---
title: "Unit 4.1 | Momentum and Impulse"
description: "Learn momentum and impulse for AP Physics 1: p = mv, J = FΔt = Δp, bouncing vs. stopping, and impulse from force vs. time graphs, with practice."
featured_image: "https://nerd-notes.com/wp-content/uploads/2023/06/kicking-a-soccer-ball-momentum-impulse.jpg"
url: "https://nerd-notes.com/unit-4-1-momentum-and-impulse/"
date_modified: "2026-10-31T09:33:49+00:00"
---

# Unit 4.1 | Momentum and Impulse

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Momentum is the second big conserved quantity in mechanics. This lesson shows what it is, how impulse changes it, and why a soft landing hurts less.

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### Unit 4 Breakdown

You are on Lesson 1 of 4

1. [Unit 4.1 | Momentum and Impulse](https://nerd-notes.com/unit-4-1-momentum-and-impulse/)
2. [Unit 4.2 | Conservation of Momentum](https://nerd-notes.com/unit-4-2-conservation-of-momentum/)
3. [Unit 4.3 | Collisions and Explosions](https://nerd-notes.com/unit-4-3-collisions-and-explosions/)
4. [Unit 4.4 | Momentum and Energy Together](https://nerd-notes.com/unit-4-4-momentum-and-energy-together/)

#### In this lesson: 

- You will learn what momentum is and why its sign matters
- How impulse changes momentum: [katex] J = F\Delta t = \Delta p [/katex]
- The 4 impulse problem types on the AP exam: finding the force or time, bounces, soft landings, and force vs. time graphs

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### Introduction

This lesson covers **momentum and impulse**: what momentum is, what changes it, and the four kinds of impulse problems you will see on the AP exam.

In [Unit 3](https://nerd-notes.com/unit-3-3-conservation-of-energy-systems-and-problem-solving/) you saw that energy is conserved. Momentum is the second big conserved quantity in mechanics, and it is the tool you need for anything that collides, explodes, or pushes off something else.

Energy asks *how much* motion an object can cause. Momentum asks *how hard it is to stop*, and in which direction it is going.

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### What Is Momentum?

**Momentum** is mass in motion:

[katex] \vec{p} = m\vec{v} [/katex]

Here [katex] m [/katex] is mass in kilograms and [katex] \vec{v} [/katex] is velocity in m/s, so momentum is measured in [katex] kg \cdot m/s [/katex].

\(10\,\text{m/s}\)

\(p = 1.5\,\text{kg·m/s}\)

\(10\,\text{m/s}\)

\(p = 30{,}000\,\text{kg·m/s}\)

*Same speed, but the truck has 20,000 times the momentum. That is why it is so much harder to stop.*

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#### Momentum is a vector

Unlike energy, momentum has a direction: the direction of the velocity. Pick a positive direction and stick with it. A ball moving right at [katex] 10\,\text{m/s} [/katex] has [katex] +mv [/katex]; the same ball moving left has [katex] -mv [/katex].

Kinetic energy can never be negative. Momentum can, because the sign is its direction. Forgetting the sign is the most common momentum mistake.

### Impulse: How Momentum Changes

Recall Newton’s 2nd law, [katex] F_{net} = ma [/katex]. Write acceleration as a change in velocity over time and rearrange:

1. Start with Newton’s 2nd law: [katex] F_{net} = m\frac{\Delta v}{\Delta t} [/katex].
2. Multiply both sides by [katex] \Delta t [/katex]: [katex] F_{net}\Delta t = m\Delta v [/katex].
3. Since [katex] m\Delta v = \Delta p [/katex], the left side is the change in momentum.

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[katex] J = F\Delta t = \Delta p [/katex]

The quantity [katex] F\Delta t [/katex] is called **impulse** ([katex] J [/katex]). A force acting over a time interval changes an object’s momentum. This is the momentum version of the work-energy theorem from Unit 3.2.

Every impulse question on the AP exam is one of the four types below. Spot the type, set it up, and the math is short.

| **Problem type** | **The clue** | **The setup** |
| --- | --- | --- |
| 1. Find the force, time or speed | “average force”, “contact time” | [katex] F\Delta t = m(v_f – v_i) [/katex] |
| 2. Bounces vs. stops | “rebounds”, “bounces back” | Opposite signs for [katex] v_i [/katex] and [katex] v_f [/katex] |
| 3. Soft landings | airbags, padding, bending knees | Same [katex] \Delta p [/katex], so [katex] F = \frac{\Delta p}{\Delta t} [/katex] |
| 4. Force vs. time graphs | a graph of [katex] F [/katex] against [katex] t [/katex] | Area = impulse = [katex] \Delta p [/katex] |

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Momentum and \(\Delta\) Momentum

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### Problem Type 1: Find the force, the time, or the speed

**How to spot it:** The question gives (or asks for) a force that acts for a time: “average force”, “contact time”, “how fast does it leave”.

**Setup:** Pick a positive direction, then write [katex] F\Delta t = m(v_f – v_i) [/katex] and solve for the one unknown.

A golf club pushes a [katex] 0.045\,\text{kg} [/katex] ball with an average force of [katex] 1000\,\text{N} [/katex] for [katex] 0.0018\,\text{s} [/katex]. How fast does the ball leave the tee?

[katex] J = F\Delta t = (1000)(0.0018) = 1.8 \, N \cdot s [/katex]. The ball started at rest, so [katex] 1.8 = (0.045)v [/katex] and [katex] v = 40 \, m/s [/katex].

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**AP trap:** Times are often in milliseconds. Convert first: [katex] 1.8\,\text{ms} [/katex] is [katex] 0.0018\,\text{s} [/katex].

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Question 1

Engineers are testing two different crash-safety systems by observing cars brought to rest from an initial speed. In Test 1, a car of mass M…

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### Problem Type 2: Bounces vs. stops

**How to spot it:** “Rebounds”, “bounces back”, “reverses direction”.

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**Setup:** Give the before and after velocities opposite signs, then [katex] \Delta p = m(v_f – v_i) [/katex].

\(p_i = \textcolor{#0072FF}{+}mv\)

\(p_f = \textcolor{#0072FF}{-}mv\)

\(\Delta p = p_f - p_i = (\textcolor{#0072FF}{-}mv) - (\textcolor{#0072FF}{+}mv) = \textcolor{#0072FF}{-2mv}\)

\(\text{twice a stop: } 0 - mv = -mv\)

*Stopping: \(\Delta p = 0 – mv = -mv\). Bouncing straight back: \(\Delta p = (-mv) – (+mv) = -2mv\), twice the impulse.*

A [katex] 0.5\,\text{kg} [/katex] ball hits a wall at [katex] 6\,\text{m/s} [/katex] and bounces straight back at [katex] 4\,\text{m/s} [/katex]. What impulse did the wall give the ball?

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Right is positive: [katex] \Delta p = (0.5)(-4 – 6) = -5 \, N \cdot s [/katex]. The impulse is [katex] 5 \, N \cdot s [/katex], pointing away from the wall.

**AP trap:** Subtracting the speeds (6 − 4) instead of the velocities. On a bounce the speeds *add*, so a bounce always takes a bigger impulse than a stop.

Question 4 needs a little Unit 3 energy first, to find the speeds just before and after the bounce.

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Question 2

A 0.10 kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the magnitude of the change in the momentum of the…

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Question 3

A 0.17 kg baseball is thrown with a speed of 38 m/s and it is hit straight back toward the pitcher with a speed of 62 m/s. What is the magnitude of…

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Question 4

A ball of mass m is moving horizontally with speed v toward a vertical racket face. The ball strikes the racket and remains in contact for a time…

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Question 5

A small rubber ball of mass 0.25 kg is dropped from rest from a height of 5.0 m above a hard floor. The ball collides with the floor, remains in…

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### Problem Type 3: Soft landings: same change, different time

**How to spot it:** Airbags, helmets, padding, bending your knees, or two trials where the same object stops over different times.

**Setup:** The change in momentum is the same either way, so [katex] F = \frac{\Delta p}{\Delta t} [/katex]: double the stopping time and the force halves.

Same impulse, very different force

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*Both areas are \(30\,\text{N·s}\), the same change in momentum. The airbag stretches the stop over \(100\,\text{ms}\) instead of \(20\,\text{ms}\), so the peak force is 5 times smaller.*

This is the idea behind airbags, helmets, bending your knees when you land, and catching an egg by pulling your hands back.

**AP trap:** Saying the cushion “reduces the impulse” or “reduces the change in momentum”. It does neither. Both stay the same; only the time gets longer, so the force gets smaller.

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Question 6

A moderate force will break an egg. However, an egg dropped on the road usually breaks, while one dropped on the grass usually does not break because…

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Question 7

A tennis ball of mass m is moving horizontally with speed v toward a player. The player hits the ball with a racket, and the ball leaves the racket…

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Question 8

Why do seat belts and air bags save lives?

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### Problem Type 4: Impulse from a force vs. time graph

**How to spot it:** A graph of force against time.

**Setup:** The area under the curve is the impulse, which is the change in momentum. Then [katex] v_f = v_i + \frac{\text{area}}{m} [/katex].

Triangle pulses, rectangles, curves: find the area, and you have the change in momentum. The airbag graph above is exactly this.

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**AP trap:** Area below the time axis counts as negative impulse: the force pushed the other way.

▶ **Watch:** a worked force vs. time graph problem (starts at 17:53).

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Question 9

A block of mass m = 2.0 kg is initially at rest on a horizontal, frictionless surface. A time-dependent horizontal force F is applied to the block…

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Impulse and Force-Time Graphs

Area under the curve, bounces versus sticks, and why airbags work.

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### Lesson 4.2 Preview

In the [next lesson](https://nerd-notes.com/unit-4-2-conservation-of-momentum/) we look at what happens when momentum does *not* change: a whole system of objects with no outside force keeps its total momentum. ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-4-1-momentum-and-impulse/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-4-1-momentum-and-impulse/
