Why is the conservation of momentum not equivalent to the conservation of energy in Newtonian mechanics? Conservation of momentum is Newton's third law. Conservation of energy is attained in theory by defining work, deducing the work-energy principle and specifying that if a force or system is conservative, then that which changes when work is done (energy) is always matched with a change of opposite sign at the same time. There can be no acceleration without a change in $ \frac{1}{2} m v^2$. This is so, both obviously because acceleration means a change in velocity, but also theoretically because a change in that quantity denotes work, and work denotes where force has acted (and force denotes where there is acceleration). We call that quantity energy. In a conservative system, no particle can accelerate while another decelerates by an equal amount in some sense. (This is a non-punctilious definition because it uses "in some sense", but as I will formally ask later, my entire question is asking whether it matters what that "in some sense" is, because in the end both momentum and energy only have reference to $F=ma$ and therefore acceleration). For if that were not the case, the system could accelerate forever and there would be no conservation of anything, only a surfeit, an excess. A conservation of excess, perhaps, haha.
Nevertheless, KE seems a good way to formalise this conservation, because a change in KE (i.e. work) is necessary for acceleration to take place. To make the conservation law work, we must make KE denote this "thing" (measured in joules) which "thing" does not change in quantity over time when it is summed across the entirety of the system. Hence, we can deduce, given particular conservative laws like gravity, a kind of energy which is different to KE but nonetheless denotes the same kind of "thing", and which sums with KE always in equal amounts of opposite sign. For gravity, given the Law of Universal Gravitation, we see what the work done is by integrating that law with respect to displacement. Subsequently, we flip the sign of the result of the integral and call it "the potential energy", because when you add it to the "kinetic energy" it adds to zero. In other words, the amount of "thing" is conserved with time. In more theoretically meaningful words, there is always a net work of zero. (In other words inferred, but not deduced from those words there is always a net momentum of zero; there is always a net acceleration (weighted according to momentum) of zero. This is a claim which is not deeply explored by me, but sounds simultaneously intuitive and counterintuitive. This is the claim that I am asking you to either corroborate or completely butcher: is conservation of energy not equivalent to conservation of momentum? (To keep some reputation before the claim is butchered I will say that I think that if it is false it would be because momentum and energy are related to acceleration in different quantitative ways or degrees)).
Changes in energy are by definition necessary for acceleration. If conservative forces cause an increase in kinetic energy and thereby acceleration, by the same amount that they cause a simultaneous decrease in energy and thereby a deceleration of another particle, since we are only talking about accelerations in the end, why is it that we require the extra layer of abstraction over $ F=ma $ and momentum in order to explain the finite of the system's ability to accelerate? The conservation of energy in the end only means that when particle A collides with particle B, it is not the case that nothing is lost between the particles such that the particles do not ever decelerate. Is this not precisely what is achieved by the conservation of momentum? Thanks.
 A: 
is conservation of energy not equivalent to conservation of momentum?

The conservation of energy is not equivalent to the conservation of momentum.
Per Noether’s theorem momentum is conserved whenever the Lagrangian is symmetric under spatial translations and energy is conserved whenever the Lagrangian is symmetric under time translations. Since it is possible to write down a Lagrangian with time translation symmetry but not space translation symmetry and vice versa you can show that they are not equivalent.
Of course, the actual physical laws show both symmetries, so they both conservation laws are always present*. But they are not equivalent because they stem from different symmetries.
*Except for cosmology where the cosmos lacks time translation symmetry and therefore does not conserve energy at cosmological scales

Changes in energy are by definition necessary for acceleration.

You said this several times but it is not true. Uniform circular motion is a counter example. In uniform circular motion energy is constant despite the continuous acceleration.
A: The other answer is very good, but not useful if we don't understand Noether's theorem (or Lagrangians). Instead, we can reason about this in a less difficult but more roundabout way.
You ask if Conservation of Energy is equivalent to Conservation of Momentum, i.e.
$$CoE \iff CoP$$
First, we will disprove
$$CoE \rightarrow CoP$$
We can do this by finding a situation where $\Delta 1/2 mv^2 = 0$, but $\Delta m\vec{v} \neq 0$. The example of circular motion indicates this, because while $1/2 mv^2$ does not change, and neither does $mv$, but the vector quantity $m\vec{v}$ changes in direction, and therefore momentum is not conserved in this example. We see that there must be some external force that accelerates the particle (centripetal), but this force does $0$ work, and therefore conserves energy and not momentum.
$$CoP \rightarrow CoE$$
This statement is also false, we can disprove it by finding the opposite scenario, where momentum is conserved, but total energy is not. Take for example two skaters side by side. They push off of one another, and begin to move in opposite directions. Conservation of momentum ensures that they move in equal speed and opposite directions, so center of mass remains still, and the momentum of the center of mass remains $0$. (total momentum is also 0) However, the total energy of the system is not zero. There are internal forces which act on the two skaters that do work, but are not relevant to the motion of the center of mass.
Why can this be? Well, we know that in physics there are independent variables (for example, $x$ and $v$). Being independent, there is no specific function $f$ that relates $x$ and $v$, even though they are related through a derivative, $\frac{d}{dt} x = v$. This means that there is guarantee that $\Delta x = 0$ implies $\Delta v = 0$, or vice versa. This is a mathematical truth.
Similarly, Energy and Momentum are also related by a derivative, $F = dP/dt$, and $E = \int F dx$. Intuitively, there can be no way to guarantee that $\Delta P = 0$ implies $\delta E = 0$. Of course, there may be special cases where both are true simultaneously, but nonetheless, one is not causing the other. In physics, we generally take both conservations.
A: Let me first discuss a reduced case, for the purpose of focus.
(The concluding statement - all the way down - explains the non-equivalence.)

I will first discuss the case of motion in a space with 1 spatial dimension. In classroom demonstrations this is of course the demonstration of two or more carts moving over an air track
In this reduced environment, let's look at momentum and kinetic energy. The expression for momentum carries directional information. Motion to the left or motion to the right are different cases, you have to keep track of that. Of course, the efficient way of expressing that directional information is to use a minus sign. If we express velocity to the right with a positive number then velocity to the left is expressed with a negative number.
If Newton's second law holds good then it follows that when two objects undergo elastic collision the velocity of the common-center-of-mass will not change.
(Both Newton and Huygens describe they did extensive experiments with colliding pendulum bobs (both equal and unequal mass), to corroborate this theoretical property.)
Of course, this velocity-of-the-common-center-of-mass-does-not-change is a way of expressing conservation of momentum.

Kinetic energy
People such as Leibniz pointed out another implication of Newton's second law: in elastic collisions another quantity is conserved: a quantity proportional to the square of the velocity. The conserved quantity: 'kinetic energy' $\tfrac{1}{2} m v^2$
This conservation of a quantity that is obtained by squaring is quite surprising, of course.
Squaring the value of the velocity means you lose the directional information.

Space with three spatial dimensions
Our actual space has three spatial dimensions. Does conservation of kinetic energy still hold good? It does, because as we know Pythagoras' theorem relates the squares of coordinates. This means that no matter the number of spatial dimensions, the concept of kinetic energy is always consistently defined.
Information loss
Just as in a space with 1 spatial dimension we have that using the square of the velocity means you lose the directional information.

Concluding:
Momentum and kinetic energy are both conserved (in elastic collisions), and yet fundamentally distinct because momentum carries directional information and kinetic energy does not.
