Two balls of masses `2 m` and `m` are moving with speed `2 v_(0)` and `v_(0)` towards each other. If the coefficient of restitution `e = 1//3` , the speed of balls , if collision is head - on

Options

A. `(v_(0))/(3) ,(4 v_(0))/(3)`
B. `(2 v_(0))/(3) , ( 5 v_(0))/(3)`
C. `(2 v_(0))/(3) ,( 4 v_(0))/(3)`
D. `(v_(0))/(3) ,(2 v_(0))/(3)`

Explanation

To solve the problem step by step, we will use the principles of conservation of momentum and the coefficient of restitution.

### Step 1: Define the masses and velocities
Let:
- Mass of ball 1, \( m_1 = 2m \) with initial velocity \( u_1 = 2v_0 \) (moving to the right)
- Mass of ball 2, \( m_2 = m \) with initial velocity \( u_2 = -v_0 \) (moving to the left)

### Step 2: Write the conservation of momentum equation
Since the collision is head-on and there are no external forces acting on the system, the momentum before the collision is equal to the momentum after the collision.

Initial momentum:
\[ p_{initial} = m_1 u_1 + m_2 u_2 = (2m)(2v_0) + (m)(-v_0) = 4mv_0 - mv_0 = 3mv_0 \]

Let \( v_1 \) be the final velocity of ball 1 and \( v_2 \) be the final velocity of ball 2. The final momentum is:
\[ p_{final} = m_1 v_1 + m_2 v_2 = (2m)v_1 + (m)v_2 \]

Setting initial momentum equal to final momentum:
\[ 3mv_0 = 2mv_1 + mv_2 \]

Dividing through by \( m \):
\[ 3v_0 = 2v_1 + v_2 \quad \text{(Equation 1)} \]

### Step 3: Use the coefficient of restitution
The coefficient of restitution \( e \) is given as \( \frac{1}{3} \). The formula for the coefficient of restitution is:
\[ e = \frac{v_2 - v_1}{u_1 - u_2} \]

Substituting the values:
\[ \frac{1}{3} = \frac{v_2 - v_1}{(2v_0) - (-v_0)} = \frac{v_2 - v_1}{3v_0} \]

Cross-multiplying gives:
\[ v_2 - v_1 = \frac{1}{3} \cdot 3v_0 = v_0 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
Now we have two equations:
1. \( 3v_0 = 2v_1 + v_2 \) (Equation 1)
2. \( v_2 - v_1 = v_0 \) (Equation 2)

From Equation 2, we can express \( v_2 \):
\[ v_2 = v_1 + v_0 \]

Substituting this expression for \( v_2 \) into Equation 1:
\[ 3v_0 = 2v_1 + (v_1 + v_0) \]
\[ 3v_0 = 3v_1 + v_0 \]
\[ 3v_0 - v_0 = 3v_1 \]
\[ 2v_0 = 3v_1 \]
\[ v_1 = \frac{2v_0}{3} \]

### Step 5: Find \( v_2 \)
Now substitute \( v_1 \) back into Equation 2 to find \( v_2 \):
\[ v_2 = v_1 + v_0 = \frac{2v_0}{3} + v_0 = \frac{2v_0}{3} + \frac{3v_0}{3} = \frac{5v_0}{3} \]

### Final Answer
The final velocities after the collision are:
- \( v_1 = \frac{2v_0}{3} \) (for mass \( 2m \))
- \( v_2 = \frac{5v_0}{3} \) (for mass \( m \))