Collision Course: Newsletter Challenge (02/02/10)

To solve this problem we would need to use the principle of conservation of momentum. Before and after the collision, the momentum would be conserved. Accordingly,

(3*0) + (6*100) = 9*x

x = 200/3

which is approximately equal to 66.667

This is an example of an inelastic collision. Energy is not conserved --- it is lost in the stickiness' deformation. However, momentum is conserved, so the following applies:

Momentum = mv. The 3-lb object has no momentum before the collision (v =0)

Momentum after collision = momentum before collision. m_{after collision} = 6 + 3 lb = 9 lb

9 lb x [Final velocity] = 6 lb x 100 cm/min

Final velocity = 6/9 x 100cm/min = 66.67 cm/min

(Odd choice of velocity ---- barely moving)

I don't see where the question mentions weightlessness. If the 3# object was motionless on a coffee table, maybe I will get this one correct.

My goodness, when do you chaps ever get any proper work done?

You are right....wastefull to discuss such a simple first grade leaver topic with just anybody who pretend to be able to do so without thinking correctly or learning their lessons...

Thank you for the advice.

The speed of the combined mass would be 1.12 meters per second. Mass of the moving object would be 2.72 Kg times 1.7 meters per second for a momentum of 4.62. 4.62 divided by 4.1Kg gives the speed of 1.12 meters per second after collision

Hey. Why complicate the problem with back-and-forth units conversions? Your answer is not correct because you made a units conversion error in there somewhere Plus, it is given in the wrong units. The original units are cm/min. Why give your answer in M/Sec? Your answer must match the original units without the reader having to convert, otherwise you did not answer his question completely.

Yeah caught the error, thanks for the Kind Feedback to a first time poster.

Yeah, we have some sticklers here, including myself.
It is alot of fun, though.

1.11 m/sec ... small conversion error

81,65 cm/min

Since the object at rest is half the mass of the moving object, it will absorb half the kinetic energy. This means that if friction is considered to be insignificant, the nine pound combined object should travel on at 50 cm/min.

I hope you´ll have better luck with Icelandic then with this collision.

Hreifiorkan (massi*hraði*hraði) varðveitist öll því áreksturinn er alófjaðrandi (sticky) en deilist eftir áreksturinn á 9 pund í stað 6 punda áður.
Þannig að:Kvaðratrótin af (( 6*100*100)/9)er:81,65cm/min

en vá samt kvað þetta er búið að flækjast fyrir liðinu. hehehe.

I guess your bankers listened to financial warnings about as avidly as you followed post#1 or tried to understand the implications of post #35? On the other hand, I don't understand a word of Icelandic

the web translator gives the following as a "translation". Even without a dictionary I managed mass, speed and pounds and didn't screw up the cm.

Hreifiorkan massihraðihraði varðveitist öll því áreksturinn is alófjaðrandi sticky ) while dividend after áreksturinn á 9 quid í stað 6 spring balance aðKvaðratrótin : with min.

You seem to be getting more screwed up as you go, depending upon whether you are the Guest who posted #78, #84, #88, #91, #94, or #95, all of which present the incorrect answer! but seem to be at least 2 or 3 different quests. Just as a side note, #94 had the right idea to use proportions, but used 6/3 as the proportion instead of 6/9.

Lawyers have nothing to do with it until you supply an incorrect answer to a client that then uses your data that results in a catastrophic failure.

As far as your reference to posts #1 and #35, I have no idea what you are trying to say. Post #1 is the first correct answer, and is the best stated, while post #35 is totally wrong. How can your put them together as being similar.

As stated in Post #1 the correct answer is 67 cm/min, rounded to two significant figures, which is probably too precise for the actual data given but then, that is another topic and "can of worms" as you can see in most of the thread to this point.

Once you identify yourself as to which guest you are, you might have some intelligent discussions in this thread. As for me, I am now bowing out of this thread.

why would u tell people to 'click here' to get answer, - then tell them to wait til march??

There are two potentially applicable physical principles which are useful in solving collision problems, Conservation of Energy and Conservation of Momentum. Since we are told that the two bodies stick together, we know that we cannot apply Conservation of Energy. Fortunately we do not have to, given that the we know that the two bodies end up moving together.

Conservation of Momentum states that the sum of the product of the velocity and the mass for each of the two bodies must be the same before and after the collision, since there are no external forces acting on the two-body system.

The math is as follows:

100 cm/min x 6 lb + 0 cm/min x 3 lb = new velocity x 6 lb + new velocity x 3 lb.

v x 9 lb = 600 cm-lb/min ==> v = 66.6... cm/min.

Dave Uggla

Not enough information. The 3 lb sticky object could be at rest on a table, etc. which would bring adhesion and friction forces into play. So there would be some conversion to heat and potential energy.

This is classic motion physics. m1v1 = m2v2; v2 = m1v1/m2; Answer 66.66 cm/min

At rest on what? ... and how well is it stuck to whatever it is resting on? In other words, how much energy is lost in breaking that bond. I's have to say the answer is indeterminent for lack of this information.

It is a reasonable assumption that the information you need to arrive at the 'correct' answer is contained within the question, otherwise the exercise is trivial.

Perhaps the collision occurs in vacuum, outside a gravity field?

We are told the weight of the objects (in a standard earth gravity) rather than their mass, but this is irrelevant. It would be polite to assume the 1 significant figures given are exact for this purpose, but again irrelevant if the only point is that the increase in mass is 50%.

At a simplistic level, the kinetic energy of the moving mass is conserved. It's impossible to arrive at a numerical answer if unquantifiable amounts of energy are absorbed / converted to heat and crunching noises etc. So 0.5 M V^2 is the same in and out. In an ideal collision the combined mass moves at 81.65 cm/s.

A reasonable footnote might suggest that the moving mass had kinetic energy, whilst the stationary mass had inertia, thus it is inevitable that for any real materials, some kinetic energy would be converted to heat, thus the actual velocity would be less than the calculated value.

Respectfully, there is no such thing as an exact measurement of weight and velocity. In a world that uses specialized instruments that measures beyond "nano" and "giga" quantities, can we continue to ignore appropriate significance and recordings of measurements?

Agreed. The most repeatable measurements we can make are of (possibly the most obscure physical phenomenon) time. Currently** to parts in 10¹⁷. The next step to parts in 10¹⁸ is curiously important. The reason is that there is a theoretical possibility that our velocity relative to the universe will introduce an asymmetry that means that special relativity breaks down - and the level of the effect would be precisely these few parts in 10¹⁸. Confirmation or denial will allow us to eliminate some of the theoretical cosmological options.

Unlike Eddington's experimental confirmation of GR, we can't yet see applications to the differences between the theories - but it will still be interesting.

**BTW, this is a specific example where rounding individual measurements to levels comparable with their accuracies before averaging not only degrades the statistical variability, but also introduces systematic offsets.

Please keep in mind that conservation of energy in a closed system and conservation of momentum in a closed system are two entirely different principles.

Energy is still consererved overall if either:

a. Some of it is converted from mechanical (kinetic) energy to potential energy, thermal energy, acoustic energy, etc. within the closed system or

b. Some of it is transferred in one way or another OUTSIDE the closed system (friction with surface, air resistance, etc.)

You can even have a situation in which the two particles move together after the impact without converting any of the energy to heat, etc. Imagine that when the two bodies collide it compresses a spring attached to one of the bodies. At the moment of closest approach, which is also the moment of zero relative velocity if the impact is head-on to the spring, Maxwell's demon or an a simple mechanical system closes a latch which binds the two bodies to each other and keeps the spring compressed. We now have potential energy stored in the spring and the sum of that potential energy and the kinetic energy of the combined mass is equal to the combined kinetic energies of the two incoming bodies. No heat, light, or noise need result.

Conservation of momentum, on the other hand, depends only on the absence of external forces acting on either body. You can have as complicated an array of forces acting between the two bodies as you want (ranging from the spring to a wad of gum to an electromagnetic damping field causing current to flow though a resistor) and conservation of momentum will still be valid. The velocity which is the solution to the problem as posed is entirely the consequence of conservation of momentum and does not depend in any way on how the excess energy is disposed of.

Dave

(combining the awesome powers of physicist, mathematician, teacher and tech-support professional)

I don't think you mean "closed" in the generally accepted sense of the term, as the definition of a closed system is that nothing leaves it or enters it. Closed systems are therefore by definition conservative of energy - although conversion between forms is to be expected. (Unfortunately, entropy increases with time - ouch).

So Energy is always conserved in a closed system - period. If you include nuclear reactions, then you have to allow nuclear energy as a form of energy storage.

As you say, in the "universe" of classical physics you could theoretically arrange a collision in which the kinetic energy is converted to potential energy. Implementation would of course be tricky.
One you accept this level of trickiness, you could also institute a situation where kinetic energy is conserved - some of what was in the linear motion is "converted" into rotational motion. The "simplest" theoretical case would be where the objects have point masses that would pass each-other rather than colliding directly, but join just as their approach velocity is zero (and maintain the separation between original centres thereafter). The most general case for which KE can be maintained is when the objects join just as the approach velocity of the Cs of G is zero, and the objects are spinning at identical angular velocities and so that the points of joining are moving at the same velocity.

Pardon my inexact terminology. You are right.

Energy is necessarily conserved in a closed system even though linear kinetic energy is not. The term inelastic collision usually refers to a collision in which kinetic energy (either linear kinetic or total kinetic) is not conserved. And if we use the introductory physics approximation of point masses rather than extended bodies, then linear kinetic energy is all we need to consider.

Have you ever seen or heard the tape of the Feynman Lecture on Physics in which he illustrates conservation of energy with "conservation of blocks" examples in which the blocks keep getting more ingeniously hidden, but their existence somewhere is determined indirectly and the result is still consistent with conservation of blocks? It is a classic!

I missed that one. Feynman was a great conjurer.

While you make some interesting points, you manage to confuse the crucial issue. Your answer implies that linear kinetic energy is conserved. There are special circumstances under which total KE could be maintained (even though the objects move as one after the collision)*. However, what is always conserved (even where KE is maintained) is momentum - which is why the combined speed after collision is the same as the mass-averaged speed before collision.

*Total KE is maintained only if linear KE is converted to rotational KE - which places very special requirements on the collision (that I have described elsewhere).

67 cm/min

answer: 0 cm/min

If the object was moving at 100 cm/min, it is only moving at 1.66 cm per second. In other words, it's barely moving.

If it hits another object while barely moving, and it is sticky, then the objects would stick together at the time of the collision forming a 9 lb mass of odd shaped sticky stuff. With such low energy at impact, and being sticky, how would it continue to move. It couldn't unless there is a constant force being applied to move it which the problem didn't state.

(not an engineer)

If it was stuck....wouldn't it stop?

(I'm just a girl)

But a three pound sticky doesn't rest unless its stuck....and stopped.

(I'm just a girl)

Don't worry about the comment made to you....

In the context, stuck together meant that the 2 object are now One and moving together as one object. Otherwise, the problem will get complicated and involve sharing the energy while again involving conservation of momentum for the 2 new object moving apart etc... Many scenarios will be involved...

Conservation of momentum between the two individual bodies and then the two stuck together: p=mv

p1=m1v1=600;

p2=m2v2=0;

p3=p1+p2=600=(m1+m2)v3;

v3=66.67 cm/min

The entity p=mv is Quantity of mouvement and is not an energy or momentum to be conserved. The Quantity of mouvement is conserved in a different situation case.

The Energy of the mouving object is conserved allways whether part of it is converted to heat sound or other. Therefore, if you use the quantity of mouvement in this solution and state that it is conserved, and consider that there are no other losses, then your resulting speed and mass will not have he same kinetic energy as when the issue started. Since we start with no loss when we study a case, and then include the losses when we can identify them, both cases must make sure that all the energy is accounted for.

Thing a little bit more and try you calculations to find out that there is a missing amount of energy

Efinal = 66.67^2 x 9/2 =20002.00 in you case

While we started with 100^2 x 6/2=30000

If there are no losses, then where is the difference of ~10000 gone? { Introducing losses due to anything you want, will have to only explain this difference if experimentally you got less than 30000}

Please consult your physics lessons etc..

"Please consult your physics lessons etc.."

Excellent advice that you would do well to heed.

As so many others have correctly pointed out, the relevant properties conserved for this problem are mass and momentum.

Nowhere does the problem (nor do the laws of physics) state that there must be no loss of kinetic energy. In fact, the statement that the two masses become one means that there MUST be loss of linear KE. (as Fyz pointed out, in very specific cases linear motion can be converted to rotation without loss of total KE, but whether the combined mass is vibrating, rotating, a bit warmer or whatever has no impact on the velocity of its center of gravity.)

You acknowledge that kinetic energy can be converted to other forms but go on to question what happened to the "missing" energy. The point is, that without a lot of additional information about material properties, exact trajectories, etc, etc, we don't know the fate of this energy, but we don't need to. Lesson: don't try to solve the problem by energy balance. Let me say it once more: conservation of momentum!

Sorry to say it but you need much more....

Energy of a system does not disppaer in thin air. It must be conserved whether it is accounted for by losses or no losses:

You will aggree that E1 (initial total energy ) is provided by the Ke of the 1st object since no other form of energy is mentioned and the 2nd object is in stand still

after collision, E2 is = to E1 - E lost ( that is transformed into other forms of energy, like heat, vibration if you like, or radiation if you want etc)

Failing to make that equation will render your reasoning invalid since thin air cannot account for the non conservation of the initial energy present (Not only Kinetic energy, all the energy present at both ends)

Therefore, You can be happy and dogmatic about momentum etc but if you consider a system without losses (as presented...), then by working as you did, you end up with v2 = 66.66 and a new mass of 9 which will give you a final Ke2 of much less that what was before,(as I calculated for you). Since we considered no losses, how do you account for the disappearance of such energy ?

If you want to include losses, then your reasoning with Quantity of mouvement (mV) must include the losses (or are you saying that you have found a genius way of calculating the outcome regardless of any losses, thus NASA should be after you to give you a medal for saving the future of space travel etc...).

I think that this topic is by far exhausted UNLESS you go and SEEK advice nearby your residence from some competent person. Or wait for the answer(and I hope it will be competent and not misleading).

Thank you

Assuming that the energy is not lost from the system, it is converted to other forms within the system. Much of this could initially be vibration (a mixture of kinetic energy and strain), but material losses will eventually convert this to heat. Some of the individual object's linear kinetic energy could be converted to kinetic energy of rotation* (the subject of some of my other posts which curiously now appear horribly relevant to your dilemma).

Your sarcasm is ill-directed. There are several competent people on this site (some of whom I know to have designed dynamic mechanical systems that performed as intended), and without exception these people have supported the answer given in post #1.

It really is a good idea to recognise what you do and don't understand - and even better not to be rude to those who know what they are about and are trying (mostly with admirable restraint) to help you.

Fyz

*If the system is closed, the rotation will not decay.

A very amusing exchange, but, since somebody asked what material the objects were made of, I'm left wondering what would happen to the importance of significant digits if the objects were made of a slippery amalgam of teflon and Plutonium-239. At least the wait for the correct answer will seem very brief compared to the half-life of 24,000 years!

Thanks for the entertainment folks! It brought back some of the horrors of waiting for the Physics final to begin!

There is absolutely no evidence of that...so why assume? Did you provide an answer?

I absolutely disagree with this answer.

After collision, some energy is spent to overcome the inertia of the 2nd object which was at rest. This energy comes from the moving object which has, for the purpose of this exercise, only a kinetic energy from which to dispense. The conservation of total energy is primordial and since no losses were considered, the kinetic energy must be reflected in the final outcome.

The momentum is generated by an energy inputat the start. It can only be used if the two object started to move at the same time due to an energy input...

Sorry, not convinced with this answer: Total energy must be conserved, nothing disappears and nothing is created from thin air (exclude the quantum theori...).

I should really learn not to rise to the bait, but one more attempt at explaining this:

As you have acknowledged elsewhere, total energy can take forms other than kinetic, so kinetic energy can indeed be lost. Your statement that "since no losses were considered" is demonstrably untrue: if the masses stick together, the collision is inelastic with respect to linear KE.

I think you are suggesting that there is insufficient information to solve the problem for a "lossy" collision and that therefore, by implication, the collision must be perfectly elastic. If the problem had indicated that the masses separated after the collision, there would indeed be too many variables to solve by conservation of momentum alone, so it would be reasonable to assume a perfectly elastic collision (you'd also have to assume a co-linear collision).

However, in this case,

1. having only a single mass post-collision means that there is sufficient information to solve by conservation of momentum alone; and
2. a perfectly-elastic (lossless) collision cannot result in a single mass - the masses must separate to preserve linear KE.
   

What great precision...(100 cm/min)(6/9) = 66.67 cm/min or

Answer is only correct if given: 100.0 cm/min, 6.000 lb, and 9.000 lb

Can someone explain how we can get an answer to the tenth of a millimeter from data like this. This is insane math!

You are assuming that the initial data was presented using the significant figures convention. There is nothing in the initial challenge to indicate this, and clearly the poser of the question knows better than we what was intended. In hindsight we must regard the situation as "being created to match the stated conditions" rather than being the result of approximate measurements. This sort of thing does happen...

Not in my world...I would be out of a job!

Why not 66.6666666667 cm/min

This is an engineering (not 9th grade) forum

This type of math may help fix a lawn mower or plant potatoes, but it will not get you into space and back safely.

This site is not for me......goodbye and good luck! I'm off to see the wizard.

When accuracy is biased there is no precision!

Even in 9th grade I would have hesitated before giving the rounded answer that corresponds to the input data: i.e. 100-cm/minute (= 1-SF at the same accuracy is the input information, which is what I had just been taught). That would have been before my maths master laboured the point that you should not assume numbers the presentation of numbers corresponds to the S.F. convention unless you have reason to do so. Working in standards labs I learned a bit more - including that no standards or measurements professional would ever use the significant figure convention. This is because of the combination of coarseness of meaning (factors of 10 in precision, for goodness sakes) and ambiguity.
And I think I've learned a bit more since then: if no tolerances are given I check. If I get a mixed bag of information I check. If additional information arrives before the answer to my checks arrives, I make a provisional interpretation. Plus that: in engineering it is generally better to retain spurious precision in the numbers than to risk throwing real precision away. This is particularly true of fabrications - if you really can't find out what is needed (it only happened once to me - a job for a customer who was temporarily not contactable), you should err towards the tolerance that you believe is sure to be good enough (it was - 'phew). From the aspect of pure data, you can always add a note that the tolerance is unknown, and/or add it in whenever you discover what it is.

Fyz

Precision index or tolerance for any data was not given. As professionals, we should be above providing data and answers at this level. Most of us are aware that modifying of SF convention (precision index/tolerance/standard deviation etc.) is commonly done in engineering, and the conventional SF does not always work with unit conversion, averaging multiply trials/population, and radians etc. But the basic principles of SF are still commonly used in science and engineering, and that is especially true for this challenge.

That was indeed approximately the case until the answer was given. I say approximately because the high-school SF interpretation of the data would not support an answer to better than +/-51%, which would make any calculation almost pointless.

I still maintain that reliance on any form of SF (even the "optimal" version using "scientific" notation to indicate precision) is inadequate for any serious usage - and that anyone teaching SF (even at high-school level) ought to warn their students of the limitations.

BTW, I've worked for about half a dozen organisations (and with a large number of others in joint projects), and in fields ranging from standards development through mechanics and fluidics to analogue electronics. Nowhere in my working experience has anyone (other than one school-leaver apprentice) used significant figures to indicate precision. Nor am I aware of any standards (IEC or equivalent) that allow for this.

My assumption from your self-description was that you had been a high-school science teacher. But you imply that you are or have been a practising professional - please elucidate.

Fyz

You assume too much.

...born and raised in Northampton, England.

Significant figures is taught to engineering students, and I did not put it in their text books. Engineers are taught many other methods which may (or may not) complement SF, but basic principles of measuring with SF is a skill set that is still taught to pre-engineering students. There are numerous facets of engineering that apply a variety of methods and techniques. Few were applicable in this challenge...what was your answer anyway?

Engineers are the back-bone of any industrial country, and I am so proud of this profession and admire and enjoy their comments (constant arrogance and disrespect appears to be applied too much) on this forum.

Cheers my friend!

Regarding teaching significant figures to engineering students: like many scholastic habitudes, this is something that has to be unlearned at the earliest possible opportunity, and for two reasons:

First, lack of generally understood meaning: there is no general standard for what tolerance expressed as S.F. actually means (see below for two cases where the meaning is clear), in addition that information that only indicates a value to a resolution of +/- half an order of magnitude is rarely useful, and
Second that if you average results that are presented to the same number of significant digits as the accuracy allows the final value will have a systematic built-in error that can be larger than the statistical uncertainty.
However, teaching SF can leave something useful behind - a preliminary awareness of the importance of tolerance. And unlearning it can leave another useful idea behind - that it is dangerous to use anything whose meaning is not both clear to you and transparent to other workers.

"My" answer? So far as I was concerned, post #1 (by Anonymous Hero) was adequate in this context, so I gave it a G.A. when I was able to post properly.
(I would not have considered tolerance for this class of problem, so I would have left it at that - other than trying to help those who seemed to have misapplied conservation of energy and lost sight of the fundamental nature of momentum).
However, the idea of tolerances was introduced, and in a way that would not fit into either physical standards work or normal engineering production. So I tried to express this problem - evidently not to the satisfaction of anyone who is pushing SF as a viable convention.

So, if I were to consider tolerances for this problem, what would my answer be? Well...

If I were to try to express the result including indicating tolerance via significant figures, there would be two options (units are cm/minute, omitted below in all cases):
70, or
100
Unfortunately, neither of these is remotely satisfactory, as the first indicates a much tighter tolerance (~7%) than the SF interpretation of the challenge would allow, and the second does not even allow you to discriminate between the answer that correctly uses conservation of momentum and a spurious answer that incorrectly uses conservation of linear kinetic energy.

So far as I am concerned, the above completely rules out the possibility of presenting the tolerance of the answer using significant figures. To my mind this directly implies that the challenge did not use the basic significant figures convention (and we can note that this was supported by the challenger's answer). But suppose we are in a situation where we know that the most basic interpretation of significant figures was intended; in my view the "correct" answer would be
69.44+/-56% (tolerance meaning similar to that used in the original presentation of significant figures)*

Failing that, if I wanted to make a statement about the tolerance (given that we can see by now that the challenge is intended to be purely about physical principles, the question arises as to "why?"), my answer would have to be expressed as
66⅔, tolerance unknown
Which is taking first-stage pedantry too far - even for me..

*In case you are wondering why this is not centred on 66⅔: it is based on the high-school standard interpretation of unassigned significant figures - which in this case would mean that all data is is presented to 1-S.F. Treating all figures on this basis would give an answer that is between 50*5.5/(3.5+5.5) =30.56, and 150*6.5/(6.5+3.5) = 108.33, which we might represent as above (69.44+/-56%). Adopting a 1-SF presentation would either give an answer of 70 (which implies about 7% accuracy) or 100, which implies an appropriate tolerance but introduces substantial systematic error and makes it impossible to judge whether the internals of the problem were correctly handled.

All this is before we even consider whether we know what significant figures actually mean; the problem here is that there are SFIK only two places where the significant figures convention has a unique and well defined meaning.
The first is the situation where it originates: the presentation of tables (log tables being the best known example) where the meaning is "the closest number to the exact value that can be presented within the assigned space".
The other place that SF has an agreed meaning is instructions for preparing pharmaceutical preparations. The convention here is that the instructions shall be expressed to sufficient S.F that variations of +/-0.5 the final significant unit are unimportant.
Although S.F. have been used for other purposes, so far as I know it has no standard meaning, and the first rule of communications (which is what this should be about) is that there should be a standard interpretation. By way of example, the +/-0.5 of last significant unit could mean any of σ, σ-sigma, σ-sigma, 90% of distribution, 99% of distribution. Note that for the pharmacists and log-table case, the meaning is 100% of distribution, an interpretation that is not normally sensible for engineering measurements.

Fyz

Maybe we'll communicate eventually...

In the meantime, I assume you would reword the question in terms of the speeds of the two object separately if they were to separate?

Fyz

What I meant is that Energy cannot disappear into thin air. A system in motion has a Kinetic energy. If all other forms of energy associated to the Object are neglected (Heat, rotation or any...) then we are left with only the Kinetic energy (motion energy) associated to the moving object.

In this context, the total energy existing is Kinetic and this must be accounted for before and after collision: Any losses will have to be added to the final Kinetic energy of the single mass in motion. The losses are not disappearances into thin air! They are energy in different forms, but emanating from the original energy that was there.

Since we ignored the losses, then the total energy in Kinetic form that was at the start, before collision, will remain after collision and involves the resulting single mass.

The equation of momentum m.v is not an energy value. It's dimensional value is kg.metre/sec while energy is kg.meter.meter/sec/sec. If you want to use the momentum equation, you must still be able to account for all the energy that is lost or not lost.

in this context, the final mass of 9 travelling at a speed of v, must have a kinetic energy of 0.5 x 9 x v^2

if you agree with this last statement, then you must be able to explain to me where is the missing energy gone since we ignored the losses.

if you bring in (again) any mention of losses, whether kinetic or else, then you must also bring them into the momentum equation! Momentum is energy converted to movement...

Finally, there is no motion or momentum without energy to account for the movement (Kinetic). At standstill status, a mass will be having 0 kinetic energy. To move the object we need to apply energy which will become Kinetic energy.

Once again: Total Energy is Conserved. It does not go into oblivion. some of it or all of will transform into different forms in the course of time and actions. All energies have the same dimensional equation value. Therefore, an account of all the transformations must be made to balance. Even if you add energy or remove it from the system in consideration, you must account it in the equation. Here, Kinetic energy is the only form we start with and the only form we consider after collision. No losses were mentioned or considered. Quantity of movement is not an energy value. It derives fromthe existing energy at play....

You really are discounting what we try to tell you. I'll try not to shout.

You don't actually need to know what form the kinetic energy of linear motion became - only that there is a mechanism that could convert it to some other from of energy. (As my history lesson should have shown, Newton was able to successfully solve this sort of problem without any idea of where the energy went)

The point is that, if the system is isolated, any answer that does not conserve momentum simply cannot be correct; and as there is only answer that both has the objects joined and also conserves linear momentum it has to be the correct one.

Now, the following may appear rude - but it is meant seriously and constructively - and the analogy is quite relevant:
To my mind what you are doing is equivalent to the man with a hole in his pocket who loses a quarter, but reasons that "I cannot identify where my quarter went, so I must still have it".
[To labour the point: the quarter represents the lost energy, the hole represents the stickiness of the objects, and the fate of the quarter represents whether the energy is converted to heat (no-one else finds it), or converted to rotation (found by a little boy who spends it - money making the world go around) or... ]

Fyz

Please clarify:

1a) Are you saying that kinetic energy cannot be converted into heat in a collision? or
1b) That there is some feature of this particular collision that prevents the kinetic energy being converted to heat?

2a) Are you saying that momentum does not need to be conserved?
2b) If the answer to 2a is yes, do you regard "conservation of momentum" as not being required in general, or are you proposing that some mechanism exists here that allows additional momentum to be imparted to the joined object?

My view is that the statement that the particles join implies that kinetic energy of linear motion is converted to some other form - presumably via the objects being "sticky". But this is something you do not actually need to consider, as momentum is conserved under all circumstances, whereas mechanical energy is readily converted to other forms (basic thermodynamics tells us that it is not so readily converted back...)

However, so far as I am concerned the major point of a simple mechanics problem such as this is to illustrate the difference between mechanical energy and momentum; i.e. that:
Momentum is conserved under all Newtonian circumstances, and there is nothing else that momentum can be converted into: and
Linear kinetic energy is in general converted to other forms of energy during collisions.

Maybe a little history is relevant here:
As early as about 1680, Leibniz observed that "kinetic energy is conserved so long as the masses do not interact". Playfair elucidated this further as "Kinetic Energy is clearly not conserved".
Attempts to use Leibniz's "equivalence" principle to define a general conservation of energy were foundering until the recognition that mechanical energy could be converted to heat - quantified by Joule as late as 1843.

Fyz