Why is kinetic energy 1/2 mv^2?

[CuriousBanker]

This is probably a stupid question because the answer is likely "it just is".
I am reading Calculus: The elements by Michael Comenetz, and when talking about mQ, he goes on to say "there happens to be advantage in working with 1/2 mQ rather than mQ itself", and then gives the equation k = 1/2 mv^2, but does not explain where the 1/2 came from, he just claims there is an advantage to using it. Is there a reason for this?

 

[SteamKing]

Here is one derivation of the KE formula:

http://hyperphysics.phy-astr.gsu.edu/hbase/ke.html#c2

 

[CuriousBanker]

That doesn't explain why mQ all of a sudden became 1/2mQ without explaining why...is this something I should just accept and move on?

 

[vanhees71]

I don't know what mQ should be, but the energy-conservation law can be directly derived from Newton's equation of motion for a point particle in a conservative force field:
m\ddot{\vec{x}}=-\vec{\nabla} V(\vec{x}).
Multiply this by \dot{\vec{x}} and use the chain rule, and you get
\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right ) = -\frac{\mathrm{d}}{\mathrm{d} t} V(\vec{x}).
Defining the total energy as
E=T+V=\frac{m}{2} \dot{\vec{x}}^2+V(\vec{x}),
the former equation is written as the conservation law
\frac{\mathrm{d} E}{\mathrm{d} t}=0.
This shows that it is convenient to define the kinetic energy as
T=\frac{m}{2} \dot{\vec{x}}^2
with the factor 1/2.

Does this answer your question?

 

[Vagn]

That doesn't explain why mQ all of a sudden became 1/2mQ without explaining why...is this something I should just accept and move on?

It comes about when considering the Work-Energy theorem and using the calculus form of Newton's second law. You can have a look at the derivation in the lower part here if the derivation by Vanhees is hard to follow.

 

[arildno]

While vanhees71 has provided you with an elegant derivation that suggests why the quantity 1/2*mv^2 might be worth giving its own name, that was certainly not clear to previous generations of physicists, and Our modern concept of kinetic energy thus defined gained ascendancy in the Scientific community in the first decades of the 19th Century.

For some of the history, you can check up on the "vis viva"-concept that Leibniz advocated, "vis viva" being, in essence, m*v^2, rather than 1/2*m*v^2:
http://en.wikipedia.org/wiki/Vis_viva
-----------------
A major reason for the rather slow acceptance of/interest in the kinetic energy concept was that it was easily shown by experiments that, in general, this quantity is NOT conserved. The latter half of the 18th century gradually developed the idea that heat was a "type" of "vis viva"/(kinetic) energy, and that the energy of the system was therefore a truly fundamental quantity, on par with momentum (which had been acknowledged a fundamental place in physics since the time of Newton). Without understanding heat as a type of energy, the energy concept itself isn't very interesting.

 

[CuriousBanker]

Ah, see the problem is that this comes up on page 14 of a calc book...which is before derivatives are taught. So I guess for now I should just accept it, and then come back to it later.
Also, one more question if I may, why is q=v^2...how is it determined that v is squared?

 

[Andrew Mason]

Ah, see the problem is that this comes up on page 14 of a calc book...which is before derivatives are taught. So I guess for now I should just accept it, and then come back to it later.
Also, one more question if I may, why is q=v^2...how is it determined that v is squared?

This has to do with the concept of work. Kinetic energy is defined at the ability of a body to do work by virtue of its motion. The amount of work that an object moving at speed v can do is proportional to the square of its speed.

If the force it applies is constant and the body goes from speed v to 0, it is easy to work out: KE = W_v→0 = Fd = mad =m(Δv/Δt)d = m(Δv/Δt)v_avg Δt = mΔv(v_avg )

Since Δv = v - 0 = v and v_avg = (v + 0)/2 you have:

KE = W_v→0 = mv²/2

AM

 

[WannabeNewton]

This is all quite natural using the Lagrangian approach. Let us consider a system in which we just have a free particle moving with constant velocity $\mathbf{v}$ relative to an inertial frame. The homogeneity of space ad time as well as the isotropy of space, for our mechanical system (the free particle), implies that the Lagrangian can only depend on the speed $v$ i.e. $L = L(v^{2})$.

The key tool now is the principle of Galilean relativity because if we have another frame moving with constant velocity $\mathbf{u}$ relative to our inertial frame then by the aforementioned principle, $\mathbf{v}' = \mathbf{v} + \mathbf{u}$. However we also know that then the Lagrangian, under this Galilean boost, must transform in such a way so as to keep the equations of motion the same in both frames therefore the two Lagrangians can differ by at most a total time derivative of a smooth function of time and the configuration space coordinates. We have $L((v')^{2}) = L(v^{2}) + 2\frac{\partial L}{\partial v^{2}}\mathbf{v}\cdot \mathbf{u}$ after a first order Taylor expansion in powers of $\mathbf{u}$ which means $2\frac{\partial L}{\partial v^{2}}\mathbf{v}\cdot \mathbf{u}$ must be a total time derivative as mentioned before which can only be possible if $\frac{\partial L}{\partial v^{2}} = \text{const.} = \alpha$ i.e. $L = \alpha v^{2}$. The constant $\alpha$ is chosen to be $\alpha = \frac{1}{2}m$ ($m$ is called the mass of the particle) and the physical reason for this is given by vanhees' link (the one about $E$ being a first integral of the equations of motion).

If you want to read more, see Landau and Lifgarbagez "Mechanics" if you can get access to it.

 

[Vanadium 50]

WannabeNewton, the OP says this is "before derivatives are taught". Do you believe that pointing someone at this stage to calculus of variations, Lagrangian mechanics and Landau & Lifgarbagez is likely to be helpful?

 

[CuriousBanker]

I appreciate all the responses, but I guess maybe it is best to just accept it as a given, go through the rest of the book, and come back to it later right?
I am self-teaching, that is why I post here and not in a classroom. I took calc I years ago but forgot it. I am self teaching because I want to learn calculus to improve my financial trading skills but also mostly just because I want to learn as much as possible in life...I hope I am able to self-teach most things...I wish I didn't always need to get things intuitively to accept them...I guess I should just accept things as a given until it all comes together, seems to be the best way to get through a subject

 

[WannabeNewton]

WannabeNewton, the OP says this is "before derivatives are taught". Do you believe that pointing someone at this stage to calculus of variations, Lagrangian mechanics and Landau & Lifgarbagez is likely to be helpful?

I'm not seeing how there is any simpler way to derive the expression for kinetic energy without going into that and then adding it on to what vanhees said. How would you do it without derivatives?

 

[Redbelly98]

How would you do it without derivatives?

Algebra-based physics courses teach that acceleration is the slope of a velocity vs. time graph, and that distance traveled is the area under a velocity vs. time graph. Students who have not taken calculus are still taught those concepts in non-calc physics.

We can use the slope and area of a v-vs.-t curve, to look at a constant force F applied to an object that is initially at rest.

The kinetic energy K of the object is equal to the work W done by the (constant) force F:
K=W= F \cdot \Delta x
K = m a \cdot \Delta x
K = m \cdot \frac{\Delta v}{\Delta t} \cdot (\frac{1}{2} \Delta v \cdot \Delta t)
This is where the 1/2 comes in. Δx is the area under the v-vs.-t curve, which is a straight line going from v=0 initially to v=Δv at some time Δt. I have used the formula for the area of a triangle to express Δx in terms of this area.

Now cancel the Δt terms, and we get
K = m \cdot \Delta v \cdot \frac{1}{2} \Delta v = \frac{1}{2} m \Delta v ^2

Hope that helps. I imagine this could also be derived using the kinematic equations for uniform acceleration.

 

[WannabeNewton]

Hi Redbelly! I had interpreted the question as asking "how do we get the expression for kinetic energy without pre-defining it in terms of work etc. i.e. does it come naturally out of some other quantity without actually defining something to be the kinetic energy a priori" but if what was wanted was just to see how it comes out of the definition in terms of work done then I do apologize for the post. I don't suppose it is possible to remove it without removing the context of the consecutive posts?

 

[Dale]

This is probably a stupid question because the answer is likely "it just is".

Kinetic energy is a defined quantity, so we certainly could have chosen to define it with a factor of 1 instead of 1/2. However, we want kinetic energy to have the property that the change in KE is equal to the work done. Therefore you need the factor of 1/2 (the details of why are in the "work-energy theorem" mentioned by Vagn).

 

[MisterX]

We can start with change in velocity equals acceleration times the change in time
<br /> \Delta v = \frac{F}{m} \Delta t<br />

Now multiply both sides by velocity v = \frac{\Delta x}{\Delta t}. On the left I am leaving this as v, but on the right I am going to express it in terms of \Delta x and \Delta t

<br /> v\Delta v = v\frac{F}{m} \Delta t <br />
<br /> v\Delta v = \frac{F}{m} \Delta t \frac{\Delta x}{\Delta t} = \frac{F}{m} \Delta x<br />
<br /> v\Delta v = \frac{F}{m} \Delta x<br />

Now here comes the crucial step, where we get the 1/2.

v\Delta v = \frac{1}{2}\Delta (v^2)

This is a result from calculus. The change in v^2 is equal to twice v times the change in v.

<br /> \frac{1}{2}\Delta (v^2) = \frac{F}{m} \Delta x<br />
<br /> \Delta (\frac{1}{2}m v^2) = F \Delta x<br />

Notice that the delta applies to the entire expression \frac{1}{2}m v^2. This means the change in \frac{1}{2}m v^2 is equal to F \Delta x. Also note that

<br /> \Delta (\frac{1}{2}m v^2) \neq \frac{1}{2}m (\Delta v) ^2 <br />

Redbelly98's derivation is wrong. He used the wrong thing for \Delta x.

 

[MisterX]

Here's an explanation of why
\Delta(v^2) = 2v \Delta v

First we express the change in v^2
\Delta(v^2) = (v + \Delta v)^2 - v^2

= v^2 +2 v\Delta v + (\Delta v)^2 - v^2 = 2 v\Delta v + (\Delta v)^2

Now something we do sometimes when doing calculus is to ignore anything like (\Delta v)^2. We do this because we are only interested in what happens as \Delta v becomes very small. Of course, if \Delta v becomes small then \Delta(v^2) also becomes small. But the two parts, 2v \Delta v and (\Delta v)^2, get small with different orders. For example, we might be interested with the rate that \Delta(v^2) is changing in time.

\frac{\Delta (v^2)}{\Delta t} = 2 v\frac{\Delta v}{\Delta t} + \frac{(\Delta v)^2}{\Delta t}

= 2 v\frac{\Delta v}{\Delta t} + \frac{\Delta v}{\Delta t}\Delta v

\frac{\Delta v}{\Delta t} is the acceleration, and even when \Delta v and \Delta t get very small, a = \frac{\Delta v}{\Delta t} does not have to get small. Just think of \frac{\Delta v}{\Delta t} as being some number. But the second term still has that extra \Delta v, so it goes away when we make \Delta v and \Delta t small.

 

[jtbell]

Here's yet another version that assumes constant force and acceleration, so we don't have to use calculus.

We push on an object with constant force and give it constant acceleration (in a straight line) via F = ma. The object starts at position x₀ with speed v₀. Some time later it ends up at position x with speed v.

Define the work done on the object as W = FΔx = F(x - x₀). Before people started talking about energy in the modern sense, they already knew about work in connection with simple machines such as levers, in which the output work equals the input work.

Let's see what happens if we rewrite the work equation in terms of the initial and final speeds, not positions. One of the standard kinematic equations for constant acceleration which you can find in any introductory physics textbook is

v² = v₀² + 2a(x - x₀)

from which

x - x₀ = (v² - v₀²) / (2a)

Substitute this equation, and F = ma, into the work equation. After cancelling out a, you end up with

W = (1/2)mv² - (1/2)mv₀²

Clearly the quantity (1/2)mv² (for any arbitrary speed) is "interesting". Let's give it a name: (kinetic) energy. Now we can say that doing work on an object changes its energy by an amount equal to the work done.

It turns out you get the same result even if the force and acceleration aren't constant, but you have to use calculus to derive it.

 

[OmCheeto]

This has to do with the concept of work. Kinetic energy is defined at the ability of a body to do work by virtue of its motion. The amount of work that an object moving at speed v can do is proportional to the square of its speed.

If the force it applies is constant and the body goes from speed v to 0, it is easy to work out: KE = W_v→0 = Fd = mad =m(Δv/Δt)d = m(Δv/Δt)v_avg Δt = mΔv(v_avg )

Since Δv = v - 0 = v and v_avg = (v + 0)/2 you have:

KE = W_v→0 = mv²/2

AM

I liked your explanation from 2005.

Here's your equation:
Fd = mad = \frac{mvd}{t} = \frac{mv(vt)}{2t} = \frac{1}{2}mv^2
Where "Fd" = work done on the system.

But like DB, it was still unclear to me where the "1/2" came from:

why is it that distance=vt/2, why divided by 2?

To which you answered:
d = \frac{v_f + v_i}{2}t = vt/2

Looking at it, I realized that the "1/2" can be added right from the start:

Fd = F \frac{v_f + v_i}{2}t = \frac{1}{2} Fvt

Which of course expands out and simplifies to:
1/2(ma)vt = 1/2 mavt = 1/2 m(d/t²)vt = 1/2 m(d/t)v = 1/2 mv² = Ke

So the 1/2 comes from the fact that the distance is the average of the velocities times time. To get an average, you have to divide by 2.

hmmm... Simple...

I never knew where that 1/2 came from either.
My apologies for not understanding everyone's explanations, but I am no longer able to comprehend calculus notations.

--------------------
I was going to be a smarty pants and just tell the OP if he doesn't like the 1/2, then he should just multiply both sides of the following equation by 2.
1/2 mv² = mgh
yielding
mv² = 2mgh
(arildno's formula!)
But then people would ask; "Where did the 2 come from?"
So I decided not to.

 

[CuriousBanker]

omcheeto, you have mvd/t=mvvt/2t
Why is there a 2t in the second denominator? Isnt d=vt? So wouldn't mvd/t=mvvt/t?

I can accept the fact that "it just is", I was just more confused by how we got from mQ to 1/2mQ with no explanation...wasn't sure if the 1/2 was supposed to be obvious to me.

On a separate note (sorry for so many questions), I know f=ma. How is velocity not part of this? Something traveling with a constant velocity and 0 acceleration towards an object...when the two objects collide, there is no force upon impact since there is no acceleration?

 

[CuriousBanker]

Ah, in reading an old post I finally get it...two posts actually

AM wrote:
Quote by fetmar
I have always plugged numbers into the following forumula:

KE = 1/2m*v^2 and it works perfectly.

I don't understand why. Can anybody please tell me why it is one half mass, and not all of the mass? And even more perplexing, is why the velocity needs to be squared.
The Kinetic Energy of a body represents the ability of that body to do work due to its motion. To do work is to apply a force through a distance. So ask yourself how much force x distance a body with mass m and velocity v can apply.

1. Force is the time rate of change of momentum.

2. Since speed is the time rate of change of distance, distance traveled is the average velocity x time.

3. So Force x distance is the time rate of change of momentum x average velocity x time, or (since the time cancels out): change of momentum x average velocity.

Since in going from v to 0, the average velocity is v/2 and the change in momentum is mv. So the work done is v/2 x mv = 12mv2.

That explains 1/2

Then moose wrote: If you understand that Energy is Force * Distance, then it should start to become clear. Imagine you are driving. Keep in mind that the energy required to brake is equal to the kinetic energy of the vehicle while it was in motion... If you are driving at 25 m/s and slow down with a force that causes an acceleration of -5m/s/s, you can do the math and figure out you had a braking distance of 62.5 m. Now, if you were driving at 50 m/s and were to apply the same force, not only would the TIME for you to brake double, your average velocity while slowing down would also double. Therefore the distance you brake for at double the speed, is four times that of the original. In this case, conceptually, the 1/2 is to get the average velocity. If you slow down at an even rate from 50 m/s to a stop, your average velocity was 25 m/s. I really hope this made some kind of sense.

Which explains the squared

I finally get it! woohoo!

Still confused why there is zero force upon impact if velocity is high but no acceleration though

 

[Andrew Mason]

omcheeto, you have mvd/t=mvvt/2t
Why is there a 2t in the second denominator? Isnt d=vt? So wouldn't mvd/t=mvvt/t?

d = vt only if v is constant. We are trying to see how much work a body traveling at speed v can do by applying a force through a distance until it stops. To avoid calculus, we assume that the rate of deceleration is constant so that the average speed is (v_i + v_f)/2. That means the distance is d = v_avgt = [(v + 0)/2]t.

On a separate note (sorry for so many questions), I know f=ma. How is velocity not part of this? Something traveling with a constant velocity and 0 acceleration towards an object...when the two objects collide, there is no force upon impact since there is no acceleration?

The acceleration does not occur before impact. It begins upon impact. The magnitude of the change in velocity depends on how fast it was going immediately before impact (ie. its kinetic energy). The amount of damage caused by the impact will be proportional to v² not v.

AM

 

[CuriousBanker]

So is the damage caused the acceleration of the non moving object that was just impacted, or the deceleration of the moving object? Or is that just the same thing?

 

[Andrew Mason]

So is the damage caused the acceleration of the non moving object that was just impacted, or the deceleration of the moving object? Or is that just the same thing?

A collision between bodies moving at different velocities results in rapid changes in velocity of the bodies. Newton's third law tells you that the forces on each body are equal and opposite: that is, the force of one body on the other is equal to the force of the other body on the first. Those forces cause the damage to the bodies involved in the collision. The forces arise because the electrical forces between atoms in the first body interact with the electrical forces in the second body when they get close. These forces cause the rapid changes in motion to occur.

Damage requires the application of force to the body over a distance ie. work. If energy is dissipated in the collision, work must done on the bodies involved in the collision. The greater the energy that is dissipated in the collision, the greater the work that is done on the bodies, hence, the greater the damage.

AM

 

[CuriousBanker]

Work is force x distance though. What if the thing being impacted does not move? Doesn't it still have damage done to it?

By the way thank you very much for all of your help

 

[Vanadium 50]

If it's dented, something moved, right? Otherwise, there wouldn't be a dent.

 

[nitsuj]

This has to do with the concept of work. Kinetic energy is defined at the ability of a body to do work by virtue of its motion. The amount of work that an object moving at speed v can do is proportional to the square of its speed.

If the force it applies is constant and the body goes from speed v to 0, it is easy to work out: KE = W_v→0 = Fd = mad =m(Δv/Δt)d = m(Δv/Δt)v_avg Δt = mΔv(v_avg )

Since Δv = v - 0 = v and v_avg = (v + 0)/2 you have:

KE = W_v→0 = mv²/2

AM

I vote this best answer, and not just because it's easy to understand but because it's simple.

got me thinking the square of it's speed, this must be "converting" the comparable (frame dependent) measures of time/length units into energy as "applied" to an invariant physical property, mass. It's the only difference I see, between the objects. Talk about equivalence principal right in your face lol, fictitious force pffft, it is what it is

below matter is referred to as a measurable rest mass, hopefully bh holes wouldn't be included in that; that's the nonsense of a photons rest frame lol. In fact there's my retort for the next post asking about a photons perspective. "What's your perspective from a black hole?"

Makes me wonder what it is about mass. why does the "effect" of gravity resemble this speed/mass relationship (kinetic energy) even though the matter is motionless. What is the "speed" part within matter that "creates" this kinda "opposite" kinetic energy? From a dimensional perspective I see EM and mass almost as reverse concepts. Does all that motion of EM within the matter cause mass and in turn the opposite effect of kinetic energy generated by EM transferred to mass (as opposed to being "within" the matter, gravity)

so am thinking matter dissipates "kinetic" energy as gravity ala continuum, measured "invariantly" as mass. I wonder if all those measurements do correlate. Both kinetic "energy" and gravitational potential "energy" seem cut of the same cloth; perhaps just in opposite ways With this perspective it's specifically not gravity that "causes" a bh to form, since KE doesn't either. And the conversion factor of speed to dilation/contraction, seems just as ridiculous as gravity is "weak" compared to mass, it's just "forever extending" and not instantly "expending" (KE)

Long and short of it I wonder if gravity is opposite speed, Now how obvious is that to the space shuttle pilot.
Sorry for posting my musings

 

[CraigHeile]

Energy defines velocity and isn't merely a derivative of it

The one-half becomes an awful lot easier to understand if you start with energy and trace it through as a conserved value. Today we have the wondrous advantange of knowing that energy is always conserved, while Newton had no such understanding, or even a word to describe energy, to my knowledge. Leibniz, a seventeenth century peer of Newton's (they each "invented" calculus independently), was so much closer to understanding this, but by the time his words gained credence, the view of looking at kinetic energy as merely a derivative of velocity had pretty much cemented in place.

When you start an object in motion or make one go faster, you are actually doing this to two objects, in opposite directions. Conservation of momentum requires this. So any energy source applied to create motion is actually being applied to two objects, and therefore kinetic energy can only be a halved quantity. The beauty of momentum conservation, however, is that for the same object to be decelerated to its original stationary position (relative to its initial inertial frame), a similar deceleration must also occur elsewhere, and the associate energy associate with accelerating the object returns to its original value. The offsetting object that is the accomplice to the deceleration is a different object from the one that helped propel the acceleration, yet total energy is conserved as it always must be.

The one-half does not always come into play when considering the relationship between velocity and kinetic energy. It exists for a third party observer (particularly important for one who is trying to do work) but disappears from the perspective of the participants (objects, particles) when they reorient to their new inertial frame. Even for the observer, the relationship grows in straight quadratic fashion (squared without the one-half, or quadrupling for every doubling) as velocities are incrementally changed.

Also, on a completely unrelated note, here is the link to the Landau and Lifgarbagez work that I have seen frequently referenced: http://ia701205.us.archive.org/11/items/Mechanics_541/LandauLifgarbagez-Mechanics.pdf

 

[Andrew Mason]

The one-half becomes an awful lot easier to understand if you start with energy and trace it through as a conserved value. Today we have the wondrous advantange of knowing that energy is always conserved, while Newton had no such understanding, or even a word to describe energy, to my knowledge. Leibniz, a seventeenth century peer of Newton's (they each "invented" calculus independently), was so much closer to understanding this, but by the time his words gained credence, the view of looking at kinetic energy as merely a derivative of velocity had pretty much cemented in place.

Welcome to PF Craig!

I hate to be picky but KE is not really a derivative of velocity: force is the time derivative of mv and it is also the rate of change of kinetic energy with respect to distance. That does not exactly make KE a derivative of velocity.

When you start an object in motion or make one go faster, you are actually doing this to two objects, in opposite directions. Conservation of momentum requires this. So any energy source applied to create motion is actually being applied to two objects, and therefore kinetic energy can only be a halved quantity. The beauty of momentum conservation, however, is that for the same object to be decelerated to its original stationary position (relative to its initial inertial frame), a similar deceleration must also occur elsewhere, and the associate energy associate with accelerating the object returns to its original value. The offsetting object that is the accomplice to the deceleration is a different object from the one that helped propel the acceleration, yet total energy is conserved as it always must be.

Be careful here. The 1/2 term in KE has nothing to do with the interaction of two objects. This sounds like you are saying that when a super ball hits a wall it can only rebound with half of its original KE because it has imparted half of its KE to the wall (earth). Not so. The amount of energy imparted to the wall/earth is miniscule. Or perhaps I have misunderstood what you were trying to say.

Also, Newton's third law says that when two objects interact so that one body applies force to another, there is an equal and opposite force of the other on the first. But the accelerations can be quite different. If one object is the Earth and the other is a ball, the acceleration/deceleration of the Earth is not similar to the deceleration/acceleration of the ball. And kinetic energy will not be conserved as there will be some heat losses/friction. Energy is conserved, but not kinetic energy of the bodies.

AM

 

[CraigHeile]

Welcome to PF Craig!

Thanks so much! I have no idea how to drive in here. This may come out looking really awful.

I hate to be picky but KE is not really a derivative of velocity: force is the time derivative of mv and it is also the rate of change of kinetic energy with respect to distance. That does not exactly make KE a derivative of velocity.

Oh, please do be picky. But here it looks like you at least we're able to capture my meaning, so that's good, although I may change it anyway. After I wrote this I felt that I was imprecise and loose with my words in places, and this was one of those places. It didn't look like anyone was paying real attention anyway, so I didn't come back into it. Perhaps I'll revise tomorrow.

Be careful here. The 1/2 term in KE has nothing to do with the interaction of two objects.

I'm afraid here, Andrew, you may have understood me exactly correctly and I'd like to hone in on this with you if you are willing.

This sounds like you are saying that when a super ball hits a wall it can only rebound with half of its original KE because it has imparted half of its KE to the wall (earth). Not so. The amount of energy imparted to the wall/earth is miniscule. Or perhaps I have misunderstood what you were trying to say.

No, I wasn't trying to say this at all. Mostly I want to get through this initial response here, and once I see it looks like it should then maybe tomorrow I can amend my original post, and/or this one so that I am at least conveying what I intend to.

Also, Newton's third law says that when two objects interact so that one body applies force to another, there is an equal and opposite force of the other on the first. But the accelerations can be quite different. If one object is the Earth and the other is a ball, the acceleration/deceleration of the Earth is not similar to the deceleration/acceleration of the ball. And kinetic energy will not be conserved as there will be some heat losses/friction. Energy is conserved, but not kinetic energy of the bodies.

Yes, totally agree.