Is mass important in calculations when two particles are collide? or is only the charge that matters? is collision in microscopic world the same with macroscopic world?

  • $\begingroup$ please note that at the level of nuclear and particle physics it is special relativity and quantum mechanics that hold. One uses the four vectors of special relativity to define masses.hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html $\endgroup$ – anna v Jan 20 '19 at 17:01

Momentum is important in the calculations involving collisions/interactions. For isolated systems, i.e., systems which you can ignore any impulse transfers into or out of the system during the time frame of the collision, the total momentum of the system will remain constant at every instant. For matter particles (not photons), the momentum depends on the mass.

If there is no outside influence on the system, i.e., nothing is doing work on the system, then the total energy will remain constant, but the kinetic energy may change if there is permanent deformation of any of the bodies (among other possible things such as internal work changing the temperature of one of the bodies). Changes in temperature or kinetic energy are also mass dependent.

For both macroscopic and microscopic collisions, momentum and energy are important and are conserved and possibly constant within the particle system, but the interactions of the bodies are different. For near-elementary particle or nucleus collisions (alphas, heavy ions), the interaction may be electromagnetic or nuclear or weak. For macroscopic collisions, the interaction is electromagnetic but we simply say "contact."

Bottom line: Yes, mass is important in collision calculations.

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  • $\begingroup$ thanks for the answer...does this apply also on gold foil experiment ? $\endgroup$ – Antonios Sarikas Jan 19 '19 at 19:09
  • $\begingroup$ @adosar Yes. The forces in Geiger-Marsden experiment are Coulomb forces. But the gold atoms are trapped in place because gold is a crystal, so effectively the momentum looks like it's not conserved (but it really is). Angular momentum remains constant and KE+PE remains constant. $\endgroup$ – Bill N Jan 20 '19 at 0:40

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