#AP Physics C: Mechanics - Momentum Review 🚀

Hey! Let's get you totally prepped for the exam. We're diving into momentum, a core concept that ties into so much of what we've learned. Remember, momentum is all about how much 'oomph' an object has when it's moving. Let's break it down and make sure you're ready to rock this!

#Conservation of Linear Momentum

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• Think of a system of objects as a single entity with a collective center-of-mass velocity. It's like finding the average motion of the whole group. 🏃‍♂️
• Formula: $$\vec{v}_{\mathrm{cm}}=\frac{\sum \vec{p}_{i}}{\sum m_{i}}=\frac{\sum\left(m_{i} \vec{v}_{i}\right)}{\sum m_{i}}$$
• This velocity stays constant when no net external force acts on the system. It's like a group of friends moving together without anyone pushing them around from the outside.

#Sum of Momenta

• The total momentum of a system is simply the sum of all individual momenta. It's like adding up all the 'oomph' in the system.

#Changes in Momentum

• In an isolated system, momentum changes are always balanced. If one object gains momentum, another must lose an equal amount. It's a perfect give-and-take! 🤝
• This is directly related to Newton's third law: every action has an equal and opposite reaction. The impulse one object exerts on another is met with an equal and opposite reaction impulse.
• Key Idea: Choosing the right system boundaries is crucial. A smart choice can make the total momentum constant.
• Impulse-Momentum Theorem: Any change in a system's total momentum equals the impulse exerted on it from outside the system. It's like giving the system a push (impulse) which changes its motion (momentum).
• Formula: $$\vec{J}=\Delta \vec{p}$$

#Velocity Before and After Collisions

• Conservation of momentum is your best friend for finding velocities right before and after collisions or explosions. It's like predicting the aftermath of a car crash or the spread of an explosion. 💥

#System Selection for Momentum

#Conservation in Interactions

• Momentum is always conserved, no matter what kind of interaction is happening. Whether it's a collision, explosion, or anything in between, momentum remains constant. It's a universal rule!

#Zero Net External Force

• If you choose a system with no net external force, the total momentum stays constant. It's like choosing a group of objects that aren't being pushed around by anything outside of the group.

#Nonzero Net External Force

• If there's a net external force acting on your system, momentum can transfer between the system and its surroundings. It's like a system being pushed or pulled by something outside of it.
• The amount of momentum transferred is equal to the impulse from the environment. It's like the push or pull from the outside directly affects the system's motion.

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• On the exam, expect to quantitatively analyze 1D and 2D collisions and interactions. 3D collisions may appear for qualitative analysis only.

#Final Exam Focus

• Highest Priority Topics:
  • Conservation of momentum in 1D and 2D collisions
  • Understanding the impulse-momentum theorem
  • Correctly selecting systems for analysis
  • Center of mass calculations
• Common Question Types:
  • Multiple-choice questions involving simple collisions and explosions
  • Free-response questions requiring detailed analysis of complex collision scenarios
  • Questions that combine momentum with energy conservation
• Time Management Tips:
  • Quickly identify the system and whether external forces are present
  • Set up equations using conservation of momentum
  • Check your work for vector directions and units
• Common Pitfalls:
  • Forgetting that momentum is a vector
  • Incorrectly applying conservation principles to non-isolated systems
  • Making algebraic errors in calculations

Practice Question

#Multiple Choice Questions

1. Two carts of masses m and 2m are at rest on a horizontal frictionless surface. A compressed spring is placed between them. When the spring is released, the cart of mass m moves to the left with a speed of v. What is the speed of the cart of mass 2m? (A) v/4 (B) v/2 (C) v (D) 2v

2. A ball of mass m moving horizontally with speed v collides with a stationary ball of mass 2m. After the collision, the two balls stick together. What is the speed of the combined mass after the collision? (A) v/3 (B) v/2 (C) v (D) 2v/3

#Free Response Question

A 2.0 kg block is sliding on a frictionless horizontal surface with a velocity of 5.0 m/s to the right. It collides with a 3.0 kg block that is initially at rest. After the collision, the 2.0 kg block moves to the left with a velocity of 1.0 m/s.

(a) Calculate the velocity of the 3.0 kg block after the collision. (b) Calculate the change in the total kinetic energy of the system as a result of the collision. (c) Is this collision elastic or inelastic? Justify your answer.

Scoring Guide:

(a) Calculate the velocity of the 3.0 kg block after the collision. (4 points)

• 1 point: Correctly stating the conservation of momentum equation: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$
• 1 point: Plugging in the correct values: $(2.0)(5.0) + (3.0)(0) = (2.0)(-1.0) + (3.0)v_{2f}$
• 1 point: Correctly solving for $v_{2f}$: $10 = -2 + 3v_{2f}$
• 1 point: Correct answer with units: $v_{2f} = 4.0 m/s$

(b) Calculate the change in the total kinetic energy of the system as a result of the collision. (3 points)

• 1 point: Calculating the initial kinetic energy: $KE_i = \frac{1}{2}(2.0)(5.0)^2 = 25 J$
• 1 point: Calculating the final kinetic energy: $KE_f = \frac{1}{2}(2.0)(-1.0)^2 + \frac{1}{2}(3.0)(4.0)^2 = 1 + 24 = 25J$
• 1 point: Calculating the change in kinetic energy: $\Delta KE = KE_f - KE_i = 25 - 25 = 0 J$

(c) Is this collision elastic or inelastic? Justify your answer. (2 points)

• 1 point: Correctly identifying the collision as elastic.
• 1 point: Justification: In an elastic collision, kinetic energy is conserved. $\Delta KE = 0$

Remember, you've got this! Keep practicing, stay confident, and you'll do great on the exam!