#AP Physics 1: Momentum & Impulse - The Night Before

Hey! Let's get you prepped and confident for tomorrow's exam. We're going to break down momentum and impulse, making sure everything sticks. Let's do this!

#1. Momentum: The Motion Measurer

#What is Momentum?

• Definition: Momentum ($\vec{p}$) is a measure of how much 'oomph' an object has in its motion. It's the product of an object's mass (m) and its velocity ($\vec{v}$).
  $$\vec{p} = m\vec{v}$$
• Key Point: Momentum is a vector, meaning it has both magnitude and direction. The direction of momentum is the same as the direction of the velocity.
• Units: kg⋅m/s

#Total Momentum

• To find the total momentum of a system, simply add up the individual momentums of each object. $$\vec{p}_{total} = \vec{p}_1 + \vec{p}_2 + \vec{p}_3 + ...$$

#Linear vs. Angular Momentum

• Linear Momentum: What we're focusing on here – for objects moving in a straight line.
• Angular Momentum: For rotating objects (you'll see this specified in questions if it's relevant).

#Example Problems

Example Problem 1:

A car (mass 1000 kg) is traveling at 30 m/s and then stops. What's the change in momentum?

• Initial momentum: $p_i = (1000 \text{ kg}) \times (30 \text{ m/s}) = 30000 \text{ kg m/s}$
• Final momentum: $p_f = (1000 \text{ kg}) \times (0 \text{ m/s}) = 0 \text{ kg m/s}$
• Change in momentum: $Δp = p_f - p_i = 0 - 30000 = -30000 \text{ kg m/s}$ (The negative sign indicates the direction of the change)

Example Problem 2:

A 0.2 kg ball is thrown at 20 m/s. What's its momentum?

• $p = (0.2 \text{ kg}) \times (20 \text{ m/s}) = 4 \text{ kg m/s}$

Example Problem 3:

A 2000 kg car at 10 m/s hits a stationary 3000 kg truck. They move together at 5 m/s. What's the total momentum before and after?

• Before: $p_{car} = (2000 \text{ kg}) \times (10 \text{ m/s}) = 20000 \text{ kg m/s}$, $p_{truck} = 0$. Total before: $20000 \text{ kg m/s}$
• After: $p_{total} = (2000 + 3000 \text{ kg}) \times (5 \text{ m/s}) = 25000 \text{ kg m/s}$

#2. Impulse: The Force of Change

#What is Impulse?

• Definition: Impulse ($\vec{J}$) is the change in momentum of an object. It's also the product of the average force ($\vec{F}_{avg}$) acting on an object and the time interval ($\Delta t$) during which the force acts. $$\vec{J} = \vec{F}_{avg} \Delta t = \Delta \vec{p}$$
• Units: N⋅s (which is equivalent to kg⋅m/s)
• Key Point: Impulse is a vector and points in the same direction as the average force.

#Impulse and Force

• Impulse is the area under a force vs. time graph. 📈

#Rocket Thrust

• Rocket thrust is related to the rate of change of mass and the velocity of expelled fuel. If fuel is expelled with speed $v$ at rate $\frac{\Delta m}{\Delta t}$, the rocket experiences a thrust force.

#Example Problems

Example Problem 1:

A 6 kg bowling ball going 10 m/s stops in 0.2 s. What's the impulse?

• Change in velocity: $Δv = 0 - 10 = -10 \text{ m/s}$
• Acceleration: $a = \frac{Δv}{Δt} = \frac{-10}{0.2} = -50 \text{ m/s}^2$
• Force: $F = ma = 6 \times -50 = -300 \text{ N}$
• Impulse: $J = FΔt = -300 \times 0.2 = -60 \text{ Ns}$

Example Problem 2:

A 0.5 kg baseball going 50 m/s stops in 0.01 s. What's the impulse?

• Change in velocity: $Δv = 0 - 50 = -50 \text{ m/s}$
• Acceleration: $a = \frac{Δv}{Δt} = \frac{-50}{0.01} = -5000 \text{ m/s}^2$
• Force: $F = ma = 0.5 \times -5000 = -2500 \text{ N}$
• Impulse: $J = FΔt = -2500 \times 0.01 = -25 \text{ Ns}$

Example Problem 3:

A 0.045 kg golf ball goes from rest to 100 m/s in 0.005 s. What's the impulse?

• Change in velocity: $Δv = 100 - 0 = 100 \text{ m/s}$
• Acceleration: $a = \frac{Δv}{Δt} = \frac{100}{0.005} = 20000 \text{ m/s}^2$
• Force: $F = ma = 0.045 \times 20000 = 900 \text{ N}$
• Impulse: $J = FΔt = 900 \times 0.005 = 4.5 \text{ Ns}$

#3. Conservation of Momentum

#The Big Idea

• In a closed system (no external forces), the total momentum remains constant. This is a HUGE concept for collisions and explosions. 💥

• Equation: $\vec{p}_{initial} = \vec{p}_{final}$

#Types of Collisions

• Elastic Collisions: Kinetic energy is conserved (objects bounce off each other). Momentum is always conserved in all collisions.
• Inelastic Collisions: Kinetic energy is NOT conserved (objects stick together or deform). Momentum is still conserved.

#4. Final Exam Focus

#High-Priority Topics

• Momentum and Impulse: Be ready to calculate momentum, impulse, and changes in momentum.
• Conservation of Momentum: Master applying this principle to collision problems (both elastic and inelastic).
• Force vs. Time Graphs: Be able to find impulse from these graphs.

Collisions: These are a staple on the AP exam. Practice problems involving both elastic and inelastic collisions. Remember to apply the conservation of momentum principle.

#Common Question Types

• Multiple Choice: Conceptual questions about momentum and impulse, calculations of momentum/impulse, and graph interpretation.
• Free Response: Multi-step problems involving collisions, often requiring you to use conservation of momentum and/or impulse-momentum theorem.

#Last-Minute Tips

• Time Management: Don't spend too much time on one question. If you're stuck, move on and come back later.
• Units: Always include units in your answers.
• Direction: Pay close attention to the direction of velocities and forces.
• Draw Diagrams: Visualizing the problem can help you set up the equations correctly.

#5. Practice Questions

Practice Question

Multiple Choice Questions:

1. A 2 kg object moving at 3 m/s collides with a 1 kg object at rest. They stick together. What is their final velocity? (A) 1 m/s (B) 2 m/s (C) 3 m/s (D) 4 m/s

2. A force of 10 N acts on an object for 2 seconds. What is the impulse? (A) 5 Ns (B) 10 Ns (C) 20 Ns (D) 40 Ns

3. A ball is dropped and bounces off the floor. Which of the following is true about the momentum of the ball-Earth system? (A) The momentum of the system increases. (B) The momentum of the system decreases. (C) The momentum of the system remains constant. (D) The momentum of the system changes direction but remains constant in magnitude.

Free Response Question:

A 0.5 kg cart is moving on a frictionless horizontal track with a velocity of 2 m/s to the right. It collides with a 1.0 kg cart initially at rest. After the collision, the 0.5 kg cart moves to the left with a velocity of 0.5 m/s.

(a) Calculate the momentum of the 0.5 kg cart before the collision. (2 points) (b) Calculate the momentum of the 0.5 kg cart after the collision. (2 points) (c) Calculate the velocity of the 1.0 kg cart after the collision. (3 points) (d) Is the collision elastic or inelastic? Justify your answer. (3 points)

Scoring Breakdown:

(a) Momentum before collision = mass * velocity = (0.5 kg) * (2 m/s) = 1 kg m/s (2 points: 1 for correct formula, 1 for correct answer) (b) Momentum after collision = mass * velocity = (0.5 kg) * (-0.5 m/s) = -0.25 kg m/s (2 points: 1 for correct formula, 1 for correct answer) (c) Use conservation of momentum: Initial total momentum = Final total momentum 1 kg m/s + 0 = -0.25 kg m/s + (1 kg) * v v = 1.25 m/s (3 points: 1 for applying conservation of momentum, 1 for correct equation, 1 for correct answer) (d) Calculate kinetic energy before and after collision. Initial KE = 0.5 * 0.5 * 2^2 = 1 J Final KE = 0.5 * 0.5 * 0.5^2 + 0.5 * 1 * 1.25^2 = 0.125 + 0.78 = 0.905 J Since kinetic energy is not conserved, the collision is inelastic. (3 points: 1 for correct initial KE, 1 for correct final KE, 1 for correct justification)

Alright, you've got this! Review these notes, take a deep breath, and go ace that exam! 🚀