Conserving Momentum in an Inelastic Collision: Explained

In summary, momentum is conserved in an inelastic collision while kinetic energy is not. This is because momentum is a vector quantity that can be transferred from one object to another, while kinetic energy is a scalar quantity that can be dissipated in the form of heat or sound. This conservation of momentum can be proven mathematically at the molecular level, as it follows from the invariance of the laws of physics under translations. Both momentum and kinetic energy were first observed and then derived from these observations, with the modern view being more fundamental.
  • #1
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Can you explain how momentum is conserved while KE isn't in an inelastic collision??

I've though about this for a while... If KE from object X is lost during the collision it means that the object will slow down. If it does, how is momentum in the system conserved? Could you explain this by means of an example? Imagine that Object X is moving and Object Y is stationary.
Thanks in advance.
 
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  • #2


Okay: say X has a mass of 1 kg and Y has a mass of 3 kg. X is moving toward the right at 15 m/s and Y is stationary. The total momentum of X and Y before the collision is 15 kg.m/s, and the total energy is 225 J.

After the collision, say X is stationary and Y is moving to the right. If momentum is conserved, then Y's speed is 5 m/s - but in that case, its kinetic energy is 75 kg.m/s. So kinetic energy obviously is not conserved in this case, but momentum is. Basically, when the momentum is transferred from object X to object Y, more of it "goes into" the mass and less of it "goes into" the velocity. But because kinetic energy is more strongly dependent on velocity, the "transfer" of momentum from mass to velocity (I'm using such loose terminology here I'm almost ashamed of myself) "affects" the kinetic energy more than the momentum - end result, kinetic energy goes down while momentum doesn't.

Speaking more generally (and more correctly now), a formula that is very common in higher-level physics is
[tex]E = \frac{p^2}{2m}[/tex]
(check it if you like, using [tex]E = mv^2/2[/tex] and [tex]p = mv[/tex]). This means that two objects can have the same momentum but different energies, if they have different masses. The lost kinetic energy generally turns into heat and sound waves.
 
  • #3


mahela007 said:
I've though about this for a while... If KE from object X is lost during the collision it means that the object will slow down.

And object Y will speed up.

mahela007 said:
If it does, how is momentum in the system conserved? Could you explain this by means of an example? Imagine that Object X is moving and Object Y is stationary. Thanks in advance.

If Object Y starts out as stationary, it won't end up as stationary.
 
  • #4


diazona said:
(check it if you like, using [tex]E = mv^2/2[/tex] and [tex]p = mv[/tex]).
Apologies for an unrelated question for someone: Why is the second Latex eq. positioned vertically up too high and what governs vertical position? Are there ways or tricks for controlling vert. alignment?

Thanx.
 
  • #5


When you put an equation inside a line of text, use "itex" and "/itex" tags, not "tex" and "/tex".
 
  • #6


Total energy and total momentum is always conserved. Now, momentum is linear in the velocity and this then implies that the velocity of the center of mass times the total mass of an object is also the total momentum of the object. So, if we keep track of only the center of mass and velocities of the objects involved, we will have accounted for all of the momentum in the system.


Kinetic energy is different. If you have N particles then the total energy can be written as the kinetic energy of the center of mass plus the sum of 1/2 m_i v_i^2 where v_i is the velocity of the ith particle relative to the center of mass and m_i the mass of the ith particle.


So, unlike momentum, kinetic energy can get lost in the motion of the parts of the system. We then say that energy has been dissipated in the form of heat.
 
  • #8


Is there a way to "prove" momentum is conserved through some type of math at the molecular level, or is it just an observed experience?
 
  • #9


Jeff Reid said:
Is there a way to "prove" momentum is conserved through some type of math at the molecular level, or is it just an observed experience?

It follows from the fact that the laws of physics are invariant under translations. Of course, this invariance is itself something that is based on observations.
 
  • #10


Here is a simpler example which may help you see the distinction.
Imagine two balls of putty of equal mass heading toward one another head on at the same speed (and thus opposite velocity).

They hit, they stick, they warm up due to conversion of kinetic energy to heat.

The total momentum was zero to begin with since momentum is a vector and the two ball's momentum was equal and opposite in direction they add to zero.

Before the collision the total momentum was (mV) + (-mV). Afterwards it is 0+0.
 
  • #11


Thanks for all the replies... I think I understand momentum and KE now.
 
  • #12


Momentum is a vector. Energy is a scalar. Makes a huge lot of difference. Two bodies moving at a frenetic pace in opposite directions would give a total momentum zero. Unfortunate!
 
  • #13


Count Iblis said:
It follows from the fact that the laws of physics are invariant under translations.

I don't understand how invariance can lead to conservation while conservation is in itself a law. A.P French in his "Newtonian Mechanics" says conservation of momentum was found first by observations. He goes on to say that the third law was derived from this and the second law.
 
  • #14


sganesh88 said:
I don't understand how invariance can lead to conservation while conservation is in itself a law. A.P French in his "Newtonian Mechanics" says conservation of momentum was found first by observations. He goes on to say that the third law was derived from this and the second law.


This is how people centuries ago who did not have the deep understanding of the laws of physics as we have today, found out that momentum is conserved.

The modern view, based on "[URL [Broken] theorem[/URL] is more fundamental, because it relates conservation of momentum to something that is far more fundamental, i.e. that the laws of physics operate in all places in the universe in exactly the same way. In experiments one can set limits to any violatons of this invariance under translations that then imply limits to violations to momentum conseervations. These limits are then much sharper than what you could get out of any direct measurement of momenta.
 
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  • #15


One last thought on the matter. Remember that Force changes momentum, in particular force is the time rate of change of momentum. Total momentum conservation is then built into Newton's third law: To every action (force) there is an equal and opposite reaction.

Newton's third law is simply the statement that momentum is not created nor destroyed it is just exchanged between systems which interact with one another. Total momentum is conserved unless Newton's third law is violated.
 
  • #16


jambaugh said:
Total momentum is conserved unless Newton's third law is violated.

But that Noether theorem page of wikipedia says
... interestingly, conservation of momentum still holds even in situations when Newton's third law is incorrect.
How is that? :uhh:
 
  • #17


Newton's third law implies conservation of momentum but conservation of momentum does not imply Newton's third law.
 
  • #18


Count Iblis said:
Newton's third law implies conservation of momentum but conservation of momentum does not imply Newton's third law.

Let's have a two particle system. m1, u1, m2,u2, v1,v2 are all corresponding parameters. What all these mean needn't be explained. :D
A force F12 is exerted by 1 on 2; and F21 exerted by 2 on 1. Both forces act for a small interval of time t. Conservation of momentum and Second law are 'known' are taken as the basis.
Initial momentum:
m1u1+ m2u2
Final momentum:
m1u1+ m2u2 + (F12+F21)t

But by conservation of momentum,
m1u1+m2u2 = m1u1+m2u2+ (F12+F21)t

So (F12+F21)t = 0
implying F12= - F21
Equal and opposite. :)
Whats wrong with this procedure?
I read this in A.P.French's book i mentioned in my earlier post. But what i don't understand about this procedure is how they defined "mass" and subsequently momentum in the 'pre-Newton's-laws period'.
 
  • #19


Kinetic energy IS conserved in an inelastic collision if you take into account the resulting increase in molecular kinetic energy after the collision and the radiation emitted by the molecules after the collision (aka, heat). The reason the momentum is conserved is that momentum is a vector quantity, and if Newton's second law holds (and it does) then Ft = mv. The "t" is always the same in all three dimensions for any collision. Energy, however, is a scalar, so you don't get this cancellation.
 
  • #20


worldrimroamr said:
Kinetic energy IS conserved in an inelastic collision if you take into account the resulting increase in molecular kinetic energy after the collision and the radiation emitted by the molecules after the collision (aka, heat).

Total energy is conserved. Kinetic energy is not.
 
  • #21


jambaugh said:
Newton's third law is simply the statement that momentum is not created nor destroyed it is just exchanged between systems which interact with one another. Total momentum is conserved unless Newton's third law is violated.

It's not quite so clear to me that Newton's 3rd law is simply that. Momentum is conserved in an inertial reference frame. The 3rd law is the means, in principle, of telling whether you're in an inertial reference frame.
 
  • #22


sganesh88 said:
Let's have a two particle system. m1, u1, m2,u2, v1,v2 are all corresponding parameters. What all these mean needn't be explained. :D
A force F12 is exerted by 1 on 2; and F21 exerted by 2 on 1. Both forces act for a small interval of time t. Conservation of momentum and Second law are 'known' are taken as the basis.
Initial momentum:
m1u1+ m2u2
Final momentum:
m1u1+ m2u2 + (F12+F21)t

But by conservation of momentum,
m1u1+m2u2 = m1u1+m2u2+ (F12+F21)t

So (F12+F21)t = 0
implying F12= - F21
Equal and opposite. :)
Whats wrong with this procedure?
I read this in A.P.French's book i mentioned in my earlier post. But what i don't understand about this procedure is how they defined "mass" and subsequently momentum in the 'pre-Newton's-laws period'.

Special relativity. If particle 1 loses momentum then particle 2 cannot gain momentum at the same time. Suppose you somehow do have conservation of momentum in one reference frame with the two particles separated by some distance d, then you don't have conservation of momentum in another frame, because events that happen at the same time in one reference frame don't happen at the same time in another frame.

Interactions are mediated by fields. A particle can lose momentum to e.g. the electromagnetic field which can then transfer that momentum to another particle.
 
  • #23


Count Iblis said:
Special relativity. Suppose you somehow do have conservation of momentum in one reference frame with the two particles separated by some distance d, then you don't have conservation of momentum in another frame, because events that happen at the same time in one reference frame don't happen at the same time in another frame.
But SR says laws of physics are valid in all inertial frames. So conservation of momentum should be valid in the other frame too- though the distance between the particles and time of collision might not be the same-.
 
  • #24


sganesh88 said:
But SR says laws of physics are valid in all inertial frames. So conservation of momentum should be valid in the other frame too- though the distance between the particles and time of collision might not be the same-.

Yes, conservation of mometum is certainly valid, but you cannot have any instantaneous transfer of momentum in any particular frame, because then you will have violation of conservation of momentum in another frame.

The momentum that is transferred from one particle to another particle must be transferred via "events" that are within the light cone of the point where the particle lost its momentum.
 

1. What is momentum and why is it important?

Momentum is a physical property of an object that describes its motion. It is defined as the product of an object's mass and velocity. In an inelastic collision, the total momentum of the system is conserved, meaning that the total momentum before and after the collision remains the same. This is important because it helps us understand and predict the motion of objects after a collision.

2. How is momentum conserved in an inelastic collision?

In an inelastic collision, momentum is conserved through the transfer of kinetic energy from one object to another. This means that the total momentum before the collision is equal to the total momentum after the collision. However, unlike in an elastic collision where kinetic energy is conserved, some kinetic energy is lost in an inelastic collision due to the deformation or sticking together of objects.

3. What is the difference between an inelastic and elastic collision?

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy before the collision are equal to the total momentum and total kinetic energy after the collision. In an inelastic collision, only momentum is conserved, and some kinetic energy is lost due to the deformation or sticking together of objects.

4. How do you calculate the final velocity of objects in an inelastic collision?

To calculate the final velocity of objects in an inelastic collision, you can use the conservation of momentum equation: m1v1i + m2v2i = (m1 + m2)vf. Here, m1 and m2 are the masses of the objects, v1i and v2i are the initial velocities, and vf is the final velocity of the objects after the collision.

5. Can momentum be conserved in an inelastic collision if objects stick together?

Yes, momentum is still conserved in an inelastic collision even if objects stick together. This is because the total mass and velocity of the objects before and after the collision remains the same. However, in this case, the final velocity of the objects will be less than the initial velocity due to the loss of kinetic energy in the collision.

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