For this next example,
what we're going to do is look
at a very similar situation,
where we also have a 200-kg car
hitting another car that's initially stationary,
also going 30 meters per second.
But now we're going to assume that in the collision,
these two stick together.
We want to know the final velocity of each car.
And then from the sticking together,
we would like to know how much energy
was lost from this collision.
Fortunately, this is a much quicker
and simpler problem.
It won't be quadratic.
It won't have the squared terms
that we have to deal with in the prior problem,
and so we should do it a little more quickly.
I recommend you give this a shot on your own,
using again the laws that we introduced
for inelastic collisions
in which the two objects stick together
and then we'll go over it here.
So let's see if we can find
the final velocity of each of these cars,
and then how much energy was lost
in the collision.
In this case, again, we have the same situation
where we have one mass is moving
at 30 meters per second of velocity
and then another mass M2 is initially not moving.
But now, our final scenario,
is that these two objects are stuck together
and they have a total mass of M1 + M2
and some new velocity
which we could call the prime.
We don't need a number on this, a 1 or 2,
because this is just the velocity
of our entire combined object.
So again, we know momentum is always conserved.
So the initial momentum is M1 x V1.
We could also say M2 x V2 for the second object,
but this object is initially not moving,
so the velocity is 0.
The final momentum is a nice, simple expression.
It's just mass x velocity.
The mass, in this case, is this big mass M1 + M2
and its velocity is V prime,
and that is the only momentum that we have to deal with
in the final analysis.
In this problem, we do not get to use conservation of energy.
We might think,
"Well, does that not mean that I have
one less equation to use to solve?"
"How can I possibly solve for everything that I need
if I only have one equation instead of two?"
The good news is, that's all right
because we only have one variable.
We only have one thing to find, which is V prime.
So we only have one variable, one unknown,
so we only need one equation.
So we'll just use conservation of momentum here,
P initial = P final.
In this case, we can just say:
M1 V1 = (M1 + M2) V prime.
And so we can solve for V prime, the final velocity, very easily.
The final velocity will be:
the ratio of the mass of the first object (M1) to the total mass (M1 + M2)
times the initial velocity of your first object (V1).
So we can very quickly plug in some numbers
and see what this is.
We have M1 which is 200 kilograms
over the total (200 + 400) for two objects
times V1 -- and the initial velocity was 30.
So this is 200/600 x 30.
We can cancel some of these zeroes and we get 2/6 x 30.
Canceling the 3 with the 6,
we get a nice simple answer of 2/2 which is 1,
and we get 1 x 10 now,
which is 10 meters per second.
So a lot of cancellations here.
Don't get lost in all the numbers.
We just had really 2/6 x 30
which is one-third of 30
which is 10 meters per second
as the final velocity of the cars
that are now stuck together.
The second part of this question will be just as easy.
They're just asking:
"How much energy was lost in this collision?"
So let's do that very quickly.
The final kinetic energy of this object
is one-half its mass.
Don't forget that's the total mass
(M1 + M2) times its velocity squared,
whereas the initial kinetic energy
is simply the kinetic energy of the first object
-- this object here.
And so this is 1/2 (M1 V1 squared)
and this is because the second object was not moving,
and so it doesn't have any kinetic energy.
So the total change in kinetic energy
for this problem
is K final, the final kinetic energy,
minus K initial, the initial kinetic energy
which is simply:
1/2 [M1 + M2] V squared - 1/2 [M1 V1 squared]
And so this is something
that is quite easy to solve
since we have all these numbers now.
We can factor out the 1/2,
[M1 + M2] is the [200 + 400]
so that's 600-kg mass.
So this is the final velocity (V) which is 10 squared minus--
and I factor out the one-half,
so that's already out here --
times [M1 V1 squared], so this [200 x 30]
and that is also going to be squared.
Finishing this up, we have:
1/2 [600 x 100 - 200 x 900].
So lots of zeroes but scientific notation saves us here.
This is 6 x 10 to the 2nd
1 x 10 to the 2nd
2 x 10 to the 2nd
and 9 x 10 to the 2nd.
So let's just focus on the numbers.
This is equal to: 1/2 [6 x 1 - 2 x 9]
[6 - 18]
10 to the 2nd x 10 to the 2nd = 10 to the 4rth.
So this is a little bit easier to think about, I think.
6 - 18 = -12
We divide that by 2, we get -6 x 10 to the 4th
and this is the kinetic energy change.
So be careful here -- the units of this is joules.
We have -6 x 10 to the 4th joules.
The reason we have a negative sign here
is because we lost kinetic energy in this collision.
It went into the compression of the cars and in places that
cause these two cars to be able to stick together and become one object.
So when you're calculating the changing kinetic energy
for totally inelastic collision where two objects stick together,
you can expect to get a negative answer,
as we expect to have lost kinetic energy in the collision.
This wraps up our summary of momentum,
conservation of momentum,
the derived quantities having to do with momentum,
including impulse and force,
as well as two different types of collision problems,
the elastic and inelastic,
and how to deal with solving these problems,
whether very long or very short.
Thanks for watching.