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. `sqrt(2/3) m u^(2)`
B. `3/4m u^(2)`
C. `1/3m u^(2)`
D. `1/8 m u^(2)`

Answer

To find the loss in kinetic energy during the inelastic collision of two particles of the same mass 'm', we can follow these steps:

### Step 1: Determine the initial velocities of the particles
The velocities of the two particles are given as:
- \( \vec{v_1} = v \hat{i} \)
- \( \vec{v_2} = \frac{v}{2} \hat{i} + \frac{v}{2} \hat{j} \)

### Step 2: Calculate the initial kinetic energy
The initial kinetic energy (KE_initial) of the system can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 \]
Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 + \frac{1}{2} m \left( \frac{v}{2} \right)^2 + \frac{1}{2} m \left( \frac{v}{2} \right)^2 \]
Calculating \( v_2^2 \):
\[ v_2^2 = \left( \frac{v}{2} \right)^2 + \left( \frac{v}{2} \right)^2 = \frac{v^2}{4} + \frac{v^2}{4} = \frac{v^2}{2} \]
Thus,
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 + \frac{1}{2} m \cdot \frac{v^2}{2} = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2 \]

### Step 3: Calculate the final velocity after the collision
In an inelastic collision, the two particles stick together and move with a common velocity \( \vec{v_0} \). Using conservation of momentum:
\[ m \vec{v_1} + m \vec{v_2} = (m + m) \vec{v_0} \]
\[ \vec{v_0} = \frac{\vec{v_1} + \vec{v_2}}{2} \]
Substituting the values:
\[ \vec{v_0} = \frac{v \hat{i} + \left( \frac{v}{2} \hat{i} + \frac{v}{2} \hat{j} \right)}{2} = \frac{(v + \frac{v}{2}) \hat{i} + \frac{v}{2} \hat{j}}{2} = \frac{\frac{3v}{2} \hat{i} + \frac{v}{2} \hat{j}}{2} = \frac{3v}{4} \hat{i} + \frac{v}{4} \hat{j} \]

### Step 4: Calculate the final kinetic energy
The final kinetic energy (KE_final) of the combined mass is:
\[ KE_{\text{final}} = \frac{1}{2} (2m) v_0^2 \]
Calculating \( v_0^2 \):
\[ v_0^2 = \left( \frac{3v}{4} \right)^2 + \left( \frac{v}{4} \right)^2 = \frac{9v^2}{16} + \frac{v^2}{16} = \frac{10v^2}{16} = \frac{5v^2}{8} \]
Thus,
\[ KE_{\text{final}} = m v_0^2 = m \cdot \frac{5v^2}{8} = \frac{5}{8} m v^2 \]

### Step 5: Calculate the loss in kinetic energy
The loss in kinetic energy is given by:
\[ \text{Loss in KE} = KE_{\text{initial}} - KE_{\text{final}} = \frac{3}{4} m v^2 - \frac{5}{8} m v^2 \]
Converting \( \frac{3}{4} \) to eighths:
\[ \frac{3}{4} = \frac{6}{8} \]
Thus,
\[ \text{Loss in KE} = \frac{6}{8} m v^2 - \frac{5}{8} m v^2 = \frac{1}{8} m v^2 \]

### Final Answer
The loss in kinetic energy during the inelastic collision is:
\[ \text{Loss in KE} = \frac{1}{8} m v^2 \]