Two particles of equal mass m are moving with velocities `vhati` and `v((hati+hatj)/(2))` respectively. If the particles collide completely inelastically, then the energy lost in the process is

Options

A. `(3mv^(2))/(4)`
B. `(2mv^(2))/(3)`
C. `(mv^(2))/(3)`
D. `(mv^(2))/(8)`

Answer

To solve the problem of two particles colliding completely inelastically, we will follow these steps:

### Step 1: Define the initial velocities
Let the velocities of the two particles be:
- Particle 1: \( \vec{v_1} = v \hat{i} \)
- Particle 2: \( \vec{v_2} = \frac{v}{2} \hat{i} + \frac{v}{2} \hat{j} \)

### Step 2: Calculate the initial momentum
The momentum of each particle can be calculated as follows:
- Momentum of Particle 1:
\[ \vec{p_1} = m \vec{v_1} = m v \hat{i} \]
- Momentum of Particle 2:
\[ \vec{p_2} = m \vec{v_2} = m \left( \frac{v}{2} \hat{i} + \frac{v}{2} \hat{j} \right) = \frac{mv}{2} \hat{i} + \frac{mv}{2} \hat{j} \]

### Step 3: Calculate the total initial momentum
The total initial momentum \( \vec{p_{initial}} \) is the vector sum of the momenta of both particles:
\[ \vec{p_{initial}} = \vec{p_1} + \vec{p_2} = m v \hat{i} + \left( \frac{mv}{2} \hat{i} + \frac{mv}{2} \hat{j} \right) = \left( mv + \frac{mv}{2} \right) \hat{i} + \frac{mv}{2} \hat{j} \]
\[ \vec{p_{initial}} = \frac{3mv}{2} \hat{i} + \frac{mv}{2} \hat{j} \]

### Step 4: Calculate the final velocity after collision
Since the collision is completely inelastic, the two particles stick together after the collision. The total mass after collision is \( 2m \). The final momentum \( \vec{p_{final}} \) is equal to the total initial momentum:
\[ \vec{p_{final}} = (2m) \vec{v_f} \]
Setting the initial momentum equal to the final momentum:
\[ \frac{3mv}{2} \hat{i} + \frac{mv}{2} \hat{j} = (2m) \vec{v_f} \]
Dividing through by \( 2m \):
\[ \vec{v_f} = \frac{3v}{4} \hat{i} + \frac{v}{4} \hat{j} \]

### Step 5: Calculate the initial kinetic energy
The initial kinetic energy \( KE_{initial} \) is the sum of the kinetic energies of both particles:
\[ KE_{initial} = \frac{1}{2} m v^2 + \frac{1}{2} m \left( \frac{v}{2} \right)^2 + \frac{1}{2} m \left( \frac{v}{2} \right)^2 \]
Calculating the second particle's kinetic energy:
\[ KE_{initial} = \frac{1}{2} m v^2 + \frac{1}{2} m \left( \frac{v^2}{4} \right) + \frac{1}{2} m \left( \frac{v^2}{4} \right) = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2 \]

### Step 6: Calculate the final kinetic energy
The final kinetic energy \( KE_{final} \) after the collision is:
\[ KE_{final} = \frac{1}{2} (2m) \left( \frac{3v}{4} \right)^2 + \frac{1}{2} (2m) \left( \frac{v}{4} \right)^2 \]
Calculating:
\[ KE_{final} = m \left( \frac{9v^2}{16} + \frac{v^2}{16} \right) = m \left( \frac{10v^2}{16} \right) = \frac{5}{8} m v^2 \]

### Step 7: Calculate the energy lost
The energy lost \( \Delta KE \) during the collision is:
\[ \Delta KE = KE_{initial} - KE_{final} \]
Substituting the values:
\[ \Delta KE = \frac{3}{4} m v^2 - \frac{5}{8} m v^2 \]
Finding a common denominator:
\[ \Delta KE = \frac{6}{8} m v^2 - \frac{5}{8} m v^2 = \frac{1}{8} m v^2 \]

### Final Answer
The energy lost in the process is:
\[ \Delta KE = \frac{1}{8} m v^2 \]