A ball moving with velocity `2 ms^(-1)` collides head on with another stationary ball of double the mass. If the coefficient of restitution is `0.5`, then their velocities (in `ms^(-1)`) after collision will be

Answer

To solve the problem, we need to find the velocities of two balls after a head-on collision. We have the following information:

- A ball of mass \( m \) is moving with a velocity \( u_1 = 2 \, \text{m/s} \).
- A stationary ball of mass \( 2m \) has an initial velocity \( u_2 = 0 \, \text{m/s} \).
- The coefficient of restitution \( e = 0.5 \).

### Step 1: Apply the Conservation of Momentum

The total momentum before the collision must equal the total momentum after the collision.

**Before Collision:**
\[ \text{Initial Momentum} = m \cdot u_1 + 2m \cdot u_2 = m \cdot 2 + 2m \cdot 0 = 2m \]

**After Collision:**
Let \( v_1 \) be the velocity of the first ball (mass \( m \)) after the collision, and \( v_2 \) be the velocity of the second ball (mass \( 2m \)) after the collision.

\[ \text{Final Momentum} = m \cdot v_1 + 2m \cdot v_2 \]

Setting the initial momentum equal to the final momentum:
\[ 2m = m \cdot v_1 + 2m \cdot v_2 \]

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

### Step 2: Apply the Coefficient of Restitution

The coefficient of restitution \( e \) is defined as the ratio of the relative velocity of separation to the relative velocity of approach.

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

The velocity of separation is given by \( v_2 - v_1 \) (since \( v_2 \) is in the direction of the positive velocity after the collision and \( v_1 \) is in the opposite direction).

The velocity of approach is given by \( u_1 - u_2 = 2 - 0 = 2 \).

Using the coefficient of restitution:
\[ 0.5 = \frac{v_2 - v_1}{2} \]

Multiplying both sides by 2:
\[ v_2 - v_1 = 1 \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations

Now we have two equations:
1. \( v_1 + 2v_2 = 2 \) (Equation 1)
2. \( v_2 - v_1 = 1 \) (Equation 2)

From Equation 2, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v_1 + 1 \]

### Step 4: Substitute \( v_2 \) into Equation 1

Substituting \( v_2 \) into Equation 1:
\[ v_1 + 2(v_1 + 1) = 2 \]
\[ v_1 + 2v_1 + 2 = 2 \]
\[ 3v_1 + 2 = 2 \]
\[ 3v_1 = 0 \]
\[ v_1 = 0 \, \text{m/s} \]

### Step 5: Find \( v_2 \)

Now substituting \( v_1 \) back into the equation for \( v_2 \):
\[ v_2 = 0 + 1 = 1 \, \text{m/s} \]

### Final Result

Thus, the velocities after the collision are:
- \( v_1 = 0 \, \text{m/s} \) (the first ball comes to rest)
- \( v_2 = 1 \, \text{m/s} \) (the second ball moves with this velocity)