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 energy in 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 Initial and Final States
In a perfectly inelastic collision, the two particles stick together after the collision. Therefore, the final mass of the combined system is \( 2m \).

### Step 2: Apply Conservation of Momentum
The momentum before the collision must equal the momentum after the collision. The initial momentum \( p_i \) is given by:
\[ p_i = m u_1 + m u_2 = m (u_1 + u_2) \]
The final momentum \( p_f \) after the collision is:
\[ p_f = (2m) v \]
Setting the initial and final momentum equal gives:
\[ m (u_1 + u_2) = 2m v \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ u_1 + u_2 = 2v \]
Solving for \( v \):
\[ v = \frac{u_1 + u_2}{2} \]

### Step 3: Calculate Initial Kinetic Energy
The initial kinetic energy \( KE_i \) of the system is the sum of the kinetic energies of both particles:
\[ KE_i = \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2 \]

### Step 4: Calculate Final Kinetic Energy
The final kinetic energy \( KE_f \) of the combined mass after the collision is:
\[ KE_f = \frac{1}{2} (2m) v^2 = m v^2 \]
Substituting \( v \) from Step 2:
\[ KE_f = m \left(\frac{u_1 + u_2}{2}\right)^2 = m \frac{(u_1 + u_2)^2}{4} \]

### Step 5: Expand the Final Kinetic Energy
Expanding \( (u_1 + u_2)^2 \):
\[ KE_f = m \frac{u_1^2 + 2u_1 u_2 + u_2^2}{4} \]

### Step 6: Calculate the Loss of Energy
The loss of kinetic energy \( \Delta KE \) is given by:
\[ \Delta KE = KE_i - KE_f \]
Substituting the expressions for \( KE_i \) and \( KE_f \):
\[ \Delta KE = \left(\frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2\right) - \left(m \frac{u_1^2 + 2u_1 u_2 + u_2^2}{4}\right) \]
Combining the terms:
\[ \Delta KE = \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2 - \frac{1}{4} m (u_1^2 + 2u_1 u_2 + u_2^2) \]

### Step 7: Simplify the Expression
Factoring out \( \frac{m}{4} \):
\[ \Delta KE = \frac{m}{4} \left(2u_1^2 + 2u_2^2 - (u_1^2 + 2u_1 u_2 + u_2^2)\right) \]
This simplifies to:
\[ \Delta KE = \frac{m}{4} \left(u_1^2 + u_2^2 - 2u_1 u_2\right) = \frac{m}{4} (u_1 - u_2)^2 \]

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