B Why is KE not conserved when momentum is?

[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.

 

[mark2142]

Extremely sorry if that annoyed you or anybody. I didn’t mean it. I was just trying to understand the difference. I did not find that in the textbook I am reading. Maybe I need to read more books. Can you suggest me one where I can find the difference. It doesn’t mean I don’t understand it already. It’s just I live in remote area and there is lack of good teachers and good resources. I have to do it all by myself. I get a lot of help from you guys. I try to understand what I see and I am beginning to understand there are a lot of wrong things I have catched up by trusting my eyes only. I hope I am not saying too much. Thank you guys. You guys are great. Thank you @jbriggs444 for writing it in so much detail and everybody else who kept up with my slow pace.

 

[sophiecentaur]

Maybe I need to read more books.

Absolutely essential. You can get hung up on the wording of just one version of a description.
There are many videos on these subjects. The Kahn Academy is recommended by many people although I find it's presentation a bit boring. They do a full course, so I believe so you could access help that way.

Also, find the YouTube videos by Walter Levin. Loads of friendly presentation of hard topics. Learn to develop your Search Skills and be prepared to spend some time finding what you need. An hour's searching can often give you just what you want.

 

[malawi_glenn]

find the YouTube videos by Walter Levin

The first post is a walter lewin video. EDIT: my bad, was another dude in that video :)

 

[N1206]

"I was just trying to understand the difference."
I get where you are coming from.
On the surface, you look and say "if mv is conserved, and mass doesn't change, doesn't it follow that 1/2 mv² should be conserved, too?"
The critical thing is that v (velocity) is a vector, having both magnitude and direction. v² is a scalar, having only magnitude.

Take the head-on collision of two 50 g ball bearings (assumed to be a completely elastic collision) each moving a 1 m/s. The total momentum of this system is zero. The masses are equal and the magnitude of the velocities are the same but the velocities are in opposite directions, so one velocity is positive and the other is negative and the sum comes to zero ( (50g * + 1m/s) + (50g * -1m/s) = 0) . After the collision, the ball bearings will go back toward where they started at a velocity with a direction completely opposite of what they had originally--and momentum of the system will still be zero.

The total kinetic energy of this system before collision is not zero, as the scalar values do not have direction and therefore do not offset each other, but rather sum up ((1/2 * 50g * (1 m/s)²) + (1/2 * 50g * (-1m/s)² = 50 gm²/s²) The definition of a perfectly elastic collision is one in which no kinetic energy is transformed--and so if the collision is perfectly elastic, KE will be perfectly conserved. Note that while energy is always conserved, it usually is not conserved entirely as kinetic energy in collisions, but transformed into heat, chemical energy, potential energy and other forms.

Change the ball bearing to eggs. Nothing changes on the momentum end of things. But now, the kinetic energy is going get transformed completely. Some of it into the energy required to break the shell, and most of it into heat, some of it into cooking the proteins, etc. The total amount of energy in the system will remain the same, but none of that energy will be kinetic in nature.

Do you see why the vector nature of v and the scalar nature of v² make the critical difference?

 

[sophiecentaur]

The critical thing is that v (velocity) is a vector, having both magnitude and direction. v2 is a scalar, having only magnitude.

I see you are trying to help. But you have to spend a lot of time and effort on such a justifying argument but the maths shows it instantly. I don't know why using the maths for an argument is always so hard for people to accept. If we argued with our Bank Manager that his calculations are not as reasonable as our 'feelings' about the state of our account, we wouldn't get far.
I previously used the phrase "can't or won't" about using maths but people just have to, if they want to get anywhere. Even the Greeks had sussed that out and realized that Science has to hang on mathematics in the end.

 

[sophiecentaur]

On the surface, you look and say "if mv is conserved, and mass doesn't change, doesn't it follow that 1/2 mv2 should be conserved, too?"

It's a bit like saying that Pythagoras' Theorem could be replaced by a+b=c, without squaring the terms.

 

[jbriggs444]

if mv is conserved, and mass doesn't change, doesn't it follow that 1/2 mv2 should be conserved, too?"

I know that you are not actually saying this. You are just quoting a straw man so that you can attack the argument.

But that straw man argument as it stands is perfectly correct. You are wrong to attack it.

If we have only one object, alone in the universe and we are given conservation of mass and of momentum, we actually do get conservation of energy for free. If we have only one object and if both mass and momentum are conserved then velocity is a constant vector and energy is a constant scalar.

But conservation of momentum also holds for isolated collections of more than one body. One can conserve total momentum for a collection of bodies without conserving total kinetic energy for the same set of bodies.

It is not correct to conclude from the conservation of $\sum_i m_i\vec{v_i}$ that $\frac{1}{2}\sum_i m_i \vec{v_i} \cdot \vec{v_i}$ is also conserved.

[As you go on to point out with the example of an inelastic collision between a pair of approaching eggs with zero total momentum]

 

[sophiecentaur]

But that straw man argument as it stands is perfectly correct.

There are circumstances where this is true but not in general.

 

[jbriggs444]

There are circumstances where this is true but not in general.

It is true in general, as stated -- for a system containing one body. Which, since the statement mentions only one m and only one v, must be the only sort of system being contemplated.

 

[sophiecentaur]

It is true in general, as stated

Hmm. That's starting with an exceptional case though.

 

[vanhees71]

Well, if you don't want to learn mathematics to understand the physics, then it's a waste of time for you to try to study physics and a waste of time trying to explain it to you.

 

[sophiecentaur]

It can get conserved or it can not.

When could it possibly not be? Did you get nothing from the past dozens of posts?

 

[mark2142]

Well, if you don't want to learn mathematics to understand the physics, then it's a waste of time for you to try to study physics and a waste of time trying to explain it to you.

I just said I am trying to grasp some information. That doesn't mean I am going to ignore maths.

 

[mark2142]

Take the head-on collision of two 50 g ball bearings (assumed to be a completely elastic collision) each moving a 1 m/s. The total momentum of this system is zero. The masses are equal and the magnitude of the velocities are the same but the velocities are in opposite directions, so one velocity is positive and the other is negative and the sum comes to zero ( (50g * + 1m/s) + (50g * -1m/s) = 0) . After the collision, the ball bearings will go back toward where they started at a velocity with a direction completely opposite of what they had originally--and momentum of the system will still be zero.

So if there is no head on collision final momentums will differ( direction change)from the head on collision because the forces will differ from the head on collision but the transfer of Kinetic energy will be same no matter how they collide. Yes?

 

[Orodruin]

So if there is no head on collision final momentums will differ( direction change)from the head on collision because the forces will differ from the head on collision but the transfer of Kinetic energy will be same no matter how they collide. Yes?

No.

 

[mark2142]

No.

Okay. If we throw a elastic ball to a wall its momentum will change but its kinetic energy will not. Force will affect p but to change the KE force need to do work.
I think Kinetic energy is like a scared running dog with its tail on fire but the momentum is like a trained horse which always move in one direction.

 

[jbriggs444]

Okay. If we throw a elastic ball to a wall its momentum will change but its kinetic energy will not.

If you are considering the elastic ball alone as your "system", then the momentum of the system changes. It is not conserved. That is because you are not considering a closed system. There is an interaction with something external: the wall.

Force will affect p

Yes. The external force affects momentum.

Force multiplied by (or integrated over) time yields change in momentum (impulse).

but to change the KE force need to do work.

Yes. The external force can affect kinetic energy.

Force multiplied by (or integrated over) displacement yields change in kinetic energy (work).

If we looked closely, we would discover that during the collision, the net displacement was zero during the collision. There was a distortion of the ball and a rebound, but the result left the ball the same as it began. Further, we would find that the force pattern was identical (though reversed) during the deflection and the rebound. The work done over the positive displacement during the deflection was equal but opposite to the work done over the negative displacement during the rebound. Zero net work done. Zero kinetic energy change.

If we looked closely during that same collision, the force would have been the same direction the whole time. Non-zero momentum change.

I think Kinetic energy is like a scared running dog with its tail on fire but the momentum is like a trained horse which always move in one direction.

I do not see that this mental image is of any great help when making predictions or performing calculations. But whatever floats your boat.

 

[mark2142]

I do not see that this mental image is of any great help when making predictions or performing calculations. But whatever floats your boat.

Ok :)

 

[mark2142]

Change the ball bearing to eggs. Nothing changes on the momentum end of things. But now, the kinetic energy is going get transformed completely. Some of it into the energy required to break the shell, and most of it into heat, some of it into cooking the proteins, etc. The total amount of energy in the system will remain the same, but none of that energy will be kinetic in nature.

Yeah! The total momentum cannot change until and unless there is an external force. We can see that the COM is at rest all the time. But the egg breaks and it does not rebound. The total KE mv^2 is lost to zero. That must have gone into breaking the eggs.