Two balls `A` & `B` of mass `m_(1)` and `m_(2)` are kept on a horizontal smooth surface. A is given a velocity towards `B` so that they perform head on collision

Options

A. If `m_(1)=m_(2)` and collision is elastic. A stops and `B` moves with the velocity of `A` after collision
B. Impulses on `A` and `B` during collision are same in magnitude whether it is elastic or inelastic collision
C. Impulses on `A` and `B` collision are equal in magnitude only if `m_(1)=m_(2)`
D. If `m_(1)gtm_(2)`, though impulses are same in magnitude velocity of `A` is still the same direction.

Explanation

To solve the problem of the head-on collision between two balls A and B with masses \( m_1 \) and \( m_2 \) respectively, we will analyze the situation step by step.

### Step 1: Identify Initial Conditions
Let:
- Mass of ball A = \( m_1 \)
- Mass of ball B = \( m_2 \)
- Initial velocity of ball A = \( u_1 = v \) (given)
- Initial velocity of ball B = \( u_2 = 0 \) (at rest)

### Step 2: Understand the Type of Collision
The problem states that the collision is elastic. In an elastic collision, both momentum and kinetic energy are conserved.

### Step 3: Apply Conservation of Momentum
The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the values of \( u_1 \) and \( u_2 \):
\[ m_1 v + m_2 \cdot 0 = m_1 v_1 + m_2 v_2 \]
This simplifies to:
\[ m_1 v = m_1 v_1 + m_2 v_2 \quad \text{(Equation 1)} \]

### Step 4: Apply Conservation of Kinetic Energy
For an elastic collision, the kinetic energy before and after the collision is also conserved:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the values of \( u_1 \) and \( u_2 \):
\[ \frac{1}{2} m_1 v^2 + 0 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
This simplifies to:
\[ m_1 v^2 = m_1 v_1^2 + m_2 v_2^2 \quad \text{(Equation 2)} \]

### Step 5: Analyze the Case When \( m_1 = m_2 \)
If \( m_1 = m_2 = m \):
From Equation 1:
\[ m v = m v_1 + m v_2 \implies v = v_1 + v_2 \]
From Equation 2:
\[ m v^2 = m v_1^2 + m v_2^2 \implies v^2 = v_1^2 + v_2^2 \]

### Step 6: Solve the Equations
We have two equations:
1. \( v = v_1 + v_2 \)
2. \( v^2 = v_1^2 + v_2^2 \)

From the first equation, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v - v_1 \]
Substituting this into the second equation:
\[ v^2 = v_1^2 + (v - v_1)^2 \]
Expanding the equation:
\[ v^2 = v_1^2 + (v^2 - 2vv_1 + v_1^2) \]
Combining like terms:
\[ v^2 = 2v_1^2 - 2vv_1 + v^2 \]
Cancelling \( v^2 \) from both sides:
\[ 0 = 2v_1^2 - 2vv_1 \]
Factoring out \( 2v_1 \):
\[ 0 = 2v_1(v_1 - v) \]
This gives us two solutions:
1. \( v_1 = 0 \) (ball A stops)
2. \( v_1 = v \) (ball B moves with the velocity of A)

Since ball A collides with ball B, the first case is valid, meaning:
- \( v_1 = 0 \) (ball A stops)
- \( v_2 = v \) (ball B moves with the velocity of A)

### Conclusion
If \( m_1 = m_2 \) and the collision is elastic, ball A stops, and ball B moves with the velocity of A after the collision.