A sphere `A` moving with speed `u` and rotating With angular velocity `omega` makes a head-on elastic collision with an identical stationary sphere `B`. There is no friction between the surfaces of `A` and `B`. Choose the conrrect alternative(s). Disregard gravity.

Options

A. `A` will stop moving but continue to rotate with an angular velocity `omega`
B. `A` will come to rest and stop rotating
C. `B` will move with speed `u` without rotating
D. `B` will move with speed `u` and rotate with an angular velocity `omega`

Explanation

To solve the problem of the elastic collision between two identical spheres, we will follow these steps:

### Step 1: Understand the Initial Conditions
- Sphere A is moving with an initial speed \( u \) and has an angular velocity \( \omega \).
- Sphere B is stationary (initial speed = 0).
- There is no friction between the surfaces of A and B.

### Step 2: Apply Conservation of Linear Momentum
In an elastic collision, the total linear momentum before the collision is equal to the total linear momentum after the collision.

Let:
- \( m \) = mass of each sphere
- \( v_1 \) = final velocity of sphere A after the collision
- \( v_2 \) = final velocity of sphere B after the collision

Using the conservation of linear momentum:
\[ m \cdot u + m \cdot 0 = m \cdot v_1 + m \cdot v_2 \]
This simplifies to:
\[ u = v_1 + v_2 \quad \text{(Equation 1)} \]

### Step 3: Apply Conservation of Kinetic Energy
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The initial kinetic energy is:
\[ KE_{initial} = \frac{1}{2} m u^2 + \frac{1}{2} I_A \omega^2 \]
where \( I_A \) is the moment of inertia of sphere A, given by \( I_A = \frac{2}{5} m r^2 \) for a solid sphere.

The final kinetic energy is:
\[ KE_{final} = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 + \frac{1}{2} I_B \omega_B^2 \]
Since sphere B is initially stationary and has no rotation, its angular velocity \( \omega_B \) after the collision will be 0.

Setting the initial and final kinetic energies equal gives us:
\[ \frac{1}{2} m u^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \omega^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 \]
This can be simplified to:
\[ u^2 + \frac{2}{5} r^2 \omega^2 = v_1^2 + v_2^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
From Equation 1, we have:
\[ v_1 = u - v_2 \]
Substituting \( v_1 \) into Equation 2:
\[ u^2 + \frac{2}{5} r^2 \omega^2 = (u - v_2)^2 + v_2^2 \]
Expanding and simplifying gives:
\[ u^2 + \frac{2}{5} r^2 \omega^2 = u^2 - 2uv_2 + 2v_2^2 \]
This leads to:
\[ \frac{2}{5} r^2 \omega^2 = -2uv_2 + 2v_2^2 \]
Rearranging gives:
\[ 2v_2^2 - 2uv_2 - \frac{2}{5} r^2 \omega^2 = 0 \]
Dividing by 2:
\[ v_2^2 - uv_2 - \frac{1}{5} r^2 \omega^2 = 0 \]

### Step 5: Solve for \( v_2 \)
Using the quadratic formula:
\[ v_2 = \frac{u \pm \sqrt{u^2 + \frac{4}{5} r^2 \omega^2}}{2} \]

### Step 6: Analyze the Results
1. Since there is no friction, sphere A will continue to rotate with the same angular velocity \( \omega \).
2. Sphere B will move with speed \( v_2 \) after the collision, but it will not rotate.

### Conclusion
- Sphere A will stop moving linearly but continue to rotate with angular velocity \( \omega \).
- Sphere B will move with speed \( u \) without any rotation.