I would like to know the complete equations for a collision of two perfectly elastic/rigid balls. I have seen something along these lines:

$$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$$

but I think there is missing one parameter, namely how much one ball touches the other.

In other words, if the collision is direct or on the other hand one ball touches the other only very slightly/indirectly. The equation above doesn't seem to incorporate this parameter. Are any items in this equation vectors rather than scalars ?

Can here the tensors be used ? They really have enough indices to be useful, but I'm not sure.

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    $\begingroup$ momentum is conserved in a closed system,including during a collision, the details do not matter as they are related to internal forces that do not contribute to the change in total momentum $\endgroup$ – Wolphram jonny Jul 31 '20 at 19:22
  • $\begingroup$ @Wolphramjonny I have read your answer several times, but I do not think I've got it. How is it that they do not contribute ? $\endgroup$ – user2925716 Jul 31 '20 at 19:31
  • $\begingroup$ there is theorem proving that internal forces do not change the moment of the CM. you cam find it on Wikipedia $\endgroup$ – Wolphram jonny Jul 31 '20 at 19:47
  • $\begingroup$ @Wolphramjonny What is CM ? Can you provide me with the appropriate link to wiki ? $\endgroup$ – user2925716 Jul 31 '20 at 19:50
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    $\begingroup$ center of mass.I could not find the demonstration, but basically says that the momentum of a system does not change in absence of external forces $\endgroup$ – Wolphram jonny Jul 31 '20 at 20:12

Here, we are talking about conservation of momentum for 2 rigid/elastic spherical bodies.

Conservation of momentum means that if there is no net external force acting on the system then the total momentum of the system remains constant.

In the case of rigid spherical bodies, the spheres come in contact with each other only at one point. This is because rigid bodies do not deform and two spheres (or even circles) can touch each other only at one point. Hence there is no question of how they touch each other.

But that was the case for perfectly rigid bodies, the case is not exactly same for perfectly elastic bodies. The important property of perfectly elastic bodies is that, they do not store/absorb any energy during deformation when they come in contact with each other. hence effectively, they can be treated as rigid bodies as they do not deform, hence don't store/absorb energy due to deformation. Alternatively, perfectly elastic bodies can be imagined as two rigid bodies, one of which connected by a spring (figure) elastic bodies

The important thing to understand here is that if there is no external force acting on the system, Total Momentum Of The System Is Conserved. But that is not always the case with Total Energy Of The System.

When there is collision between two not perfectly rigid/elastic bodies, Momentum of the system is conserved if there is no external forces acting, same as elastic collisions. But, if we look at the total energy of the system, before and after the collision, the energy before collision may not be equal to the energy after collision as some energy is lost in deformation of the bodies due to contact or sound or heat, etc. Hence, total mechanical energy is not conserved in this case, but momentum is still conserved.

I think,by 'how much one body touches the other' you mean, how much deformation is happening to each of the body and how do we account for the energy lost in the equations. To account the energy lost in the deformation, we use something called as coefficient of restitution. it is the ratio of initial relative velocity if the bodies to final relative velocities of the bodies. $$V_1 - V_2/V_1' - V_2'$$ You can see more about centre of mass and collisions from these lectures, there are really good experiments and explanations in these. COM/Momentum Collisions

  • $\begingroup$ What is a physical example of Completely elastic collision ($Q=0$) in the time 4:08 in this video ? $\endgroup$ – user2925716 Aug 8 '20 at 20:16
  • $\begingroup$ @user2925716 Cases where change in kinetic energy is 0, is not naturally possible if we take collision of two balls and collide them, since some energy will always be lost in various forms like deformation of balls, sound, heat, etc. But, at a bsic level to ignore all that, we assube change in ke = 0 most of the time $\endgroup$ – Harshavardhan Hajeri Aug 9 '20 at 2:13

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