Note that energy is not conserved when there is kinetic friction. In this scenario it is better to analyse the impulse on infinitesimal elements to get force and power. Energy is dependent on reference frame, but energy dissipated is not dependent on the reference frame.
$$ m\ddot x-F=0 \implies \int (m \ddot x-F)\,dx=0 $$
$$ \int (m \frac{d\dot x}{dx}\dot x -F)\,dx=0 $$
$$ \frac{1}{2}m\dot x^2-\int F \, dx=E ,\quad E \in \R $$
This is essentially proving the work-energy theorem.
Generally, energy methods are superior to force methods. The only exception is, well, when energy isn’t conserved. Energy methods also have the benefit that they usually only depend on the macroscopic state of the system, and insensitive to small details that may not be easily found. E.g. efpho 2007 t2, eotvos 2018 t1.
Total linear and angular momentum about any point is always conserved Angular momentum of each object is always conserved about the collision point. Total energy is only conserved in an elastic collision
Note that very massive objects can absorb momentum, but not energy, so it may appear that angular/linear momentum is not conserved (but it is).
By Newton’s third law, the impulse on each object is equal and opposite.
This is equivalent to momentum conservation, but the important thing to note is the direction of the impulse (along the normal, without friction). Conservation of energy has to be used to find the magnitude of the impulse exerted. Alternatively, if the final relative velocity is known (through the coefficient of restitution), then we can use this to solve for the final state. It is also useful when there is friction in the collision, as the ratio of the tangential impulse to the normal impulse is $\mu_k$.
In other words, where energy conservation fails, we analyse the specifics of the collision to figure out the final state. This is similar to how we can solve fluid flows even if energy is not conserved, by making use of additional information given.
CM frame is very useful when dealing with point mass collisions.
In the CM frame, since linear momentum is conserved (sum to 0), the final speed of each mass must be an equal multiple of the initial speed, and the direction of the velocity of one mass is opposite to the other. For elastic collisions, energy is conserved so the speed of each mass is conserved.
This means that for a system constrained to a line, the final velocity of each mass is equal and opposite to what it began with. (multiple must be $\pm 1$, but it can’t be $+1$ as that means no collision occurred)
$$ v_1'=-v_1+2v_{CM} \qquad v_2'=-v_2+2v_{CM} $$
If $m_1=m_2$, their velocities will simply swap.