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 absording enegry. If final velocities of particles be `v_(1)` and `v_(2)` then we must have

Options

A. `m_(1)^(2)u_(1)+m_(2)^(2)u_(2)^(2)-epsi=m_(1)^(2)v_(1)+m_(2)^(2)v_(2)`
B. `1/2m_(1)u_(1)^(2)+1/2m_(2)u_(2)^(2)=1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)-epsi`
C. `1/2m_(1)u_(1)^(2)+1/2m_(2)u_(2)^(2)-epsi=1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)`
D. `1/2m_(1)u_(1)^(2)+1/2m_(2)u_(2)^(2)+epsi=1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)`

Explanation

To solve the problem, we need to analyze the situation using the principles of conservation of energy and the concept of kinetic energy.

### Step-by-Step Solution:

1. **Identify the Initial Kinetic Energy:**
The initial kinetic energy (KE_initial) of the two particles before the collision can be expressed as:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]

2. **Identify the Final Kinetic Energy:**
After the collision, the final kinetic energy (KE_final) of the two particles can be expressed as:
\[ KE_{\text{final}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

3. **Energy Absorbed by the Excited Particle:**
Let \( E \) be the energy absorbed by one of the particles during the collision that causes it to get excited to a higher energy level. According to the conservation of energy, the initial kinetic energy is equal to the final kinetic energy plus the energy absorbed:
\[ KE_{\text{initial}} = KE_{\text{final}} + E \]

4. **Set Up the Equation:**
Substituting the expressions for initial and final kinetic energy into the equation gives us:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + E \]

5. **Rearranging the Equation:**
Rearranging the equation to isolate the energy \( E \) yields:
\[ E = \left( \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \right) - \left( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \right) \]

6. **Final Expression:**
Thus, we can conclude that the energy absorbed during the collision, which causes one of the particles to become excited, can be expressed as:
\[ E = \frac{1}{2} m_1 (u_1^2 - v_1^2) + \frac{1}{2} m_2 (u_2^2 - v_2^2) \]