects to Newton’s laws

Newton’s 2nd law can be written in momentum form:

\vec{F}_{net} = \frac{d\vec{p}}{dt}

If the mass of the object is constant, then \vec{p} = m\vec{v} gives

\frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}

so you recover \vec{F}_{net} = m\vec{a}.

For a system of particles, the key idea is that internal forces occur in equal-and-opposite pairs. When you sum forces over the whole system, internal forces cancel (as long as they’re all included inside the system boundary). That sets up the conservation laws you’ll use constantly.

Example 1: Comparing momenta

A 0.15\,\text{kg} baseball moves at 40\,\text{m/s} and a 1.5\,\text{kg} brick moves at 4\,\text{m/s}. Compare the magnitudes of their momenta.

Compute each magnitude:

p_{ball} = (0.15)(40) = 6.0\,\text{kg}\cdot\text{m/s}

p_{brick} = (1.5)(4) = 6.0\,\text{kg}\cdot\text{m/s}

They have the same momentum magnitude even though their masses and speeds differ—momentum balances both.

Example 2: Total momentum of a system (1D)

Two carts on a track:

• Cart A: m_A = 2.0\,\text{kg}, v_A = +3.0\,\text{m/s}
• Cart B: m_B = 1.0\,\text{kg}, v_B = -2.0\,\text{m/s}

Total system momentum:

P = m_A v_A + m_B v_B

P = (2.0)(3.0) + (1.0)(-2.0) = 4.0\,\text{kg}\cdot\text{m/s}

The positive result means the system’s net momentum points in the positive direction.

Exam Focus

• Typical question patterns:
  • Compute individual and total momentum vectors from mass and velocity (often with signs or components).
  • Use \vec{F}_{net} = d\vec{p}/dt to connect force graphs or time intervals to changes in momentum.
  • Define a system and identify which momenta belong in \vec{P} = \sum \vec{p}.
• Common mistakes:
  • Treating momentum as scalar in 2D problems (forgetting x- and y-components).
  • Dropping negative signs in 1D when objects move opposite your chosen positive direction.
  • Using \vec{F}_{net} = m\vec{a} blindly in situations better handled by momentum/impulse (short collisions, large forces over small times).

Impulse and Momentum Change

What impulse is

Impulse is a measure of how much a force changes an object’s momentum over a time interval. If momentum tells you “how hard it is to stop,” impulse tells you “how much stopping (or speeding up) the force actually accomplishes.”

For a constant net force acting over a time interval \Delta t, impulse is

\vec{J} = \vec{F}_{net}\Delta t

The impulse has the same direction as the net force.

The impulse-momentum theorem (the core relationship)

The most important idea in this section is the impulse-momentum theorem:

\vec{J} = \Delta\vec{p}

meaning the impulse delivered to an object equals the change in its momentum.

To see why this is true, start from Newton’s momentum form:

\vec{F}_{net} = \frac{d\vec{p}}{dt}

Integrate both sides over time from t_i to t_f:

\int_{t_i}^{t_f} \vec{F}_{net}\,dt = \int_{t_i}^{t_f} \frac{d\vec{p}}{dt}\,dt

The right side becomes the net change in momentum:

\int_{t_i}^{t_f} \frac{d\vec{p}}{dt}\,dt = \vec{p}_f - \vec{p}_i = \Delta\vec{p}

So the general definition of impulse is

\vec{J} = \int_{t_i}^{t_f} \vec{F}_{net}\,dt

This integral form is essential in AP Physics C because forces in collisions often vary strongly with time.

Average force vs peak force

In many collision problems, you’re asked for an average force over the interaction time. Define average net force by

\vec{F}_{avg} = \frac{\Delta\vec{p}}{\Delta t}

This does not mean the force was constant—only that a constant force of magnitude F_{avg} over the same time would produce the same impulse.

A very common misconception: students treat F_{avg} as if it were the maximum force. Real collision forces often have sharp peaks, so F_{max} can be much larger than F_{avg}.

Impulse as area under a force-time graph

Because

\vec{J} = \int \vec{F}_{net}\,dt

impulse equals the area under the force vs time curve (including sign/direction). In 1D, you can treat this as signed area.

• A rectangular force pulse: area is F\Delta t.
• A triangular pulse: area is \frac{1}{2}F_{peak}\Delta t.

This is a frequent AP question format: you’re given a graph and asked for \Delta p or final speed.

Why “increase collision time” reduces force (safety physics)

From

\vec{F}_{avg} = \frac{\Delta\vec{p}}{\Delta t}

if you need the same momentum change (for example, bringing a head or a car to rest), increasing the stopping time decreases the average force. That’s the physics behind airbags, crumple zones, padded gym mats, and follow-through when catching a ball.

Be careful: increasing time reduces average force, but the momentum change depends on initial and final velocities, not on how “gentle” the collision feels.

Worked example 1: Impulse from a force-time graph (triangular pulse)

A force on a cart acts in the positive direction, rising linearly from 0 to 800\,\text{N} over 0.020\,\text{s}, then falling linearly back to 0 by 0.050\,\text{s}. The cart starts from rest with mass m = 4.0\,\text{kg}. Find its final speed.

Step 1: Compute impulse as area. The graph is a triangle with base 0.050\,\text{s} and height 800\,\text{N}.

J = \frac{1}{2}(800)(0.050) = 20\,\text{N}\cdot\text{s}

Step 2: Use impulse-momentum theorem. Starting from rest means p_i = 0, so

\Delta p = p_f = J = 20\,\text{kg}\cdot\text{m/s}

Step 3: Convert momentum to speed.

p_f = mv_f

v_f = \frac{p_f}{m} = \frac{20}{4.0} = 5.0\,\text{m/s}

Worked example 2: Average force during a stop

A 0.20\,\text{kg} ball moving at +15\,\text{m/s} is caught and brought to rest in 0.030\,\text{s}. Find the average force on the ball.

Step 1: Momentum change.

\Delta p = m(v_f - v_i) = (0.20)(0 - 15) = -3.0\,\text{kg}\cdot\text{m/s}

Step 2: Average force.

F_{avg} = \frac{\Delta p}{\Delta t} = \frac{-3.0}{0.030} = -100\,\text{N}

The negative sign means the force is opposite the ball’s initial direction, as expected for a stopping force.

Exam Focus

• Typical question patterns:
  • Use force-time graphs to compute impulse (area) and then find \Delta p or v_f.
  • Compute average force given stopping time and change in velocity.
  • Use integral form \vec{J} = \int \vec{F}\,dt when force varies with time.
• Common mistakes:
  • Using speed instead of velocity in \Delta\vec{p} = m(\vec{v}_f - \vec{v}_i) (sign errors).
  • Confusing impulse with force: a big force for a tiny time can give the same impulse as a smaller force for longer.
  • Forgetting that the impulse theorem requires net force (or being clear about which force you mean if multiple forces act).

Conservation of Linear Momentum

The big idea: when total momentum stays constant

Conservation of linear momentum says that the total momentum of a system remains constant if the net external impulse on the system is zero.

A precise way to state it:

• If \vec{J}_{ext} = \int \vec{F}_{ext}\,dt = \vec{0} over an interaction, then

\Delta\vec{P} = \vec{0}

so

\vec{P}_i = \vec{P}_f

This is not “always true”—it has a condition: the system must be sufficiently isolated so that external forces either are zero or produce negligible impulse during the interaction time.

Why internal forces don’t change total momentum

In collisions and explosions, forces between objects in the system are internal. By Newton’s 3rd law, internal forces come in equal-and-opposite pairs. Over the same time interval, these equal-and-opposite forces produce equal-and-opposite impulses on the two objects, so they cancel in the total.

That’s why two ice skaters pushing off each other can dramatically change their individual momenta, but the total momentum of the pair stays fixed (assuming negligible external horizontal forces).

The system choice is everything

Momentum conservation is only as good as your system definition.

• If you choose “two colliding carts” as the system on a low-friction track, external horizontal impulse is small, so momentum is conserved horizontally.
• If you choose “just one cart,” the other cart’s force is external to your system, so that one-cart momentum is not conserved.

A typical AP skill is stating (explicitly) why momentum is conserved: “Net external impulse is negligible in the horizontal direction” is the kind of justification graders like.

Component form (2D conservation)

Because momentum is a vector, conservation applies independently in each direction where external impulse is negligible.

In two dimensions:

P_{x,i} = P_{x,f}

P_{y,i} = P_{y,f}

This is why 2D collision problems usually split into x and y component equations.

Collisions: elastic vs inelastic (what’s conserved?)

Momentum conservation and kinetic energy conservation are different ideas.

• Momentum is conserved (for an isolated system) in all collisions.
• Kinetic energy is conserved only in elastic collisions.

A quick comparison:

A frequent misconception is thinking “inelastic means momentum not conserved.” Inelastic means kinetic energy is not conserved; momentum still is (if external impulse is negligible).

Strategy for collision/explosion problems

When you see “collision,” “sticks together,” “explodes,” “recoil,” or “push off,” momentum conservation is usually the starting point.

1. Choose the system so internal forces dominate (often both objects together).
2. Decide the direction(s) where external impulse is negligible (often horizontal only).
3. Write momentum conservation in those directions.
4. If needed, add other relationships (like “they stick” implies same final velocity, or geometry for components).

Worked example 1: Perfectly inelastic collision (stick together)

A 0.050\,\text{kg} bullet moving at 400\,\text{m/s} embeds in a 2.0\,\text{kg} block initially at rest on a frictionless surface. Find the speed of the bullet-block system immediately after impact.

Step 1: Choose system and apply momentum conservation. Take bullet + block as the system. External horizontal impulse is negligible during the short collision.

Initial momentum:

P_i = (0.050)(400) + (2.0)(0) = 20\,\text{kg}\cdot\text{m/s}

Final: they stick, so they share a common speed v_f with total mass 2.05\,\text{kg}.

P_f = (2.05)v_f

Set P_i = P_f:

20 = (2.05)v_f

v_f = \frac{20}{2.05} = 9.76\,\text{m/s}

Important note: Kinetic energy is not conserved here; a lot becomes internal energy (heat, deformation). If you incorrectly set kinetic energy equal before and after, you’ll get the wrong answer.

Worked example 2: Explosion / push-off from rest

Two skaters start at rest on frictionless ice. Skater A has mass 60\,\text{kg} and skater B has mass 40\,\text{kg}. They push off; afterward A moves at -2.0\,\text{m/s}. Find B’s velocity.

Step 1: Total initial momentum. They start from rest, so

P_i = 0

Step 2: Conserve momentum.

P_f = m_A v_A + m_B v_B = 0

Plug in:

60(-2.0) + 40v_B = 0

-120 + 40v_B = 0

v_B = 3.0\,\text{m/s}

B moves opposite A and faster because B has smaller mass.

Worked example 3: 2D collision (components)

A puck of mass 0.20\,\text{kg} slides east at 5.0\,\text{m/s} and collides with a stationary puck of mass 0.30\,\text{kg}. After the collision, the 0.20\,\text{kg} puck moves at 3.0\,\text{m/s} at 30^\circ north of east. Find the velocity components of the 0.30\,\text{kg} puck.

Step 1: Initial momentum components. Choose east as +x, north as +y.

P_{x,i} = (0.20)(5.0) = 1.0\,\text{kg}\cdot\text{m/s}

P_{y,i} = 0

Step 2: Final momentum of puck 1 (given speed and angle).

v_{1x,f} = (3.0)\cos(30^\circ) = 2.60\,\text{m/s}

v_{1y,f} = (3.0)\sin(30^\circ) = 1.50\,\text{m/s}

So its final momentum components:

p_{1x,f} = (0.20)(2.60) = 0.520\,\text{kg}\cdot\text{m/s}

p_{1y,f} = (0.20)(1.50) = 0.300\,\text{kg}\cdot\text{m/s}

Step 3: Conserve momentum in x and y to get puck 2 momentum.

p_{2x,f} = P_{x,i} - p_{1x,f} = 1.0 - 0.520 = 0.480\,\text{kg}\cdot\text{m/s}

p_{2y,f} = P_{y,i} - p_{1y,f} = 0 - 0.300 = -0.300\,\text{kg}\cdot\text{m/s}

Step 4: Convert puck 2 momentum to velocity.

v_{2x,f} = \frac{p_{2x,f}}{m_2} = \frac{0.480}{0.30} = 1.60\,\text{m/s}

v_{2y,f} = \frac{p_{2y,f}}{m_2} = \frac{-0.300}{0.30} = -1.00\,\text{m/s}

So puck 2 moves east and south.

When momentum conservation is only approximate

In real situations, external forces like friction and gravity exist. Momentum can still be conserved approximately if:

• The collision time is very short (so external impulses are small).
• You consider only components where external forces don’t act significantly (for example, horizontal momentum in a tabletop collision, since gravity is vertical).

A typical AP reasoning step is comparing external impulse to internal collision impulse: if the collision force is huge compared to friction during the brief impact, then friction’s impulse is negligible.

Exam Focus

• Typical question patterns:
  • “Two objects collide/explode; find final velocity(ies)” using \vec{P}_i = \vec{P}_f (often 1D carts, bullet-block, recoil).
  • 2D collisions requiring separate conservation equations in x and y.
  • Conceptual prompts asking you to justify why momentum is conserved (system choice, negligible external impulse).
• Common mistakes:
  • Assuming kinetic energy is conserved in all collisions (it is not).
  • Writing conservation of momentum for a non-isolated system (for example, including only one object when the other exerts an external force on it).
  • In 2D, conserving momentum magnitude instead of conserving x and y components separately.