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Particle in a box, find the frequency of collisions

  1. Jul 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Given a particle of mass m in box of length L, how would you find the frequency of collisions on a given wall and the momentum of each collisions?

    2. Relevant equations

    3. The attempt at a solution

    Should I start with the equation for the root-mean-square-velocity? How can one derive the freqeuncy of collisions from that?
  2. jcsd
  3. Jul 20, 2007 #2
    Well, if initially the particle has velocity [tex]\vec{v}[/tex], then after the collision, its velocity should be [tex]-e\vec{v}[/tex], where e is the coefficient of restitution. Assuming that there is no other force acting on the particle (air resistance etc...), the particle would take time [tex]t=\frac{L}{ev}[/tex]. Hence, the frequency of collision is [tex]f=\frac{ev}{L}[/tex].

    After the first collision, the velocity was ev, after the second one it will be [tex]e^2v[/tex], and hence, after the nth one it will be [tex]\vec{v(n)}=(-1)^ne^n\vec{v}[/tex]. Therefore, the momentum after each collision will be mev, [tex]-me^2\vec{v}[/tex], and after the nth collision [tex]\vec{p(n)}=(-1)^nme^n\vec{v}[/tex].
  4. Jul 20, 2007 #3
    Hmmm. I asked the first part of the question hoping the second half would be obvious after that. But since it is not, let me ask the whole question:

    We are given a classical particle with energy E (from which we know its velocity [tex]\sqrt{2E/m}[/tex]. We want the frequency with which it collides with a given wall, the momentum transfer per collision, and finally the average force.

    So the frequency as you said is [tex]f=\frac{ev}{L}[/tex] and the momentum transfer (I think) would be -e, where e is the coefficient of restitution. So would the average force be:

    momentum transfer per collision * collisions / time


    I'm not sure but I do not think the answer should have a coefficient of restitution in it?
  5. Jul 21, 2007 #4
    I guess the momentum transfer per collision will be |p(n)|-|p(n-1)|. This would give you [tex]\del p(n)=m\vec{v}e^{n-1}(e-1)[/tex].

    The average force per collision should be momentum transfer per collision*frequency of collision.

    Perhaps there's another way to do this, but I think by this method, you WILL have the coefficient of restitution present in the solution expression.
  6. Jul 21, 2007 #5
    to get the answer in my book, I think you have to pretend the particle is bouncing back and forth with no vertical velocity component and that the collisions are perfectly elastic. Then,

    [tex] p = \sqrt{2Em} [/tex]

    and the change in momentum per collision is 2p and the frequency of collisions are 2Lm/p. So the average force would be

    [tex] \Delta p / \Delta t = 2E/L [/tex]

    Am I right about the need to make those assumptions?
  7. Jul 21, 2007 #6


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    I think you are right. This is probably a question intended to compare a quantum situation with classical, right?
  8. Jul 22, 2007 #7
    Yeah. Probably. But then I dont think the momentum transfer will be with the mod values, but with the vector values.
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