# Particle in a box, find the frequency of collisions

1. Jul 19, 2007

### ehrenfest

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. Jul 20, 2007

### 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 $$\vec{p(n)}=(-1)^nme^n\vec{v}$$.

3. Jul 20, 2007

### 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?

4. Jul 21, 2007

### 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.

5. Jul 21, 2007

### 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?

6. Jul 21, 2007

### Dick

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

7. Jul 22, 2007

### chaoseverlasting

Yeah. Probably. But then I dont think the momentum transfer will be with the mod values, but with the vector values.