Conservation of Momentum Overview

• Topic: Conservation of Momentum in the context of physics, focusing on the principle governing collisions.

Law of Conservation of Momentum

• Definition: In a closed (isolated) system with no external forces, the total momentum before a collision is equal to the total momentum after the collision.
• Mathematical Representation: $\text{TOTAL MOMENTUM BEFORE} = \text{TOTAL MOMENTUM AFTER}$$M<em>1V</em>1 + M<em>2V</em>2 = M<em>1V</em>1' + M<em>2V</em>2'$
  • Where:
  • $M1$, $M2$ are the masses of the objects involved.
  • $V1$, $V2$ are the initial velocities of the objects.
  • $V1'$, $V2'$ are the final velocities of the objects after the collision.

Sample Problem

• Problem Statement: An 11 kg stone moving at 33 m/s strikes a second stone at rest. After the collision, the 11 kg stone moves with a velocity of 13 m/s and the second stone moves with a velocity of 8 m/s.
• Task: What is the mass of the second stone?
• Known Values:
  • Mass of Stone 1 ($M_1$) = 11 kg
  • Initial Velocity of Stone 1 ($V_1$) = 33 m/s
  • Final Velocity of Stone 1 ($V_1'$) = 13 m/s
  • Final Velocity of Stone 2 ($V_2'$) = 8 m/s

Formula & Solution

• Formula used:
  $P = M \cdot V$
• Total momentum before collision:
  $P<em>{\text{before}} = m</em>{\text{stone1}} \cdot v<em>{\text{stone1}} + m</em>{\text{stone2}} \cdot v<em>{\text{stone2}} = 11 \cdot 33 + m</em>{\text{stone2}} \cdot 0$
• Total momentum after collision:
  $P<em>{\text{after}} = m</em>{\text{stone1}} \cdot v<em>{\text{stone1}}' + m</em>{\text{stone2}} \cdot v<em>{\text{stone2}}'$$= 11 \cdot 13 + m</em>{\text{stone2}} \cdot 8$
• Equating the two:
  $11 \cdot 33 + 0 = 11 \cdot 13 + m_{\text{stone2}} \cdot 8$
• Solving gives:
  $363 = 143 + 8m<em>{\text{stone2}}$$220 = 8m</em>{\text{stone2}}$
  $m_{\text{stone2}} = 27.5 \text{ kg}$
• Result: Mass of the second stone is 27.5 kg.

Types of Collision

Elastic Collision

• Definition: In an elastic collision, the total kinetic energy of the system remains constant, and the colliding objects rebound off each other without losing kinetic energy.
• Characteristics:
  • Objects bounce apart.
  • No kinetic energy is transformed into other forms of energy.

Inelastic Collision

• Definition: In inelastic collisions, the total kinetic energy is not conserved and is transformed into other forms of energy (e.g., heat, sound).
• Characteristics:
  • Objects may stick together or deform after the collision.
  • Perfectly inelastic collisions are a subset where colliding objects stick together post-collision.

Conservation of Momentum Experiment

• Overview:
  • Procedure involves a simple system of two balls colliding.
  • Principle established: The total momentum before the collision equals the total momentum after the collision.

Group Presentation Assignments

• Group 1: Part 1 Elastic Collision (Simulation 1 & questions 1-3)
• Group 2: Part 1 Elastic Collision (Simulation 2 & questions 4-7)
• Group 3: Part 2 Inelastic Collision (Simulation 3 & questions 8-10)
• Group 4: Part 2 Inelastic Collision (Simulation 3 & questions 8-10)

The "Great Escape" Collision Scenario

• Background: Two astronauts in zero gravity, A (80 kg) and B (120 kg), pushing off against each other to move apart, both initially at rest.
• Initial Conditions:
  • Both at rest: $v = 0$
  • Astronaut A moves to the left at 3.0 m/s after the push.

Learner Tasks

1. Recoil Calculation:
   • Calculate the final velocity of Astronaut B using conservation of momentum.
2. Directional Logic:
   • Explain why Astronaut B’s velocity must be opposite to Astronaut A's using vectors.
3. Force Inquiry:
   • Determine which astronaut felt a greater force during the 0.5 seconds of pushing, referencing Newton’s 3rd Law.

Given & Missing Values for Astronauts

Application of Momentum Equation

• Utilizing the momentum formula:
  $P = M \cdot V$
• Before Collision:
  $(80 kg)(0 m/s) + (120 kg)(0 m/s) = 0$
• After Collision:
  $(80 kg)(-3.0 m/s) + (120 kg)(v_B) = 0$
• Rearranging:
  $-240 + 120v<em>B = 0$$120v</em>B = 240$
  $v_B = 2.0 \text{ m/s}$
• Result: Astronaut B moves at 2.0 m/s to the right.

Directional Logic Explanation

• Explanation: Astronaut B’s movement is opposite to Astronaut A’s to comply with the conservation of momentum principles and directional vectors.

Force Inquiry Discussion

• According to Newton’s 3rd Law, forces experienced are equal in magnitude but opposite in direction.
• Since the duration of the push is equal for both astronauts (0.5 s), the conclusion is that neither astronaut felt a greater force.

Quiz Questions

1. When a moving ball hits a stationary ball of the same mass in a Newton's cradle, what usually happens?

   • A: Both balls stop moving.
   • B: The first ball stops, and the second ball moves forward.
   • C: Both balls move backward.
   • D: The second ball remains at rest.

2. A 2 kg ball moves at 4 m/s toward a 3 kg ball at rest. What is the total momentum of the system before collision?

   • A: 6 kg·m/s
   • B: 8 kg·m/s
   • C: 12 kg·m/s
   • D: 14 kg·m/s

3. Two objects collide in a closed system. The total momentum before collision is 15 kg·m/s. What is the total momentum after collision?

   • A: 0 kg·m/s
   • B: 7.5 kg·m/s
   • C: 15 kg·m/s
   • D: It depends on the masses only.

4. A 1 kg cart moving at 6 m/s collides with a 2 kg cart at rest. After collision, the 1 kg cart moves backward at 2 m/s. What is the velocity of the 2 kg cart?

   • A: 2 m/s
   • B: 3 m/s
   • C: 4 m/s
   • D: 5 m/s

Quiz Solutions

1. Solution to Question 2:

   • Calculations:
     $\text{Total Momentum} = (2 \text{ kg}) \times (4 \text{ m/s}) + (3 \text{ kg}) \times (0 \text{ m/s})$
     = 8 \text{ kg·m/s}
   • Correct Answer: B (8 kg·m/s)

2. Solution for Question 4:

   • Applying momentum conservation:
     $mv<em>1 = mv</em>2$
   • Setup:
     $(1 \text{ kg}) \cdot (6 \text{ m/s}) + (2 \text{ kg}) \cdot (0 \text{ m/s}) = (1 \text{ kg}) \cdot (-2 \text{ m/s}) + (2 \text{ kg}) \cdot v<em>2$ 6 \text{ kg·m/s} + 0 = -2 \text{ kg·m/s} + 2 \text{ kg} \cdot v2
     6 \text{ kg·m/s} + 2 \text{ kg·m/s} = 2 \text{ kg} \cdot v2 8 \text{ kg·m/s} = 2 \text{ kg} \cdot v2
     $v_2 = 4 \text{ m/s}$
   • Correct Answer: C (4 m/s)

3. Question about closed system example:

   • A: Soccer ball slowing down due to friction on grass
   • B: Two ice skaters pushing off one another on a frictionless surface
   • C: A car accelerating due to its engine
   • D: A falling object undergoing air resistance

Practical Applications

• Concept of momentum in vehicle bumpers: Understanding how conservation of momentum applies in crash safety designs.
  • Notable aspects involve mass and velocity considerations during collisions to ensure safety.

Conclusion

• Final Note: Importance of understanding the Conservation of Momentum in practical scenarios and physics applications.