AP Physics 1 Unit 4 Notes: Understanding Momentum and Impulse

Linear Momentum

What momentum is (and what it isn’t)

Linear momentum is a measure of “motion quantity” for an object moving in a straight line. It combines two ideas you already understand from dynamics:

• how much stuff is moving (mass)
• how fast and in what direction it’s moving (velocity)

The key point is that momentum is a vector—it has direction. In one dimension, you often represent direction with a positive or negative sign; in two dimensions, you track components (like $p_x$ and $p_y$).

Momentum is defined as

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

Here:

• $\vec{p}$ is the momentum vector
• $m$ is mass (a scalar)
• $\vec{v}$ is velocity (a vector)

Because velocity is a vector, momentum points in the same direction as velocity.

A common confusion is to treat momentum like kinetic energy. They are related to motion, but they behave very differently:

• Kinetic energy is scalar and depends on $v^2$.
• Momentum is vector and depends on $v$.

That difference matters a lot in collisions and recoil: opposite directions can cancel momentum (vector addition), but energies never cancel that way.

Why momentum matters

Momentum is the central quantity in situations where objects interact over a short time—especially collisions, explosions, recoil, and push-offs. In those events, the forces can be huge and complicated, but the total momentum often changes in a clean, predictable way when external influences are limited.

Even before you formally use conservation of momentum, thinking in momentum helps you describe “how hard it is to stop” something that’s moving. For example:

• A truck and a car moving at the same speed do not have the same momentum. The truck’s larger mass means much larger momentum, so it typically requires a larger impulse (a longer time, a bigger force, or both) to stop.

Units and interpreting magnitude

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

$\mathrm{kg\cdot m/s}$

Bigger momentum can come from bigger mass, bigger speed, or both.

In one dimension, you’ll often use sign to represent direction. For instance, if you choose “to the right” as positive, then an object moving left has negative velocity and therefore negative momentum.

Momentum in systems (internal vs external influence)

Momentum becomes especially powerful when you consider a system—a chosen collection of objects you analyze together.

• Internal forces are forces objects in the system exert on each other.
• External forces are forces exerted on the system by objects outside the system.

Why this distinction matters: internal forces can change individual momenta dramatically, but they tend to cancel in pairs (Newton’s third law) when you look at the total momentum of the system. External forces are what can change the system’s total momentum.

Even in this “Momentum and Impulse” section (before full collision conservation problems), you should already be thinking: “What is my system, and what external forces act during the interaction time?” That is the gateway to deciding whether momentum will be approximately conserved later on.

Worked example 1: computing momentum (including direction)

A $0.150\ \mathrm{kg}$ baseball moves horizontally at $40\ \mathrm{m/s}$ to the right.

Momentum magnitude and direction:

$p = mv$

$p = (0.150)(40) = 6.0\ \mathrm{kg\cdot m/s}$

So $\vec{p}$ points to the right. If “right” is positive, you might write

$p = +6.0\ \mathrm{kg\cdot m/s}$

If instead the ball were moving left at the same speed, you would write

$p = -6.0\ \mathrm{kg\cdot m/s}$

The physics is the same; the sign just tracks direction.

Worked example 2: same speed does not mean same momentum

A $1200\ \mathrm{kg}$ car and a $12\ \mathrm{kg}$ bicycle both move at $8\ \mathrm{m/s}$.

Car:

$p = (1200)(8) = 9600\ \mathrm{kg\cdot m/s}$

Bicycle:

$p = (12)(8) = 96\ \mathrm{kg\cdot m/s}$

The car has $100$ times the momentum because it has $100$ times the mass. That’s why stopping distances and required braking impulses differ so much.

What typically goes wrong with momentum

Students often make these conceptual slips:

• Forgetting momentum is a vector. If an object reverses direction, its momentum changes more than you might expect because the sign flips.
• Mixing up momentum and force. Momentum describes motion; force describes interaction that changes motion.
• Assuming mass cancels or is irrelevant. In many momentum situations (recoil, collisions), mass strongly shapes the outcome.

Exam Focus

• Typical question patterns:
  • Compute momentum from mass and velocity, including sign/direction.
  • Compare momenta for different objects or different velocities.
  • Identify momentum components (conceptually) from a velocity direction.
• Common mistakes:
  • Using speed instead of velocity (dropping direction).
  • Treating momentum like energy (forgetting momentum can be negative in 1D).
  • Unit confusion (writing $\mathrm{N}$ or $\mathrm{J}$ instead of $\mathrm{kg\cdot m/s}$).

Change in Momentum and Impulse

The big idea: forces change momentum

If momentum is the “amount of motion,” then a force is what changes it. In AP Physics 1, the central connection is the impulse-momentum theorem:

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

Here:

• $\vec{J}$ is impulse
• $\Delta \vec{p}$ is the change in momentum

A very common student mistake is to think impulse is “a big force.” Impulse is not a force—it is a measure of a force acting over a time interval.

Defining change in momentum

Change in momentum is final minus initial momentum:

$\Delta \vec{p} = \vec{p}_f - \vec{p}_i$

In one dimension:

$\Delta p = p_f - p_i$

Because momentum is a vector, a reversal in direction can make $\Delta \vec{p}$ large even if speed stays the same.

Example of why direction matters: if a ball comes in at $+10\ \mathrm{m/s}$ and bounces back at $-10\ \mathrm{m/s}$, then the velocity changed by $-20\ \mathrm{m/s}$, not $0$.

Defining impulse in an AP Physics 1 way

Impulse is defined as the effect of a force over a time interval. For a constant (or average) net force acting over a time interval $\Delta t$:

$\vec{J} = \vec{F}_{\mathrm{net,avg}}\Delta t$

Where:

• $\vec{F}_{\mathrm{net,avg}}$ is the average net force during the interval
• $\Delta t$ is the interaction time

Units:

$\mathrm{N\cdot s}$

Using base SI units, you can show the unit equivalence:

$\mathrm{N\cdot s} = \mathrm{kg\cdot m/s}$

That equivalence is not a coincidence—it’s exactly what $\vec{J} = \Delta \vec{p}$ says.

Connecting impulse to Newton’s second law (conceptual bridge)

Newton’s second law is often introduced as $\vec{F}_{\mathrm{net}} = m\vec{a}$ for constant mass. Momentum form emphasizes the “rate of change of momentum” idea:

$\vec{F}_{\mathrm{net}} = \frac{\Delta \vec{p}}{\Delta t}$

Rearranging gives

$\Delta \vec{p} = \vec{F}_{\mathrm{net,avg}}\Delta t$

This is the impulse-momentum theorem again.

What this means physically:

• A given change in momentum can happen with a large force for a short time or a smaller force for a longer time.
• Safety devices exploit this by increasing collision time to reduce average force.

Force-time graphs: impulse as area

On AP Physics 1 exams, impulse is often extracted from a force vs time graph. The key interpretation:

• The impulse equals the area under the net force vs time curve over the time interval.

For a constant force, the area is a rectangle.
For a linearly changing force, it’s often a triangle or trapezoid.

If the force changes with time, you can still compute impulse by area (counting geometric areas). The sign of the area matters: force opposite your positive direction produces negative impulse.

Why “increasing collision time reduces force” is true

Suppose you must change an object’s momentum by a fixed amount $\Delta p$ (for instance, bring it to rest). From

$\Delta p = F_{\mathrm{avg}}\Delta t$

you get

$F_{\mathrm{avg}} = \frac{\Delta p}{\Delta t}$

So for the same $\Delta p$:

• Larger $\Delta t$ means smaller $F_{\mathrm{avg}}$.

This is the physics behind:

• airbags and crumple zones (increase stopping time)
• catching a ball by “giving” with your hands (increase stopping time)
• bending your knees when landing (increase stopping time)

A subtle but important point: the theorem uses net force. If multiple forces act (like gravity and a contact force), the net force determines momentum change.

Notation reference (common symbols you’ll see)

Worked example 1: impulse from stopping time (safety reasoning)

A $0.060\ \mathrm{kg}$ tennis ball traveling at $25\ \mathrm{m/s}$ to the right is brought to rest.

1) Find the change in momentum.

Initial momentum:

$p_i = mv_i = (0.060)(25) = 1.5\ \mathrm{kg\cdot m/s}$

Final momentum:

$p_f = 0$

Change in momentum:

$\Delta p = p_f - p_i = 0 - 1.5 = -1.5\ \mathrm{kg\cdot m/s}$

The negative sign means the momentum changed in the leftward direction (opposite the initial motion).

2) Compare average force if the stopping time is $0.020\ \mathrm{s}$ vs $0.10\ \mathrm{s}$.

Use

$F_{\mathrm{avg}} = \frac{\Delta p}{\Delta t}$

Case A: $\Delta t = 0.020\ \mathrm{s}$

$F_{\mathrm{avg}} = \frac{-1.5}{0.020} = -75\ \mathrm{N}$

Case B: $\Delta t = 0.10\ \mathrm{s}$

$F_{\mathrm{avg}} = \frac{-1.5}{0.10} = -15\ \mathrm{N}$

The longer stopping time reduces the average net force by a factor of $5$. This is the same momentum change in both cases—the difference is how quickly you make it happen.

Worked example 2: impulse from a force-time graph (area)

A cart experiences a net force that increases linearly from $0\ \mathrm{N}$ at $t = 0\ \mathrm{s}$ to $40\ \mathrm{N}$ at $t = 0.50\ \mathrm{s}$, then returns linearly to $0\ \mathrm{N}$ at $t = 1.0\ \mathrm{s}$ (a symmetric triangle). Find the impulse.

Impulse equals the area under the force-time graph. The shape is a triangle with base $1.0\ \mathrm{s}$ and height $40\ \mathrm{N}$.

Triangle area:

$J = \frac{1}{2}(\mathrm{base})(\mathrm{height})$

$J = \frac{1}{2}(1.0)(40) = 20\ \mathrm{N\cdot s}$

Direction: if the force is defined positive over that interval, then the impulse is positive.

If the cart’s initial momentum were $5\ \mathrm{kg\cdot m/s}$ in the positive direction, its final momentum would be

$p_f = p_i + J = 5 + 20 = 25\ \mathrm{kg\cdot m/s}$

Worked example 3: bounce vs stop (why bouncing hurts more)

A $0.20\ \mathrm{kg}$ rubber ball hits a wall moving at $+12\ \mathrm{m/s}$ and rebounds at $-12\ \mathrm{m/s}$. Compare the magnitude of $\Delta p$ to the case where it simply stops (final velocity $0$).

First compute initial and final momenta for the bounce:

$p_i = (0.20)(+12) = +2.4\ \mathrm{kg\cdot m/s}$

$p_f = (0.20)(-12) = -2.4\ \mathrm{kg\cdot m/s}$

$\Delta p = p_f - p_i = -2.4 - 2.4 = -4.8\ \mathrm{kg\cdot m/s}$

Magnitude is $4.8\ \mathrm{kg\cdot m/s}$.

If it stops instead:

$p_f = 0$

$\Delta p = 0 - 2.4 = -2.4\ \mathrm{kg\cdot m/s}$

Magnitude is $2.4\ \mathrm{kg\cdot m/s}$.

So bouncing produces twice the momentum change magnitude compared to stopping from the same incoming speed. If the collision time is similar, bouncing implies a larger average force magnitude. This is why “softening” a collision (reducing rebound or increasing contact time) can reduce forces.

Common conceptual pitfalls (and how to fix them)

1) Confusing impulse with impact force.
Impulse is not just “force.” Two collisions can have the same impulse with very different peak forces if their contact times differ.

2) Using only speeds when direction matters.
If an object reverses direction, you must use signed velocities (or vectors). The change in momentum depends on direction.

3) Forgetting that it’s net force that matters.
During a hit, gravity is still acting. Often the contact force is so large that gravity’s impulse during the short collision is negligible, but you should justify that idea conceptually rather than assuming it blindly.

4) Mixing up which quantity is “area.”
Area under an $F$ vs $t$ graph is impulse (units $\mathrm{N\cdot s}$). Area under a $v$ vs $t$ graph is displacement (different topic). Area under an $a$ vs $t$ graph is change in velocity.

Real-world applications that AP questions love

• Airbags and crumple zones: increase $\Delta t$ for the same $\Delta p$, reducing average force.
• Catching a ball: pulling your hands back increases stopping time.
• Sports collisions: “follow-through” changes how long a force acts and therefore the impulse delivered.
• Hammering a nail: a large momentum change over a short time implies a large average force.

Exam Focus

• Typical question patterns:
  • Use $\vec{J} = \Delta \vec{p}$ to find impulse, final velocity, or average force.
  • Interpret an $F$ vs $t$ graph: compute impulse as area (rectangles, triangles, trapezoids) and connect it to momentum change.
  • Compare scenarios (stop vs bounce, short vs long collision time) qualitatively and quantitatively.
• Common mistakes:
  • Using $\Delta p = p_i - p_f$ instead of $p_f - p_i$ (sign errors).
  • Calculating area under a graph incorrectly (forgetting triangle factor $\frac{1}{2}$ or wrong base/height units).
  • Treating impulse as always positive (ignoring direction of force and momentum change).