Impulse and Momentum

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Systems of Particles and Linear Momentum· 14 min read

1. Core Definitions of Momentum and Impulse★★☆☆☆⏱ 3 min

Impulse and linear momentum are core concepts for particle and system motion, and this subtopic makes up roughly one-third of AP Physics C: Mechanics Unit 4, which counts for 14-18% of your total exam score. It appears regularly in both multiple-choice and free-response sections.

📘 Definition

Linear Momentum

$p$

A vector quantity describing the inertia of motion of a moving point mass, pointing in the same direction as the object's velocity.

Example:

A 2 kg mass moving at 3 m/s right has $p = + 6 kg \cdotp m/s$ .

Impulse (denoted $J$ ) is the change in momentum caused by a net force acting over a finite time interval. It captures the cumulative effect of a force over time to change an object's motion. Unlike work (which relates force over displacement to kinetic energy change), impulse relates force over time to momentum change, making it ideal for analyzing short-duration interactions like collisions, where forces vary rapidly and are hard to model with constant acceleration kinematics. Units for both momentum and impulse are $kg \cdotp m/s$ , which is equivalent to $N \cdotp s$ .

2. The Impulse-Momentum Theorem for Single Particles★★☆☆☆⏱ 4 min

🔬 Derivation

Goal:

Derive the impulse-momentum theorem from Newton's second law

Starting from:

Newton's second law in momentum form: $F_{n e t} = \frac{d p}{d t}$

1
Rearrange to separate variables:
2
$d\vec{p} = \vec{F}_{net} dt$
3
Integrate both sides over the interval from initial time $t_{i}$ to final time $t_{f}$ :
4
$\int_{p_i}^{p_f} d\vec{p} = \int_{t_i}^{t_f} \vec{F}_{net} dt$
5
The left-hand side simplifies to total change in momentum, and the right-hand side is defined as total impulse.

Result:

The core impulse-momentum theorem is: $J = Δ p = p_{f} - p_{i}$

For a variable force, impulse equals the signed area under a $F (t)$ vs $t$ graph between the start and end times of the interaction. Because impulse and momentum are vectors, direction always matters: forces opposite to motion decrease momentum, while forces in the direction of motion increase momentum.

📐 Worked Example

A 2.0 kg object moving along the x-axis has an initial velocity $v_{i} = + 3.0 m/s$ . A net force acts on the object, varying with time to form three segments: a linear increase from 0 N to 4 N between $t = 0$ and $t = 2$ s, constant 4 N between $t = 2$ s and $t = 3$ s, and a linear decrease from 4 N back to 0 N between $t = 3$ s and $t = 5$ s. Find the final velocity of the object.

1
1. Total impulse equals the signed area under the $F (t)$ graph; all force values are positive here, so sum the area of each segment.
2
1. Calculate area for each segment: Area 1 (0–2 s, triangle):
3
$A_1 = \frac{1}{2} \times 2 \times 4 = 4 \text{ N·s}$
4
Area 2 (2–3 s, rectangle):
5
$A_2 = 1 \times 4 = 4 \text{ N·s}$
6
Area 3 (3–5 s, triangle):
7
$A_3 = \frac{1}{2} \times 2 \times 4 = 4 \text{ N·s}$
8
1. Total impulse is the sum of areas: $J = 4 + 4 + 4 = 12 N \cdotp s$ . By the impulse-momentum theorem, $J = Δ p = m (v_{f} - v_{i})$ .
9
1. Solve for final velocity:
10
$v_f = v_i + \frac{J}{m} = 3.0 + \frac{12}{2.0} = 9.0 \text{ m/s}$

Exam tip:

Always mark the sign of force on your F(t) graph before calculating area; any segment below the time axis contributes negative impulse, which subtracts from total momentum change.

3. Average Force and Impulse★★☆☆☆⏱ 3 min

For short-duration interactions like collisions, we rarely need the full time dependence of the force. Instead, we use average force: the constant force that produces the same total impulse (and thus same momentum change) as the actual variable force over the same time interval $Δ t$ .

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

Rearranged, this gives the common form used for collision problems: $F_{a v g} = \frac{Δ p}{Δ t}$ . This aligns with Newton's third law: for two colliding objects, the average force on object 1 from object 2 is equal and opposite to the force on object 2 from object 1, so their impulses are also equal and opposite.

📐 Worked Example

A 0.15 kg baseball travels horizontally toward a bat at 35 m/s. After being hit, it travels at 45 m/s away from the bat along the same line. If the collision lasts 0.002 s, what is the magnitude of the average force exerted on the ball by the bat?

1
1. Set a coordinate system where positive x is away from the bat. Initial velocity is $v_{i} = - 35 m/s$ , final velocity is $v_{f} = + 45 m/s$ .
2
1. Calculate initial and final momentum:
3
$p_i = m v_i = (0.15)(-35) = -5.25 \text{ kg·m/s}, \quad p_f = m v_f = (0.15)(45) = 6.75 \text{ kg·m/s}$
4
1. Find momentum change:
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$\Delta p = p_f - p_i = 6.75 - (-5.25) = 12 \text{ kg·m/s}$
6
1. Solve for average force:
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$F_{avg} = \frac{\Delta p}{\Delta t} = \frac{12}{0.002} = 6000 \text{ N}$

Exam tip:

When an object reverses direction, do not drop the negative sign on initial velocity. Failing to do this cuts your calculated momentum change (and thus average force) in half, which is a very common exam error.

4. Impulse for Calculus-Based Variable Force Problems★★★☆☆⏱ 4 min

AP Physics C: Mechanics regularly tests your ability to calculate impulse for a force given as an explicit function of time, using the definition of impulse as a definite integral. For 1D motion, the formula is:

J = \int_{t_1}^{t_2} F_{net}(t) dt = \Delta p

For 2D motion, the impulse-momentum theorem holds independently for each vector component: you calculate $J_{x} = Δ p_{x}$ and $J_{y} = Δ p_{y}$ separately, since x and y components of motion do not mix. This is a core calculus application that appears on almost every AP exam.

📐 Worked Example

A variable force $F (t) = 10 t^{2} - 5 t$ (in N, t in s) acts on a 4.0 kg stationary object along the x-axis from $t = 0$ s to $t = 2.0$ s. No other forces act on the object. Find the speed of the object at $t = 2.0$ s.

1
1. The object starts from rest, so initial momentum $p_{i} = 0$ , meaning $J = Δ p = p_{f} = m v_{f}$ , so $v_{f} = \frac{J}{m}$ .
2
1. Set up the definite integral for impulse:
3
$J = \int_0^2 (10t^2 - 5t) dt$
4
1. Evaluate the integral. The antiderivative is $\frac{10}{3} t^{3} - \frac{5}{2} t^{2}$ :
5
$J = \left(\frac{10}{3}(2)^3 - \frac{5}{2}(2)^2\right) - 0 = \frac{80}{3} - 10 = \frac{50}{3} \approx 16.67 \text{ N·s}$
6
1. Solve for final speed:
7
$v_f = \frac{50/3}{4.0} \approx 4.17 \text{ m/s}$

✓ Quick check

Test your understanding of impulse from F(t) graphs:

A 0.50 kg cart moves right at 2.0 m/s. A net force acts on it: 0-2 s forms a positive triangle of total area +1 N·s, 2-4 s forms a negative triangle of total area -1 N·s. What is the speed at t=4 s?
- 0 m/s
- 2.0 m/s
- 3.0 m/s
- 4.0 m/s
Reveal answer
1 —
Total impulse is +1 - 1 = 0, so no change in momentum, so speed remains 2.0 m/s. The common mistake is adding absolute areas to get 4 m/s.

Exam tip:

If the force function changes form at different time boundaries, split your integral into separate intervals matching each function, then sum the results to get total impulse.

5. Common Pitfalls

Wrong move:

Calculating momentum change for a bouncing object as $Δ p = m (v_{f} - v_{i})$ where $v_{i}$ and $v_{f}$ are both positive magnitudes.

Why:

Momentum is a vector, so reversing direction changes the sign of velocity, leading to a momentum change half the correct size.

Correct move:

Always set a coordinate system before starting, assign positive/negative signs to velocity based on direction, then calculate $Δ p = m v_{f} - m v_{i}$ with signed velocities.

Wrong move:

Treating impulse as force times distance instead of force times time.

Why:

Students confuse impulse with work, which is force times displacement.

Correct move:

Label each quantity by its units on your paper: impulse has units $N \cdotp s$ , work has units $N \cdotp m$ , so mismatched units will immediately flag a mistake.

Wrong move:

Counting negative area (force below the time axis) as positive impulse when calculating area under F(t).

Why:

Students calculate total absolute area regardless of force direction, especially for graphs that cross the time axis.

Correct move:

Split the area into positive regions (above the axis) and negative regions (below the axis), add positive areas and subtract negative areas to get total impulse.

Wrong move:

When asked for the impulse exerted on object A by object B, use the change in momentum of object B instead of object A.

Why:

Students mix up which object the impulse acts on when applying Newton's third law.

Correct move:

Explicitly write that impulse on X equals the change in momentum of X, then match your calculation to the correct object.

Wrong move:

Calculate impulse using only the applied force, ignoring other forces like gravity or friction.

Why:

Problems often only mention the applied force, leading students to forget the impulse-momentum theorem uses net force.

Correct move:

Always check for other forces acting on the object, and add them to get net force before calculating impulse.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Linear Momentum	$p = m v$	Vector, same direction as velocity; units: $kg \cdotp m/s = N \cdotp s$
Impulse (variable force)	$J = \int_{t_{i}}^{t_{f}} F_{n e t} (t) d t$	Equal to the signed area under the $F (t)$ vs $t$ graph
Impulse (constant force)	$J = F_{n e t} Δ t$	Only applies when net force is constant over the interval
Average Force	$F_{a v g} = \frac{Δ p}{Δ t}$	Constant force that gives the same impulse as a variable force over $Δ t$
Impulse-Momentum Theorem	$J = Δ p = p_{f} - p_{i}$	Holds for each vector component independently for single particles
Impulse (average force)	$J = F_{a v g} Δ t$	Works for any force, variable or constant
2D Momentum Components	$J_{x} = Δ p_{x}, J_{y} = Δ p_{y}$	X and y components are calculated separately, no mixing

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Impulse calculation from F(t) graph
2022 · FRQ
Variable force impulse problem

What's Next

Mastering impulse and momentum for single particles is an absolute prerequisite for the rest of Unit 4, starting with conservation of linear momentum for multi-particle systems. This topic gives you the foundation to prove that for an isolated system (no net external force), total momentum is conserved, the core tool for solving all collision and interaction problems. Without being able to calculate impulse and momentum change correctly, you will not be able to analyze elastic and inelastic collisions, center of mass motion, or variable-mass systems like rockets, all of which are heavily tested on the AP exam. Impulse and momentum also connect to energy conservation, as you will combine momentum and energy analysis to solve collision problems that cannot be solved with force or energy alone.