Two balls of equal mass have a head-on collision with speed `6 m//s`. If the coefficient of restitution is `(1)/(3)`, find the speed of each ball after impact in `m//s`

Explanation

To solve the problem, we need to determine the speeds of two balls after a head-on collision, given that they have equal masses and a coefficient of restitution of \( \frac{1}{3} \). Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Let the mass of each ball be \( m \).
- The initial speed of ball A (moving towards ball B) is \( u_1 = 6 \, \text{m/s} \).
- The initial speed of ball B (moving towards ball A) is \( u_2 = -6 \, \text{m/s} \) (negative because it's in the opposite direction).

2. **Coefficient of Restitution:**
- The coefficient of restitution \( e \) is given as \( \frac{1}{3} \).
- The formula for the coefficient of restitution is:
\[ e = \frac{\text{Relative speed after collision}}{\text{Relative speed before collision}} \]
- This can be expressed as:
\[ e = \frac{v_2 - v_1}{u_1 - u_2} \]
where \( v_1 \) and \( v_2 \) are the final velocities of balls A and B, respectively.

3. **Calculate Relative Speeds:**
- The relative speed before the collision:
\[ u_1 - u_2 = 6 - (-6) = 6 + 6 = 12 \, \text{m/s} \]
- Substitute this into the equation for \( e \):
\[ \frac{1}{3} = \frac{v_2 - v_1}{12} \]

4. **Express the Equation:**
- Rearranging gives:
\[ v_2 - v_1 = \frac{12}{3} = 4 \quad \text{(1)} \]

5. **Conservation of Momentum:**
- Since the masses are equal, we can use the conservation of momentum:
\[ m u_1 + m u_2 = m v_1 + m v_2 \]
- Simplifying (dividing by \( m \)):
\[ 6 + (-6) = v_1 + v_2 \]
- This simplifies to:
\[ 0 = v_1 + v_2 \quad \text{(2)} \]

6. **Solve the Equations:**
- From equation (2), we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = -v_1 \]
- Substitute \( v_2 = -v_1 \) into equation (1):
\[ -v_1 - v_1 = 4 \]
\[ -2v_1 = 4 \implies v_1 = -2 \, \text{m/s} \]
- Now substitute back to find \( v_2 \):
\[ v_2 = -(-2) = 2 \, \text{m/s} \]

7. **Final Speeds:**
- The final speeds of the balls after the collision are:
- Ball A (v1): \( -2 \, \text{m/s} \) (moving in the opposite direction)
- Ball B (v2): \( 2 \, \text{m/s} \)

### Conclusion:
The speed of ball A after impact is \( 2 \, \text{m/s} \) in the opposite direction, and the speed of ball B after impact is \( 2 \, \text{m/s} \).