Lecture 11.4 – Example: Dropping Stacked Balls
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Lecture 11.4 – Example: Dropping Stacked Balls

November 19, 2019

SPEAKER 1: Ah, yes. So there’s a reason. It’s not because I am a
fan of a certain big box retailer that shall be nameless. All right. So what I’m going to do,
I have the little ball and I have the allegedly very
elastic liquid metal ball here. And I’m going to drop them. Probably very poorly, so we’re
going to think about it a lot before I actually screw up the demo. So I’m going to drop them, and
I think they’re going to fall, pretty confident on that score. So down they go. And if they fall the same
distance, then they’re probably going about the same speed. But the bottom ball will
encounter the floor first. And we’re going to say
for simplicity that it undergoes an elastic
collision with the floor. It’s cement. It’s pretty heavy. So it bounces up. If it was elastic and
it went down at v0, then does that imply that
it needs to come up at v0? Unless the earth is recoiling
with great velocity, right? OK. So we’ll pretend that the
earth doesn’t recoil a lot. Then the yellow one is still going down
at v0, and so they’re going to collide. And after the collision, well
maybe I don’t know for sure that the lower ball is
still going up, but I’m pretty confident that the
upper ball will be going up. And my question for you if all
the collisions are elastic, and if the blue ball’s mass is way big
compared to the yellow ball’s mass, then when the yellow ball
rebounds from the blue ball its speed will be most nearly which? All right. I’m terrible at this demo. So it may take a couple of
tries, and then I’ll give up. That’s about as good as
I’m ever going to do. So hang on one sec
before we explore that. So this one has four balls of decreasing
mass, and the top one can fly off. So let’s see– SPEAKER 2: But can the others? SPEAKER 1: No. Well they ain’t supposed to,
Yeah, that was really good. Let’s try this again, never fails. Get back there. All right, far away, here we go. Ah, missed her. All right. Thank you so much. So I will let you play with
these in recitation section. But I hope you noticed that
the little grape at the top actually got a lot of the kinetic
energy from the falling masses. So let’s now see if we can
understand the answer to this one. If fact, I think I heard
it right over here. Do you want to give it a shot? You’ve got this thing. Yeah, I know you do. I heard it. I heard the right answer. Go for it. SPEAKER 3: C, because since the
blue ball’s mass is a lot bigger than the yellow ball’s mass, the yellow
ball won’t slow the blue ball down. But the difference in speed
still needs to be the same. So– SPEAKER 1: So she says, OK. Correct me if I don’t get it right. So since the mass of the
blue ball is way big compared to the mass of the yellow, its velocity
isn’t really going to be changed much. It’s sort of like when the blue
ball collides with the earth, and the earth is effectively infinitely
massive compared to the blue ball, so the blue ball’s
velocity just reverses. So the blue ball is
going up at speed v0. And the yellow ball is coming
into this collision at speed v0. So what’s the relative speed? To v0, right. And then if we can approximate, because
this is so much bigger than that that the blue ball frankly doesn’t
care about the yellow ball. So it continues to move
up at speed about v0. Then in order for this to
be an elastic collision and preserve the relative
speed, then the yellow ball will have to be moving
up at 2v0 with respect to the blue ball, which
is moving up at v0. So therefore the yellow ball will
go up at 3v0, or thereabouts. Do you follow? I think she deserves
a round of applause. That was awesome. Any question on that? SPEAKER 4: Can we do
that a little bit slower? SPEAKER 1: Yes. Ian says, I want more chocolate,
or can you do that slower? All right. So here’s the large ball going up at v0. That’s too big. And the small one is coming down at v0. Now what has to be
conserved in the collision? Certainly momentum. So it is true that M, I’m going to
pick up as my positive direction. So initial momentum
would be Mv0 minus mv0. And the final momentum
will be M times, I guess I’ve got to make this a
big V going up, and a small v. Now I want to simplify this taking
into account that little m is puny compared to big M. So what if I
were to divide through by big M? Try to illustrate the little terms. No cheating yet, right? I didn’t divide by zero,
and then the world explodes. So in the limit that this is
huge compared to that, then I could maybe neglect this term. And I could neglect that term. And that implies that the
final v for the blue ball is the same as the initial v0. It’s not exact, but it’s close. And so if that’s the case, now
this one is moving up at v0. But the relative speed is
preserved in the collision. So this one has to go out
at some speed, v, such that v minus v0 is equal
well to the relative speed that we had before the collision, 2v0. Because here the relative speed is 2v0. Here the relative speed is 2v0. So therefore the outgoing speed of the
light ball, the ball with small mass, must be 3 times v0. About. OK now, I fudged things, right? I shouldn’t be putting–
this is a good equal sign. But by the time I get
down to here, it’s going to be relying on my ability to neglect
the difference between– that I can set the mass ratio to go to zero. There was a question. Yeah? SPEAKER 5: How is
kinetic energy conserved? SPEAKER 1: Ah. So she says, say it loud and proud. But then how is kinetic
energy conserved? Right? Because what would be the
kinetic energy before? Well it would be 1/2Mv0
squared, plus 1/2mv0 squared. And if we say that the big mass didn’t
change its kinetic energy, because it didn’t change its speed, then
we somehow manufactured energy. And that bugs you. Yeah. She says, yeah, that bugs me. So it can somebody help me out? Did we? Yeah? SPEAKER 6: So and then assuming that
big M is definitely [INAUDIBLE]. SPEAKER 1: So the
answer is, yeah we did. And he’s saying, well but in
the limit that big M is so much bigger than little m, all of the kinetic
energy in essence, is residing where? In the blue ball. And so even if we were to shave a
tiny little bit of its kinetic energy, it would still be moving
almost at v0, right? Because the top one is really light. So this is an approximate equals,
because I erased some terms in my momentum conservation. I did this. So my momentum conservation isn’t exact. But the approximation captures
the spirit of what’s going on. Because I mean if it’s true that big M
is this huge compared to little m, then almost all the kinetic
energy– so if you steal just a little
bit from the big one, you’ll hardly notice it’s slowing down. But you can shoot the little one up. So a little subtle
reasoning approximation.

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