# Confusing Elastic and Inelastic Collisions :S

1. Jul 12, 2009

### rice1am

Can someone please 'stupify' the definitions of an elastic and inelastic collision?

Do objects bouncing off of each other undergo elastic collisions, while objects which stick together undergo completely inelastic collisions? This concept is very confusing to me :S

thank you!

2. Jul 12, 2009

If there is no loss of kinetic energy it is a completely elastic collision.Such collisions can occur with atomic scale objects such as gas molecules but not with macroscopic objects.If there is a total loss of kinetic energy it is a completely inelastic collision.Total energy and momentum are conserved in all collisions.
If a ball is dropped from rest and bounces to 90 percent of its original height it is an approximately elastic collision.If an egg is dropped and splatters on impact it is an approximately inelastic collision.

3. Jul 12, 2009

### tiny-tim

Hi rice1am!

Yes, a completely inelastic collision is where the two bodies have the same final velocity, in other words they stick together.

(It's really not a very sensible name for it )

But no, bouncing is usually not elastic … some energy is almost always lost, in practice.

General rule: a collision is not (completely) elastic unles the exam question says it is!

4. Jul 12, 2009

### rice1am

Thanks everyone! I'm starting to understand the concepts. :D

5. Jul 12, 2009

### RoyalCat

Just to add onto that, a completely inelastic collision (Sometimes called a plastic collision) isn't necessarily one where where ALL kinetic energy is 'lost' (Note that this means that it has gone into deforming the objects in question, or into heat energy. On an atomic scale, energy is conserved), but just one where the objects stick together.

So if you had an object of mass $$m$$ traveling at a velocity $$v$$ colliding with an object of mass $$M$$ that is at rest, a completely inelastic collision would mean that the two objects would combine into one with a shared mass of $$m+M$$.
Momentum is conserved regardless of the elasticity of the collision. See if you can derive what the shared velocity of the new $$m+M$$ object is, and what the kinetic energy of the system is after the collision, and how that compares with its initial energy.

That should be a good exercise to see if you understand the subject well.

As a counter to that, see if you can derive a simple condition about the velocities of the objects in question from preservation of kinetic energy and momentum in the case of a completely elastic collision.

6. Jul 12, 2009

Hello tiny tim and Royal cat.There can be inelastic collisions where all the kinetic energy is lost so what is the correct terminology to use for such collisions?I have always called them completely inelastic but should they be called 100 percent inelastic or something similar.Thank you.

7. Jul 12, 2009

### RoyalCat

Whoopsie, only just noticed I was replying to a quote by Dadface.
Everything I posted here, with the exception of the first paragraph, is addressed at rice1am. :)

Using the definitions I'm familiar with, such a collision would be defined by two factors.
The first would be that the two colliding objects fuse into one. The second would be that the final velocity of the resulting object is 0.

I think I can see where Dadface was coming from now:
If you work in the center of mass frame (The frame of reference for which the center of mass is at rest), then a plastic collision would result in the loss of ALL the kinetic energy.

To explain further, there are two cases I can think of from the top of my head where a plastic collision would lead to the total loss of kinetic energy : $$E_{k_f}=0$$

The first is the case where the initial momentum of the system is 0. The requirement it be preserved coupled with the fact you're only dealing with one mass in the final state, means that your final speed would have to be 0 as well.
Choosing an appropriate frame of reference will bring you to this case. Note that this is a useful transition to make in some cases as it cuts down on the number of expressions you would have to keep track of in an equation.

The second is where you have a single particle of mass $$m$$ traveling at some velocity $$v$$ prior to a collision with a particle of mass $$M$$ which is at rest.
A quick rundown using conservation of momentum would lead to the following conclusion (Try and do this for yourself if you haven't already):
The velocity of the combined particle of mass $$m+M$$ would be:
$$\bar V = \frac{m}{m+M}v$$
Consider the private case where $$m<<M$$ (Where the mass $$m$$ is negligible when compared with $$M$$), for instance, when a ball of putty hits a wall.
Mathematically, you would have to look at the case where $$m\rightarrow 0$$
Taking the limit, you see that the final velocity of the ball of putty+wall will approach 0, meaning the same goes for the kinetic energy of the system.

8. Jul 13, 2009

### tiny-tim

Forget loss of kinetic energy … it's frame-dependent, and it will only confuse you.

I think the official term is perfectly inelastic … but "completely" is also used, and is fine, though I've never seen "100 per cent".

9. Jul 13, 2009

Thanks tiny-tim.I like to think of perfectly elastic and perfectly inelastic collisions as being at two extremes and that most collisions with macroscopic objects lie somewhere between those extremes.Although macroscopic object cannot collide perfectly elastically they can collide perfectly inelastically,although such collisions may be rare.Just one example would be when, in the rest frame of the observer,a bullet with a certain momentum in one direction collided with and became embedded in a block of wood with a numerically equal momentum but travelling in the opposite direction.In fact, I see perfectly inelastic collisions as being like explosions but running in reverse.