What is the relation between Newton's third law and law of conservation of energy? Using newton's third law we can find law of conservation of momentum.

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*$F_{12} = - F_{21}$ $\hspace{2cm}$  Newton's 3rd law


*$F_{12} t = - F_{21} t$ $\hspace{1.8cm}$ Multiplying by "t" o.b.s , it is impulse (this step is important for question)


*$\Delta{P_{12}} = - \Delta{P_{21}}$ $\hspace{1.4cm}$ Change in momentum


*$m_1\Delta{V} = - \space m_2\Delta{V'}$


*$m_1(V_1 - U_1) = - \space m_2(V_2 - U_2)$ $\hspace{1cm}$ Rearrange we get law of conservation of momentum


*$m_1U_1 + m_2U_2 = m_1V_1 + m_2V_2$


*$\sum P_i = \sum P_f$ $\hspace{1.5cm}$ Total momentum before and after collision conserved (remain the same).
Now if in steps-2 we multiply newton's third law with displacement then we can find law of conservation of energy in terms of kinetic energy. Pls confirm the below steps are correct or not , If not why.
$F_{12} \times s_{12} = - F_{21} \times s_{21}$
$\frac{1}{2}m_1 \Delta V^2 = - \frac{1}{2} m_2 \Delta V'\space^2  $
$\frac{1}{2}m_1(V_1^2 - U_1^2) = - \frac{1}{2}m_2(V_2^2 - U_2^2)$
$\frac{1}{2}m_1U_1^2 + \frac{1}{2}m_2U_2^2 = \frac{1}{2}m_1V_1^2 + \frac{1}{2}m_2V_2^2$ $\hspace{1.5cm}$ Law of conservation of energy
Does above equation correct for driving conservation of energy. The total kinetic energy before and after collision remain the same (conserved).
 A: The internal forces can change the internal energy of a system. The absence of external forces does not grant the conservation of KE but it does so for momentum. So, the third law does not imply the conservation of kinetic energy. KE is conserved in some special cases but it is not a consequence of N's 3rd law which is valid in general. And the displacements are equal only in these special cases. You cannot assume they are equal in general.
Also, you call it "conservation of energy" but you are dealing with KE only. Conservation of energy usually refers to total energy or at least mechanical energy (when all forces are conservative). There is no general "conservation of kinetic energy" so there is nothing to prove, really.
In your derivation, you assume that the two displacements have the same sign on both sides of the equation. This is not true in general. The work on the two sides of the equation may have the same sign so the net work is not zero. This is where the difference resides, mathematically, when compared with the momentum case. The time interval has the same sign for both objects but the displacement does not have to.<
For the case of a perfectly elastic collision, we assume, by definition, that the kinetic energy before and after the collision is the same. If this happens, the collision is called "perfectly elastic". What makes this to be so is not some general principle but the elastic properties of the colliding bodies. During any collision, some of the original KE is converted into other forms of energy (related to plastic and elastic deformation, sound, thermal etc.). In the last stage of the collision, when the two bodies start to move apart, some of this energy may be converted back into KE. For very special conditions, related to the properties of the bodies but also on some idealization of reality, all the conversion of the initial KE is into elastic energy of deformation and then this elastic energy is re-converted back fully into KE.
You can see here a difference when you compare this behavior (of the KE) with that of the total energy and total momentum. Their value is the same at any time during a collision (not just before and after), as long as we neglect interactions of the colliding bodies with the outside world. The KE is not like this even for the ideal elastic collision. And there is no way to prove from basic principles that for a given collision the final KE will be the same as the initial one. You need to know in detail the specific properties of the system.
When we solve collision problems we may say that we use conservation of momentum and "conservation of KE" but this does not mean that there is a fundamental law of conservation of KE. It is just a short way to say that we assume that the KE will be completely restored after the collision, with good approximation.
A: The third law does not imply conservation of energy. Suppose the potential energy of a two particle system is $V(x_1,x_2,t)=\frac{t}{x_1-x_2}$.
Then, the forces on the particles are:
$$F_1(t)=-\frac{\partial V}{\partial x_1}=\frac{t}{(x_1-x_2)^2}$$
$$F_2(t)=-\frac{\partial V}{\partial x_2}=\frac{-t}{(x_1-x_2) ^2}$$
The forces are equal and opposite, but the total energy, given by kinetic plus potential energy, is not constant in time because of the lack of time-translational symmetry of the dynamics determined by $V$,
In Lagrangian or Hamiltonian mechanics, it is derived that energy conservation is a consequence of the time symmetry of the dynamics. Newton's third law does not guarantee that.
A simple derivation can look like this:
$$\frac{dE}{dt}=\frac{d}{dt} (\frac{1}{2}m_1 v_1^2+\frac{1}{2}m_2v_2^2 +V(x_1, x_2, t)) $$
$$=m_1a_1v_1+m_2a_2v_2+\frac{\partial V}{\partial x_1} \frac{dx_1}{dt}+ \frac{\partial V}{\partial x_2} \frac{dx_2}{dt}+\frac{\partial V}{\partial t}$$
The first four terms sum to $0$ after using $\frac{\partial V}{\partial x_i}=-m_ia_i$ and $\frac{dx_i}{dt}=v_i$
So we have:
$$\frac{dE}{dt}=\frac{\partial V}{\partial t}$$
You can see that, if $\frac{\partial V}{\partial t}=0$, then $E$ is conserved. The proof in Hamiltonian mechanics is a bit more general than this.
