Impact Parameter: Solving Collision Problems

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Collision problems can be analyzed using conservation of momentum and energy, impulsive forces, and the impact parameter, which is the perpendicular distance between the centers of mass of colliding particles. When given the impact parameter, angular momentum conservation allows for separating radial and angular motion, with angular momentum treated as an effective potential in the radial direction. The effective potential acts like a fictitious force that pushes particles away from the center of mass, particularly in non-colliding scenarios. For forces following a 1/r^2 relationship, such as in Rutherford scattering, existing solutions can be applied. Understanding how to calculate the impact parameter or scattered angle is crucial for solving these problems effectively.
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Any problem of collision can be solved using four equations: Conservation of momentum and energy, impulsive forces and impact parameter (distance between center of masses of the particle taken perpendicular to velocity). I can solve problems using first three cases. But how to analyse a problem when impact parameter is given as data. ??
 
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First of all, angular momentum is conserved, so you can separate the motion into radial motion and angular motion. You can calculate the angular momentum from the impact parameter. You can solve the radial motion separately by treating the angular momentum as an effective potential acting in the radial direction. Find the minimum distance, and integrate the forces from infinity to the minimum distance. It's just some messy math.

Of course, the solution is already available for 1/r^2 forces (Rutherford scattering).
 
Ok. I got the first part. But can you please elaborate "effective potential" stuff a little more.
 
The effective potential is basically another way of writing the centrifugal force. It is a fictitious force which pushes stuff away from the center where r=0. If you consider two particles which pass beside each other without colliding and you write the equations of motion for one particle with respect to the center of mass frame, you see
dx/dt = dy/dt = dz/dt = 0, because there are no forces because these particles don't collide. But if you look at dr/dt, dr/dt > 0. There is an effective potential that pushes the particle away from the center of mass. see http://en.wikipedia.org/wiki/Effective_potential
 
See Section titled Determining the Impact Parameter in

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html

Plugging values of nuclear Z, the scattered angle, and the incident alpha particle energy into the empty boxes is sufficient to determine the dependent variable impact parameter b. So either the impact parameter or the scattered angle can be used as an input parameter.
 
Thanks
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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