Two-Dimensional Momentum: Vector Analysis and Component Methods

Introduction to Two-Dimensional Momentum

• Conceptual Shift from 1D to 2D:
  • Previously, momentum problems involved objects moving parallel to each other (e.g., side-to-side or east and west).
  • In 1D, vectors can be treated with simple addition or subtraction using positive and negative signs to indicate direction.
  • Example of 1D Math: If Object 1 has a momentum of $10\,\text{kg}\cdot\text{m/s}$ East and Object 2 has a momentum of $5\,\text{kg}\cdot\text{m/s}$ West, the resultant momentum ($P_R$) is simply $10 - 5 = 5\,\text{kg}\cdot\text{m/s}$.
  • In 2D, vectors are no longer parallel (e.g., perpendicular or at varying angles), meaning simple arithmetic is insufficient. Vector addition rules must be applied.

Collision at 90 Degrees (Workbook Page 224)

• General Collision Equation:

  • The conservation of momentum dictates that the total initial momentum equals the total final momentum.
  • For two objects: $P_1 + P_2 = P'$

• Problem Scenario:

  • A southerly moving object collides with an easterly moving object.
  • The objects stick together after the collision.

• Given Data:

  • Object 1 (Moving South):
    • Mass ($m_1$) = $4\,\text{kg}$
    • Velocity ($v_1$) = $2.8\,\text{m/s}$
    • Momentum ($P_1$) = $4 \times 2.8 = 11.2\,\text{kg}\cdot\text{m/s}$ [South]
  • Object 2 (Moving East):
    • Mass ($m_2$) = $6\,\text{kg}$
    • Velocity ($v_2$) = $3\,\text{m/s}$
    • Momentum ($P_2$) = $6 \times 3 = 18\,\text{kg}\cdot\text{m/s}$ [East]

• Mathematical Approach:

  • Since the vectors are perpendicular, they form a right-angled triangle when added tip-to-tail.
  • Finding Resultant Momentum ($P'$): Use the Pythagorean theorem.
    • $(P')^2 = (11.2)^2 + (18)^2$
    • $P' = \sqrt{125.44 + 324} = \sqrt{449.44}$
    • $P' \approx 21.2\,\text{kg}\cdot\text{m/s}$
  • Finding Final Velocity ($v'$):
    • Since the objects stick together, the total mass is $4\,\text{kg} + 6\,\text{kg} = 10\,\text{kg}$.
    • $v' = \frac{P'}{m_{total}} = \frac{21.2}{10} = 2.12\,\text{m/s}$
    • Rounded to significant figures: $2.1\,\text{m/s}$

• Determining Direction (Angle):

  • Use the tangent function based on the vector diagram.
  • $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{18}{11.2}$
  • $\theta = \tan^{-1}(\frac{18}{11.2}) \approx 58^{\circ}$
  • Directional Description: The angle is tilted from the South toward the East. Therefore, it is described as $58^{\circ}$ East of South.

Non-90 Degree Collisions and Component Method (Workbook Page 225)

• Complex Scenarios:

  • If a collision does not occur at a perfect 90-degree angle, the Pythagorean theorem cannot be used directly on the initial vectors.
  • The preferred method is to break all momentum vectors into horizontal ($x$) and vertical ($y$) components.

• General Equations for Components:

  • The conservation of momentum must hold true for both dimensions independently:
    1. $P_{1x} + P_{2x} = P_{1x}' + P_{2x}'$
    2. $P_{1y} + P_{2y} = P_{1y}' + P_{2y}'$

• Example Problem Analysis:

  • Initial Conditions:
    • Object 1 ($4\,\text{kg}$) is moving generally East.
    • Object 2 ($6.1\,\text{kg}$) is at rest ($P_2 = 0$).
  • Post-Collision (Prime) Conditions:
    • Object 1 ($P_1'$):
      • Mass = $4\,\text{kg}$, Velocity = $2.8\,\text{m/s}$.
      • $P_1' = 11.2\,\text{kg}\cdot\text{m/s}$ at $32^{\circ}$ North of East.
    • Object 2 ($P_2'$):
      • Mass = $6.1\,\text{kg}$, Velocity = $1.5\,\text{m/s}$.
      • $P_2' = 6.1 \times 1.5 = 9.15\,\text{kg}\cdot\text{m/s}$ at $41^{\circ}$.

• Step 1: Calculate X-Components (Yellow Vectors):

  • $P_{1x}' = 11.2 \times \cos(32^{\circ}) \approx 9.5$
  • $P_{2x}' = 9.15 \times \cos(41^{\circ}) \approx 6.91$
  • Total Initial X-Momentum ($P_{1x}$):
    • Since these two are parallel and in the same direction:
    • $P_{1x} = 9.5 + 6.91 = 16.4\,\text{kg}\cdot\text{m/s}$

• Step 2: Calculate Y-Components (Green Vectors):

  • $P_{1y}' = 11.2 \times \sin(32^{\circ}) \approx 5.94$ [Up/North]
  • $P_{2y}' = -(9.15 \times \sin(41^{\circ})) \approx -6.0$ [Down/South]
  • Total Initial Y-Momentum ($P_{1y}$):
    • $P_{1y} = 5.94 - 6.0 = -0.06\,\text{kg}\cdot\text{m/s}$

• Step 3: Solve for the Initial Vector ($P_1$):

  • Now that we have the aggregate $x$ and $y$ components of the initial state, we can use the Pythagorean theorem because $x$ and $y$ are perpendicular.
  • $(P_1)^2 = (16.4)^2 + (-0.06)^2$
  • $P_1 \approx 16.4\,\text{kg}\cdot\text{m/s}$
  • Initial Angle ($\theta$):
    • $\tan(\theta) = \frac{0.06}{16.4}$
    • $\theta \approx 0.21^{\circ}$
    • Direction: South of East (due to the negative sign in the y-component calculation).

Summary of Principles

• Vector Decomposition: You are essentially turning one vector equation into two equations (one for $x$, one for $y$).
• Independence of Dimensions: Calculations for the horizontal plane do not affect calculations for the vertical plane.
• Tip-to-Tail Visualization:
  • $x$-components (yellow arrows) add tip-to-tail to reach the final horizontal position of the resultant.
  • $y$-components (green arrows) add tip-to-tail to reach the final vertical position of the resultant.
• Final Step Requirement: Pythagorean theorem and trigonometric functions (tan) are only appropriate after components have been consolidated into single resultant $x$ and $y$ values.

Questions & Discussion

• Question from Student: Does my name look like a girl? / Did you post this?
• Mr. C4's Response: I don't care… let me think about that… don't care what you say. Yes, I already posted this one. You like the grid system better? Okay.
• Question/Comment regarding Login/Workbook: "Mr. C4 what lowercase all one word we're all one mr c4 no everyone else is joined so no space look there we go okay that's probably why"
• Mr. C4 Remark on Sleep: "I didn't get any sleep last night… could have been because I had some espresso before… it usually doesn't bother me… but it's a coffee definitely."

Assignments and Next Steps

• Reading: Review the example problems on Workbook pages 224 and 225.
• Homework Practice:
  • Attempt problems 1, 2, and 3 on page 228.
  • These involve 90-degree angles (the "simpler" type similar to Example 1).
  • Students may optionally attempt problems 4 and 5 if they feel confident with non-90-degree calculations.
• Upcoming Schedule: Mr. C4 intends to go over homework problems 1-3 the next class day and provide extra practice specifically for non-90-degree collisions, which are considered more difficult.