Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How do air molecules rebound?

  1. Aug 25, 2008 #1
    Hi, so glad to be here. This is my first posting. I'm just a 61 year-old dummy, so please help!

    If a rubber ball is dropped to a hard surface the impact will compress the rubber, storing some of the kinetic energy as potential elastic energy. The rebound will come from this stored potential elastic energy. If the ball is dropped straight down I presume this elastic force is stored in the ball mostly in an axis normal to the ground, ie. the ball is smushed out making it just a little fatter at its' "equater" than at its' poles (the axis of impact).

    If I've got this bouncing ball thing right so far, can this idea of storing potential energy mostly along the axis of impact be carried over to air molecules? Do air molecules impacting a surface, like the walls of a containing vessel, hold potential energy (for the rebound) mostly in the axis of impact (normal to the impact surface)?

    Let's change the world.....does anybody out there know the answer?

    Thank you in advance!
  2. jcsd
  3. Aug 25, 2008 #2


    User Avatar
    Science Advisor

    If a ball rebounds from the ground perfectly (ie. rebounds to the same height from which it was dropped), then all the energy of ball is completely kinetic just before and just after it hits the ground. The point at which it hits the ground is complex, and perhaps potential energy is stored in the material of the ball for a very short time. So if there is potential energy stored in the material of the ball, it is only at the point of impact.

    Just after impact, the energy of a perfectly rebounding ball entirely kinetic. As it rises in height, the kinetic energy is converted to potential energy - but this potential energy is not stored in the material of the ball, it is stored in the ball by virtue of its height in the gravitational field.

    So it's the same for an air molecule, any potential energy stored by "compressing" it occurs only at the point of impact with the walls of the container or with other air molecules. It's a good first approximation to think of the energy of air as entirely kinetic. A tall column of air has potential energy, but again, that's just because things held high above the ground have gravitational potential energy.

    Actually, it's not clear that the smushing (away from the axis of impact) of the material at the point of impact aids a perfect rebound. If the compression occurs in the wrong way, then some of the kinetic energy of the ball is converted into heat, so the ball will rebound to a lesser height, but will feel slightly warmer after the rebound. Heat is the vibration of the material of the ball in random directions, so for there to be no loss of energy as heat, you are right that the compression should be along the axis of impact.

    Another way to get a perfect rebound is to make the ball of material so hard that it cannot have any internal compressions or vibrations at all, so it can never be heated up. You should also make the ball very small, because internal rotations will take away from the rebound. If you use a perfectly hard material for your ball, you still may not get the perfect rebound, because the ground must also be perfectly hard for that to work. So this is very impractical, and I think only theorists use such prefectly rebounding balls.
  4. Aug 25, 2008 #3


    User Avatar
    Homework Helper

    A better analogy might be the fully elastic interation of two opposing magnets approaching each other on a firctionless surface. They will not physically collide, but there will be a collision like reaction, with preservation of momentum and of kinetic energy. Even without a collision, the opposing forces will result in a compression at the inner surfaces of the two magnets.

    Regarding the relastionship between speed and temperature:


    There's a really neat mathematical equation based on a theorem called
    the "equipartition theorem" which states that the energy of a gas system
    (equal to 1/2*mv^2) is equal to the temperature of the gas (equal to 3/2*kT).
    If we rewrite this equation to solve for velocity we get:

    sqrt(3*T*k/m) = v

    where T is the temperature in Kelvin, k is the Boltzman constant = 1.3805*10^-
    23 J/K and m is the mass of the gas particle.

    If we assume that the average mass of air (since it is a mixture of different
    gases) is 28.9 g/mol (or each gas particle is around 4.799*10^-26), and room-
    temperature is 27C or 300K, we find that the average velocity of a single air
    particle is around 500 m/s or 1100 miles per hour!


    Note that this average speed of air molecule is much faster than the speed of sound
    at a specific temperature, since the speed of sound is based on the speed of propgation
    of collisions, instead of the speed of the molecules. At 27C, the speed of sound in
    the air is 347 mps or 777mph.
    Last edited: Aug 25, 2008
  5. Aug 25, 2008 #4
    If i have understood your question you are trying to apply the mechanism of rebound you have just analyzed at macroscopic scale to a microscopic scale, for example air molecule instead of a rubber ball. I think the main difference for microscopic object, such as molecule, atom or particle, is that you cannot always speak of "compression" because it may happens that the object dosent allow any state "more compressed" that one in which is. However the mechanism is similar: you have exchange of energy from kinetic to an interaction's potential and then release.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook