I Baked Cookies For My Students

A while back, I wrote an article on how to impress upon the students of the need have units in most of the numbers that they write in physics. I gave them a recipe for a banana bread, but I left out all the units of measure. It was the students themselves who noticed what was wrong with the recipe, so in the process, I managed to convey to them that (i) without units, these numbers are meaningless and (ii) this is not just something in physics (or science) but rather something common that we encounter and take for granted.

Over the Summer, I did the same thing but I showed them a recipe for my often-requested Chewy Oatmeal Cranberry cookies. Same reaction. But the difference happened at the end of the arduous and intense 8-week summer session. On the 2nd to last day of the class (last day was the final exam), after we did our review, I showed them again the cookie recipe and asked them if they remembered why I was showing them the recipe. All of them did.

I then whipped out a container that had the very same cookies, from the recipe, that I had baked the day before. Oh yeah, they were pleasantly surprised! We basically came full circle, and had a lovely time the last 15 minutes of class time as we sat around chatting and munching on the cookies. Even the coffee machine was nearby and a few of us got some coffee to go along with the cookies.

It was a wonderful end to the class, in my opinion. I am considering this Fall semester if I want to do that again. I just might, if I can find the time.

Zz.

The Unseen Impact of Physics In Healthcare

This is a nice news article that provides a basic summary of the applications of physics in healthcare and medicine. It's another one of those where if someone thinks physics only deals with esoteric and useless ideas, show him/her this. I've mentioned many examples of similar medical/health/etc. applications and concepts that came directly from physics, such as this one.

As someone who often teaches general physics to life science/premed/bio/kinesiology major, this is definitely another useful evidence to get them to realize that the physics class they are taking has a direct relevance to their area of study.

Zz.

Since When Does A "Proposal" Become "Evidence"?

It's one of the reasons why I groan at popular media's reporting of science.

This article is reporting on a paper that proposes a possibility of finding evidence of large-scale symmetry breaking from the  data ".. in current and upcoming surveys such as those undertaken by Dark Energy Spectroscopic Instrument, the Euclid satellite, and the Vera C. Rubin Observatory.... ". Yet, the article is trumpeting the "Incredible new evidence...." as if it has been found. This is similar to accepting speculation as the truth.

Over-selling and over-hyping science does no one any good, other than making it a click-bait.

Z.

The First 12 Months of the James Webb Space Telescope

Many of us knew that it would be a significant instrument. We just didn't know that in its early days, it would make this many discoveries.

In case you were asleep for most of the past 12 months, here is an article that will highlight some of the groundbreaking discoveries made by the JWST. It will not be hard to guess that there will be more earth-shattering (universe-shattering?) discoveries to be made in the next 12 months.

Zz.

ChatGPT and Projectile Motion, Part 2

If you missed my previous physics questions to ChatGPT, check them out here and here. 

In this particular case, I asked ChatGPT a rather common question, and ended up having an argument with it. My questions are in red while ChatGPT's responses are in blue.

Three projectiles of the same mass are launched from the top of a cliff with the same initial speed. The first projectile is launched at an angle of 20 degrees above the horizontal. The second is launched horizontally. The third is launched at an angle 20 degrees below the horizontal. Which projectile hit the ground with the largest speed?

All three projectiles are launched with the same initial speed, which means they all have the same initial kinetic energy. The only difference is the angle at which they are launched.

The maximum range of a projectile launched at a given speed is achieved when it is launched at an angle of 45 degrees above the horizontal. This means that the first and third projectiles will have the same range, but the second projectile, launched horizontally, will have a shorter range.
However, the question asks which projectile hits the ground with the largest speed, not which has the greatest range. To answer this question, we need to consider the vertical component of each projectile's velocity at impact.

The first projectile is launched at an angle of 20 degrees above the horizontal, so it will have a significant vertical component of velocity at impact, causing it to hit the ground with a larger speed than the third projectile, which is launched at an angle of 20 degrees below the horizontal and will have a smaller vertical component of velocity at impact.

The second projectile, launched horizontally, will have no vertical component of velocity at impact, so it will hit the ground with the smallest speed of the three projectiles.

Therefore, the first projectile, launched at an angle of 20 degrees above the horizontal, will hit the ground with the largest speed.

It started off so well, realizing that all three have the same KE. But then, after it answered something that wasn't asked (the range), it went on to be obsessed with the vertical component of the velocity.

I decided to challenge its answer.

But they all have the same initial energy in the beginning. Shouldn't they have the same energy at the bottom when they hit the ground?

You're correct that all three projectiles have the same initial kinetic energy, and conservation of energy tells us that the total energy of the system must be conserved throughout the motion. Therefore, at the bottom of the cliff, all three projectiles will have the same total energy as they did at the top of the cliff.

However, the question specifically asks which projectile hits the ground with the largest speed, not which has the most kinetic energy. The speed of a projectile at any point during its motion depends not only on its kinetic energy but also on its direction and the effects of external forces such as gravity and air resistance.

In the case of these three projectiles, they all have different launch angles, which means they will follow different trajectories and experience different vertical and horizontal components of acceleration due to gravity. These differences will cause the projectiles to hit the ground with different speeds, even though they all have the same initial kinetic energy.

It's a bit puzzling what it is thinking at this point. If they have the same KE in the end, via conservation of energy, then should they also have the same speed?

But kinetic energy is 1/2*m*v^2. If they have the same kinetic energy, they should have the same speed.

Yes, you are correct that the kinetic energy of an object is given by the formula KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity. If three projectiles of the same mass are launched from the top of a cliff with the same initial speed, they will all have the same initial kinetic energy, and therefore the same total energy throughout their motion.
 

However, as the projectiles follow their different trajectories, they will experience different amounts of air resistance and vertical and horizontal components of acceleration due to gravity. These differences in the forces acting on the projectiles will affect their velocities as they reach the ground.
The projectile launched horizontally will experience no vertical acceleration due to gravity, so it will not gain any speed in the vertical direction as it falls, only in the horizontal direction. The first and third projectiles, on the other hand, will experience significant vertical accelerations due to gravity, which will affect their speeds at impact.
 

Therefore, even though all three projectiles have the same initial kinetic energy, the projectile launched at an angle of 20 degrees above the horizontal will hit the ground with the largest speed due to the combination of its initial horizontal velocity and the vertical velocity gained as it falls.

How can the projectile launched horizontally not experience vertically acceleration due to gravity? After all, it is falling to the ground!

You are correct that a projectile launched horizontally will experience vertical acceleration due to gravity, and it will also fall to the ground just like the other two projectiles.

What I meant to say is that the horizontal component of the velocity of the projectile launched horizontally will remain constant throughout its motion, since there are no external forces acting on it in the horizontal direction. This means that the horizontal velocity of the projectile will be the same at the top of the cliff and at the moment it hits the ground.

However, the vertical component of the velocity of the projectile launched horizontally will increase due to the acceleration of gravity. At the top of the cliff, the vertical component of the velocity is zero, but as the projectile falls, the vertical velocity will increase, and the projectile will hit the ground with a non-zero vertical speed.
 

Therefore, while the horizontal velocity of the horizontally launched projectile remains constant, the vertical component of its velocity will increase due to gravity, resulting in a non-zero speed at impact. However, since the horizontal velocity of the projectile remains constant, it will hit the ground with the smallest speed of the three projectiles.
 

The question asks about the speed, not the velocity. All three gain in speed as it falls to the ground. If their kinetic energy is the same, their speeds must be the same.

You are correct that the question asks about the speed of the projectiles, not the velocity. The speed of an object is the magnitude of its velocity, and it is calculated by taking the absolute value of its velocity vector.
 

If all three projectiles have the same initial kinetic energy, then they must have the same initial speed. As the projectiles fall, they will gain speed due to the force of gravity.
 

Therefore, at the moment of impact, all three projectiles will have the same kinetic energy, which means they will have the same speed. The angle at which the projectiles are launched and the resulting trajectories they follow may affect their final velocities, but their final speeds will be the same since they all have the same kinetic energy at impact.

So, to answer the original question, all three projectiles will hit the ground with the same speed, even though they were launched at different angles.

Phew!! That felt like pulling teeth!

It was as if it doesn't know the difference between speed and velocity, and I'm wondering if it was arguing with me about the vertical component of the velocity. It still makes no sense because clearly I explicitly mentioned "speed" at the very beginning.

Once again, if you don't know what you are looking for, there is no guarantee that you'll get the correct answer and consistent explanation.

Zz.

ChatGPT and Buoyant Force

I posted previously about ChatGPT's response to a projectile motion question. It got the answer correct, but with a wrong starting point, which made the whole response rather puzzling, inconsistent, and self-contradictory.

This time, I decided to ask a question that I had asked my students in a poll during a lesson on buoyancy and Archimedes principle.

A block of Styrofoam floats on water while a lead block of the same size block lies submerged in the water. Which one has the larger buoyant force?

The buoyant force on an object is equal to the weight of the fluid displaced by the object. The Styrofoam block is less dense than water and floats because it displaces an amount of water equal to its own weight, which is equal to the buoyant force acting on it. On the other hand, the lead block is more dense than water and sinks because it displaces an amount of water that is less than its own weight.

Therefore, the Styrofoam block has a larger buoyant force acting on it than the lead block of the same size because it displaces more water.

This time, ChatGPT got the answer wrong. Unlike the first time, it started out with the correct idea, which is that the buoyant force is equal to the weight of the displaced fluid.

However, in this case, the Styrofoam is displacing a very small amount of fluid because it is floating above the surface, while the lead block is completely submerged in the water, displacing its entire volume. This means that the lead block is displacing a larger volume of the fluid, and thus, having a larger buoyant force, despite it sinking in the water.

So in this case, ChatGPT once again exhibits its inconsistent and self-contradictory behavior.

I followed up trying to correct this error.

This is incorrect. The lead block displaces more water than the Styrofoam.

You are correct, I apologize for the mistake in my previous answer. The buoyant force acting on an object is equal to the weight of the fluid displaced by the object, and the amount of fluid displaced depends on the volume of the object. Since the lead block and the Styrofoam block have the same volume, the lead block will displace more water than the Styrofoam block because it is denser.

Therefore, the lead block has a larger buoyant force acting on it than the Styrofoam block of the same size, even though it is submerged in the water while the Styrofoam block floats on the surface.

It gets it now, but it is arguing that the lead block displaces more water than the Stryrofoam because it is "denser". While that is somewhat correct, it is not the reason why it displaces more water. It is because it is completely submerged in the water and displaces a volume of water equal to its volume.

If I reshape the same mass of lead into the shape of a bowl, the object itself will have the same "density", but this time, it can float on the water. It now displaces enough of the water such that the weight of the displaced water is equal to its weight, and thus, it will float. Granted that the whole "object" now includes the air that it encompasses so that its "total" density is less than the density of water, but the density of the lead itself hasn't changed. It is still denser than water but this time it floats!

Once again, if you don't know the material, it is hard to know whether you are being given not just the correct answer, but whether the whole response is consistent and not self-contradictory.

More to come.....

Zz.

ChatGPT and Projectile Motion

In my previous post, I mentioned an exercise that I often give to my General Physics students that are learning about projectile motion. Using a PhET application, they found out that two different scenarios produced a projectile motion that has the same maximum height and same time of flight. I asked the students to figure out why.

I decided to post the same question to ChatGPT. The result was a bit of an eye-opener. I will show my question (in red) and its response below:

Why does a projectile shot with initial velocity of 20 m/s and angle 30 degrees with respect to the horizontal reach the same height as the projectile shot with initial velocity of 10 m/s and angle 90 degrees?

The key to understanding this phenomenon is to recognize that the vertical motion of the projectile is independent of its horizontal motion.
For both projectiles, the initial vertical velocity is zero. The only force acting on the projectile in the vertical direction is gravity, which causes the projectile to accelerate downward at a rate of 9.8 m/s^2. The vertical displacement of the projectile can be determined using the formula:
Δy = v0y*t + (1/2)at^2
where Δy is the vertical displacement, v0y is the initial vertical velocity, a is the acceleration due to gravity, and t is the time elapsed.
Since the projectile that was shot with an initial velocity of 10 m/s at an angle of 90 degrees goes straight up, it reaches its maximum height when its vertical velocity becomes zero. At this point, the time elapsed is given by:
t = v0y/a
where v0y is the initial vertical velocity. Substituting in the values, we get:
t = (10 m/s) / (9.8 m/s^2) = 1.02 s
Using this value of t in the above formula, we get:
Δy = (0 m/s)(1.02 s) + (1/2)(9.8 m/s^2)*(1.02 s)^2 = 5.1 m
So, the projectile reaches a maximum height of 5.1 m before falling back down to the ground.
For the projectile that was shot with an initial velocity of 20 m/s at an angle of 30 degrees, we can break down its initial velocity into its horizontal and vertical components:
v0x = v0cos(30) = 17.3 m/s
v0y = v0sin(30) = 10 m/s
The horizontal motion of the projectile is uniform, meaning it moves with a constant velocity of 17.3 m/s. The time taken for the projectile to reach its maximum height is the same as the time taken for it to reach half the range of its motion (i.e., the distance it travels horizontally before hitting the ground). The formula for the time of flight is:
t = 2*v0y/a
Substituting in the values, we get:
t = (2*10 m/s) / (9.8 m/s^2) = 2.04 s
The horizontal distance traveled by the projectile can be calculated using the formula:
R = v0x*t
Substituting in the values, we get:
R = (17.3 m/s)*(2.04 s) = 35.3 m
The maximum height reached by the projectile can be determined using the formula for vertical displacement:
Δy = v0y*t + (1/2)at^2
Substituting in the values, we get:
Δy = (10 m/s)(2.04 s) + (1/2)(9.8 m/s^2)*(2.04 s)^2 = 5.1 m
So, the projectile that was shot with an initial velocity of 20 m/s at an angle of 30 degrees reaches the same maximum height of 5.1 m as the projectile that was shot with an initial velocity of 10 m/s at an angle of 90 degrees, despite traveling much farther horizontally before hitting the ground.

Now, the answer is correct. In fact, it gave me more than what I asked for, i.e. it included the actual calculation of even the range, which I did not ask for.

However, its starting point is incorrect, and in fact, it is self-contradictory. It starts of by saying that the initial vertical velocity for both projectile is ZERO. This is clearly wrong.

Yet, a few lines down, it then mentioned and even found the vertical component of the initial vertical velocity, all while it was oblivious to what it just said in the beginning. So while the answer in the end may be correct, the reasoning, if we were to go by what it said, is puzzling and inconsistent.

I decided to challenge it by pointing out its mistake:

You are mistaken about the initial vertical velocity being zero. For both projectiles, the initial vertical velocity is not zero.

You are correct, I apologize for the mistake. The initial vertical velocity for the projectile shot with an initial velocity of 20 m/s and angle 30 degrees with respect to the horizontal is:
$v_{0y} = v_0 \sin \theta = 20 \text{ m/s} \sin 30^\circ = 10 \text{ m/s}$
And the initial vertical velocity for the projectile shot with an initial velocity of 10 m/s and angle 90 degrees with respect to the horizontal is:
$v_{0y} = v_0 \sin \theta = 10 \text{ m/s} \sin 90^\circ = 10 \text{ m/s}$
So, both projectiles have an initial vertical velocity of 10 m/s. The rest of the explanation and calculations remain the same.

This is now correct. Unfortunately, it requires someone who actually understands projectile motion to spot this. A student learning about it may not notice the inconsistency of the reasoning.

This seems to be the most common issue that I've come across and read about as far as scientific questions being asked to ChatGPT. This was similarly observed in another post that I wrote about. ChatGPT response has correct parts, but there are also parts that are puzzling, contradictory, and often downright wrong, all within one response.

There's more. In my next blog on ChatGPT, I asked it about buoyant forces. You don't want to miss that!

Zz.

My Favorite Web Applications - Part 7

Previous posts:

My favorite web applications - Part 1

My favorite web applications - Part 2

My favorite web applications - Part 3

My favorite web applications - Part 4

My favorite web applications - Part 5

My favorite web application - Part 6

This one is an obvious one. It is from PhET, and it is on projectile motion (the "Lab" option).

I have used this web app in many different situations and for many different purposes, including using it as a virtual lab when we went remote. However, even in my face-to-face classes, I continue to use this during our lessons on projectile motion.

One of the most difficult concepts for students to understand with projectile motion is that the maximum height and the time-of-flight of the projectile depends only on the vertical component of the motion. If the vertical component of the motion remains the same, then regardless of what the horizontal component is doing, the maximum height and time-of-flight will be the same as well.

What I typically have the student do with the app is the following:

• Set the canon to an angle of 30 degrees with respect to the horizontal and an initial speed of 20 m/s.
• Fire away!
• Measure the maximum height and the time of flight using the tools available in the app.
• Then change the angle to 90 degrees and an initial speed of 10 m/s.
• Fire away!
• Again, measure the maximum height and time of flight.
• Compare the two situations.

The students will find that for these two different situations, the maximum height and the time of flight are the same. I ask them to discuss this with their partner/s and figure out why these values come out this way. Then I ask them to find another angle and initial speed where the projectile gets to the same height and has the same time of flight.

Of course, the reason for this is that the vertical component of the initial velocity is the same for both situations. The is the only thing that the two motion has in common. Thus, since the maximum height and time of flight depends only on the vertical motion, the two different situations will naturally produce the same values for each of those two quantities. If they understand this, then they will be able to quickly find another angle and initial speed via simple calculation rather than by trial-and-error.

BTW, watch this space as I will be posting a link to an upcoming blog post of my interaction with ChatGPT on this same question that I ask my students.

Edit: This is my blog post on what happened when I asked this projectile motion question to ChatGPT.

Zz.

Physics Can Be So Distracting!

I've been the internet for a very, very long time, longer than a lot of people have been alive. In all those years, I've had battle scars from my battle with weirdos and cranks of all kinds, especially during the early wild, wild west days of unmoderated Usenet. Even now, I have to deal with them frequently, both online at various forums, and of course, the occasional stuff that tried to appear on this blog.

So you'll understand when I say that dealing with physics cranks is such a major distraction that, often, I see something and the first thing that comes to mind is just that!

I was at a wonderful Turkish restaurant last week, and I knew about this dish called Testi Kebab. I've seen it being served the last time I was at the same restaurant. I inquired about it, and our waiter described it and told us that if we wish to order that, they require a reservation a day in advance. This is because they bake the dish in this clay pot that is sealed. When they serve it, they literally break the pot and pour the content out. The broken clay pot is then tossed away, so they make a brand new clay pot each time this dish is ordered.

I immediately wanted to order that dish, and sure enough, we made a reservation for a group of people, and I ordered two of the dish, one large and one small, since there were 3 of us who wanted it. It was wonderful. The meat was lamb, and it came with vegetables stewed slowly in the sealed clay pot. After they broke the pots at our table, I asked if I could keep the broken pots, and they said yes.

I took them home, washed them, and they are now on display somewhere in my living room. But I have to tell you that as soon as I saw them when I got them home, the first thing that came across my mind was that I now have a couple of cracked pots in my house!

 

 

Oh well, I guess I just have to live with them! 😁

Zz.

Room-Temperature Superconductor?

Here we go again!

Big news with the new publication out of Nature this week. A report on an observation of room-temperature superconductivity on a sample that is under pressure at only 1 GPa. That pressure is exceedingly low considering that most of the other superconductors that that has a high transition temperatures tend to be under hundreds of GPa.

Superconductivity has been observed at 20 °C (294 K) in a nitrogen-doped lutetium hydride under a pressure of 1 GPa (10 kbar). The material was made and studied by Ranga Dias and colleagues at the University of Rochester in the US, who claim that the finding raises hopes that a material that superconducts at ambient conditions may soon be found.

Not only that, this thing changes color as pressure is increased, with it turning from blue to pink at the onset of superconductivity. I'm sure doing a reflectivity measurement such as UV-VIS to look at the phonon modes would be very interesting here. 

But as with anything here, this needs to be independently verified, meaning that another group must be able to replicate the recipe and observe the same thing, before this is widely accepted. We will just have to wait.

Z.

Blackholes The Source Of Dark Energy?

Can blackholes at the center of galaxies be the source of the dark energy that we have been detecting?

That seems to be the conclusion based on two recently published papers [1,2]. Both of these are open access papers, so the full papers are available to everyone.

You may read an explanation and review of the papers at the AAS news website. The implication here is that if this is true, then dark energy is not something exotic or new since it can already be explained with General Relativity.

Now, if only we can find those pesky dark matter.... if they exist.

Zz.

[1] D. Farrah et al., Astrophy. J. Lett., v.944, p.L31 (2023).

[2] D. Farrah et al., Astrophy. J., v.943, p.133 (2023).

ChatGPT Does Physics

I'm guessing that most of you who are reading this have heard of ChatGPT or maybe have even tried it. I have. I had to, because I need to know what it can and cannot do in case my students are also using it. I am still playing with it and trying several different things, so I will have a lot more to say about it.

Still, when I came across this video on YouTube, I had to post it here, because it seems that we have similar general observations (not including the writing the Python code for the infinite square well problem - I've never asked ChatGPT to write a code).

It is true that ChatGPT still can't handle graphs and figures as of now. While this may be a way out for instructors to prevent students from using ChatGPT effectively, the extensive use of figures and graphs is also a hindrance to students with certain disabilities. In many cases, I have tried to make my exams and questions to be more "friendly" toward such students, especially in the spirit of making all my material accessible and moving toward conforming to the idea of Universal Design in Learning (UDL).

Of course, if my questions, including the figures, can be deciphered by text or document readers, then it should, in principle, be capable to be fully understood by ChatGPT, which then removes that barrier of using it to solve those problems. Sigh....

Like I said, there's more to come about this.

Z.