This might be a very basic concept but I have never been able to understand it.

Lets say a object with speed $v$ collides with another of same mass at rest in an perfectly inelastic collision in which the momentum. Why does then the both object join and move with speed $v/2$ ? If all the kinetic energy is lost then where does the energy to move the two blocks come from ? Also if I throw wet clay at my bedroom wall it does not move with the clay ?

I understand, plugging values in conservation of momentum equation gives this result.


In a perfectly inelastic collision, all the kinetic energy in the center of mass frame is lost. This is the key fact you are missing.

Problem 1: If you compute the problem of 2 equal masses, one a rest and the other moving to collision at speed $v$, in the center of mass frame, those 2 masses are moving at $v/2$ and $-v/2$ towards each other. The final state the 2 masses are "stuck" together and there's no kinetic energy. All the initial kinetic energy of $${1 \over 2} m ({v \over 2})^2+ {1 \over 2} m ({v \over 2})^2$$

is lost in the CM frame. You were analyzing the collision in a non-CM frame!

Problem 2 In the wet clay problem, you can model your wall as an infinite mass (ok, technically, the wall is glued to the earth, and so you could use the earth as the 2nd mass). Then for all intents and purpose, the center of mass frame is the earth, coz, well the earth is pretty heavy and dominates the CM. All the kinetic energy is lost in the frame of the wall/earth as you observe!

  • $\begingroup$ Can I determine how much is lost in non-CM frame ? $\endgroup$ – A---B Jun 16 '17 at 23:04

To slightly reformulate physicsdude's answer: a perfectly inelastic collision doesn't mean that all kinetic energy is lost - it means that the maximum possible amount of kinetic energy is lost that is compatible with the conservation of momentum.

Note that defining it to mean "all kinetic energy is lost" wouldn't be as nice of a definition, because the kinetic energy depends on your frame of reference, so whether or not a given collision is perfectly inelastic would also be frame-dependent under this definition.

  • $\begingroup$ Oh, it makes sense now. Don't know how I missed this even after attending classes on this topic. $\endgroup$ – A---B Jun 16 '17 at 23:02

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