Conservation of momentum/kinetic energy in inelastic collisions.by Sweeney Tags: inelastic collision, kinetic energy, momentum 

#1
Dec1912, 01:09 PM

P: 5

I'm finding it difficult to understand how momentum is conserved in an inelastic collision, but kinetic energy isn't. I understand some energy leaves the system in the form of sound and heat.
If we look at the momentum of the system before and after the collision, P=mV, they are equal, due to CoM. Assuming a constant mass we can also say that the velocity of the system is constant. Now looking at kinetic energy, E=(1/2)mv^2 we know that it changes. Again assuming a constant mass we can say that v^2 changes and thus v changes. Obviously these two statements contradict each other, but I don't understand where I went wrong. 



#2
Dec1912, 01:27 PM

P: 647

I'll try and think of a good way to put it... but in response to your first line, inelastic collisions are defined by a nonconservation of kinetic energy. That being said, is your question really, "why is kinetic energy conserved in some, but not all, collisions?"




#3
Dec1912, 01:40 PM

Mentor
P: 40,890





#4
Dec1912, 01:41 PM

P: 749

Conservation of momentum/kinetic energy in inelastic collisions.It would seem that you are taking m to denote the sum of the masses of the colliding objects and V to denote the velocity of the centerofmass of the colliding objects. For momentum, that works out because P(total) = m(total) * V(average) For energy, this does not work because E(total) != 1/2 m(total) * V(average)^2 Let's take an more familiar example... Suppose that you have a pile of building blocks. They are all shaped like cubes. Some are big and some are little. If you pile all the blocks on top of one another you can determine the height of the tower by multiplying the average size of each block by the number of blocks. But you cannot correctly determine the total volume of the blocks by taking the average size the blocks, cubing that and multiplying by the number of blocks. The average of the cubes is not equal to the cube of the average. The average of the squares is not equal to the square of the average. The average of the inverse is not equal to the inverse of the average. Averages don't commute in general. That's why there are things like the "harmonic mean" or the "root mean square". 



#5
Dec1912, 02:21 PM

P: 5

Absolutely perfect answer! I finally can understand it.



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