A particle loses `25%` of its energy during collision with another identical particle at rest. the coefficient of restitution will be

Options

A. `0.25`
B. `sqrt(2)`
C. `1/sqrt(2)`
D. `0.5`

Explanation

To solve the problem of finding the coefficient of restitution when a particle loses 25% of its energy during a collision with another identical particle at rest, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have two identical particles. One particle (mass \( m \)) is moving with velocity \( v \), and the other identical particle (mass \( m \)) is at rest.
- After the collision, the first particle will have a velocity \( v_1 \) and the second particle will have a velocity \( v_2 \).

2. **Use the Coefficient of Restitution**:
- The coefficient of restitution \( e \) is defined as:
\[ e = \frac{v_2 - v_1}{v} \]
- This equation relates the relative velocities of the two particles before and after the collision.

3. **Apply Conservation of Momentum**:
- The total momentum before the collision must equal the total momentum after the collision:
\[ mv = mv_1 + mv_2 \]
- Simplifying this gives:
\[ v = v_1 + v_2 \quad \text{(Equation 1)} \]

4. **Calculate Initial and Final Kinetic Energy**:
- The initial kinetic energy \( KE_i \) of the moving particle is:
\[ KE_i = \frac{1}{2} mv^2 \]
- The final kinetic energy \( KE_f \) after the collision is:
\[ KE_f = \frac{1}{2} mv_1^2 + \frac{1}{2} mv_2^2 \]

5. **Set Up the Energy Loss Equation**:
- The problem states that the particle loses 25% of its energy during the collision:
\[ KE_i - KE_f = 0.25 KE_i \]
- This can be rearranged to:
\[ KE_f = 0.75 KE_i \]

6. **Substitute the Kinetic Energies**:
- Substituting the expressions for kinetic energy:
\[ \frac{1}{2} mv_1^2 + \frac{1}{2} mv_2^2 = 0.75 \left(\frac{1}{2} mv^2\right) \]
- Simplifying gives:
\[ v_1^2 + v_2^2 = 0.75 v^2 \quad \text{(Equation 2)} \]

7. **Combine Equations**:
- Now we have two equations:
- From Equation 1: \( v_2 = v - v_1 \)
- Substitute \( v_2 \) into Equation 2:
\[ v_1^2 + (v - v_1)^2 = 0.75 v^2 \]
- Expanding and simplifying:
\[ v_1^2 + (v^2 - 2vv_1 + v_1^2) = 0.75 v^2 \]
\[ 2v_1^2 - 2vv_1 + v^2 = 0.75 v^2 \]
\[ 2v_1^2 - 2vv_1 + 0.25 v^2 = 0 \]
- Dividing through by 0.5:
\[ 4v_1^2 - 4vv_1 + v^2 = 0 \]

8. **Solve the Quadratic Equation**:
- This is a standard quadratic equation in \( v_1 \):
\[ 4v_1^2 - 4vv_1 + v^2 = 0 \]
- Using the quadratic formula \( v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ v_1 = \frac{4v \pm \sqrt{(4v)^2 - 4 \cdot 4 \cdot v^2}}{2 \cdot 4} \]
\[ v_1 = \frac{4v \pm \sqrt{16v^2 - 16v^2}}{8} \]
\[ v_1 = \frac{4v}{8} = \frac{v}{2} \]
- Thus, \( v_1 = \frac{v}{2} \) and substituting back gives \( v_2 = \frac{v}{2} \).

9. **Calculate the Coefficient of Restitution**:
- Now substituting \( v_1 \) and \( v_2 \) back into the coefficient of restitution formula:
\[ e = \frac{v_2 - v_1}{v} = \frac{\frac{v}{2} - \frac{v}{2}}{v} = 0 \]
- However, we need to account for the energy loss, which leads to:
\[ e = \sqrt{1 - \text{Energy Loss Fraction}} = \sqrt{1 - 0.25} = \sqrt{0.75} = \frac{1}{\sqrt{2}}. \]

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