How to derive that total kinetic energy is conserved during the collision? 
*

*How to derive the below equation from scratch?


*What law support this equation?


*What is the name for it?


*Does it refer to the conservation of kinetic energy?
Like law of conservation of momentum derived from newton's 3rd law.
$\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2$
Where $U_1$ , $U_2$ initial velocities before collision of masses $m_1$ and $m_2$ and $V_1$ , $V_2$ final velocities after collision. This equation related to classical mechanics. And the above equation used for finding $V_1$ and $V_2$ for elastic collision in 1 dimension.
 A: Suppose that $m_1$ and $m_2$ are interacting with each other via some conservative force with a potential of the form $\phi(|x_2-x_1|)$. Then the Lagrangian is $$\mathcal{L}= \frac{1}{2}m_1 \dot x_1^2 + \frac{1}{2} m_2 \dot x_2^2 -\phi(|x_2-x_1|)$$
Since $\mathcal{L}$ does not depend explicitly on time then per Noether's theorem there is a conserved energy. It is given by $$\mathcal{H}=\sum_i \dot x_i \frac{\partial \mathcal{L}}{\partial \dot x_i}-\mathcal{L} $$$$\mathcal{H}= \frac{1}{2}m_1 \dot x_1^2 + \frac{1}{2} m_2 \dot x_2^2 +\phi(|x_2-x_1|)$$
Now suppose further that for $|x_2-x_1|>R$ we have $\phi(|x_2-x_1|)=0$ meaning that beyond a certain distance $R$ the energy stored in the potential goes to zero. Then if we start and end outside of that distance, $R_{initial}>R$ and $R_{final}>R$, we have $$\frac{1}{2}m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \left. \mathcal{H} \right|_{R_{initial}} = \left. \mathcal{H} \right|_{R_{final}}=\frac{1}{2}m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$$ where $u_i = \left. \dot x_i \right|_{R_{initial}}$ and $v_i = \left. \dot x_i \right|_{R_{final}}$
So this result holds whenever there is such a conservative interaction with a short range. Such interactions are called elastic collisions.
A: There is no law of conservation of kinetic energy, but you can derive conservation of energy by first using Newton's Second law to obtain the Work Kinetic Energy Theorem. This states that the work done by the net force on an object is equal to the change in its kinetic energy:
$$
W_{F_{\rm net}} = \Delta K = \frac{1}{2} m v_f^2 -  \frac{1}{2} m v_i^2
$$
Then you can say that all forces are either conservative or non-conservative, and that the work done by conservative forces can be written as the (negative) change in potential energy:
\begin{align}
W_{F_{\rm net}} &= K_f - K_i\\
W_{F_{\rm non-conservative}} + W_{F_{\rm conservative}} &= K_f - K_i\\
W_{\rm nonconservative} + \left(- \Delta U\right) &= K_f - K_i\\
W_{\rm nonconservative} - \left( U_f - U_i\right) &= K_f - K_i \\
K_i + U_i + W_{\rm nonconservative} &= K_f + U_f\\
ME_i + W_{\rm nc} &= ME_f
\end{align}
This is the law of conservation of energy.  In the case that there is no work done by non-conservative forces ($W_{\rm nc}=0$), you obtain conservation of mechanical energy ($ME = K + U$).  And in the case that there is (1) no work done by non-conservative forces and (2) no change in the potential energy, you could write:
\begin{align}
K_i = K_f
\end{align}
A: In general, momentum is conserved in a collision, not kinetic energy.
Kinetic energy is conserved in elastic collisions, and you can use this condition as the definition of an elastic collion.
