A ball of mass 1 kg moving with velocity `vecv = 3hati + 4 hatj` collides with a wall and after collision velocity of the ball is `-2 hati + 6 hatj` . Which of the following unit vectors is perpendicular to wall?

Options

A. `1/5 (-3hati - 4hatj)`
B. `1/(sqrt(29))(5 hati + 2 hatj)`
C. `1/(sqrt(40)) -(2 hati + 6 hatj)`
D. `1/(sqrt(40))(5 hati - 2hatj)`

Explanation

To find the unit vector that is perpendicular to the wall after a ball collides with it, we can follow these steps:

### Step 1: Identify the initial and final velocities of the ball
The initial velocity of the ball before the collision is given as:
\[ \vec{v_i} = 3 \hat{i} + 4 \hat{j} \]
The final velocity of the ball after the collision is:
\[ \vec{v_f} = -2 \hat{i} + 6 \hat{j} \]

### Step 2: Determine the change in velocity
The change in velocity (\(\Delta \vec{v}\)) can be calculated as:
\[ \Delta \vec{v} = \vec{v_f} - \vec{v_i} = (-2 \hat{i} + 6 \hat{j}) - (3 \hat{i} + 4 \hat{j}) = (-2 - 3) \hat{i} + (6 - 4) \hat{j} = -5 \hat{i} + 2 \hat{j} \]

### Step 3: Analyze the components of the velocity
The component of the velocity that is parallel to the wall will remain unchanged, while the component perpendicular to the wall will change direction.

### Step 4: Define the unit normal vector to the wall
Let the unit vector perpendicular to the wall be:
\[ \hat{n} = x \hat{i} + y \hat{j} \]
Since it is a unit vector, it must satisfy:
\[ \sqrt{x^2 + y^2} = 1 \quad \Rightarrow \quad x^2 + y^2 = 1 \]

### Step 5: Set up the equations based on the reflection
The component of the initial velocity perpendicular to the wall will change direction, while the parallel component will remain the same. The reflection principle gives us:
\[ \text{Initial component along } \hat{n} = \text{Final component along } \hat{n} \]
This can be expressed as:
\[ \vec{v_i} \cdot \hat{n} = -\vec{v_f} \cdot \hat{n} \]

### Step 6: Calculate the dot products
Calculating the dot products:
\[ \vec{v_i} \cdot \hat{n} = (3 \hat{i} + 4 \hat{j}) \cdot (x \hat{i} + y \hat{j}) = 3x + 4y \]
\[ -\vec{v_f} \cdot \hat{n} = -(-2 \hat{i} + 6 \hat{j}) \cdot (x \hat{i} + y \hat{j}) = (2 \hat{i} - 6 \hat{j}) \cdot (x \hat{i} + y \hat{j}) = 2x - 6y \]

### Step 7: Set the equations equal
Setting the two expressions equal gives us:
\[ 3x + 4y = 2x - 6y \]
Rearranging this gives:
\[ x + 10y = 0 \quad \Rightarrow \quad x = -10y \]

### Step 8: Substitute into the unit vector equation
Substituting \(x = -10y\) into the unit vector equation:
\[ (-10y)^2 + y^2 = 1 \quad \Rightarrow \quad 100y^2 + y^2 = 1 \quad \Rightarrow \quad 101y^2 = 1 \quad \Rightarrow \quad y^2 = \frac{1}{101} \quad \Rightarrow \quad y = \pm \frac{1}{\sqrt{101}} \]
Thus,
\[ x = -10y = \mp \frac{10}{\sqrt{101}} \]

### Step 9: Form the unit vector
The unit vector perpendicular to the wall is:
\[ \hat{n} = \left(-\frac{10}{\sqrt{101}} \hat{i} + \frac{1}{\sqrt{101}} \hat{j}\right) \text{ or } \left(\frac{10}{\sqrt{101}} \hat{i} - \frac{1}{\sqrt{101}} \hat{j}\right) \]

### Step 10: Conclusion
Thus, the unit vector perpendicular to the wall can be expressed as:
\[ \hat{n} = \frac{1}{\sqrt{101}} \left(-10 \hat{i} + 1 \hat{j}\right) \text{ or } \frac{1}{\sqrt{101}} \left(10 \hat{i} - 1 \hat{j}\right) \]