Linear Momentum — AP Physics 1 Phys 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Linear momentum definition, the impulse-momentum theorem, 1D and 2D conservation of momentum, elastic and inelastic collisions, and center of mass per the 2024-25 AP Physics 1 CED.

You should already know: Algebra 2, basic trig, no calculus required.

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 papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Linear Momentum?

Linear momentum is a vector quantity that describes the "quantity of motion" of a moving object, quantifying how difficult it is to stop the object’s translational motion. It is a core topic in the 2024-25 AP Physics 1 CED, accounting for 10-15% of your total exam score across both multiple-choice and free-response sections. Often shortened to just "momentum" (when discussing translational motion), it follows strict conservation rules that let you predict the outcome of collisions, explosions, and other interactions between objects without needing to calculate complex internal forces.

2. Linear Momentum $p=mv$

The mathematical definition of linear momentum is given by the formula: $$\vec{p}=m\vec{v}$$ Where:

• $\vec{p}$ = linear momentum (vector, units of $\text{kgcdotpm/s}$)
• $m$ = inertial mass of the object (scalar, units of $\text{kg}$)
• $\vec{v}$ = instantaneous velocity of the object (vector, units of $\text{m/s}$) Momentum’s direction always matches the direction of the object’s velocity, so you must account for sign when working in one dimension, or x/y components for 2D problems.

Worked Example

A 1200 kg car travels east at 18 m/s, then reverses direction and travels west at 12 m/s. Calculate its momentum in both scenarios, using east as the positive direction.

1. Eastbound momentum: $p=1200\ast 18=21600\text{ kgcdotpm/s}$ (positive, matches east direction)
2. Westbound momentum: $p=1200\ast (-12)=-14400\text{ kgcdotpm/s}$ (negative, matches west direction) Examiner tip: A common 1-point MCQ question tests that momentum depends on velocity, not speed: an object moving in a circle at constant speed has constantly changing momentum, because its velocity direction changes.

3. Impulse-Momentum Theorem

The impulse-momentum theorem is derived directly from Newton’s second law, which was originally written as $F_{net}=\frac{\Delta p}{\Delta t}$ rather than the more common $F=ma$. Rearranging this definition gives the theorem: $$\vec{J}={\vec{F}}_{avg}\Delta t=\Delta \vec{p}$$ Where $\vec{J}$ = impulse (vector, units of $\text{Ncdotps}$, equivalent to $\text{kgcdotpm/s}$), ${\vec{F}}_{avg}$ = average net force applied over time interval $\Delta t$, and $\Delta \vec{p}=m({\vec{v}}_f-{\vec{v}}_i)$ = change in momentum. This theorem is particularly useful for short-duration interactions like bat hits, car crashes, and ball bounces, where the peak force is too large to measure directly, but the change in momentum and contact time are easy to calculate.

Worked Example

A 0.14 kg baseball is thrown west at 35 m/s, then hit by a bat and travels east at 45 m/s. The bat is in contact with the ball for 0.002 s. Calculate the average net force applied to the ball by the bat.

1. Assign east as positive: $v_i=-35\text{ m/s}$, $v_f=45\text{ m/s}$
2. Calculate change in momentum: $\Delta p=0.14\ast (45-(-35))=0.14\ast 80=11.2\text{ kgcdotpm/s}$
3. Solve for average force: $F_{avg}=\frac{\Delta p}{\Delta t}=\frac{11.2}{0.002}=5600\text{ N east}$ Exam note: FRQs frequently ask you to use this theorem to explain safety features like airbags and padded gym floors: increasing the contact time $\Delta t$ for a fixed change in momentum reduces the average force applied to the person, lowering injury risk.

4. Conservation of Momentum in 1D and 2D Collisions

Momentum is conserved for any closed, isolated system: a system with no net external force acting on it, and no mass entering or leaving the system. For collisions between two or more objects, internal forces between the objects are equal and opposite (Newton’s third law), so they cancel out, leaving total system momentum unchanged: $$\sum {\vec{p}}_{initial}=\sum {\vec{p}}_{final}$$ For 1D collisions, you only need to track sign to account for direction. For 2D collisions, momentum is conserved separately for the x and y components of motion, as perpendicular vector components do not affect each other.

Worked 1D Example

A 2 kg cart travels right at 3 m/s on a frictionless track, collides with a stationary 3 kg cart, and the two stick together. Calculate their final velocity.

1. Total initial momentum: $p_{initial}=(2\ast 3)+(3\ast 0)=6\text{ kgcdotpm/s right}$
2. Total final momentum: $p_{final}=(2+3)v_f=5v_f$
3. Equate and solve: $5v_f=6\to v_f=1.2\text{ m/s right}$

Worked 2D Example

A 0.5 kg ice puck travels east at 4 m/s, collides with an identical stationary puck. After collision, the first puck moves 30° north of east at 2 m/s. Find the velocity of the second puck.

1. X-direction (east positive): Initial $p_x=0.5\ast 4=2\text{ kgcdotpm/s}$ Final $p_x=0.5\ast 2\ast \cos (30^{\circ })+0.5v_x=0.866+0.5v_x=2\to v_x=2.27\text{ m/s east}$
2. Y-direction (north positive): Initial $p_y=0$ Final $p_y=0.5\ast 2\ast \sin (30^{\circ })+0.5v_y=0.5+0.5v_y=0\to v_y=-1\text{ m/s (south)}$
3. Combine components: Magnitude $v=\sqrt{(2.27{)}^2+(1{)}^2}\approx 2.48\text{ m/s}$, direction $23.8^{\circ }$ south of east.

5. Elastic vs Inelastic Collisions

While momentum is conserved for all isolated collisions, kinetic energy (KE) is only conserved for a small subset of collisions, classified by their energy behavior:

• Elastic collision: Both momentum and kinetic energy are conserved. No permanent deformation occurs, and no energy is lost to heat, sound, or friction. This only occurs for subatomic particle collisions, or near-perfect macroscopic interactions like billiard ball collisions or low-friction air track collisions.
• Inelastic collision: Momentum is conserved, but kinetic energy is not. Some kinetic energy is converted to other forms (heat, sound, deformation energy). Most real-world collisions are inelastic.
• Perfectly inelastic collision: A special case of inelastic collision where the colliding objects stick together after impact, resulting in the maximum possible loss of kinetic energy.

Worked Example

Classify the 1D cart collision from Section 4 as elastic, inelastic, or perfectly inelastic using kinetic energy calculations.

1. Initial KE: $K{E}_{initial}=\frac{1}{2}\ast 2\ast (3{)}^2+\frac{1}{2}\ast 3\ast (0{)}^2=9\text{ J}$
2. Final KE: $K{E}_{final}=\frac{1}{2}\ast 5\ast (1.2{)}^2=3.6\text{ J}$
3. 5.4 J of KE was lost, and the carts stuck together, so this is a perfectly inelastic collision. Exam trap: Never assume KE is conserved unless the problem explicitly states the collision is elastic. Examiners regularly include distractor options that use incorrect KE conservation for inelastic collision problems.

6. Centre of Mass

The centre of mass (COM) of a system is the single point where you can consider the entire mass of the system to be concentrated for the purpose of analyzing translational motion. For any isolated system with no net external force, the velocity of the centre of mass remains constant, even if individual objects in the system collide or move relative to each other. For a 1D system of particles, the COM position is calculated as: $$x_{cm}=\frac{m_1{x}_1+m_2{x}_2+...+m_n{x}_n}{M_{total}}$$ Where $m_i$ is the mass of the i-th particle, $x_i$ is its position, and $M_{total}$ is the total mass of the system. The COM velocity follows the same structure: $v_{cm}=\frac{\sum m_i{v}_i}{M_{total}}=\frac{p_{total}}{M_{total}}$, so constant total momentum means constant COM velocity.

Worked Example

A 2 kg mass is placed at x=1 m, and a 3 kg mass is placed at x=6 m on a frictionless track. The 2 kg mass moves right at 3 m/s, while the 3 kg mass remains stationary. Calculate the initial COM position and COM velocity.

1. COM position: $x_{cm}=\frac{(2\ast 1)+(3\ast 6)}{2+3}=\frac{20}{5}=4\text{ m}$
2. COM velocity: $v_{cm}=\frac{(2\ast 3)+(3\ast 0)}{5}=1.2\text{ m/s right}$ Note that this $v_{cm}$ matches the final velocity of the combined carts after the collision in Section 4, as expected for an isolated system.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Treating momentum as a scalar, ignoring direction when calculating change in momentum for rebounding objects. Why students do it: They confuse momentum with kinetic energy, which is a scalar. Correct move: Always assign a positive direction first, use negative values for velocity/momentum in the opposite direction, especially for objects that bounce off surfaces.
• Wrong move: Applying conservation of momentum to systems with net external forces (e.g., a ball rolling down a hill colliding with a rock, where gravity is an unbalanced external force). Why students do it: They forget the closed, isolated system requirement. Correct move: For very short collisions, external forces like gravity have negligible impulse, so momentum can be approximated as conserved only during the collision interval.
• Wrong move: Using kinetic energy conservation for inelastic collisions. Why students do it: They mix up momentum conservation (always true for isolated systems) with KE conservation (only true for elastic collisions). Correct move: Only use KE conservation if the problem explicitly labels the collision as elastic.
• Wrong move: Adding 2D momentum vectors as scalars instead of splitting into x and y components. Why students do it: They skip setting up a coordinate system to save time. Correct move: Always separate total initial momentum into x and y components, set each equal to total final x and y momentum, then combine final components to find magnitude and direction.
• Wrong move: Forgetting that impulse on two colliding objects is equal in magnitude and opposite in direction. Why students do it: They assume the heavier object experiences a larger impulse. Correct move: Newton’s third law guarantees equal force magnitude and equal contact time for both objects, so their impulse (and change in momentum) are equal and opposite, which is why total system momentum is conserved.

8. Practice Questions (AP Physics 1 Style)

Question 1 (MCQ)

A 0.05 kg tennis ball is thrown downward at 10 m/s, bounces off a concrete floor, and travels upward at 8 m/s. The contact time between the ball and floor is 0.01 s. What is the magnitude of the average force exerted on the ball by the floor? A) 10 N B) 40 N C) 90 N D) 140 N

Solution

1. Assign upward as positive: $v_i=-10\text{ m/s}$, $v_f=+8\text{ m/s}$
2. Calculate change in momentum: $\Delta p=0.05\ast (8-(-10))=0.9\text{ kgcdotpm/s}$
3. The net impulse equals change in momentum: $(F_{floor}-mg)\Delta t=\Delta p$. The weight force $mg=0.49\text{ N}$ is negligible compared to the floor force, so $F_{floor}\approx \frac{0.9}{0.01}=90\text{ N}$. Correct answer: C.

Question 2 (FRQ Part)

Two ice skaters on frictionless ice are initially stationary. Skater A has a mass of 60 kg, Skater B has a mass of 80 kg. They push off each other, and Skater A moves left at 4 m/s. a) Calculate the velocity of Skater B after the push. b) Classify this interaction as elastic, inelastic, or perfectly inelastic. Justify your answer with calculations.

Solution

a) Total initial momentum = 0, so total final momentum = 0. Assign right as positive: $60\ast (-4)+80v_B=0\to -240+80v_B=0\to v_B=3\text{ m/s right}$ b) Initial KE = 0 J. Final KE = $\frac{1}{2}\ast 60\ast (4{)}^2+\frac{1}{2}\ast 80\ast (3{)}^2=480+360=840\text{ J}$. KE is not conserved (it increased from chemical energy in the skaters’ muscles), and the skaters do not stick together, so the interaction is inelastic.

Question 3 (2D Collision FRQ)

A 2 kg puck travels east at 5 m/s, collides with a 3 kg puck traveling north at 2 m/s on frictionless ice. After collision, the 2 kg puck moves 30° north of east at 3 m/s. Find the magnitude and direction of the 3 kg puck’s final velocity.

Solution

Set east as positive x, north as positive y.

1. Initial x momentum: $2\ast 5+3\ast 0=10\text{ kgcdotpm/s}$ Final x momentum: $2\ast 3\ast \cos (30^{\circ })+3v_x=5.196+3v_x=10\to v_x=1.60\text{ m/s east}$
2. Initial y momentum: $2\ast 0+3\ast 2=6\text{ kgcdotpm/s}$ Final y momentum: $2\ast 3\ast \sin (30^{\circ })+3v_y=3+3v_y=6\to v_y=1\text{ m/s north}$
3. Magnitude: $v=\sqrt{(1.60{)}^2+1^2}\approx 1.89\text{ m/s}$
4. Direction: $\theta =\arctan (\frac{1}{1.60})\approx 32^{\circ }$ north of east.

9. Quick Reference Cheatsheet

10. What's Next

Linear momentum is a foundational topic that connects directly to later AP Physics 1 units, including rotational motion, where you will learn about angular momentum: the rotational analog of linear momentum that follows nearly identical conservation rules. Understanding collision dynamics also prepares you to analyze energy transfer in complex systems, a cross-cutting theme that appears on every AP Physics 1 exam, across both multiple-choice and free-response sections. Mastering momentum rules will also give you a head start on AP Physics 2 and AP Physics C mechanics content if you plan to take those exams later. If you struggle with any of the concepts, worked examples, or practice questions in this guide, you can ask Ollie for step-by-step explanations, additional practice problems, or targeted review tips at any time. Head to the homepage, type in your question, and get personalized support tailored to your AP Physics 1 exam prep needs.