2026 AP Physics 1 Free Response

Added: 2026-05-09

[00:00:00]Hi everybody, Joel pal Mr. Friendsley here and I'm sure nobody else on the internet is piling on to upload a video of them solving these glorious free responses from AP Physics 1 2026. So I guess it's got to be me that does it. So anyway, uh before I jump in, uh let me just tell you that this is a service of my little company, World's Finest Physics. There's my website right there.

[00:00:28]And we provide two major services.

[00:00:30]Tutoring for students, not just physics, but math, chemistry, I need to learn a little bit more statistics, you know, uh but otherwise calculus all the way up through college math, uh chemistry, computer science also. And then we have two uh experiences for teachers, professional development experiences for physics teachers. They are called semester 1 and semester 2. And you can see kind of the topics there. uh uh a full asynchronous virtual experience for semester 1 is like AP physics 1 and C all in one package uh first semester physics for most high school honors regulars onramps physics one it's basically all of mechanics including some calculusbased bits but if you're a teacher and you don't teach calculus based physics you just skip over those uh and then the same for semester 2 right and if you teach AP physics CG you would skip the waves and thermos stuff and if you were an AP physics 2 teacher you would skip the calculusbased EMA stuff, right? Um, and this could go with uh anyone else on ramps, physics 2, anyone else who teaches those kind of topics and wants to see uh my instructional methods. Okay, so I'm more than happy to have your school pay for it and not you. All right, so here we go. I have not seen these before. This is going to be my uh disclaimer. I literally just ran them off site unseen.

[00:01:51]So you're seeing me do this fresh for the first time. That means I may actually make some mistakes. Okay. Um, okay. So, let's see what's going on here. Up. Looks like a fluids question.

[00:02:02]So, we've got a fountain and the water shooting out an angle theta kn. At time zero, a droplet of water exits the nozzle and follows the path shown in figure one. The droplet reaches a maximum height h1 above the nozzle. At time tf final, the droplet returns the same height from which it came from the nozzle. Okay. All right. So on the axis shown in figure two, sketch graphs the horizontal and vertical components of velocity of the water droplet. I feel like I want to do this with a straight edge. I want to do this with a straight edge.

[00:02:36]Where's my straight edge? I guess this is it right here. Not the straight edge I was looking for. Okay. So horizontal velocity, it's going to be a big fat flat. The horizontal velocity does not change. And it doesn't say I need to label anything. the vertical velocity component should and I'm going to be super careful here. I don't think the AP people care how how careful you are, but I'm going to be super careful here. And I'm going to make sure that this has the exact same initial and final velocity, but in opposite sign, opposite directions. All right, so there you go.

[00:03:11]So that is my vertical velocity component. decreases on the way up, zero vertical velocity at the highest point, increases on the way down, and uh constant horizontal velocity. Anybody who knows projectile motion knows that stuff. Okay. Dive an expression for the speed and I guess I got to get separate paper nowadays for the speed of the water exiting the nozzle. Okay, express your answer in terms of theta and h1.

[00:03:37]So, here's what I'm thinking right here.

[00:03:39]I think what I want to say is I want to say this. I want to pull out a a kinematics equation. And the kinematics equation I'm going to pull out I think is going to be the no time equation. So I'm going to say vf final 2 = v initial^ 2 + 2 acceleration. Now at this point I'm going to only take the y component the y component and and so that means that the acceleration is down gravity and I'm going to call this deltay. Well deltay is the same thing as h1. That's the same thing as h1. And I am going to be very careful to make sure I call it h1 and not just h because I need to do exactly the same uh subscripts that it gives me. All right. Minus 2 g. Now when you reach the highest point, your height is h1, but your y velocity is zero. You only have vx. Your v y is zero at the highest point. So I'm gonna say v initial y^2 equals zero. Okay. Now with that being said, I'm going to say, okay, that means that V initial Y^2 is equal to 2 GH1. Okay, now I got to figure out this speed right here. I got to figure out the hypotenuse right there. Oh, I know. V initial Y is equal to V. Did they give a name to this? Did they call it anything? They didn't call it like V KN or anything like that. Okay, well, whatever. I'm just going to call it V initial, but I'm going to call V KN sine of theta KN. Okay, I've got to use the exact same subscripts they say. All right, so uh if I substitute this in, I get ah heck, let's just take the square root. V initial y= radical 2 g h initial h1 I mean h1 I mean it's h1. Okay. And then so this is v kn sin thet= radle 2 gh1. So the v kn will be square root of 2 gh1 over sin theta kn. So to review, I only considered the vertical and nobody asked me about time. They say time and it didn't give me time. They didn't ask for time. So I think that I should use the no time equation.

[00:05:43]But the rule is that you have to do the same components. So it had to be a vertical, vertical, vertical, vertical, vertical, vertical, vertical, vertical, vertical, vertical, vertical, vertical.

[00:05:52]Had to be all vertical. Okay? I couldn't have any horizontal in there. So then once I got the v initial y by himself, then I convert them to uh uh magnitude and and sign of the direction and divide. Okay, the nozzle has a circular cross-section with radius r. Dive an expression for the volume flow rate.

[00:06:11]Okay, so volume flow rate and my symbol for volume flow rate is fi. Okay, volume flow rate, but I think the AP physics exam uses Q. That's weird to me. I would use fi if it were me. Okay. Now what is that? That is area time velocity. So the area is pi r squared. And the velocity is 2 gh h1 over sin theta kn. Okay. Now I'm allowed to have theta somewhere in my answer.

[00:06:42]I'm allowed to have h somewhere in my answer. I'm allowed to have r somewhere in my answer. And physical constants like uh pi as a physical constant. So there's your flow rate right there.

[00:06:51]Okay. I'm pretty sure that this worth three points and this worth two. Part B.

[00:06:57]The fountain nozzle is replaced with a new nozzle with a radius smaller. The water exits the new nozzle at the same angle above the horizontal.

[00:07:06]Excuse me. I'm allergic to all this fluid dynamics. The volume flow rate of the water is equal to before. A water droplet exiting a new nozzle reaches a maximum height H2. Is H2 greater? Yeah, it sure is. All right. So, we're going to say h2 is greater than h2 is greater than h1. Here's why. Flow rate equals area times the flow speed. The flow speed less uh radius less radius less area. So the water comes out at a faster or with a faster maybe I should say with a faster speed than V knot. Okay, this makes the Vy at launch greater.

[00:08:05]So the droplet the droplet goes higher before reaching the top where vy equals zero. I wonder how much detail they really need from this. I I don't know and I don't think anybody knows until they actually get to the AP reading. All right, so there you go. Oh, justify include qualitative reasoning beyond mathematical derivational expressions. Yeah, I didn't really use any of that. I guess I did say flow rate equals area time speed, but I I'm sure I said everything I needed to say. All right. Two. Two small discs R and S are on straight horizontal track. At time zero, disc R of mass M KN is located position X knot. Disc R is initially moving with speed V KN in the positive X direction. Disc has mass 3 KN. 3M KN is initially at rest as shown in the top view. Figure one. Frictional forces are negligible. At time t1, the discs collide. Immediately after the collision, disc R moves at the speed 1/2 v knot in the other way. Okay, so here's figure one. This is the before picture.

[00:09:10]So I'm going to draw the after picture.

[00:09:12]Mot is going to be that way at 1/2 v knot and 3 m uh v kn is going to be going that way, but I don't know how fast. Okay. The momentum vector diagram in figure two represents the momentums of disc r and s before the collision.

[00:09:29]Okay, so disc R has one, two, three, four, and disc S has nothing for momentum.

[00:09:38]Draw arrows on figure three to represent the momentum vectors of disc R and disc S. Okay. Now, if you reverse direction, the arrow has to go the other way. And if it's half of V KN, this arrow has to be half as long in the other direction.

[00:09:52]A one, a two. Okay. Okay. Now, this is important. This is important. This momentum, which I'm going to say is four, and this momentum, which is zero, adds up to a net momentum of four.

[00:10:08]That's just basic addition. That net momentum has to be the same here. So, in other words, this -2 and this number has to add up to four. So, what number do I put here with negative? Six. One, two, three, four, five, six.

[00:10:27]There you go. That's what the momentum vectors look like. OH MY GOSH, MR. FRIENDS, you didn't conserve momentum.

[00:10:33]You have more momentum after than before. No, you don't. You don't have momentum after than before. You have the same momentum after as before. The momentum before is four because momentum is a vector. And you have momentum four because momentum is a vector. Okay? No, you don't have more momentum after.

[00:10:49]Don't say 2 plus 6. It's -2 plus 6.

[00:10:52]Okay. Starting with conservation of linear momentum. Dive an expression for the kinetic energy of disc S.

[00:10:57]Immediately after the collision, express your final answer in terms of that crap.

[00:11:01]Okay. So, I'm going to say this. Here's conservation of momentum. It's going to be m v. That's dis S R. That's dis R.

[00:11:10]And that's going to equal M * -2 V KN.

[00:11:14]That's dis R. And then disc S is going to be 3M time his final velocity. I'm going to call him VF final. Okay. So now let's see mv KN equals - one2 mv KN + 3 MV final add.

[00:11:31]So 1 andgative - 1/2. Now you're going to add 1/2 is three halves mv knot equals 3 m vfal. Okay we're going to cancel out the m's. We're going to cancel out the threes. And it looks like vfal is just 1/2 v kn. Vfal is just 1/2 v kn. Okay. Now what is the kinetic energy? So now now we're going to say the kinetic energy of disc S is 12 mv ^2. So that's going to be 12 3 m KN uh 12 VN^ 2. So you get 12 * 3 m * 1/4 V kn that is the kinetic energy of disc S.

[00:12:18]Okay. Um let's see the graph shown in figure 4 represents the positions x as functions of time t until time t1 for each of the following disc one disc s and the center mass two block system.

[00:12:31]This is position versus time on the graph uh shown in figure four. Draw three lines that represent the positions x of disc r disc s and the center mass uh during that time. Clearly label each line. Okay. So let's see. Oh, okay.

[00:12:48]Well, disc S has half the speed. So, what I'm looking at for R is I see a slope of two over one up two. But they told me that R now has half as much speed. So, that's half as much slope.

[00:13:04]That would be a slope of one, but in the opposite direction. This is R. Hey, I don't know if you're paying attention.

[00:13:10]Anybody who's a teacher, tell your students that when they scan these, because remember, we don't handle their papers anymore because of COVID and now the Hanta virus. Okay? So, when they scan them and they show up on our computer screen, they are scanned in in black and white. So, if you say red, if they bring colored pencils or pens into any AP exam and say red is R, uh, blue is S, and black is the center of mass, well, we don't see any of that, okay? We we don't see those colors. It's all black and white. Okay. Disc S is now going to be again half the speed that R had half the speed that R had. So R has a speed of V KN. Now half V KN. So half the slope. S is also half V not. That's what I got before. So that's S. And then the center mass does not change its motion during any collision. Thank you for playing the end. The center of mass does not change its motion during any center of mass keeps the same motion throughout any collision anywhere. During the collision, the magnitudes of the change in momentum of R and S are those.

[00:14:20]Indicate whether they are the same.

[00:14:22]Yeah, they're the same. Okay. Briefly justify your response by referencing a fundamental conservation.

[00:14:30]conservation of momentum.

[00:14:36]Okay, don't actually call the reader an And by the way, the readers, if you hate the question, they hate the question, too. The person reading it is not the person who wrote it. And the person reading it probably hates the question as much as you, right? which says that the net momentum stays the same during the collision.

[00:15:03]So to do this, the momentum lost by r must equal this is embarrassing. the momentum gain by S. I feel like everybody should have gotten that. Uh, shame on you. I'm not even lying to you. Shame on you if you didn't get it. Oh, it's the old block going down and it's frictionless and then it's rough. I can tell already. I don't have to read it. All right. The group of students uh investigating friction. The students release a block unknown mass at the top of a curved ramp. The block slides down the ramp.

[00:15:42]The block transitions to the horizontal surface. frictional forces are uh are negligible on the ramp but there's definitely friction on the horizontal surface. Okay. The students are asked to perform an experiment in which they a single quantity is varied in order to collect data that could be graphed in order to determine the value of mu kinetic the coefficient kinetic friction on the horizontal bit. Students have access to only a meter stick. Awesome.

[00:16:03]That means I'm only going to measure distances. Okay. indicate quantities that could be measured by the students that would allow them to determine okay the height the block is released from H and the distance along the flat frictiony bit D the block slides okay describe A method to reduce experimental uncertainty. Repeat trials with different release height h. Okay. Indicate what quantities the students could graph on the horizontal and vertical axis to create a linear graph. I've got to do some work and I'm going to do it up here. So here if this is h and this distance that it slid is d then potential energy mgh turns into friction energy dissipated energy uh force friction times distance that is the energy dissipated by friction friction is mu force normal and when you're sliding which you are and on a flat piece force normal is mg because there's no other up or down motion or forces or anything like that and it's flat so you get mgh equals mu mg GD. The mgs cancel out and you just get H equals mud. So, I'm going to do is I'm going to plot H on the vertical and D on did I call it capital D here?

[00:17:41]Let's be consistent. Okay. Uh capital D.

[00:17:44]Not that they look up here. All right.

[00:17:47]Uh capital D. Okay. Uh D on the horizontal.

[00:17:52]Describe the relationship between mu K and a feature of the graph from part of the feature is going to be the slope.

[00:17:56]So, mu k is the slope of h versus d.

[00:18:02]Okay, so I'm even going to do hd. It's going to be a line and the slope is mu.

[00:18:07]Okay, it it really is that simple, folks. There there's not a lot of writing that needs to go on here. Okay.

[00:18:12]Um there. Okay. So, clearly state what I All right. Now, a new story. In a different experiment, the students released a block from the rest of the rough ramp and the block is released distance d from a photo gate. Okay. The ramp is incline at angle theta. There's friction everywhere. Uh pass through the photo gate which is positioned near the bottom of the ramp. The photogate determines the speed v. It really doesn't folks. Come on. You can the photo gate only measures time. You just have to do width over time. Can you at least say that the photogate determines speed v of the block as it passes through the photo gate. An unknown coefficient kinetic friction between the block lands that. Okay. Um the experiments repeat several times with different release distances D for the same block. Table one is what they measured. The students correctly determined that this is the relationship between D and V is given by this. Okay, where theta equals 30°. The students want to determine mu k. The students create a graph with dpl plotted on the horizontal axis. V ^2 is going to be on the vertical. So I'm just going to tell you right now if this is the x and this is the y, then this is going to be the uh the slope right here. Now I'm going to take a quick moment. You you don't need this for the AP exam, but I'm just going to take a quick moment and tell you where did this equation come from.

[00:19:22]And here's where it came from. It came from this uh you had to say uh mgh initially becomes 12 mv^2 and friction force times distance. Okay, friction force times the distance d. Right? Do you see that? Okay. Now um okay so friction force is mu force normal and on an incline force normal is mg cosine theta. Okay. 12 mv ^2. And what is mg?

[00:19:49]Well, what is h? Well, let's look here.

[00:19:51]Okay. So this is theta and this is h.

[00:19:54]Okay, theta h. So h over d is sin theta.

[00:19:59]So h is d sin theta. So I'm going to write here that this is going to be mg d sin theta. Okay. Now we're going to cancel out m. It's in every term. Okay.

[00:20:10]Uh I am going to bring this over. So it's g d sin theta uh minus mu g cosine theta d = 12 v ^2. So then here's v ^2, here's g and d. g and d. So g and d are going to factor out. So um and I'm going to multiply by two. So everyone's going to get multiplied by two. So two and g and d get canceled out. Factored out.

[00:20:34]Sin theta minus mu k. It should have been mu k this whole time. Okay. We should use the same subscript. Cossine theta= v ^2. So you see this is equivalent to this. But they gave it to you and it was correct. So this the y, this is the slope, this the x. So what's happening on the next side? Okay, the quantity here is going to be v ^2 in me squar/s squared. Okay, all good. Now I guess I'm going to need a calculator here. Here we go. Okay, so I'm going to do this. You could do this on your Desmos if you want, but I'm doing it on a regular calculator. 2.3 point4.5 6.

[00:21:08]Okay. Uh 29 384149.52.

[00:21:14]Okay. And then we have to square these. Sorry. Okay. So, we're going to square these. All right. All good. So, v squared in me squared per second squared. Hello.

[00:21:28]>> Thank you so much for keeping my room clean. I really appreciate it.

[00:21:34]We would be nowhere without y'all. Thank you so much. This is greatly appreciated. Uh, okay. So, uh, all good.

[00:21:42]270. Okay. So, that's what I'm going to be plotting.

[00:21:46]All right. Uh, what was my highest value? 027. 1 2 3 4 5 6. Oh, they love me. Okay. So, I'm going to do nickel, dime, 15, 20, 25 cents, and 30 cents. Okay.

[00:22:02]All good. Um, okay. So, I'm going to go to 0 2. It's going to be84, which will be about right there. Okay. 3 and.144.

[00:22:13]It's about right there. And then 0.168.

[00:22:16]Oh, that's uh that's kind of ugly right there. Okay. Well, you know what? They give you buggy uh nonlinear, not a perfectly lined data on purpose so that they can test to see whether you can draw a best line. Don't force your line to go through the the uh the origin.

[00:22:33]Okay. So, that's good enough for me.

[00:22:35]Using the best fit line you drew in calculate experimental value. I am using my TI 84 to get the slope of the linear regression. Okay. So, let's just go ahead and do that right now. Okay. Um then right L1 L2. Okay. So, the slope the slope is4683 and that is equal what the devil was that equal to? uh that was equal to sin oh no all of it 2 g sin theta minus mu k cosine theta okay do y'all see that right okay now let me just tell you that if you want I mean Desmos is built into the testing platform use a linear regression on Desmos they make it so easy now to do it but if you really want you can choose two points on the line on the line on the line not two data POINTS NOT TAKING DO THE DATA POINTS BUT TWO POINTS ON THE LINE. The readers are specifically trained to check whether you are taking two data points to make the slope. And you can do uh let's see 23 minus uh 006 over.5 minus.1. Okay, those are points on the line which I show with open rings u not uh not points from the data table. Okay.

[00:24:02]All right. So that's the slope. All right. So, what I'm going to do now is I'm going to say 4683 is equal to 2 * 9.8. Okay. In here, the s of 30. It said 30, didn't it? Theta equals 30°. Okay. So, uh sine of 30 is 12 minus mu. The cosine of 30 is some I don't know what it is. The cosine of 30. I should know it's 866. I should know that. Shame on me for forgetting radical 3 over two. All right, so let's divide.

[00:24:36]Okay, so we're going to take 4683 and we're going to divide that by 2 and 9.8. Okay, then I am going to uh do.5 minus that result and I get 866 mu =476.

[00:24:56]So if I divide that by 866, I get a mu of 0.55.

[00:25:02]Yay. All right. Almost there. And now it's time for the qualitative quantity translation. All right. So, qualitative quantitative translation. Toys X and Y are free to rotate uh with no friction about a vertical axle through the center. They look like a top. The upper portion of each toy has a radius R not as shown in the side view in figure one.

[00:25:25]Great. A string of length Ln sure is completely wrapped around the upper portion of each toy. The rotational inertial toy X is ix and toy I Y is half as much. That's the only difference. Okay. Toys X and Y are initially at rest. A constant horizontal force F not is exerted on the string of each toy which causes toys to rotate as shown overhead view. The axles of the toys remain vertical as the force is exerted on the strings. The strings don't slip. When the string loses contact with the axle of toy X, the angular speed of toy X is omega X and then omega Y. Okay. So, what do we want from me? All right. Um, did it did it ever tell me that there was a uh uh Oh, okay. Okay. In the overhead view in figure two. Okay, fine. All right. Uh, indicate whether omega y is great. Are you serious? Is this for real? Is that all you have to do? Y is easier to rotate. Did I read that? Yeah, y is easier to rotate. Are you for real? Like this is pathetic. So why am I am I being pumped right now? Is this is this okay?

[00:26:32]Uh justify your answer using qualitative reason referencing equation. Okay. So I'm going to say this. Uh both uh both toys have the same kinetic energy since the same force Fnot acted through O U G the same uh distance. What do they call the distance? Do they call it lnot? Okay. Yeah. So force and distance makes work and that becomes kinetic energy. But if y has less rotational inertia, then y must rotate faster to compensate compensate for its lower omega uh lower rotational inertia. I meant to say it's lower rotational inertia. All right, starting with the work energy theorem. Awesome. F* L. I should have read part B before I jumped into part A. F* L = 12 I omega^2. Okay.

[00:27:46]Uh, omega X^2. Uh, what was the IIX?

[00:27:49]This is called ix. Okay, so this is a work energy theorem. Work is equal to change in kinetic energy. Work is equal to change in kinetic energy. Oh or or Newton second law rotational form. Man, they're really helping you out on this exam nowadays. They must think everybody's stupid now. Uh, drive an expression for the angular speed omega x toy x the instant the string losses contact with the axle. Express your final answer in terms of all this good stuff. So I'm going to 2 fn l= iix omega x^2 fn ln over ix= omega^ 2. So I'm going to square root it. Okay. I'm going to try it with Newton second law for rotation.

[00:28:26]So the Newton second law for rotation is going to be net torque equals ix alpha of x and the net torque is f * r * f okay equals ix okay and um alpha x okay so now at this point what am I going to do I don't even have omega I think I have to use this equation v ^2 = 2 a delta shoot I didn't mean that I meant omega^2= 2 alpha delta theta. Okay. So I'm going to just uh say that alpha is omega^2 over 2 delta theta. That's what alpha is. So r f= ix * this rigomearroll omega^2 / 2 delta theta. And then if you bring the delta theta over r * delta theta radius * radians that's length. So you get fot equals two in the denominator. OKAY. SO IT'S JUST MORE STEPS. IT'S JUST MORE STEPS.

[00:29:32]So Newton second law of rotational form was just more steps. Just fire your derived equation part B is or is not consistent with your reasoning. The ix is in the denominator.

[00:29:46]So since IY is less, a lesser denominator makes a le a a greater fraction and a greater omega final something like that. Okay.

[00:30:11]I don't know man. I think my test might be too difficult that I'm giving to my students because this was sad.