Two identical balls `A and B` having velocity of `0.5 m//s and -0.3 m//s` respectively collide elastically in one dimension. The velocities of `B and A` after the collision respectively will be

Options

A. `-0.3 m//s and 0.5 m//s`
B. `0.3 m//s and 0.5 m//s`
C. `-0.5 m//s and 0.3 m//s`
D. `0.5 m//s and -0.3 m//s`

Explanation

To solve the problem of the elastic collision between two identical balls A and B, we will use the principles of conservation of momentum and conservation of kinetic energy.

### Step 1: Define the initial conditions
Let:
- Mass of ball A, \( m_A = m \)
- Mass of ball B, \( m_B = m \)
- Initial velocity of ball A, \( u_A = 0.5 \, \text{m/s} \)
- Initial velocity of ball B, \( u_B = -0.3 \, \text{m/s} \)

### Step 2: Apply the conservation of momentum
The total momentum before the collision must equal the total momentum after the collision. This can be expressed as:
\[ m u_A + m u_B = m v_A + m v_B \]
Where \( v_A \) and \( v_B \) are the final velocities of balls A and B after the collision.

Since the masses are identical, we can simplify the equation by dividing through by \( m \):
\[ u_A + u_B = v_A + v_B \]
Substituting the values:
\[ 0.5 + (-0.3) = v_A + v_B \]
\[ 0.2 = v_A + v_B \quad \text{(Equation 1)} \]

### Step 3: Apply the conservation of kinetic energy
In an elastic collision, the total kinetic energy before the collision equals the total kinetic energy after the collision:
\[ \frac{1}{2} m u_A^2 + \frac{1}{2} m u_B^2 = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2 \]
Again, we can simplify by dividing through by \( \frac{1}{2} m \):
\[ u_A^2 + u_B^2 = v_A^2 + v_B^2 \]
Substituting the values:
\[ (0.5)^2 + (-0.3)^2 = v_A^2 + v_B^2 \]
Calculating the left side:
\[ 0.25 + 0.09 = v_A^2 + v_B^2 \]
\[ 0.34 = v_A^2 + v_B^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( v_A + v_B = 0.2 \) (Equation 1)
2. \( v_A^2 + v_B^2 = 0.34 \) (Equation 2)

From Equation 1, we can express \( v_B \) in terms of \( v_A \):
\[ v_B = 0.2 - v_A \]
Substituting this into Equation 2:
\[ v_A^2 + (0.2 - v_A)^2 = 0.34 \]
Expanding the equation:
\[ v_A^2 + (0.04 - 0.4v_A + v_A^2) = 0.34 \]
Combining like terms:
\[ 2v_A^2 - 0.4v_A + 0.04 = 0.34 \]
\[ 2v_A^2 - 0.4v_A - 0.3 = 0 \]
Dividing the entire equation by 2:
\[ v_A^2 - 0.2v_A - 0.15 = 0 \]

### Step 5: Use the quadratic formula to find \( v_A \)
Using the quadratic formula \( v_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -0.2, c = -0.15 \):
\[ v_A = \frac{-(-0.2) \pm \sqrt{(-0.2)^2 - 4 \cdot 1 \cdot (-0.15)}}{2 \cdot 1} \]
\[ v_A = \frac{0.2 \pm \sqrt{0.04 + 0.6}}{2} \]
\[ v_A = \frac{0.2 \pm \sqrt{0.64}}{2} \]
\[ v_A = \frac{0.2 \pm 0.8}{2} \]

Calculating the two possible values:
1. \( v_A = \frac{1.0}{2} = 0.5 \, \text{m/s} \)
2. \( v_A = \frac{-0.6}{2} = -0.3 \, \text{m/s} \)

### Step 6: Find \( v_B \)
Using \( v_B = 0.2 - v_A \):
1. If \( v_A = 0.5 \, \text{m/s} \), then \( v_B = 0.2 - 0.5 = -0.3 \, \text{m/s} \)
2. If \( v_A = -0.3 \, \text{m/s} \), then \( v_B = 0.2 - (-0.3) = 0.5 \, \text{m/s} \)

### Final Answer
Thus, the final velocities of balls A and B after the collision are:
- \( v_A = -0.3 \, \text{m/s} \)
- \( v_B = 0.5 \, \text{m/s} \)