Two particles of mass m & 2m have velocities `u hat( j) ` & `u hat( i ) ` respectively . They collide completely inelastically. Find the loss in kinetic energy of the system.

Options

A. `"mu"^(2)`
B. `2/3"mu"^(2)`
C. `("mu"^(2))/4`
D. `("mu"^(2))/4`

Explanation

To solve the problem of finding the loss in kinetic energy during a completely inelastic collision between two particles of masses \( m \) and \( 2m \) with initial velocities \( \mathbf{u} \hat{j} \) and \( \mathbf{u} \hat{i} \) respectively, we can follow these steps:

### Step 1: Calculate the Initial Kinetic Energy
The initial kinetic energy (\( KE_{\text{initial}} \)) of the system can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
For the first particle (mass \( m \) with velocity \( \mathbf{u} \hat{j} \)):
\[ KE_1 = \frac{1}{2} m (u^2) = \frac{1}{2} m u^2 \]
For the second particle (mass \( 2m \) with velocity \( \mathbf{u} \hat{i} \)):
\[ KE_2 = \frac{1}{2} (2m) (u^2) = mu^2 \]
Thus, the total initial kinetic energy is:
\[ KE_{\text{initial}} = KE_1 + KE_2 = \frac{1}{2} m u^2 + mu^2 = \frac{1}{2} m u^2 + \frac{2}{2} m u^2 = \frac{3}{2} m u^2 \]

### Step 2: Calculate the Final Velocity After Collision
In a completely inelastic collision, the two particles stick together. We can use conservation of momentum to find the final velocity (\( \mathbf{v} \)) of the combined mass.

The initial momentum (\( \mathbf{p}_{\text{initial}} \)) is:
\[ \mathbf{p}_{\text{initial}} = m \mathbf{u} \hat{j} + 2m \mathbf{u} \hat{i} = 2m \mathbf{u} \hat{i} + m \mathbf{u} \hat{j} \]
The total mass after collision is:
\[ M = m + 2m = 3m \]
Using conservation of momentum:
\[ \mathbf{p}_{\text{initial}} = \mathbf{p}_{\text{final}} \Rightarrow 2m \mathbf{u} \hat{i} + m \mathbf{u} \hat{j} = 3m \mathbf{v} \]
Thus, we can solve for \( \mathbf{v} \):
\[ \mathbf{v} = \frac{2m \mathbf{u} \hat{i} + m \mathbf{u} \hat{j}}{3m} = \frac{2\mathbf{u} \hat{i} + \mathbf{u} \hat{j}}{3} \]

### Step 3: Calculate the Final Kinetic Energy
The final kinetic energy (\( KE_{\text{final}} \)) can be calculated using the final velocity:
\[ KE_{\text{final}} = \frac{1}{2} M v^2 = \frac{1}{2} (3m) v^2 \]
First, we need to find \( v^2 \):
\[ v^2 = \left(\frac{2\mathbf{u} \hat{i} + \mathbf{u} \hat{j}}{3}\right) \cdot \left(\frac{2\mathbf{u} \hat{i} + \mathbf{u} \hat{j}}{3}\right) = \frac{1}{9} \left( (2u)^2 + (u)^2 \right) = \frac{1}{9} (4u^2 + u^2) = \frac{5u^2}{9} \]
Now substituting back:
\[ KE_{\text{final}} = \frac{1}{2} (3m) \left(\frac{5u^2}{9}\right) = \frac{15mu^2}{18} = \frac{5mu^2}{6} \]

### Step 4: Calculate the Loss in Kinetic Energy
The loss in kinetic energy (\( \Delta KE \)) is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = \frac{3}{2} mu^2 - \frac{5}{6} mu^2 \]
Converting \( \frac{3}{2} mu^2 \) to a common denominator:
\[ \frac{3}{2} mu^2 = \frac{9}{6} mu^2 \]
Thus:
\[ \Delta KE = \frac{9}{6} mu^2 - \frac{5}{6} mu^2 = \frac{4}{6} mu^2 = \frac{2}{3} mu^2 \]

### Final Answer
The loss in kinetic energy of the system is:
\[ \Delta KE = \frac{2}{3} mu^2 \]