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I am familiar with the notion of an elastic and inelastic collision. However, I have an issue. For a combined collision:

For 2 masses, $m_1, m_2$, with known velocities $v_1,v_2$, the initial and final momenta are: $m_1u_1+m_2u_2=(m_1+m_2)v$. Therefore the initial and final energies are: $\frac{1}{2}(m_1u_1^2+m_2u_2^2), \frac{1}{2}(m_1+m_2)v^2$. These are not always equal.

This is because due to the C.O.M, $v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$

Hence doing some rearranging with both equations, the energy is conserved iff, $m_1u_1+m_2u_2=m_1u_1^2+m_2u_2^2$ In other words, not very often.

How is it that these equations inherently produce inelastic collisions? Do objects have a theoretical energy loss based on velocity and mass alone? This doesn’t seem to make sense as these are just theoretical equations based on ideal scenarios, so I do not know why there is an energy loss incorporated into these equations.

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    $\begingroup$ Your momentum equation is correct only if the masses stick together after the collision, which is of course is inelastic. More often the masses separate after collision and have different different velocities which allows an elastic collision theoretically. Elastic collisions are not realistic, though, but are mathematically possible if you use the correct equations. $\endgroup$
    – Bill Watts
    Commented Feb 5, 2020 at 8:43
  • $\begingroup$ @BillWatts Elastic collisions are not realistic Nope, it's pretty much realistic, take a look at collisions between billiard balls. $\endgroup$
    – Agnius Vasiliauskas
    Commented Feb 5, 2020 at 10:59
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    $\begingroup$ Your equation after the "iff" is dimensionally inconsistent. You must have made a mistake somewhere when deriving it. $\endgroup$
    – Ruslan
    Commented Feb 5, 2020 at 11:27
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    $\begingroup$ In an elastic collision both energy and momentum are conserved. In the uni dimensional case, this means that given the initial velocities, the final velocities are determined by the two equations. If you put an additional constraint, such as the two masses move together or the final speed of one of the masses, then only conservation of momentum would be satisfied, and the amount of energy loss can be automatically determined $\endgroup$
    – user65081
    Commented Feb 5, 2020 at 14:37
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    $\begingroup$ @AgniusVasiliauskas I know that, and you know that, but does a novice young physics student know that? (they do now! :-) ) $\endgroup$
    – garyp
    Commented Feb 5, 2020 at 18:13

2 Answers 2

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Kinetic energy is a frame-dependent quantity. However, due to momentum conservation, the center-of-mass frame is the same before and after the collision, and we can use it to analyze the situation.

Before the collision, we have two masses moving towards the center, ie non-zero kinetic energy. After the collision, the two masses rest at the center, ie zero kinetic energy. That energy must have been transferred somewhere (deformation of the bodies, heat, ...).

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Momentum conservation law applies to both - elastic & inelastic collisions as : $$ \frac {d}{dt}\sum p_{i}=0 $$ Where $p_i$ is i-th body momentum. It shows that total system's momentum is $\textrm{const}$ in a closed system.

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