Download Conservation of Momentum Video Script

Physic 602 – Conservation of Momentum
(Read objectives on screen.)
Instructor
Good. You’re back. We’re just about ready to start this lab on conservation of momentum during
collisions and explosions. In the lab, we’ll show you four different examples of elastic collisions, two
inelastic collisions, and two explosions. At first, we’ll help you show, both mathematically and with a
diagram, how the momentum of each individual cart changes during the collision and how the total
momentum of the system is conserved. Later, you’ll do this on your own.
The first thing you want to do when you observe the collision or explosion is to draw arrows over the
carts that are moving to show the direction and magnitude of the velocity.
If the cart is at rest, don’t draw an arrow.
If it’s moving to the right, draw an arrow pointing right.
And if it’s moving faster to the left, draw a longer arrow pointing left. This will make it easier for you
to remember what happened when we start plugging in numbers and calculating. Now, watch the first
collision and draw your arrows.
(student on screen)
VO
In case one, cart B is at rest in the middle of the track. Watch what happens when cart A is given a
little push toward cart B. When they collide, B stops and A starts moving at the same speed B had
before the collision. Draw your arrows and then we’ll help you plug in some numbers.
(diagram of carts on screen)
VO
Your arrows for case one should look like this.
To make calculations very easy, we’re using one kilogram for the mass of the empty carts. We’ll also
use one meter per second for the velocity of the moving cart. You’ll fill in other possible velocities,
before and after the collision, to show how momentum is conserved. Let’s do the first one together.
In case number one, cart A was moving, so let’s draw a medium length arrow to represent its velocity.
Cart B was not moving, so we don’t draw an arrow for it. After the collision, cart A stops and cart B
starts moving at the same speed as A was moving before, so we draw the same length arrow.
Now let’s use simple math to show how momentum is conserved. We’re using one kilogram for the
masses of both carts. Before the collision, we’re assigning one m/s as the velocity of the moving cart
A and zero for cart B, which is at rest. So cart A has a momentum of 1 kg·m/s, and cart B has a
momentum of zero. After the collision, the two carts simply trade velocities and momentum. A’s
momentum is now zero and B’s is 1 kg·m/s. One kg·m/s plus 0 equals 0 plus 1 kg·m/s. So we see
that the total momentum before and after the collision are the same. Momentum has been conserved.
Now watch collision two and draw arrows as before.
(student on screen)
VO
In case number two, a one kilogram mass is placed on top of cart B, doubling its mass. Watch what
happens when cart A is given the same push as before and collides with B.
(diagrams on screen)
VO
In this case, cart A is moving at 1 m/s, just like before, and cart B is not moving, so its momentum is
zero. Now you fill in the masses and velocities of the carts after the collision. When you do the math,
you should see that the momentum before the collision is the same as the momentum after the
collision. When you’ve finished, we’ll move on to collision three.
(Pause Tape Now graphic)
(student on screen)
VO
In number three, both empty carts are placed at the same end of the track. Cart B is given a small
push to give a low velocity, and cart A is given a greater velocity so that it will catch B. Watch what
happens when they collide. Cart A slows down and B speeds up.
(diagrams on screen)
VO
For this situation, give cart A a velocity of 2 m/s and cart B a velocity of 1 m/s. Now, you fill in the
velocities after the collision and show how momentum is conserved.
(Pause Tape Now graphic)
VO
In case four, the empty carts are started at opposite ends of the track and given the same velocity so
that they meet in the center. What do you expect to happen when they collide? Tell your teacher.
The carts bounce off each other and go in the opposite directions at the same speed as before the
collision.
VO
Because velocity is a vector quantity, we’ll need to use positive 1 m/s for the left to right motion of
cart A and negative 1m/s for right to left. You take it from here.
(Pause Tape Now graphic)
Instructor
In Part Two of the lab, we’ll see how momentum is conserved during inelastic collisions. Before we
start, tell your teacher what an inelastic collision is.
Did you say that during an inelastic collision, the objects stick together or are deformed and energy is
lost in the form of heat?
That leads me to a question. How many of you have seen movie scenes where cars drive off cliffs and
explode on the way down? Now that you know some physics, does that make any sense to you? An
explosion usually occurs when the car collides with the ground and the heat generated ignites the gas
fumes. But I guess it’s more dramatic showing the car explode in mid-air.
Woops! I got off the subject for a minute, didn’t I?
Let’s get back to the inelastic collisions in the lab. You won’t notice any heat being given off, but the
carts will stick together.
As you watch the two inelastic collisions, draw arrows, as before. Then your teacher will pause the
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tape while you plug in numbers and calculate.
(student on screen)
VO
In case number five, we’ll repeat what we did in number one, but this time the carts will stick
together. Watch what happens when they collide.
The joined carts move forward at half the original speed of cart A.
(student on screen)
VO
For our last collision, case six, we’ll start the empty carts from opposite ends of the track and give
them the same velocities, but in opposite directions.
The carts collide and stop.
(Pause Tape Now graphic)
Instructor
All right. You’ve seen six cases in which our carts collide. Now, in Part Three, comes more fun.
You’ll see two explosions and show how the momentum of the system is conserved. For this, we’ll
place the carts in the middle of the track, with the plunger on the exploder cart touching the other cart.
To release the plunger, all we have to do is tap the release at the top, like this. Now some of you
physics types may think that I’m exerting a net external force on the system when I tap, but remember
that I’m not exerting any force in the horizontal direction that the carts will be moving. Tapping the
release rod is like lighting the fuse of a firecracker. The system is still isolated, so momentum should
be conserved. We’ll show you two explosions in the lab
Draw arrows as you observe the carts before and after the explosions.
(student on screen)
VO
In case number seven, we place two carts in the middle of the track, so that the plunger of cart A
touches cart B. Watch what happens when the plunger is released.
The two carts move apart in opposite directions at the same speed.
(student on screen)
VO
In our last case, we’ll repeat case number seven, but this time, we’ll double the mass of cart A. Tell
your teacher what you expect to happen when the release rod is tapped.
The carts separate as before, but cart A moves with half the speed of cart B. Did you get it right?
Instructor
Your teacher will pause the tape now and give you time to finish your calculations and answer the
questions in the conclusion section of your lab report. You’ll go over the lab results in class. When
you come back, we’ll go over the questions.
(Pause Tape Now graphic)
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Instructor
I hope that the theoretical numbers we used to describe the collisions and explosions in the lab helped
you to see how the momentum of the system is conserved. Remember that the momentum of each
cart is changed because an outside force acts on it, but when you consider the two carts together, no
outside forces act on the system, so the total momentum does not change.
Now let’s go over the answers to the questions.
(toy truck on screen)
VO
Question number one was “If a truck runs into a wall and stops, the truck loses momentum. Since
momentum cannot be created or destroyed, where does it go?” The answer is that the momentum is
transferred to the wall. This means that the wall must move a little, but it is so massive compared to
the cart that you probably can’t see the small movement.
(student on screen)
VO
Here’s question two. Someone throws a heavy ball to you when you are standing on roller skates.
You catch it and roll backwards. How does your speed compare to the speed of the ball and why?
After the collision, you and the ball move together at a much slower speed. This decrease in velocity
is necessary to conserve momentum because your total mass is greater than the mass of the ball alone.
(prospector on screen)
VO
Here’s the situation in question three. A prospector finds himself holding his bag of gold and
standing in the middle of a large pond of frictionless ice. How can he get to the side before he
freezes? Now don’t ask how he got himself into this mess. That’s not the point of the question.
Since the ice is frictionless, he can’t push backward on the ice to move himself forward. How can he
get to the side before he gets frostbite or freezes to death?
His only choice is to use the Law of Conservation of Momentum and throw the heavy gold
backwards. That will make him move forward fast enough to make it to the other side before he
freezes.
Instructor
Easy come, easy go! Now that you know how momentum is conserved, let’s put some real numbers
together and solve some problems. We’ll start with a couple of example problems.
(text on screen)
VO
The first problem involves an inelastic collision. A 1,200 kilogram car is stopped at a traffic light
when a 3,500 kilogram truck moving at 8.4 (Eight point four) meters per second hits it from behind.
If the bumpers lock, how fast will the two vehicles move?
We start with two objects and they hook together to make one. The initial momentum of the car is
zero, but filling in the numbers may help. The mass of the car is 1,200 kilograms and its velocity is
zero. And the mass of the truck is 3,500 kilograms and its velocity is 8.4 (eight point four) meters per
second. After the collision, the mass of the two vehicles hooked together is 1,200 plus 3,500
kilograms. We don’t know their velocity.
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When we do the math, “v” turns out to be 6.3 (six point three) meters per second.
(text on screen)
VO
The second problem involves the kick of a gun. A 0.0050kg bullet is fired at a velocity of 310 m/s
from a 2.0 kg gun. What is the recoil velocity of the gun?
All the conservation of momentum problems will involve a before and an after or initial and final
momentum. And since momentum is conserved in all situations, the initial and final must be equal.
In this case, the gun and bullet together are at rest before the gun is fired. So the total initial
momentum is zero. After the internal explosion, the bullet goes one way and the gun recoils or kicks
in the other direction. So we’ll have to use a negative sign for one velocity. I choose the velocity of
the gun to be negative.
So the final momentum is .0050kg times 320 m/s plus 2.0kg times negative v. We can move the 2.0
times “v” to the other side of the equation to make the velocity positive. The rest is easy, and the
answer is 0.80 m/s.
(physics challenge on screen)
VO
Now here’s a challenge question for you. If the gun in the last problem is being held tightly against
the shoulder of a marksman who is braced for the shot, will it still kick? And how will the recoil
velocity be affected? Tell your teacher.
Instructor
The gun will still kick because the momentum of the gun in one direction must equal that of the bullet
in the other direction.
But the recoil velocity will be much less because the mass of the gun and man together is much
greater than the gun alone. And if the man were braced against a tree, you’d have to add that mass,
too. There must be some change in velocity or some motion, but the greater the mass, the less
noticeable it will be.
Now you try two problems and we’ll go over the answers together.
(text on screen)
VO
Local Teachers, turn off the tape and give students problem set number one from facilitator's guide.
(Pause Tape Now graphic)
(text on screen)
VO
The initial momentum of the astronaut and her tank is zero. So the sum of the momenta of two after
the tank is thrown must equal zero. This means that the tank will move in one direction. Let’s make
that negative. And Shirley will move in the opposite or positive direction. The equation becomes “0
= 12 kg times negative 2.2m/s plus 65 kg times “v.” We move this term to the left of the equals to
change its sign and then do the math. The answer is 0.41 m/s forward. I hope Shirley has something
else she can throw to get back to the ship.
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In our second problem, we have two objects before the collision and one after. And the directions will
all be positive. The momentum of the car number one is 14,000 kg times 5.2m/s. And the momentum
of car number two is zero, since it is at rest. After the collision, the momentum is the total mass of the
joined cars is 14,000 plus “m2” times 3.6 m/s. And the mass of car number two turns out to be
6,200kg.
Instructor
We’re almost finished with our study of momentum, so we need to review. I want you to put together
the concepts of momentum, impulse, and conservation of momentum to analyze this situation. A bug
hits the windshield of a moving car.
(car and table on screen)
VO
Which object, the car or the bug, experiences the greater force? Does the car or the bug experience
more time of contact when the force is applied? Which experiences the greater impulse, and which
has the greater change in momentum? And finally, which has the greater mass and change in
velocity?
Instructor
Your teacher will give you the questions again and time to answer them. Now don’t surf. Think
about everything you’ve learned in this unit and previous units. When you’ve made you decisions,
talk it over in class and then we’ll come back and go over the right answers.
(Pause Tape Now graphic)
(table on screen)
VO
Did you all agree on the answers? Here are the correct ones. First of all, I hope you remembered
Newton’s Third Law. A force is an interaction between the bug and car, in this case. And the car
cannot hit the bug harder than the bug hits the car. The forces are equal in magnitude and opposite in
direction?
I’ll bet you got the second one right. The time of contact between the bug and car have to be equal. I
certainly can’t push or pull on you for a longer period of time than you’re pushing or pulling back,
can I?
Now, here’s where your bad answer could snowball, since impulse is force times time. Since force
and time are equal, the car and bug experience equal impulses.
And if the impulses are equal, then their changes in momentum must be equal. That’s because an
impulse produces a change in momentum and vice versa.
So far, everything is equal, which may surprise you. But this collision just obeys the Law of
Conservation of Momentum. If the car and bug are a system, then any momentum lost by one object
must be gained by the other. So the changes in momentum must be equal.
But you know that the bug and car are very different, so something must be different about the way
they behave during this collision. That’s where mass and change in velocity enter the picture. The
car’s mass is much greater than the bug, so the bug’s change in velocity must be much greater than
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the car’s. Got it?
Instructor
I’d say that not only has our bug’s velocity changed a great deal, but it’s day has been pretty well
ruined by this run in with a car. Rest in peace, little bug.
So much for our study of momentum. It’s time to…SHOW WHAT YOU KNOW!!
Jot down your choice for each question. Your local teacher will go over the correct answers with you.
(Read Show What You Know questions on screen)
Instructor
Before you go, I have one more thing for you to think about. It has to do with the toy I showed you
earlier. The steel balls have elastic collisions with each other, and of course, obey the Law of
Conservation of Momentum. When I let one ball hit the others, momentum is transferred from one
ball to the other until the last ball flies out with equal velocity, just like the collisions between carts in
our lab. Now watch when I pull two balls back and let them hit. Two fly out together. But think
about this.
Wouldn’t momentum still be conserved if two balls moving at one meter per second hit and then one
ball, with half the mass, were to fly out at two meters per second?
Momentum would be the same if the mass is doubled and the velocity is cut in half.
But this never happens. There must be more physics involved here than just conservation of
momentum. We’ll have to study the concept of energy before we can explain. That’s the topic we’ll
begin to study in our next program. See you then.
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