Unit 4.1 | Momentum and Impulse

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.

[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]

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].

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

Problem Type 2: Bounces vs. stops

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

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

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

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

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

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

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?

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.

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

nerd-notes.com

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.

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.

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).

Lesson 4.2 Preview

In the next lesson 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/