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`

Options

A. 18 m/s
B. 2 m/s
C. 6 m/s
D. 4 m/s

Answer

To solve the problem of two balls colliding head-on, we can follow these steps:

### Step 1: Understand the initial conditions
- Both balls have equal mass (let's denote the mass as `m`).
- Both balls are moving towards each other with a speed of `6 m/s`.

### Step 2: Define the coefficient of restitution
- The coefficient of restitution (e) is given as \( \frac{1}{3} \).
- The formula for the coefficient of restitution is:
\[ e = \frac{V_2 - V_1}{U_1 - U_2} \]
where:
- \( V_1 \) and \( V_2 \) are the final velocities of ball 1 and ball 2, respectively.
- \( U_1 \) and \( U_2 \) are the initial velocities of ball 1 and ball 2, respectively.

### Step 3: Assign initial velocities
- Let’s assume ball 1 is moving towards the right with a velocity of \( U_1 = 6 \, m/s \).
- Ball 2 is moving towards the left with a velocity of \( U_2 = -6 \, m/s \) (negative because it’s in the opposite direction).

### Step 4: Set up the equation using the coefficient of restitution
- Plugging the values into the coefficient of restitution formula:
\[ \frac{V_2 - V_1}{6 - (-6)} = \frac{1}{3} \]
Simplifying the denominator:
\[ \frac{V_2 - V_1}{12} = \frac{1}{3} \]

### Step 5: Solve for the final velocities
- Cross-multiplying gives:
\[ V_2 - V_1 = \frac{12}{3} = 4 \]
Thus, we have:
\[ V_2 - V_1 = 4 \quad \text{(1)} \]

### Step 6: Use conservation of momentum
- The total momentum before the collision must equal the total momentum after the collision:
\[ m \cdot 6 + m \cdot (-6) = m \cdot V_1 + m \cdot V_2 \]
This simplifies to:
\[ 0 = V_1 + V_2 \quad \text{(2)} \]

### Step 7: Solve the system of equations
- From equation (2), we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = -V_1 \]
- Substitute \( V_2 \) into equation (1):
\[ -V_1 - V_1 = 4 \]
This simplifies to:
\[ -2V_1 = 4 \quad \Rightarrow \quad V_1 = -2 \, m/s \]
- Now, substituting \( V_1 \) back into equation (2) to find \( V_2 \):
\[ V_2 = -(-2) = 2 \, m/s \]

### Final Result
- The final speeds of the balls after the collision are:
- \( V_1 = -2 \, m/s \) (Ball 1 moves to the left)
- \( V_2 = 2 \, m/s \) (Ball 2 moves to the right)

### Summary
- The speed of each ball after impact is:
- Ball 1: \( 2 \, m/s \) (moving in the opposite direction)
- Ball 2: \( 2 \, m/s \) (moving in the original direction)