2026 AP Physics II SOLUTIONS Free Response Exam

Welcome. We are going to walk through the solutions to the AP Physics 2 2026 free response section. So, whether you are looking to just, you know, calm yourself down and see whether you got that passing score or you're looking to study, I've got you here today. So, we're going to go through the problems starting with number one. In part we have sample of moles of monatomic ideal gas in large sealed thermally conducting container with fixed volume. The small sphere has mass ms in the container and the volume of the sphere is much less than the volume of the container. So, we got little bit in this container here. The gas is initially at state with pressure and volume as shown here. So, we've got We're initially in thermal equilibrium. The sphere is the same temperature of the gas. That's what that means. The gas undergoes heating process until the gas is in state with pressure 3P. So, our pressure goes from to 3P. And the sphere is again in thermal equilibrium with the gas. So, we let the system to come into an equilibrium. The sphere is at the same temperature. The total energy transferred to the sphere in the heating process is Qs. Heat to sphere. The graph in figure two represents the number of atoms per unit speed as function of atom speed for the gas when the gas is in state So, initially, we have Maxwell-Boltzmann distribution showing the velocity speed of each atom and then sort of the number of atoms found at each speed. So, here right in the middle, we've got whole bunch moving at that speed. We've got some that tail off as we get faster and faster. And same thing over here. There's bunch of slow molecules down on this side. We want to sketch curve that represents the number of atoms per unit speed as function of atom speed when the gas is in state So, we are at constant volume. We increased the pressure to 3 And that means at constant volume, is also constant. We're not leaking any molecules from the container. That means the temperature has to increase as well. So, we're probably going to go to temperature of 3 Temperature, we should know correlates to the average kinetic energy of the molecules in the gas, which means that the the speed on average should go higher. so, this kind of like this peak here should shift to the right. And the curve should flatten out essentially. So, we have the same number of molecules under the curve. The area should be the same. If this peak goes over here, we're kind of accessing more atoms at higher speeds, and so the curve should flatten out as we do that. If the temperature got colder, we would have higher peak towards the left, but we have the opposite going here. Our peak should shift noticeably down. probably could have exaggerated that little more. and the the peak shifts to the right. So, that's the key features there. We want the peak to shift to the right to let us know that the the molecules are reaching higher temperature, and therefore higher average speed. And then, kind of in order to access these higher speeds, the curve needs to flatten out to maintain that same area. So, peak shifts right and down for higher temperature. If you show that, you should get full credit there. Moving on. Now, we want to derive an expression for the change delta in temperature for the gas for the heating process the gas undergoes as this gas goes from to Express your answer in terms of NPV and physical constants as appropriate. Begin your derivation by writing fundamental physics principle. so, ideal gas law would probably be the way to go here. We need that equals This is true, but it's also true that for any state, because and are all constants here. If we kind of find that ratio, over should be equal to over This is constant. So, we can see whatever happens to the same thing will have to happen to So, really, think we can just write, you know, minus is as by definition our change in temperature. Let me push it over so you can actually see what I'm doing here. We can use our ideal gas law over here to rewrite each and in terms of the pressure at that time and this constant ratio over So, is going to be over minus over equals delta So, this is good. Our equation is in terms of so far we have and We'll get that in terms of just regular in the second. and and physical constants are. So this is in terms of everything we're allowed to have. All we need to do now is remember that our pressure goes from to 3 So at PY we have 3 over minus just at state over Three of the same quantity. This PV over NR is just kind of like variable 3 minus gets us 2 and writing keeping in terms of the right variables we get that delta is 2 PV over NR. And that does that there. Now we're asked to derive an expression for the specific heat CS of the sphere. Begin your answer Express your answer in terms of MS PV and QS and physical constants as appropriate. So we want to remember that the specific heat The main equation you see that in is this you know equals MC delta always think of this as MCAT if you were planning to go pre-med in the future. You have the heat exchanged to something is equal to the mass of that object. So this should be an MS for the mass of the sphere. This is CS for the mass of the the sphere and this is delta the change in temperature. Now the trick here is that the change in temperature of the sphere because it is in thermal equilibrium with the gas is the change in temperature of the gas which hey we have over here. So all we need to do is recognize that this equation exists for the heat transferred to something for the sphere for example MC delta and we're going to plug in the delta for the gas to make that connect there. So we get MSCS parentheses 2 PV over NR. If you like to pull your constant values out, you can go ahead and pull the two out. And there we have the heat transferred to the sphere. It has the same change in temperature as the gas, so we just substitute our answer from part two over in here to finish that off. Now, we get to part An insulated container is filled with liquid of mass ML, so mass of liquid is ML and it has specific heat CL. The original sphere is submerged in the liquid. The sphere has mass where the mass of the sphere is less than the mass of the liquid. Specific heat is also less than the liquid. So, if we're comparing the two, the sphere has both less mass and less specific heat. Initial temperature of the sphere is greater than the initial temperature of the liquid. Later, the liquid of the and the sphere reach thermal equilibrium. The absolute values of the changes in the temperatures of the liquid are delta TL and delta TS, respectively. So, here we're going to use the fact that if these two are going to come into thermal equilibrium, the heat lost by the sphere, so negative QS represents that is the hotter object, it's going to lose heat in this transaction, and the liquid is going to gain that heat. They must be equal to each other. So, we're going to use our MCAT equation again. MSCS delta equals LCL delta TL. We know the liquid has both greater mass and greater specific heat, so this is large and this is small. In order for this equation to to remain true, for both sides to be equal, we need the change in temperature of the sphere to be proportionally large and the change in temperature of the liquid to be proportionally small. So, we're just kind of setting up this equation, knowing that the stuff on the left has to equal the stuff on the right. We can identify what pieces of this are large and small, and then what needs to proportion out. So, this is going to tell us that the delta TL is going to be less than delta TS. The sphere being smaller, less stuff to heat up, and is more readily changing in its temperature because its specific heat is smaller. need less energy to change it in 1°. Means that it's going to be able to transfer it's going to have its temperature change the most as that heat is exchanged between the two objects. So, less heat or less mass and specific heat means temperature will change more. And that ends question one. We move on to question two. It looks like we have some energy level diagrams here for hypothetical atom. It's never real atom. It's always some made-up one. but we have three states here, one, two, and three with corresponding energies -5 E0, -3, and -2. We're asked to draw arrows to represent all possible atomic transitions that could result in the emission of photon. So, an emission means that we are at higher energy state, and we are going down to lower state. So, we could either go from start at equals three, and we're just going to work our way down. We could transition directly to state two. That's possibility. Or we could go at state three, and we can go all the way down to state one. Alternatively, we could start at state two and move down to state one. So, we have three arrows representing all the possible transitions from higher state down to lower state, three to two, three to one, or just two to one. If we want to label these, we can. Delta 2 1 Delta 3 1 This guy over here is Delta 3 2. Check. Got that one. Now, deriving. Derive an expression for the wavelength of the highest energy photon that can be emitted from the atom. Express your answer in terms of E0 and physical constants. So, higher energy level. We want to remember that the energy of photon is directly proportional to its frequency. that means we want the largest energy is going to have the largest frequency, and thereby is also going to have the lowest wavelength. So, the largest energy transition we can see here is going to correspond to the biggest gap. so, that's going to be from = 3 to = 1. So, we are looking at, you know, essentially delta 3 1, which is 2 E0, 5 E0, there's difference of 3 E0 between those two states, 3 E0 is equal to hf. However, we're asked to find the wavelength. We've not done that yet. We're not allowed to have in our equation at all. E0 is fine, is physical constant, so that's good, but should not be there. So, we're going to use the fact that the frequency can be related to the wavelength by the speed of light over the wavelength there. We're going to make that substitution, hc over lambda, and now we have 3 E0 = hc over lambda. Usually they like it when your derivations, like it's fine to have them just algebraically equivalent. So, we basically have function of the wavelength with just E0 and some physical constants. But, if you're purist like am, you'll actually want to solve for that lambda. So, lambda's stuck under the HC here. We can pick that HC up, drop it on top of the left-hand side, and we'll end up with lambda equals HC over 3 epsilon not. So, not not too bad there. Now, we're going to get to part and We have new information. device can emit monochromatic electromagnetic radiation. The wavelength lambda of the radiation can be varied over continuous wave of range of wavelengths. On figure two, which is guess it's on the next page. We're going to sketch the energy of emitted photons from the device as function of lambda for the range of lambda not to four times that. photons with lambda not have an energy 4 E0. So, if we're going back to the idea here, energy is proportional or is equal to HF. That means it's equal to HC over lambda. So, when lambda equals lambda zero, we're told that is equal to 4 E0. That means that when lambda is four times its value here, the maximum value of our range, we're going to multiply our denominator by four, and that's going to cut our energy in half by four. So, we get just an energy equal to not. So, we're going to use that probably in our diagram. We're starting with highest energy for the lowest wavelength at 4e and then we get down to eventually just once we get to the highest wavelength. So, we are sketching the energy of emitted photons from the devices function of lambda. So, we're starting at 4e. We are getting down to we just said we're going to get to at that highest wavelength and our relationship again between and lambda is inverse. So, we should draw kind of smooth downward curve between the two. It should be concave up as 1/x type function. So, we got the right end points at you know, the the lowest wavelength and the highest wavelength and we just want to kind of nicely connect that with smooth curve. Want this is goes like 1 over lambda. Now, the student claims that based on the energy states shown in figure one, the atom can emit photon of wavelength lambda naught and energy 4e. indicate whether this claim is correct. So, we're looking for on our diagram change in 4e, which don't think we can get. We have three emissions here. One has an energy level of only 1e. The next one has an energy drop of 3e. That's what we found in part in our derivation and then this one here from three to five only has an energy gap of two. So, we would say no, the claim is wrong. Atoms can only emit frequencies or wavelengths rather, guess that's what we're talking about. Same difference, but we should be more answering the direct question. Atoms can only emit wavelengths that correspond to energy levels. Energy We'll say energy level differences between states. The atom has emissions of Now, we said E0, 2E0, and 3E0. Not 4E0. So, we've mentioned that atoms can only emit wavelengths that have energies equal to the gaps in our energy diagram. gap of 4E is not found in our diagram in figure one. So, sorry student, but you need to read your diagram bit more carefully. Hopefully, you read yours the right way. All right, that does it for question two. And we're moving on to three. In experiment one, scientists want to collect data that can be graphed to determine the magnitude B0 of an external uniform magnetic field. The field is directed in to into into figure one. So, it's going into the page. So, those X's mean the kind of that's an arrow, like literal archery arrow pointing away from us, and we're looking at the little feather fletching on the back. it's into the page between identical parallel conducting plates. They're connected to power supply of variable EMF or voltage, as we sometimes call it. the plates are known distance apart, where is much smaller than the dimensions of the plate. So, the area is really, really big. We have nice little shh shh small distance between them. In each trial, device emits small charged spheres with speed through the region between the plates. So, this little device here, this rectangle, this magic charge shooting rectangle, is shooting them through the plates and towards this motion detector here. We can vary between the trials. Gravitational effects are negl- negligible. So, we're only looking at the effects of the electric and magnetic fields. We're going to indicate quantities that could be measured using the available equipment that would allow scientists to determine the magnetic field by using linear graph. So, it seems the idea is that we are varying the electric field from the power supply. So, the the the voltage here is going to control the electric field between the plates, and eventually the magnitude of that electric field will equal will be large enough to give you magnet or sorry, an electric force on this the moving charge that is equal to the magnetic force from the field there. So, the charge is going to move through in straight line, and we should be able to measure its velocity with the motion detector. So, we're going to measure with motion detector. We're going to measure the EMF with the power supply reading. And that should be enough. So, we're going to vary until the velocity coming out of the device is the right speed to be undeflected. From that, we should be able to build an equality between the electric and magnetic fields and solve for or extract Briefly describe method to reduce experimental uncertainty. Here we go. This is where you just say do this. do multiple trials. So, do the experiment multiple times and take an average. Make sure that no single data point is fluke. You take kind of what all the data says together. You can also use wide range of velocities. So, we're going to use wide range of velocities to you know, just get more data. We don't just want to do one velocity and one EMF to go with it. We want to see full trend between the the velocities that we put in from the device and the resulting EMFs. It gives us the right feel. So, use wide range of velocities to see trend of and epsilon. Want to indicate what quantities the scientist could graph on the horizontal and vertical axes to determine B0. So, our idea here is we want the electric force provided by the the the plates to be equal to the magnetic force. The magnetic force on charged particle is QVB. So, there's our That's where we're going to get that. the force provided by our power supply here. These capacitor plates are going to charge up and there's going to be an electric force that goes straight up or down between them. the electric force is just the charge times the electric field. And then the electric field in the And then we can re-express this electric field as the voltage, the EMF supplied by the battery, divided by the distance between the plates. We should get constant electric field, and then we can use the relationship, you know, usually it's equals ED. Ed is the way always remember it. So that is going to be delta over that means that we can change delta into the terms of the problem here, EMF epsilon, and we should get the Q's will cancel out. we get epsilon over equals VB. So we can look at how epsilon and times all relate to each other. So we basically solve for epsilon. So what we're going to plot, we're going to graph on the vertical axis, the EMF. We're going to graph the velocity of the charge that we measure with the motion detector on the horizontal axis. And that means we're going to How are we going to get So that we already did part one. Part two says briefly describe the relationship between B0 and one feature of the graph. So if we're modeling this in equals MX plus linear model, plays the role of the EMF. plays the role of our independent variable, which means the coefficient of that BD, is going to represent the slope of that line. So the slope is going to be equal to BD. We can divide our slope by to get the magnetic field. If you want, you could also plot like times and then your slope would just be but kind of like to have my clean and as as my variables. It's kind of personal preference thing. So, we're going to get the slope from the graph of epsilon EMF versus the velocity of the particles. Once we extract that slope, we're going to divide by to get the magnetic field we're looking for. All right. We have the second experiment. there's always an experiment two. scientists want to use graph to determine the mass of an identical charged particles that are emitted from device. Each has charge up here. In each trial, particle is emitted with speed 3 * 10 ^ 6, about 1/100 of the speed of light, toward region of the external uniform magnetic field with variable magnitude So, this field is going out of the page. The little tips of the arrows are coming out of at us with the dots. And we have the ability to change the field in here. The radius of the path along the particle moves is shown here. In bunch of trials, we get relationship between the magnetic field the scientist used and the radius of that circular motion. We want to find the mass. So, what we would do here is note that while the charge is going through the magnetic field, it is moving in circle, which means that the net force is also the centripetal force mv ^ 2 / So, our centripetal force is there. The force that's supplying that centripetal force is the magnetic force, which we can write as we did earlier as qvb. We're going to rearrange this thing to get relationship between and and so, let's see. We're changing the field. We want to see what the turns into. So, we can rewrite this as / qr = will be our variable. is going to be inversely proportional to this radius And so, that means to linearize this graph and be able to get from this. So, we can see we can plot versus 1/R. That means the slope of our graph is going to be the coefficient of 1/R, which is going to be MV/Q. And we know and we know we're just going to be able to get from that constant. To do this, we need to make another column in our data table here, 1/R. Make sure that when you transform your variable, 1 over the radius is going to get us 1 over meters as well for those units. They're going to want to see you understand that everything changes when we linearize. We'll have to do 1 over each of these values. So, we can carefully put that into our calculators. And let's see. 1 over 1.8 is 0.556. 1 over 1.2 0.8333. 1 over half is just two. That one's nice and clean. 1 over 1.4 or sorry, 0.4 is 2.5. And then 1 over 0.3 is 3.333. So, we are going to ultimately plot this on our axis and this on our axis. Now, got to do the actual graphing. We want to label the axes of this grid. and so we said we're going to have one over here and one over meters. We're going to have the field here in Tesla. Our field values go kind of pretty nicely up .04 up to .2. So, we can maybe go in increments of .04, 0.08, 0.12, 0.16, 0.2. 0.24 at the top. Our one over radii go from, you know, zero up to about 3.5. we only have 1 2 3 4 5 6 there. So, maybe we go up in 0.75 increments. 1.5, 2.25, 3, 3.75. And that should capture all of our data there. So, our first data point is at one over equals .556. The way did this, maybe this wasn't the best way, each tick mark here is .15. so, .15, .3, .45. We're kind of in between here, so plot there for .556. the next one's at .883. So, that is going to be kind of right here at .6. So, right in the center at this point here. Our next one is at two. So, that jumps way up, but it's at .14. So, like right in the center here. there. Next one's at 2.5. So, this gets us to 2.4. We're kind of in between these two tick marks here. You go up to 0.16 right on the line. These two. And then lastly, 0.2 3.33. So, like two tick marks over and little bit more. Something like that. Now, we're going to draw line of best fit. Here's where you use your straight edge. I'm just going to draw it straight through trying to get an equal number of points above and below my line. So, something like that is good enough. Now, we're going to use our slope to calculate So, I'm going to find This point luckily ended up on the line, so can use that. I'm going to find another one that intersects grid line. kind of like this point here. At 0.9 and whatever that ends up being. So, our first value is 0.16. Our next one is it going to be 0.04 to 0.08. So, each of those tick marks is 0.04 / 5. So, that lands us, you know, three tick marks at about 0.064. Our values are that one was at 2.5. Minus this one's right on like 0.9. So, we subtract those values. 0.16 - 0.064 is 0.096. We're going to divide that by 2.5 - 0.9. 1.6 Slope is equal to about 0.06. So, now that we have that, we can go back to our relationship that we wrote earlier. slope is going to equal MV over So, we know was 3 * 10 ^ 6. Our charge was 6.4 * 10 ^ -19. And this is all times which is equal to the slope we found of 0.06. So, now we got to do some calculator fun. We're going to multiply our charge 6.4 * 10 ^ -19 * 0.06. And then we're going to divide that by 3 * 10 ^ 6. So, mass is 1.28 * 10 ^ -26 kg, which is very small, but is not you know, it's it's about 10 times heavier than proton, so it's not completely unusual mass for particle. This seems to make sense given the, you know, the speed and the charge that it can reach. this might actually be helium nucleus. Now that I'm looking at it, this looks about four times the electron charge. Or no. That'd be too many. Yeah, doesn't matter. Point is we found the slope using the relationship we got from setting the centripetal force equal to the magnetic force. And now we're on to question four, the final question. We've got small sphere, not shown. Great, thanks. with positive charge plus is initial speed VS in the direction and kinetic energy KS at point So, we're starting here and then we move to point The force from the electric field is the only force exerted on the sphere. So, we're ignoring things like gravity and such as the sphere moves between those two points. Has final speed VT and kinetic energy KT when it finally arrives at We're going to indicate whether VT is greater than, less than, or equal to VS. So, is the velocity of the charge greater than or less than or maybe even equal to it was when it started at There's couple ways to think about this. One, we have this dividing zero line here with negative potentials on the left and positive potentials on the right. So, we can kind of imagine our positive charge at moving away from negative charge closer to positive charge over here. That's why we have this negative potential and positive potential. We're between positive and negative charges. Positive charge is naturally going to be repelled by positive charge. So, as we go from to we would expect the the object to slow down as that repulsion happens. So, we're going to, you know, move toward positive potential. That's going to have more repulsive force for the positive charge. That repulsion is going to decelerate the charge and cause it to slow down. So, we predict that VT will be less than VS. Another way to think about this is with conservation of energy. We can look at what energy is going on at So, we have the potential energy times -3 V0. We're going to add that to the kinetic energy at And then we're going to equal that to the potential energy it has at point Sorry, this is KS. My fault. Potential energy at kinetic energy at We're going to set it equal to the potential energy at 2 V0 plus KT. So, if we know we start with negative potential energy and we're going to get positive potential energy at the end of the day. So, we have negative thing plus positive thing needs to equal the same value. So, we have very negative thing here. As this thing gets more positive we need KT to decrease compared to the KS we started out with. So, if this is very negative guess if these are both positive, this is positive potential energy and this is you know, kinetic energy is always positive. So, our right-hand side tells us the final energy is positive, which means the initial energy had to be positive as well. so, we have very large kinetic energy with you know, very large negative potential energy taken away with that to get us positive number. Here to get that same positive number, we have positive potential plus small positive kinetic energy. So, to kind of balance out very negative this is positive this needs to be small to get the same value. So that agrees that VT is less than VS. We want to derive an expression for KT, the kinetic energy at in terms of V0, this kind of potential unit we're using, the charge, and KS, physical constant. So can kind of do what we just did up here. We have VS plus KS equals VT plus KT. We can leave KS as it is. We're allowed to just have kinetic energy in our answer. We need to rewrite these potentials potential energy. Like said, should be used, not V's. Rewrite these potential energies in terms of the potential and the charge. Well, that's an easy transformation. is just going to be times at given point. So we can replace that with So it's the same charge, so it's going to be here. We have the potential at point which is -3 V0. We're going to add to that KS, and we're going to do similar thing on the right-hand side. UT is going to become times 2 V0 plus KT. We're trying to solve for KT, so we're going to subtract Q2V from both sides. We get -3 -2 more is going to get us -5 QV0 plus KS equals KT. And there we have it. This is all, you know, 5's fine to have. It's guess it's fundamental constant. and are both allowed. KS is allowed, so we are have satisfied the rules of the derivation here. Finally, we come to part We have new small sphere with the same mass but with negative charge at the same speed as before in the positive direction. The force in the electric field is the only force. So, every everything is the same except it's negative charge. The sphere has kinetic energy new when we get to point We want to indicate whether new is greater than, less than, or equal to the original thing. So, we go back to our derivation. V0 + KS = KT. So, it tells us to go back to your derivation and we're going to justify our answer. the only thing that's different here is that is now negative. This term here for KT is already quite negative with positive charge. When we have negative charge, negative is going to give us double negative. The negative from the negative charge is going to cancel out with the negative five here. So, this whole term is going to be positive. And we're going to add that to an already positive thing and we're going to get big KT. Here we have positive KS. That's the same in both situations and then we subtract from that just by virtue of having negative term over here. So, the negative charge is going to make this term positive. Two positives added together is larger than positive and negative. If they're the same terms at least. and that's what we have here. We have the same terms. It's just, you know, originally one was positive, one was negative. Now we're adding those same two numbers together but they're both positive. So, we're naturally going to get larger thing. So, KT should be greater Sorry, new should be greater than the original KT. And there we have it. We have completed the exam. hope you found this helpful. if I've made mistake, again, this is the first time I've seen these questions, feel free to complain about it in the comments. best of luck with any other exams you have, and have great rest of your day. Thanks for watching.