#AP Physics 1: Momentum - The Night Before 🚀

Hey! Let's get you totally prepped for the AP Physics 1 exam. We're going to break down momentum, systems, and how it all connects. You've got this! 💪

#Systems: Open vs. Closed

Understanding systems is KEY for momentum problems. Let's clarify:

• Closed System: Think of it like a sealed box. No mass, energy, or charge can get in or out. Whatever's inside stays inside. 🔒

  • Key Point: Total momentum within a closed system is always conserved.

• Open System: Imagine a fish tank. Stuff can be exchanged with the outside world. Water, fish food, etc. can go in and out. 🐠

  • Key Point: Momentum is not necessarily conserved because the system can gain or lose momentum.

Image Credit: physics.usyd.edu

#Momentum of a System

Let's get into the nitty-gritty of momentum within a system:

• Internal Forces: These are forces within the system (like your hand pushing a cart). They always come in action-reaction pairs (Newton's 3rd Law).

  • Key Point: Internal forces always cancel each other out. They don't change the total momentum of the system. 💡

• External Forces: Forces acting on the system from outside (like friction or gravity). These do affect the total momentum.

• Net External Force and Momentum:

  • If the net external force is zero, the total momentum of the system is conserved. $p_{initial} = p_{final}$

  • If there is a net external force, the momentum of the system changes. $\Delta p = F_{net} \Delta t$

#Calculating Momentum

Remember the basic formula:

$$p = mv$$

Where:

• $p$ is momentum (kg⋅m/s)
• $m$ is mass (kg)
• $v$ is velocity (m/s)

Let's break down the example problems:

#Example Problem #1: Car and Passenger

• Car:
  • Mass ($m_c$) = 1000 kg
  • Velocity ($v_c$) = 50 m/s
  • Momentum ($p_c$) = $m_c v_c$ = 1000 kg * 50 m/s = 50000 kg⋅m/s
• Passenger:
  • Mass ($m_p$) = 75 kg
  • Velocity ($v_p$) = 50 m/s
  • Momentum ($p_p$) = $m_p v_p$ = 75 kg * 50 m/s = 3750 kg⋅m/s
• Total Momentum:
  • $p_{total}$ = $p_c$ + $p_p$ = 50000 kg⋅m/s + 3750 kg⋅m/s = 53750 kg⋅m/s

#Example Problem #2: Spaceship and Cargo

• Spaceship:
  • Mass ($m_s$) = 1000 kg
  • Velocity ($v_s$) = 50 m/s
  • Momentum ($p_s$) = $m_s v_s$ = 1000 kg * 50 m/s = 50000 kg⋅m/s
• Cargo:
  • Mass ($m_{cargo}$) = 500 kg
  • Velocity ($v_{cargo}$) = 25 m/s (relative to spaceship)
  • Momentum ($p_{cargo}$) = $m_{cargo} v_{cargo}$ = 500 kg * 25 m/s = 12500 kg⋅m/s
• Total Momentum:
  • $p_{total}$ = $p_s$ + $p_{cargo}$ = 50000 kg⋅m/s + 12500 kg⋅m/s = 62500 kg⋅m/s

#Example Problem #3: Train and Cars

• Train:
  • Mass ($m_{train}$) = 10000 kg
  • Velocity ($v_{train}$) = 50 m/s
  • Momentum ($p_{train}$) = $m_{train} v_{train}$ = 10000 kg * 50 m/s = 500000 kg⋅m/s
• Cars:
  • Mass per car ($m_{car}$) = 1000 kg
  • Velocity per car ($v_{car}$) = 50 m/s
  • Number of cars = 20
  • Total mass of cars ($m_{cars}$) = 20 * 1000 kg = 20000 kg
  • Total momentum of cars ($p_{cars}$) = $m_{cars} v_{car}$ = 20000 kg * 50 m/s = 1000000 kg⋅m/s
• Total Momentum:
  • $p_{total}$ = $p_{train}$ + $p_{cars}$ = 500000 kg⋅m/s + 1000000 kg⋅m/s = 1500000 kg⋅m/s

#Final Exam Focus

Okay, let's talk strategy for the exam:

• High-Priority Topics:

  • Conservation of Momentum in closed systems
  • Impulse and its relation to momentum change
  • Analyzing collisions (elastic and inelastic)
  • Understanding the difference between internal and external forces

• Common Question Types:

  • Multiple-choice questions testing conceptual understanding of momentum conservation
  • Free-response questions involving calculations of momentum and impulse in various scenarios
  • Questions that combine momentum with other concepts like energy and work

• Time Management:

  • Don't spend too long on a single problem. If you're stuck, move on and come back later.
  • Practice with timed tests to get a feel for the pacing.

• Common Pitfalls:

  • Forgetting to consider external forces
  • Mixing up mass and velocity units
  • Not treating momentum as a vector quantity

#

Practice Question

Practice Questions

Alright, let's solidify your understanding with some practice questions. These are designed to mimic what you'll see on the exam.

#Multiple Choice Questions

1. A 2 kg ball is moving to the right at 5 m/s. It collides head-on with a 3 kg ball moving to the left at 2 m/s. If the collision is perfectly elastic, what is the total momentum of the system after the collision? (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

2. A rocket expels hot gas to move forward. This is best explained by: (A) Conservation of energy (B) Conservation of momentum (C) Newton's First Law (D) Newton's Second Law

3. A system consists of two objects. Object A has a mass of 2 kg and is moving at 3 m/s, and object B has a mass of 3 kg and is at rest. If the two objects collide and stick together, what is the final velocity of the combined mass? (A) 1.2 m/s (B) 1.8 m/s (C) 2.0 m/s (D) 3.0 m/s

#Free Response Question

A 0.5 kg cart is moving on a frictionless track at 2 m/s towards a stationary 1.5 kg cart. After the collision, the 0.5 kg cart bounces back with a velocity of 0.5 m/s.

(a) What is the velocity of the 1.5 kg cart after the collision? (3 points) (b) Is the collision elastic or inelastic? Justify your answer. (3 points) (c) If the collision took 0.1 seconds, what was the average force exerted on the 0.5 kg cart during the collision? (3 points)

#Scoring Breakdown:

(a) Velocity of the 1.5 kg cart (3 points)

• 1 point: Correctly applying conservation of momentum: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$
• 1 point: Plugging in the correct values: $(0.5 kg)(2 m/s) + (1.5 kg)(0 m/s) = (0.5 kg)(-0.5 m/s) + (1.5 kg)v_{2f}$
• 1 point: Correctly solving for $v_{2f}$: $v_{2f} = 0.75 m/s$

(b) Elastic or Inelastic? (3 points)

• 1 point: Calculating the initial kinetic energy: $KE_i = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}(0.5 kg)(2 m/s)^2 + 0 = 1 J$
• 1 point: Calculating the final kinetic energy: $KE_f = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 = \frac{1}{2}(0.5 kg)(-0.5 m/s)^2 + \frac{1}{2}(1.5 kg)(0.75 m/s)^2 = 0.5 J$
• 1 point: Correctly stating that the collision is inelastic and justifying it by showing that kinetic energy is not conserved ( $KE_i$ ≠ $KE_f$)

(c) Average Force on the 0.5 kg cart (3 points)

• 1 point: Using the impulse-momentum theorem: $F\Delta t = \Delta p$
• 1 point: Calculating the change in momentum: $\Delta p = m(v_f - v_i) = (0.5 kg)(-0.5 m/s - 2 m/s) = -1.25 kg⋅m/s$
• 1 point: Solving for the force: $F = \frac{\Delta p}{\Delta t} = \frac{-1.25 kg⋅m/s}{0.1 s} = -12.5 N$

You've made it through! Remember to stay calm, read each question carefully, and trust your preparation. You've got this! Go get that 5! 🎉