A spherical ball of mass `m_(1)` collides head on with another ball of mass `m_(2)` at rest . The collision is elastic . The fraction of kinetic energy lost by `m_(1)` is `:`

Options

A. `(4m_1m_2)/((m_1+m_2)^2)`
B. `m_1/(m_1+m_2)`
C. `m_2/(m_1+m_2)`
D. `(m_1m_2)/((m_1+m_2)^2)`

Explanation

To find the fraction of kinetic energy lost by the mass \( m_1 \) during an elastic collision with another mass \( m_2 \) that is initially at rest, we can follow these steps:

### Step 1: Understand the Problem
We have two masses, \( m_1 \) (moving) and \( m_2 \) (at rest). The collision is elastic, meaning both momentum and kinetic energy are conserved.

### Step 2: Apply Conservation of Momentum
The conservation of momentum before and after the collision can be expressed as:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Since \( m_2 \) is at rest initially, \( u_2 = 0 \). Thus, the equation simplifies to:
\[ m_1 u_1 = m_1 v_1 + m_2 v_2 \]
This can be rearranged to:
\[ u_1 = v_1 + \frac{m_2}{m_1} v_2 \quad \text{(Equation 1)} \]

### Step 3: Apply Conservation of Kinetic Energy
For elastic collisions, kinetic energy is also conserved:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Again, since \( u_2 = 0 \), we have:
\[ \frac{1}{2} m_1 u_1^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
This can be rearranged to:
\[ m_1 u_1^2 = m_1 v_1^2 + m_2 v_2^2 \quad \text{(Equation 2)} \]

### Step 4: Use Coefficient of Restitution
For elastic collisions, the coefficient of restitution \( e = 1 \):
\[ e = \frac{v_2 - v_1}{u_1 - u_2} = 1 \]
Since \( u_2 = 0 \), we have:
\[ v_2 - v_1 = u_1 \quad \Rightarrow \quad v_2 = u_1 + v_1 \quad \text{(Equation 3)} \]

### Step 5: Substitute Equation 3 into Equation 1
Substituting \( v_2 \) from Equation 3 into Equation 1 gives:
\[ u_1 = v_1 + \frac{m_2}{m_1}(u_1 + v_1) \]
Rearranging this leads to:
\[ u_1(1 - \frac{m_2}{m_1}) = v_1(1 + \frac{m_2}{m_1}) \]
This can be simplified to:
\[ v_1 = u_1 \frac{1 - \frac{m_2}{m_1}}{1 + \frac{m_2}{m_1}} \quad \text{(Equation 4)} \]

### Step 6: Calculate the Kinetic Energy Loss
The initial kinetic energy of \( m_1 \) is:
\[ KE_{initial} = \frac{1}{2} m_1 u_1^2 \]
The final kinetic energy of \( m_1 \) is:
\[ KE_{final} = \frac{1}{2} m_1 v_1^2 \]
The kinetic energy lost is:
\[ KE_{lost} = KE_{initial} - KE_{final} = \frac{1}{2} m_1 u_1^2 - \frac{1}{2} m_1 v_1^2 \]
The fraction of kinetic energy lost is:
\[ \text{Fraction lost} = \frac{KE_{lost}}{KE_{initial}} = \frac{\frac{1}{2} m_1 u_1^2 - \frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_1 u_1^2} = 1 - \frac{v_1^2}{u_1^2} \]

### Step 7: Substitute \( v_1 \) from Equation 4
Substituting \( v_1 \) from Equation 4 into the fraction gives:
\[ \text{Fraction lost} = 1 - \left( \frac{m_1 - m_2}{m_1 + m_2} \right)^2 \]

### Final Result
After simplifying, we find the fraction of kinetic energy lost by \( m_1 \):
\[ \text{Fraction lost} = \frac{4 m_1 m_2}{(m_1 + m_2)^2} \]