Two identical balls A and B moving with velocities + 0.5 m/s and `-0.3`.m/s respectively collide head-on elastically. The velocities of the ball A and B after collision will be respectively

Options

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

Explanation

To solve the problem of two identical balls A and B colliding elastically, we will use the principles of conservation of momentum and conservation of kinetic energy. Here’s a step-by-step breakdown:

### Step 1: Define the Variables
- Let the mass of both balls be \( m \) (since they are identical).
- The initial velocity of ball A, \( u_A = +0.5 \, \text{m/s} \).
- The initial velocity of ball B, \( u_B = -0.3 \, \text{m/s} \).
- Let the final velocity of ball A after the collision be \( v_A \).
- Let the final velocity of ball B after the collision be \( v_B \).

### Step 2: Apply Conservation of Momentum
According to the conservation of momentum:
\[ m u_A + m u_B = m v_A + m v_B \]
Since the masses are equal, we can cancel \( m \):
\[ u_A + u_B = v_A + v_B \]
Substituting the values:
\[ 0.5 + (-0.3) = v_A + v_B \]
This simplifies to:
\[ 0.2 = v_A + v_B \quad \text{(Equation 1)} \]

### Step 3: Apply Conservation of Kinetic Energy
For elastic collisions, kinetic energy is also conserved:
\[ \frac{1}{2} m u_A^2 + \frac{1}{2} m u_B^2 = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2 \]
Again, canceling \( \frac{1}{2} m \):
\[ u_A^2 + u_B^2 = v_A^2 + v_B^2 \]
Substituting the values:
\[ (0.5)^2 + (-0.3)^2 = v_A^2 + v_B^2 \]
Calculating the left side:
\[ 0.25 + 0.09 = v_A^2 + v_B^2 \]
This simplifies to:
\[ 0.34 = v_A^2 + v_B^2 \quad \text{(Equation 2)} \]

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

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:
\[ v_A^2 + (0.04 - 0.4 v_A + v_A^2) = 0.34 \]
Combining like terms:
\[ 2v_A^2 - 0.4v_A + 0.04 = 0.34 \]
Rearranging gives:
\[ 2v_A^2 - 0.4v_A - 0.3 = 0 \]

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \( v_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 2, b = -0.4, c = -0.3 \):
\[ b^2 - 4ac = (-0.4)^2 - 4 \cdot 2 \cdot (-0.3) = 0.16 + 2.4 = 2.56 \]
Calculating the roots:
\[ v_A = \frac{0.4 \pm \sqrt{2.56}}{4} = \frac{0.4 \pm 1.6}{4} \]
This gives two possible values:
1. \( v_A = \frac{2}{4} = 0.5 \, \text{m/s} \)
2. \( v_A = \frac{-1.2}{4} = -0.3 \, \text{m/s} \)

### Step 6: Determine Final Velocities
From \( v_A + v_B = 0.2 \):
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 \( v_B = 0.2 - (-0.3) = 0.5 \, \text{m/s} \).

Thus, the final velocities after the collision are:
- \( v_A = -0.3 \, \text{m/s} \)
- \( v_B = 0.5 \, \text{m/s} \)

### Final Answer
The velocities of ball A and B after the collision will be:
- \( v_A = -0.3 \, \text{m/s} \)
- \( v_B = 0.5 \, \text{m/s} \)