The Emperor
Harmless
Posts: 4
Registered: 27-4-2015
Member Is Offline
Mood: No Mood
|
|
Kinetics Homework Help
Hey folks, I've had a little problem that I needed some help with.
1. As two particles approach each other, what happens to their kinetic energy?
What about their potential energy?
I thought that the particles kinetic energy would increase because of the rebounding motion.
As for potential energy, I think that it would decrease.
Thanks for any help
|
|
|
j_sum1
Administrator
Posts: 6326
Registered: 4-10-2014
Location: At home
Member Is Offline
Mood: Most of the ducks are in a row
|
|
You did that well violet sin.
I wasn't even sure what the question was.
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Assuming elastic and frictionless collisions, basically two Laws govern collisions: Conservation of Energy and Conservation of Momentrum (a vector).
The sum of the potential and kinetic energies of the particles is the same before and after the collision:
Σ(E<sub>k</sub> + E<sub>p</sub> <sub>before</sub> = Σ(E<sub>k</sub> + E<sub>p</sub><sub>after</sub> (sum over all particles)
Similarly the sum of the momenta (product of mass and velocity, m x v) of both particles is the same before and after the collision:
Σ(m x v)<sub>before</sub> = Σ(m x v)<sub>after</sub> (sum over all particles)
In principle these equations are sufficient to determine the post-collision trajectories and all energies of the particles taking part.
In reality the complexity of the solution depends on the specific geometry of the initial trajectories, whether potential energy plays a part or not
(in billiards e.g. it does not) etc and each problem needs to be looked at case-by-case. In 2 or 3 dimensional problems, speeds and momenta have to be
treated as vectors.
A head-on, two-particle, one dimensional collision on an axis parallel to the Earth's surface is more or less the simplest case and boils down to a
system of two simultaneous equations with two unknowns (the post-collision velocities of both particles - this simple case can be found here: http://en.wikipedia.org/wiki/Elastic_collision#One-dimension...). Even simpler is the scenario in which one particle is stationary, prior to the
collision.
I'm not keen on the pendulum analogy because it's quite a different problem, although it does illustrate conservation of energy (potential to kinetic
and vice versa) very well. But there's no conservation of momentum.
If the OP specifies the problem in greater detail, I'm willing to try and treat it specifically.
[Edited on 28-4-2015 by blogfast25]
|
|
|
Hexavalent
International Hazard
Posts: 1564
Registered: 29-12-2011
Location: Wales, UK
Member Is Offline
Mood: Pericyclic
|
|
I must firstly confirm that you are indeed talking about particles in a chemical sense; many mathematics problems model macroscale objects as
"particles" - smooth, spherical objects - which obviously behave differently to particles in a chemical sense.
If you are referring to the latter case, the kinetic energy changes associated with the particles moving towards one another may depend on the nature
of the particles themselves, i.e. if there is any electrostatic attraction or repulsion between them. For example, even in species where there is no
permanent electrostatic force between particles, such as in diatomic element molecules such as Cl2 or lone noble gas atoms, temporary
dipoles can cause neighbouring particles to accelerate towards them, by the formation of an electrostatically attractive induced dipole.
If the problem is simpler than this and assumes no such interactions, then blogfast is correct. If you are required to work out kinetic energy values,
if you can determine velocities and have knowledge of the particles involved, you can use the equation Ek=0.5mv<sup>2</sup> to
find them.
With a knowledge of the coefficient of restitution of the particles, you can determine velocities using Newton's Law of Recovery alongside the
principle of conservation of momentum.
[Edited on 28-4-2015 by Hexavalent]
"Success is going from failure to failure without loss of enthusiasm." Winston Churchill
|
|
|
The Emperor
Harmless
Posts: 4
Registered: 27-4-2015
Member Is Offline
Mood: No Mood
|
|
I thank you a lot for that great explanation, but could you kind of simplify it down a little?
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Can you specify what you mean by 'particles'? And what kind of potential energy are you referring to? Gravitational? Coulombic (electrostatic
repulsion/attraction)? Magnetic repulsion/attraction?
These problems are never simple.
For macroscopic, non-charged, non-magnetic rigid objects, the explanation I gave above holds.
[Edited on 29-4-2015 by blogfast25]
|
|
|
The Emperor
Harmless
Posts: 4
Registered: 27-4-2015
Member Is Offline
Mood: No Mood
|
|
By particles I mean molecules.
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
For non-charged molecules in non-relativistic conditions (v <<< c) collisions are roughly ruled by what I wrote higher up.
Molecules contain electron clouds, which when the molecules get very, very close, cause temporary polarisation of these molecules, in turn leading to
'sticky' collisions.
Some molecules are permanent dipoles, increasing that 'stickiness'.
But these electrostatic interactions only take place at very close range, because electrostatic repulsions/attractions decrease/increase with the
inverse of the square of the distance (1/r<sup>2</sup> between the
charged particles.
|
|
|
j_sum1
Administrator
Posts: 6326
Registered: 4-10-2014
Location: At home
Member Is Offline
Mood: Most of the ducks are in a row
|
|
Quote: Originally posted by blogfast25  |
For non-charged molecules in non-relativistic conditions (v <<< c) collisions are roughly ruled by what I wrote higher up.
Molecules contain electron clouds, which when the molecules get very, very close, cause temporary polarisation of these molecules, in turn leading to
'sticky' collisions.
Some molecules are permanent dipoles, increasing that 'stickiness'.
But these electrostatic interactions only take place at very close range, because electrostatic repulsions/attractions decrease/increase with the
inverse of the square of the distance (1/r<sup>2</sup> between the
charged particles. |
It's worse than inverse square. That applies to charges spheres. But molecules have dipoles -- a region of localised positive charge and a region of
negative charge. So, while a positively charged region of one molecule may be attracted to the negative of another, it is simultaneously repelled by
the corresponding positive acting at a slightly greater distance. Only at very close proximities is there a significant difference between the
attraction and repulsion forces.
For a first approximation the relationship works out to be an inverse cube law although it obviously depends on orientation. Magnetic dipole is also
often cited as following an inverse cube law.
|
|
|
The Emperor
Harmless
Posts: 4
Registered: 27-4-2015
Member Is Offline
Mood: No Mood
|
|
I was aiming for a concise answer to my question, as I'm a little confused now.
My original question was what happened to the kinetic and potential energy when two molecules approached eachother?
p.s what happened to violent sin's comment?
|
|
|
j_sum1
Administrator
Posts: 6326
Registered: 4-10-2014
Location: At home
Member Is Offline
Mood: Most of the ducks are in a row
|
|
Without knowing what potential energy you are referring to, it is difficult to answer. If you are just thinking of theoretical particles that behave
in (ideal) billiard ball fashion then there really is no potential energy to speak of. (On a micro scale, the mechanism of collision is caused by a
very quick compression of the billiard ball material but it is not stored for any great length of time.)
In an ideal gas, the particles are assumed to have perfectly elastic collisions -- ie, no energy los, no build up of any kind of potential energy.
So, if I understand you correctly,
Kinetic energy is conserved
Potential energy -- there is none.
Beyond that we are talking about specific scenarios or contexts which you haven't provided.
|
|
|
alexleyenda
Hazard to Others
Posts: 277
Registered: 17-12-2013
Location: Québec, Canada
Member Is Offline
Mood: Busy studying chemistry at the University
|
|
I would complete by suggesting that you read about the Lennard-Jones potential which sums up the variation of potential energy depending of the
distance between 2 not charged atoms. When the total energy (kinetic + potential) is lower than the 0, the sum of the potential and kinetic energy
will not allow the molecule to get out of the well. Let's say the particules come close together and have a kinnetic energy of less than the 0 on the
graphic, they are going to be stuck in that well and oscillate from one side to the other of the well, there is a chemical bond between the two. If
the kinnetic energy is higher than the 0, it will transform into potential energy equal to the quantity of kinetic energy there was, getting as close
as the graphic says the distance between the atoms/molecules is with that potential energy, then all the potential energy gets transformed back into
kinetic energy and the atoms/molecule bounce away from each other.
Here is a clear link with the graphic i'm talking about.
http://chemwiki.ucdavis.edu/Physical_Chemistry/Intermolecula...
[Edited on 29-4-2015 by alexleyenda]
Help us build the Sciencemadness Wiki! Every question and tips about amateur chemistry two clicks away, wouldn't that be awesome?!
sciencemadness.org/smwiki
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
