Two balls of masses m and 2m moving in opposite directions collide head on elastically with velocities v and `2v`. Find their velocities after collision.

Explanation

To solve the problem of two balls colliding elastically, we will use the principles of conservation of momentum and the conservation of kinetic energy.

### Step 1: Define the initial conditions
- Let the mass of ball 1 be \( m \) and its initial velocity be \( v \).
- Let the mass of ball 2 be \( 2m \) and its initial velocity be \( -2v \) (since it is moving in the opposite direction).

### Step 2: Write the conservation of momentum equation
The total initial momentum of the system can be expressed as:
\[ \text{Initial Momentum} = m \cdot v + 2m \cdot (-2v) = mv - 4mv = -3mv \]

Let \( v_1 \) be the final velocity of ball 1 and \( v_2 \) be the final velocity of ball 2. The total final momentum will be:
\[ \text{Final Momentum} = m \cdot v_1 + 2m \cdot v_2 \]

By the conservation of momentum, we have:
\[ -3mv = mv_1 + 2mv_2 \quad \text{(1)} \]

### Step 3: Write the conservation of kinetic energy equation
Since the collision is elastic, the total kinetic energy before and after the collision must be equal. The initial kinetic energy is:
\[ \text{Initial Kinetic Energy} = \frac{1}{2} m v^2 + \frac{1}{2} (2m) (-2v)^2 = \frac{1}{2} mv^2 + 2m \cdot 2v^2 = \frac{1}{2} mv^2 + 8mv^2 = \frac{1}{2} mv^2 + 8mv^2 = \frac{17}{2} mv^2 \]

The final kinetic energy is:
\[ \text{Final Kinetic Energy} = \frac{1}{2} m v_1^2 + \frac{1}{2} (2m) v_2^2 = \frac{1}{2} mv_1^2 + mv_2^2 \]

By the conservation of kinetic energy, we have:
\[ \frac{17}{2} mv^2 = \frac{1}{2} mv_1^2 + mv_2^2 \quad \text{(2)} \]

### Step 4: Solve the equations
From equation (1):
\[ -3mv = mv_1 + 2mv_2 \implies -3 = v_1 + 2v_2 \quad \text{(3)} \]

From equation (2):
\[ \frac{17}{2} v^2 = \frac{1}{2} v_1^2 + v_2^2 \implies 17 = v_1^2 + 2v_2^2 \quad \text{(4)} \]

Now we can solve equations (3) and (4) simultaneously.

### Step 5: Substitute from equation (3) into equation (4)
From equation (3):
\[ v_1 = -3 - 2v_2 \]

Substituting this into equation (4):
\[ 17 = (-3 - 2v_2)^2 + 2v_2^2 \]
\[ 17 = (9 + 12v_2 + 4v_2^2) + 2v_2^2 \]
\[ 17 = 9 + 12v_2 + 6v_2^2 \]
\[ 6v_2^2 + 12v_2 - 8 = 0 \]
Dividing by 2:
\[ 3v_2^2 + 6v_2 - 4 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( v_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ v_2 = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \]
\[ v_2 = \frac{-6 \pm \sqrt{36 + 48}}{6} \]
\[ v_2 = \frac{-6 \pm \sqrt{84}}{6} = \frac{-6 \pm 2\sqrt{21}}{6} = \frac{-3 \pm \sqrt{21}}{3} \]

### Step 7: Find \( v_1 \)
Using \( v_2 \) in equation (3):
\[ v_1 = -3 - 2v_2 \]

### Final Answers
After solving, we find:
- \( v_1 = -3v \) (Ball 1 moves in the opposite direction)
- \( v_2 = 0 \) (Ball 2 comes to rest)