A ball of mass m moving with velocity v collides head-on which the second ball of mass m at rest. I the coefficient of restitution is e and velocity of first ball after collision is `v_(1)` and velocity of second ball after collision is `v_(2)` then

Options

A. `v_(1)=((1-e)u)/(2), v_(2)=((1+e)u)/(2)`
B. `v_(1)=((1+e)u)/(2), v_(2)=((1-e)u)/(2)`
C. `v_(1)=(u)/(2), v_(2)=-(u)/(2)`
D. `v_(1)=(1+e)u, v_(2)=(1-e)u`

Answer

To solve the problem of two balls colliding, we can follow these steps:

### Step 1: Understand the Problem
We have two balls of mass \( m \). The first ball is moving with an initial velocity \( u \), and the second ball is at rest. After the collision, we need to find the velocities \( v_1 \) and \( v_2 \) of the first and second balls, respectively. The coefficient of restitution is given as \( e \).

### Step 2: Apply the Law of Conservation of Momentum
The total momentum before the collision must equal the total momentum after the collision.

**Initial Momentum:**
- The first ball has momentum \( mu \) (since it has mass \( m \) and velocity \( u \)).
- The second ball is at rest, so its momentum is \( 0 \).

Thus, the total initial momentum is:
\[ \text{Initial Momentum} = mu + 0 = mu \]

**Final Momentum:**
- After the collision, the first ball has momentum \( mv_1 \) and the second ball has momentum \( mv_2 \).

Thus, the total final momentum is:
\[ \text{Final Momentum} = mv_1 + mv_2 \]

Setting the initial momentum equal to the final momentum gives us:
\[ mu = mv_1 + mv_2 \quad \text{(Equation 1)} \]

### Step 3: Apply the Coefficient of Restitution
The coefficient of restitution \( e \) relates the relative velocities of separation and approach. The formula is:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]

**Velocity of Approach:**
- The approach velocity before the collision is \( u \) (the first ball moving towards the second ball at rest).

**Velocity of Separation:**
- After the collision, the velocities of separation are \( v_2 - v_1 \) (the second ball moving away minus the first ball moving away).

Thus, we can write:
\[ e = \frac{v_2 - v_1}{u} \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have two equations:
1. \( mu = mv_1 + mv_2 \)
2. \( e = \frac{v_2 - v_1}{u} \)

From Equation 1, we can simplify to:
\[ u = v_1 + v_2 \quad \text{(Equation 1')} \]

From Equation 2, we can rearrange to find:
\[ v_2 - v_1 = eu \quad \text{(Equation 2')} \]

### Step 5: Add the Two Equations
Adding Equation 1' and Equation 2':
\[ u + (v_2 - v_1) = v_1 + v_2 + eu \]
This simplifies to:
\[ u + eu = 2v_2 \]
Thus, we can solve for \( v_2 \):
\[ v_2 = \frac{u(1 + e)}{2} \]

### Step 6: Substitute \( v_2 \) Back to Find \( v_1 \)
Now substitute \( v_2 \) back into Equation 1':
\[ u = v_1 + \frac{u(1 + e)}{2} \]
Rearranging gives:
\[ v_1 = u - \frac{u(1 + e)}{2} \]
Factoring out \( u \):
\[ v_1 = u \left(1 - \frac{1 + e}{2}\right) = u \left(\frac{2 - (1 + e)}{2}\right) = u \left(\frac{1 - e}{2}\right) \]

### Final Results
Thus, the final velocities after the collision are:
\[ v_1 = \frac{u(1 - e)}{2} \]
\[ v_2 = \frac{u(1 + e)}{2} \]