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 a step-by-step solution:

### Step 1: Identify the Given Data
- Mass of ball 1, \( M_1 = 1 \, \text{kg} \)
- Initial velocity of ball 1, \( v_1 = 4 \, \text{m/s} \) (moving in the positive x-direction)
- Mass of ball 2, \( M_2 = 2 \, \text{kg} \)
- Initial velocity of ball 2, \( v_2 = -6 \, \text{m/s} \) (moving in the negative x-direction)

### Step 2: Apply Conservation of Momentum
The total momentum before the collision must equal the total momentum after the collision. The equation for conservation of momentum is:

\[ M_1 v_1 + M_2 v_2 = M_1 v_1' + M_2 v_2' \]

Substituting the known values:

\[ (1 \, \text{kg} \cdot 4 \, \text{m/s}) + (2 \, \text{kg} \cdot -6 \, \text{m/s}) = (1 \, \text{kg} \cdot v_1') + (2 \, \text{kg} \cdot v_2') \]

Calculating the left side:

\[ 4 - 12 = v_1' + 2v_2' \]

This simplifies to:

\[ -8 = v_1' + 2v_2' \quad \text{(Equation 1)} \]

### Step 3: Apply the Coefficient of Restitution
For an elastic collision, the coefficient of restitution \( e \) is defined as:

\[ e = \frac{\text{Relative velocity after collision}}{\text{Relative velocity before collision}} \]

Since the collision is elastic, \( e = 1 \). Thus, we have:

\[ 1 = \frac{v_2' - v_1'}{v_1 - v_2} \]

Substituting the initial velocities:

\[ 1 = \frac{v_2' - v_1'}{4 - (-6)} = \frac{v_2' - v_1'}{10} \]

This leads to:

\[ v_2' - v_1' = 10 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations Simultaneously
Now we have two equations:

1. \( -8 = v_1' + 2v_2' \) (Equation 1)
2. \( v_2' - v_1' = 10 \) (Equation 2)

From Equation 2, we can express \( v_2' \) in terms of \( v_1' \):

\[ v_2' = v_1' + 10 \]

Substituting this into Equation 1:

\[ -8 = v_1' + 2(v_1' + 10) \]

Expanding this gives:

\[ -8 = v_1' + 2v_1' + 20 \]

Combining like terms:

\[ -8 = 3v_1' + 20 \]

Rearranging gives:

\[ 3v_1' = -28 \quad \Rightarrow \quad v_1' = -\frac{28}{3} \, \text{m/s} \]

### Step 5: Find \( v_2' \)
Now substituting \( v_1' \) back into Equation 2 to find \( v_2' \):

\[ v_2' = -\frac{28}{3} + 10 = -\frac{28}{3} + \frac{30}{3} = \frac{2}{3} \, \text{m/s} \]

### Final Result
The velocities after the collision are:
- \( v_1' = -\frac{28}{3} \, \text{m/s} \) (Ball 1)
- \( v_2' = \frac{2}{3} \, \text{m/s} \) (Ball 2)