A ball A moving with a velocity 5 m/s collides elastically with another identical ball at rest such that the velocity of A makes an angle of `30^(@)` with the line joining the centres of the balls. Then consider following statements :
(A) Balls A an B move at right angles after collision
(B) kinetic energy is conserved
(C) speed of A after collision is (5/2) m/s
(D) speed of B after collision is (`5sqrt(3)//2`) m/s
Select correct alternative :-

Options

A. Only (A) and (B)
B. Only (A), (B) and (C)
C. Only (A), (B) and (D)
D. All are correct

Answer

To solve the problem, we will analyze the elastic collision between the two identical balls, A and B.

### Step 1: Understand the Initial Conditions
- Ball A is moving with a velocity of \(5 \, \text{m/s}\) at an angle of \(30^\circ\) to the line joining the centers of the balls.
- Ball B is initially at rest.

### Step 2: Break Down the Velocity of Ball A
We can resolve the velocity of Ball A into two components:
- **Along the line of impact (x-direction)**:
\[ V_{A_x} = V_A \cos(30^\circ) = 5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \, \text{m/s} \]
- **Perpendicular to the line of impact (y-direction)**:
\[ V_{A_y} = V_A \sin(30^\circ) = 5 \cdot \frac{1}{2} = \frac{5}{2} \, \text{m/s} \]

### Step 3: Analyze the Collision
In an elastic collision between identical balls:
- The components of the velocities along the line of impact are exchanged.
- The perpendicular components remain unchanged.

### Step 4: Determine the Final Velocities
- After the collision:
- The velocity of Ball A along the line of impact becomes \(0\) (it transfers its momentum to Ball B).
- The velocity of Ball B along the line of impact becomes \(V_{A_x}\):
\[ V_{B_x} = V_{A_x} = \frac{5\sqrt{3}}{2} \, \text{m/s} \]
- The perpendicular component of Ball A remains the same:
\[ V_{A_y} = \frac{5}{2} \, \text{m/s} \]
- The perpendicular component of Ball B remains \(0\).

### Step 5: Calculate the Final Speeds
- The speed of Ball A after the collision is:
\[ V_A = \sqrt{V_{A_x}^2 + V_{A_y}^2} = \sqrt{0^2 + \left(\frac{5}{2}\right)^2} = \frac{5}{2} \, \text{m/s} \]
- The speed of Ball B after the collision is:
\[ V_B = \sqrt{V_{B_x}^2 + 0^2} = \frac{5\sqrt{3}}{2} \, \text{m/s} \]

### Summary of Results
- Speed of A after collision: \( \frac{5}{2} \, \text{m/s} \)
- Speed of B after collision: \( \frac{5\sqrt{3}}{2} \, \text{m/s} \)

### Step 6: Evaluate the Statements
- (A) Balls A and B move at right angles after collision: **True** (since one moves along the line of impact and the other perpendicular).
- (B) Kinetic energy is conserved: **True** (in elastic collisions, kinetic energy is conserved).
- (C) Speed of A after collision is \( \frac{5}{2} \, \text{m/s} \): **True**.
- (D) Speed of B after collision is \( \frac{5\sqrt{3}}{2} \, \text{m/s} \): **True**.

### Conclusion
All statements (A), (B), (C), and (D) are correct.