DEE DEE MESSER: Hi.
Welcome to another session of AP Daily Practice
Sessions for Mechanics.
I'm Dee Dee Messer, physics C teacher at William Mason High
School in Mason, Ohio.
In this Daily session, we are going
to look at a free response question that
deals with linear mechanics.
On the screen you should see a PDF link
if you want to follow along or have a paper copy.
Or as I change the slides and I put a new question on there,
you can pause the video, answer it on your own,
and then resume to see how well you did with my answers.
So let's get started.
All right.
Here's the scenario.
You usually get a graphic that shows you what's happening
and then a paragraph that explains
the initial conditions.
In this one, I have a block of mass
to 2M on rest at a horizontal table,
and it's attached to a relaxed spring.
So that's my equilibrium position.
The spring is nonlinear and obeys
F equals negative Bx cubed, where B is a positive constant
and x is the displacement from equilibrium.
A block of mass 3M has an initial velocity
of v0 moving to the left.
There is negligible friction, so we do not
need to worry about that in the initial conditions.
Now that we have the initial conditions,
let's look at part a.
On the dots below, which represent
both blocks 2M and 3M, draw and label
the forces, not the components, that act on each block
before they collide.
Each force must be represented by a distinct arrow
starting on and pointing away from the appropriate dot.
It is important to understand that a free-body diagram is
an accurate depiction of what is happening in the scenario.
They are not meant to be drawn fast or sloppy.
If your arrows don't start on the dot,
we typically won't count them.
If they're pointing towards the dot,
they also are not going to be counted.
Make sure that you follow along with the directions.
So let's start with block of mass 2M.
I know the force of gravity is pulling it
down with a quantity of 2Mg.
And the table's exerting a normal force upwards
equal in magnitude of FN1.
It is at rest, so there are no horizontal forces.
Let's move on to block 3M.
It also has a gravitational force pulling down
with a quantity of 3Mg.
And the table's exerting upward with an equal magnitude,
this time of normal force FN2.
Because there is no friction, there is no horizontal forces,
so those are my free-body diagrams.
Let's check the scoring to see how many points we
would have earned.
Part a is worth two points.
The first point is for correctly drawing and labeling
the vectors on the block of mass 2M.
So you needed both Fg and FN correct
to get that first point.
The second point is for correctly drawing and labeling
the vectors on the block of 3M and for using symbols
for the vectors that are physically
correct and different from those of the block of 2M.
For example, if you had labeled Fg for both block 2M and 3M,
you would not have earned the second point.
Because the gravitational forces have
different numerical values, therefore they
should have different notations.
All right.
Let's move on to the next scenario.
Some time has passed and now the blocks
have collided and are moving together as one system.
The two--block system compresses the spring a maximum distance
of D. Express my answers in terms of b, c, and d--
in terms of big M, big B, and v0 and physical constants.
Remember, the only physical constant
we need to worry about in mechanics
is little g, the gravitational field.
It is important to remember the variables that you're allowed
to express your final answers.
If you need to, underline them, circle them, or even write them
beside parts b, c, and d to remind you
when you can stop substituting.
All right.
So there's my initial conditions and the important information,
since everything couldn't fit on the same screen.
Let's look at part b.
Derive an expression for the speed of the blocks
immediately after the collision.
Derive means equation, substitute,
solve with variables.
And the collision lets me know that linear momentum is always
conserved.
So I'm going to start with rho 0 equals rho.
Initially, only block 3M was moving with a speed of v0.
And after the collision, the blocks
stuck together as one system.
This gives me 3/5 v0 equals my new velocity.
Let's check the scoring guideline.
Part b was worth two points.
The first point was for using a conservation
of momentum expression.
"Using" is the important word.
If all you write down on the mechanics exam
is an equation and nothing else, you're
not going to earn any points.
You have to at least try to make one substitution
within that equation to earn the "using" point.
The second point was for the correct substitutions
into the correct conservation of momentum equation.
So the second line is worth the second point.
All right.
We're still having the blocks stuck together.
We want to now determine an expression
for the kinetic energy of the two-block system.
Kinetic energy-- 1/2 mv squared.
I substitute in the mass of the system, my velocity from part
b, and I get 9/10 Mv0 squared.
For part c, it was only worth one point.
And the verb "determine" is what allows
you to understand that you do not need to show your work.
Determine questions are usually one point.
And they mean that it is a simple equation or something
that is shown on a graphic that you can just write down.
So here, the one point was for the correct answer,
consistent with part b.
That means if you were not sure how to answer part b,
write down anything as the answer,
so that you can substitute it in for the rest of the questions
and still earn that point.
All right.
Finally, part d.
Derive an expression for the maximum distance D
that the spring is compressed.
Since friction is negligible, I know energy is conserved.
All of the initial energy was kinetic.
And at maximum compression, all of my energy is potential.
The initial kinetic energy I have.
That is my answer from part c.
And I know that the change of potential energy
is equal to negative integral of F dx
because spring force is a conservative force.
Making the appropriate substitutions
and setting my bounds from 0 to D for maximum compression,
I can integrate and find that my final answer is 18M v0 squared,
all over 5B, and then all over square root
to the fourth power, which is maybe not what you consider
a pretty answer, but it nevertheless
follows the rules of making sure that your answer's in terms
of M, B, v0 and physical constants.
So let's check the scoring guideline.
For part d, this is worth four points.
"Derive" again being the important verb
to make sure that you show all of your work and substitutions.
It was one point for using conservation
of energy expression.
I highlighted the first two lines
to let you know anything that showed us
that you're using a conservation of energy equation
would have been awarded that first point.
But again, it has to be using.
So if you just wrote E0 equals E, for example,
you would earn zero points.
But if you then went and tried to make
any substitution consistent with the problem,
you would get that first point.
The second point is for integrating the spring force
equation.
So I needed to remember that the spring was non ideal
and substitute in that correct expression.
The third point was for the correct limits
or a constant of integration.
Here, I wanted to make sure I integrated
from it being compressed from 0 meters to some distance D.
And the last point was from using the kinetic energy
consistent with part b.
Notice how the final answer is not worth
any points because it relied on information
from a previous part.
So don't give up, and don't get nervous.
If you miss one part of the question
and you're not sure what to do, write down an answer,
and then just keep substituting that into the correct spot.
And you won't lose any additional points.
All right.
More time has passed, and now I want
to look at away from expressions and talk into explains
and justifies.
For the first part of e, in which direction
is the net force, if any, on the block of mass 2M
when the spring is at maximum compression?
You have three choices to check from-- left, right, or zero.
And then justify your answer.
So I'm going to make a check mark
on right for the direction of the net force
and my justification.
These should be short, sweet, to the point.
Be very careful.
At the AP reading when we are grading
exams, what we sometimes run into
is that students talk too much.
The more they write, the more wrong
their answer becomes, and that's not what we want.
So initially, the blocks were moving to the left.
At maximum compression, the system is momentarily at rest.
Therefore, the spring needed to exert a force
to the right to cause that acceleration of slowing down.
This rightward force causes the rightward acceleration.
Part e(i) is worth two points.
The first point was for selecting
"right" as the direction of the force and an attempt
at a relevant justification.
Once again, if it's a check mark question
and you just select a check mark with nothing else,
you're not going to earn any points on the exam.
You have to select a check and attempt
at a relevant justification.
So the check mark with something about the system
gives you the first point.
The second point is making sure that your justification
is correct and valid for that scenario.
Let's keep going.
The second part of e, which of the following
correctly describes the magnitude
of the net force on each of the two blocks
when the spring is at maximum compression?
There are four checks and a justify.
So I'm going to select the net force is
greater on the block of 3M.
And my justification-- the blocks
move with the same acceleration.
Therefore, the 3M block has more mass
and will require more force to cause the same acceleration.
Short, sweet, and to the point.
It's worth two points.
One point for stating that both blocks have
the same acceleration.
So this is an example where the check mark was not
worth any points.
The check mark there is to help you organize your thoughts,
and it's to help the person grading your exam
to see where you're going in your explanation.
Finally, the second point--
for a correct justification why the force needed to be greater.
When you are relating different variables,
make sure that you state which ones are constant.
For example, Newton's second law, F equals ma.
In this scenario, acceleration, or a, was constant.
That's why force and mass have the direct relation.
The more massive object, the more force required.
Last part.
Do the two blocks, which remain stuck together and attached
to the spring, exhibit simple harmonic motion
after the collision?
Again, there's a check and a justification.
I'm going to select No with a justification
that the restoring force exerted by the spring on the blocks
is not a linear force.
Therefore, the force is not proportional
to the displacement from equilibrium.
So motion after the collision is not simple harmonic.
And this is also worth two points.
Where again, the check mark and an attempt
at a relevant justification is the first point.
A check by itself, not worth anything.
The second point is the correct justification.
Understanding that this is not an ideal spring,
therefore it does not obey the rules of simple harmonics.
And that concludes this AP Daily Practice Session.
The more you practice, the better
you're going to see the results on the day of the AP exam.
So continue watching the practice sessions
and continue practicing on your own.
Good luck, and I'll see you next time.