A ball of mass `1kg` moving with `4m^-1` along +x-axis collides elastically with an another ball of mass `2kg` moving with `6m//s` is opposite direction. Find their velocities after collision.

Explanation

To solve the problem of two colliding balls, we will use the principles of conservation of momentum and the properties of elastic collisions. Here’s the step-by-step solution:

### Step 1: Identify the Given Data
- Mass of ball A (m1) = 1 kg
- Initial velocity of ball A (v1) = 4 m/s (moving in the +x direction)
- Mass of ball B (m2) = 2 kg
- Initial velocity of ball B (v2) = -6 m/s (moving in the -x direction)

### Step 2: Write the Conservation of Momentum Equation
The total momentum before the collision must equal the total momentum after the collision. This can be expressed as:
\[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \]
Substituting the values:
\[ (1 \, \text{kg} \cdot 4 \, \text{m/s}) + (2 \, \text{kg} \cdot -6 \, \text{m/s}) = (1 \, \text{kg} \cdot v_1') + (2 \, \text{kg} \cdot v_2') \]
This simplifies to:
\[ 4 - 12 = v_1' + 2v_2' \]
\[ -8 = v_1' + 2v_2' \] (Equation 1)

### Step 3: Write the Elastic Collision Equation
For an elastic collision, the relative velocity of approach equals the relative velocity of separation:
\[ v_1 - v_2 = -(v_1' - v_2') \]
Substituting the values:
\[ 4 - (-6) = -(v_1' - v_2') \]
This simplifies to:
\[ 4 + 6 = -v_1' + v_2' \]
\[ 10 = -v_1' + v_2' \] (Equation 2)

### Step 4: Solve the Equations Simultaneously
Now we have two equations:
1. \( -8 = v_1' + 2v_2' \) (Equation 1)
2. \( 10 = -v_1' + v_2' \) (Equation 2)

From Equation 2, we can express \( v_1' \):
\[ v_1' = v_2' - 10 \]

### Step 5: Substitute \( v_1' \) in Equation 1
Substituting \( v_1' \) in Equation 1:
\[ -8 = (v_2' - 10) + 2v_2' \]
\[ -8 = v_2' - 10 + 2v_2' \]
\[ -8 = 3v_2' - 10 \]
\[ 3v_2' = 2 \]
\[ v_2' = \frac{2}{3} \, \text{m/s} \]

### Step 6: Find \( v_1' \)
Now substitute \( v_2' \) back into the expression for \( v_1' \):
\[ v_1' = \frac{2}{3} - 10 \]
\[ v_1' = \frac{2}{3} - \frac{30}{3} \]
\[ v_1' = -\frac{28}{3} \, \text{m/s} \]

### Final Results
- Final velocity of ball A (v1') = \(-\frac{28}{3} \, \text{m/s}\) (moving in the -x direction)
- Final velocity of ball B (v2') = \(\frac{2}{3} \, \text{m/s}\) (moving in the +x direction)