Car accident

Let's consider the kinetic energy of the cars and the work wich dameges all of them. $K_A=\frac{1}{2}m_Av_A^2; \quad K_B=\frac{1}{2}m_Bv_B^2; \quad K_C=\frac{1}{2}m_Cv_C^2; \quad \text {with}\,\, v_A=v_B=v, \quad m_A=m_B=m_C=m.$ The text of the problem says that we can write $K_A+K_B=mv^2=W_{AB}=W_A+W_B; \,\,\,\,\,\,\, W_C=K_C=\frac{1}{2}mv_C^2=W_A=W_B=\frac{W_{AB}}{2}$ where $W_{AB}$ is the total energy in the first scenario transformed into work equally reparted between $A$ and $B$ so that $W_A=W_B=(1/2)W_{AB}$ while the work $W_C$ is equal to both $W_A$ and $W_B$ because the car $C$ reports the same damage as $A$ and $B$ taken separately. Now, considering $v$ and $v_C$ being positive since they are the moduli of two vectors, we conclude that $\frac{1}{2}mv^2=\frac{1}{2}mv_C^2 \Leftrightarrow v^2=v_C^2$ so, by taking the positive solution, we have $v_C=v$ .