Odds of Collision between Incident Object and Target Objects

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AI Thread Summary
The discussion focuses on calculating the odds of an incident object colliding with target objects based on their sizes and density. The collision cross-section is defined as σ = π(r1 + r2)², and the probability of collision with a single target is given by P1 = σ/A. The probability of no collision is expressed as (1 - P1), and for multiple targets over a distance L, the total probability of no collision becomes (1 - P1)^(Ln). The mean free path is introduced as a relevant concept, indicating that the probability of no collision follows an exponential distribution, simplifying the calculations. The conversation emphasizes the need for careful consideration of units and limits in the calculations.
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Homework Statement



The problem is what are the odds of an incident object of radius r1 colliding with any of a collection of target objects of radius r2, where the r2 objects have a number density N / m^3 = n and the incident object travels a distance L. Incident object is moving much faster than the other objects so they can be considered still.


Homework Equations



Collision cross-section for collision between incident object and a single target is:

σ = \pi (r1 + r2)^2

Probability of collision for a single target object is
P1 = σ/A
were A is the total area of the domain in question.

Probability no collision for a single target object is 1-P1

Maybe relevant, the mean free path is
λ = 1 / nσ

The Attempt at a Solution



My thinking is, if probability of no collision for a single target is (1-p1), then if the incident object travels a distance L, the number of targets to consider is Ln. So the total probability for no collision is

(1-p1)^(Ln)

And probability of colliding with a single one of these is 1 minus this answer.

Is this correct?

I was also trying to use the mean free path but I wasn't sure how.

Thanks
 
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Your exponent has units, that needs another factor. And then you need the limit for an infinite area.

The probability to have no collision before length L is an exponential distribution, where the mean free path gives the factor in the exponent. That is easier to set up.
 
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