Let's analyze a one-dimensional collision,
where we're in the laboratory reference frame.
And we have V1 initial and V2 initial is 0.
It's a frictionless surface.
But there's a collision here.
And now we're going to make this collision totally inelastic.
Now what that means is that the two particles stick together.
So if this is our initial state--
and let's, again, choose a direction i hat-- then
our final state will just stick those two particles together.
And they're going to move.
We only need one velocity for the final state, i hat.
Now again, let's assume that there's no external forces.
There are clearly internal forces
that are acting between the two particles as they collide.
But, remember, we know that internal forces
cancel in pairs.
And if there's no external forces,
the momentum of the system is constant.
So we can write down our momentum condition very simply.
And we'll write it again in terms of components.
We have some incoming momentum into the system.
And all of that incoming momentum is going out.
And now, is energy constant?
Well, a totally inelastic collision,
energy cannot be constant.
And the reason is that these internal forces
will cause deformations that are irreversible.
So the objects might be deformed,
which is a loss of-- some of that kinetic energy will
go into deformation.
There could be noise.
There could be heat generated in the collision, lots of sources.
When objects stick together, kinetic energy is not constant.
In fact, we can even figure out the change in kinetic energy,
because we can see here that Vx final is m1 over m1 plus m2 V1
x initial.
And now let's ask ourselves, in a totally inelastic collision
what is the change in the kinetic energy?
So our final kinetic energy is 1/2 m1 plus m2 V-- here,
we can, again, right in terms of components-- Vx final,
because the component's squared.
That's no problem.
And we just have this initial kinetic energy coming in.
Now we have a solution for our problems.
So let's just put that in.
m1 plus m2 times m1 over m1 plus m2 times V1
x initial squared minus 1/2 m1 V1 x initial squared.
And now let's pull out the 1/2 m1
V1 x initial squared-- 1/2 one factor of m1 V1 x
initial squared.
But what we're left with here is another factor of m1.
And downstairs, we have an m1 plus m2
outside but an m1 plus m2.
Now remember, we didn't square that.
We have an m1 plus m2 inside.
So there's that m1 plus m2 minus 1.
Now this is the initial kinetic energy.
And when we subtract these terms,
we have negative m2 over m1 plus m2.
And so what we see here is-- I'm just
going to rewrite this again-- that the kinetic energy,
the change in kinetic energy, just
depends on the initial kinetic energy and this mass ratio.
In fact, if we asked ourselves, what
is the ratio of that loss of kinetic energy
to the initial kinetic energy, that just
depends on the mass ratio of our two objects.
In particular, it should be a minus sign,
because the kinetic energy is decreasing.
And so this is the fact that it's very transparent now
that in a totally inelastic collision
the kinetic energy is not constant.
Where did that energy go?
It went into other forms of energy.