Assertion : Object A of mass collides elastically with another object B of same mass at rest and if found that A and B move perpendicular to each other after collision .
Reason : Collision between A and B is not head on .

Options

A. If both assertion and reason are corret and reason is a correct explanation of the assertion .
B. If both assertion and reason are correct but reason is ot the correct explanation of assertion .
C. If assertion is correct but reason is incorrect .
D. If both assertion and reason are incorrect .

Explanation

To solve the problem, we need to analyze the assertion and the reason provided.

### Step 1: Understanding the Assertion
The assertion states that object A of mass collides elastically with another object B of the same mass at rest, and after the collision, A and B move perpendicular to each other.

### Step 2: Understanding Elastic Collision
In an elastic collision, both momentum and kinetic energy are conserved. If two objects of the same mass collide elastically, the velocities after the collision can be determined based on their initial velocities.

### Step 3: Analyzing the Collision
1. **Initial Conditions**:
- Let the mass of object A be \( m \) and its initial velocity be \( u_1 \).
- Let the mass of object B be \( m \) (at rest), so its initial velocity \( u_2 = 0 \).

2. **Final Velocities**:
- After the collision, let the velocity of A be \( v_A \) and the velocity of B be \( v_B \).
- According to the problem, \( v_A \) and \( v_B \) are perpendicular to each other.

### Step 4: Conservation of Momentum
The conservation of momentum before and after the collision can be expressed as:
\[ m u_1 + m u_2 = m v_A + m v_B \]
Since \( u_2 = 0 \), this simplifies to:
\[ m u_1 = m v_A + m v_B \]
Dividing through by \( m \) gives:
\[ u_1 = v_A + v_B \]

### Step 5: Conservation of Kinetic Energy
The conservation of kinetic energy in an elastic collision can be expressed as:
\[ \frac{1}{2} m u_1^2 = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2 \]
Again, dividing through by \( \frac{1}{2} m \) gives:
\[ u_1^2 = v_A^2 + v_B^2 \]

### Step 6: Relationship Between Velocities
Since \( v_A \) and \( v_B \) are perpendicular, we can use the Pythagorean theorem:
\[ v_A^2 + v_B^2 = u_1^2 \]
This confirms that the velocities can satisfy both the momentum and kinetic energy conservation equations.

### Step 7: Conclusion
Since A and B move perpendicular to each other after the collision, it indicates that the collision is not head-on. In a head-on elastic collision, the velocities would simply exchange if the masses are equal, which does not happen here.

### Final Answer
- **Assertion**: True
- **Reason**: True
- The reason correctly explains the assertion.