Two particles of masses `m_1,m_2` move with initial velocities `u_1 and u_2` On collision, one of the particles get excited to higher level, after absorbing energy `epsilon` if final velocities of particles be `v_1 and v_2` then we must have

Options

A. `1/2m_1u_1^2+1/2m_2u_2^2=1/2m_1v_1^2+1/2m_2v_2^2-epsilon`
B. `1/2m_1u_1^2+1/2m_2u_2^2-epsilon=1/2m_1v_1^2+1/2m_2v_2^2`
C. `1/2m_1^2u_1^2+1/2m_2^2u_2^2+epsilon=1/2m_1^2v_1^2+1/2m_2^2v_2^2`
D. `m_1^2u_1+m_2^2u_2+epsilon=1/2m_1^2v_1^2+1/2m_2^2v_2^2`

Explanation

To solve the problem, we need to apply the principle of conservation of energy. Here's a step-by-step breakdown:

### Step 1: Understand the Initial Conditions
We have two particles with masses \( m_1 \) and \( m_2 \) moving with initial velocities \( u_1 \) and \( u_2 \) respectively.

### Step 2: Identify the Energy Before Collision
The total kinetic energy before the collision can be expressed as:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]

### Step 3: Identify the Energy After Collision
After the collision, one of the particles (let's say particle 1) absorbs energy \( \epsilon \) and moves with a final velocity \( v_1 \). The other particle moves with a final velocity \( v_2 \). The total kinetic energy after the collision can be expressed as:
\[ KE_{\text{final}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + \epsilon \]

### Step 4: Apply Conservation of Energy
According to the conservation of energy, the total energy before the collision must equal the total energy after the collision:
\[ KE_{\text{initial}} = KE_{\text{final}} \]
Substituting the expressions from Steps 2 and 3, we get:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + \epsilon \]

### Step 5: Rearranging the Equation
We can rearrange the equation to isolate the absorbed energy \( \epsilon \):
\[ \epsilon = \left( \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \right) - \left( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \right) \]

### Step 6: Final Expression
Thus, we have the final expression that relates the initial and final kinetic energies along with the absorbed energy \( \epsilon \):
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 - \epsilon = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]