Kinetic Energy and Inelastic Collisions

[protonman]

Why is KE not conserved in inelastic collisions?

 

[jcsd]

Imagine two objects of masses m_1 and m_2, traveling with velocities of v_1 and v_2 which after collding inelastically form a new object of m_3, traveling at velocity m_3, which due to the conservations of mass must equal m_1 + m_2

We can say this due to the conservation of momentum:

m_3v_3 = m_1v_1 + m_2v_2

therefore:

v_3 = \frac{m_1v_1 + m_2v_2}{m_3}

We can also say that due to the conservation of energy:

\frac{1}{2}m_3{v_3}^2 = \frac{1}{2}m_1{v_1}^2 + \frac{1}{2}m_2{v_2}^2

therefore:

v_3 = \sqrt{\frac{m_1{v_1}^2 + m_2{v_2}^2}{m_3}}

combing the equations we get:

\frac{m_1v_1 + m_2v_2}{m_3} = \sqrt{\frac{m_1{v_1}^2 + m_2{v_2}^2}{m_3}}

square and mutiply {m_3}^2 byboth sides,substitue in m_3 = m_1 + m_2 and mutiply out:

{m_1}^2{v_1}^2 + {m_2}^2{v_2}^2 + 2m_1m_2v_1v_2 = {m_1}^2{v_1}^2 + {m_2}^2{v_2}^2 + m_1m_2{v_1}^2 + m_1m_2v_2^2

Simply eliminate and you get:

2v_1v_2 = v_1^2 + v_2^2

Which can be re-arranged as:

v_1^2 - 2v_1v_2 + v_2^2 = 0

using the quadratic formula we can solve for v_1

And we find that:

v_1 = v_2


So for an inelastic collision the intial velcoties of the two colliding objects must be the same, hence no collision.

QED

 

[jcsd]

I'm sure there must be a quicker way to prove it, but though I have seen the proof in textbooks, it's completely gone from my mind so I derived it myself.

 

[robphy]

Why is KE not conserved in inelastic collisions?

I'm going to read this as
"Why is total-KE not conserved in inelastic collisions?"

Strictly speaking, an Elastic Collision, by definition, is one in which total-KE is conserved... otherwise the collision is, by definition, Inelastic.

Now, if you were to ask, "Why is the total-KE not conserved when, say, two objects collide and stick?", this can be demonstrated with a short calculation.

To make the calculation trivial, consider the special case of two objects colliding and sticking together with final velocity zero [the total-inelastic or perfectly-inelastic case]. Clearly, the initial total-KE is positive and the final total-KE is zero. (Of course, total-momentum is still conserved.)

 

[protonman]

These are all obvious trivial calculations. What I am asking is why they turn out this way. Simply going through the derivations is not an explanation.

I don't think the people who write on this site think very much.

 

[protonman]

Actually robphy made a good suggestion of looking at the problem in the center of mass frame. This at least simplifies the calculations.

The problem here is that energy is not conserved. This is a big problem.

 

[ophecleide]

The remaining energy turns into heat. When two pieces of playdough collide, they stick together and all the remaining KE is transferred to heat through the internal friction of the playdough when it is deformed. With a perfectly elastic rubber ball, for example, the energy is stored as potential energy for the time during which the ball is stressed, then it is released in such a manner as to accelerate the ball and none of the kinetic energy is turned into heat. There can, of course, be all cases in between.

 

[Integral]

The problem here is that energy is not conserved. This is a big problem

Not true.

KINETIC ENERGY is not conserved. An inelastic collision is usually accompanied by deformation of one or both bodies. This requires energy. Thus TOTAL energy is conserved but not necessarily KINETIC ENERGY.

 

[protonman]

Not true.

KINETIC ENERGY is not conserved. An inelastic collision is usually accompanied by deformation of one or both bodies. This requires energy. Thus TOTAL energy is conserved but not necessarily KINETIC ENERGY.

But the equation is only considering kinetic energy there is no mention of deformation or internal potential energies.

 

[krab]

But the equation is only considering kinetic energy there is no mention of deformation or internal potential energies.

What equation? Kinetic energy is not conserved, so there is no equation of the form: total kinetic energy before = total kinetic energy after. Momentum, however is conserved, so for two particles,
m_1v_{1i}+m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

 

[jcsd]

These are all obvious trivial calculations. What I am asking is why they turn out this way. Simply going through the derivations is not an explanation.

I don't think the people who write on this site think very much.

It's a silly question then, I only went for the one-dimensional example but the whole point is that by the defintion of momentum and by the defintion of kinetic enrgy some energy must be transferred to other forms, kinetic energy can't be conserved in an inelastic collsion which is the question you asked.

If you wanted something deeper, I could've of proved that 4-momentum isn't conserved (which is a useful result as it illustrates why a free electron can't absorb a photon for example), but it amounts to the same thing.

 

[protonman]

What equation? Kinetic energy is not conserved, so there is no equation of the form: total kinetic energy before = total kinetic energy after. Momentum, however is conserved, so for two particles,
m_1v_{1i}+m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

Kinetic energy is conservered in an elastic collision but using the same equation for an inelastic collision it is not conserved.

 

[protonman]

It's a silly question then, I only went for the one-dimensional example but the whole point is that by the defintion of momentum and by the defintion of kinetic enrgy some energy, kientic enrgy can't be conserved which is the question you asked.

If you wanted something deeper, I could've of proved that 4-momentum isn't conserved (which is a useful result as it illustrates why a free electron can't absorb a photon for example), but it amounts to the same thing.

4-momentum is conserved. Maybe you should read an intro relativity text.

 

[jcsd]

4-momentum is conserved. Maybe you should read an intro relativity text.

Not in a totally inelastic collison, hence a photon may not collide inelastically with a free electron, though it may collide elastically.

 

[protonman]

Not in a totally inelastic collison, hence a photon may not collide inelastically with a free electron, though it may collide elastically.

Both photons and electrons have momentum. If they have an equal amount of momentum in opposite directions they will collide and both have a final velocity of zero.

 

[jcsd]

Both photons and electrons have momentum. If they have an equal amount of momentum in opposite directions they will collide and both have a final velocity of zero.

In this case the answer is clear, we find that the E_TOT is more than the rest mass of the electron, so the electron must have kinetic energy, but if it did have kientic energy then it must have velocuiity and momentum can't be conserved so the situation is impossible.

 

[protonman]

In this case the answer is clear, we find that the E_TOT is more than the rest mass of the electron, so the electron must have kinetic energy, but if it did have kientic energy then it must have velocuiity and momentum can't be conserved so the situation is impossible.

You need to start thinking real soon. You are using what is in question to answer the question. I am saying that you can have an inelastic collision between a photon and an electron because you can have to momenta in opposite directions whose sum is zero. Therefore after they collide they will both be at rest.

 

[Alkatran]

Both photons and electrons have momentum. If they have an equal amount of momentum in opposite directions they will collide and both have a final velocity of zero.

No. Assume that two identical objects are colliding, with opposite speeds (thus opposite momentums).

Now, total momentum needs to be conserved, so logicly they will have equal speeds (but opposite) at the end of the collision as well. The only limit on the new speeds is that they musn't use more kinetic energy than the original speeds and they must be above or equal to 0. If they have less kinetic energy than before, some of the energy was transformed (into things like the sound of them smashing together, and the heat created by friction), if they have the same amount none was (and, there was no sound to the collision).

The point is, if two things with opposite momentums collide, they don't have to have a final velocity of 0.
m1=1
m2=2
v1=2
v2=-1

1*2+2*-1 = 0 = 1*x + 2*y
x = -2y
Nothing says x or y have to be 0.

 

[protonman]

No. Assume that two identical objects are colliding, with opposite speeds (thus opposite momentums).

Now, total momentum needs to be conserved, so logicly they will have equal speeds (but opposite) at the end of the collision as well. The only limit on the new speeds is that they musn't use more kinetic energy than the original speeds and they must be above or equal to 0. If they have less kinetic energy than before, some of the energy was transformed (into things like the sound of them smashing together, and the heat created by friction), if they have the same amount none was (and, there was no sound to the collision).

The point is, if two things with opposite momentums collide, they don't have to have a final velocity of 0.
m1=1
m2=2
v1=2
v2=-1

1*2+2*-1 = 0 = 1*x + 2*y
x = -2y
Nothing says x or y have to be 0.

You need to read the post before you write.

They do have zero velocity because we are talking about inelastic collisions.

Is there anyone here who can think?

BTW, if you do the calculation out for a photon and electron moving with opposite momentums towards one another it does work out. Using the relativistic momenta.

(Energy of photon)/c - (gamma)(mass of electron)(velocity of electron) = 0

Just solve for the velocity of the electron in this case and see for yourself.

 

[ophecleide]

Is there anyone here who can think?

If you're such a thinker, why didn't you think your way out of having to ask a question on this forum? We already told you the energy you're looking for goes to heat , sound, etc. What's the problem?

 

[protonman]

If you're such a thinker, why didn't you think your way out of having to ask a question on this forum? We already told you the energy you're looking for goes to heat , sound, etc. What's the problem?

You may have told me that but you did not support it. You need to provide a reason. I am not here to get answers.

What is happening here is that people who have never thought about this stuff before are being asked questions they have never considered. In this model we are neglecting friction.

 

[jcsd]

You need to start thinking real soon. You are using what is in question to answer the question. I am saying that you can have an inelastic collision between a photon and an electron because you can have to momenta in opposite directions whose sum is zero. Therefore after they collide they will both be at rest.

God you're obnoxious. I didn't say momenta isn't conserved I said 4-momenta isn't conserved. For 4-momentum to be conserved, energy (which can be seen as the time coponent of 4-momentum) must also be conserved in all reference frames.

 

[protonman]

God you're obnoxious. I didn't say momenta isn't conserved I said 4-momenta isn't conserved. For 4-momentum to be conserved, energy (which can be seen as the scalar of the time coponent of 4-momentum) must also be conserved in all reference frames.

Energy is conserved in all reference frames. That is one of the fundamental postulates of SR. Are you sure you know you are on a physics site?

 

[Gokul43201]

Alkatran: when Protonman said 'inelastic' he meant that the objects stick to each other.

Protonman: Inelastic does not necessarily mean that the objects need to stick. I believe that's referred to as 'perfectly inelastic'. An inelastic collision is anything that does not conserve KE.

In general, the only requirement on the energy that can be written down as an equation is that the total energy is conserved.
i.e : E1(i) + E2(i) + E3(i) +...+ En(i) = E1(f) + E2(f) + ...+ En(f)
where E = tot. energy + KE + PE + other forms (sound, heat, etc - which are also really forms of KEs and PEs of different things)

In the case of an elastic collision, all the KE before the collision is regained after it (through an intermediate conversion into PE and back). So the energy conservation can be wirtten purely in terms of KEs.
ie. KE1(i) + KE2(i) +...+ KEn(i) = KE1(f) + KE2(f) + ...+ KEn(f), since no other forms of energy are involved.

However, for inelastic collisions, we have
KE1(i) + OE1(i) +...+ KEn(i) + OEn(i) = KE1(f) + OE1(f) + ...+ KEn(f) + OEn(f), where OEs are other energies, whose values we may not know, unless a more complete description of the event is given. Since we know not what the OEs are - or even if we do, but find them to be different before and after - we can not write the equation that we write for elastic collisions.

P.S: A free electron can not absorb a photon, because there's no OE possible to put the lost KEs into. However, in an atom, the lost KEs can go into increasing PE, so that's cool, but still a little tricky, because KEs and PEs are quantized.

 

[protonman]

Physics Lesson One:

What is called the 4-momentum is not really "momentum" in the classical sense. If you differential the space-time 4-vector with respect to proper time you get what is called the momentum 4-vector. This is no more momentum than the space-time 4-vector is actually a spatial displacement. The first component of space-time 4-vector is time multipled by a constant to produce units of distance. Parallel, although not completely similar, thinking applies to the momentum 4-vector.

The reason 4-vectors are defined is that if something can be written in 4-vecotr notation its norm is a Lorentz invariant.

 

[jcsd]

That energy is conserved in all referbce frames, is a result of the fundamental postualtes of SR.

Yes I know what 4-momentum and for a free electron to absorb a phton is a violation of 4-momenta.

 

[protonman]

God you're obnoxious. I didn't say momenta isn't conserved I said 4-momenta isn't conserved. For 4-momentum to be conserved, energy (which can be seen as the time coponent of 4-momentum) must also be conserved in all reference frames.

Here you imply that energy is not conserved in all reference frames.

That energy is conserved in all referbce frames, is a result of the fundamental postualtes of SR.

Yes I know what 4-momentum and for a free electron to absorb a phton is a violation of 4-momenta.

Here you say energy is conserved in all reference frames.

 

[protonman]

Yes I know what 4-momentum [is] and for a free electron to absorb a ph[o]ton is a violation of 4-momenta.

Can you violate space-time 4-vector?

 

[Alkatran]

My apologies, I thought that all sub-atomic collisions weren't (perfectly) inelastic?

jcsd: I don't get what you're arguing. You've been told that the energy goes to heat (from friction) and sound (as well as others, I'm sure...). Do you think the energy just disappears? Why aren't you accepting the answers you wanted?

 

[protonman]

My apologies, I thought that all sub-atomic collisions weren't (perfectly) inelastic?

jcsd: I don't get what you're arguing. You've been told that the energy goes to heat (from friction) and sound (as well as others, I'm sure...). Do you think the energy just disappears? Why aren't you accepting the answers you wanted?

Lets get back on track. In the center of momentum frame an inelastic collision can be shown to violate conservation of energy. More directly mechanical energy. Trying to explain this by introducing sound and heat is not admissable. We are neglecting friction. The problem still stands.

 

[protonman]

Here is my understanding of the problem. What is being overlooked is the object's internal elastic potential energy. The incorrect assumption is that the objects involved in the collision are incompressable. If the objects were in compressable then the instant the blocks collide the force at the point of contact would instantaneously arise at all points along the object [in the direction of motion]. This would violate SR because the information [that there was contact] would travel faster that the speed of light [i.e. instantaneously].

The situation is like a compressed spring that once compressed does not return to equilibrium. The initial kinetic energy of both objects is stored in internal potential energy. In reality a perfect inelastic or perfectly elastic collision is impossible because it neglects friction. But in our model we were not concerned with the effects of friction. Even in [other] models that ignore friction KE is conserved. The key issue here is that a perfectly rigid [i.e. incompressable] body is in violation of SR.

Newtonian mechanics is not complete. It is only an approximation and therefore where it does not consider the postulates of SR it should not be assumed to be correct [in all cases].

 

[jcsd]

Here you imply that energy is not conserved in all reference frames.

Here you say energy is conserved in all reference frames.

Where did I imply that energy is not conserved in all reference frames? 4-momenta is always conserved, so energy is always conserved in all refernce frames, the two go hand-in-hand.

 

[jcsd]

My apologies, I thought that all sub-atomic collisions weren't (perfectly) inelastic?

jcsd: I don't get what you're arguing. You've been told that the energy goes to heat (from friction) and sound (as well as others, I'm sure...). Do you think the energy just disappears? Why aren't you accepting the answers you wanted?

Again where did I see the energy can't go to heat. I never asked any questions either.

 

[protonman]

Where did I imply that energy is not conserved in all reference frames? 4-momenta is always conserved, so energy is always conserved in all refernce frames, the two go hand-in-hand.

On Post #22

 

[robphy]

In my opinion, I think the original problem
"Why is KE not conserved in inelastic collisions?"
is not clearly posed since the original poster is obviously seeking for something deeper than the textbook definition [that an "inelastic collision" is one in which the total-KE is not conserved],

Can you please pose a clear question (possibly followed by what features you are looking for, or not looking for)?

[If "inelastic collision" means something other than the textbook definition, it would be helpful if you precisely define what YOU mean.]

 

[protonman]

In my opinion, I think the original problem
"Why is KE not conserved in inelastic collisions?"
is not clearly posed since the original poster is obviously seeking for something deeper than the textbook definition [that an "inelastic collision" is one in which the total-KE is not conserved],

Can you please pose a clear question (possibly followed by what features you are looking for, or not looking for)?

[If "inelastic collision" means something other than the textbook definition, it would be helpful if you precisely define what YOU mean.]

The textbook answer is wrong and/or incomplete.

A bit of advice. You don't know how much physics I have studied. I ask questions here to inspire discussion, not to get answers. This is the last place I would look for answers. If you want to test my understanding of the physics I have been discussing look at what I have said about SR. The best most people can do here is regurgitate their textbooks which is useless because anyone who can read can do the same. The whole point is to understand what the books are saying and then check if what they say is valid.

 

[robphy]

Physics Lesson One:

What is called the 4-momentum is not really "momentum" in the classical sense. If you differential the space-time 4-vector with respect to proper time you get what is called the momentum 4-vector.

I think you need to multiply by the rest mass to get what is called the momentum 4-vector.

Both photons and electrons have momentum. If they have an equal amount of momentum in opposite directions they will collide and both have a final velocity of zero.

Photons can't be brought to rest.

 

[protonman]

I think you need to multiply by the rest mass to get what is called the momentum 4-vector.



Photons can't be brought to rest.

WRT the first statement what I said initially is correct.

But they can be absorbed. The whole problem is that physicsts don't even understand what a photon is.

 

[robphy]

You need to read the post before you write.

They do have zero velocity because we are talking about inelastic collisions.

Is there anyone here who can think?

BTW, if you do the calculation out for a photon and electron moving with opposite momentums towards one another it does work out. Using the relativistic momenta.

(Energy of photon)/c - (gamma)(mass of electron)(velocity of electron) = 0

Just solve for the velocity of the electron in this case and see for yourself.

But they can be absorbed. The whole problem is that physicsts don't even understand what a photon is.

Let k be the null 4-momentum of the incident photon.
Let p be the timelike 4-momentum of the electron before the collision.
Let p' be the timelike 4-momentum of the electron after the collision.
On the assumption that the photon was "absorbed", there is no k'.

4-momentum conservation implies
k + p = p'

squaring
<br /> \begin{align*}<br /> k\cdot k + 2 k \cdot p + p \cdot p &amp;= p&#039; \cdot p&#039; \\<br /> 0 + 2 k \cdot p + m^2 &amp;= m^2 <br /> \end{align*}<br />
where m is the rest mass of the electron.

So,
k \cdot p = 0
Since k is null and p is timelike and nonzero, then k must be zero [that is, there is no such incident photon... the proposed collision is not possible.] (This supports Gokul43201's statement that "A free electron can not absorb a photon".)

Alternatively,
evaluate the invariant k \cdot p = 0 in the rest frame of the electron (so, p = \left( \begin{array}{cc} m \\ \vec 0\end{array}\right) and k = \left( \begin{array}{cc} k_0 \\ \vec k \end{array}\right), where k_0=|\vec k|). Then 0=k \cdot p = m(k_0)-\vec 0\cdot \vec k=m(k_0). Since m \neq 0, then k_0=0 (and hence \vec k=\vec 0 and so the 4-vector k=0).

 

[protonman]

Let k be the null 4-momentum of the incident photon.
Let p be the timelike 4-momentum of the electron before the collision.
Let p&#039; be the timelike 4-momentum of the electron after the collision.
On the assumption that the photon was "absorbed", there is no k&#039;.

4-momentum conservation implies
k + p = p&#039;

squaring
<br /> \begin{align*}<br /> k\cdot k + 2 k \cdot p + p \cdot p &amp;= p&#039; \cdot p&#039; \\<br /> 0 + 2 k \cdot p + m^2 &amp;= m^2 <br /> \end{align*}<br />
where m is the rest mass of the electron.

So,
k \cdot p = 0
Since k is null and p is timelike and nonzero, then k must be zero [that is, there is no such incident photon... the proposed collision is not possible.] (This supports Gokul43201's statement that "A free electron can not absorb a photon".)

Alternatively,
evaluate the invariant k \cdot p = 0 in the rest frame of the electron (so, p = \left( \begin{array}{cc} m \\ \vec 0\end{array}\right) and k = \left( \begin{array}{cc} k_0 \\ \vec k \end{array}\right), where k_0=|\vec k|). Then 0=k \cdot p = m(k_0)-\vec 0\cdot \vec k=m(k_0). Since m \neq 0, then k_0=0 (and hence \vec k=\vec 0 and so the 4-vector k=0).

Your mathematics don't impress me. If you want to debate with adults you need to think for yourself. Just because the math works does not mean you have proved anything.

 

[jcsd]

On Post #22

No I didn't, read what I wrote. 4-momanta is always conserved so:

1) Free electronns can't absorb photons

2) energy is conserved in all referbce frames

 

[Gokul43201]

It looks like protonman has figured out where the lost KE of the inelastic collision goes - "the object's internal elastic potential energy". So, pending the question of photon absorption by a free electron and doubts about 'who said what, and when', I believe this question has reached a resolution.

 

[jcsd]

It looks like protonman has figured out where the lost KE of the inelastic collision goes - "the object's internal elastic potential energy". So, pending the question of photon absorption by a free electron and doubts about 'who said what, and when', I believe this question has reached a resolution.

There's nothing to resolve about whthere or not a free photon can absorb an electron it's basic physics, and it's hardly a big secret as it appears in enough physics textbooks.

 

[electronman]

No I didn't, read what I wrote. 4-momanta is always conserved so:

1) Free electronns can't absorb photons

2) energy is conserved in all referbce frames

Recall from post #22

God you're obnoxious. I didn't say momenta isn't conserved I said 4-momenta isn't conserved. For 4-momentum to be conserved, energy (which can be seen as the time coponent of 4-momentum) must also be conserved in all reference frames.

 

[electronman]

It looks like protonman has figured out where the lost KE of the inelastic collision goes - "the object's internal elastic potential energy". So, pending the question of photon absorption by a free electron and doubts about 'who said what, and when', I believe this question has reached a resolution.

And the interesting part is that no one even acknowledged my post. It seems here that when the bad guy is the right guy they ignore and eventually ban users.

 

[jcsd]

Recall from post #22

And if you recall the context, it's apparent that I meant 4-momentum isn't conserved in the example that he gave and hence the example is physically impossible.

 

[electronman]

And if you recall the context, it's apparent that I meant 4-momentum isn't conserved in the example that he gave and hence the example is physically impossible.

Ha Ha you lose!

 

[jcsd]

Ha Ha you lose!

What!? quit your babbling, what was the point to this thread? have you just come here to troll?

 

[electronman]

What!? quit your babbling, what was the point to this thread? have you just come here to troll?

This post ended at post #31 as far as I am concerned since no one raised any objection to my answer.

 

[jcsd]

The question was answered befopre then. You don't need to consider SR at all and the compressibilty of objects is irrelevant as even in SR objects can be point particles.