Two particles of same mass m moving with velocities `u_(1)` and `u_(2)` collide perfectly inelastically. The loss of energy would be :

Options

A. `1//2m(u_(1)-u_(2))^(2)`
B. `1//4m(u_(1)-u_(2))^(2)`
C. `m(u_(1)-u_(2))^(2)`
D. `2m (u_(1)-u_(2))^(2)`

Answer

To find the loss of kinetic energy during a perfectly inelastic collision between two particles of the same mass \( m \) moving with velocities \( u_1 \) and \( u_2 \), we can follow these steps:

### Step 1: Understand the scenario
In a perfectly inelastic collision, the two particles stick together after the collision. The initial kinetic energy of the system can be calculated using the velocities of both particles before the collision.

### Step 2: Calculate the initial kinetic energy
The initial kinetic energy (\( KE_{initial} \)) of the two particles is given by:
\[ KE_{initial} = \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2 \]
This can be simplified to:
\[ KE_{initial} = \frac{m}{2} (u_1^2 + u_2^2) \]

### Step 3: Calculate the final velocity after collision
Since the collision is perfectly inelastic, the final velocity (\( v_f \)) of the combined mass after the collision can be calculated using the conservation of momentum:
\[ m u_1 + m u_2 = (m + m) v_f \]
\[ v_f = \frac{u_1 + u_2}{2} \]

### Step 4: Calculate the final kinetic energy
The final kinetic energy (\( KE_{final} \)) after the collision is:
\[ KE_{final} = \frac{1}{2} (2m) v_f^2 = m v_f^2 \]
Substituting \( v_f \):
\[ KE_{final} = m \left(\frac{u_1 + u_2}{2}\right)^2 = m \cdot \frac{(u_1 + u_2)^2}{4} \]

### Step 5: Calculate the loss of kinetic energy
The loss of kinetic energy (\( \Delta KE \)) during the collision is given by:
\[ \Delta KE = KE_{initial} - KE_{final} \]
Substituting the expressions we derived:
\[ \Delta KE = \frac{m}{2} (u_1^2 + u_2^2) - m \cdot \frac{(u_1 + u_2)^2}{4} \]

### Step 6: Simplify the expression
To simplify \( \Delta KE \):
1. Expand \( (u_1 + u_2)^2 \):
\[ (u_1 + u_2)^2 = u_1^2 + 2u_1u_2 + u_2^2 \]
2. Substitute back into the equation:
\[ \Delta KE = \frac{m}{2} (u_1^2 + u_2^2) - m \cdot \frac{(u_1^2 + 2u_1u_2 + u_2^2)}{4} \]
3. Combine the terms:
\[ \Delta KE = \frac{2m}{4} (u_1^2 + u_2^2) - \frac{m}{4} (u_1^2 + 2u_1u_2 + u_2^2) \]
\[ = \frac{m}{4} (2u_1^2 + 2u_2^2 - u_1^2 - 2u_1u_2 - u_2^2) \]
\[ = \frac{m}{4} (u_1^2 - 2u_1u_2 + u_2^2) \]
\[ = \frac{m}{4} (u_1 - u_2)^2 \]

### Final Answer
Thus, the loss of kinetic energy during the perfectly inelastic collision is:
\[ \Delta KE = \frac{m}{4} (u_1 - u_2)^2 \]