#AP Physics 1: Linear Momentum Study Guide 🚀

Hey there, future physics pro! Let's break down linear momentum for your AP exam. This guide is designed to be your go-to resource the night before the test. Let's make sure you're feeling confident and ready to ace it! 💪

#What is Linear Momentum?

#Definition of Linear Momentum

Key Concept

Linear momentum ( $\vec{p}$ ) is the measure of an object's motion, combining its mass and velocity. Think of it as the "oomph" behind a moving object.

Formula: $\vec{p} = m\vec{v}$ where:
- $\vec{p}$ is the linear momentum (kg⋅m/s)
- $m$ is the mass (kg)
- $\vec{v}$ is the velocity (m/s)
Key Insights:
- Momentum depends on both mass and velocity. Double either, and you double the momentum.
- SI units: kilogram-meters per second (kg⋅m/s)

Quick Fact

Momentum is a vector, so don't forget about direction!

#Vector Nature of Momentum

Direction Matters: Momentum is a vector with both magnitude (size) and direction. It points in the same direction as the object's velocity.
Positive vs. Negative:
- Positive momentum: motion to the right or upwards
- Negative momentum: motion to the left or downwards
Vector Addition: You can add or subtract momentum components just like any other vectors to find the net momentum of a system.
Changing Momentum: Change in velocity (speed or direction) means a change in momentum.

Memory Aid

Think of momentum as the "push" of an object. A heavier object or a faster object has more "push".

#Momentum in Collisions and Explosions 💥

Conservation of momentum is HUGE for collisions and explosions! This is a core concept you'll see again and again.

Conservation Principle: In an isolated system (no external forces), the total momentum remains constant before and after a collision or explosion. 💡
Collisions:
- Objects exert large forces on each other over a short time.
- We analyze the initial and final states using the object model, ignoring the collision details.
- Momentum is always conserved, but kinetic energy might not be (inelastic collisions).
Explosions:
- Internal forces push objects apart.
- Total momentum is conserved, but individual momenta change.
- Kinetic energy increases as potential energy is converted.
Types of Collisions:
- Elastic Collisions: Both momentum and kinetic energy are conserved.
- Inelastic Collisions: Momentum is conserved, but kinetic energy is not (some energy is lost to heat, sound, etc.).

Common Mistake

Don't confuse conservation of momentum with conservation of kinetic energy. Momentum is ALWAYS conserved in an isolated system, but kinetic energy is only conserved in elastic collisions.

Exam Tip

When solving collision problems, always start by stating the conservation of momentum equation. This will often get you partial credit even if you can't solve the whole problem.

#Final Exam Focus

High-Priority Topics:
- Conservation of momentum in collisions (elastic and inelastic).
- Vector nature of momentum and how to add/subtract components.
- Understanding the difference between elastic and inelastic collisions.
Common Question Types:
- Calculating final velocities after collisions or explosions.
- Determining if a collision is elastic or inelastic.
- Analyzing momentum changes in systems.
Last-Minute Tips:
- Time Management: Don't spend too long on a single question. Move on and come back if you have time.
- Common Pitfalls: Watch out for direction (positive/negative) when dealing with momentum vectors. Always check your units!
- Strategies: Draw diagrams to visualize the problem. Write out the conservation of momentum equation before plugging in numbers.

#Practice Questions

Practice Question

Multiple Choice Questions

A 2 kg object moving at 3 m/s collides head-on with a 1 kg object moving at -4 m/s. If the collision is perfectly inelastic, what is the final velocity of the combined mass? (A) -1 m/s (B) 2/3 m/s (C) 1 m/s (D) 10/3 m/s
A system consists of two objects. Object A has a mass of 2 kg and a velocity of 5 m/s to the right. Object B has a mass of 3 kg and a velocity of 2 m/s to the left. What is the total momentum of the system? (A) 4 kg⋅m/s to the right (B) 4 kg⋅m/s to the left (C) 16 kg⋅m/s to the right (D) 16 kg⋅m/s to the left
In an isolated system, two objects collide. Which of the following is always conserved during the collision? (A) Kinetic energy (B) Potential energy (C) Momentum (D) Both kinetic and potential energy

Free Response Question

Two blocks are on a frictionless, horizontal surface. Block A has a mass of 2.0 kg and is moving to the right with a speed of 5.0 m/s. Block B has a mass of 3.0 kg and is initially at rest. The blocks collide. After the collision, Block A moves to the left with a speed of 1.0 m/s.

(a) Calculate the momentum of Block A before the collision. (2 points) (b) Calculate the momentum of Block A after the collision. (2 points) (c) Calculate the momentum of Block B after the collision. (2 points) (d) Calculate the velocity of Block B after the collision. (2 points) (e) Is the collision elastic or inelastic? Justify your answer. (2 points)

Scoring Breakdown:

(a) Momentum of Block A before collision: * $\vec{p}_{A} = m_A \vec{v}_{A} = (2.0 \text{ kg})(5.0 \text{ m/s}) = 10 \text{ kg m/s}$ to the right. (2 points) * 1 point for correct formula/setup * 1 point for correct answer with units

(b) Momentum of Block A after collision: * $\vec{p}_{A}' = m_A \vec{v}_{A}' = (2.0 \text{ kg})(-1.0 \text{ m/s}) = -2.0 \text{ kg m/s}$ (or 2.0 kg m/s to the left). (2 points) * 1 point for correct formula/setup * 1 point for correct answer with units

(c) Momentum of Block B after collision: * Conservation of momentum: $\vec{p}_{A} + \vec{p}_{B} = \vec{p}_{A}' + \vec{p}_{B}'$ * $10 + 0 = -2 + \vec{p}_{B}'$ => $\vec{p}_{B}' = 12 \text{ kg m/s}$ to the right (2 points) * 1 point for correct application of conservation of momentum * 1 point for correct answer with units

(d) Velocity of Block B after collision: * $\vec{v}_{B}' = \frac{\vec{p}_{B}'}{m_B} = \frac{12 \text{ kg m/s}}{3.0 \text{ kg}} = 4.0 \text{ m/s}$ to the right. (2 points) * 1 point for correct formula/setup * 1 point for correct answer with units

(e) Type of collision: * Calculate initial kinetic energy: $KE_i = \frac{1}{2} m_A v_A^2 = \frac{1}{2} (2.0)(5.0)^2 = 25 \text{ J}$ * Calculate final kinetic energy: $KE_f = \frac{1}{2} m_A v_A'^2 + \frac{1}{2} m_B v_B'^2 = \frac{1}{2} (2.0)(1.0)^2 + \frac{1}{2} (3.0)(4.0)^2 = 1 + 24 = 25 \text{ J}$ * Since $KE_i = KE_f$ , the collision is elastic. (2 points) * 1 point for correct calculation of initial and final KE * 1 point for correct conclusion with justification