Physics 11 Final Exam Review Part 3 (Forces Part 2, Momentum)

So looking at forces, remember for forces, what were our tools? Free body diagrams and Newton's laws. We only had really only had three equations. It was Fnet equals MA. That became winner minus loser equals MA. We had weight. No, now the force of gravity was mg, which technically is ma in disguise where g is the acceleration due to gravity. And we had an equation for friction. It was mu times the normal force. We introduced the concept of the normal force. How hard the ground pushes up. Also, what a scale measures. So, example three says a 75 Newton. Ooh, what did I just give you when I said a 75 Newton sled? Did I give you its mass? What did I give you? No. Now, I gave you its weight. So, this is mg, which means I can find m. I might make a little note here. M equals it's going to be 75. How do I turn newtons into kilograms? Yeah. In other words, to get the m by itself, divide the g over because the g is multiplying. So, I'll divide by 9.8. And I'll probably store that on my calculator because I don't think this is going to work out to a nice number. Maybe it will, but I'm doubtful. 75 / 9.8. I'll make a note. Yeah. Uh 7.653. Maybe I don't want to round off too much. I wouldn't say 8 kg even 7.7. That's going to introduce a lot of error into the rest of my calculations. Uh, a 75 Newton sled is pulled across the snow at a constant speed. A horizontal force of 85 newtons is exerted to keep it moving. What is the coefficient of friction? This is already a pretty difficult question. I would consider this probably a C plus B. I call this a B-level question. Um, there's a key phrase in here that's going to make this slightly easier. Ryan, what do you see as the key phrase that's going to make things slightly easier? If it's at a constant speed, what's my acceleration? And that means forces are everybody. I'm looking for the stress letter B. Okay, this is going to be a job for a free body diagram. So here's my sled. What are the forces acting on it? Martin, get the obvious one. Is the sled sinking into the ground like quicket? No. Is it flying through like Superman? No. You know, and in this case, I think the normal force is the same size as mg. So later on, I'm probably going to say I don't know the normal force. Oh, but look, look, look, look, look, look. I don't know the force size as the normal force. What else? We have this horizontal force of 86 Newton. Who remembers what did we call the mystery force coming from off the page quite often because we don't want to make up a backstory. Yeah, we called it F applied. That equals 86. By the way, I guess I could even put the 75 right there, which means the normal force is also 75. It's got to be. Why can't this be correct? Why is this impossible based on this question? That can't be going at a constant speed because forces aren't balanced. Jacobe, which way do I need to add an arrow? How big? Yes, there's friction. That doesn't quite look the same size, but it should be the same size. In fact, I can even say friction is 86 newtons in this case. Now, if there was an acceleration, that would make raise this to like a B plus A minus level question that I'd have to do a full winner minus loser. But in this case, because it's a constant speed, it's a tie. I could go, who's winning? Who's No one's winning. What I can really say is horizontally friction equals F applied, where F applied is 86 newtons. They told me that. Friction is what times what? So I could say on the next line, mu times the normal force. And they want me to get the coefficient of friction by itself. You know what they want me to get by itself, Rat? The mu. What's the normal force doing to the mu, Renat? Adding, subtracting, multiplying, or dividing. How move it over then? And so I'm going to divide by the normal force. I don't know. Oh yes, I do. The normal force is 75 newtons. It's going to be 86 divided by 75 newtons. I've made up a terrible question with terrible numbers. I'm going have to change this next time because we're going to get a coefficient of friction bigger than one on snow. It should be like 0.1. We're going to get a huge nonsense coefficient of friction. I'll change that for next year. This is I made up really dumb numbers. To keep this going at a constant speed, I probably only need to pull with 8 newtons, not 86 newtons in real life because it's easy to pull stuff on snow. So, I apologize for that. But it would be 1.15, which h I hate making up dumb numbers, but clearly I did this time. This irritates me. And and that that's what they wanted us to find. Normally coefficients of frictions again are between zero and one. And if the closer to zero, the slipperier it is. Snow is slippery. Should be like 0.152 maybe. So I feel bad for this one. Sorry. An object with a mass of 32 kg has a weight of 528 newtons on a certain planet. What's the acceleration due to gravity on this planet? Okay, they gave me the mass. They gave me the weight. Oh, my equation for weight is mg. Now, we're on a different planet. G is not 9.8. Eve, they want me to get the acceleration due to gravity by itself. How would I do that? How would I get the G by itself? Because that's the acceleration due to its gravity. This mystery planet is gravity. I'll move the m over. What's what? Yep. Yeah. We call this gravitational field strength. So, it's going to be the force of gravity on the planet divided by the mass. Remember, mass is the one that doesn't change. It's constant anywhere in the universe. Wait, no. Now, that's the one that changes cuz it depends on what planet you're on because of the G. What's G on Earth? 9.8. What's G on the moon? 1.6. G on Mars about 3.3. So, this is a mystery planet. It says the weight is 528 newtons. It says the mass is 32. Let's find out what the acceleration due to gravity is on this mystery planet. I get 16.5. Probably a bigger planet than Earth. Am I wrong? Not Newton's Mr. Dick. Thank you, Mason, for giving me the stink eye there. It would be m/s squared cuz it's an acceleration. Or you could say newtons per kilogram cuz we did go force divided by mass. That is newtons per kilogram. These two units actually work out to the same thing. B. Mason, what does B want me to find? Oh, that's easy. The weight is mg. So, it's going to be uh 125 * 9.8, right? Oh, we're on a different planet. It's going to be 125 times new gravitational field, which is 16.5. Let's crunch that and see what we get. Pause the video. Got someone coming in late. So, it's going to be 125 * 16.5. Uh, do you get 2,62.5? I'll go 260 newtons, but if you wrote the whole thing, that's whatever. Your test is multiple choice, so that's the one you would pick. I'd probably go to three sigfigs. Turn the page. This I would consider a solid A level question. And the reason is there's an extravertical force besides mg. I see friction. I see an F applied. So it says given the following situation, find the normal force. Find the acceleration. You know what? Before I even start, what might this be a good job for? >> Yeah, I'm going to do a free body diagram. So, did they give me a picture? I'm going to label the picture right on here with my forces. What are the forces acting on this C? And I get the obvious one. Which way? Yep. So, I'll put that there. MG. What else? Are we sinking into the ground like Quickset? No. Are we flying like Superman? No. So, there is a normal force, but you notice there's an extra vertical force lifting up. That's not the normal force because someone's lifting up. The ground isn't having to push up as hard. The normal force is going to be smaller than mg. If someone was pushing down, did I do pushing down? Nope. If someone was pushing down, the ground would have to push up harder. Then we have F applied. Oh, there's a coefficient of friction. So, I should also add Sienna. There's friction. So, A said, find the normal force. Did I give you lots of space? Maybe. I'm going to do part A. I think right over here in the margin over here. Sienna, back to you. Are we sinking to the ground like quick sand? Are we flying here like Superman? Normally, I would say I don't know the normal force. Oh, but look, look, look, look. Except here, it's a bit more complicated because they're extravertical forces. What I can say is everything down has to equal everything. So, everything down in this case is mg. Everything up is the normal force plus and the 36. If I was pushing down, then Jamie, I would have mg plus something equals the normal force, which is kind of easier because the normal force is already by itself. See, I got to get the FN by itself. How? I'll move the plus 36 over. Yeah. How? Yeah. The normal force is going to be mgus 36. And I'm going to go straight to numbers. M is 12. I'm going to go 12 * 9.8 - 36. And I'll carry all the sigfigs. I get, double check me, 81.6. So there's part A. I would have to do that if on a test I gave you this one. I don't think I gave you one this tough. Oh, sorry. That's a lie. On your final, I did ask you to find a normal force that isn't mg. Probably half of them will have something pushing down. Half of them will have something lifting up. But I don't think I added the part B. Part B says find the acceleration. I did this just to review the fact that we can handle it. Uh if we are accelerating, we're probably accelerating to the right. So Claire, who's winning? Losing you think friction's winning and f is that correct? I think maybe now we'll find out if we get a negative acceleration. You might have been right. Minus friction equals ma. Was it Ryan's theorem? Are you the one I and you barely showed up on time? Wow. We can get the A by itself. Divide by M. I'm going to move sideways to try and make sure I fit this all in. F applied is going to be 43 minus mu * the normal force divided by m. You know what that's going to be? 43 minus what's mu? I don't know what's mu.36 uh the normal force 81.6 divided by 12. That'll get us the acceleration. Once you have that written out, go to your calculator. Uh, a a solid A or A+ level question would be me giving you the acceleration, asking you to get the coefficient of friction, the mu by itself. Let's see. Mr. Dick brackets around the top. 43 -.36 * haha answer button close bracket divided by 12 sadie did you get or cla sorry did you get 1.14 yeah m/s squared it's a positive answer so f applied is winning if it wasn't then friction would be winning I'd do a quick change I'd make that a positive that a negative in a more complicated question I might have hadrict pointing the wrong way. But here we're pretty straightforward. I definitely put an Atwood machine on your test. Okay, this is probably the longest question in terms of the most amount of writing. Jacobe, what's it asking me to find? I can't. I first of all, I'm going to find the acceleration of everything. Sorry, I'm going to write an equation for everything to find the acceleration of everything. Why can't I find tension from that equation? Help me out, folks. I lose tension. Wow. Right. I did the stupid joke about the tension deficit disorder. Then I'll find the tension. So, to start out, you know, this is going to be a good job for Yeah. So, let's label what are the forces on M1. Jacobe, get the obvious one. M1G because there's more than one mass. What else? Yep. On the second mass, it was M2G. And tension. The tugofwar is really between the two gravitational forces. Which one's bigger? Which is the bigger mass? So, which way is m2 accelerating? Which is the winning force? Don't say m2 because that's not on my free body diagram. Which is the winning force? Help me out, folks. Yes. Good. Now, I often, just so I didn't forget, once I figured out the direction of the acceleration, I often just drew that on there because jazz, it's easy to lose track and forget. So now we got our equation. We're going to start out walking along the rope. Who's winning? M2G. Anything that ends up pointing down once it gets to the right hand side is going to be winner plus. So then we have minus tension. What about this tension? Is it a winner because it's pointing up? No. Follow it. Follow it. Follow it. Follow it. Follow it. When it gets to this side, it ends up pointing. So is it a loser because it's pointing up? I said that backwards. It's a winner because it ends up pointing down once it gets to the right hand side. This one, Lachland, you might say, is a winner because it's pointing down. But again, if you follow it, I don't want to draw on there. If you follow it, follow it. When it gets to this side, I think it ends up pointing loser or minus. What do I have to change the right hand side to if I have forces from more than one mass? M all a tensions cancel. And then we can still use Ryan's [clears throat] theorem. [snorts] Get the acceleration. And it's going to be 23 * 9.8 - 16 * 9 9.8 all divided by 16 + 23. These masses are fairly close. I don't think we're going to get a big acceleration. If it was like 23 and 22, they would barely accelerate at all cuz they'd almost be balanced. I know it's going to be less than 9.8. I can tell you that. Uh 23 * 9.8 - 16 * 9.8 8 close bracket divided by a bracket 16 + 23. Let me know what you get, Jacobe. So 1.75897. I'm going to write 1.76, but you know I'm storing this on my calculator, right? Or if I knew I was going to do be doing more than one step, I'd write this down elsewhere. So 1.76 m/s squared. Not what they asked us to find. What do they ask us to find? Look at an individual mass to find an individual force. So which mass do you want to use? Okay. Which way is m1 accelerating? Which way is the winning direction? Same answer. Which force is winning then on m1? Yes. Which one is losing? M1 g. And that's going to be m1. If you used m2, it would be m2g minus tension equals m2a. You'd have to do a swappy dance to get the t by itself. Here we don't need to do a swappy dance. We just need to plus the M1G over, which is actually a little easier. So, it's going to be M1 A plus M1G. It's going to be M16 A button plus M116 G9.8. Oh, I don't need to put in brackets. 16 * answer button plus 16 * 9.8 and I get 185 newtons. So you are going to see an Atwood machine on your test. What I'm not going to give you on your final is a missing mass question. I don't if you remember we also did Atwood machines with missing masses. I put that on your test. I didn't put it on your final. That was where you got to use that clever math 9 operation. How many m's do I have? One. How many? Two. How many would I prefer? One. And you have to use the GCF trick. We'll revisit that if you take physics 12. And I think this is the last forces question. Yes, it is. I have several different versions of the test. Some have an Atwood machine, some have this one. These are the two multimask questions. Find the acceleration. Find the tension. Brooklyn spoiler alert. What am I going to do first? What might this be a good job for? And again, I'm going to write find the equation of the acceleration of everything by writing an equation for everything. And then I'll find an individual force by looking at an individual mass. Probably this one, the 12, because it's going to be less cluttered. Brooklyn, what are the forces acting on the 4.3? Get the obvious one. I'll call that M1G. And I think there's going to be a normal force the same size as M1G. Renaut, I'll call that for normal force number one to go with mass number one. Is there a rope? We have tension. Is there friction in this question? And this is going to be pulling down, sliding to the right. So, I'm pretty sure friction is pointing to the left. I'll call it friction force number one to go with mass number one. What are the forces acting on the 12 kg mass? Get the obvious one. Yep. What else? Yep. A says, find the acceleration. Okay. Who's winning? m2g. And then the most common mistake people wanted to include m1g. Remember, we're only going to include forces along the rope. So, I'm going to walk along the rope, which is going to give me m2g minus tension. This tension here becomes a winner plus when it gets over this side. And friction force number one is a loser. M1G hasn't showed up in the equation. It'll show up in the friction portion later, but it doesn't show up in the first section. that equals m all a. Once again, I would stretch it out losing tension. Once again, Ryan's theorem. So, a is going to be m2g minus friction is what times what, Brooklyn? I don't know the normal force. Oh, but look, look, look, look, look. I know the force the same size as the normal force. And this is where the m1g makes its appearance inside the friction equation. Uh, divided by m1 + m2. I've scrolled down. What was it? 12 and 4.3. 12 * 9.8us what's mu? 28 * 4.3 * 9.8 8 all divided by 4.3 + 12. Oh, I made a typo. Glad I caught that. Insert a time sign. 12 * 9.8 minus mu * the normal force / Brooklyn. Did you get 6.49? No, I see people nodding. Yeah, I wouldn't round off to 6.5. I don't need to. Three sigfigs is pretty good. If I know I'm going to be using an answer to find another answer. I usually carry four or five just to be paranoid. But 6.49. I'm going to store this on my calculator. If I gave you this question, just like the Atwood machine, I said find the tension. If you get this one, I'm just gonna say find the tension and I'll expect you to clue in Nikki. You have to find the acceleration first. So Nikki, let's find the uh tension. Which mass would you go with? Yeah, less cluttered. Uh which way is m2 accelerating? Which way is the winning direction? Same answer. Who's winning? M2G minus tension equals M2A. That's the equation just for that mass. What's right in front of the T that I don't like? I'll swap you dance. It's going to end up being m2 g minus m2 a that equals tension. It's going to be m212 g 9.8 8 minus M212 times answer button. You get 39.7. Newtons. So, from last class and today, that's a quick review of forces. That's what I'm going to expect you to know. Okay? If you're looking for a better idea of how I turned these into multiple choice questions, look at the final practice final exam that I gave you today. You we're going to work on it Wednesday. And then I would also work on it before you write your final as part of your practice. Pause for a second. Give me one second if you're okay. I want to try and plow through all of momentum for the remainder of class so I can give you a complete review period on Wednesday. That's the goal. So really quickly, key concepts for momentum. Remember momentum was a vector. So if there's more than one direction, we needed to let one way be something and one way be something. Okay? plus V. That's my abbrev minus V. You can write POSOS or NEG. You need to know the difference between just plain old momentum and change in momentum. So if you look at your formula sheet, plain old momentum was just P= MV. Oh yeah, we had to use P because uh M was lowerase M is mass uppercase M something else already taken. So it's from a Latin word. What's change in? Well, what's the symbol for change in? So a lot of people kept calling either this momentum. It's the change in momentum, which means the only way that occurs is if momentum is changing. If there's a difference. Uh what's another word for change in momentum? Starts with an I. And what's changing anything? I'm going to abbrev F minus I. And so that gave us our impulse equation which ended up becoming MV final minus MV initial and force time. Uh collisions and explosions. If you see a collision or an explosion, there's two masses. Uh that you want to use is the law of conservation of momentum. And then I'm dropping some blatant hints here because I turned these into multiple choice questions. Forces come in opposite. So in a collision between something big and something small, are the forces the same? Yes, in opposite directions, but they come in opposite pairs. What about impulses? In a collision between something big and something small, impulses also come in opposite pairs. Do accelerations come in opposite pairs? Why not? F equals what? What? The only way the accelerations will be in pairs is if the masses are the same. And this goes my to my big rant of why a small car is often so much worse off in a collision with a big truck. Same force, small car, way bigger acceleration. And I've tried to teach you what does damage. Acceleration is what does damage. There was several of those questions on your momentum test and I li as multiple choice and I lifted those and put them right on the final. All righty. A force of 12 Newtons acts on an object for 8 seconds. The mass is 3 kg. What's the object's change in momentum? Remember change in momentum is delta P. It's impulse. So I'm going to look at that equation on my formula sheet. I notice they gave me a force. They gave me a time. I think I'm going to use the version of impulse that says force times time. The third version Eve is looking. Yeah. Make sure you see where it is on your formula sheet. Right. It's that that middle one that had three equations in one that we said really it was a nice shortcut. Uh it's going to be 12 * 8 which is 96. Double check me. And the units for momentum, oh 96, Mr. Dick, were kilogram meters/s. The units for change of momentum were kilogram meters/s. But the same units are also Newton seconds because force times time. If you simplify force newtons time seconds, you will end up with kilogram meters/s. So you can use them interchangeably. B. Jess, what did B want me to find? What did we find in part A? There's a word in front of it. We didn't find the momentum. I got to be nitpicky. Change in momentum. Okay, so we're not doing P= MV. We're in that middle equation. Hey, that middle equation also said that change in momentum is M delta V. How would I get that? That's the change in velocity. That looks like an A, Mr. Do it. Make it a delta. Jess, how would I get the delta V by itself? how I move the m over. Yep. The change in velocity is going to be the impulse, the change in momentum divided by the mass. It's going to be 96 / 3, which is I should be able 32. Double check me. Yeah. And it's a veloc a change in velocity. So, it's just meters/s. That means that its velocity changed by 32 meters/s. If it was originally going two, now it's going 34. It was going 10, now it's going 10 plus 32, 42. Example two. What will the recoil velocity, magnitude, and direction be if a rifle fires a bullet with a velocity of 410 m/s north? When you fire a rifle, that's an explosion. There's more than one mass. Harit, this is going to be conservation of momentum. What did we say for that? Well, we said your the sum of your initial momentum had to equal the sum of your final momentum. That was the summation notation that I taught you, which was the shortest way to write that. But really, we need to visualize firing a rifle. Uh, I said be lazy but organized. I'm going to use R for rifle. I'm going to use B for bullet. You could also use mass A and mass B, but if I could, I tried to use words that started with it. Easier to remember. So, here is my rifle analog. Before I pull the trigger, what's moving? The rifle, the bullet, both or neither? Be stupid obvious. Neither. So, what's my initial momentum? Yeah. After I pull the trigger, what will be moving? Well, definitely the bullet. And it wants me to fire find the recoil velocity which does have to be there because forces come in pairs. There's going to be a force on the gun. I'm going to have momentum sorry mass of the rifle velocity of the rifle. There's the final momentum of the rifle plus mass of the bullet velocity of the bullet because momentum is mass times velocity. They want us to find this. Okay, Lachlan, I need to move this term over. How will I move this term over? What's it doing to the velocity of the rifle? So, and there's a zero over here. 0 minus anything is just negative that thing. So, I'm going to say, you know what we end up with over here is negative mass bullet v bullet. That takes care of this term. Now I need to take care of this mass in front of the rifle. What's this mass doing to the V? So, and if you don't mind, I'm going to put the velocity of the rifle over here because that's where we're used to seeing the thing by itself. And I'm putting in the vector signs to remind myself I better pay attention to direction. Mass of the bullet it says is 036. Oh, don't forget the negative. Mr. What's the velocity of the bullet, Lachland? And by saying that, you've just decided north is positive. You could have said negative 410 and decided north was negative and I would have gone along with it. But if we get a positive answer, that'll be north. What if we get a negative answer with confidence and authority in your voice? Yes. Uh divided by the mass of the rifle, 2.2. We are going to get a negative answer. I'm not going to say negative south. After I get an answer, I'll say equals and then south. So, it's going to be 036 * 410. No, Mr. Dick, it's going to be.6 * 410 / 2.2. Do you all get -6.71? I need nods. Yeah. So, -6.71, which is 6.71 m/s south. That was the explosion question that we looked at for momentum. Sometimes we had an explosion where the initial momentum wasn't zero. We did a spacecraft capsule that blew up into two pieces and actually it worked out well because the probe got extra velocity for free. That's why we have multi-stage rockets. The engines lose momentum. Well, then the probe has to gain or the capsule has to gain momentum. We're okay. So, everyone looking out in the hallway. We're back now. Squirrel distraction. Uh, I think this is the last one or two more car crashes. I put one car crash on your final. Either in the car crash they stick together or they bounced apart. Ooh. What do we call a collision where they stick together? Inelastic. Bounce apart. Elastic. Rubber band. Bounce apart. Elastic. So car A has a mass of that traveling west hits a stationary car B. If it's stationary, what's car B's initial velocity as a number exactly? What's car B's initial momentum as a number exactly? I made the question easier for you. The collision is inelastic. That means they stick together. Find the final velocity. Okay. Is this a collision? Yes, we're going to use conservation of momentum. Before the collision, what was moving, Sadi? Car A, car B, both or neither? So, I'm going to have the mass of car A, velocity of car A initial. And what do I always do here? Do you remember? Bam. They collide after the collision stuck together or separate inelastic. So, in brackets, car A plus car B val. This is the trickiest one to set up but the easiest one to solve because so many things are either stuck together or zero because jazz this wants me to get the final velocity magnitude and direction. What's the bracket doing to the V jazz? Adding, subtracting, multiplying or dividing. So how I move the whole bracket over? What's the opposite of multiplying? Yeah, even though this starts out complicated, this is probably the easiest one mathematically I can give you. It ends up being the mass of car A, velocity of car A initial divided by the mass of both of them. That's going to be Vfal. And now I can crunch my numbers. 2,350. I've just let west be positive. If I get a negative answer, then that'll be east. and stuck together. 2,350 + 1,400. This is the same physics. Instead of a car, I could make it a football player tackling and wrapping someone up when that person was standing still. That would be say if you watch football, if the receiver was standing still when they caught the ball and then right when they caught it, the defender went plowing through them. That happens fairly often. uh 2,350 * 26 / bracket 2350 +,400 close bracket 16.3 direction. It's got to be west cuz I let west be positive. By the way, that W since it's after an at symbol is west. Otherwise, it could be work. It could be watts. I mentioned we don't have enough. Hopefully, I mentioned we don't have enough letters going around. Example four. This is probably the toughest type of collision. Here we have I was still nice and I think on the final exam I always let one of the objects be stationary at the beginning just so you had a little fewer terms. But here afterwards they're both moving. It says it's elastic. They bounced apart. So before the collision, what's moving? Car A. So we're going to have the mass of car A, velocity of car A initial. They collide. Sorry for those of you on YouTube with headphones. After the collision, Eveuna, read the question. Car B is traveling south and it wants me to find the velocity of car A. They must not be stuck together. Otherwise, the answer to car A would also be 24 meters/s south. Uh, I see the word south. I see the word south. I don't see the word north. I'm going to make a little note cuz this one's more complicated. I'm going to let south be positive. If I get a negative answer because it's possible car A bounced off and reversed direction. It might have, then I'll know. Uh, mass A V A final mass B val final momentum of car A final momentum of car B not stuck together Eugen what's this asking me to find which one there's three which one there's two of car What car? What? Yes. Wants me to get that by itself. Okay. Hey, this is formula manipulation. Complicated but still doable. What's going to drop down like a domino? The left side. Yes. Mass a va initial. And then I'm going to do my adding and subtracting. You I need to move this over. How will I move this term over? Yep. How will I move the mass of A over? Yep. That equals VA final. I'm not sure whether I'm going to get a positive or a negative answer. Let's see. Mass A 2250 car a ma velocity A positive 32 minus mass B 1,400 car B 24. You know what? It's getting rear ended. Car B is going south at 24. It's getting rear ended by car A. So, car B is going to speed up, which it Okay. So, no, car A, car B was stuck still. Yeah. Yeah. Yeah. Sorry, I'm visualizing the question here. Uh, divided by mass A. Mass A 2250. I think you do end up with a positive answer. I think I think I think I hope maybe maybe what'd you get? 17.1. Anybody else? Yep. So car A slowed down because it gave some of its momentum to car B, but it's still going forwards. We visualize that. Car B was parked. Car A came in. Boom. And they both kept going, but car B kept going away faster. 17.1 meters/s direction positive. So I'm not doing a review of work and energy. I'm kind of hoping that's recent enough. Uh on Wednesday, if you want, you can also ask me to look at any of your unit tests if you're thinking about doing a rewrite. I won't do that today, but I'll sign those out and collect them at the end of class. any flex block this week. If you want to come in, you can also sign out a unit test. Folks, that is a really quick review of the course. Let me just hit stop here.