## Wednesday, March 29, 2023

### 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.

## Monday, March 27, 2023

### 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.

## Saturday, March 25, 2023

### 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.

## Thursday, March 16, 2023

### 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.

## Thursday, March 09, 2023

### 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.