Two particles of masses $m_{1},m_{2}$ move with initial velocities $u_{1},u_{2}$. On collision, one particle gets excited to a higher level, after absorbing energy $\epsilon$. If final velocities of particles be $v_{1},v_{2}$ then we have
\[\begin{align}
  & A.\dfrac{1}{2}{{m}_{1}}u_{1}^{2}+\dfrac{1}{2}{{m}_{2}}u_{2}^{2}-\varepsilon =\dfrac{1}{2}{{m}_{1}}v_{1}^{2}+\dfrac{1}{2}{{m}_{2}}v_{2}^{2} \\
 & B.\dfrac{1}{2}m_{1}^{2}u_{1}^{2}+\dfrac{1}{2}m_{2}^{2}u_{2}^{2}-\varepsilon =\dfrac{1}{2}m_{1}^{2}v_{1}^{2}+\dfrac{1}{2}m_{2}^{2}v_{2}^{2} \\
 & C.m_{1}^{2}u_{1}^{2}+m_{2}^{2}u_{2}^{2}-\varepsilon =m_{1}^{2}v_{1}^{2}+m_{2}^{2}v_{2}^{2} \\
 & D.\dfrac{1}{2}{{m}_{1}}u_{1}^{2}+\dfrac{1}{2}{{m}_{2}}u_{2}^{2}=\dfrac{1}{2}{{m}_{1}}v_{1}^{2}+\dfrac{1}{2}{{m}_{2}}v_{2}^{2} \\
\end{align}\]

Hint: A elastic collision occurs when the total kinetic energy of the two bodies interacting bodies is conserved. Here, we can consider the interaction of the molecules with each other to be in one direction and solve the following.

Formula used:
 $\Delta KE=0$

Complete step-by-step answer:
We know that collision is the interaction between two bodies which are moving with some velocities for a short duration of time. We also know that when two identical bodies undergo collision there is a transfer of momentum and energy between the bodies. Generally, collision results in the change in velocity of the interacting bodies. Since velocity is a vector, we can say that collision changes the direction and speed of the interacting bodies.
We know from the conservation of momentum, that the total momentum of the system remains unchanged. However, the kinetic energy of the system depends on the type of collision occurring.
Let us assume that the given particles undergo elastic collision, then we know that the sum of the initial kinetic energies is equal to the sum of the final kinetic energies.
Here, the initial kinetic energy is given as $\dfrac{1}{2}m_{1}u_{1}+\dfrac{1}{2}m_{2}u_{2}$
Also, the final kinetic energy contains the absorbed energy, then it is given as $\dfrac{1}{2}m_{1}v_{1}+\dfrac{1}{2}m_{2}v_{2}+\epsilon$
$\implies \dfrac{1}{2}m_{1}u_{1}+\dfrac{1}{2}m_{2}u_{2}=\dfrac{1}{2}m_{1}v_{1}+\dfrac{1}{2}m_{2}v_{2}+\epsilon$
$\implies \dfrac{1}{2}m_{1}u_{1}+\dfrac{1}{2}m_{2}u_{2}-\epsilon
=\dfrac{1}{2}m_{1}v_{1}+\dfrac{1}{2}m_{2}v_{2}$
Hence the correct option is \[A.\dfrac{1}{2}{{m}_{1}}u_{1}^{2}+\dfrac{1}{2}{{m}_{2}}u_{2}^{2}-\varepsilon =\dfrac{1}{2}{{m}_{1}}v_{1}^{2}+\dfrac{1}{2}{{m}_{2}}v_{2}^{2}\]

So, the correct answer is “Option A”.

Note: Collision can be broadly classified into two types of collision namely the elastic collision, and the inelastic collision. In elastic collision, both the momentum and the kinetic energy of the system is conserved. In inelastic collisions, the momentum of the system is conserved while the kinetic energy is not conserved.