What is the fractional decrease in kinetic energy of a body of mass `m_1` when it makes a head on elastic collision with another body of mass `m_2` kept at rest?

Explanation

To find the fractional decrease in kinetic energy of a body of mass \( m_1 \) when it makes a head-on elastic collision with another body of mass \( m_2 \) that is at rest, we can follow these steps:

### Step 1: Understand the Initial and Final Kinetic Energy
The initial kinetic energy (\( K_i \)) of the body of mass \( m_1 \) before the collision is given by:
\[ K_i = \frac{1}{2} m_1 v_1^2 \]
where \( v_1 \) is the initial velocity of mass \( m_1 \).

The body of mass \( m_2 \) is at rest, so its initial kinetic energy is:
\[ K_{i2} = 0 \]

### Step 2: Determine the Final Velocities After Collision
In an elastic collision, both momentum and kinetic energy are conserved. The final velocity of mass \( m_1 \) after the collision (\( v_1' \)) and the final velocity of mass \( m_2 \) after the collision (\( v_2' \)) can be calculated using the following formulas:
\[ v_1' = \frac{(m_1 - m_2)}{(m_1 + m_2)} v_1 \]
\[ v_2' = \frac{2 m_1}{(m_1 + m_2)} v_1 \]

### Step 3: Calculate the Final Kinetic Energy
The final kinetic energy (\( K_f \)) of mass \( m_1 \) after the collision is:
\[ K_{f1} = \frac{1}{2} m_1 (v_1')^2 = \frac{1}{2} m_1 \left(\frac{(m_1 - m_2)}{(m_1 + m_2)} v_1\right)^2 \]
The final kinetic energy of mass \( m_2 \) after the collision is:
\[ K_{f2} = \frac{1}{2} m_2 (v_2')^2 = \frac{1}{2} m_2 \left(\frac{2 m_1}{(m_1 + m_2)} v_1\right)^2 \]

### Step 4: Combine the Final Kinetic Energies
The total final kinetic energy after the collision is:
\[ K_f = K_{f1} + K_{f2} = \frac{1}{2} m_1 \left(\frac{(m_1 - m_2)}{(m_1 + m_2)} v_1\right)^2 + \frac{1}{2} m_2 \left(\frac{2 m_1}{(m_1 + m_2)} v_1\right)^2 \]

### Step 5: Calculate the Fractional Decrease in Kinetic Energy
The fractional decrease in kinetic energy is given by:
\[ \text{Fractional Decrease} = \frac{K_i - K_f}{K_i} \]

Substituting the values of \( K_i \) and \( K_f \):
\[ \text{Fractional Decrease} = \frac{\frac{1}{2} m_1 v_1^2 - K_f}{\frac{1}{2} m_1 v_1^2} \]

### Step 6: Simplify the Expression
After simplifying the expression, we find:
\[ \text{Fractional Decrease} = 1 - \frac{m_2}{m_1 + m_2} \left(\frac{v_1'}{v_1}\right)^2 \]
Substituting \( v_1' \) from earlier, we can derive the final expression:
\[ \text{Fractional Decrease} = \frac{4 m_1 m_2}{(m_1 + m_2)^2} \]

### Final Answer
Thus, the fractional decrease in kinetic energy of the body of mass \( m_1 \) when it makes a head-on elastic collision with another body of mass \( m_2 \) at rest is:
\[ \frac{4 m_1 m_2}{(m_1 + m_2)^2} \]