Homework Help: Concept of Elastic/Inelastic Collision ?

1. Nov 12, 2012

Lo.Lee.Ta.

Hi, you guys! :)

I am studying Elastic/Inelastic Collision my Physics class right now.

So I know:
Elastic Collision: P(initial) = P(final) ; KE(initial) = KE(final)

Inelastic Collison: P(initial) = P(final) ; KE(initial) ≠ KE(final)

I'm wondering how I am supposed to know right when I see a type of problem, whether elastic or inelastic collision is taking place?

There was a pool ball example problem that I was working, and it required that you know it is elastic collion (and therefore, KE(i) and KE(f) are equal) taking place.
....In that particular problem, the collision was not head-on, so the two balls moved in different directions after the collision.

But Newton's Cradle is an example of elastic collision also, right? If one ball hits the other balls, the middle ones stay stationary, but the end ball swings out and continues that pattern.

So if the pool balls don't have a head-on collision, resulting in the striking ball to become stationary, and to transfer all of it's kinetic energy to the other ball, how could an elastic collision be taking place???

2. Nov 12, 2012

Zayin

In inelastic collisions, momentum is conserved but kinetic energy is not.
In elastic collisions, momentum and kinetic energy are conserved.

Therefore Newton's Cradle is an example of an elastic collision because, as you said, all kinetic energy is being transferred from one ball to the other.

--

Another good way to think about inelastic and elastic collisions is like this. When two objects collide and stick together, the collision is called 'perfectly inelastic'. For example, when two pieces of putty collide, they stick together and move with a common velocity after the collision. If either object in a collision is deformed at all, the collision is inelastic. If they aren't, or are very barely deformed (e.g. pool balls), the collision is elastic.

3. Nov 12, 2012

Lo.Lee.Ta.

Thank you for your response! :)
That makes sense.

Now to summarize:

Inelastic Collision:
*takes place when two objects collide, stick together, and behave like a single mass.
*Will very likely look deformed after collision
*Kinetic energy initial is different from Kinetic energy final

Examples: car collision where both cars stick together, putty colliding and sticking together to form one mass, grandma on roller skates zoomes in and scoops up her grandkid

Elastic Collision:
*Objects DON'T stick together after collision.
*They retain their shape
*Kinetic Energy initial is equal to kinetic energy final

Examples: Pool ball collision, grocery carts bump into each other, Newton's Cradle

4. Nov 12, 2012

SammyS

Staff Emeritus
What you have described above is a Perfectly Inelastic Collision. As Zayin pointed out, if the objects stick together, the collision is perfectly inelastic.

For a general inelastic collision, all that's specified is that the total Kinetic Energy is reduced after the collision, compared to the case before the collision.

It's interesting to look at such collisions in the Center of Mass reference frame. In this reference frame, the total momentum of the system is zero. If the objects stick together, then after the collision, the objects will both be at rest, so that the total Kinetic Energy after the collision is zero. That's as inelastic as you can get, thus the terminology, perfectly inelastic collision.

5. Nov 12, 2012

Lo.Lee.Ta.

Okay, now I'm confused... =_=

I am working on a problem that says:
A VW Beetle (m= 840kg) traveling at 9.6m/s collides with a stationary SUV (m= 1800kg) on a icy road. The beetle rebounds at 2.5m/s. Find:
a. What is the SUV'S velocity right after collision?
b. Is the final KE equal to, less than, or greater then the initial KE?

I am saying object 1 is the VW Beetle and object 2 is the SUV.

So the P1(initial)= (840)(9.6) = 8064

P1(final)= (840)(-2.5) = -2100

P2(initial) = 0

P2(final)= (1800)(vf)

Pi = Pf
8064 + 0 = -2100 + (1800)(vf)
vf = 5.65m/s <--- So this is the final velocity of the SUV

So to determine part B... It seems that just by looking at what happens in this problem, there is no object sticking together, therefore there is no inelastic collision.
It made me think there was elastic collision here...
But, obviously, that does not make sense if you work it out:

KE(initial) = KE(final)
1/2(840)(9.6^2) + 1/2(1800)(0^2) = 1/2(840)(-2.5^2) + 1/2(1800)(5.65^2)
38707.2J ≠ 30447.24J

It looks like the final kinetic energy is less than the initial kinetic energy...

So, clearly, the initial and final kinetic energies are not the same... and so it could not have been an elastic collision...

I guess it really was an inelastic collision, but how am I supposed to know beforehand?
I thought is was REQUIRED that the two objects stick together for an inelastic collision to occur.
I mean, these two objects are likely deformed, but I didn't think that would be enough to qualify it as an inelastic collision.... I thought they HAD to stick together...

6. Nov 12, 2012

SammyS

Staff Emeritus
Incorrect. If they stick together that would be a perfectly inelastic collision -- that's one in which the maximum amount of Kinetic Energy ia lost.
A collision can be inelastic without being perfectly inelastic. (In fact this one is just such a collision, and that may be part of what this problem was intended to demonstrate.)
Since the final KE is less than the initiel KE, that means that the collision is inelastic !

Did you not read the earlier posts?

7. Nov 12, 2012

Lo.Lee.Ta.

Thank you SammyS! :)

I gather that only in COMPLETELY inelastic collision do the objects colliding stick together. In completely inelastic collision, the maximum amount of kinetic energy is lost. The actual act of the two objects bonding together is mostly what uses up the maximum kinetic energy.

But for just PARTIALLY inelastic collision, the objects do not stick together but still lose some of their kinetic energy.

8. Nov 13, 2012

SammyS

Staff Emeritus
Yes. That's the idea.

As a technical term, I have only seen "perfectly inelastic collision" ,a not "COMPLETELY inelastic collision", even though the latter is a good description.