Velocity of a particle of mass `2 kg` change from `vecv_(1) =-2hati-2hatjm/s` to `vecv_(2)=(hati-hatj)m//s` after colliding with as plane surface.

Options

A. The angle made by the plane surface with the positive x-axis is `tan^(-1)(1/3)`
B. The angle made by the plane surface with the positive `x`-axis is `tan^(-1)(1/3)`
C. The direction of change in momentum makes an angle `tan^(-1)(1/3)` with the positive `x`-axis.
D. The direction of change in momentum makes an angle `90^(@) + tan^(-1)(1/3)` with the plane surface.

Explanation

To solve the problem, we need to analyze the change in velocity of a particle after it collides with a plane surface. We will calculate the change in momentum and determine the angles involved.

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities:**
- Initial velocity \( \vec{v_1} = -2 \hat{i} - 2 \hat{j} \, \text{m/s} \)
- Final velocity \( \vec{v_2} = \hat{i} - \hat{j} \, \text{m/s} \)

2. **Calculate Change in Velocity:**
- Change in velocity \( \Delta \vec{v} = \vec{v_2} - \vec{v_1} \)
- Substitute the values:
\[ \Delta \vec{v} = (\hat{i} - \hat{j}) - (-2 \hat{i} - 2 \hat{j}) = \hat{i} - \hat{j} + 2 \hat{i} + 2 \hat{j} \]
- Simplifying gives:
\[ \Delta \vec{v} = (1 + 2) \hat{i} + (-1 + 2) \hat{j} = 3 \hat{i} + 1 \hat{j} \]

3. **Calculate Change in Momentum:**
- Mass \( m = 2 \, \text{kg} \)
- Change in momentum \( \Delta \vec{p} = m \Delta \vec{v} \)
- Substitute the values:
\[ \Delta \vec{p} = 2 \cdot (3 \hat{i} + 1 \hat{j}) = 6 \hat{i} + 2 \hat{j} \]

4. **Determine the Angle of Change in Momentum:**
- The angle \( \theta \) made by the change in momentum with the positive x-axis can be calculated using the tangent function:
\[ \tan \theta = \frac{\text{y-component}}{\text{x-component}} = \frac{2}{6} = \frac{1}{3} \]
- Therefore,
\[ \theta = \tan^{-1} \left( \frac{1}{3} \right) \]

5. **Analyze the Plane Surface Angle:**
- The angle made by the plane surface with the positive x-axis is \( 90^\circ + \theta \) because the change in momentum is perpendicular to the plane surface.
- Hence, the angle made by the plane surface is:
\[ 90^\circ + \tan^{-1} \left( \frac{1}{3} \right) \]

6. **Conclusion:**
- The direction of change in momentum makes an angle of \( \tan^{-1} \left( \frac{1}{3} \right) \) with the positive x-axis.
- The angle made by the plane surface with the positive x-axis is \( 90^\circ + \tan^{-1} \left( \frac{1}{3} \right) \).

### Final Answer:
The correct statement is that the direction of change in momentum makes an angle of \( \tan^{-1} \left( \frac{1}{3} \right) \) with the positive x-axis.

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