# Why Momentum is Conserved in Inelastic Collisions

1. Apr 3, 2008

### daveyman

This is the way I understand inelastic collisions. Please correct me if I'm wrong on this.

In an inelastic collision, kinetic energy is converted into internal energy and is lost (often in the form of heat). So, one cannot say that energy is conserved. Momentum, however, is conserved in this case.

Here's the problem I have with this explanation:
Momentum is dependent upon mass and velocity (p=mv).
Kinetic energy is also dependent upon mass and velocity (K=1/2(mv$$^{2}$$)).

So, if both kinetic energy and momentum are dependent upon the same variables, how can one be conserved while the other isn't?

2. Apr 3, 2008

### tiny-tim

Welcome to PF!

Hi daveyman! Welcome to PF!

Very simply: if a - b = c - d is true, then that usually means that a^2 - b^2 = c^2 - d^2 isn't true!

(If you're not happy with that, just try a few examples until you're convinced!)

For a collection of variables to successfully obey two different equations is actually very unusual!

3. Apr 4, 2008

### pam

Conservation of momentum is a direct consequence of Newton's third law.

4. Apr 4, 2008

### manjuvenamma

Yes, like Pam said law of conservation of momentum is a direct consequence of newton's law about action - reaction pair. Please go through any standard book giving the derivation of this. It is a beatuiful proof - very useful, yet very simple. In the process, you will also be able to calculate the forces involved.

Kinetic energy is also conserved, but kinetic energy is conserved but in general, this some of this energy is lost/converted into other forms of energy such as heat/light/deformation energy etc. so the kinetic energies before and after the collision don't equal each other.

5. Apr 4, 2008

### manjuvenamma

"Kinetic energy is also conserved, but kinetic ",

no, sorry, I meant energy is also conserved, but kinetic ......

6. Apr 7, 2008

### Volcano

You are wrong, momentum is not conserving.

7. Apr 7, 2008

### Volcano

İn collisions total momentum conserving, not each one. Also don't forget that one is vector other one is scalar, comparison is wrong. By the way, if focus on everything, total energy is conserving too. Missing energy turning to heat and vibration energy.

8. Nov 20, 2010

### markem

Sorry to bump this thread but I've been digging around on the web trying to find an answer to the same worry.

Suppose as a conceptual "limiting case" we have an inelastic collision where the initial speed of something is, say, 1 m/s, followed by an inelastic collision with another object where ALL the kinetic energy is dissipated as heat, thus both objects have 0 velocity. Thus we have m*(1m/s) = 0 (since both velocities are 0 and thus masses are irrelevant, momenta = 0). Clearly this violates p1+p2 = p1f+p2f... In "usual" scenarios I presume the same thing happens but to a lesser degree...

What am I not getting?

9. Nov 20, 2010

### dgOnPhys

The situation you are presenting cannot happen unless it preserves total momentum: so the only way you can get all kinetic energies to be zero is if the initial total momentum was already zero.

A completely inelastic interaction between two bodies is one in which the two bodies are stuck or lumped together. After this collision the resulting body will still move with the initial velocity of the center of mass

10. Nov 20, 2010

### fzero

Well momentum conservation makes such a scenario impossible as long as the two objects are considered as a closed system. One could however contrive viable examples with an external force or an external medium. For example, gas molecules in an atmosphere would carry momentum away from the objects. This is what happens to the dissipated momentum when an object experiences air resistance.

11. Nov 20, 2010

### marsdk

@Markem:

What you say could happen if both objects had the same mass and where travelling with the same speed, but in opposite directions along the same line. Then the total momentum before would be 0, just as it is after.

Remember that momentum is a vector so the direction of the velocities matters. In the case stated above, p would be:

p = m*v + m*(-v) = 0

12. Nov 20, 2010

### Ambidext

Kinetic energy isn't conserved- it's energy that is conserved.

Etotal = Epotential + Ekinetic + Einternal

"Momentum", like "force" are merely concepts if you think about it. We can't see or feel a force. These concepts are experimental and yet have been predicting motion in inertial frames for the past few centuries.

13. Dec 13, 2010

### pinsky

So what you're saying is that the conservation of momentum prevents all the kinetic energy to be converted (in this situation).

What is the amount of energy that can be converted then? (given in a ratio of the starting kinetic energy of the system)

14. Dec 13, 2010

### lalbatros

It is useful to consider the Noether theorem to answer this question.
(see http://en.wikipedia.org/wiki/Noether's_theorem)

This theorem shows that conserved quantities are related to symmetries.

When a system doesn't change by rotation, it has a rotational symmetry.
When a system doesn't change by translation, it has translational symmetry.
When a system doesn't change in time, it has time-translation symmetry or time invariance symmetry.

("when a system doesn't change" means: the hamiltonian (energy) doesn't change)

The Noether theorem states that:

rotational symmetry implies the conservation of angular momentum,
translational symmetry implies the conservation of momentum,
time invariance implies the conservation of energy.

Talking about "Inelastic Collision" is actually hiding a part of the story.

There can be different kind of inelastic collisions.

Consider a collision between two rubber spheres.
The total energy (hamiltonian) of the system contains the usual kenetic and potential energy, but it should be complemented for all the internal energy that both spheres can accomodate, including vibrations and even heat. Taking these extra energy terms into account, would restore energy conservation. But of course the "simplified energy that neglects internal energy" would not be conserved. Note further that the total energy (including internal effects) would be time invariant as well as translation invariant, leading to total energy and momentum conservation. Note also that the internal degrees would modify both the expression for the total energy as well as the expression for the total momentum. The "simplified" momentum like the "simpified" energy need not be conserved since the internal degrees of freedom need to be taken into account. We need to view each sphere as the more complex object they are!

Consider a collision between two electrons.
In this case, electromagnetic radiation will be emitted.
The total energy of the two electrons is not conserved since electromagnetic radiation will take energy away.
The total momentum of the two electron will also not be conserved since the electromagnetic radiation wil take momentum away.

15. Jan 24, 2011

### markem

lalabtros, thank you so much for your reply. Finally elucidated something I was stuck on for a long time, namely, how could momentum be conserved when energy (supposedly) wasn't. Thanks a ton!