Inelastic Head-On Collision: Understanding Momentum and Kinetic Energy

[bobie]

Homework Statement

I could not find anywhere an example of inelastic head-on collision, all examples are of bodies crashing into an object at rest.
I do not have a specific question, if you have, I'd appreciate it. I just want to know in general:

suppose a ball A (of mass 100 kg rolling at speed 1 m/s, so, p = 100, KE = 50 J) collides head on with another ball B . In order to stop A I know we need to do 50J, right? so, if B has mass 4 and speed 5 m/s (p= 20 . E = 50J) will it stop A in its track even if it has 1/5 of its momentum?

Suppose now that B has speed 1 m/s (p=4, E =2J) and they collide and stick together without any deformation, should we subtract 50-2 = 48 to find The kinetic energy and then speed and momentum of A-B?

Thanks for your help. A link will do

Homework Equations

E = mv^2/2 p = mv

 

[HallsofIvy]

"Suppose a ball A (of mass 100 kg rolling at speed 1 m/s, so, p = 100, KE = 50 J) collides head on with another ball B . In order to stop A I know we need to do 50J, right? so, if B has mass 4 and speed 5 m/s (p= 20 . E = 50J) will it stop A in its track even if it has 1/5 of its momentum?[/quote]

How inelastic? An "elastic collision" is one in which kinetic energy is conserved. A "perfectly inelastic" collision is one in which the two colliding objects stick together and move off as one object. Kinetic energy is not conserved. But just an "inelastic" collision can have kinetic energy after the collision anywhere from 0 up the initial total kinetic energy. Are you assuming "perfectly inelastic" so that both objects come to a complete stop?

Even in an inelastic collision, momentum is conserved. Initially, ball A has momentum 100(1)= 100 kg m/s. In order to bring it to a stop in a perfectly inelastic collision, in which both A and B have 0 speed, the total momentum after the collision must be 0 so the total collision before the collision must be 0: B must have momentum -100 kg m/s, where the negative indicates that B is moving in the opposite direction to A- they are moving toward one another. If B has mass 4 kg it must have speed 100/4= 25 m/s, not 5 m/s.

Energy has nothing to do with it- in an inelastic collision, kinetic energy is NOT conserved.

 

[bobie]

. Are you assuming "perfectly inelastic" so that both objects come to a complete stop?
.

I called my thread crash as I was not sure if it is an inelastic collision or what.
A and B are two bodies that have same KE but different momentum. They collide and stick together without any deformation. That is why I assumed speed 3.6 Km/h.

I know that work is the energy required to stop a body so each body can do enough work (50J) to stop the other body, but
I know also that If A has momentum 100, B cannot bring it to a halt with its meagre 20 Kg m/s

In conclusion if A (v = +1) and B ( v = -5) crash into each other and by any device are bound together, what is the result of the crash? the body AB (26 kg) will stop dead or will continue in the direction +x at reduced speed (.77 m/s ??)

 

[haruspex]

Are you assuming "perfectly inelastic" so that both objects come to a complete stop?

I think you mean "so that the two objects coalesce, i.e. finish with the same velocity"

 

[bobie]

I think you mean "so that the two objects coalesce, i.e. finish with the same velocity"

But is that final same velocity = 0 or not, is the KE of B (50J) enough to do the necessary work to stop A? Of course in this case their KE are mutually canceled out.

 

[haruspex]

They collide and stick together without any deformation.

That's not a physically possible combination. If they coalesce, work is lost. If there's no deformation, no work is lost.

In conclusion if A (v = +1) and B ( v = -5) crash into each other and by any device are bound together, what is the result of the crash? the body AB (26 kg) will stop dead or will continue in the direction +x at reduced speed (.77 m/s ??)

Total momentum is conserved. Add up the momenta before crash (taking care with the signs) and divide by the total mass.

 

[bobie]

That's not a physically possible combination. If they coalesce, work is lost. If there's no deformation, no work is lost.Total momentum is conserved. Add up the momenta before crash (taking care with the signs) and divide by the total mass.

Could you explain that, please? If a crash makes you doubtful, imagine this situation:
we join the two bodies by a cable and give them an equal amount of KE in opposite directions. When the cable is tight why shouldn't mass B employ its 50 J to do the necessary work to stop A, even though it has 1/5 momentum? If the two joined bodies were to proceed in the +x direction the necessary Ke should be supplied by A, but it has been drained of all 50 J it possesed.

If we consider momentum A has + 100, B - 20, net momentum = +80, M would be 104 Kg so 80/104 is $\approx .77 m/s$ as I said, that means that Ke of AB is 30.83 J, right?
That would mean that it took A only 19 J to do the work necessary to stop B (to annihilate 50J), this seems impossible: if the car crashed into a tree or any other object the energy to stop is $- mv^2/2 = - 4*25/2 = 50 J$ se here at hyperphysics http://hyperphysics.phy-astr.gsu.edu/hbase/carcr.html#cc1

 

[haruspex]

Could you explain that, please? If a crash makes you doubtful, imagine this situation:
we join the two bodies by a cable and give them an equal amount of KE in opposite directions. When the cable is tight why shouldn't mass B employ its 50 J to do the necessary work to stop A, even though it has 1/5 momentum? If the two joined bodies were to proceed in the +x direction the necessary Ke should be supplied by A, but it has been drained of all 50 J it possesed.

If we consider momentum A has + 100, B - 20, net momentum = +80, M would be 104 Kg so 80/104 is $\approx .77 m/s$ as I said, that means that Ke of AB is 30.83 J, right?
That would mean that it took A only 19 J to do the work necessary to stop B (to annihilate 50J), this seems impossible: if the car crashed into a tree or any other object the energy to stop is $- mv^2/2 = - 4*25/2 = 50 J$ se here at hyperphysics http://hyperphysics.phy-astr.gsu.edu/hbase/carcr.html#cc1

Positive work does not cancel positive work. Both lose work as other forms of energy, such as heat. There is therefore no requirement for each to perform/lose the same amount of work.

 

[bobie]

Positive work does not cancel positive work. Both lose work as other forms of energy, such as heat. There is therefore no requirement for each to perform/lose the same amount of work.

B is going to the left at -5m/s and has 50 J of energy in that direction. A does work on B in the opposite direction and must make it stop before giving it acceleration to the right and v +.77m/s. I am not saying that they should lose same amount. I quoted a reliable link that says it is necessary -mv^2/2 energy to stop a body. so to stop A any other body , be it a tree , a wall or a moving body must do work in the opposite direction.
Are you refuting hyperphysics? If you are not:
body A must spend -50 J work to stop B, then, from where does it get the +20J necessary to make itself plus B travel at +.77 m/s?, if you accept the link, A had 50J but, in all, spent 77J if the body AB continued to move.
On the other hand, if AB moves to the right, that might imply that A used only 29J to bring B to a halt. How do you explain that?
Is this incorrect?

 

[haruspex]

B is going to the left at -5m/s and has 50 J of energy in that direction. A does work on B in the opposite direction and must make it stop before giving it acceleration to the right and v +.77m/s. I am not saying that they should lose same amount. I quoted a reliable link that says it is necessary -mv^2/2 energy to stop a body. so to stop A any other body , be it a tree , a wall or a moving body must do work in the opposite direction.
Are you refuting hyperphysics? If you are not:
body A must spend -50 J work to stop B, then, from where does it get the +20J necessary to make itself plus B travel at +.77 m/s?, if you accept the link, A had 50J but, in all, spent 77J if the body AB continued to move.
On the other hand, if AB moves to the right, that might imply that A used only 29J to bring B to a halt. How do you explain that?
Is this incorrect?

No, it doesn't need to do that amount of work. Consider a box slid along a rough floor, starting at some speed and coming to rest. It lost some amount x of KE. The floor did not do work x on the box. The box did work x on the two surfaces in contact, heating them up. When two bodies collide inelastically, each does work that goes into heat, sound etc. They don't need to do work in order to stop the other body. Something has to absorb the work.
E.g. Consider a completely elastic collision between two identical bodies, one starting at rest. The initially moving one stops. The other did no work to achieve that, rather, work was done on it.

 

[BvU]

Bobie: pipe down (*). Haru isn't refuting physics. You are not listening (or rather: reading) carefully enough. A reliable link is no authority if you misquote/misinterpret. Nowhere does hyperphysics say the tree does any work at all: the literal wording is: work required to stop the car. If you don't want to know, that's OK. But if you do: all the work is done by the car itself; that's why there is so much damage to the thing. The tree does ZERO work if it is hard enough.

Momentum conservation holds in all cases. Energy is conserved too. But kinetic energy can be converted into deformation.
Your AB collision momentum balance is Ok. Energy balance isn't. Energy has no direction. Pre-collision there is 100 J, post 30.8 J.
What about the 69 J in between ? Deformation in case of fully inelastic. Case "by any device are bound together" it depends: if you think of little hooks, kinetic energy can be fully conserved, but then that energy is distributed over kinetic energy of the center of mass (which is the same pre- and post collision thanks to momentum conservation! -- very important concept; work it out!) and kinetic energy of the rotation movement around that center of mass.

(*) Just to make sure: I can understand you feel a little frustrated if you think you get it and then you don't. But don't take it out on the folks that try to help you get it right with the best of intentions.

 

[bobie]

Momentum conservation holds in all cases. Energy is conserved too. But kinetic energy can be converted into deformation.

(*) Just to make sure: I can understand you feel a little frustrated if you think you get it and then you don't. But don't take it out on the folks that try to help you get it right with the best of intentions.

I am sorry if I gave that impression, BvU. I was not 'high', nor frustrated.
Now coming to the issue, you say energy has no direction, right, but v has direction.
- If I throw a bowl attached to a string and give it v =5 -y direction and energy 50 J, if I want to stop it and pull whe string, how much energy must I spend /force exert in the +x direction? There is no deformation here, I suppose the bowl does no work. Please, answer directly this specific question and if your answer is not 50 J add an explanation.
I promise I'll listen and read carefully. :)

 

[haruspex]

I am sorry if I gave that impression, BvU. I was not 'high', nor frustrated.
Now coming to the issue, you say energy has no direction, right, but v has direction.
- If I throw a bowl attached to a string and give it v =5 -y direction and energy 50 J, if I want to stop it and pull whe string, how much energy must I spend /force exert in the +x direction? There is no deformation here, I suppose the bowl does no work. Please, answer directly this specific question and if your answer is not 50 J add an explanation.
I promise I'll listen and read carefully. :)

You don't need to do any work to stop it. You have to have work done on you. But it can be very confusing using muscular scenarios because of the way muscles work. Muscles burn energy just staying tense, even though in a mechanical sense they are not doing work.

 

[bobie]

. Energy has no direction. .

That is true, probably for other forms of energy, BvU, but surely not for KE, you have E =m/2 * v^2 and not by squared speed, to begin with.

But formula is a formality, the point is that if you want to stop a body you must do work, exert a force in the opposite direction to that energy.

The effect of a force acting on a body depends not only on the magnitude & direction but also on its point of application & line of action.

E → ← F, if you do work in another, different direction you will do less or no work, right? If you do work in the same direction you do not stop it , you increase its v and KE

Now correct me if I am wrong, and please acknowledge you are mistaken if I am right:

In order to stop a moving body you must do work in the opposite direction equal to the amount of its KE. I hope there is no argument on this.
When the baseball catcher stops the ball ( E = k) in his glove he spends -k J of energy on the ball, and the ball does k J of work on his hand.
If the ball is not solid and is , say, an egg the egg will crush and explode. The work has been done by the rear part of the egg that doesn't discharge its energy on the hand because the shell is not able to transmit it to the hand and collapses. Right?

That happens in a flimsy car that crashes into a tree or a concrete wall. But that doesn't change the physics behind it. Work = -k J has been done by the hand/tree/wall. The work on the hand, tree, wall doesn't reach its final destination and destryes the egg.

Can we agree on this, BvU?

 

[haruspex]

In order to stop a moving body you must do work in the opposite direction equal to the amount of its KE.

You keep asserting this but it is not true, it is not based on Newton's laws, and the only supporting material you supply is the hyperphysics link.
At that link, the work "done by the tree to stop the car" is negative. That is, the work is actually done by the car on the tree. That said, the hyperphysics text is really misleading. Work is done on the tree only to the extent that the tree yields. Most yielding is done by the car body, so that's where the work goes.

 

[BvU]

If I throw a bowl attached to a string and give it v =5 -y direction and energy 50 J, if I want to stop it and pull whe string, how much energy must I spend /force exert in the +x direction? There is no deformation here, I suppose the bowl does no work. Please, answer directly this specific question and if your answer is not 50 J add an explanation.

"Can't be done" is the right answer here. Either there is deformation (of the plate in case it has to be stopped instantaneously) or of the string (if it isn't stiffer/stronger than the plate) or of the arm. That's the hard one, as Haru points out. Too difficult in this stage.

What you will have to learn to accept is that if $\vec F \cdot d\vec s=0$ that means that the work is zero. Period.

And it's not really that hard, if you bear in mind that $\vec F = {d\vec p\over dt}$ it becomes clear that instantaneous (dt = 0) processes can not occur.

And energy has no direction. It is a number, a scalar, not a vector. No work is done when a ball reverses direction and heads off with the same speed. Even your hyperphysics says so (here). Since we're into thought experiments: Throw up the bowl and let go of the string. Little hook on the bowl gets caught on a rigid steel bar higher up and the bowl describes a half circle until the hook let's go. No work done.

I promise I'll listen and read carefully. :)

Good. I believe you are sincere in trying to master physics. Accepting everything straightaway is not a good quality in this field, so in that respect you are well underway. I bumped into another thread you originated and I see some similarities. You don't give up, which is another good attribute in physics.

if you want to stop a body you must do work, exert a force in the opposite direction to that energy.

No, Yes and: energy has no direction. It really doesn't. It's momentum you want to think of.

Formulas are a formality is a tautonomy.

If you exert a force without moving the point where the force is applied, you do no work. The famous brick wall or large tree does no work. Holding a book on a shelf: the shelf does no work (but it exerts a force on the books to counteract gravity and result in acceleration = 0). Holding a book in your hand: no work. Holding a book in front of you, arms extended: a lot of sweat, but no work.

And, if it makes you feel better: hyperphysics doesn't delve into these aspects too deeply, also not in this link . Even a good student might get confused ...

[edit]this in response to post #12. Have to read later posts carefully in order to find the minimum resources path towards greater understanding. After all, you are getting quite a bit out of this free PF deal, and there are many others who would appreciate to get some attention too!

 

[bobie]

[edit]this in response to post #12.!

If you want to answer, you should cosider post #14.
I insist when my specific questions are ignored. Just say true or false and if I do not agree I'll provide evidence.
1) is work the energy required to stop a body and equale to its KE?
2) if yes, if we want to do work to stop a body must we apply a force in the direction opposite to its motion?

If you do not justify your no to these 2 questions, I cannot understand

 

[BvU]

Please, answer directly this specific question and if your answer is not 50 J add an explanation.
I promise I'll listen and read carefully. :)

One thing at a time. I do what you ask, YOU promise to read.

Any further questions instead of demands ? I don't have infinite time, and I start to think I can use what I do have a lot better than reading repeats of your misconceptions, let alone answer them on demand.

Jumping to 17 (might save time): 1) yes 2) yes

Does not mean you understand now. 2) is a yes because if you want to do work, you must... But you don't have to do work. Read posts. If you still cannot understand, ask yourself what is blocking you.

 

[bobie]

I don't have infinite time,
Jumping to 17 (might save time): 1) yes 2) yes
Does not mean you understand now. 2) is a yes because if you want to do work, you must... But you don't have to do work. .

That is a huge step toward my understanding, BvU, I thank you for the time devoted to this thread, I hope someone will continue these explanations.

So, I have learned that if I want to stop a ball (E = k) I must spend k J of energy doing work in the direction opposite to its v:
I hold up my hand and stop it . I have done negative work on the ball and spent k J

3) can I stop it without doing any work as BvU states? how?
I cannot find an explanation in previous posts. I already said that if it is an egg some energy will go into its distruction, but if the body is solid and indeformable (like a steel ball or a stone) I supposes my hand will have to take k J in whatsover way I stop it. Any example?

 

[NTW]

May I participate in the conversation...?

bobie: A steel ball may be hard to deform permanently, but can be elastically deformed. If you throw a steel ball against an absolutely rigid wall, part of its KE will be converted, in the impact, into elastic energy and some will be used up in in heat, rising the temperature of the ball, the wall, and the surrounding air. But that elastic energy will be almost instantly converted again, as the ball recovers its original shape, to KE (again with some looses as heat), and the ball will rebound. With a lower velocity, it's true. The difference, times the ball's mass and divided by two, is the energy lost as heat...

 

[BvU]

On Saturday, OP started with inelastic head-on collisions and had difficulty with kinetic energy and work considerations. As I start to believe, this thread is a continuation of this one . Rather tedious with mixing up of fact and supposition and hopping from one question to another. Extended in this thread . As Dale said: at some point you have to make up your mind what problem you want to solve and describe it consistently.

Here, again, we are hopping from one thing to another and disregarding sound responses in favor of demanding an answer to yet another question. Catching a ball is definitely not a collision process.

Somehow, my impression is that it is difficult to sell to bob the idea that a brick wall does no work on a car that crashes into it and for sure ends up with zero velocity.

 

[bobie]

May I participate in the conversation...?
.

Hi NTW, I'd appreciate your contribution. But please, I beg you, as I am the one who does not understand, follow the step I need , do not hop as we did so far from one case to another. I hope you are more patient.

Now I ascertained 2 basic points, I need to fill in another pidgeon hole
3) how can I stop a ball/stone etc with my baseball glove without doing any work? is it true what BvU stated?
If you want to use the steel ball , please follow my logic, let me catch it with my hand. Is there a way in which I can stop a ball/stone (k J) etc with my hand doing zero or less than k J of work?

I thank you in advance, if you look at the other thread you'll get a clear picture of my difficulties, yes, I am not ashamed to admit that I am trying to understand the issue through more, many posts. I suppose that is what PF is all about, starting many threads, giving kind answers.

 

[NTW]

I think that the explanations have been clear. I'll try my luck...

Stopping a moving object means reducing to zero its KE. But that KE can't simply disappear. It converts to work (for example, the work made by the force of your hand for the distance needed to stop the ball, plus some heating of ball, hand, and surroundings...).

 

[HallsofIvy]

That is true, probably for other forms of energy, BvU, but surely not for KE, you have E =m/2 * v^2 and not by squared speed, to begin with.

No, "energy", all forms of energy, since they are all equivalent, is a scalar, not a vector. And for a vector, v, v^2 is the dot product of the velocity vector with itself so is scalar.

But formula is a formality, the point is that if you want to stop a body you must do work, exert a force in the opposite direction to that energy.
E → ← F, if you do work in another, different direction you will do less or no work, right? If you do work in the same direction you do not stop it , you increase its v and KEi

Exerting force is NOT the same as doing work and does not necessarily require any energy transfer. When you sit in a chair, the chair is constantly exerting force (mg) on you but is doing no work.

Now correct me if I am wrong, and please acknowledge you are mistaken if I am right:

In order to stop a moving body you must do work in the opposite direction equal to the amount of its KE. I hope there is no argument on this.

I wouldn't call it an "argument", just that you are wrong. If a moving body hits a wall, the wall stops it without doing any work.

When the baseball catcher stops the ball ( E = k) in his glove he spends -k J of energy on the ball, and the ball does k J of work on his hand.
If the ball is not solid and is , say, an egg the egg will crush and explode. The work has been done by the rear part of the egg that doesn't discharge its energy on the hand because the shell is not able to transmit it to the hand and collapses. Right?

That happens in a flimsy car that crashes into a tree or a concrete wall. But that doesn't change the physics behind it. Work = -k J has been done by the hand/tree/wall. The work on the hand, tree, wall doesn't reach its final destination and destryes the egg.

Can we agree on this, BvU?

No, we cannot. What you are saying is simply not true. You seem to have serious misunderstandings about "work" and "energy". You also seem to have some misunderstanding about vectors operations. There are two different ways to multiply two vectors, the dot product, which results in a scalar and the cross product which results in a vector. But the cross product of two parallel vectors, and, in particular, the cross product of a vector with itself, is always 0. When we talk about "$v^2$", for v a vector, we can only mean the dot product. $v^2$ is the same as $|v|^2$, a scalar.

 

[bobie]

I think that the explanations have been clear. I'll try my luck...

Stopping a moving object means reducing to zero its KE. But that KE can't simply disappear. It converts to work (for example, the work made by the force of your hand for the distance needed to stop the ball, plus some heating of ball, hand, and surroundings...).

I never said that, did I? and that's why I said I am wearing a baseball catcher's glove. The energy goes 99% to my poor hand. Sure
In this case the ball is douing no work on the ball. And I am doing negative work on the ball.
3) has no answer yet, can you tell me a case when I stop the ball with my hand doing no (or less) work?

Thanks a lot

 

[NTW]

I never said that, did I? and that's why I said I am wearing a baseball catcher's glove. The energy goes 99% to my poor hand. Sure
In this case the ball is douing no work on the ball. And I am doing negative work on the ball.
3) has no answer yet, can you tell me a case when I stop the ball with my hand doing no (or less) work?

Thanks a lot

In the first case, that of the glove, let's suppose that your hand doesn't move at all. Well, the KE of the ball will transform in work done by the glove and your hand tissues, resisting the braking ball and deforming elastically. The ball will also deform elastically, and if it does not rebound, when the deformations disappear and everything regains its former shape, all the KE the ball had will have turned into heat, rising the temperature of the glove, the hand, the ball, and the surrounding air...

If your hand were indeformable, and didn't move at all when struck by the ball, all the elastic deformations mentioned above will take place in the ball only...

 

[bobie]

In the first case, that of the glove, let's suppose that your hand doesn't move at all. Well, the KE of the ball will transform in work done by the glove and your hand tissues, resisting the braking ball and deforming elastically. The ball will also deform elastically, and if it does not rebound, when the deformations disappear and everything regains its former shape, all the KE the ball had will have turned into heat, rising the temperature of the glove, the hand, the ball, and the surrounding air...

If your hand were indeformable, and didn't move at all when struck by the ball, all the elastic deformations mentioned above will take place in the ball only...

Ok i know all that, but all in all I suppose my hand (and glove) are still doing 95% of the work and it it is a stone even 99. When can I do zero work?

 

[NTW]

Ok i know all that, but all in all I suppose my hand (and glove) are still doing 95% of the work and it it is a stone even 99. When can I do zero work?

I'm not sure of that 95%, but all the KE will go into the ball, in other energy form, and possibly ending up in heat, if you have a fully indeformable and immobile hand and glove...

 

[haruspex]

So, I have learned that if I want to stop a ball (E = k) I must spend k J of energy doing work in the direction opposite to its v:
I hold up my hand and stop it . I have done negative work on the ball and spent k J

No, if you have done negative work then you have not spent k J, you have absorbed k J. The ball does k J, you do -k J (i.e. absorb k J), and the total work done is 0 (which is correct).

 

[Nugatory]

I do not have a specific question, if you have, I'd appreciate it. I just want to know in general:

A thread that starts this way isn't going to respond well the homework help process, and this one is no exception.