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

Hi Hero.

All I can quickly think of is that the collision is not completely inelastic and hence some energy is lost (not immediately as heat, because that would have momentum as well). The sound of the collision may be one way of momentum loss?

In such a case, the combined speed must be marginally less than 66.67 cm/min.

-J

Conversion of energy to heat and sound is why conservation of energy can't be used for inelastic collisions. However, momentum should always be conserved so I think the answer is correct (albeit with some funky momentum units - lb·cm/min, but they divide out). I agree that this seems to be rather simple for a monthly challenge.

" I agree that this seems to be rather simple for a monthly challenge."

And then have the drama of waiting over one month to confirm the answer. :)

But wait - if (non-relativistic) momentum is conserved, while there is energy loss, then must the final velocity not be marginally higher than the 66.67 cm/min?

If p = mv = constant, with m=E/c², then lesser E means higher v, not so?

-J

I don't think so. The total sum of the net energy of the system will be same before and after the collision.

If 100% of the energy is transferred into the final sticky mass, then the calculations are correct.

If some fraction of that collision is liberated in heat and sound, then something less than 100% of the total system energy is imparted as kinetic energy in the final mass. Therefore, the velocity must be less or the mass would need to change to a slightly less mass.

Ke = (m * V^2)/2

The kinetic energy must be less after the collision (lost to heat sound etc when the sticky object grabs the other one) which can be verified by looking at the 1/2 times velocity squared of the moving objects before and after. There is no way to conserve both kinetic energy and momentum if the objects stick together. The only answer that only uses the info given implies conservation of momentum must be used as in your original response.

For a uniformly-coloured object radiated heat would not affect the velocity. The reason is that the radiation would be front-back symmetrical in the frame of reference of the object. From another viewpoint, the asymmetry of radiation in the frame of reference of the observer (higher-energy radiated forward) will exactly compensate for the mass-loss from the object.

Sound is another matter - that will to an extent depend on the shape of the objects. Given the sizes, it is likely that significantly more sound would be emitted forward. But even for an identical pair of objects (not this challenge) the higher acoustic impedance in the forward direction would slow the pair below the half-velocity.

However, there would be a far larger effect in the presence of a sound-conducting medium - fluid expelled during the collision. The chances (no certainty) are that this fluid would have forward momentum - also reducing the final velocity.

Please don't consider gravitational waves...

Good reply -- The medium in which this collision occurs is very important!

Think about the difference between a collision of this sort in air versus the same collision in water. BIG difference. The water being squeezed straight out to the sides makes a great damper! It takes a lot of energy to accelerate that water out, compared with the energy required to move the air out.

I would have thought that any dissipated energy (be it sound, heat, or whatever) would result in less energy in the congealed mass that what was initially present at T0 when the experiment started.

Under ideal theoretical conditions if 100% of the energy was transferred to the congealed mass upon impact, then the initial posted answer I gave would be the correct one.

However, any dissipation of the T0 system energy after the collision would and must result in a lower net velocity of the congealed mass. At least, that is what the equations tell us (non-relativisticly).

The radiation was of course relativistic... Non-relativistic calculations ignoring Doppler but allowing momentum=mc=hν/c would give a mean forward momentum for radiated light of zero - so they would "predict" that radiation losses caused the lighter objects to speed up. Non-relativistic calculations including Doppler would give an average momentum for the light of frequency ν of hν(V/c)/c, or mV; unsurprisingly, Doppler plus light momentum gives the same result as full relativity.

If after collision the conjoined object is hotter than its radiative environment, we could make the front reflective and blacken the rear - radiation pressure will then result in a net forward force. We might do something similar with sound - by making the front surface interfere destructively at the joint object's resonant frequency, but allowing emission from the rear

And I could shape a cavity in the 3 lb object and a point in the 6 lb object so that material is ejected backwards, increasing the overall momentum of the pair immediately after the collision. Artificial, certainly, but showing that there is still a theoretical possibility of acceleration following impact.

Of course, for non-contrived situations I entirely agree with your viewpoint - even if you did allow that other guest to lure you into using the term "energy" when I think you really mean momentum.

"And I could shape a cavity in the 3 lb object and a point in the 6 lb object so that material is ejected backwards..."

Then it would not be a net 9 lb mass as stated in the challenge.

The context here included the possibility of sound emission (this was in response to your post #34). That requires a medium. It was of course this medium that was ejected backwards, not material from the objects themselves.

Some of the sticky mass could be dispersed out from between the two objects on impact.

There are no inelastic collisions. If there were such a thing, it would release infinite energy due to division by zero, neither of which is valid.

Not sure what you have on your denominator that goes to zero, but the total energy in the system is most assuredly finite, so there will certainly not be a release of infinite energy.

Perhaps you mean the RATE of energy release? For sure a real-word collision won't be instantaneous (elastic or inelastic), and the agglomerated mass might squish or wobble for a bit, but eventually all the energy will be dissipated, so a perfectly inelastic collision is the result. The finite duration for the collision makes no difference to basic conservation of momentum and the final velocity will still be 2/3 of the original.

(The only exception to that is addressed by those answers that allow for directional transfer of energy outside of the two bodies, but since the question contained no information to allow calculation of such effects, we can assume that they were not part of the questioner's intent.)

No actual division by zero because there are no inelastic collisions, none, they don't exist. There are no inelastic substances, they don't exist, hence, no need to divide by zero whether it is rate or quantity, it is invalid because inelastic materials don't exist.

Many calculations (in the industrial sector) deal with the idealistic approach/equations, and AH was justified to drop zero appropriately in his solution. This appears to be a hypothetical problem (closed system): evidenced by the lack of specificity and absence of variables. KE might have some roll in the solution, but unless this factor can be calculated and would influence the velocity by more than 1cm/min only then would it be significant...I see no evidence of that. Therefore, AH solved the problem out of the gate using the conservation of momentum. Correcting the specificity of his results, the answer is 7cm/min. I do agree that the simplicity of this problem make ones suspicious of the KE factor.

Senior moment (allow me to correct my own math): 66.667cm/min should be rounded off to 70cm/min when the answer (by AH) is limited to one significant digit, thus there is a 65cm/min to 75cm/min or 10cm/min variance. Sorry, some neurons are obviously not firing. Happens, when you reach 70years old ...well I will be on March 2 the day they release the answer to this problem (70cm/min).

Inelastic here means lossy, not ideally rigid. If you did have ideally rigid and totally lossy materials you would get infinite power conversion for zero time - so finite energy and no conflict with physical laws. The reason that perfectly rigid bodies can't exist (and you are correct that they can't) lies elsewhere.

Well how about that!

Further ---- Key assumptions are zero friction, massless media, and direct hit, which are ideal conditions one might duplicate with objects flying in space (at least to an acceptable level of precision for okiescienceprof to buy off on it). Then, as stated, the collision causes the masses to adhere themselves to each other with no loss or gain in total mass and no bounce.

Check out HarryBurt's comment #41 about the bullet and ball of clay, which points out that this is a conservation of momentum question-- not a conservation of energy question. This is also a very good example of the "sticky stuff".

If we agree that upon initial contact of two bodies where one is moving at a different rate than the other, and the change in velocities due to acceleration and deceleration constitutes work, then plugging t into the formula for power: P = W / t we would have division by zero since without elasticity (deformation) the change in velocity would have to be instantaneous and any amount of work performed in zero time infers infinite power which would likely annihilate everything. Also implied would be infinite acceleration which would create infinite force: F = ma which is of course also a bogus notion. Come to think of it, maybe this is why matter is not solid. Anything perfectly solid contacting something else perfectly solid would be somewhat of a problem. Maybe that's what happened at the "big bang", two perfectly solid objects collided, hahaha. However, if we could create such a substance we would have weapons more powerful than hydrogen bombs. Good thing it's impossible, we are dangerous enough as it is.

See #60 for the fault in your arguments

Absolutely correct, but in this challenge those terms were intangible. In other words, there were no quantitative clues as to amount of energy transferred to heat or sound, etc.

The best I could do was to work in the theoretical theater.

Values for some variables were not given (nor can be calculated). Thus, they must be either considered zero, a standard value, or insignificant within the boundaries of the data provided. The first two factors do not apply so they must be insignificant within the boundaries of the answer. Therefore, the answer should be stated as 7cm/min (or some value between 6.5cm/min to 7.5cm/min).

I have a problem with the data supplied. What shape is the "sticky stuff" objects? Is the stick stuff against the surface of the platform? What initial force will be required to get the 2nd object of sticky stuff moving? What are the drag constants on each object from the sticky stuff? Apparently not frictionless motion. I'd like to see a discussion re are there any achievable conditions that would alter the half-life of isotopes, not including neutron etc bombardment that would change the isotope into a different species.

This method (conservation of momentum) is essentially correct, but the answer is technically incorrect. Each value (measurements) in the problem only had one significant digit thus the answer can only have one significant digit. The answer should be 7cm/min. Since the value 7cm/min ranges somewhere between from 6.5cm/min to 7.5cm/min, the Gibbs free energy for this collision would not be significant (the data was not given anyway). The answer is 7 cm/min.

How about a round number - say 40-m/hour?

Have phun

You missed that kinetic energy is proportional to velocity squared not simply velocity,

ergo K.E.= 1/2MV(squared). If KE(1) = KE(2) then

1/2 M(1)*V(1)squared = 1/2M(2)*V(2)squared.

6*100(squared) = 9*V(2)squared

{root}60,000/9 = V(2)
V(2) = 81.65 cm/min

Not so. DAC was quite correct

You have tried to conserve kinetic energy. But this is only conserved in a perfectly elastic collision - in which event the two objects could not stick together. In other words, kinetic energy is not conserved in an inelastic collision - it is converted to vibration and eventually heat.

What is conserved is momentum, and this is what DAC calculated (and he even explained that this was what he was doing).

Yes, I believe you did miss something. They never stated if the objects were on wheels or sliding along the ground. In any event, you will need to have the coeficient of friction (or rolling resistance) to accurately determine the final velocity of the combined body.

If it is not stated or derivable it is safe to assume those variables as zero. That's pretty much standard practice for all school/university problems.

As I stated earlier AH, omissions can also be a standard condition (e.g. pressure = one atmosphere), but usually (not always) are stated correctly as STP or SRC etc. in the problem. But, more often (you are essentially correct) they should be considered zero or insignificant. This is a hypothetical/idealistic problem, and the answer is 70cm/min. Realistically, the speed of the residual mass would be slower than this value, but the difference would have to be greater than +/- 5cm/min to affect the answer. Data in the problem are at the best weak estimates and ambiguous, suggesting an easy idealistic solution. I think you nailed the answer out of the gate (except for SF).

Conversions...Mass of blocks are in IP units, velocities are in SI units. One of them need to be normalized.

100 cm/min = ≈0.0547 ft/sec

.5*6*(0.0547^2) + 0 = .5*9*vf^2

vf ≈ 0.0446 ft/sec -0R- 81.6 cm/min

Seems to me there is insufficient data to render a meaningful answer. Most (if not all) of you seem to assume that the center of mass of the "projectile" is moving directly at the C.O.M. of the "target", otherwise some of the energy that would go toward its combined final vector would be translated into rotation of the merged body. This would also assume that the resulting centripital forces are not stronger than that of the "stickiness", otherwise the two original bodies would fly apart after impact. But this is all crazy talk on my part anyway... ^¿^

So long as we make the usual assumption of "no external action", this is a challenge where you really do not need any additional information. The point is that both momentum and angular momentum will be conserved.
This means that (ignoring external actions) the centre of gravity of the pair will be moving at the speed stated by Anonymous_Hero before, during and after the collision. This would be true regardless of whether or how the objects join - though you are explicitly told that they do form a "joint object".
Similarly, the angular momentum of the pair will be the same before during and after the collision.

Energy, on the other hand is not conserved* during this collision - but that does not affect the answer to the challenge.
*As described elsewhere, if the objects are to join to make a single object, kinetic energy would only be conserved if both objects were rotating with the same angular velocity, the edges where they first touch are moving at the same velocity, and the objects are sticky-rigid. As we are told that the second object is not rotating, we know that kinetic energy can only be conserved if the second object has zero moment of inertia.

Fyz

You're right, the second part of my comment is silly because we are explicitly told the two objects become one after the collision. While we are NOT explicitly told that there is no post-collision rotation, its omission suggests that there is none for the purpose of the challenge. I stand corrected. ^¿^

I've obviously still not been quite clear: there is no need to assume that there is no post-collision rotation. All we need is to maintain the convention that the speed of an object is the speed of its centre of gravity - in which case the speed of the combined object is uniquely specified by the challenge. This should not be problematical, as we use this convention all the time when describing movement - whether it be aeroplanes with propellors, spinning baseballs (basketballs or footballs or whatever) or even planets.

Fyz

This is a lot of gibberish... The first object is stated as moving and no mention of rotations! Therefore, simply assume the data as given and don't add IFs... The fact is that if any amount of energy is required to overcome the Inertia of the fixed object, this will be deducted from the object that has energy to give... Kinetic. Therefore, the final system is stated as forming one mass, will now contain all the initial energy which was kinetic... conservation of kinetic energy is appropriate here since all the mass is now moving together... no difformation or anything else is mentioned and no parts of any object is left out to carry energy somewhere else...

How rude! But it might (just) be acceptable if you were had taken the trouble to find out what was going on.

Energy will of course be conserved - but the sticky collision will convert some of the kinetic energy in the original linear motion to some other form of energy - be it sound, heat, rotational kinetic energy or whatever. So kinetic energy will not help you answer the challenge. What is needed to "overcome inertia" (horrible misleading expression - no wonder people get confused) is not energy, but momentum. To answer the challenge you need to conserve momentum, not energy, as was correctly done by AnonymousHero* in the first posted answer to the challenge.

The stuff you are objecting to does not conflict with AnonymousHero's solution. It is simply addressed to erroneous conclusions that are often attached to such solutions - an extension to understanding rather than part of the original challenge as posed.

Note that if the collision had been co-linear (neither stated in the challenge, nor necessary to solve it), 1/3 of the original kinetic energy would have been converted to other forms.

*Using only the standard simplifying assumptions such as "you can ignore interactions with the environment).

I am sorry if my expression was rude.....

But on the subject: The problem was stated simply and there is no way you could evaluate just by working out on paper all the extra variables and conditions that you are adding to solve a simply stated problem. You would need to guess what the author wanted to assume etc... and then conduct some experiments to define the values...

Therefore, simply put, the moving object collides with a stationary object and being sticky (just for this exercise), they start moving together in whatever direction. Linearity or any other consideration is irrelevant here. Just assume in space, the collision will be direct on, and not sideways since not mentioned... Therefore, since moving in unison, the conservation of momentum cannot be equated since the centre of gravity is moving for this system!

Thanks..LAA_LUke

Subject to "standard assumptions", the problem was fully and satisfactorily solved way back in post #1, which I repeatedly acknowledged - including in a response to one of your postings. Nothing I'm saying changes this result.

Regarding your statement about a "moving center of gravity", this is precisely what conservation of momentum is about. We can rewrite conservation of momentum as "the centre of gravity of a system will continue to move at a constant velocity unless that system is acted on by an external force" - which is obviously the expression of Newton's first law of motion in terms of a multi-bodied system. And it is the only constraint that can be used to answer this challenge.

Once you have absorbed and understood this, you may be in a position to follow some of the more subtle aspects (not part of the original challenge) that I was considering.

I'm exploring these related (additional) areas because others have raised them, and they appeared interesting. One of these (though not necessarily expressed in these terms) was "is 1/3 of the kinetic energy inevitably converted into other forms?". And I was showing that the answer to this would be that, depending on the details of the collision, anywhere between none and 1/3 of the KE would be converted.

Thank you, Dave. I WAS complicating the exercise unnecessarily, which on reflection was meant to be straightforward (and unexpectedly easy).

C-M ^¿^

You have considered the conservation of mouvement, whereby the center of gravity remains in its place. Your response would be correct if you were looking at jet propulsion systems for example.

Here, the conservation of kinetic energy is in question. The moving object has a Ke = (6 x 100^2)2 and the stationary object is = 0. After collision, the problem states that they form ONE new moving mass of 9lb. No elasticity or any other factor is mentioned and therefore no need to agravate the problem for no reason.

Therefore, resolving Ke1 = Ke2 will give V2 = V1 x sqrt(6/9)= 81.65cm/min

m1 x V1 = m2 x V2 will only apply if both objects start going in differen directions after collision, but still you will then need to work out how the Ke energy will split to each object....

Have you ever played rackets? And noticed that the ball gets hot? This is because kinetic energy is not necessarily conserved - in this case some of it is converted to heat. Momentum, on the other hand is ALWAYS conserved.

It's very simple: any supposed solution that does not satisfy conservation of momentum cannot be correct. The corollary is that if conservation of momentum is sufficient to provide a result, then it is the only thing you need to know about.

You are correct for the tennis balls and any such type of action involving rackets etc... the reason is that the racket stays in your hand while the ball is redirected with the same mass...

In this case, it is stated that the 2 masses become as ONE and the will be moving together... In the case of the tennis ball, being a deformable and elastic material, when it hits, let us say another object similar or not, and if they become one, they will re-absorb the energy of deformation and forget about the heat loss since that was not debated by the initial statement of the problem to solve, and you cannot calculate unless you make soooo many experiments and define soooo many variables( constraints) complicating unnecessary the original question....LAA_Luke

Is the material elastic, and does it matter ? From the wording of the question, I guess not. Still, I've said it now, so I'm stuck with it . I'd roll with the 2/3 approach of #1, though (as ever) we have to make a zillion assuptions about conditions (distortion/shape/aerodynamic loss pre-collision etc). The 6 lb object might even have back-spin, bit like playing pool - some redneck is going to be well unhapy when it all ends up in his beer glass.

If I've inspired folk to further thought, then I consider it a job well done . It matters not, because if previous is anything to go on, we'll all be basting Turkey for Thanksgiving/Christmas before the 'official' answer is given

I don't think rotational affects from colliding off centre will make any difference to the motion of the centre of mass of the combined object. This can be illustrated by considering the motion of the centre of mass of the two body system which must be the same before and after the collission as there are no external forces being applied. If the straight line distance between the centre of mass (COM) of the 6 lb object and the 3 lb object is 3 length units at a certain instance before the collission, the COM of the combined 2 body system is 1 length unit from 6 lb object and 2 length units from 3 lb object along the same line at the same instance of time. This means that the COM of the 2 body system is travelling at 2/3 of the speed of the 6 lb object (ie, 66.67 cm/min) before and after the collision (ie, it is unchanged by the collission). This off course assumes that no mass is lost from the 2 objects (I think we can safely ignore relativistic effects at these low speeds and energies).

The total kenetic energy before collision is 378x10-6 J. The total kenetic energy after collision is 252x10-6 J. Therefore, the energy lost in the two bodies is 126x10-6 J which would be lost as heat.

I like the 'spinning' idea.

No shape is given, so it is possible, if colliding offset at an angle, the combined objects would spin.

I've no idea how much. But I am sure the velocity with a spin would be somewhat less than the velocity of a direct head on hit with no spin.

And by some considerable amount I guess.

Conservation of linear momentum means that the initial rate of movement of the CofG is independent of the spin.

Absolutely right. A spinning combined mass will have more kinetic energy and less conversion to thermal/whatever than a non-spinning mass. The angular momentum will NOT come at the expense of linear momentum.

Other than resulting in an increase in aerodynamic drag - but that's only a finite deceleration, so it shouldn't affect the velocity immediately after the collision (whatever that means)

Despite the distractions of the endless discussion on significant figures, I think I have gained some useful mechanical insight from this thread.

Suppose two bodies collide with glancing impact and some ingenious hooking device were to prevent them from separating, then the energy which would otherwise be lost (in plastic deformation during a head-on collision) gets transformed into rotational energy of the combined body. The linear momentum is conserved, and angular momentum appears to get created as they start 'spinning', but I suppose it exists already with respect to the combined centre of mass and the original trajectories.

Is that correct? I've never had to dabble in this kind of dynamics, and would like some confirmation from gurus on this thread, since it is still somewhat counter-intuitive.

Maybe using a simple numerical example, =TeeSquare=

Almost: it will only be exact if both objects are spinning at the final angular velocity before the collision, otherwise kinetic energy will be converted (to vibration then heat) in synchronising the rotation.

If the objects are separated so that each can oscillate about a(n assumed inextensible and infinitely flexible) joining cord, then there will be no loss of kinetic energy if the points of attachment have zero approach/separation velocity at the time of attachment. [The dynamics of the coupled resonances can be quite amusing...]

Fyz

I possibly misconstrued at least one phrase of Tee-Square's note, so (although my intended meaning was correct) I need to re-write the above (unfortunately in mind-numbing detail):

Angular momentum is conserved absolutely in all cases. It is easiest to see this if we define it wrt the joint centre of gravity.

What I was writing about was kinetic energy. For objects joining at a surface this can be conserved if the differential velocity between the joining surfaces is zero at the initial moment of contact, and the angular rotation rates are identical before and after joining. (The rotational constraint could equally be written as the angular rotation rate of each object being the same at the moment of joining as the rotation of the line joining the centres of gravity).
Note that the above constraints implies that the joining surface is orthogonal to the direction of motion.
I should also mention the "motherhood" constraint that the change in shape of the objects (due to increased centripetal effect after joining) is negligible.

For objects joining via a cord we require either that the cord be either losslessly elastic, or if inextensible and lossy, the approach/separation rate at the moment of joining be zero. Other assumed constraints were that the cord would stick to the objects wherever it touched them, and that the separation was sufficient that the objects would not roll so far up the cord that they touched.

Apologies that the original did not quite say what was necessary

Fyz (temporarily at his own PC)

Although presenting facts not in evidence (a common occurance on this board) drives me nuts, I'll propose that, due to the low velocity of the 6 pound object, there may not be enough energy to dislodge the sticky 3 pound object. We don't know the coefficient of friction or any other useful details but that's a very low-speed collision.

On that basis, why wouldn't the 6-lb object stop before it hits the 3-lb one?

Because we are told it is traveling at speed when it hits. A fact that is in evidence.

The 6 pound object may be decelerating when it collides and will continue to decelerate.

Assuming zero friction and conservation of mass momentum (M1*V1=M2*V2), the velocity of the combined mass will be 66.67 cm/min

I believe atmospheric conditions also must be known to achieve a correct answer. (temp, humidity ext.)

Mixing up ones units was a no-no when I was at college. ie lbs and cms. Spacecraft can go off course that way!

In a vaccuum it would be moving at 66 cm/min.

Doesn't anyone remember their first physics problems that were always prefaced with the conditions of the experiment being performed in a closed room or box with a vacuum and no extraneous forces would come into play?

I haven't read all of the comments, but the ones I have read all talk about energy being radiated. Imagine a large ball of clay being hit by a bullet. Assuming the clay is not blown apart, it will move at some rate determined by their relative masses and pre-impact velocities. The clay will be greatly deformed by the bullet impact; this deformation will heat the clay. It doesn't have to transfer this heat to the environment instantly so discussion of how to dissipate the energy is irrelevant. The final velocity will be determined by conservation of momentum.

The energy of a moving object (assuming no losses due to friction, heat of collision or aerodynamic drag) is predicted by the simple formula:

KE = 0.5 * M * v^2

Therefore the moving object has a kinetic energy of 30,000 lb - cm^2/min^2. The object at rest has a kinetic energy of "0". Therefore after collision, assuming kinetic energy is conserved, the velocity of the 9 pound object will be:

v = (KE (total)/ (0.5 * (6 + 3))^0.5 = 81.6 cm/min.

"assuming kinetic energy is conserved"

Wrong assumption - for KE to be conserved the objects would have to bounce off each other.

I agree conservation of kinetic energy is the correct means to the solution. The exact answer is 81.64966 cm/sec

Regardless, your answer can't have seven significant digits unless each of the initial measurements had an equal (or greater ~ limit based on smallest) number of SD. Thus the answer would be 8cm/min.

Kinetic energy is only conserved in elastic collisions (for which this is not) so conservation of momentum is the correct principle to use. Also keep in mind that the 6 lb object is moving at an extremely slow speed of 100 cm/min (0.0167 m/s or 0.037 mph) so I'm not sure that you can even consider this a collision at all. A collision is an interaction between two objects that exerts strong forces between the objects over a short period of time, but at this velocity the interaction will probably be longer than normally considered as a collision.

Elastic collisions ~ momentum is conserved and kinetic energy is conserved

Inelastic collisions ~ momentum is conserved but kinetic energy is not conserved

Since solving the problem with the conservation of kinetic energy and momentum give different answers, the problem is assumed to be an inelastic collision (as implied) and the method you are using can not be correct. When in doubt, always use conservation of momentum method because it is valid for both types of collisions.

Those who use the fact that a perfect inelastic collision does not exist are correct, but need to teach for a while. Then you will understand why it is important to assume ideal conditions and work with hypothetical problems for the simplification of learning. Even if the collision was not perfectly inelastic, the variance in the problem is +/- 5cm/min. This would allow for some elasticity.

The correct answer is 70cm/min (any answer with more than one significant digit is mathematically incorrect)

We are told that the two objects become a single object after the collision. The combined object may vibrate or spin after the collision, but this does not change the velocity of the centre of gravity, which is what we must use for its velocity. This means that immediate losses and/or subsequent vibration have no bearing on the variance of the answer.

So the variance in the answer is the same as the variance in the values that set the problem. Although the inputs are presented with only a single digit value, we have no way of knowing whether the values were set by rounding or in the creation of the objects (e.g. bags of molasses + arbitrary but accurate velocity settings ?).

Even assuming the given values are the result of rounding, I think that some explanation of standard deviation is required.
The 3lbs could be anywhere between 2.5 lb and 3.5 lb; knowing nothing about the origins of the object we could then assign a S.D. of about 0.5/√3 (~0.3 lb) to the mass of the object (this is accurate for a uniform distribution). The expected S.D. on the 6-lb object would also be about 0.3-lbs, and the variance on the velocity (range presumably 95...105-cm/sec) could be about 3-cm/sec. In percentage terms these would contribute about 3%, 2%, and 3% respectively - giving a S.D. of about 4.5% or 3 cm/sec.
Junior high-school tradition should be to present the resulting value to two digits; however, even this practice should be discouraged at higher levels, as it can result in systematic errors when averaging multiple measurements of consistent phenomena.

If, as you desire, we are to be precise about this, the result can be presented to a substantially larger number of significant figures together with the S.D. - e.g. 66.7 cm/sec (σ ≈ 3 cm/sec).In this case the basis for the calculation of the error should probably be stated.
The resolution at which the mean is presented would depend on the context; I believe that our lack of knowledge as to context would mean we should err slightly on the side of over-presentation, as the inclusion of the S.D. means that this is not misleading.

Regards

Fyz

So the "correct" answer is 67 cm/min +or- 3cm/min.

I would hate to think that someone could read the 70 cm/min to mean that there is a very good likelihood that the object could be traveling at 75 cm/min! 65 cm/min, OK. But not 75, which is actually 8 cm/min faster than the calculated value, assuming no losses. That is 12 percent! Hey, that's not as bad as I thought it would be!

The "correct" answer could easily be argued to be either 67 cm/min or 70 cm/min. Quote 67 cm/min if the answer is to be used for downstream calcs. Use 70 cm/min if this is a quoted number in a report, but maintain the 67 cm/min value for further calculations in the report to limit the accumulated effect of the in-built (possibly 12 percent) error.

With all due respect to both of you ~ my understanding is that keeping/carrying a two SF value calculated (using multiplication or division) from two or more values with only one SF is not valid. The answer should be rounded to one SF.

20 X 3.2 = 64 = 60

Want more significance then measure more accurately

I don't know where those numbers come from.

But if I understand you correctly, you are proposing that we discard precision midway through the calculations? If this is what you mean, I know of no basis for it; worse yet, the answer can depend on the order in which you perform the calculations, though in this case the single-digit rounding does at least result in the same value.
I.e.:
100*6/9 = 600/9 = 66.66 rounds to 70, and
100/9*6 = 11.11*6 = 11*6 = 66 still rounds to 70

Fyz

I disagree only slightly. Your 20 should be taken to be accurate to the nearest 5 units, even though it happens to be specified only to the nearest 10 units. What if the original number specified is 18, 19, 21 or 22, but supplied to you as the rounded number of 20, the specifier not wanting the answer to be returned to the nearest 1 unit. Or what if the original number is, itself, 20 units (say marbles) why would you expect that he wants the answer only to the nearest 10, which could actually produce a number as far off as 14? That is a lot of marbles!

I truly hope you are not teaching your students that this is the correct way to do things for when they get out into the real working world. Teach them to use their common sense, if only for just a little bit. No wonder I see so many graduates these days that cannot think for themselves.

In your example your answer of 64 should be rounded to 65. By your definition, your value of 60 could be taken to include 50 units, which is in error by 22 percent! Not acceptable in most circles of practical application.

For example, would you take 16 as being accurate only to the nearest 4 or 8 units just because it is evenly divisible by 4 or 8? Not likely. Perhaps one could take it to be accurate to the nearest 2 units (12 percent), but never 4 or 8 units.

For much larger items, say 1,000,000 pounds, the magnitude of the number is very significant (pun intended). I normally take such a number to be accurate to the nearest 1000, 10,000 or 100,000. Never to the nearest 10. It generally cannot be measured to the nearest 10 pounds, and such accuracy would be totally useless.

First - apologies for the trypo (not to be confused with tripos). I've no good excuse for failing to catch this myself.

And even less excuse for dividing the range of a uniform distribution by √3 (I don't know what's come over me today). Obviously the divisor should have been by √2 - which gives a one-sigma error of about 5.5% or 3.7-cm/minute (managed to type it this time - I think). This is somewhat closer to the prof's numbers.

Returning to your comment: yes, it depends on further use of the data. If you don't know the application I still think that it's a good idea to give more resolution and specific error data.

It may also be worth reminding ourselves that for a normal distribution the probability of the actual value lying outside the 1-sigma used for error bars would be slightly more than 32% - though this is of course not exact for the convolution of only 3 uniform distributions.

Indeed, for this case (unlike for the normal distribution) the extreme range would be between 0.55*95/(0.55+0.35) and 0.65*1.05/(0.65+0.25), or between 58.1 and 75.9, so not much chance of exceeding 75 cm/minute.

Apologies

Fyz

100cm/min is also a measurement with one significant digit

My schooling puts the tolerance for "100 units" at +or- 5 units unless specified otherwise, which is in line with the thought process in reply #69.

However, specifying 99 or 101 would put the intended tolerance at +or- 1 unit. I have always been bothered by the fact that "100" could actually have two totally different inferences.

Now, all that said, I do agree with you for most applications. For most of my applications, a tolerance of 1 or 5 does not make much difference if the error is less than 5 or 10 percent. Therefore, I would use +or- 5 units for any specified number that is evenly divisible by 5 and greater than or equal to 100; but I would use +or- 1 unit for any whole number less than 100.

Hey. We got way off the original topic way back there somewhere. However, we are now right on the current topic. Right?

I believe so...maybe the author will ignore SF...I could live with 67cm/min

Wouldn't 989 or 102 normally imply +/-0.5 unit? Bur the ambiguity would force 99 to be -0.5/+1 unit, which appears even less satisfactory than

Of course, thissimply illustrates the intrinsic ambiguity of whole number presentation. A useful convention if you wish to indicate rounding errors is to write the number in so-called scientific notation - e.g. 1.0E2 for a rounding error in the region of +/-5, 1.00E2 to indicate a rounding error of +/-0.5.

Does "one significant digit" here mean that all we know is that it would lie between 50 and 150?

Please! This is not the forum to teach the methods for applying significant digits nor consequences of not using these rules so I will try to keep this simple and in the context of this topic. Several posts have questioned and/or made comments about the application of significant digits to arrive at the correct answer. Rules for S.D.'s are in most general text, and I did not invent them as someone implied.

The original data was given as 6lb, 3lb, 9lb, and 100cm/min. Did the author mean the number for 100cm/min was:

A. 1 X 10^2 (range 50 to 150) one significant digit

B. 1.0 X 10^2 (range 95 to 105) two significant digits

C. 1.00 X 10^2 (range 99.5 to 100.5) three significant digits

It does not matter as far as the answer (when multiplication or division is used) is concerned because it is also limited by the other variables which only have one SD.

Graphics in this forum does not allow a hyphen to be placed above the last significant digit (when it is represented by "0"). When feasible (as someone pointed out), the number should be expressed as an exponential number where the mantissa may have only significant digits (thus all "0's" in the mantissa are significant). In engineering, the exponent is commonly written with a number divisible by thee coinciding with units and numbers (e.g. nano, milli, and billion etc.). Thus, 15.4 X 10^6 is sometimes preferred over 1.54 X 10^7 (chemists text typically don't do this).

Answer of 66.667cm/min is 70cm/min or 7 X 10^1 cm/min

Well, that takes the cake. You write that "this is not the forum for", having yourself first introduced the issue - and you then proceed to lay down the law on the matter.

I see that you concur my statement of the school-level 'rules' (using 1.0 x 10^2 instead of the equivalent but more compact 1.0E2), so I can hardly complain about that part.

However, once the subject is raised it becomes important not to mislead, and it is well known that the use of rounding to match accuracy can in many cases lead both to substantial increases in the errors of data derived from those results, as well as to systematic bias when results from multiple sources are combined.
For measurements of physical processes, the resolution should always be at a level that allows variable errors from other sources to substantially exceed the resolution error - combined with error "measures" and preferably an indication of the type of the error. This last is because measurements that are dominated by rounding errors (e.g. equipment resolution) can directly result in consistent offsets - unless other non-systematic errors are large enough (= either "very much larger" or "uniform distribution of related size") to allow these to average out. (Electronic engineers may be familiar with the use of dither to improve the effective resolution of DACs and ADCs, which is equivalent to non-systematic error in its effects).

Finally, to return to the presentation of solutions to the challenge: the challenge does not specify precision in this way, and there is no secondary information that would allow us to assess this, so it's an aspect we cannot properly address. Short of adding this to an already long list of uncertainties about the problem, I think that Anonymous_Hero got it as right as possible by giving a pair of answers - a recognisably truncated answer (66.67) combined with the statement "2/3 of the original velocity". I would vote him yet another G.A. were this practical from here.

Fyz.

66 2/3 cm per min. Conservation of momentum. Total mass times velocity initial = total mass times velocity final 6x100 + 3x0 = 9 x 66 2/3