DEE DEE MESSER: Hi.
Welcome to AP Physics Practice Sessions for Mechanics.
I'm Dee Dee Messer, AP Physics C teacher at William Mason High
School in Mason, Ohio.
This video, we're getting to practice multiple choice
questions over the topics of gravitation and simple
harmonic motion.
You have two choices.
You can download the PDF on the link that is shown
and have a paper copy and follow along or do the questions ahead
of time, or every time I switch the screen
and show a new question, you can pause the video,
try to answer the question by yourself,
and then resume to see what the final answer is.
Either way, it's up to you.
But let's get started.
For question 1, we have a mass of a Planet X
that is 1/10 that of Earth.
Its diameter is 1/2 that of Earth.
And what we want to know is, what is the acceleration
due to gravity at the surface of Planet X?
So let's start with analyzing planet Earth.
From our equation sheet, our gravitational force equation--
and I know that Fg on the surface of the planet Earth
is mg.
The mass of the object cancels out.
And I get the acceleration equation is big M big G
all over R squared.
Starting with that equation, now for Planet X,
I'm going to substitute in 1/10 of the mass of planet
Earth, 1/2 of the radius of planet Earth,
and I derive this equation.
Here, I can see that big G big M over R squared
is also equal to the little g.
So I make that substitution, put in 10 for little g.
And I find that the acceleration is 4 meters per second squared.
That is choice B.
Let's try the next question.
The radius of Earth is approximately 6,000 kilometers.
I have an astronaut that is 300 kilometers above the Earth.
And we would like to know what that astronaut's acceleration
is in a perfectly circular orbit.
I know that if they are in orbit,
then there is an acceleration, because the direction
is changing.
So it can't be 0.
And 300 kilometers above the surface of the Earth
is not far enough away to be essentially 0.
So choices A and B can be removed.
Now we're down to three choices left.
Starting with that gravitational force equation,
making the substitution for the surface of the Earth equation,
I get g equals big G big M over R squared.
This time, however, the radius of the Earth is changing.
So I'm going to move that over to the other side,
and I'm going to leave my constants big G and big M.
Starting with the equation now, for the radius that's
300 kilometers above the surface of the Earth,
then I get this equation.
And I can see that, again, big G big M in the equation
make that substitution for little g R squared.
Now I have all of my variables where
I know their numerical values.
Make those substitutions, and I get
that we are approximately 9 meters per second squared,
which is choice B.
All right, let's try question 3.
Once again, a satellite is moving
in a circular orbit with a speed V0 at a distance R
from the center of a planet.
Now the satellite is moved, still in a circular orbit,
but has a new distance of 2R from the center of the planet.
What we want to do is calculate what is the speed required
to make that true?
Starting with my gravitational force equation,
this time, since I want speed, I know
that the acceleration is centripetal acceleration
because the orbit is circular.
And centripetal acceleration is v
squared over R. The mass of the satellite cancels out.
And solving for v, I get v squared equals big G
big M over R.
Now I'm going to substitute in my new radius
or my new distance of 2R.
And I see big G big M over R in the equation from my original.
Substitute that in for v0 squared,
take the square root of both sides,
and I get the speed of v0 all over radical 2,
which is choice B.
All right, let's try question 4--
here, I have a newly discovered planet
with twice the mass of the Earth.
But the acceleration is exactly the same as it is on Earth.
I want to know, what is the radius, then,
of this new planet, in terms of R
being the radius of the Earth?
Once again, I'm going to start with Newton's
universal gravitational force equation.
And on the surface of planet Earth, Fg is equal to mg.
Little m cancels out.
And I'm solving for the radius, so I'm
going to get R squared equals big G big M all over little g.
Now I want this to be for the radius of the new planet.
And I get big G times 2M all over little g.
I know that big G big M over little g
is equal to R squared of planet Earth.
So I can make that substitution.
Taking the square root of both sides,
I find the answer to be radical 2 times R, which is choice C.
All right, let's look at a different type of question.
Sometimes on the multiple choice,
we have a stem that is similar or the exact same
for multiple questions.
So here's a situation where I have an elliptical orbit
of a comet around the Sun.
Positions 1 and 2 are shown.
And position 1 is 10 times as far away as position 2.
So what I'm going to do is say that from the Sun to position 2
is a distance of x, and from the Sun to position 1
is a distance of 10x.
The first question, what is the ratio
of v1 over v2 of the speed of the comment at position
1 to its speed at position 2?
It is important to remember which variable is on top
and which one's on bottom when they ask for ratios.
So here, make sure that you're paying attention to v1
is on top and v2 is on the bottom.
If the orbit is elliptical, then I
know angular momentum is conserved because there
are no external torques.
If I want linear speed ratio, then my equation for angular
momentum, I'm going to use the one that
has linear speed, which is r cross rho.
Making the substitutions, I know that the distance
from the Sun to position 1 is perpendicular to the velocity.
So the sine of 90 is 1.
So the cross product on both sides goes to 1.
And then linear momentum is mass times velocity.
So I make that substitution.
Mass of the comet cancels out.
Again, make sure that you put v1 on top of v2 for the ratio.
The distance for position 2 was x.
The distance for position 1 was 10x.
And I get 1/10 as my ratio, which is choice B.
Now we're going to stay with the same stem.
But this time, we're going to ask a different question.
Here, at position 2, we're going to calculate or figure out what
the comet's kinetic energy is.
Well, I've already figured out that it
has a velocity at position 2.
So I know it can't be equal to 0.
So choice E is removed.
I also know that there is a ratio that was not equal to 1.
So the velocities can't be the same.
Therefore, the kinetic energies can't be the same.
So choice A is also removed.
Now we're down to three choices.
Kepler's second law talks about how
satellites sweep equal areas in equal times as they orbit.
So the farther away that the comet is,
the slower it is moving, and the closer it is to the sun,
the faster it is moving.
So at position 2, I am the closest I will be.
So I am moving the fastest.
Therefore, the kinetic energy is a maximum at position 2.
That is choice C.
All right, let's switch it up from gravitation
to simple harmonic.
In this situation, I have both a graphic
and a stem paragraph that's going
to talk about a situation.
And then multiple questions ask about that same situation.
So, here we go.
We have an unstretched ideal spring hanging vertically
from a fixed support.
There is a 0.4 kilogram object attached
to the lower end of the spring.
The object is pulled down to 0.35 meters
below the stretched position and released from rest at time t
equals 0.
Therefore, at the graph, you can see
that the spring is already stretched as soon as we
start collecting data.
The graph shows the vertical position
of the lower end of the spring as a function of time t,
where y equals 0 when the spring was initially unstretched.
The first question-- at which of the following times
is the upward velocity of the object the greatest?
I know that the slope of a position time graph
is equal to the velocity of the object.
So at times 0, 0.5, and 1 second,
I am either at a maximum or a minimum value.
Therefore, my velocity is 0, and can't be the choice.
So just knowing that piece of information
gets me down to choices B and D. At least
you get a 50% chance of guessing if you're not
sure what to do next.
Here, though, I know that I want upward velocity.
That means I need a positive slope.
And between 0.25 and 0.75, the only one
that shows a positive slope is 0.25 seconds, choice B.
Staying with that same situation of an ideal spring,
the next question asks, what is the spring constant?
Well, knowing that it's ideal, I can
start with the equation from the equation sheet 2 pi
equals the square root of m over k.
k is what I am solving for for the spring constant.
What I need is mass, which is given to me in the problem,
of 0.4 kilograms.
And I need the period t, which I can get from the graph,
of 1 second.
Now that I know I have all of my numerical values,
take the equation and solve it for k.
Then make the numerical substitutions
for both mass and period.
And we find that k is approximately 16 Newton
per meter.
One thing to notice with this one
is that when you read the question
and you know you're solving for k, and pi is in the answer
or is in the equation, look at the answer choices
and make sure you see if they are in terms of pi
or not so that when you do the calculation, what constants
you can leave and which ones you need to multiply through.
New scenario-- a certain one-dimensional conservative
force is given as a function of x by the expression F
equals negative kx cubed, where F is in newtons and x
is in meters.
What we're asked to find is a possible potential energy
function for the force.
Because it's conservative, I know
that force is the negative of the slope of a potential energy
curve, or negative of the derivative potential energy
with respect to x.
Taking that equation and solving for a potential energy,
we get U equals negative integral F dx.
I substitute in my expression for the force.
And I want to integrate from 0 to x.
That gets me x to the fourth over 4.
And when I evaluate with respect to my bounds,
I get positive 1/4 kx to the fourth, which is choice D.
All right, last one--
which of the following is a differential equation
that correctly describes Newton's second law
for a simple harmonic oscillator of mass m and force constant k?
So here, the equation of Newton's second law
lets me know where to get started, that my net force is
equal to the spring force, which is causing
the simple harmonic motion.
My net force, according to Newton,
is mass times acceleration.
And Hooke's law lets me know that the spring
force is equal to negative kx.
I know that the acceleration is the second derivative
of position with respect to time.
And that is my differential equation.
Notice, again, when you look at the choices,
you do not need to solve the differential.
You just needed to write it down.
And that is choice B.
And that is all we have today for this section of the AP
Daily Practices for Mechanics.
All the practice sessions that you watch
and any additional practices that you do
are going to make a big difference come AP exam day.
Good luck, and I'll see you next time.