What exactly is the cause/source of the energy loss in a non-elastic collision? The forces during a collision are the Normal forces: which don't do work I believe. The Discussion with FakeMod regarding the following problem motivated this question:

We have a rope (uniform, mass density $u$, length $l$) which is coiled and kept on the table and we start pulling up the rope with uniform velocity. We need to find the work done in pulling up the chain completely.

  1. Approach 1:

Suppose $x$ length of the chain is above the table, and we pull a mass element $dm$(with length $dx$) off the table.This $dm$ mass element gains momentum $dm * v$ in upward direction (which means the part of chain above the table has exerted an upward force on this element), which further means it exerted a force = $dp/dt$= $(v*dm)/dt$ on the part of the rope already above the table, in the downward direction. Now, $dm/dt=udx/dt=uv$. Therefore $vdm/dt= uv^2$. Now, the $x$ length of rope is subjected to two forces: Weight = $(ux)g$ and this "thrust" force = $uv^2$ both in the downward direction.To maintain constant velocity V, we need to Apply external force= $ugx+uv^2$.The work done in pulling the entire chain=$\int_{0}^{l}Fdx$. This comes out to be "$ugl^2/2 + uv^2l.$

  1. Approach 2:

I tried to use the work-energy theorem. Total work done by external forces=$\Delta$$KE$ + $\Delta$$PE$ = $1/2(ul)v^2 + (ul)g(l/2)$. This differs slightly from the answer in approach 1. Which means that some other force is also there in the system (Its an inelastic collision, as FakeMod pointed out), which does work= $ 1/2(ul)v^2$.

What exactly is this Force?

  • 2
    $\begingroup$ I suggest editing the question to look less like a "check my work" question. Usually questions that have sections of "Approach 1" etc. end with "where did I go wrong?" $\endgroup$ Commented Apr 30, 2020 at 14:26
  • $\begingroup$ would you suggest what exactly I should change? $\endgroup$
    – satan 29
    Commented Apr 30, 2020 at 14:49
  • $\begingroup$ Maybe put your conceptual question up front and give the supporting work later. $\endgroup$ Commented Apr 30, 2020 at 15:04
  • 2
    $\begingroup$ If you hear a sound, that is energy loss, and if heat is generated that is more energy loss. $\endgroup$ Commented May 2, 2020 at 19:35
  • 1
    $\begingroup$ @AakashSinghBais It is L written in small letters as l not 1. $\endgroup$ Commented May 4, 2020 at 16:12

7 Answers 7


Let's take it from first principles. I am starting away from the rope of the question, but I hope that I will be able to paint a good picture of what happens. This will require getting back to basics unless we want to have generalities that don't really explain anything.

The tldr is that internal forces in a system and between a system and its environment (air for example) can "absorb" energy, that is some of the work that you perform on a system using a force that you control does not get transferred to kinetic energy or gravitational potential energy. But, this is the kind of "generalities" that I want to avoid in the first place.

Conservation of momentum in systems.

If you have a system of interacting elements and two elements 1 and 2 interact through a force, we know that the force exerted by 1 on 2 is equal in magnitude and opposite in direction to the force exerted on 2 by 1 (Newton's third law). Therefore, the momentum imparted on 1 by 2 is equal and opposite to the momentum imparted by 2 on 1 as F = dp/dt. This has for consequence that in a system, internal forces cannot change the momentum of the whole system (equal to the momentum of all the elements of the system), only external forces can. This also has for consequence that if you know the external forces exerted on a system, you know how the system will change in momentum, globally, as a function of time, whatever happens inside the system as the external forces are redistributed. That is, the rate of change in the sum of the momenta of each portion of the system is equal to the total force external exerted on the system, whatever happens inside the system.

Of note, momentum is a vector and each component of the vector is conserved. If you have two identical electrons coming towards each other, they will typically not move in the same direction they were moving after the interaction as they would have to be perfectly aligned to do so. They can interact even if slightly offset from each other. However, if the electrons had the same speed before interaction, they will have the same speed after the interaction, whatever the direction they are then going. The center of mass will keep on moving at the same speed it was going before the interaction.

Conservation of energy in systems.

At the microscopic level, energy is also conserved. However, energy is tricky. Mechanics is usually concerned either with a few particles interacting together, or with substances that are approximated as continuous. However, in the real world, stuff is made out of atoms. A very large number of atoms. When you get many particles (of the order of Avogadro's number ) interacting together, a lot of things happen. Local applications of energy can be redistributed among many atoms and in the end what started as a local application of force in a specific direction results in atoms moving in all directions. To get back to my two electrons, energy is a scalar. Even if the electrons change direction, their mechanical energy does not change.

This is where we need to be very careful and where energy "losses" happen. In mechanics, all this redistribution of energy among many particles and all these changes in direction can be so drastic that the object as a whole does not move in a single direction at the speed it "should" be going anymore. Instead, all of its atoms vibrate, but in different directions, and even change direction of motion (it all depends on the type of system). This is called temperature/heat and mechanics by itself doesn't know how to deal with this. Derivation of what happens exactly in such energy redistribution is extremely complex. In mechanics, we deal with this by saying that energy was "converted" to heat, when all of the energy stays in the object/system we consider, or "lost to friction" or other things, when our system interacts with the outside world and leaks a bit of its energy, as we like to use conservation of energy.

It can happen also that an object deforms under the application of a force. In these cases, atoms that were in equilibrium at a certain position relative to their neighbors will be moved to a new equilibrium position. Moving the atom required giving is energy, to break attraction, and then accepting energy from the atom, so that it comes to its new equilibrium. This deformation is a very good way of transferring energy in directions that differ from the original force application.

As final remarks, for completeness, in solids, atoms move about an equilibrium position and transfer energy with their neighbors all the time, very rapidly. Also, this is a classical explanation. In real life, things are even more messy. For example, objects radiate energy in the form of photons all the time.


Interaction between atoms in a system can redistribute energy in all directions. Even a force applied normally at the surface of an object can result in atoms in the object moving in all kinds of directions. Note that some ways of moving may preferred, and if energy dissipation is not too quick, the atoms will want to move in some ways that is coordinated, so that, for example, a metal object will ring. Once energy has been redistributed in all directions, we don't call it "kinetic energy" of the object as such, which is usually reserved for the movement of the center of mass of the object, or of its constituents "macroscopic" objects. You did not get rid of energy, you just redistribute it.

Note that however, there is no way to get rid of momentum, due to Newton's third law, unless you interact with your environment. That is, you cannot get rid of "macroscopic" momentum without outside interaction, while you can get rid of "macroscopic" energy without outside interaction.

A simple example.

Imagine you have two balls close to each other and take a third ball in your hands to hit both balls simultaneously, moving in a direction perpendicular to the line joining the first two balls "though the middle". Since you hit the two initial balls a bit from the side, they will get momentum both in the direction in which you are moving, and in the direction perpendicular to this direction. The center of mass of the two initial balls will start moving in the direction in which you exerted a force, but if you forget that the balls are also moving away from each other in the other direction, you will "lose" energy.

Even better, now join the two balls with a spring. After you finish hitting the two balls, they will both move in the direction of the initial force, and vibrate sideways relative to this force (imagine a very fast interaction with the third ball, has if hitting a drum). If you look the system from very far, you will not see this vibration and it will seem like you dissipated energy, but that's only because you forget about the sideways movement. Now, imagine that you balls move in space and that they can sometimes hit a smaller ball. They will impart momentum to this smaller ball and slowly lose energy. This is analogous to friction.

Rope example.

In the rope example, I would not think of "collisions" though. A rope is not a chain. A rope is made of fibers that ware weaved together. The problem statement says that the rope is coiled. To start from a coiled configuration to a straight configuration, at the end, the fibers in the rope will slip on each other and exert friction on each other. Also, lifting the rope will create deformations inside the rope. In the rope case, energy is lost by friction between the fibers. Depending on the details, some energy may be dissipated in the air (friction between rope and air molecules). Depending on the exact structure of the rope, I will concede that maybe collisions occur, but I would not consider this as a main energy dissipation effect.

What is wrong with approach 2

Conservation of energy works very well in problems concerning rigid bodies, as there is a theorem stating that the work done on the rigid body is equal to the kinetic energy given by the mass of the object moving at the speed of its center of gravity ("translation" kinetic energy)plus a term including the moment of inertial of the object relative to the center of mass and its angular speed about the same point "rotation kinetic energy". However, this is a theorem that applies only to rigid bodies. The rope in our problem is not a rigid body.

To start from an horizontal coiled rope to a purely vertical straight rope, a lot of things had to happen that require deformations.This is where the "missing" energy goes. Either mechanical energy has been lost to friction, or it has been stored in stretching of the rope. This latter possibility seems to be neglected though as the length of the rope seems unaffected by the whole process, at least to first order.

How could one store mechanical energy in similar problem? Well, imagine a rigid rod lying on a table. If you pick it up at one end and move this end vertically, at the end, after leaving the table, the rod will oscillate. If you then let go of the rod when it is vertical, it will rotate on itself. Here, total mechanical energy is conserved, but not linear kinetic energy. You store energy in the rotation degree of freedom of the rod.

What is confusing about approach 1.

Approach 1 is appropriate as there is no way of "dissipating" momentum imparted by the outside force unless external friction becomes significant. The problem is stated so that is it not. However, the solution given in the question leaves a lot unsaid, which has caused confusion. I would prefer to split (even an idealization here) the rope into 3 sections: Section 1: what is on the table. Section 2: what is vertical in the air moving at speed v. Section 3: the transition between the two.

The solution leaves a lot of things unsaid that may be confusing.

Section 1: The normal force and gravity are exerted on this section, so why don't we care about them? Because by definition the normal is what balances gravity. These two forces are equal and opposite, so they do no work and impart no momentum to section 1. Also, there is no tension at the free end of 1 as it is assumed that the rope is coiled static on the table.

Section 3: This is where all the messy stuff happens and its existence is omitted. One presumes that the rope is moved from its end at constant V, but at the same time presumes that a piece of rope goes from 0 to V in a time dt, implying an acceleration. The fact is that there is a transition zone, where a piece of rope is accelerated from 0 to V. How is this possible?

If you look at a piece of rope (I mean, literally, go, take a rope and pick it up), that is supported from above with a part that lays on a table, and try to lift the rope a bit, you will see that the rope is curved between the vertical and horizontal parts, with an horizontal distance between the two. This means that a large displacement in the vertical portion (away from where the rope lays flat) can lead to a small displacement close to the table, where a small piece of rope starts moving. It's just like a lever. Speed close to the point of rotation is very small, even if speed away is large. The point of rotation here always move as the rope is unwound. As you continue lifting the rope, this part of the rope will get closer and closer to the vertical and pick up speed. Approach 1 makes the implicit assumption that section 3 is small and negligible, or that at least, section 3 has the same velocity profile at all times while the rope is lifted.

I would prefer a solution in which we state what the momenta of sections 1 and 2 is as a function of time (always zero for section 1), and say that whatever happens in section 3 is not important as it is small and stays more or less the same as we pull the rope. Then, adding the momenta of all 3 sections at time t and using F = dp/dt, we can get the force we exert and integrating over x will give work. The formula we obtain at the end will be the same, but this would end a lot of confusion as to why this works and avoid details about accelerating masses while having this mass having a constant speed and other confusing statements.

[EDIT] Storage of energy other than gravitational potential energy and translation kinetic energy in the rope

One thing I forgot to mention is that you can maybe temporarily store energy in the rope. Imagine you have a mass linked to a spring. You pull the end of the spring at constant velocity horizontally. The spring will elongate while the mass accelerates. However, once the mass reaches speed V, the spring still exerts a force and the mass overshoots. There will oscillations and energy is stored in the spring/mass system. You will need to exert a force that is larger than the force required to move the mass from 0 to its final speed in absence of a spring, so that you can do work on the spring. However, if there is friction, as there always is in real world situations, you will lose this energy eventually. My answer presumes that energy is lost quick enough that we don't accumulate much energy in the rope, so that it does not have any movement other than vertical at speed V just as the last piece of rope leaves the surface. In real life, this may not be the case. So, even conservative internal forces can in fact do work against you, not just friction. You can get more details of how this works in this simple system by looking for the treatment of harmonic oscillators. The right keywords are "forced (or driven) damped harmonic oscillator", and more particularly "step input" to such movements.

  • $\begingroup$ This is a well-written answer with a nice conceptual motivation. Enjoyed reading it. +1! $\endgroup$
    – Vivek
    Commented May 9, 2020 at 12:57

To rephrase the question, we are trying to solve the problem of why Approach 2 has a factor of 1/2 that Approach 1 does not. The problem lies in the kinetic term.

Consider the same analysis for only a single chain link with mass $dm = udx$. You have said that there is a "thrust force" when the piece $dm$ is picked up, but you also assume that $v$ is constant immediately after. Integrating, you get $uv^2l$, even if the entire rope is only 1 piece $dx$ long, whereas we would expect the energy to be just a kinetic energy term (can't have internal losses if there is only one piece).

Rather, this implies the force is some Dirac like function, since the change in momentum is nearly instantaneous. When you perform the integration $\int F \cdot dx$, one of your limits ($0$) is the peak of the Dirac delta. Integration around the Dirac delta is $1$, but having one bounds be at the peak of a Dirac delta is not really well defined. This stackexchange question suggests the integral is 1/2 (as long as we define the Dirac delta with a delta sequence), which is a natural guess and gives you the missing factor. Doing the same for each piece gives you this result.

This answer is a little bit more satisfying, because we have never assumed anything in Approach 1 that would suggest anything about the elasticity of the collisions between chains links or rope segments.

  • $\begingroup$ My point is you should not model speed as a heaviside function, as imo that is the source of our inconsistency. My example to illustrate this is to consider that the chain has only one link. Following the analysis in Approach 1, you would still see a discrepancy. $\endgroup$
    – xzkxyz
    Commented May 5, 2020 at 23:33
  • $\begingroup$ Integrating over the "first half of the next link" results in a double counting. If we assume that the links that are off the table can be approximated by a rigid body, there should be no difference between picking up the chains as-is and picking up each link one-by-one. You do not need to count the impulse on the link higher up and also count the impulse required to pick up the link. Those are the same $\endgroup$
    – xzkxyz
    Commented May 5, 2020 at 23:35
  • 1
    $\begingroup$ I misread your answer. You are basically trying to so something similar to what Charles Francis is doing. I have to admit that the way approach 1 solves the problem is a bit awkward. If you look at a piece of rope coiled on a surface and pull it up from one end, no part of the rope goes from 0 to V instantly. There is a region in which a piece accelerates. /1 $\endgroup$ Commented May 6, 2020 at 15:00
  • 1
    $\begingroup$ However, imagine that the rope is coiled in a perfect circle and we pull from the center. After we uncoiled the rope one whole turn, the "transition region" in which stuff accelerates is back to what it was in term of velocity distribution, and the rope, above the transition region, has gained velocity V, so a change in momentum 2 pi r u V, with r radius of the rope circle. There is no 1/2 factor here. /2 $\endgroup$ Commented May 6, 2020 at 15:02
  • $\begingroup$ This is becoming a discussion and we can take this elsewhere if you don't agree. My original comment was false, It is deleted. I misunderstood your point. The problem with approach 2 is that the conservation potential + kinetic energy is only valid for objects as a whole for rigid objects. Being "one piece" is not the right condition. Imagine a vibrating rope under tension. It stores plenty of energy, even if its center of mass does not move. In the rope of the question, a lot of things happen and it is not obvious that energy will be conserved. /3 (and final) $\endgroup$ Commented May 6, 2020 at 15:08

Okay I guess you need a compact and satisfactory answer without beating around the bush. Here you go:

Firstly you need to understand that internal forces are capable of doing work.

Let's take yourself and the Earth as system. Initially, approximate both to be at rest. So there is no kinetic energy. But then you jump in the air with some velocity. So now there is kinetic energy! All the forces in play were between you and earth, so it is all inside the system.(The force here is electrostatic).

So what happens in a collision?

When 2 objects collide, their atoms come so close to each other that there is really significant repulsion between them, which prevents them from going closer. Electrostatic force is a conservative force. So in an ideal(Elastic) collision, all the kinetic energy gets converted to potential energy and then back to kinetic energy. Thus kinetic energy is conserved.

But in a real life scenario this doesnt happen. There are repulsive/attractive forces between the contact electrons and other neighbouring electrons/nuclei. Thus a few of the particles tend to vibrate and lose energy as sound. If you even ignore sound,light etc, some electrons come too close to another electron, leading to a large force, that is a little,or even a lot larger than the elasric limit.

So some energy may go into deformations also instead of being converted back to kinetic energy.

Thus there you have it! You just lost some kinetic energy without caring about anything that happened!

Hope you got it :)

p.s it took a long time to type this

  • $\begingroup$ I too was thinking on the same lines, (which probably is the reason), but it seemed to imply that the rope that rose was permanently different from the one that was coiled. Though, what the difference is confuses me. $\endgroup$
    – Elendil
    Commented May 3, 2020 at 17:53
  • $\begingroup$ The tension thing takes care of itself if you assume a very large rigidity (stiffness). If something is stretched with F = -kx, the stored energy is E= .5*kx^2. At equilibrium, x = -F/k, so that at equilibrium E = .5*F^2 /k. If k is large enough, you don't store much energy. $\endgroup$ Commented May 5, 2020 at 19:27
  • $\begingroup$ Well even if rigidity is very high, doesnt it still imply a slight deformations? For example if 2 stones collide inelastically, it is still possible even if rigidity is high $\endgroup$ Commented May 6, 2020 at 5:23


Although the following answers revolves around chains, however, the same physics can even be applied to ropes, since ropes itself is made up of thousands of such tiny chains. Therefore, the object under consideration is unimportant here, what is more important is the nature of collisions taking place.


The reason why the energy approach doesn't work is because there are inelastic collisions between the chain's loops going on.

What should happen if the collisions were elastic

enter image description here

Image source

Now to understand this, imagine pulling the chain up. Now let's analyze the lowest loop of the chain which was rising up with a velocity of $v$. The next loop after this loop has not been give a velocity yet. But as this loop moves up, the next loop also gains a velocit $v$. However, if we were to compare this case with an elastic collision, then we should expect the velocities of the chains to be exchanged (since they have the same mass). So what you'd expect is that the chain initially rising up, will come to rest after the collision and the chain which was on the ground would start rising up with a velocity $v$.

What really happens

However, this is not at all the case. In reality, when the lower loop collides with the rising loop, booth the loops start moving at the same velocity. But, this is precisely the characteristic of an inelastic collision, which implies that the collision in the chain case is inelastic.

Energy loss and the factor of 2

As you'd know, energy losses are bound to happen in inelastic collisions. And in this case, we have a case of a body moving with velocity $v$ colliding inelastically with a body (of the same mass) at rest. In this particular type of case, we can compute the difference between the initial and final energies and that difference will always be equal to half the initial energy. This implies that half the energy gets lost/dissipated. That's what's happening in your cas as well and thus you are getting an unwanted factor of $2$ while comparing both the approaches. Thus you do double the work to increase the kinetic energy ($\mu v^2 l$), but only half of that work is manifested as an increase in kinetic energy ($0.5 \mu v^2 l$).


So there isn't any specific force responsible for this discrepancy. It is due to dissipative nature of inelastic collisions. The lost energy would be converted to heat and sound energy, and the temperature of the chain will rise and you might hear the clanking sound of the chain.

References supporting the ides of inelastic collisions

$[1] :$ An HTML page containing similar problems by University of Tennessee

$[2] :$ Few relevant pages of David Morin's "Introduction to Classical Mechanics" (PDF)

  • $\begingroup$ Makes a lot of sense. However, i still cant come to terms with the fact that there isnt any specific force responsible for the energy loss in an inelastic collision, $\endgroup$
    – satan 29
    Commented Apr 30, 2020 at 13:00
  • $\begingroup$ @satan29 Basically, it is the normal force which is responsible here, however I, in my experience, have never found any explicit mention of the specific force which causes the energy loss in an inelastic collision. And anyways, a change in energy does not always have a single force as its direct reason. However, in this case, the energy loss can be attributed to the normal reaction between the two loops. $\endgroup$
    – user258881
    Commented Apr 30, 2020 at 13:07
  • $\begingroup$ for instance: take the classic case of a ball with velocity "sticking" to another ball. There is a loss in KE, and I attributed this to friction since it is friction, which makes it possible for them to stick! $\endgroup$
    – satan 29
    Commented Apr 30, 2020 at 13:09
  • $\begingroup$ in that case, Wouldnt the normal force affect method 1?? Youll need to include that as a force on the "x" length of the chain we were analysing $\endgroup$
    – satan 29
    Commented Apr 30, 2020 at 13:10
  • $\begingroup$ @satan29 No! Friction doesn't make the balls stick. Friction always acts along the surface of contact (in this case, perpendicular to the line joining the centers of the balls). But the force which you are talking about is attracting the two balls together (acting along the line joining the centers), so it definitely can't be friction. $\endgroup$
    – user258881
    Commented Apr 30, 2020 at 13:12

I think other answers have pointed out the effect already, viz. - inelastic processes that cause loss of energy associated with center of mass motion, and is eventually dissipated because of damping/friction (sound, heat etc.).

Here is a way to make this quantitative. Consider a completely inelastic collision of a body of mass $m_1$, velocity $\mathbf{v}_1$ with another body of mass $m_2$, velocity $\mathbf{v}_2$. Assuming momentum is conserved, we can determine the loss of (kinetic) energy very easily. The answer is

$$ \Delta T = \frac{1}{2} \mu (\mathbf{v}_2-\mathbf{v}_1)^2, $$

where $\mu = \frac{m_1 m_2}{m_1+m_2}$ is the reduced mass of the two body system.

You ask what is the relevance to your problem? Here goes: Your chain is coiled, so the part on the ground is stationary...but as you start pulling more and more mass into the moving portion, you're accelerating some of the stationary part into the moving portion in some messy way we don't talk about (nicely pointed out by @ManuelFortin). If the rope were inextensible, this wouldn't be possible, but that's of course the whole point!

Now imagine a time interval $\mathrm{d}t$. You pull mass $\mathrm{d}m = u v \thinspace \mathrm{d}t$ on to the moving portion from rest to velocity $v$ - must be an inelastic collision. How much energy is lost, you ask yourself again? The previous formula says:

$$\mathrm{d} T = \frac{1}{2} u v^3 \mathrm{d}t$$

(You may object the formula is not applicable because of the external force acting on the system. Sure, this would slow down the rope a little bit, but it would then be the job of $F$ to counter the slowdown in time $\mathrm{d}t$. The reason why this is consistent with overall momentum conservation is because one assumes that as the infinitesimal piece latches on to the moving part, it doesn't pull the part on the ground. So, if you really calculate all the impulses on the full rope from start to end, you'd indeed see that the net impulse is equal to the change in momentum. In fact, if you take up this hypothesis as the starting point, you'd again get the same formula for the force $F$ needed to pull the rope with constant speed $v$.)

Result: You lose mechanical energy associated with center of mass motion at a rate,

$$\frac{\mathrm{d} T}{\mathrm{d}t} = \frac{1}{2} u v^3 $$

And obviously you conclude that you have to apply an extra force $\delta F$ compared to the case in which energy is conserved. The power generated by this "extra force" must exactly compensate the loss of energy in the inelastic, dirty stuff we didn't want to bother modeling (but still wanted to include). Therefore,

$$\delta F v = \frac{1}{2} u v^3,$$

which unsurprisingly gives us the extra force as,

$$\boxed {\delta F = \frac{1}{2} u v^2} $$

So, the extra force is the work that you do and you feed it into the system. But such is the system that it loves to dissipate it as heat and sound, cuz, as they say, there is no free lunch and the heat/sound is not for nothing!

Further Remarks

You may ask what sorcery is this. And indeed it is. We got away with calculating energy dissipated without even modeling how the dissipation happened in the first place.

So how does that happen here? First of all, the piece of the rope that is accelerated in time $\mathrm{d}t$ has two different velocities on two different ends. This results in elongation of the piece in the 'transition region' (as already pointed out by @ManuelFortin). Thus the system is quite complex with the top part being in some kind of steady state at velocity $v$, the part on the ground is stationary and there is a transition zone where all sorts of elongations and elastic oscillations take place. The assumption is that to a first approximation, the system doesn't lose momentum/angular momentum to any other degree of freedom (eg. to the air, or an internal degree of freedom such as the magnetization), while the elastic oscillations are damped out after a certain natural relaxation time. Thus, one implicitly assumes that the infinitesimal piece transferred to the moving portion eventually enters a steady state of the upper moving part, but in the process also undergoes elastic damping. This energy appears as heat and sound, and has to result from some kind of dissipative mechanism (the same that damps out a ringing bell etc.).

The reason we could calculate the loss exactly is the approximation that we assume momentum conservation and this gives rise to a process that can be "effectively" modeled as a completely inelastic collision of an infinitesimal mass latching on to a moving piece in which the upper bound to the energy loss is set by the momentum conservation constraint.

p.s. - 1. I recommend the even more well-written and detailed answer of @ManuelFortin for understanding what we're actually doing in this problem.

  1. References provided by @FakeMod
  • $\begingroup$ Well, that is less convoluted than my explanation and maybe a bit clearer as such. I would not talk about "collision" though. If you replace "collision" by "process", I think the answer would be clearer. In the problem, we have a rope. There is no literal collision. Also, you say "we assume momentum conservation". This is not really an assumption. It is the truth. That is F = dp/dt is always true. The problem is that delta E = delta (potential energy) + delta (kinetic energy) is not true all the time. There are other forms of energy that this equation forgers. $\endgroup$ Commented May 11, 2020 at 16:17
  • $\begingroup$ I had a chain in mind, when I was writing that. It works on the chain because of lack of rigidity. But also works on ropes because of stretchability and subsequent dissipation of the resulting oscillations. The "collision" is the whole process, indeed, as you say. But that is also how energy is lost in an inelastic collision of two bodies...the objects first first, twist, squeeze deform, oscillate and then amicably enter into thermal equilibrium. So temporarily, if you think of the upper steady part as the system, it's effectively experiencing a process that is "collisional". $\endgroup$
    – Vivek
    Commented May 11, 2020 at 17:12
  • $\begingroup$ I do stress momentum conservation and angular momentum conservation are approximations, though, and I agree with you that energy is generally more subtle in stat-mech (because it is always affected by temperature). Transferring momentum/angular-momentum to an internal degree of freedom is unusual, and only more common when macroscopically ordered states are involved, eg. Einstein-de Haas. Also, the sound waves transmitted to air carry away both energy and momentum. $\endgroup$
    – Vivek
    Commented May 11, 2020 at 17:25
  • $\begingroup$ I felt the need to stress on the momentum conservation bit because a certain reputed poster on the answer page claims momentum is not conserved and calls the consideration of inelastic losses as physics which is non-mainstream (and a personal theory of mine :)). $\endgroup$
    – Vivek
    Commented May 11, 2020 at 17:36
  • $\begingroup$ Good point about momentum in air. One could say that's an external force, but I had not thought about this! Of course, if the sound is about isotropic, then total momentum transfer is about zero. As to the Einstein-de-Haas effect, if we add electromagnetic fields, then of course we have to consider momentum of this field. $\endgroup$ Commented May 11, 2020 at 19:16

The force between the 'x' length of chain and the mass element about to be lifted off ( the normal or thrust force) is doing some net work here.

Let's see how.

Consider 'x' length of chain off the ground .After a small time interval (delta)t , a small mass element of length (delta)x starts moving with the 'x' length with speed v . Clearly , in this time interval the normal force does negative work on the 'x' length of chain and positive work on the incoming element .

The displacement of the 'x' length of chain is v .(delta)t However, the displacement of the mass element is gonna be less than this because the speed of the mass element is gonna be less than v throughout this interval ( it becomes equal to v at the end of this interval)

We can conclude that the magnitude of the negative work done on the 'x' length is more than the positive work done on the incoming mass element.

Hence , the normal force does a net negative work here. Normal force is an internal force but does net work here. The total work done by the external force F and the normal force gives us the sum of change in kinetic and potential energy which solves the problem.

We can also look at the whole process another way . As explained above, after time delta(t) , the separation between the last mass element of 'x' length and the incoming mass element would increase as they both undergo unequal displacements , and consequently some energy would be stored as elastic potential energy of this configuration.

Adding this elastic potential energy term to the kinetic and gravitational potential energy In approach 2 , we get the net work done by external force F which solves the problem . For better visualization, we can consider a spring between the 'x' length of chain and the mass element. After time interval delta(t) , the spring stretches as its ends undergo unequal displacements and some energy is stored in it .

  • $\begingroup$ To conclude , the normal force is responsible for this discrepancy $\endgroup$
    – user260622
    Commented May 5, 2020 at 10:26

There is an error in the way in which you are treating work done by the force to accelerate a chain element. We must integrate its acceleration, $a$, which takes its velocity, $w$, from $0$ to $v$ in time $dt$, distance $dx$. Notationally this is a little awkward, because we have to integrate a variable with infinitesimal bounds. I will use $D$ to denote the smaller infinitesimal (since $d$ is already taken). We have, during the acceleration of the rope/chain element in time $dt$,

$$a = \frac 1 2 \frac{Dw^2}{Dx}$$

Then the work done to accelerate the chain element $dx$ is

$$ \int_0^{dx} udx a Dx = udx \frac 1 2 v^2 $$

and you just have to integrate this to get your answers to agree.

The specific error in argument 1 arises because the force to accelerate the rope element $dx$ to velocity $v$ in time $dt$ is given by $$ F = \frac{udx v}{dt} = u v^2 $$ but the work done by this force has moved the centre of gravity of the rope element a distance $\frac 1 2 dx$. Consequently the total work done accelerating a length $l$ of rope is $$\int_0^l u v^2\frac 1 2 dx = \frac 1 2 ul v^2$$ again in agreement with answer 2.

@Fakemod has given as a reference 5.8 Inherently inelastic processes from Introduction to Classical Mechanics by David Morin, who basis his argument on a common misinterpretation of Newton's second law Force = rate of change of momentum. Morin uses the product rule

$$ {d\over dt} p = {d\over dt} mv = m{d\over dt} v + v{d\over dt} m $$

But mass is a conserved quantity in Newtonian mechanics. In all cases $${d\over dt} m = 0 $$ Thus $$ {d\over dt} p = {d\over dt} mv = m{d\over dt} v = ma$$ is correct.

In supposed "changing mass" problems, whenever mass elements are moving or accelerating differently, we must treat the motions as belonging to different bodies. Morin uses the conveyor belt example, but he ignores the acceleration of the sand as it lands on the belt, setting $$\frac{dv}{dt} = 0$$ because the belt is not accelerating, when he should consider the acceleration of the sand. He gets the right force $\sigma v$ by the wrong method, and fails to recognise that in going from a vertical flow to a layer of sand on the belt, the centre of gravity has moved $\frac 1 2 v dt$, so that the work done by the belt in accelerating the sand is precisely half what he calculates.

There can be no issue in these questions losses due to inelacticity. Even if there are losses (as there always are in practice), they are material dependent and will not give an exact factor of 2. All the forces, neglecting losses to heat, are already contained in both argument 1 and in argument 2. These can disagree because of a mathematical error.

  • $\begingroup$ Although I didn't mention it explicitly, The answer we got from approach 1 is correct. As, @FakeMod pointed out, (and another Source, which asks us to, infact, calculate the energy loss in the process), there will be some loss in energy. $\endgroup$
    – satan 29
    Commented May 3, 2020 at 14:43
  • $\begingroup$ And what exactly is wrong with approach 1? I believe its the standard treatment for variable mass systems.. $\endgroup$
    – satan 29
    Commented May 3, 2020 at 14:44
  • $\begingroup$ Why should there be a loss of energy? There need be no inelastic losses here. Your answer 2 shows precisely that there is an error in answer one. It is always wrong to differentiate momentum with the product rule, generating a term $vdm/dt$ in force. This is a common error, not a standard treatment. Mass is constant, but you have to take into account that force applies to the mass element which is accelerating. The mass element accelerates from $0$ to $v$ and the work done accelerating it is found in the usual way, by integrating force with respect to distance. Else you violate work energy. $\endgroup$ Commented May 3, 2020 at 17:22
  • $\begingroup$ @CharlesFrancis Don't think your claim here is correct. One can do the real physical experiment with a coiled rope. If the rope were inextensible, indeed, your claim would hold. However, it's not inextensible, and this allows the part on the ground to be stationary while the upper part can still be pulled...the resulting elastic oscillations are eventually damped and the loss of energy has to be accounted for. $\endgroup$
    – Vivek
    Commented May 8, 2020 at 19:40
  • $\begingroup$ @Vivek, if elasticity had anything to do with it, there would not be a factor of exactly 2, but a different factor depending on the material. On the other hand, abusing Newton's laws does give a factor of exactly 2. The mass of the rope does not change. In each infinitesimal time interval, a different mass element has to be accelerated, and the work energy relation is obeyed for that acceleration. All the question has done is demonstrate that a common error gives wrong results. $\endgroup$ Commented May 8, 2020 at 19:51

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