A ball impinges directly on another ball at rest and is reduced to rest to rest b y impact. If half the initial kinetic energy is lost by impact, then the coefficient of elasticity is

Options

A. `1//4`
B. `1//3`
C. `3//4`
D. None of these

Answer

To solve the problem, we need to analyze the situation using the concepts of kinetic energy and the coefficient of restitution.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the mass of the moving ball be \( m \) and its initial velocity be \( u \).
- The initial kinetic energy (KE) of the moving ball is given by:
\[ KE_{initial} = \frac{1}{2} m u^2 \]

2. **Kinetic Energy After Impact**:
- After the impact, the first ball comes to rest, so its final velocity is \( 0 \).
- The second ball (initially at rest) will gain some velocity \( v \) after the impact.
- The kinetic energy of the second ball after the impact is:
\[ KE_{final} = \frac{1}{2} m v^2 \]

3. **Energy Loss**:
- According to the problem, half of the initial kinetic energy is lost during the impact. Therefore, the kinetic energy after the impact is:
\[ KE_{final} = KE_{initial} - \text{Energy lost} \]
- Since half the initial kinetic energy is lost:
\[ \text{Energy lost} = \frac{1}{2} KE_{initial} = \frac{1}{2} \left(\frac{1}{2} m u^2\right) = \frac{1}{4} m u^2 \]
- Thus, we can express the final kinetic energy as:
\[ KE_{final} = \frac{1}{2} m u^2 - \frac{1}{4} m u^2 = \frac{1}{4} m u^2 \]

4. **Setting Up the Equation**:
- From the expression for \( KE_{final} \):
\[ \frac{1}{2} m v^2 = \frac{1}{4} m u^2 \]
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = \frac{1}{4} u^2 \]
- Rearranging gives:
\[ v^2 = \frac{1}{2} u^2 \]
- Taking the square root:
\[ v = \frac{u}{\sqrt{2}} \]

5. **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:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
- Before the impact, the velocity of approach is \( u \) (the first ball's velocity) and after the impact, the second ball moves with velocity \( v \):
\[ e = \frac{v}{u} \]
- Substituting \( v \):
\[ e = \frac{\frac{u}{\sqrt{2}}}{u} = \frac{1}{\sqrt{2}} \]

### Final Answer:
The coefficient of restitution \( e \) is \( \frac{1}{\sqrt{2}} \).