A sphere is moving on a smooth surface with angular velocity `omega` and linear velocity v. If collides elastically with another identical sphere B kept at rest. Neglecting friction everywhere, let `V_(A)` and `omega_(A)` be linear and angular velocities of sphere A after the impact and let `V_(B)` and `omega_(B)` be the corresponding parameters for spehre B. Then

Options

A. `V_(A) =0, omega_(A) = 0`
B. `V_(B) = v, omega_(B) = 0`
C. `V_(B) =0, omega_(B) = omega`
D. `omega_(A) = 0 , omega_(B) = omega`

Answer

To solve the problem of two identical spheres colliding elastically, we can break it down into a few steps.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- Sphere A is moving with a linear velocity \( v \) and an angular velocity \( \omega \).
- Sphere B is at rest.
- The collision is elastic, meaning both momentum and kinetic energy are conserved.

2. **Conservation of Linear Momentum**:
- Before the collision, the total linear momentum is given by:
\[ P_{\text{initial}} = m \cdot v + 0 = mv \]
- After the collision, let \( V_A \) and \( V_B \) be the linear velocities of spheres A and B respectively. The total linear momentum after the collision is:
\[ P_{\text{final}} = m \cdot V_A + m \cdot V_B \]
- Setting the initial and final momenta equal gives:
\[ mv = mV_A + mV_B \]
- Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ v = V_A + V_B \]

3. **Conservation of Kinetic Energy**:
- The initial kinetic energy is:
\[ KE_{\text{initial}} = \frac{1}{2} mv^2 + 0 = \frac{1}{2} mv^2 \]
- The final kinetic energy is:
\[ KE_{\text{final}} = \frac{1}{2} m V_A^2 + \frac{1}{2} m V_B^2 \]
- Setting the initial and final kinetic energies equal gives:
\[ \frac{1}{2} mv^2 = \frac{1}{2} m V_A^2 + \frac{1}{2} m V_B^2 \]
- Dividing through by \( \frac{1}{2} m \) (again assuming \( m \neq 0 \)):
\[ v^2 = V_A^2 + V_B^2 \]

4. **Analyzing the Collision**:
- Since the collision is elastic and the spheres are identical, we can conclude that the velocities are exchanged.
- Thus, after the collision:
\[ V_A = 0 \quad \text{(Sphere A comes to rest)} \]
\[ V_B = v \quad \text{(Sphere B moves with the initial velocity of A)} \]

5. **Angular Velocity**:
- The angular velocity of sphere A after the collision remains unchanged because the collision does not affect the angular motion. Therefore:
\[ \omega_A = \omega \quad \text{(remains the same)} \]
- Sphere B, being at rest initially and not spinning, will have:
\[ \omega_B = 0 \]

### Conclusion:
- After the collision:
- \( V_A = 0 \)
- \( \omega_A = \omega \)
- \( V_B = v \)
- \( \omega_B = 0 \)

### Final Answer:
The correct option is:
- \( V_A = 0 \), \( \omega_A = \omega \), \( V_B = v \), \( \omega_B = 0 \)