Two particles of same mass 'm' moving with velocities `vecv_1= vhati, and vecv_2 = (v/2)hati + (v/2)hatj`collide in-elastically. Find the loss in kinetic energy

Options

A. `(mv^2)/8`
B. `5(mv^2)/8`
C. `(mv^2)/4`
D. `3(mv^2)/8`

Explanation

To find the loss in kinetic energy during the inelastic collision of two particles with the given velocities, we can follow these steps:

### Step 1: Define the velocities of the particles
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 kinetic energy
The initial kinetic energy (\( KE_{initial} \)) of the system is the sum of the kinetic energies of both particles.

\[ KE_{initial} = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 \]

Calculating each term:
- For Particle 1:
\[ KE_1 = \frac{1}{2} m v^2 \]
- For Particle 2:
\[ KE_2 = \frac{1}{2} m \left( \frac{v}{2} \right)^2 + \frac{1}{2} m \left( \frac{v}{2} \right)^2 = \frac{1}{2} m \left( \frac{v^2}{4} + \frac{v^2}{4} \right) = \frac{1}{2} m \left( \frac{v^2}{2} \right) = \frac{m v^2}{4} \]

Thus, the total initial kinetic energy is:
\[ KE_{initial} = \frac{1}{2} m v^2 + \frac{m v^2}{4} = \frac{2m v^2}{4} + \frac{m v^2}{4} = \frac{3m v^2}{4} \]

### Step 3: Determine the final velocity after the collision
In an inelastic collision, the two particles stick together. We can use the conservation of momentum to find the final velocity (\( \vec{v'} \)) of the combined mass.

The total momentum before the collision is:
\[ \vec{p_{initial}} = m \vec{v_1} + m \vec{v_2} = m(v \hat{i}) + m\left(\frac{v}{2} \hat{i} + \frac{v}{2} \hat{j}\right) = mv \hat{i} + m\left(\frac{v}{2} \hat{i} + \frac{v}{2} \hat{j}\right) \]
\[ = mv \hat{i} + \frac{mv}{2} \hat{i} + \frac{mv}{2} \hat{j} = \left(mv + \frac{mv}{2}\right) \hat{i} + \frac{mv}{2} \hat{j} = \frac{3mv}{2} \hat{i} + \frac{mv}{2} \hat{j} \]

The total mass after the collision is \( 2m \). Therefore, the final velocity \( \vec{v'} \) is given by:
\[ \vec{v'} = \frac{\vec{p_{initial}}}{2m} = \frac{\frac{3mv}{2} \hat{i} + \frac{mv}{2} \hat{j}}{2m} = \frac{3v}{4} \hat{i} + \frac{v}{4} \hat{j} \]

### Step 4: Calculate the final kinetic energy
The final kinetic energy (\( KE_{final} \)) of the combined mass is:
\[ KE_{final} = \frac{1}{2} (2m) (v')^2 = m (v')^2 \]
Calculating \( (v')^2 \):
\[ (v')^2 = \left(\frac{3v}{4}\right)^2 + \left(\frac{v}{4}\right)^2 = \frac{9v^2}{16} + \frac{v^2}{16} = \frac{10v^2}{16} = \frac{5v^2}{8} \]
Thus,
\[ KE_{final} = m \cdot \frac{5v^2}{8} = \frac{5mv^2}{8} \]

### Step 5: Calculate the loss in kinetic energy
The loss in kinetic energy (\( \Delta KE \)) is given by:
\[ \Delta KE = KE_{initial} - KE_{final} \]
Substituting the values we calculated:
\[ \Delta KE = \frac{3mv^2}{4} - \frac{5mv^2}{8} \]
To subtract these fractions, we need a common denominator (which is 8):
\[ \Delta KE = \frac{6mv^2}{8} - \frac{5mv^2}{8} = \frac{1mv^2}{8} \]

Thus, the loss in kinetic energy is:
\[ \Delta KE = \frac{mv^2}{8} \]

### Final Answer
The loss in kinetic energy is \( \frac{mv^2}{8} \).

---