Why is KE not conserved when momentum is?

[mark2142]

Its clear in elastic collision that both KE and momentum is conserved. Bodies exchange their velocities. It is seen clearly in this video. There is no decrease in speed. Total KE is constant.

But in an inelastic collision momentum is conserved again but not the KE. There is loss in KE (I guess total). I don't really understand how is momentum conserved but KE is not. Both depend on v. So if there is a loss in KE or speed then that means there is a loss in momentum too.

If we look at it from different point of view. Like let's say the momentum is always conserved no matter what if external force is zero. Then if we take a Newtons cradle for example.
Then assuming $v_i= 1 m/s$ of the first ball and $m= 0.1 kg$ of all five balls.
$$p_i=0.1 kg m/s$$
That means $$p_f=0.1 kg m/s$$ too
So if the velocity of say the last ball is $v=0.8$ m/s then according to conservation of momentum $v=0.05 m/s$ of the remaining first four balls. Yes?
So is it right that the velocity is governed by the law of conservation of momentum and assuming that if the velocity of first ball is 0.8 m/s then the velocity of the 4 remaining balls should be 0.2 each(i.e the velocity of the 4 ball system will be 0.2) is wrong?
( It will be 0.2/4 each).
Is according to the momentum left v is decided to each body?
(weird!)

Now if we look at the KE then:
$KE_i= 0.05 J$
$KE_f= 0.032+0.0005=0.0325 J$
(loss of KE=0.0175 J)
Is this purely mathematical because I don't see a reason why the KE will be lost when momentum is not because both depend on m and v?

There is also a line that Feynman says: https://www.feynmanlectures.caltech.edu/I_10.html
"That the speeds before and after an elastic collision are equal is not a matter of conservation of momentum, but a matter of conservation of kinetic energy. That the velocities of the bodies rebounding after a symmetrical collision are equal to and opposite each other, however, is a matter of conservation of momentum."
Is this because of speed is scalar as well as KE whereas momentum and velocity are vectors?

 

[Orodruin]

So if there is a loss in KE or speed then that means there is a loss in momentum too.

No, this is an erroneous conclusion.

KE and momentum depend on velocity in different ways.

Consider the center of mass frame where total momentum is zero by definition. In an elastic collision of two objects, kinetic energy is conserved and both objects will move after the collision (they do not ”exchange velocities”, generally the velocity of one object after collision will not be that of the other before collision). In a completely inelastic collision, both objects stick together and overall momentum is still zero so they do not move. Hence KE is not conserved but momentum is.

 

[mark2142]

In a completely inelastic collision, both objects stick together and overall momentum is still zero so they do not move. Hence KE is not conserved but momentum is.

That is a very general statement. In order to get a feel for it(Why Ke is not conserved in inelastic collision) I am talking about subtleties in a Newtons cradle.

 

[Orodruin]

That is a very general statement. In order to get a feel for it(Why Ke is not conserved in inelastic collision) I am talking about subtleties in a Newtons cradle.

The generality still applies. There is no reason why KE would be conserved generally just because momentum is.

 

[Ibix]

The way I think of it is this:

Both momentum and energy are always conserved.

Energy, however, frequently ends up being stored in deformed bodies or dissipated as heat. This is where the difference between starting and finishing energy ends up. It's typically difficult to calculate from first principles, so we usually say "energy (or at least the bits that we are measuring) is not conserved".

The key point is that you can't convert momentum into heat or material stresses or whatever. So we can't "lose track" of momentum in our accounting, but we can lose track of energy.

 

[mark2142]

Both momentum and energy are always conserved.

So is it right that the velocity is governed by the law of conservation of momentum and assuming that if the velocity of first ball is 0.8 m/s then the velocity of the 4 remaining balls should be 0.2 each(i.e the velocity of the 4 ball system will be 0.2) is wrong?
( It will be 0.2/4 each).

So momentum governs velocity not that velocity governs momentum?

 

[hutchphd]

So momentum governs velocity not that velocity governs momentum?

Governs? Physics is egalitarian.

 

[mark2142]

Governs? Physics is egalitarian.

I mean momentum decides velocity. In any collision it looks like v is transferred to other body but actually momentum is transferred not v. yes?
That makes sense?

 

[Ibix]

I mean momentum decides velocity. In any collision it looks like v is transferred to other body but actually momentum is transferred not v. yes?
If that makes sense!

The conserved quantity is momentum, yes, and it makes a lot more sense to talk about transferring momentum than transferring velocity. You'd have to implicitly compute momentum in order to calculate a velocity transfer anyway.

Like @hutchphd, I wouldn't talk about one quantity "governing" another. I might say the whole process is governed by various conservation laws, including conservation of momentum but not "conservation of velocity" (because there ain't no such animal).

 

[mark2142]

The conserved quantity is momentum, yes, and it makes a lot more sense to talk about transferring momentum than transferring velocity. You'd have to implicitly compute momentum in order to calculate a velocity transfer anyway.

Thank you. yes.
Now can I say the velocity or speed is transferred when the collision is elastic because it looks like it?

 

[Ibix]

Thank you. yes.
Now can I say the velocity or speed is transferred when the collision is elastic because it looks like it?

You can say whatever you like. You're still going to have to compute momentum and energy transfers. The "equal and opposite velocities swap from one object to the other" thing only applies to equal mass objects - for a counter example imagine an elastic collision between a tank and a ping pong ball with equal and opposite velocities.

 

[sophiecentaur]

I don't really understand how is momentum conserved but KE is not.

You are in very good company. In Newton's day, there was a lot of argument / discussion about how to 'describe' a suitable quantity to associate with motion and 'substance' (Mass, these days). Momentum and Kinetic energy (not those words exactly) were both candidates but it was finally acknowledged that both were needed if you wanted to calculate and predict the outcomes of collisions etc.

There are many collision situations in which the Momentum starts of as Zero (equal and opposite velocities and equal masses) and ends up with zero. But the initial total KE can be very high (two fast objects) and ends up with either High total KE (steel ball bearings flying apart fast) or Zero (lumps of putty sticking together and stationary).

I mean momentum decides velocity.

Stick to the definition of momentum, P = mv. "Decides" is not maths. When you get down to it, Physics replaces (or attempts to replace) arm waving terms with mathematical relationships between quantities. You just have to be rigorous. If something just doesn't feel right then you may well be wrong if your feelings aren't supported by the maths. Everyone gets this at times.

 

[mark2142]

a counter example imagine an elastic collision between a tank and a ping pong ball with equal and opposite velocities.

That is a special case with large difference in mass.
I have a very dumb question. If velocity v is transferred during elastic collision between two bodies and we see the balls going up and as fast as they come together then why don't we describe 'v' as Kinetic energy instead of 1/2mv^2?
That would make much more sense.

 

[nasu]

Two bodies of equal mass, this is a special case. Two bodies with different masses is the general case. There is no "velocity transfer" in general.

 

[Nugatory]

why don't we describe 'v' as Kinetic energy instead of 1/2mv^2?

Because if we bring an object moving at speed $v$ to a complete stop with a friction brake, we will find that the heat energy ”created” (actually the brakes are turning the kinetic energy into heat) is equal to $mv^2/2$. Conversely, we need to “spend” (actually convert to kinetic energy from chemical, electrical, or heat energy) at least that much energy to accelerate an object initially at rest to speed $v$.

 

[vanhees71]

I'd not say "velocity is transferred". I would use this wordings only for "extensive conserved quantities" like energy and momentum.

From a modern point of view, it turns out that energy and momentum are useful quantities, because they are related to the symmetry of space and time. In Newtonian (and also in special relativistic) physics space (for an inertial observer) is Euclidean and thus homogeneous. This means that on a fundamental level the natural laws must be independent of the place where you do an experiment, and this implies, according to a beautiful theorem by Emmy Noether (1918) that momentum is conserved for a closed system. Another symmetry is "time-translation invariance", which says that the natural laws do not change with time, and this symmetry implies energy conservation. This analysis shows that energy and momentum conservation are independent conservation laws.

An elastic collision of two bodies by definition means by definition that both energy and momentum are conserved, i.e., that you can consider the system as closed, i.e., there is no energy transferred to some form of intrinsic energy due to the deformation or heating of the colliding bodies.

Even for the case of one-dimensional elastic collisions momentum conservation is not enough to determine the momenta after the collision from the given momenta before the collision, because you have only one equation,
$$m_1 v_1 +m_2 v_2=m_1 v_1'+m_2 v_2'$$
for the two unknown velocities $v_1'$ and $v_2'$. You need one more equation, and for an elastic collision that's energy conservation, i.e.,
$$ \frac{m_1}{2} v_1^2 + \frac{m_2}{2} v_2^2 = \frac{m_1}{2} v_1^{\prime 2} + \frac{m_2}{2} v_2^{\prime 2}.$$
This is now indeed sufficient to solve for $v_1'$ and $v_2'$. To set end we bring everything related to particle 1 on one and everything related to particle 2 to the other side of the two equations, leading to
$$m_1 (v_1-v_1') = m_2 (v_2'-v_2), \quad m_1 (v_1^2-v_1^{\prime 2}) = m_2 (v_2^{\prime 2}-v_2^2).$$
With the 3rd binomial formula you can rewrite the 2nd equation and use the first one
$$m_1 (v_1 -v_1')(v_1+v_1') = m_2 (v_2'-v_2) (v_2'+v_2) \; \Rightarrow \; v_1+v_1'=v_2'+v_2.$$
The latter is the 2nd linear equation in addition to the momentum-conservation equation you need to solve uniquely for $v_1'$ and $v_2'$.

To that end first multiply by $m_1$ and add the equation to the momentum-conservation equation, leading to
$$2 m_1 v_1=(m_1+m_2) v_2' +(m_1-m_2) v_2$$
or
$$v_2'=\frac{m_1 (2 v_1-v_2)+m_2 v_2}{m_1+m_2}=\frac{2 (m_1 v_1 +m_2 v_2)}{m_1+m_2}-v_2.$$
In the same way by multiplying with $m_2$ and subtracting from the momentum-conservation equation you get
$$v_1'=\frac{m_2(2v_2-v_1) + m_1 v_1}{m_1+m_2}=\frac{2 (m_1 v_1+m_2 v_2)}{m_1+m_2}-v_1.$$

 

[PeroK]

That is a special case with large difference in mass.
I have a very dumb question. If velocity v is transferred during elastic collision between two bodies and we see the balls going up and as fast as they come together then why don't we describe 'v' as Kinetic energy instead of 1/2mv^2?
That would make much more sense.

IMHO you can't learn physics thinking like that.

Newton's third law directly implies conservation of momentum.

Note that momentum is a vector. There is conservation of momentum separately in all three spatial dimensions.

 

[sophiecentaur]

That would make much more sense.

Perhaps to you but not to Physicists wot know a thing or two about this. This is very basic stuff and has been thrashed out many times by cleverer people than you and me. If you try and make up your own version of stuff like that, your system just won't work. Even people like Albert Einstein started off with the accepted ideas and then moved us on a little bit.

 

[Ibix]

That is a special case with large difference in mass.

So what? Consider a tennis ball and a ping pong ball. You won't get equal and opposite velocities in any case with unequal masses.

I have a very dumb question. If velocity v is transferred during elastic collision between two bodies and we see the balls going up and as fast as they come together then why don't we describe 'v' as Kinetic energy instead of 1/2mv^2?

All quantities in physics are defined because they're useful. Velocity is useful because it is directly related to "do I need to get out of the way of that car?" and "how much time do I have to do so?". I don't need to worry about the mass of the car to do this calculation, which makes sense - why would mass be involved in answering "is that thing coming towards me?". On the other hand, if I don't manage to get out of the way then there is a collision and momentum and energy are needed to calculate the energy transfer, which relates to how much this is going to hurt.

Don't forget that there are other types of problems than the ones you are currently working on.

 

[sophiecentaur]

So what? Consider a tennis ball and a ping pong ball. You won't get equal and opposite velocities in any case with unequal masses.

I sympathise with your attempt to make all this more approachable but, when you get down to it, the maths of the topic are basically very straightforward and there's a serious limit to what can be explained / justified without maths. Anyone who cannot or who will not use the maths will never get hold of the topic.

Analogies and examples can only take people so far. OH - I just read my signature, below. Well, at least I'm consistent.

 

[jbriggs444]

we see the balls going up and as fast as they come together then why don't we describe 'v' as Kinetic energy instead of 1/2mv^2?

This almost sounds as if you are trying to suggest the magnitude of closing velocity as a conserved quantity for elastic two-body collisions.

And indeed, for a two body collision, the magnitude of closing velocity is conserved if kinetic energy and momentum are.

But kinetic energy is useful because it tells you how fast you need to be going to coast to a stop at the top of a hill. For a fixed mass, the height that you can reach is directly proportional to kinetic energy. It is not directly proportional to velocity.

We use different words for different things because those things are different. That's how language works.

 

[mark2142]

Because if we bring an object moving at speed $v$ to a complete stop with a friction brake, we will find that the heat energy ”created” (actually the brakes are turning the kinetic energy into heat) is equal to $mv^2/2$. Conversely, we need to “spend” (actually convert to kinetic energy from chemical, electrical, or heat energy) at least that much energy to accelerate an object initially at rest to speed $v$.

Okay. That makes much more sense than just using v. There is mass involved also since we need more energy to accelerate to a velocity v if the mass is more or more K energy is converted to more heat if mass is more when stopping a car. For a given velocity more mass means more energy needed or more energy is lost in other forms.

On the other hand, if I don't manage to get out of the way then there is a collision and momentum and energy are needed to calculate the energy transfer, which relates to how much this is going to hurt.

Yeah! If we think like that then a tank and a ping pong ball has huge difference in KE due to mass and after collision the KE of the ball will increase tremendously whereas the KE of the tank will be unaffected. That is due to huge mass and velocity, actually the KE ,not due to velocity alone.We should be talking in terms of energy and momentum and not just velocity. Yes?

 

[mark2142]

Perhaps to you but not to Physicists wot know a thing or two about this. This is very basic stuff and has been thrashed out many times by cleverer people than you and me. If you try and make up your own version of stuff like that, your system just won't work. Even people like Albert Einstein started off with the accepted ideas and then moved us on a little bit.

Okay. But in books it is not properly explained. So I came here.

 

[jbriggs444]

after collision the KE of the ball will increase tremendously

Or decrease or stay the same. Head-on collisions with on-coming bodies are not the only possible situation.

In an elastic head-on collision with an equal speed tank, the ping-pong ball will have its KE increased by approximately a factor of nine. The change in kinetic energy of the tank will be equal and opposite, approximately eight times the initial kinetic energy of the ping-pong ball.

 

[mark2142]

Or decrease or stay the same. Head-on collisions with on-coming bodies are not the only possible situation.

In an elastic head-on collision with an equal speed tank, the ping-pong ball will have its KE increased by approximately a factor of four. The change in kinetic energy of the tank will be equal and opposite, approximately three times the initial kinetic energy of the ping-pong ball.

So that is the reason why we can't ignore mass or more presicely the Kinetic energy. It plays the main role in speeding up the body? (We can't just say the velocities are transferred.That would be wrong?).

 

[vanhees71]

Okay. But in books it is not properly explained. So I came here.

For the elastic central collision, I've tried to explain it in #16. If you have questions, I'm happy to help further!

Don't be afraid of math. All the many words don't help much in really understanding a physical phenomenon. The only adequate, precise enough "language" is math! Here it's not too complicated math, i.e., pure algebra.

 

[jbriggs444]

So that is the reason why we can't ignore mass or more presicely the Kinetic energy. It plays the main role in speeding up the body? (We can't just say the velocities are transferred.That would be wrong?).

You seem to want to find the "one true quantity" that explains everything. There is no such thing.

We can compute various quantities. Some of them are useful in many different ways. We give the useful quantities names. Like "mass", "time", "position", "velocity", "momentum", "kinetic energy", "potential energy", "work" and "impulse".

We do not bother giving names to less useful quantities like $v^3$ or $\sqrt{m}$.

Like any good workman, we choose the tools that are appropriate for the task at hand. We do not use a chisel to turn a screw or a screwdriver to chip out a mortise.

 

[sophiecentaur]

So that is the reason why we can't ignore mass or more presicely the Kinetic energy. It plays the main role in speeding up the body? (We can't just say the velocities are transferred.That would be wrong?).

I think you are trying very hard to see things 'your way' and will only let go when you absolutely have to. You will have much more success (with Science in general) if you start at the beginning by learning and accepting basics and work upwards. It's not our job to convince you about Physics; it's your job to learn it.

 

[mark2142]

The change in kinetic energy of the tank will be equal and opposite, approximately eight times the initial kinetic energy of the ping-pong ball.

The KE of ball will increase by factor of nine. Yes but I got little very bit decrease in KE of tank. How will it be equal ?

You seem to want to find the "one true quantity" that explains everything. There is no such thing

No No! sorry if it looks like that. I am just trying to make sense of the existing KE formula. Previously I was thinking that its the velocity that gets transferred but now I think, thanks to our discussion, that its the KE that is gets transferred not v when collision happens. Yes?

 

[mark2142]

It plays the main role in speeding up the body?

Sorry! wrong word. Not speeding.

 

[sophiecentaur]

I am just trying to make sense of the existing KE formula.

Making sense comes after accepting.
A formula is a formula - you just have to learn it and use it. KE is about Energy and Energy is all about making things happen and causing changes. Energy can be described as 'the potential to do work' (1960's School Physics) and it's like money - the KE of a body can be transferred into heat, changing shape or into Potential Energy (spring, height etc.). It's far more basic a quantity than velocity or force - neither of those will tell you you much, on their own, without more information whereas, if someone tells you there's 100MJ of Kinetic Energy, that tells you about what sort of job you could do with it. (Lifting a flea or a car - or what?)

 

[mark2142]

For the elastic central collision, I've tried to explain it in #16. If you have questions, I'm happy to help further!

Thank you for being so kind but I am trying to digest some previous information.

 

[jbriggs444]

The KE of ball will increase by factor of nine. Yes but I got little very bit decrease in KE of tank. How will it be equal ?

In an elastic collision, kinetic energy is conserved. If the ball gains a KE equal to approximately 8 times its own initial KE then the tank must lose an amount equal to approximately 8 times the ping pong ball's initial KE.

If your calculated result is otherwise, the error is in your calculations. I do not see those calculations. Perhaps I missed them in failing to keep up with this firehose of a thread :-)

Let us assume a ping pong ball with a mass of 2.7 grams colliding at 10 meters per second with an on-coming M1A1 Abrams with a mass of 54 metric tons also moving at 10 meters per second. Conveniently, this is a mass ratio of twenty million (20,000,000) to one.

So we can just say that the ping pong ball has mass 1 mumbles and the tank has mass 20,000,000 mumbles.

To a good approximation, the ping pong ball will rebound at 30 meters per second (minus a tad). It came in at 20 meters per second tank relative and departed at 20 meters per second tank relative for a total of 30 meters per second.

This represents a change in momentum of 40 mumble meters per second. A positive change if we accept the tank's direction of motion as positive.

The change in the tank's momentum is equal and opposite. It started with 200,000,000 mumble meters per second and finished with 199,999,960 mumble meters per second. Momentum is conserved.

If we divide by the tank's mass, we get that the tank's final velocity is 9.999998 meters/sec.

What about kinetic energy? $KE=\frac{1}{2}mv^2$.

The ping pong pong ball started with a kinetic energy of 50 mumble meters squared per second squared. It ended up moving 3 times faster which means 9 times the KE for a total of 450 mumble meters squared per second squared. That is a delta of 400 mumble meters squared per second squared.

The claim is that the change in the tank's KE will match this figure of 400 mumble meters squared per second squared.

The tank's initial kinetic energy is one billion (1,000,000,000) mumble meters squared per second squared.

The tank's final kinetic energy is approximately 999,999,600 mumble meters per second squared.

BINGO.

[One can do the same calculation algebraically with $m$, $M$ and variously subscripted $u$'s and $v$'s and get a completely accurate result with no approximations, but there is something about doing it with numbers that sometimes feels a little more real]

now I think, thanks to our discussion, that its the KE that is gets transferred not v when collision happens. Yes?

In an elastic collision (one that conserves kinetic energy), it is a useful to think about kinetic energy as something that gets transferred, yes.

One should be just a little bit wary of thinking of it as some kind of real physical fluid that flows from one object to another. Energy is not an "invariant" quantity. The direction in which it flows can depend on one's chosen frame of reference. If, for instance, we viewed the ping pong ball experiment from a passing car, we might decide that energy is moving from ping pong ball to tank instead of the other way around.

 

[sophiecentaur]

But in books it is not properly explained.

That is a very annoying statement to read. Is everybody out of step but you? Generations of successful Physicists have started using standard textbooks.
If you find more than one book confuses you then, at this level, it's you who has to change. There are so many textbooks that will tell you things in the right way. If you can't understand them on the first reading then go over it all again.

PLUS - there is no point in just dipping into a textbook, using the index and to expect to get your answer. Self discipline is needed and you need to work through from the start of a topic and not expect to get 'the right flavour' from a bit of speed reading in the middle of a chapter. (PS We've all tried this at times. Sometimes you get lucky but there's no point in complaining. )

 

[malawi_glenn]

Okay. But in books it is not properly explained

I learned it by using a textbook in my youth.

Try this, replace one bowling boll with a rubber ball. The momentum will be conserved because the net force on the system is 0, but the rubber ball will deform and thus some of the kinetic energy in the system is required to perform this deformation.