Two identical billiard balls undergo an oblique elastic collision. Initially, one of the balls is stationary. If the initially stationary ball after collision moves in a direction which makes an angle of `37^(@)` with direction of initial motion of the moving ball, then the angle through which initially moving ball will be deflected is

Options

A. `37^(@)`
B. `60^(@)`
C. `53^(@)`
D. `gt53^(@)`

Explanation

To solve the problem of the oblique elastic collision between two identical billiard balls, we can follow these steps:

### Step 1: Understand the Initial Conditions
- We have two identical billiard balls, A and B.
- Ball A is moving with an initial velocity \( u \) towards ball B, which is initially stationary (velocity = 0).

### Step 2: Analyze the Collision
- After the collision, ball B moves at an angle of \( 37^\circ \) with respect to the initial direction of ball A.
- We need to determine the angle through which ball A is deflected after the collision.

### Step 3: Apply Conservation of Momentum
- Since the collision is elastic and the balls are identical, we can apply the conservation of momentum in both the x and y directions.
- Let \( \theta \) be the angle through which ball A is deflected.

### Step 4: Set Up the Momentum Equations
1. **In the x-direction**:
\[ m \cdot u = m \cdot V_B \cdot \cos(37^\circ) + m \cdot V_A \cdot \cos(\theta) \]
where \( V_B \) is the velocity of ball B after the collision, and \( V_A \) is the velocity of ball A after the collision.

2. **In the y-direction**:
\[ 0 = m \cdot V_B \cdot \sin(37^\circ) - m \cdot V_A \cdot \sin(\theta) \]

### Step 5: Simplify the Equations
- Since the masses are identical, we can cancel \( m \) from all terms:
1. **X-direction**:
\[ u = V_B \cdot \cos(37^\circ) + V_A \cdot \cos(\theta) \]

2. **Y-direction**:
\[ 0 = V_B \cdot \sin(37^\circ) - V_A \cdot \sin(\theta) \]

### Step 6: Solve for the Angles
- From the y-direction equation, we can express \( V_A \) in terms of \( V_B \):
\[ V_A \cdot \sin(\theta) = V_B \cdot \sin(37^\circ) \implies V_A = \frac{V_B \cdot \sin(37^\circ)}{\sin(\theta)} \]

### Step 7: Substitute into the X-direction Equation
- Substitute \( V_A \) into the x-direction equation:
\[ u = V_B \cdot \cos(37^\circ) + \left(\frac{V_B \cdot \sin(37^\circ)}{\sin(\theta)}\right) \cdot \cos(\theta) \]

### Step 8: Factor Out \( V_B \)
- Factor \( V_B \) from the equation:
\[ u = V_B \left(\cos(37^\circ) + \frac{\sin(37^\circ) \cdot \cos(\theta)}{\sin(\theta)}\right) \]

### Step 9: Analyze the Geometry
- Since the collision is elastic and involves identical balls, the angle \( \theta \) can be determined using the geometry of the situation. The angles formed must satisfy the relationship:
\[ \theta + 37^\circ + 90^\circ = 180^\circ \]
Therefore, \( \theta = 53^\circ \).

### Final Answer
The angle through which the initially moving ball (ball A) will be deflected is \( 53^\circ \).

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