Particle in a box, find the frequency of collisions

[ehrenfest]

Homework Statement

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?

Homework Equations

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?

 

[chaoseverlasting]

Well, if initially the particle has velocity \vec{v}, then after the collision, its velocity should be -e\vec{v}, 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 t=\frac{L}{ev}. Hence, the frequency of collision is f=\frac{ev}{L}.

After the first collision, the velocity was ev, after the second one it will be e^2v, and hence, after the nth one it will be \vec{v(n)}=(-1)^ne^n\vec{v}. Therefore, the momentum after each collision will be mev, -me^2\vec{v}, and after the nth collision \vec{p(n)}=(-1)^nme^n\vec{v}.

 

[ehrenfest]

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 \sqrt{2E/m}. 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 f=\frac{ev}{L} 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?

 

[chaoseverlasting]

I guess the momentum transfer per collision will be |p(n)|-|p(n-1)|. This would give you \del p(n)=m\vec{v}e^{n-1}(e-1).

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.

 

[ehrenfest]

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,

p = \sqrt{2Em}

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

\Delta p / \Delta t = 2E/L

Am I right about the need to make those assumptions?

 

[Dick]

I think you are right. This is probably a question intended to compare a quantum situation with classical, right?

 

[chaoseverlasting]

Yeah. Probably. But then I don't think the momentum transfer will be with the mod values, but with the vector values.