In perfectly elastic collision between two masses `m_(1)and m_(2)` in one dimension energy transfer is a maximum, when

Options

A. `m_(1)=2m_(2)`
B. `m_(1) lt ltm_(2)`
C. `m_(1) gt gt m_(2)`
D. `m_(1)=m_(2)`

Answer

To solve the problem of maximum energy transfer in a perfectly elastic collision between two masses \( m_1 \) and \( m_2 \) in one dimension, we can follow these steps:

### Step 1: Understand the Scenario
In a perfectly elastic collision, both momentum and kinetic energy are conserved. We denote the initial velocity of mass \( m_1 \) as \( v \) and assume mass \( m_2 \) is initially at rest (velocity = 0).

### Step 2: Write the Conservation of Momentum Equation
The conservation of momentum before and after the collision can be expressed as:
\[ m_1 v + m_2 \cdot 0 = m_1 v_1 + m_2 v_2 \]
This simplifies to:
\[ m_1 v = m_1 v_1 + m_2 v_2 \quad \text{(1)} \]

### Step 3: Write the Conservation of Kinetic Energy Equation
For perfectly elastic collisions, the kinetic energy before and after the collision is also conserved:
\[ \frac{1}{2} m_1 v^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
This simplifies to:
\[ m_1 v^2 = m_1 v_1^2 + m_2 v_2^2 \quad \text{(2)} \]

### Step 4: Relate the Velocities Using the Coefficient of Restitution
The coefficient of restitution \( e \) for a perfectly elastic collision is equal to 1. Thus, we have:
\[ e = \frac{v_2 - v_1}{v - 0} = 1 \implies v_2 - v_1 = v \quad \text{(3)} \]

### Step 5: Solve the System of Equations
From equation (3), we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v + v_1 \]
Substituting this into equation (1):
\[ m_1 v = m_1 v_1 + m_2 (v + v_1) \]
This simplifies to:
\[ m_1 v = m_1 v_1 + m_2 v + m_2 v_1 \]
Rearranging gives:
\[ m_1 v = (m_1 + m_2) v_1 + m_2 v \]
Thus,
\[ v_1 = \frac{m_1 v - m_2 v}{m_1 + m_2} = \frac{(m_1 - m_2)v}{m_1 + m_2} \]

Now substituting \( v_1 \) back into equation (3) to find \( v_2 \):
\[ v_2 = v + \frac{(m_1 - m_2)v}{m_1 + m_2} = \frac{(m_1 + m_2)v + (m_1 - m_2)v}{m_1 + m_2} = \frac{2m_1 v}{m_1 + m_2} \]

### Step 6: Calculate the Kinetic Energy Transfer
The kinetic energy of mass \( m_2 \) after the collision is:
\[ KE_{m_2} = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_2 \left(\frac{2m_1 v}{m_1 + m_2}\right)^2 \]
This simplifies to:
\[ KE_{m_2} = \frac{2 m_2 m_1^2 v^2}{(m_1 + m_2)^2} \]

### Step 7: Find the Condition for Maximum Energy Transfer
To maximize \( KE_{m_2} \), we can differentiate it with respect to the mass ratio \( k = \frac{m_2}{m_1} \) and set the derivative to zero. After performing the differentiation and simplification, we find that maximum energy transfer occurs when:
\[ m_2 = m_1 \]

### Conclusion
Thus, the maximum energy transfer in a perfectly elastic collision occurs when the two masses are equal:
\[ \boxed{m_1 = m_2} \]