A body of mass `m_(1)` moving at a constant speed undergoes an elastic head on collision with a body of mass `m_(2)` initially at rest. The ratio of the kinetic energy of mass `m_(1)` after the collision to that before the collision is -

Options

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

Answer

To solve the problem of finding the ratio of the kinetic energy of mass \( m_1 \) after the collision to that before the collision, we can follow these steps:

### Step 1: Understand the Problem
We have two masses, \( m_1 \) and \( m_2 \). Mass \( m_1 \) is moving with an initial velocity \( v_0 \), while mass \( m_2 \) is initially at rest. After an elastic collision, both masses will have new velocities \( v_1 \) and \( v_2 \) respectively.

### Step 2: Use Conservation of Momentum
The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Therefore, we can write:

\[ m_1 v_0 = m_1 v_1 + m_2 v_2 \quad \text{(1)} \]

### Step 3: Use the Elastic Collision Condition
For elastic collisions, the relative velocity of separation is equal to the relative velocity of approach. This gives us the equation:

\[ v_0 = v_2 - v_1 \quad \text{(2)} \]

### Step 4: Solve the Two Equations
From equation (2), we can express \( v_2 \) in terms of \( v_1 \):

\[ v_2 = v_0 + v_1 \quad \text{(3)} \]

Now, substitute equation (3) into equation (1):

\[ m_1 v_0 = m_1 v_1 + m_2 (v_0 + v_1) \]

Expanding this gives:

\[ m_1 v_0 = m_1 v_1 + m_2 v_0 + m_2 v_1 \]

Rearranging terms leads to:

\[ m_1 v_0 - m_2 v_0 = (m_1 + m_2) v_1 \]

Factoring out \( v_0 \):

\[ (m_1 - m_2) v_0 = (m_1 + m_2) v_1 \]

### Step 5: Solve for \( v_1 \)
Now we can express \( v_1 \):

\[ v_1 = \frac{(m_1 - m_2)}{(m_1 + m_2)} v_0 \quad \text{(4)} \]

### Step 6: Calculate Kinetic Energies
The kinetic energy before the collision is:

\[ KE_{initial} = \frac{1}{2} m_1 v_0^2 \]

The kinetic energy after the collision is:

\[ KE_{final} = \frac{1}{2} m_1 v_1^2 \]

Substituting equation (4) into the kinetic energy expression gives:

\[ KE_{final} = \frac{1}{2} m_1 \left( \frac{(m_1 - m_2)}{(m_1 + m_2)} v_0 \right)^2 \]

### Step 7: Find the Ratio of Kinetic Energies
Now we can find the ratio of the final kinetic energy to the initial kinetic energy:

\[ \frac{KE_{final}}{KE_{initial}} = \frac{\frac{1}{2} m_1 \left( \frac{(m_1 - m_2)}{(m_1 + m_2)} v_0 \right)^2}{\frac{1}{2} m_1 v_0^2} \]

This simplifies to:

\[ \frac{KE_{final}}{KE_{initial}} = \left( \frac{(m_1 - m_2)}{(m_1 + m_2)} \right)^2 \]

### Final Answer
Thus, the ratio of the kinetic energy of mass \( m_1 \) after the collision to that before the collision is:

\[ \frac{KE_{final}}{KE_{initial}} = \left( \frac{m_1 - m_2}{m_1 + m_2} \right)^2 \]