After an head on elastic collision between two balls of equal masses, one is observed to have a speed of `3m//s` along the positive x-axis and the other has a speed of `2m//s` along the negative x-axis. What are the original velocities of the balls?

Explanation

To solve the problem of finding the original velocities of two balls after a head-on elastic collision, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( V_1 \) = initial velocity of ball 1 (before collision)
- \( V_2 \) = initial velocity of ball 2 (before collision)
- \( V_1' = 3 \, \text{m/s} \) (final velocity of ball 1 after collision, positive x-direction)
- \( V_2' = -2 \, \text{m/s} \) (final velocity of ball 2 after collision, negative x-direction)

### Step 2: Apply the Conservation of Momentum
Since no external forces are acting on the system, the total momentum before the collision is equal to the total momentum after the collision.

The equation for conservation of momentum is:
\[ m V_1 + m V_2 = m V_1' + m V_2' \]
Since the masses are equal, we can cancel \( m \) from the equation:
\[ V_1 + V_2 = V_1' + V_2' \]
Substituting the known final velocities:
\[ V_1 + V_2 = 3 + (-2) \]
\[ V_1 + V_2 = 1 \quad \text{(Equation 1)} \]

### Step 3: Apply the Conservation of Kinetic Energy
For an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The equation for conservation of kinetic energy is:
\[ \frac{1}{2} m V_1^2 + \frac{1}{2} m V_2^2 = \frac{1}{2} m (V_1'^2 + V_2'^2) \]
Again, cancel \( \frac{1}{2} m \):
\[ V_1^2 + V_2^2 = V_1'^2 + V_2'^2 \]
Substituting the known final velocities:
\[ V_1^2 + V_2^2 = 3^2 + (-2)^2 \]
\[ V_1^2 + V_2^2 = 9 + 4 \]
\[ V_1^2 + V_2^2 = 13 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have two equations:
1. \( V_1 + V_2 = 1 \) (Equation 1)
2. \( V_1^2 + V_2^2 = 13 \) (Equation 2)

From Equation 1, we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = 1 - V_1 \]

Substituting \( V_2 \) in Equation 2:
\[ V_1^2 + (1 - V_1)^2 = 13 \]
Expanding the equation:
\[ V_1^2 + (1 - 2V_1 + V_1^2) = 13 \]
\[ 2V_1^2 - 2V_1 + 1 = 13 \]
\[ 2V_1^2 - 2V_1 - 12 = 0 \]
Dividing the entire equation by 2:
\[ V_1^2 - V_1 - 6 = 0 \]

### Step 5: Factor the Quadratic Equation
Factoring the quadratic:
\[ (V_1 - 3)(V_1 + 2) = 0 \]
This gives us:
\[ V_1 = 3 \quad \text{or} \quad V_1 = -2 \]

### Step 6: Find Corresponding \( V_2 \)
Using \( V_1 + V_2 = 1 \):
1. If \( V_1 = 3 \):
\[ V_2 = 1 - 3 = -2 \]
2. If \( V_1 = -2 \):
\[ V_2 = 1 - (-2) = 3 \]

### Conclusion
The original velocities of the balls are:
- Ball 1: \( V_1 = -2 \, \text{m/s} \) (moving in the negative x-direction)
- Ball 2: \( V_2 = 3 \, \text{m/s} \) (moving in the positive x-direction)