Inelastic collisions -- how is momentum conserved but not energy?

[richengle]

Homework Statement

in an INELASTIC collision, how is momentum conserved, but not energy?

Relevant Equations

Cons of Momentum
m1v1+m2v2=m1vs'+m2v2' , if car hits small fluffy object m2, initially v2=0, and v1'=v2' ... so
m1v1=[m1+m2](v2')
but why not energy? Why is there a KElost in Cons of Energy?
Cons of Energy
.5m1v1^2+.5m2v2^2=.5m1v1'^2+.5m2v'2^2 +KElost , and again v2=0, v1'=v2'
.5m1v1^2=.5[m1+m2]v2'^2+KElost
I know you could say deformation, but do you somehow have to take the derrivative of energy [wrt what i dont know] to find a maximum, which is dependent on the initial and final masses and velocities?

m1v1+m2v2=m1vs'+m2v2' , if car hits small fluffy object m2, initially v2=0, and v1'=v2' ... so
m1v1=[m1+m2](v2')

but why not energy? Why is there a KElost?
.5m1v1^2+.5m2v2^2=.5m1v1'^2+.5m2v'2^2 +KElost , and again v2=0, v1'=v2'
.5m1v1^2=.5[m1+m2]v2'^2+KElost
using consv of momentum...
KElost=.5[m1v1^2[1-m1/(m1+m2)] ?
.. and if m1>>m2, this is zero,,, how? Some energy should still be lost?
I know you could say deformation, but do you somehow have to take the derrivative of energy [wrt what i don't know] to find a maximum, which is dependent on the initial and final masses and velocities?

 

[kuruman]

You don't have to take a derivative to find the maximum energy lost in an inelastic collision. Maximum kinetic energy is lost when the two masses are not moving relative to each other after the collision. That amount of energy is $$\Delta (KE)=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(v_2-v_1)^2$$where $v_1$ and $v_2$ are the velocities of the masses before the collision.

 

[haruspex]

Maximum kinetic energy is lost when the two masses are not moving relative to each other after the collision.

That's certainly true for perfectly inelastic collisions in one dimension (head-on collisions), but unqualified "inelastic" is anything that is not perfectly elastic.

@richengle: That momentum is conserved follows from Newton's action/reaction law. If A exerts force F(t) on B then B exerts -F(t) on A. The change in momentum due to a force is its integral over time, so the changes in momentum from the collision are ∫F.dt and -∫F.dt, which clearly add to zero.
The work done on B is ∫F.dx, where the range is the displacement of B in the direction of F. So the work they do on each other also adds to zero, but this time there is no guarantee that the work will all reappear as KE.

For oblique collisions, I'm not sure perfectly inelastic implies sticking together. I would think it only means the coefficient of restitution is zero. So if the bodies are smooth, it is only the velocity components in the direction of the line of centres at collision that will become equal.

 

[Delta2]

Just to help state the statement of the OP a bit more accurately, in inelastic collision momentum is conserved, and energy is conserved too, it is kinetic energy that is not necessarily conserved.

@richengle: That momentum is conserved follows from Newton's action/reaction law. If A exerts force F(t) on B then B exerts -F(t) on A. The change in momentum due to a force is its integral over time, so the changes in momentum from the collision are ∫F.dt and -∫F.dt, which clearly add to zero.
The work done on B is ∫F.dx, where the range is the displacement of B in the direction of F. So the work they do on each other also adds to zero, but this time there is no guarantee that the work will all reappear as KE.

So the "moral" lesson of this is that impulse always transfers momentum between bodies, but work doesn't necessarily transfer kinetic energy, some work might go to heat or potential energy of deformation .e.t.c..

 

[kuruman]

That's certainly true for perfectly inelastic collisions in one dimension (head-on collisions), but unqualified "inelastic" is anything that is not perfectly elastic.

I agree, but I also think that we are looking at OP's question about a maximum

$\dots~$ but do you somehow have to take the derrivative of energy [wrt what i don't know] to find a maximum, which is dependent on the initial and final masses and velocities?

from different standpoints. Mine is goal-oriented. Given that one is looking for maximum kinetic energy loss and since the kinetic energy of the CM is conserved, the inelastic collision must be head-on and perfectly inelastic. In order to achieve this maximum, the parameters to be varied are two: the impact parameter and the coefficient of restitution. When both are set equal to zero, one has a head-on, perfectly inelastic collision that maximizes the loss of kinetic energy.

 

[Frodo]

When I was at school I always thought the conservation of momentum was inferior, in some way, to the great conservation of energy law which applies to everything. It was only when I read Principia that I finally understood - that Aha! moment - what momentum is.

Newton defines it beautifully: Momentum is the quantity of motion a body possesses.

As momentum is mv, and kinetic energy is 1/2 mv^2 how can both be conserved when their equations are so different?

Because when two things collide we get sound , which takes energy away; the bodies deform and heat a little, which takes energy away. When you add up all those bits of energy loss it explains why momentum (the quantity of motion) is conserved and also why energy (kinetic energy of motion + sound loss + heating loss + whatever else ...) is also conserved, even though their equations are so different.

The conservation of momentum follows as Corollary III from Action and reaction are equal and opposite.