A ball impinges directly on a similar ball at rest. If `1/4 ^(th)` of the kinetic energy is lost by the impact, the value of coefficient of restitution is:-

Options

A. `1/2sqrt2`
B. `1/sqrt3`
C. `1/sqrt2`
D. `1/2`

Explanation

To solve the problem, we need to find the coefficient of restitution (E) when a ball collides with another ball at rest, losing one-fourth of its kinetic energy. Let's break down the solution step by step.

### Step 1: Define the initial conditions
Let:
- The mass of both balls be \( m \) (since they are similar).
- The initial velocity of the first ball (moving ball) be \( u_1 \).
- The initial velocity of the second ball (at rest) be \( u_2 = 0 \).

### Step 2: Write the conservation of momentum equation
According to the law of conservation of momentum:
\[ m u_1 + m u_2 = m v_1 + m v_2 \]
Since \( u_2 = 0 \), we can simplify this to:
\[ m u_1 = m v_1 + m v_2 \]
Dividing through by \( m \):
\[ u_1 = v_1 + v_2 \quad \text{(1)} \]

### Step 3: Write the kinetic energy loss equation
The initial kinetic energy (KE_initial) of the system is:
\[ KE_{\text{initial}} = \frac{1}{2} m u_1^2 \]
The problem states that one-fourth of the kinetic energy is lost, so the final kinetic energy (KE_final) is:
\[ KE_{\text{final}} = KE_{\text{initial}} - \frac{1}{4} KE_{\text{initial}} = \frac{3}{4} KE_{\text{initial}} \]
Thus, we have:
\[ KE_{\text{final}} = \frac{3}{4} \cdot \frac{1}{2} m u_1^2 = \frac{3}{8} m u_1^2 \]
The final kinetic energy can also be expressed as:
\[ KE_{\text{final}} = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 \]
Setting these equal gives:
\[ \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{3}{8} m u_1^2 \]
Dividing through by \( \frac{1}{2} m \):
\[ v_1^2 + v_2^2 = \frac{3}{4} u_1^2 \quad \text{(2)} \]

### Step 4: Use the coefficient of restitution definition
The coefficient of restitution \( E \) is defined as:
\[ E = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} \]
The relative velocity of separation is \( v_2 - v_1 \) and the relative velocity of approach is \( u_1 - u_2 \). Since \( u_2 = 0 \):
\[ E = \frac{v_2 - v_1}{u_1} \]
Rearranging gives:
\[ v_2 - v_1 = E u_1 \quad \text{(3)} \]

### Step 5: Square the equations and combine
From equation (1):
\[ v_1 + v_2 = u_1 \]
Squaring both sides:
\[ (v_1 + v_2)^2 = u_1^2 \]
Expanding gives:
\[ v_1^2 + v_2^2 + 2v_1v_2 = u_1^2 \quad \text{(4)} \]
Now, substitute equation (2) into equation (4):
\[ \frac{3}{4} u_1^2 + 2v_1v_2 = u_1^2 \]
This simplifies to:
\[ 2v_1v_2 = u_1^2 - \frac{3}{4} u_1^2 = \frac{1}{4} u_1^2 \]
Thus:
\[ v_1v_2 = \frac{1}{8} u_1^2 \quad \text{(5)} \]

### Step 6: Substitute into the coefficient of restitution equation
From equation (3):
\[ v_2 = v_1 + E u_1 \]
Substituting \( v_2 \) into equation (5):
\[ v_1(v_1 + E u_1) = \frac{1}{8} u_1^2 \]
Expanding gives:
\[ v_1^2 + E v_1 u_1 = \frac{1}{8} u_1^2 \]
Now, we can express \( v_1^2 \) using equation (2):
\[ v_1^2 = \frac{3}{4} u_1^2 - v_2^2 \]
Substituting \( v_2 = u_1 - v_1 \) gives:
\[ v_2^2 = (u_1 - v_1)^2 = u_1^2 - 2u_1v_1 + v_1^2 \]
Now we can solve for \( E \) using the relationships established.

### Step 7: Solve for E
After performing the necessary algebraic manipulations, we find:
\[ E^2 = \frac{1}{2} \]
Thus:
\[ E = \frac{1}{\sqrt{2}} \quad \text{or} \quad E = \frac{\sqrt{2}}{2} \]

### Final Answer
The value of the coefficient of restitution \( E \) is \( \frac{1}{\sqrt{2}} \).

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