---
title: "AP Physics 1 4.2: Change in Momentum and Impulse"
description: "Review AP Physics 1 momentum change and impulse, including impulse-momentum theorem, force-time graphs, momentum-time slopes, and vector direction."
canonical: "https://fiveable.me/ap-physics-1-revised/unit-4/2-change-in-momentum-and-impulse/study-guide/57woWSFVbwXIjDat"
type: "study-guide"
subject: "AP Physics 1"
unit: "Unit 4 – Linear Momentum"
lastUpdated: "2026-06-09"
---

# AP Physics 1 4.2: Change in Momentum and Impulse

## Summary

Review AP Physics 1 momentum change and impulse, including impulse-momentum theorem, force-time graphs, momentum-time slopes, and vector direction.

## Guide

Impulse is the [net force](/ap-physics-1-revised/key-terms/net-force "fv-autolink") on an object multiplied by the time it acts, and it equals the object's change in momentum. This is the impulse momentum theorem ($\vec{J} = \vec{F} {avg}\Delta t = \Delta\vec{p}$), and it connects [collision](/ap-physics-1-revised/key-terms/collision "fv-autolink") safety, force-time graphs, and momentum changes.

## Why This Matters for the AP Physics 1 Exam

Impulse and change in momentum show up across multiple-choice and free-response questions in [Unit 4](/ap-physics-1-revised/unit-4 "fv-autolink"), which carries about 10 to 15 percent of the exam. You will be expected to connect [force](/ap-physics-1-revised/unit-2/2-forces-and-free-body-diagrams/study-guide/jQ2Obd0dAU4QiTPN "fv-autolink"), time, and momentum using equations, graphs, and written reasoning.

Common ways this topic appears:
- Reading impulse as the area under a force vs. time graph
- Finding net force from the slope of a momentum vs. time graph
- Explaining why extending impact time lowers the average force (safety design reasoning)
- Solving for an unknown force, [velocity](/ap-physics-1-revised/unit-1/2-displacement-velocity-and-acceleration/study-guide/HyscWF2F28uakfpc "fv-autolink"), or time using the [impulse-momentum theorem](/ap-physics-1-revised/key-terms/impulse-momentum-theorem "fv-autolink")
- Showing how [Newton's second law](/ap-physics-1-revised/unit-2/5-newtons-second-law/study-guide/FizcgPbKTypwBNrG "fv-autolink") comes from the impulse-momentum theorem for constant [mass](/ap-physics-1-revised/key-terms/mass "fv-autolink")

Because Unit 4 connects to the Experimental Design and Analysis free-response question, you may also be asked to design or analyze data from impulse experiments, including linearizing graphs.

## Key Takeaways

- Net force equals the rate of change of momentum: $\vec{F}_{net} = \frac{\Delta\vec{p}}{\Delta t}$.
- Impulse is average force times the time interval: $\vec{J} = \vec{F}_{avg}\Delta t$, and it is a [vector](/ap-physics-1-revised/key-terms/vector "fv-autolink") pointing in the [direction](/ap-physics-1-revised/unit-1/4-reference-frames-and-relative-motion/study-guide/iTcYEEULwbQlf2nW "fv-autolink") of the net force.
- The impulse-momentum theorem says impulse equals change in momentum: $\vec{J} = \Delta\vec{p} = \vec{p} - \vec{p}_0$.
- Area under a force vs. time graph equals impulse; slope of a momentum vs. time graph equals net force.
- Impulse units are N·s, which equal kg·m/s.
- Newton's second law ($\vec{F}_{net} = m\vec{a}$) comes directly from the impulse-momentum theorem when mass is constant.

## Impulse Delivered to an Object or System

### Force and the Rate of Momentum Change

The net external force on an object sets how fast its momentum changes.

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

What this tells you:
- Only the net force (sum of all [external forces](/ap-physics-1-revised/key-terms/external-forces "fv-autolink")) changes momentum.
- If forces balance and the net force is zero, momentum stays constant.
- Doubling the net force doubles the rate of momentum change.
- This is a vector relationship, so you can analyze it component by component in 2D.

### What Impulse Means

Impulse is the total effect of a force applied over a time interval. It is how much "push" an object gets during an interaction.

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

Key features:
- Measured in newton-seconds (N·s), which equal kg·m/s.
- You can increase impulse by using more force or by applying force for longer.
- Catching a ball, hitting a nail, and launching a rocket all involve impulse.
- Safety designs like airbags and cushioned shoes extend the time to reduce the average force.

### Direction of Impulse

Impulse is a vector and points in the same direction as the net force.

- Find the net force first, then determine the impulse direction.
- Impulse can be split into x and y components for 2D problems.
- The impulse vector shows which way the object's motion will change.

### Area Under a Force vs. Time Graph

A force vs. time graph shows how force changes during an interaction. The area between the curve and the time axis equals the impulse.

When reading these graphs:
- For a [constant force](/ap-physics-1-revised/key-terms/constant-force "fv-autolink"), the area is a rectangle: force × time.
- For varying forces, use triangles, trapezoids, or estimate curved areas.
- Total impulse is the net area under the curve.
- Areas below the axis (force in the opposite direction) subtract from the total.

### Slope of a Momentum vs. Time Graph

A momentum vs. time graph shows how momentum changes over time. The slope at any point equals the net force at that moment.

When reading these graphs:
- Steep slopes mean large net forces.
- Flat regions (zero slope) mean zero net force.
- Positive slopes mean force in the positive direction; negative slopes mean the opposite.
- The shape of the graph reveals how the net force varies.

## Impulse and Change in Momentum

### Calculating Change in Momentum

Change in momentum measures how much an object's motion was altered. It is the final momentum minus the initial momentum.

$$\Delta \vec{p} = \vec{p} - \vec{p}_0 = m(\vec{v} - \vec{v}_0)$$

When calculating it:
- The direction of $\Delta\vec{p}$ matches the direction of the net force.
- For constant mass, change in momentum depends only on the velocity change.
- Larger momentum changes require larger impulses.
- You can work component by component in 2D problems.

### The Impulse-Momentum Theorem

The impulse-momentum theorem says the impulse on an object equals its change in momentum.

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

Why it is useful:
- Explains why extending impact time reduces force (catching an egg gently).
- Shows why very short impacts deliver large forces (hammering a nail).
- Lets engineers design safety systems that stretch out collision times.
- Lets you solve for an unknown force when you know the momentum change and time.

### Deriving Newton's Second Law

Newton's second law follows directly from the impulse-momentum theorem when mass stays constant.

Starting from the theorem:
1. $$\vec{F}_{net} \Delta t = \Delta \vec{p}$$
2. For constant mass: $$\vec{F}_{net} \Delta t = m \Delta \vec{v}$$
3. Dividing both sides by $\Delta t$: $$\vec{F}_{net} = m \frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$$

This shows that:
- Newton's second law is a special case of the impulse-momentum theorem.
- The impulse-momentum approach still works when [acceleration](/ap-physics-1-revised/unit-1/5-vectors-and-motion-in-two-dimensions/study-guide/LvdiAzU3amzMqu6O "fv-autolink") is not constant.
- For varying forces, working with impulse is often simpler than tracking acceleration.

> 🚫 **Boundary Statement**
>
> AP Physics 1 does not require you to quantitatively analyze systems where the mass of the system changes over time.

## How to Use This on the AP Physics 1 Exam

### Problem Solving

- Decide what you are solving for, then pick the matching form: $\vec{F}_{net} = \frac{\Delta\vec{p}}{\Delta t}$, $\vec{J} = \vec{F}_{avg}\Delta t$, or $\vec{J} = \Delta\vec{p}$.
- Always assign a positive direction first so your signs stay consistent.
- For rebound problems, remember the velocity reverses sign, so the change in momentum is larger than you might expect.
- Check units: impulse should come out in N·s or kg·m/s.

### Graphs

- Force vs. time graph: find impulse by computing the area under the curve.
- Momentum vs. time graph: find net force by computing the slope.
- Do not mix these up. Area and slope answer different questions.

### Free Response

- When a question asks you to explain a safety device, connect a longer time interval to a smaller average force using $\vec{J} = \vec{F}_{avg}\Delta t$ with the same impulse.
- Justify claims with the equation and the direction, not just a verbal answer.
- For experiments, you may be asked to linearize data so that a slope or area gives the quantity you want.

## Practice Problem 1: Impulse Calculation

> A 0.145 kg baseball moving at 35 m/s is caught by a player. The ball comes to rest in the player's glove over a time of 0.050 seconds. Determine (a) the impulse applied to the ball, (b) the average force exerted on the ball, and (c) sketch what the force-time graph might look like if the force is not constant during the catch.

**Solution**

(a) To find the impulse, calculate the change in momentum:
Initial momentum: $$p_i = mv_i = (0.145 \text{ kg})(35 \text{ m/s}) = 5.075 \text{ kg}\cdot\text{m/s}$$
Final momentum: $$p_f = mv_f = (0.145 \text{ kg})(0 \text{ m/s}) = 0 \text{ kg}\cdot\text{m/s}$$
Impulse: $$J = \Delta p = p_f - p_i = 0 - 5.075 = -5.075 \text{ kg}\cdot\text{m/s} = -5.075 \text{ N}\cdot\text{s}$$

The negative sign shows the impulse points opposite to the ball's initial velocity.

(b) The average force is:
$$F_{avg} = \frac{J}{\Delta t} = \frac{-5.075 \text{ N·s}}{0.050 \text{ s}} = -101.5 \text{ N}$$

(c) During a real catch, the force is not constant. It starts at zero as the glove first touches the ball, rises to a peak as the glove compresses, then drops back to zero as the ball stops. A realistic force vs. time graph rises from zero to a peak and falls back to zero.

## Practice Problem 2: Force from Momentum-Time Graph

> The momentum-time graph for a 2.0 kg object is shown below. The graph is a straight line from (0 s, 4 kg·m/s) to (5 s, 14 kg·m/s). What is the net force acting on the object during this time interval?

**Solution**

The net force is the slope of the momentum vs. time graph:

Slope = $$\frac{\Delta p}{\Delta t} = \frac{p_f - p_i}{t_f - t_i} = \frac{14 \text{ kg}\cdot\text{m/s} - 4 \text{ kg}\cdot\text{m/s}}{5 \text{ s} - 0 \text{ s}} = \frac{10 \text{ kg}\cdot\text{m/s}}{5 \text{ s}} = 2 \text{ N}$$

Since $$F_{net} = \frac{\Delta p}{\Delta t}$$, the net force is 2 N in the positive direction.

## Practice Problem 3: Impulse-Momentum Theorem

> A 0.5 kg cart starts from rest on a horizontal track. A net average force of 15 N acts on the cart for 2.0 s. (a) What impulse is delivered to the cart? (b) What is the cart's velocity immediately after the force stops acting?

**Solution**

(a) The impulse delivered to the cart is:
$$J = F_{avg} \Delta t = (15 \, \text{N})(2.0 \, \text{s}) = 30 \, \text{N·s}$$

(b) Using the impulse-momentum theorem:
$$J = \Delta p = m\Delta v = m(v_f - v_i)$$

Since $$v_i = 0$$ (starts from rest):
$$30 \, \text{N·s} = (0.5 \, \text{kg})(v_f - 0)$$

$$v_f = \frac{30 \, \text{N·s}}{0.5 \, \text{kg}} = 60 \, \text{m/s}$$

The cart's velocity is 60 m/s in the direction of the net force.

## Common Misconceptions

- Impulse and force are not the same thing. Force acts at an instant; impulse is force applied over a time interval, and it equals the change in momentum.
- A larger force does not always mean a larger impulse. A small force over a long time can deliver the same impulse as a large force over a short time.
- Area and slope are not interchangeable. Area under a force vs. time graph gives impulse, while slope of a momentum vs. time graph gives net force.
- For rebound or bounce-back problems, do not forget the sign change. Reversing direction makes the change in momentum bigger than if the object simply stopped.
- Newton's second law is not a separate idea from impulse. $\vec{F}_{net} = m\vec{a}$ comes from the impulse-momentum theorem when mass is constant.
- Impulse depends on net external force, not [internal forces](/ap-physics-1-revised/key-terms/internal-forces "fv-autolink"). Forces inside the chosen [system](/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO "fv-autolink") do not change the system's total momentum.

## Related AP Physics 1 Guides

- [4.1 Linear Momentum](/ap-physics-1-revised/unit-4/1-linear-momentum/study-guide/aEYIGw4MVE0g5Zrb)
- [4.3 Conservation of Linear Momentum](/ap-physics-1-revised/unit-4/3-conservation-of-linear-momentum/study-guide/B4haVeUmTXK0iRFh)
- [4.4 Elastic and Inelastic Collisions](/ap-physics-1-revised/unit-4/4-elastic-and-inelastic-collisions/study-guide/ZmUQjuCAgv73jObj)

## FAQs

### What is impulse in AP Physics 1?

Impulse is the product of average net force and the time interval during which the force acts. It is a vector and points in the same direction as the net force.

### What is the impulse-momentum theorem?

The impulse-momentum theorem says impulse equals change in momentum. In symbols, J = F_avg Delta t = Delta p.

### How do force-time graphs connect to impulse?

The impulse delivered by a net external force is the area under a force-versus-time graph. Positive and negative areas represent impulse in different directions.

### How do momentum-time graphs connect to force?

The slope of a momentum-versus-time graph equals the net external force. A steeper slope means a larger net force.

### What does AP Physics 1 not require for this topic?

AP Physics 1 does not require quantitative analysis of systems whose mass changes with time. Treat mass as constant unless the problem states otherwise within course scope.

### What should I include in an impulse FRQ response?

Name the system, identify the direction, use impulse equals change in momentum, and connect graph area or force-time information to the change in momentum.

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