Momentum and Impulse — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Definition of linear momentum, calculation of impulse for constant and variable forces, the impulse-momentum theorem, vector sign conventions for 1D problems, and impulse calculation from force-time graphs, aligned to AP Physics 1 CED.

You should already know: Newton's laws of motion, vector sign conventions for 1-dimensional motion, area calculation for basic geometric shapes.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Momentum and Impulse?

Linear momentum (commonly just called momentum) is a vector quantity that describes the "amount of motion" an object has, relating its mass and velocity. Impulse describes the change in motion caused by a net force acting over a period of time. Together, these concepts form the foundation of Unit 5: Momentum, which the AP Physics 1 CED lists as 12–18% of the total exam score. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a standalone question or combined with collision problems, energy analysis, or real-world engineering scenarios. Unlike energy, momentum is always conserved for systems with no net external force, making it a critical tool for analyzing interactions between objects that we cannot study with force alone. AP Physics 1 only assesses momentum and impulse in one dimension, so we use positive and negative signs to indicate direction rather than full vector components, simplifying calculations while still requiring careful attention to vector rules.

2. Linear Momentum

Linear momentum is defined as the product of an object’s mass and its velocity. As a vector, its direction matches the direction of the object’s velocity, which means we must account for sign in 1-dimensional problems. The formula for momentum is: $$\vec{p}=m\vec{v}$$ Common units for momentum are kilogram-meters per second ($\text{kg}\cdot \text{m/s}$). The intuition for momentum is that it measures how hard it is to stop a moving object: a slow moving semi-truck has far more momentum than a fast moving baseball, because the truck’s large mass outweighs the baseball’s higher velocity. For systems of multiple objects, total momentum is just the vector sum of the momentum of each individual object. This means we add momentum values with their correct signs (not just magnitudes) to get total system momentum, a critical step for conservation of momentum problems later.

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 $v_1=+3.0\text{m/s}$ (for the 2.0 kg cart) and $v_2=-4.0\text{m/s}$ (for the 1.0 kg cart).
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 to get total momentum: $p_{\text{total}}=p_1+p_2=6.0-4.0=+2.0\text{kg}\cdot \text{m/s}$.

The total momentum of the system is $+2.0\text{kg}\cdot \text{m/s}$, meaning it has net momentum to the right.

Exam 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.

3. Impulse

Impulse is the quantity that describes the effect of a net force acting on an object over a period of time. If the net force is constant, impulse is simply the product of force and the time interval the force acts: $J=F_{\text{net}}\Delta t$. If the force changes over time (which is almost always true for collisions, where force rises to a peak then falls back to zero), impulse equals the area under a net force vs. time graph. This is because impulse is defined as the integral of force over time: $$J={\int }_{t_1}^{t_2}{F}_{\text{net}}(t)dt$$ AP Physics 1 does not require calculus for this calculation, so the area will always be made of simple shapes (triangles, rectangles, trapezoids) that you can calculate with basic geometry. The units of impulse are Newton-seconds ($\text{N}\cdot \text{s}$), which is equivalent to $\text{kg}\cdot \text{m/s}$, matching the units for momentum. The key intuition for impulse is the impulse-force-time relation: to get the same total impulse (same change in momentum), you can have a large force over a short time or a small force over a long time. This is the physics behind airbags, padded dashboards, and crash-absorbing bumpers: increasing collision time reduces force.

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 a triangle in this case.
2. The area of a triangle is given by $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$.
3. The base of the triangle is the total contact time $\Delta t=0.005\text{s}$, and the height is the peak force $F_{\text{max}}=2000\text{N}$.
4. Calculate impulse: $J=\frac{1}{2}\times 0.005\times 2000=5.0\text{N}\cdot \text{s}$.

The magnitude of the impulse delivered to the ball is $5.0\text{N}\cdot \text{s}$.

Exam tip: Always check the x-axis label of any force graph before calculating area. Area of a force vs. position graph is work, not impulse—this is a common MCQ distractor trap.

4. The Impulse-Momentum Theorem

The impulse-momentum theorem is the core relationship between impulse and momentum that connects cause (force over time) to effect (change in motion). It can be derived directly from Newton’s second law: start with $F_{\text{net}}=ma$, and rewrite acceleration as $a=\frac{\Delta v}{\Delta t}$. Rearranging gives $F_{\text{net}}\Delta t=m\Delta v$, which is the impulse-momentum theorem: $$J_{\text{net}}=\Delta p=p_f-p_i=m(v_f-v_i)$$ This theorem states that the net impulse delivered to an object equals the total change in the object’s momentum. It works for constant and variable forces, because we already account for variable force by calculating impulse as area under the F-t graph. For systems of objects, internal impulses (forces objects in the system exert on each other) cancel out per Newton’s third law, so only external forces contribute to net impulse for the whole system. This theorem lets you solve for any unknown quantity: final velocity, average force, contact time, etc., as long as you know the other values.

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 $F_{\text{net}}=-120\text{N}$, $v_i=+5.0\text{m/s}$, $\Delta t=1.5\text{s}$, $m=60\text{kg}$.
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: $v_f=5.0+\frac{-180}{60}=5.0-3.0=2.0\text{m/s}$.

The skateboarder’s final velocity is 2.0 m/s to the right.

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

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating impulse for a triangular force vs. time graph as $F_{max}\Delta t$ instead of $\frac{1}{2}{F}_{max}\Delta t$. Why: Students confuse the shape of the graph and use the rectangle area formula, resulting in twice the correct impulse, which is a common MCQ distractor. Correct move: Always explicitly identify the shape of the F-t region and write the correct area formula before plugging in numbers.
• Wrong move: Adding magnitudes of momentum for objects moving in opposite directions, instead of adding signed values. Why: Students forget momentum is a vector and treat it like a scalar quantity. Correct move: Write your coordinate system at the start of the problem, assign signs to all velocities before calculating momentum.
• Wrong move: Equating impulse to total momentum, instead of change in momentum. Why: Students abbreviate the theorem to "impulse equals momentum" when memorizing, leading to wrong answers for final velocity. Correct move: 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 move: Calculating $\Delta p=p_i-p_f$ instead of $\Delta p=p_f-p_i$. Why: Students mix up the "final minus initial" rule for change, leading to a sign error that propagates through the whole problem. Correct move: Double-check the order of subtraction for any change in quantity, and always put final value first.
• Wrong move: Using area under force vs. position to find impulse. Why: Students confuse impulse (uses F-t) and work (uses F-x), because both are areas under force graphs. Correct move: Always check the x-axis label before calculating area: x = time → impulse, x = position → work.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

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

Worked Solution: First, define the direction toward the wall as positive, so initial velocity $v_i=+15\text{m/s}$ and final velocity after bouncing is $v_f=-10\text{m/s}$. By the impulse-momentum theorem, impulse equals change in momentum: $J=\Delta p=m(v_f-v_i)=0.2(-10-15)=-5\text{Ncdotps}$. The question asks for magnitude, so the magnitude is 5 N·s. Common errors include 1 N·s (wrong order of subtraction) and 1 N·s (wrong order of subtraction) and 3 N·s (forgetting to flip the sign of final velocity). The correct answer is C.

Question 2 (Free Response)

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.

Worked Solution: (a) Total impulse equals the area under the F-t graph. Split the graph into three regions: left triangle (0-1s): $A_1=\frac{1}{2}(1)(4)=2\text{Ncdotps}$; middle rectangle (1-3s): $A_2=(2)(4)=8\text{Ncdotps}$; right triangle (3-4s): $A_3=\frac{1}{2}(1)(4)=2\text{Ncdotps}$. Total impulse $J=2+8+2=12\text{Ncdotps}$. (b) Use the impulse-momentum theorem: $J=m(v_f-v_i)⟹v_f=v_i+\frac{J}{m}=6.0+\frac{12}{2.0}=12\text{m/s}$. The final velocity is 12 m/s to the right. (c) For a fixed total impulse, the average force is inversely proportional to the time interval ($F_{avg}=J/\Delta t$). Increasing $\Delta t$ reduces the average force, and for a similar force profile, maximum force scales with average force, so the maximum force also decreases. This is the core principle behind crash safety design.

Question 3 (Application / Real-World Style)

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 wall for both barriers.

Worked Solution:

1. Define initial direction of motion as positive, so $v_i=+15\text{m/s}$, $v_f=0\text{m/s}$, $m=1500\text{kg}$.
2. Calculate change in momentum: $\Delta p=m(v_f-v_i)=1500(0-15)=-22500\text{kgcdotpm/s}$.
3. Rearrange $F_{avg}\Delta t=\Delta p$ to get $F_{avg}=\Delta p/\Delta t$.
4. Rigid barrier: $F_{avg}=-22500/0.08=-2.8\times 10^5\text{N}$. Energy-absorbing barrier: $F_{avg}=-22500/0.30=-7.5\times 10^4\text{N}$. The negative sign indicates force acts opposite the direction of motion. The energy-absorbing barrier reduces the average force on the car by nearly 75%, making crashes far less dangerous for passengers.

7. Quick Reference Cheatsheet

8. What's Next

This chapter gives you the foundational tools for the rest of 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 to energy conservation, where you will compare when to use momentum vs. energy for collision problems. Momentum principles also extend to later topics like rotational motion, where angular impulse and angular momentum follow identical logic. Next topics to study after this chapter are: Conservation of Momentum Collisions and Explosions Energy and Momentum Comparisons