Hello Internet, And welcome to MedVIP!
Today, we are going to talk about projectile motion.
Today is the first in a series of videos I hope to make about passage-based questions.
These are passages that I own, because I made them up!
They're only going to have a couple of questions in each video.
The reason is to make sure that you guys actually have an adequate explanation of those concepts.
My main focus isn't to make questions.
There are plenty of people who do that better than me and have tons and tons of people working
on that and a huge payroll to do so.
My expertise is to focus on explaining them and explaining them in a way that is tenable
and graspable by not only undergraduates, but even those who may be taking high school
courses, or not even yet thinking about taking the MCAT.
A physiologist is studying a group of high school shotput athletes.
He begins with measurements of the average arm length and the average Force during the
throw as show in Table 1.
So, we look at table 1, read the column headings, and we see: Grade, Average Force during the
Throw, and the Wrist displacement in centimeters.
So, we have Newtons and centimeters.
That's Force and Distance, and for those of you thinking ahead, you should be thinking
that this has something to do with Work, but beyond that you don't really need to pay
attention to Table 1 at this point.
Notice that you should always read it in conjunction with the part of the paragraph that actually
mentions that table, figure, or graph.
The physiologist then leads the team through a series of exercises to determine how far
they can throw the shot to determine who makes the cutoff for their next meet.
Afterwards, to show off in an extremely dangerous way, the classes compete to see who can throw
the shot straight up the highest.
Infuriated to discover they have been so reckless as to throw an iron ball with 5 kg mass so
that it could land on their heads, he makes them run the following series of physics problems.
Given the density of the shot, he tells them to ignore air resistance.
Now of course, this is a silly passage because I made it up.
But the point is to learn how to read these passages and extract the relevant information.
Some people like to highlight; I was never a huge highlighter, unless there was a fact
or a number buried in the text, and for those, I would absolutely highlight.
Clearly this passage is going to have something to do with projectile motion because it's
a shotput team.
After highlighting all the relevant facts in the text that are hidden, the other main
goal of the passage-reading is to get an idea of what kinds of questions you might be asked,
to sort-of anticipate what's coming along.
So that table has Newtons and meters, which should give you Force and Distance, some sort
of calculation based on Work.
We also have at least one trial where the projectile's being thrown straight up and
down.
Alright, on to question 1.
What is the maximum height above the point of release that the average junior can throw
the shot straight up?
We've got a series of answers.
They're all in meters.
It's going to be a straight projectile problem, but it's one where, like I said, the projectile's
being thrown straight up.
And of course, question 2: What is the maximum distance that the average freshman should
be able to throw the shot?
Now, maximum distance implies maximum range.
So, for those of you who are thinking ahead, there's a little bit of a trick to this,
but we're going to discuss these answers in just a second.
Alright, so, the explanation of question 1: They're asking for the maximum height above
the point of release, so really, they're only looking for the change in height here.
The actual final height in some projectile problems is gonna depend upon where the initial
height was, but here we're only asking for the difference in height, so we can just apply
the formula much more simply to figure out the change in potential energy.
So, the concept here is that the Muscular Work that is done is going to give the object
it's initial Kinetic Energy, which is going to be all the mechanical energy it has upon
the point of release.
Since there is no Friction, that's also going to be exactly equal to the maximum amount
of Potential Energy it will have at the peak of the parabola, even though that's a skinny
parabola because it's just up and down.
And that of course, gives us the final or maximum height that is achieved by the projectile.
So the series of connections between the Work of the Muscles, the Kinetic Energy, the Potential
Energy, and the change in height, that is the solution path.
And the solution path is the logical chain that connects the givens to the final answer.
So, this concept of logical connections, this is a critical skill that will serve you from
now until your last day in medicine.
Now, some problems have longer solution paths.
Some ones have shorter and straighter solution paths.
Some have branching, or forking, or merging solution paths.
But the solution path is really the key to solving any problem.
And to play off the Wu Tang Clan, "A test question [is like a swordfight – you must
think first, before you move]."
Alright, so let's start with that first concept in the logical chain, the Work done
by the muscles.
Now, in order to get the work done by the muscles, we're going to need to apply that
formula, Force times distance.
And since they're co-linear, the cosine of theta is simply going to be 1.
So, because this is an average junior, we're going to have to look up that line in Table
1, which gives us the Force during the throw, and the displacement of the Wrist during the
Throw, because that's going to be the displacement of the shot.
For those of you who said, "Well, it's the displacement of the wrist, not the shot,
can we make that assumption?"
Yes.
Keep it Simple.
Don't read too much into the question.
Just make sure that you're careful and don't skip steps in your calculation.
When we read the third line in table one, we see that we get 406 newtons and 100 centimeters.
Now, the problem is, of course, that Newtons times centimeters is not a real unit, so we
better go ahead and convert that 100 centimeters into meters.
And of course, there are 100 centimeters in 1 meter, so conveniently enough, that's
a distance of 1 meter.
Now, if this sounds like an overly convenient number, it's not really an overly convenient
number.
The AAMC will generally give you numbers that are easier than almost anything you are going
to find on the average test prep question.
Once we have our givens as 406 Newtons and 1 meter, we can actually cancel out and combine
the units.
So, in this case, a Newton times a meter combine to equal 1 Joule.
So, 406 Joules is equal to the work of the muscles and is also equal to the initial kinetic
energy that the ball has, which, given that there's no air resistance, means that that's
also equal to the final Potential Energy that the ball achieves at the peak of its vertical
motion.
So that final potential energy is going to be equal to mg∆h, according to the formula.
Now, why did I say ∆h?
Because, the thing about Potential Energy you gotta remember is that there is no real
meaningful zero.
Potential Energy is always defined relative to some point where we define it to be zero.
So, we define that to be the point of release.
However, the proper formula is not "mgh," which would just imply some arbitrary height,
but rather the change in height above the zero-position.
Rearranging to get the ∆h, we get the potential energy divided by mg.
Plugging in what we know, we get 406J for the Potential Energy-final from our earlier
calculations, 5 kg for the mass of the shot because of that part of the passage where
the physiologist was angry at the students for engaging in this dangerous activity, and
g is one of the constants the MCAT wants you to know.
Now, this is not Gravity, this is "little g," the acceleration due to gravity, and
while it is 9.81 meters per seconds-squared downwards, we're going to go ahead and use
10 because we're not masochists.
So, combining the terms in the denominator, we're going to see 5 kilograms times 10
meters per seconds-squared, and a kilogram times a meter per seconds-squared is a unit
of mass times a unit of acceleration, which should give us a unit of Force, which is a
Newton.
So, we get 406 Joules divided by 50 Newtons.
Now, of course, 406 does not get evenly divided by 50, so we're going to go ahead and estimate.
And that estimation is going to get us to 400 Joules in the numerator, divided by 50
Newtons in the denominator.
The first thing we are going to do is get rid of any unnecessary zeroes when we're
trying to simplify the numerator and denominator because tens just make everything unnecessarily
complicated.
There's two zeroes in the numerator, and one in the denominator, so we can only get
rid of one, so we do that.
The next thing we do is simplify the units.
Since we have Joules in the numerator, and Newtons in the denominator, we're going
to divide through.
And a Joule divided by a Newton is a meter.
How do we remember that?
Remember, Force times the distance is Work, which is Energy.
So, a Newton times a meter is equal to a Joule.
Now that we've simplified as much as we can, we have 40 meters divided by 5, which
is around 8 meters.
So we take our prediction of around 8 meters to the answer choices, and we see one of them
matches really well.
So we pick that, and move on.
So, Question 2: What is the maximum distance that the average freshman should be able to
throw that shot?
So this is a range question, and specifically, because they're asking for the maximum distance
should be able to throw the shot, they're asking for maximum range.
Now, in order to answer that question and explain it properly, we're going to have
to diverge a little bit into a lecture on Projectile Motion.
So here you see an animated GIF of the projectile during its flight.
And one of the reasons I like this GIF is that it actually shows you how much of the
speed of the projectile is directed in the x- and y-axes during the flight at the different
points where the projectile passes.
So, notice initially that the projectile has some vertical AND horizontal components to
motion, but eventually, when it reaches its peak, in the x-axis the object continues to
move, but in the y-axis, the object is basically just going up, stopping, and turning around,
going back down towards the negative y direction.
After the projectile has reached its peak, notice it's not moving in the y-axis any
longer, so all of the velocity it has is directed in the horizontal direction.
And if you look at the horizontal component of the velocity, you'll notice that it never
changes during the entire flight, whereas the vertical vector that represents the velocity
absolutely does change, first shortening, then becoming zero, then relengthening, but
in the negative direction.
Without calculus, it is impossible to factor in air resistance, because the size of the
force resistance varies as a function of the speed, and thus would constantly getting smaller
as the projectile constantly varies its speed through the parabolic arc.
So there's no way of doing a smooth averaging or simplifiying it for the MCAT.
So, for the MCAT, and for those taking algebra-based physics, we always ignore air resistance.
Heck, even for taking calculus-based physics, it's a reasonable approximation to assume
that the force of air resistance is negligible for a dense enough projectile.
This really wouldn't work for a ping-pong ball, but you see my point.
The key to projectile motion is that both axes of motion should be considered separately
Since, for the MCAT, you always ignore the force of air resistance, the only force acting
on the projectile is gravity, which acts only in the vertical axis.
Thus, the horizontal component of the initial velocity goes unchanged, while the object's
motion in the vertical, or y, axis is an accelerated one.
So the object will continue to move in a horizontal axis at a constant speed until it connects
with the ground, at which point it will stop.
In other words, the horizontal component of the velocity is constant for the entire duration
of the time of flight.
And you get the time of flight from y-axis motion.
The steeper the angle of launch, the more that initial speed is directed straight upwards,
so the longer the projectile will spend in the air.
However, the closer that angle θ gets to 90 degrees, the less horizontal velocity there
is.
Thus, the range is a trade-off between the time of flight and the horizontal velocity.
To make that more clear, let's work on some generalized math.
So, as you can see from this formula, the range is simply the horizontal velocity times
the time of flight.
Because v-nought-x (the horizontal component of velocity) is the side of the right triangle
that is adjacent to the angle θ, it is equal to the hypoteneuse (the overall velocity vector)
times the cosine of theta.
If you didn't follow how to do that, don't worry, I'll be making a video on trig.
Let me know in the comments whether you'd like that sooner rather than later!
Also, take a moment to like and subscribe so you get notified when all my videos come
out.
To figure out the time of flight, we need to look at the motion of the y-axis.
We start by picking a kinematics equation to work with.
(For those of you who want to know how to pick and apply these equations, let me know
in the comments.
These videos will come out eventually, but I want to make them in the order that you,
my viewers, [ my tribe ;) ], need and want.
Alternatively, you can check out my website, medvipvideos.com for a brief written treatment.
Of the 3 equations you must know, VAX, VAT, and XAT, we need to know time, so you can't
use VAX, which doesn't contain it, and since we don't know the vertical displacement,
we can't use use XAT, which would really be YAT in this equation.
But, either way, it doesn't matter, since we're really going to use VAT.
So what is VAT?
Again, check the website.
Now, we have a choice in how we want to calculate: We could calculate the time it takes the projectile
to reach the peak of the motion, which has the advantage of v-final being 0, but has
the disadvantage of making YOU have to remember to multiply by 2 at the end.
And that's an easy way to lose points because it's one of the things that the AAMC is
really good at: creating distractors.
And distractors pull you away based on common mistakes of calculation and thinking that
students will encounter.
And remember, the key to testing strategy for both the MCAT, the USMLE, and beyond,
is: ERROR REDUCTION.
Leave yourself no chance to mess up, and you will consistently grab points.
So, to do this problem in one step, we actually look at the velocities will be when the projectile
finally hits the ground, on the far side of the parabola.
The horizontal component of the v-final will be the same, but the vertical component of
the v-final will be the exact opposite of the vertical component of v-initial, because
the parabola is symmetric.
The acceleration for this falling object is simply little g, the acceleration due to gravity,
directed in the negative y direction.
Rearranging for time of flight, we get 2 times the vertical component of the initial velocity
divided by little g.
The vertical component of the initial velocity is the hypoteneuse, the overall initial velocity,
times the sine of θ.
Combining the algebra we did to get the formula for the time of flight with the formula for
the horizontal velocity, we can actually combine them to get a formula for the range in terms
of the initial velocity and the angle θ.
We can see that the range depends upon the square of the initial velocity (Just like
the Kinetic Energy, right?)
Little g in the denominator is a constant, so that doesn't influence the range unless
we were to switch planets.
The only thing left is the trig terms.
Because they are in the numerator, the Range is directly proportional to the product of
the sine and cosine of θ.
Since one gets bigger as the other gets smaller in the domain 0 to 90 degrees, their product
is the biggest when they are equal.
For the geometry lovers out there, that statement is analogous to the idea that the Area of
a rectangle with a fixed perimeter is the greatest when that sides are equal, in other
words, a square, as opposed to being long and skinny in either axis, as you can see
from this little animation over here.
√3=1.7…
≈1.6 √3/4=1.7/4…
≈1.6/4≈0.4
Given that on the MCAT, there are only 5 angles whose sine and cosine you need to know, the
product is either 0, root 3 over 4, or 2 over 4, aka one-half.
Note that root 3 is roughly 1.7, which we can round down for the purposes of estimating
the division to be 1.6.
1.6 over 4 is 0.4, definitely less than one-half.
So, the largest value that the product of sine and cosine of θ can have is when θ
is equal to 45 degrees or one-half.
It is not really worth working this out EVERY time you have a projectile motion problem,
but it's totally worth knowing that the maximum range that a projectile can achieve
is when the angle of launch θ is 45 degrees.
Take a look at the very colorful picture to hammer home the idea that you exchange horizontal
velocity and time of flight to achieve range.
Now that we've discussed projectile motion problems in general, I want to return your
attention to some of the interim algebra we did so we can discuss how to do projectile
motion problems in a consistent and pragmatic way, not always when they're asking for
the maximum range.
Just keep in mind that the range is the horizontal velocity times the time of flight.
Approach this problem like others, in a top-down fashion.
You can always get the horizontal velocity from applying trig to the initial velocity.
You can get the time of flight by applying the kinematics equations to the vertical axis
of motion.
I would suggest solving for those two things separately, and always keeping your eyes on
the prize and writing down the name of variable you are solving for overall as your very first
step.
Now that we have discussed the mechanics of what we will do to arrive at the solution,
let's actually get it done!
So, that work of the muscles gets us that initial Kinetic Energy, which we can use to
calculate the initial velocity.
Using the overall initial velocity and the angle of launch, we can get the horizontal
and vertical components of the velocity.
We can use the vertical component of the velocity to get time of flight, and multiply that by
the horizontal component of the velocity to get range.
And that is the solution path for range problems.
[Meowing].
So, this is Professor Sidney, master of physics.
Unfortunately, he is a little cranky because this is past his bedtime.
But, that's the only reasonably-quiet time I get to record.
Every once in a while, he becomes a kitty projectile!
[Chuckling] Not really.
Professor Sidney disagrees with my teaching methods!
So, to do this from top-down, remember all we do is, to get range, we keep in mind that
we need the horizontal velocity times the time of flight.
Remember that we can get the horizontal velocity just using trig, and we can get the time of
flight by applying kinematics to the vertical axis of motion.
So, let's focus on the beginning of the solution path: how we get from Work to the
initial velocity.
Again, the work done by the muscles is simply going to be the force applied by them times
the displacement of the wrist during the motion.
So, the work done by the muscles will be equal to the initial kinetic energy the object has,
which is going to be equal to one-half times its mass times the initial speed it has, squared.
We can rearrange that last formula to solve for the initial speed that the object has,
which is simply going to be the square root of two times its initial kinetic energy divided
by its mass.
So, now that we've got our algebra out of the way, we can plug-n-chug.
Yay!
Looking back at the passage, we can see that the average freshman can apply 375 Newtons
of force during the throw, and that during that throw, they displace their wrist, and
thus the shot, an average of 80 centimeters.
Better yet, 0.8 m.
Better yet still, four-fifths of a meter.
Keeping numbers in fractional form is useful because it allows you to simplify and cross
out.
5 divides the numerator nicely, so we get 75 Newtons times 4 meters.
And 300 Newton-meters is the same thing as 300 Joules.
The Work done by the freshman's muscles will be the initial Kinetic Energy of the
Shot, in other words, 300 Joules.
We can use that number to calculate the initial speed.
Plugging into the formula we "algebrized" earlier, we see that we can cross out the
denominator numerically.
5 divides into 300 sixty times, and 60 times 2 equals 120.
We can also cross out the units.
Remember, a Joule is a kg times meters squared per second squared.
Rewriting Joules as its component units lets us cancel out kilograms.
What we have left is a square unit, which is great, because it's under a radical.
The number which is under the radical, 120, is actually not a square number, but it's
REALLY close to another number which is.
So, we estimate everything under the radical to be 121 meters-squared per seconds-squared.
We can take the square root of that, and we can get our estimate of v-initial to be 11
meteres per second.
Now that we have a value for the initial speed, we can calculate for the horizontal component
of the initial velocity, and the time of flight.
It doesn't matter in which order we calculate them, so I choose to calculate the time of
flight first because it's harder.
Knowing the value for the initial velocity, we can just plug n' chug here.
We can see that the denominator for the sine term is 2, which can cross out the 2 that's
already there in the overall numerator of the right-hand side of the equation.
Looking at the units, we see that we have a fractional unit in the denominator.
To eliminate it, we have to multiply by its reciprocal.
That allows us to cross out everything except a second in the overall numerator, which is
part of our logic check, since we are looking for time and getting units of time on the
other side of the equation.
This is part of what tells us that we're on the right track.
Notice that this doesn't help us with constants, but that's why the algebra step of problem-solving
is such important one.
So, our time of flight is 11 root 2 over 10 seconds.
Until we are all done, it's best to keep things in fractional form, because it makes
any further calculations easier.
Now that we have the time of flight, the last interim calculation we need to do is to get
the horizontal velocity, which, in air-resistanceless projectile motion problems, is the same thing
as the horizontal component of the initial velocity vector.
We can plug and chug, and we get that the horizontal velocity is 11 root 2 over 2 m/s.
Now that we have values for the horizontal velocity and the time of flight, we simply
multiply them together.
The seconds cross out, the root 2's in the numerator multiply together to make 2, which
then crosses out with the 2 in the denominator.
Thus, the range is 121 divided by 10, or 12.1, meters.
Taking our prediction to the answer choices, we see that the closest match is C, 12.2m.
Notice that the real answer is slightly larger than our estimate.
And it IS an estimate, since we divided by an estimate for little g, 10 meters per seconds-squared,
that was slightly larger than the actual value of 9.81 meters per seconds-squared, making
our estimate a little bit of an under-estimate, which is part of our logic check, and helps
us gain confidence in our answer.
And with that, we're out of questions, and almost out of video.
But not quite!
Do the Youtube things.
Like, comment, subscribe, and stay tuned for the rest of the videos that are coming out
this week.
I know I started this channel out in a slightly different time in my work schedule, so I wasn't
as regular as I'd liked to have been, but I do this for you guys, y'know.
Check out my website, and see what else I can do for you: medvipvideos.com.
Until next time, this is MedVIP, signing off and reminding you: Keep It Simple!
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