Exploring the Concept of Conservation in Elastic and Inelastic Collisions

In summary, the problem is that you cannot determine whether or not a collision between a ball and a wall is elastic or inelastic. The only option is to consider it to be completely inelastic.
  • #1
Alem2000
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I had a conceptual question. Say I have a ball and I throw it against the wall and it bounces back. I have initial values for its velocity but not for the final velocity using the [tex]\vec{P}=m\vec{v}[/tex] I noticed that momentum is not conserved. Well if momentum is not conserved then that means it is not inelastic or elastic and the only option left is completely inelastic but..the wall can't move with the ball...whats going on? Is it just that you can't determine these kinds of collisions? Or do I have some misunderstandings of the whole concept of the elastic or inelastic collisions?
 
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  • #2
Alem2000 said:
I noticed that momentum is not conserved. Well if momentum is not conserved then that means it is not inelastic or elastic and the only option left is completely inelastic
I'm not sure about the wording, but you basically have 3 choices: completely elastec, completely inelastic, and somewhere in between.
but..the wall can't move with the ball...whats going on?
Can't it? Why not? There is no such thing as a perfectly rigid object.

Also think about what else is happening: is there any other energy that you can account for during the impact?
 
  • #3
The wall exherts a force back on the ball, so would that mean that the wall moves a tad bit? All I know is that there is a very small [tex]\Delta{t}[/tex] where there is contact btween the ball and the wall there is a mass and an initial velocity [tex]\vec{v_0}[/tex].Also I know the impulse, using the [tex]\vec{J}\int\sum\vec{F}dt[/tex] I found all the values that I needed. I saw that the initial [tex]KE[/tex] does not equal the final...and the same with the initial [tex]\vec{P}[/tex] it can't be elastic or inelastic but there would only leave 2 other option 1) I can't know 2)completely inelastic

completely inelastic would mean that [tex]m\vec{v_0}=(M+m)\vec{v_f}[/tex]
looking at it in terms of the equations I doesn't make sense..but it would if the the wall moved a littele...you said that there is no perfectly rigid body so a completely inelastic collision would have to be the safest answer..? Is that the correct reason or answer?
 
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  • #4
you'll get a small amount of deformation in the wall at the time of impact. think about what happens to the ball at impact though.
 
  • #5
It will experience a force from the wall, an equal force in magnitude.
 
  • #6
isn't that where e, the coefficient of restitution comes into play?
 
  • #7
Alem2000 said:
It will experience a force from the wall, an equal force in magnitude.
might that force cause the ball to deform as well?

as russ said earlier, any time you're talking about collisions in a problem, you have to specify what type of collision you are going to be working with. if your problem deals with elastic collisions, or if you have defined your objects to be completely rigid and non-deformable, then momentum and energy are conserved for your collision problem.

if your problem deals with inelastic collisions, or if one or more of your objects are deformable, then momentum is not conserved (energy is always conserved). some of that momentum/energy is taken up deforming your object(s).

you cannot talk about collisions without specifying the type of collision you want to work with, or the properties of your colliding objects.
 
  • #8
Yeah, the problem of approaching the collision problem from the momentum side of things is that you need to now the characteristics of the collision and can't, at least as I see it, link it to the behavior of the problem itself... directly at least if you don't have any information on the actual event. If you'd want to incorporate the real deal what is going on in the collision you would have to do a dynamic contact analysis, where the deformations subjected to both the wall and the ball would be accounted for. That, however gets really easy really complex and at this point in time the usual way to solve it is to numerically solve Newton's 2nd incorporating stiffness of both parties (in this case if the ball is as soft as they usually are, in nonlinear sense), in some cases even damping would come to play.
 
  • #9
For an efficient ball (like a superball or a tennis ball) and a sufficiently smooth and dense, sturdy wall, I think you could approximate conservation of momentum to be true. There will be less afterwards but it is a small change compared to the total energy, if you throw it hard.
 
  • #10
All I know is this! I was given a problem I solved it and then I was asked what kind of collision it is. I am not making this problem up for "fun" it was given to me. The final and initial momentum are different (by a significant amount) and so is the kinetic energy :cry: what to do?
 
  • #11
Alem2000 said:
The final and initial momentum are different (by a significant amount) and so is the kinetic energy :cry: what to do?
if momentum is not conserved, then it is an inelastic collision.
 
  • #12
imabug said:
if momentum is not conserved, then it is an inelastic collision.

Er... no. If the kinetic energy is not conserved, then it is an inelastic collision. If the momentum is not conserved, then there is an external force acting on the system.

Zz.
 
  • #13
ZapperZ said:
Er... no. If the kinetic energy is not conserved, then it is an inelastic collision. If the momentum is not conserved, then there is an external force acting on the system.

Zz.
duh, that's right. my mistake. thanks for correcting me.
 

1. What is conservation of momentum in elastic and inelastic collisions?

Conservation of momentum states that the total momentum of a closed system remains constant before and after a collision. In elastic collisions, both momentum and kinetic energy are conserved, while in inelastic collisions, only momentum is conserved.

2. How does the conservation of momentum apply to elastic and inelastic collisions?

In elastic collisions, the total momentum of the system remains constant, meaning that the sum of the momenta of the objects before the collision is equal to the sum of the momenta after the collision. In inelastic collisions, the total momentum of the system is also conserved, but some kinetic energy is lost due to the objects sticking together or undergoing deformation.

3. What is an example of an elastic collision?

An example of an elastic collision is when two billiard balls collide on a pool table. The total momentum of the system remains the same before and after the collision, and both balls bounce off each other without any energy loss.

4. What is an example of an inelastic collision?

An example of an inelastic collision is when a car crashes into a wall. The total momentum of the system is conserved, but some kinetic energy is lost due to the deformation of the car and the energy used to crush the wall.

5. How do you calculate the velocities of objects after a collision?

To calculate the velocities of objects after a collision, you can use the conservation of momentum equation: m1v1i + m2v2i = m1v1f + m2v2f, where m1 and m2 are the masses of the objects, v1i and v2i are the initial velocities, and v1f and v2f are the final velocities. This equation can be solved for the final velocities by knowing the initial velocities and the masses of the objects.

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