Ball 1 collides directly with another identical ball 2 at rest. Velocity of second ball becomes two times that of 1 after collison. Find the coefficient of restitution between the two balls?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have two identical balls, Ball 1 and Ball 2. Ball 1 is moving with an initial velocity \( U \) and collides with Ball 2, which is at rest. After the collision, the velocity of Ball 2 becomes twice that of Ball 1.

### Step 2: Define Variables
- Let the mass of each ball be \( m \).
- Let the initial velocity of Ball 1 be \( U \).
- Let the final velocity of Ball 1 after the collision be \( V_1 \).
- Let the final velocity of Ball 2 after the collision be \( V_2 \).

Given that \( V_2 = 2V_1 \).

### Step 3: Apply Conservation of Momentum
Since no external forces are acting on the system, we can apply the law of conservation of momentum:

\[ \text{Initial Momentum} = \text{Final Momentum} \]

The initial momentum is:

\[ mU + 0 = mU \]

The final momentum is:

\[ mV_1 + mV_2 = mV_1 + m(2V_1) = mV_1 + 2mV_1 = 3mV_1 \]

Setting the initial momentum equal to the final momentum:

\[ mU = 3mV_1 \]

Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):

\[ U = 3V_1 \]

Thus, we can express \( V_1 \) in terms of \( U \):

\[ V_1 = \frac{U}{3} \]

### Step 4: Calculate the Coefficient of Restitution
The coefficient of restitution \( e \) is defined as:

\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]

**Velocity of separation** after the collision:

\[ \text{Velocity of separation} = V_2 - V_1 = 2V_1 - V_1 = V_1 \]

**Velocity of approach** before the collision:

\[ \text{Velocity of approach} = U - 0 = U \]

Substituting the values into the coefficient of restitution formula:

\[ e = \frac{V_1}{U} \]

Now, substituting \( V_1 = \frac{U}{3} \):

\[ e = \frac{\frac{U}{3}}{U} = \frac{1}{3} \]

### Conclusion
The coefficient of restitution between the two balls is:

\[ \boxed{\frac{1}{3}} \]