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### AP Practice Exam: Free Response, Part 2

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Question 3 0:14
• Question 4 4:34

### Transcription: AP Practice Exam: Free Response, Part 2

Hi everyone and welcome back to Educator.com.0000

I am Dan Fullerton and now we are going to continue our work through the 1998 AP Physics B examination by doing free response questions 3 and 4, so let us dive right in with number 3.0002

As we start number 3, one thing to note, the AP curriculum has changed a little bit since they released this practice exam, so not all of this question has items that are covered in the curriculum currently and I will point those out when we get to them.0016

To begin with, we have this very interesting apparatus that is designed to explore the conversion of mechanical energy into thermal energy.0030

In Question A, it is asking us what happens if you close a lower valve, flip this upside down so that you have all of these beads on top -- What amount of work must you do in order to do that?0038

That is a pretty fair question for them to ask us, so work, if you recall, is force × (r) × cos(θ) or in this case it is a conservative force so we really do not even care about that cos(θ), it is just going to be Fr, which is going to be Mg, the force required to overcome gravity times (h).0049

Or you could have done that by conservation of energy, realizing that you are going from what we are calling 0 down here all the way to gravitational potential energy Mgh, so there is our answer to part (A).0069

Now in part (B), it says we are going to open the valve, let the leads tumble down into the hollow sphere and as that gravitational potential energy gets converted into thermal energy, what is the temperature increase of the lead?0082

This would not be on the current test or no longer asked, but we will do it anyhow, since we are here for those of you who might have an interest.0096

In this case, that energy that we had, (Mgh), is going to be equal to the mass times a specific heat, times the change in temperature, therefore, δt will be Mgh/MC or just gh/C.0103

Again, if that looks unfamiliar, that is fine, it is probably supposed to.0122

For section (C), again this is another section that really probably should look unfamiliar to you, but I will do it because we are here anyhow.0128

It is asking us to calculate the cumulative temperature increase after you do this 100 times, so first thing, let us figure out what the temperature increase after doing it once.0138

That is just going to be gh/C and it gives us these values of 10 m/s2, our height is 2 m, and our specific heat is 128 J/kg-K, so that is going to be 0.156 K for one repetition.0148

If you do that 100 times or 100 repetitions, you are going to get an increase of 15.6 K for all of those repetitions.0167

Part (D) is saying what if you did this with a mass that was 1/10 as large? How would your answers change?0182

Well, if you look at (B), we did not have any mass dependents, so since there are no mass dependents, our answer would not change.0191

In part (C), also, had no mass dependents, so the exact same answer -- no mass dependents, therefore, no change.0208

The last part of the question (E) -- When the experiment was actually done, you had less of a temperature increase than you predicted.0217

Explain what could have happened there?0224

Well, I would say something about the friction between the beads and the container could lead to less than 100% efficient energy transfer, something about friction between beads and container leading to less than 100% efficient energy transfer.0227

That should put you in good stead on that problem, but again, if that one gave you trouble, really most of this problem is not covered anymore on today's current AP exam.0261

Let us move on to one that is.0271

Number 4 -- We are given a circuit with a bunch of light bulbs, (A), (B), (C), and (D) and we know the battery gives us a constant potential difference, so it is going to be an ideal battery -- we are going to neglect any effects of internal resistance.0275

The first thing we are asked to do is draw a circuit diagram here and that is probably going to be pretty useful to us because looking at it in its current form is kind of tough to read.0287

Let us draw it with a voltage source, (V), and as we go from the top of our voltage source...0296

...we see resistor (B) and we will call that (B) there and from (B) we go to (C) and from (C) we head back through (A) to the negative terminal of our battery.0306

But then we also have another loop here, where we have resistor (D), so that would be my diagram for part (A).0323

Now for (B), it asks us to list the bulbs in order of their brightness, from brightest to least.0336

The first thing we need to know is that brightness is proportional to the power.0344

Really what we are looking for are the bulbs that are dissipating the most power.0350

Without a doubt, (A) is going to have all of the current and most of the voltage, so pretty easy to see that (A) must come first.0355

Why? It is going to have total current of the circuit going through it and the most voltage, therefore, the most power.0363

Next up, (D) has the same voltage as (B) and (C), but because there is only one of these it is going to have lower resistance, so it is going to have more current flow, so (D) must come next.0376

It has the same (V), same potential drop as the (BC) branch, but lower resistance, so that is more current, and therefore, more power.0396

Finally, (B) and (C) have the same current and the same voltage drop across them, so they would come last -- (B) and (C) -- same current, same voltage drop, same power, so equal brightness.0420

Great! Moving on to Part (C), we are going to remove bulb (D) from the socket and describe what changes happen to bulb (A).0443

Let us redraw that where we pull (D) out of here, so now we are going to have our voltage source and our circuit is going to have (B), (C), and (A), a simple series circuit.0453

What happens here is our total current, or our total (r) is going to increase because we do not have this parallel section, so if our total current goes up, that means that the current through (A), which was the total current must go down...0471

...our power is therefore going down -- Power = I2R -- so our power is going down, so the brightness of (A) must decrease.0487

In (C2), we are going to describe what happens to the brightness of bulb (B) when this happens, so bulb (B) -- What is going to happen to bulb (B)?0505

Well, across bulb (B), we are going to have the potential drop go up, so the voltage drop across (B) is increasing because we do not have this parallel element anymore...0520

...so the voltage is going up and power equals V2/R, so that is going to be a very strong term and we are going to see the power through (B) go up, therefore, (B) is going to get brighter.0530

That will do it for questions 3 and 4 on this practice exam.0550