# Momentum and Impulse

> AP Physics 1 · AP Physics 1 CED Unit 5
> Source: https://www.owlsprep.com/study/ap-physics-1-u5-momentum-and-impulse/

This guide covers core definitions of linear momentum and impulse, impulse calculation for constant/variable forces, the impulse-momentum theorem, and 1D vector sign conventions, aligned to AP Physics 1 CED for Unit 5.

**Prerequisites:** [Newton's laws of motion](https://www.owlsprep.com/study/ap-physics-1-u3-newtons-laws/); [1D vector sign conventions](https://www.owlsprep.com/study/ap-physics-1-1d-kinematics/); Area calculation for basic geometric shapes

## Learning objectives

- Define linear momentum and impulse as vector quantities
- Calculate impulse for constant and variable forces from force-time graphs
- Apply the impulse-momentum theorem to solve 1D motion problems
- Correctly use vector sign conventions for 1D momentum problems

## Linear Momentum

**Linear Momentum** — Vector quantity describing the amount of motion of an object, equal to the product of the object's mass and velocity. Direction matches velocity, so use positive/negative signs for direction in 1D problems. Units are $\text{kg} \cdot \text{m/s}$.

*Notation:* \vec{p} = m\vec{v}

*Example:* A 1000 kg car moving at 20 m/s right has momentum $+20000 \, \text{kg} \cdot \text{m/s}$.

Momentum measures how hard it is to stop a moving object: a slow-moving semi-truck has more momentum than a fast-moving baseball, because mass has a larger effect than velocity here. For systems of multiple objects, total momentum is the vector sum of individual momentum values: always add signed values, not just magnitudes, to get the correct total.

**Worked example:** A 2.0 kg cart moving to the right at 3.0 m/s collides head-on with a 1.0 kg cart moving to the left at 4.0 m/s. What is the total momentum of the two-cart system before the collision, taking right as the positive direction?

1. Define the coordinate system (right = positive), so velocities are:

   $$v_1 = +3.0 \, \text{m/s}, \quad v_2 = -4.0 \, \text{m/s}$$
2. Calculate momentum of the first cart:

   $$p_1 = m_1 v_1 = (2.0)(+3.0) = +6.0 \, \text{kg} \cdot \text{m/s}$$
3. Calculate momentum of the second cart:

   $$p_2 = m_2 v_2 = (1.0)(-4.0) = -4.0 \, \text{kg} \cdot \text{m/s}$$
4. Sum the signed momentum values to get total system momentum:

   $$p_{\text{total}} = p_1 + p_2 = 6.0 - 4.0 = +2.0 \, \text{kg} \cdot \text{m/s}$$

> **tip**
>
> Always explicitly define your positive direction at the start of every momentum problem. AP graders require this for FRQ credit, and it eliminates 90% of common sign errors.

## Impulse

**Impulse** — Quantity describing the effect of a net force acting over a time interval. For constant net force, $J = F_{\text{net}} \Delta t$; for variable force, $J$ equals the area under a net force vs. time graph. Units are $\text{N} \cdot \text{s}$, which is equivalent to $\text{kg} \cdot \text{m/s}$.

*Notation:* J

Impulse follows the key force-time relationship: to get the same total impulse (same change in momentum), you can apply a large force over a short time or a small force over a long time. This principle explains the function of airbags, padded dashboards, and crash-absorbing bumpers: increasing collision time reduces the peak force experienced during impact.

AP Physics 1 does not require calculus for impulse calculation: the area under a force-time graph will always be made of simple geometric shapes (triangles, rectangles, trapezoids) that can be calculated with basic geometry.

> **warning**
>
> The area under a force vs. position graph equals work, not impulse. This is a very common multiple-choice distractor trap.

**Worked example:** A student hits a 0.05 kg golf ball with a club. The force exerted by the club on the ball as a function of time forms a triangle with a peak force of 2000 N and total contact time of 0.005 s. What is the magnitude of the impulse delivered to the golf ball?

1. Impulse equals the area under the force vs. time graph, which is triangular in this case.
2. Area of a triangle is given by:

   $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$
3. Base = contact time $\Delta t = 0.005$ s, height = peak force $F_{\text{max}} = 2000$ N. Substitute to find impulse:

   $$J = \frac{1}{2} \times 0.005 \times 2000 = 5.0 \, \text{N} \cdot \text{s}$$

## The Impulse-Momentum Theorem

**Impulse-Momentum Theorem** — The net impulse delivered to an object (or system) equals the total change in the object's momentum. This is derived directly from Newton's second law.

*Notation:* J_{\text{net}} = \Delta p

*Example:* A 1 kg object that speeds up from 2 m/s to 5 m/s has $\Delta p = 3 \, \text{kg} \cdot \text{m/s}$, so net impulse $J = 3 \, \text{N} \cdot \text{s}$.

The full form of the theorem is:

$$J_{\text{net}} = \Delta p = p_f - p_i = m(v_f - v_i)$$

For systems of multiple objects, internal impulses (forces that objects within the system exert on each other) cancel out per Newton's third law, so only external forces contribute to the net impulse of the whole system. This relationship works for both constant and variable forces, since variable force impulse is calculated as area under the F-t graph.

**Worked example:** A 60 kg skateboarder moving right at 5.0 m/s hits a rough patch of ground that exerts an average net force of 120 N to the left on the skateboard for 1.5 s. What is the skateboarder’s final velocity?

1. Define right as positive, so all values are:

   $$F_{\text{net}} = -120 \, \text{N}, \, v_i = +5.0 \, \text{m/s}, \, \Delta t = 1.5 \, \text{s}, \, m = 60 \, \text{kg}$$
2. Calculate net impulse:

   $$J_{\text{net}} = F_{\text{net}} \Delta t = (-120)(1.5) = -180 \, \text{N} \cdot \text{s}$$
3. Rearrange the impulse-momentum theorem to solve for final velocity:

   $$v_f = v_i + \frac{J_{\text{net}}}{m}$$
4. Substitute values and solve:

   $$v_f = 5.0 + \frac{-180}{60} = 2.0 \, \text{m/s}$$

> **tip**
>
> When asked for average force, always use net impulse (sum of impulse from all forces acting over the interval), not just the impulse from the applied force. Friction or gravity often contribute a non-negligible impulse that changes the result.

## AP-Style Worked Practice Problems

**Check your understanding**

Try this multiple-choice question to test your understanding before looking at the solution:

1. A 0.2 kg ball is thrown straight toward a wall at 15 m/s, and bounces straight back at 10 m/s. What is the magnitude of the impulse delivered to the ball by the wall?

   - A) 1 N·s
   - B) 3 N·s
   - C) 5 N·s
   - D) 25 N·s

   *Why:* You must account for the change in direction of the ball, so the velocity flips sign. Correct calculation: $J = \Delta p = 0.2(-10 - 15) = -5$ N·s, magnitude 5 N·s.

**Worked example:** A 2.0 kg block slides right along a frictionless horizontal surface at 6.0 m/s. A variable force (positive when pointing right) is applied to the block over 4.0 seconds, with the following force vs. time shape: from t=0 to t=1 s, force increases linearly from 0 to 4 N; from t=1 s to t=3 s, force is constant at 4 N; from t=3 s to t=4 s, force decreases linearly back to 0. (a) Calculate the total impulse delivered to the block over 4.0 seconds. (b) Calculate the final velocity of the block after 4.0 seconds. (c) Explain why increasing the time over which a fixed total impulse is applied reduces the maximum force on the block.

1. (a) Split the F-t graph into three regions and calculate area for each:
- Left triangle (0-1 s): $A_1 = \frac{1}{2}(1)(4) = 2$ N·s
- Middle rectangle (1-3 s): $A_2 = (2)(4) = 8$ N·s
- Right triangle (3-4 s): $A_3 = \frac{1}{2}(1)(4) = 2$ N·s
2. Total impulse is the sum of all areas:

   $$J = 2 + 8 + 2 = 12 \, \text{N·s}$$
3. (b) Use the impulse-momentum theorem to solve for final velocity:

   $$J = m(v_f - v_i) \implies v_f = v_i + \frac{J}{m} = 6.0 + \frac{12}{2.0} = 12 \, \text{m/s}$$
4. (c) For a fixed total impulse, average force is inversely proportional to the time interval ($F_{avg} = J/\Delta t$). Increasing $\Delta t$ reduces the average force, and maximum force scales with average force for a similar force profile. This is the core principle behind automotive crash safety design.

**Worked example:** A car manufacturer tests crash safety by running a 1500 kg car into a barrier at 15 m/s. The car comes to a complete stop after impact. Two setups are tested: a rigid barrier stops the car in 0.08 s, and an energy-absorbing barrier stops the car in 0.30 s. Calculate the average force exerted on the car by the barrier for both setups.

1. Define initial direction of motion as positive, so $v_i = +15 \, \text{m/s}$, $v_f = 0$, $m = 1500 \, \text{kg}$.
2. Calculate total change in momentum:

   $$\Delta p = m(v_f - v_i) = 1500(0 - 15) = -22500 \, \text{kg·m/s}$$
3. Rearrange the impulse-momentum theorem to solve for average force:

   $$F_{\text{avg}} = \frac{\Delta p}{\Delta t}$$
4. Rigid barrier average force:

   $$F_{\text{avg, rigid}} = \frac{-22500}{0.08} = -2.8 \times 10^5 \, \text{N}$$
5. Energy-absorbing barrier average force:

   $$F_{\text{avg, absorb}} = \frac{-22500}{0.30} = -7.5 \times 10^4 \, \text{N}$$
6. The negative sign indicates force acts opposite the car's initial direction. The energy-absorbing barrier reduces average force by nearly 75%.

## Common pitfalls

- **Wrong:** Calculating impulse for a triangular force vs. time graph as $F_{max}\Delta t$ instead of $\frac{1}{2}F_{max}\Delta t$
  - Why it fails: Students confuse the graph shape and use the rectangle area formula, resulting in twice the correct impulse, which is a common MCQ distractor.
  - Correct: Always explicitly identify the shape of the F-t region and write the correct area formula before plugging in numbers.
- **Wrong:** Adding magnitudes of momentum for objects moving in opposite directions, instead of adding signed values
  - Why it fails: Students forget momentum is a vector and treat it like a scalar quantity, leading to wrong total momentum.
  - Correct: Write your coordinate system at the start of the problem, assign signs to all velocities before calculating momentum.
- **Wrong:** Equating impulse to total momentum, instead of change in momentum
  - Why it fails: Students abbreviate the theorem to 'impulse equals momentum' when memorizing, leading to wrong answers for final velocity.
  - Correct: Always write the full theorem $J_{net} = \Delta p = p_f - p_i$ at the start of your calculation to remind yourself it is a change.
- **Wrong:** Calculating $\Delta p = p_i - p_f$ instead of $\Delta p = p_f - p_i$
  - Why it fails: Students mix up the 'final minus initial' rule for change, leading to a sign error that propagates through the whole problem.
  - Correct: Double-check the order of subtraction for any change in quantity, and always put the final value first.
- **Wrong:** Using area under force vs. position to find impulse
  - Why it fails: Students confuse impulse (uses F-t graphs) and work (uses F-x graphs), because both are areas under force graphs.
  - Correct: Always check the x-axis label before calculating area: x = time → impulse, x = position → work.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Linear Momentum | $p = mv$ | Vector, direction matches velocity; use signs for direction in 1D |
| Total System Momentum | $p_{\text{total}} = p_1 + p_2 + ... + p_n$ | Add signed momentum values, not just magnitudes |
| Impulse (Constant Force) | $J = F_{\text{net}} \Delta t$ | Units: $\text{N·s} = \text{kg·m/s}$, same as momentum |
| Impulse (Variable Force) | $J = \text{Area under } F \text{ vs. } t \text{ graph}$ | Use triangle area $A=\frac{1}{2}bh$, rectangle area $A=bh$ |
| Impulse-Momentum Theorem | $J_{\text{net}} = \Delta p = p_f - p_i = m(v_f - v_i)$ | Core relation, applies to all constant-mass problems |
| Average Force | $F_{\text{avg}} = \frac{J_{\text{net}}}{\Delta t}$ | Used to find average force for variable impulse over time |

## What's next

This sub-topic gives you the foundational tools for the rest of AP Physics 1 Unit 5: Momentum. The most immediate application is conservation of momentum for closed systems, where net external impulse is zero, so total momentum remains constant. Without mastering the impulse-momentum theorem and consistent sign conventions for vector momentum, you will not be able to correctly solve collision and explosion problems, which make up the majority of Unit 5 FRQ points on the AP exam. This topic also connects directly to energy conservation, where you will learn when to use momentum vs. energy to solve different types of collision problems, and core logic extends to angular momentum in rotational motion.

- [AP Physics 1 Impulse-Momentum Theorem](https://www.owlsprep.com/study/ap-physics-1-u5-impulse-momentum-theorem/)
- [Conservation of Momentum for Isolated Systems](https://www.owlsprep.com/study/ap-physics-1-u5-conservation-of-momentum-for-isolated/)
- [One-Dimensional Collisions](https://www.owlsprep.com/study/ap-physics-1-u5-one-dimensional-collisions/)

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