How momentum is conserved in an inelastic collision

In summary, according to my current understanding, energy can be transferred by either work out heat or by the transfer of momentum. If energy is transferred from our system (two masses), then some of the energy is transferred by the means of work which involves force and if mass exerts force on surrounding then equal and opposite force is exerted to mass according to Newton's third law. If there is an external force, then momentum should not be conserved.
  • #1
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In inelastic collision their is loss of energy and according to my current knowledge energy can be transferred by either work out heat...Now The problem is that we use law of conservation of momentum in problems related with inelastic and if energy is transferred from our system (two masses) then some of the energy is transferred by the means of work which involves force and if mass exerts force on surrounding then equal and opposite force is exerted to mass according to Newton's third law, now if their is some external force then momentum should not be conserved.
 
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  • #2
Hemant said:
In inelastic collision their is loss of energy and according to my current knowledge energy can be transferred by either work out heat...Now The problem is that we use law of conservation of momentum in problems related with inelastic and if energy is transferred from our system (two masses) then some of the energy is transferred by the means of work which involves force and if mass exerts force on surrounding then equal and opposite force is exerted to mass according to Newton's third law, now if their is some external force then momentum should not be conserved.

If an external force acts during the collision, then momentum may not be conserved. E.g. if there is friction on a surface and the collision is not considered instantaneous, then friction may change the momentum of the system during the time of the collision.
 
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  • #3
Hemant said:
... if mass exerts force on surrounding then equal and opposite force is exerted to mass according to Newton's third law, now if their is some external force then momentum should not be conserved.
The collision does not result in external forces on the system, only internal forces.
 
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  • #4
PeroK said:
If an external force acts during the collision, then momentum may not be conserved. E.g. if there is friction on a surface and the collision is not considered instantaneous, then friction may change the momentum of the system during the time of the collision.
Okay,but my another argument is that if we say kinetic energy is ##\frac {p²} {2m}## and if kinetic energy of centre of mass changes then also momentum should be changed.
 
  • #5
Hemant said:
Okay,but my another argument is that if we say kinetic energy is ##\frac {p²} {2m}## and if kinetic energy of centre of mass changes then also momentum should be changed.
That's not a valid argument. Note that in the centre of momentum frame the total momentum is zero. That can't actually go down any lower!
 
  • #6
The system consists of two objects. No external forces act on the objects. The NET moment of the WHOLE SYSTEM does not change by Newton’s first law. However, each of the objects has its own momentum. The momentum of each object can and does change and their kinetic energy may be lost.
 
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  • #7
Hemant said:
Now The problem is that we use law of conservation of momentum in problems related with inelastic and if energy is transferred from our system (two masses) then some of the energy is transferred by the means of work which involves force and if mass exerts force on surrounding then equal and opposite force is exerted to mass according to Newton's third law, now if their is some external force then momentum should not be conserved.
That all is one sentence. And I still don't know what the problem is.
 
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  • #8
Hemant said:
Okay,but my another argument is that if we say kinetic energy is ##\frac {p²} {2m}## and if kinetic energy of centre of mass changes then also momentum should be changed.
Momentum is a vector, energy is a scalar. If 2 balls in opposite directions collide, they may end up at rest if it's a squishy collision, but momentum is conserved since one ball started out + and the other -. 0 total momentum before, 0 total momentum after.
 
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  • #9
I think the OP may be worried that energy can leave the system and this means external forces are somehow acting . But the momentum of the system will change only if there is a net external force... those that involve heat transfer will average to zero over a very short time frame. Indeed if there is energy dissipated there will always be small fluctuations which average to zero.
 
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  • #10
I got it.
Even if energy is lost from system the velocity will somehow adjust to conserve momentum.
 
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  • #11
rude man said:
Momentum is a vector, energy is a scalar. If 2 balls in opposite directions collide, they may end up at rest if it's a squishy collision, but momentum is conserved since one ball started out + and the other -. 0 total momentum
Sir but isn't this is general case.
I was thinking that if kinetic energy is lost by system then we should be able to find it out by the formula ##\frac {p²} {2m}## but problem using this formula is that momentum doesn't changes but kinetic energy changes so even then if we apply this formula we will find that their is no change in kinetic energy of system.
 
  • #12
Hemant said:
Sir but isn't this is general case.
I was thinking that if kinetic energy is lost by system then we should be able to find it out by the formula ##\frac {p²} {2m}## but problem using this formula is that momentum doesn't changes but kinetic energy changes so even then if we apply this formula we will find that their is no change in kinetic energy of system.

For a system of particles:

##E \ne \frac{p^2}{2m}##

That formula only holds for a single particle.

To see this simply consider the centre-of-momentum frame. The momentum is zero, but the kinetic energy is zero only in the special case where all particles are at rest.
 
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  • #13
PeroK said:
For a system of particles:

##E \ne \frac{p^2}{2m}##

That formula only holds for a single particle.

To see this simply consider the centre-of-momentum frame. The momentum is zero, but the kinetic energy is zero only in the special case where all particles are at rest.
Got it!
 
  • #14
Hemant said:
the velocity will somehow adjust to conserve momentum.
"Somehow" seems to imply a 'little brain', taking care of things. Momentum IS conserved. When you introduce losses in a collision, the result is that the resulting Kinetic Energy is less and the share of the KE is different. If Net Work is done during the collision, due to distortion then the parting velocity will be less than the approach velocity. A fair approximation to this is often covered by the 'Coefficient of Restitution' but it may not work for all velocities is the materials are not linear. Energy that's permanently absorbed will turn up as heat and sound so total Energy is also (of course) conserved.

I remember the way C.O.R. was taught at A level and it was not really 'explained - just presented as a number that applied to each of the problems we were given. Not very satisfactory for me because I wasted a lot of worry time on it when the teacher could have dealt with it in terms of Work Done and I / we would have accepted it and moved on.
 
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  • #15
Hemant said:
I got it.
Even if energy is lost from system the velocity will somehow adjust to conserve momentum.

One more comment, generally when we say energy is lost, it doesn’t actually leave the system (or at least most of it doesn’t). What we are saying is that it is converted into a form we aren’t counting. For example if your carts have putty between them and the putty deforms and the carts stick together we say that energy is not conserved. However, really it is conserved. The kinetic energy changes, but the missing energy went into deforming and heating the putty. If we included some thermometers and measured the thermal energy we would find that energy was actually conserved. At least mostly.

Now, when I say “mostly”, I make that distinction only because we generally can’t (or at least don’t) make truly isolated systems. We try to minimize your cart’s interaction with the track by using wheels, but there is still some friction, and there is air resistance, and when the carts collide the make sound which is energy being lost to the air. These interactions that lose energy to the surroundings will keep energy from being conserved in the system. They also violate the concept of “no external forces” and can keep momentum from being conserved. When we say momentum and energy are conserved we are claiming we have made all of the external interactions negligibly small.
 
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  • #16
Conservation of Momentum is one of the very few things in Science that are treated with almost religious observance. It just works every time and I don't think there is even a hint that it won't work next time.
Momentum is defined differently under relativistic situations so it just keeps working and fits seamlessly into classical mechanics.
 
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  • #17
Hemant said:
Sir but isn't this is general case.
...
Momentum is directional, energy is not.

:cool:

The key is in the velcro:

 
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  • #18
sophiecentaur said:
Conservation of Momentum is one of the very few things in Science that are treated with almost religious observance. It just works every time and I don't think there is even a hint that it won't work next time.
Momentum is defined differently under relativistic situations so it just keeps working and fits seamlessly into classical mechanics.
😮,so no one knows how it works.
 
  • #19
Hemant said:
😮,so no one knows how it works.
That reads as if it's just a trick of some sort.
Does anyone 'know' how anything works? All you can say is that you can rely on it being conserved. Same as 1+1 = 2 . That's just an axiom, I know but it's never been known to fail so we work with it.
 
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  • #20
Hemant said:
so no one knows how it works.
We know it works well.
 
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  • #22
A.T. said:
We know it works well.
I was not questioning about if it works or not but I wanted to knew how does it work but i understand what sophiecentaur meant to say.

sophiecentaur said:
Does anyone 'know' how anything works?
When I read this post I didn't liked it but after thinking about it I came to conclusion that whenever we ask 'why' and start finding for an answer we get another question like it on every step and the number of questions grow exponentially and we can't even answer one of them and ended up with nothing rather than existential crisis.
PeroK said:
Try reading Feynman's take on conservation of energy here.

https://www.feynmanlectures.caltech.edu/I_04.html
Thanks for the great help.i again get the line which I get in many forms(energy is not description of some mechanism or something concrete but an mathematical number which remains conserve)and that is the true reality.
 
  • #23
Hemant said:
Sir but isn't this is general case.
I was thinking that if kinetic energy is lost by system then we should be able to find it out by the formula ##\frac {p²} {2m}## but problem using this formula is that momentum doesn't changes but kinetic energy changes so even then if we apply this formula we will find that their is no change in kinetic energy of system.
You're forgetting the energy lost in an inelastic collision including physical distortion forces, heat generation, sound generation ... lots of ways energy is dissipated other than via momentum change! Momentum change if and only if EXTERNAL forces applied to system, such as friction.

You realize of course that, the Earth taken as the sysrem, momentum is conserved under all circumstances!
 
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  • #24
Hemant said:
When I read this post I didn't liked it but after thinking about it I came to conclusion that whenever we ask 'why' and start finding for an answer we get another question like it on every step and the number of questions grow exponentially and we can't even answer one of them and ended up with nothing rather than existential crisis.
While you are in a Feynman mood, I get to recommend his comments about "why" questions...perhaps my favorite:
 
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  • #25
In any collision, there is an equal and opposite transfer of momentum via the collision impulse. Since ##-\Delta P+\Delta P=0 ##, there is no net change in momentum.
 
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  • #26
Hemant said:
Okay,but my another argument is that if we say kinetic energy is ##\frac {p²} {2m}## and if kinetic energy of centre of mass changes then also momentum should be changed.
The kinetic energy of a two body system is not determined by the "kinetic energy of the center of mass" (whatever that means). The kinetic energy of a two body system is the sum of the kinetic energies of each of the bodies separately.
 
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