Let's now get to problem number 5B.3, and I will get
back to the first problem later.
It is a one dimensional collision.
I started out with object mass m 1, it has velocity v 1, and
it collides with mass m 2, which has velocity v 2.
This is before the collision, and here we see the situation
after the collision.
There is [UNINTELLIGIBLE], m 1, v 1 prime,
m 2, and v 2 prime.
If there is no external force on the system, only internal,
momentum is conserved.
The conversation of momentum: m 1 v 1 plus m 2 v 2 equals m
1 v 1 prime plus m 2 v 2 prime.
That is equation number one, the conservation of momentum.
I could do delete these bars since I'm dealing with a one
dimensional situation, keeping in mind that if the velocities
in this direction would then, for instance, assign it a plus
sign, but if the velocity is in this direction, it would
then have a minus sign.
I can delete the bars, not forgetting that this is now my
convention, so think of the bars as not being there.
It is given, or I'm telling you that this is also an
elastic collision, so you also have that the kinetic energy
is conserved: 1/2 m 1 v 1 squared plus 1/2 m 2 v 2
squared equals 1/2 m 1 v 1 prime squared plus 1/2 m 2 v 2
prime squared.
That is equation number two, and that is the conservation
of kinetic energy, which only holds if this
is an elastic collision.
This is the conservation of momentum, which in this one
big case, observing our sign convention, can then even
delete the bars.
If I massage this a little further--
and I will let you do that, since we have little time
today, so you will have to do some massaging of this--
you can then show that v 1 minus v 2 equals v 2 prime
minus v 1 prime.
That is quite remarkable: this is the relative velocity
before the collision between the two objects, and this is
the relative velocity after the collision
between the two objects.
It's not so obvious that that would be the same.
Of course, it's only true if kinetic energy is conserved--
but what is not required is that m 1 is m 2.
That is not required: when you derive that, you will see that
that is not a must.
Let's take an example.
For instance, I have m 1 equals 5 kilograms, and I have
m 2 that equals 22 kilograms.
This object is v 1, magnitude 3, and this object is v 2 in
the opposite direction with the magnitude of
6 meters per second.
Be very careful and think of our sign convention: v 1 minus
v 2, and this is now positive 3, plus 3.
v 2, itself, is negative--
minus 6, so minus minus 6 becomes plus six, so this is
plus 9 meters per second.
This now also has to be v 2 prime minus v 1 prime,
observing the same sign conventions.
This, of course, is quite remarkable when you come to
think about it.
You may say, great--
I know the difference in velocity, but what is now v 2
prime, and what is now v 1 prime?
Now you have to go back to your equation number one and
equation of number two.
If you know v 1, you know m 1, you know v 2, and you know m
2, then you have in principle two equations with two
unknowns, and you should be able to solve for v 1 prime
and v 2 prime.
It takes a little bit of hassling, a little bit of
massaging, a little bit of algebra: that you can find in
books, or you can work on it yourself.
It would be a waste of your time if I used my precious 57
minutes right now, so I'll give you the answer.
If you do it correctly, you'll find then that v one prime
equals m 1 minus m 2 divided m one plus m 2 times v 1 plus 2
m 2 divided by m 1 plus m 2 times the v 2.
Then we have v 2 prime equals 2 m 1 divided by m 1 m 2 times
v 1 plus m 2 minus m 1 divided by m 1 plus m 2 times v 2.
You can check whether I did that correctly--
if you're changing the first equation with the 1 to a 2 and
the 2 to a 1, you obviously should get this equation,
because the problem is completely symmetric.
Let's take this term, and the one becomes a 2, the 2 becomes
a 1, and indeed this looks good.
This 2 becomes a 1 here, so I think we are great--
this v 1 is v 2, and this v 2 is v 1, so this looks good.
Remember, this is only true if it is elastic.
Remember, that a positive sign means going to the right for
the velocity, and a negative sign means going to the left.
Now let's take a very special case--
let's take the case that m 1 equals m 2.
Here is m 1 with velocity v 1, here is m 2 with velocity v 2,
and if you now take m 1 equals m 2, this is
gone, and this is gone.
This is 1, this is 1, and what do you find
to your great surprise?
That v 1 prime equals v 2, and v 2 prime equals v 1.
What does this mean?
It means that after the collision, the velocity of 1
is now the velocity of 2, and the velocity of 2 is now the
velocity of 1--
that is not so obvious.
They exchange their velocity--
it's by no means intuitively so obvious.
If we take a very special case that v 2 equals 0, it would
follow then from what I just told you, that v 1 prime would
be 0-- they exchange, and that v 2 prime would be v 1.
I would like to demonstrate this to you, and the best way
I could do that is to have two pucks of equal mass: this one
bangs into this one, and this one will be standing still.
There has to be no friction, and I could show you, then,
after the impact that this one gets this velocity, and this
one stands still--
but I don't have a frictionless horizontal table,
so I can't do that.
You may say, do it with marbles: if I do it with
marbles, there is another problem.
I can take two marbles of equal mass, but when I do this
collision process on my pad, then they begin to roll.
Rolling adds a complication that we have not at all
discussed, and that is very special.
We can't do it with rolling marbles-- that's a little
tricky, and if you slide the marbles,
then there is friction.
If there is friction, there's external forces-- so let's not
do it with marbles either.
I will do it with a pendulum.
I have here a pendulum, and I have here a pendulum of equal
length and equal mass.
I will swing this one towards this one.
You will say, there are external forces-- there's
gravity, which is very true.
There is a gravitational force here, mg, and this accelerates
this object, so clearly--
if there's an external force, how can
possibly momentum be conserved?
At the moment that they collide, in the x direction,
there is no external force.
They collide, there are internal forces, and action
equals minus reaction, but that's OK--
there are no external forces in the x direction.
As they collide in this direction, I have in the x
direction-- in the horizontal direction--
an ideal situation: kinetic energy is conserved, because
these metal balls are very flexible.
They bounce back very nicely, and in addition, I have no
external force, so momentum is conserved.
So when this object hits this one, this one should stand
still, and this one should pick up this speed.
This is something that I want to--
see, I will do it with this famous device, which is called
a Newton's Cradle.
You see here the two pendulums of equal mass, and we will
show it to you from above: that's easier, because I can
also zero in then, and then you see it better.
In fact, when I zero in, you don't even see the vertical
motion, you only see the horizontal motion, which makes
it even more convincing.
I move this one out, and at the moment of impact, I have
kinetic energy conserved, and I have momentum conserved in
the x direction--
there we go.
Watch as this hits--
this one will stand still, and this one will
pick up that speed.
Are you ready-- there we go.
You see?
It stands still--
I go back, and it stands still.
Of course, if you do this, then you get this alternation,
which you have seen, no doubt, many times.
A great toy--
can I have the overhead, because I want to be able to
adjust the screen?
A great toy, and it makes this point very convincing.
Let's now take a situation that m 1 is not m 2, but let's
for simplicity still take v 2 equals 0, so our equations
becomes a little simple.
It's the same equation that I wrote before, but I substitute
v 2 equals 0, so this is v 1, and v 2 equals 2 m 1, divided
by m 1 plus m 2 times v 1.
It's the same equations as before, but I substitute in v
2 equals 0.
Let's assume v 1 is in this directions, which we have
called positive--
of course, this holds only if kinetic energy is conserved,
because that's the way we derive them.
It holds only if there is no external force on the system,
so the momentum is also conserved--
that's very important.
Notice that v 1 has--
that v 1 prime--
has three options: it can be larger than 0, if m 1 is
larger than m 2.
It can be equal to 0, if 1 equals m 2--
we saw that before, that's not new--
but it can also be smaller than 0 if m 1 is
smaller than m 2.
This is intuitively pleasing, because what it means is if a
small, low mass ball, like a ping pong ball collides with a
baseball, or with a billiard ball, which
may be a better example--
then it can bounce back.
It's intuitively pleasing, and you see that.
If m 1 is smaller than m 2, it will bounce back and will go
in this direction.
However, v 2 prime is always larger then 0.
Notice no matter how you choose here--
m 1, m 2, or v 1--
if v 1 is the positive direction, then v 2 prime is
also in a positive election.
That is also very intuitive, because if I have here an
object, and I hit it, obviously it
must go in this direction.
It would be absurd if I hit this object,
that it would go back.
That would be absurd.
It's completely consistent with what we find here: v 1
prime, you have three possibilities.
The v 2 prime always goes into the direction of v 1, and that
is exactly what you expect.
Now we'll take an extreme case that m 2 goes to infinity, and
so we have a wall.
Obviously, if v 2 is 0, the wall is not moving--
let's assume the wall is not moving.
If the wall were moving, the kinetic energy of the wall
would be infinitely large, and that would be a bit of a
problem-- so v 2 better be 0.
I find it if I collide an object with m 1--
it goes with the velocity v 1 I'm against the wall--
I find that v 1 prime equals minus v 1.
I find that v 2 prime equals 0--
kinetic energy is conserved, and momentum is conserved.
This is quite obvious, you would say: this is the wall,
and an object comes in here with a certain velocity, and
it bounces back with exactly the same speed, but it
reverses its velocity.
Yes, this is very intuitive, but I want you
to think about something.
The kinetic energy is all in here: it's not in the wall.
The kinetic energy of the wall, before and
after, equals 0.
The kinetic energy here is 1/2 m 1 v 1 squared, and it is
also 1/2 m 1 v 1 squared when it bounces
back, because look--
so that's fine.
Let's now look at the situation of the momentum, and
I'll give you some numbers, because it's
easier to deal with--
m 1, let's say, is 1 kilogram.
That's the ball that bounces off the wall.
It comes in with 5 meters per second, and it bounces back
with five meters per second.
In other words, p 1 has decreased by 10 kilogram
meters per second.
It's first in the positive direction, and now it's in the
negative direction, so it has decreased by 10.
However, I've stated that momentum must be conserved.
How can it be--
momentum is changing.
Is it perhaps true that the momentum of the wall is also
changing, so that the net momentum of the wall
plus the ball is 0?
It better be, but it's kind of puzzling, isn't it, because
the wall has infinite mass, and it has no speed or
velocity for the impact or after the impact.
You think about that, and it's very important that you get
the answer to that, because it's very fundamental.
Conservation of momentum must hold: so yes, indeed, the wall
must have changed its momentum, in spite of the fact
that it never moved.
Isn't that cute?
Think about it.