Two balls of equal masses m each undergo oblique collision. If colision is perfectly elastic, then angle between their velocities after collision is

Options

A. `(pi)/(4)`
B. `(pi)/(3)`
C. `(pi)/(6)`
D. `(pi)/(2)`

Answer

To solve the problem of finding the angle between the velocities of two balls of equal mass after a perfectly elastic oblique collision, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two balls of equal mass \( m \) that collide obliquely.
- The collision is perfectly elastic, meaning both momentum and kinetic energy are conserved.

2. **Initial Setup**:
- Let the initial velocity of ball 1 be \( u \) and assume it is moving towards ball 2, which is initially at rest.
- The angle of incidence (the angle at which ball 1 approaches ball 2) is denoted as \( \theta \).

3. **Conservation of Momentum**:
- Since the collision is perfectly elastic and there are no external forces acting along the line of impact, we can apply the conservation of momentum.
- The momentum before the collision must equal the momentum after the collision.
- Let \( v_1 \) and \( v_2 \) be the velocities of ball 1 and ball 2 after the collision, respectively.
- The momentum conservation equation along the line of impact can be written as:
\[ m u \sin(\theta) = m v_1 + m v_2 \]
- Simplifying gives:
\[ u \sin(\theta) = v_1 + v_2 \]

4. **Conservation of Kinetic Energy**:
- For a perfectly elastic collision, the kinetic energy before and after the collision is also conserved:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 \]
- Simplifying gives:
\[ u^2 = v_1^2 + v_2^2 \]

5. **Solving the Equations**:
- We now have two equations:
1. \( u \sin(\theta) = v_1 + v_2 \) (1)
2. \( u^2 = v_1^2 + v_2^2 \) (2)

- From equation (1), we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = u \sin(\theta) - v_1 \]

- Substitute \( v_2 \) into equation (2):
\[ u^2 = v_1^2 + (u \sin(\theta) - v_1)^2 \]

- Expanding this gives:
\[ u^2 = v_1^2 + (u^2 \sin^2(\theta) - 2u \sin(\theta) v_1 + v_1^2) \]
\[ u^2 = 2v_1^2 - 2u \sin(\theta) v_1 + u^2 \sin^2(\theta) \]

- Rearranging leads to a quadratic equation in \( v_1 \):
\[ 2v_1^2 - 2u \sin(\theta) v_1 + (u^2 \sin^2(\theta) - u^2) = 0 \]

6. **Finding the Angle**:
- The angle between the velocities after the collision can be derived from the relationship between \( v_1 \) and \( v_2 \).
- In a perfectly elastic collision, the angle between the two velocities after the collision is always \( 90^\circ \) (or \( \frac{\pi}{2} \) radians).

### Conclusion:
The angle between the velocities of the two balls after the perfectly elastic collision is \( \frac{\pi}{2} \) radians (90 degrees).