Conservation of kinetic energy in perfectly Elastic collision Suppose two bodies of mass $m_1$ and $m_2$ moving with velocity $u_1$ and $u_2$ and they collide (perfectly elastic collision) and after collision their velocities are $v_1$ and $v_2$.
Then by law of conservation of linear momentum.
$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$
And if I am given only this information is it possible to prove that the initial kinetic energy of the system is equal to final kinetic energy of the system, or the information is insufficient?
Any help would be appreciated.
 A: By the very definition of an elastic collision, both kinetic energy and momentum are conserved. If it were any collision, you are certain that if no net external forces are acting on the system, the initial momentum equals the final momentum. However, only in perfectly elastic collisions is the kinetic energy conserved.
A: Conservation of linear momentum doesn't equate to collision being elastic ,so no you can't prove energy conservation using that equation
A: The equation for the conservation of momentum does not imply that kinetic energy is equal before and after a collision - oftentimes, it is not. For any input parameters $m_1, m_2, u_1, u_2$, there are an infinite number of possible $v_1$ and $v_2$ that will satisfy the equation and the conservation of momentum. There are few solutions that will also satisfy the conservation of kinetic energy. The fact that a collision is elastic by definition implies conservation of kinetic energy, but the mere conservation of momentum does not imply conservation of kinetic energy (inelastic collisions conserve momentum but not KE).
A: Let's suppose two bodies collide the total kinetic energy from the encounter is equal because of the elastic net conversion K = 1/2 m v2
which is adequate and satisfies the law of conservation in kinematics as well as in non-linear dynamics hence I would say the total energy is conserved throughout unless it were to be acted upon otherwise, so in technicality this would be a perfect frame that we discuss of to have it be conserved. I hope I gave you a bit more of a grasp on the coefficient and those parameters!
A: If for a system:$$\sum F_{external}=0$$
Then  $$\sum_i(m_iv_i)=constant$$
Or simply $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ; for a two body system. But we cant conclude about $KE=constant$, simply from this equation.
So here we use concept of elasticity.
In elastic collisions, kinetic energy remains constant while in inelastic collision it doesn't remains constant.
For checking nature of collision we use Newton's coefficient of restitution ($e$), which is defined as:
$$e=\left|\frac{\Delta v}{\Delta u}\right|$$
Here, $\Delta v$ is velocity of seperation (after collsion) and $\Delta u$ is velocity of approach (before collision).
So we can clearly see that $e\in [0,1]$.
So for, $e=1$, collision is perfectly elastic and kinetic energy remains conserved.
For other values of $e$, we have inelastic collsion, and energy is wasted in form of friction, heat, sound or other forms of energy. So, for our real non-idealistic world, $e\in (0,1)$.
