Ball B, moving in the positive direction of an x axis at speed v, collides with stationary ball A at the origin. A and B have different masses. After the collision, B moves in the negative direction of the y axis at speed v/3. In what direction does A move, as an angle with respect to the x axis?

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Chapter1: Units, Trigonometry. And Vectors
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### Problem Statement

Ball B, moving in the positive direction of an x-axis at speed \( v \), collides with stationary ball A at the origin. A and B have different masses. After the collision, B moves in the negative direction of the y-axis at speed \( v/3 \). In what direction does A move, as an angle with respect to the x-axis?

### Approach

To solve this problem, we'll use the principles of conservation of momentum. Since there's no external force acting on the system, the total momentum before and after the collision must be the same.

1. **Initial Momentum:**

   - Ball B: Momentum in x-direction = \( m_B \cdot v \)
   - Ball A: Momentum = 0 (since it's stationary)
   
   Total Initial Momentum = \( m_B \cdot v \)

2. **Final Momentum:**

   - Ball B: 
     - Momentum in x-direction = 0 (since it moves along y-axis)
     - Momentum in y-direction = \( m_B \cdot (-v/3) \)
   
   - Ball A:
     - Let \( v_A \) be the final speed of Ball A and \(\theta\) be the angle with respect to the x-axis.
     - Momentum in x-direction = \( m_A \cdot v_A \cdot \cos(\theta) \)
     - Momentum in y-direction = \( m_A \cdot v_A \cdot \sin(\theta) \)
   
3. **Conservation Equations:**

   a. **In x-direction:**
   \[
   m_B \cdot v = m_A \cdot v_A \cdot \cos(\theta)
   \]

   b. **In y-direction:**
   \[
   0 = m_A \cdot v_A \cdot \sin(\theta) + m_B \cdot \left(-v/3\right)
   \]

### Solving the Equations
From the above equations, you can determine both \(\theta\) and \(v_A\), using algebraic manipulation and substitution.

### Conclusion
By solving these equations, you will find the angle \(\theta\) that Ball A makes with the x-axis, indicating the direction of its movement post-collision.

**Note:** Please ensure all steps are clearly shown with any necessary assumptions or approximations highlighted.

### Reminder
Transcribed Image Text:### Problem Statement Ball B, moving in the positive direction of an x-axis at speed \( v \), collides with stationary ball A at the origin. A and B have different masses. After the collision, B moves in the negative direction of the y-axis at speed \( v/3 \). In what direction does A move, as an angle with respect to the x-axis? ### Approach To solve this problem, we'll use the principles of conservation of momentum. Since there's no external force acting on the system, the total momentum before and after the collision must be the same. 1. **Initial Momentum:** - Ball B: Momentum in x-direction = \( m_B \cdot v \) - Ball A: Momentum = 0 (since it's stationary) Total Initial Momentum = \( m_B \cdot v \) 2. **Final Momentum:** - Ball B: - Momentum in x-direction = 0 (since it moves along y-axis) - Momentum in y-direction = \( m_B \cdot (-v/3) \) - Ball A: - Let \( v_A \) be the final speed of Ball A and \(\theta\) be the angle with respect to the x-axis. - Momentum in x-direction = \( m_A \cdot v_A \cdot \cos(\theta) \) - Momentum in y-direction = \( m_A \cdot v_A \cdot \sin(\theta) \) 3. **Conservation Equations:** a. **In x-direction:** \[ m_B \cdot v = m_A \cdot v_A \cdot \cos(\theta) \] b. **In y-direction:** \[ 0 = m_A \cdot v_A \cdot \sin(\theta) + m_B \cdot \left(-v/3\right) \] ### Solving the Equations From the above equations, you can determine both \(\theta\) and \(v_A\), using algebraic manipulation and substitution. ### Conclusion By solving these equations, you will find the angle \(\theta\) that Ball A makes with the x-axis, indicating the direction of its movement post-collision. **Note:** Please ensure all steps are clearly shown with any necessary assumptions or approximations highlighted. ### Reminder
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