inelastic collisions and potential energy

[Clarphimous]

Okay, the problem is this. An object runs into a wall and bounces off at a slower speed than it began with. The floor is assumed frictionless. So some of the kinetic energy is dissipated. I need to find out how much of each form of energy (kinetic, potential, dissipated) there is when the object is stopped against the wall. In order to know that, I need a knowledge of when during a collision the energy is dissipated.

Of course, there is no kinetic energy at that point. Just potential energy and possibly dissipated energy.

Help?

[Mantikore]

i just did this today in mechanics. though i still need time to absorb it.


anyway, we know kinetic energy is 0.

theoretically, IF the collistion were perfectly elastic, dissipated energy would also = 0, meaning the ball will bounce back at the same speed as if it were dropped.

since it dropped slower, could you somehow work out the potential energy of the ball when it is released, and subtracted it by the potential energy at the apex of the bounce, then you can get the dissipated energy.

by the way, did they give you any values? such as the initial height

[Clarphimous]

Alright, here's the exact problem:

http://i36.tinypic.com/1hdlcz.png

Part (b) (3).

[spelling_bee]

Mantikore's right.
Since energy isn't lost by friction, the energy at each step would be:
E=0.5*0.2*(2.5^2)
=0.625J

So if it only has KE of 0.4J after the wall, the dissipated energy would be 0.225J.
At the moment of hitting the wall, the 0.4J would all be PE.

[Clarphimous]

Mantikore's right.
Since energy isn't lost by friction, the energy at each step would be:
E=0.5*0.2*(2.5^2)
=0.625J

So if it only has KE of 0.4J after the wall, the dissipated energy would be 0.225J.
At the moment of hitting the wall, the 0.4J would all be PE.

It isn't the moment of hitting the wall that I need, it's the moment at which it has stopped. The object doesn't stop instantaneously; rather, the object and wall compress like springs until they come to a stop. So it's still mostly kinetic energy at the instant it touches the wall.

What I need to find out is when during this compression/extension of the collision is the energy dissipated, specifically the amount dissipated when the object is stopped.

edit: if you were measuring the heat/sound dissipation of an actual inefficient spring, it would be throughout both compression and extension, right?

[Mantikore]

a) kinetic energy should be dead easy. just plug in the values.
E=|0.5mv^2|, where E is in J, m is in Kg and v is in m/s

so initial KE = |0.5*0.2*2.5^2| = 0.625 J
final KE = |0.5*0.2*2^2| = 0.4 J

KE at the moment of impact is 0 since it changes direction
Most of the KE is transformed into potential energy. The rest is lost as heat, sound, damage to the wall etc

b)
1)
K
P=====
D
T=====

2)
K=====
P
D
T=====

3)
K
P====
D=
T=====

4)
K====
P
D=
T=====

where K = kinetic, P = potential, D = dissipated, T = total


e)impulse is just change of energy isnt it?

so 0.625 - 0.4 = 0.225

ill do the other two questions later

edit; actually, im reading the above post, the doesnt tell you if the objects are rigid. if they arent, you have to take the compression of the block into consideration, which i think we dont have enough information for

[Clarphimous]

a) kinetic energy should be dead easy. just plug in the values.
E=|0.5mv^2|, where E is in J, m is in Kg and v is in m/s

so initial KE = |0.5*0.2*2.5^2| = 0.625 J
final KE = |0.5*0.2*2^2| = 0.4 J

KE at the moment of impact is 0 since it changes direction
Most of the KE is transformed into potential energy. The rest is lost as heat, sound, damage to the wall etc

b)
1)
K
P=====
D
T=====

2)
K=====
P
D
T=====

3)
K
P====
D=
T=====

4)
K====
P
D=
T=====

where K = kinetic, P = potential, D = dissipated, T = total


e)impulse is just change of energy isnt it?

so 0.625 - 0.4 = 0.225

ill do the other two questions later

edit; actually, im reading the above post, the doesnt tell you if the objects are rigid. if they arent, you have to take the compression of the block into consideration, which i think we dont have enough information for

Impulse can be defined in two ways. It is the change in momentum. It is also the integral of the force with respect to time, which can be estimated by multiplying the average force times the amount of time.

If the object and wall were "perfectly rigid", the object would never stop but rather suddenly be going in the opposite direction after touching the wall. For anything less, the collision takes a certain amount of time to accelerate the object in the opposite direction. It may be a very, very short period of time, but any real collision with macroscopic objects takes time to reverse directions (or stop, if the collision is perfectly inelastic).

It turned out that my teacher didn't really care about (b) (3), as long as all the energy were both in potential and dissipated. Still, I'm thinking that since collisions act as springs, and inefficient springs would dissipate energy throughout both stretching and compressing, that part of it would be dissipated as the object came to a stop, and the rest dissipated as it accelerates away from the wall. But it probably wouldn't be in equal amounts since the velocities are different in the two different parts, and it would involve a whole bunch of other mechanics that I don't want to get into yet.