A LEVEL PHYSICS REVISION: PROJECTILE MOTION EXAM QUESTIONS

hello and welcome back to another one of the a level physics mechanics videos So today we're going to be doing the much requested projectile motion so projectile motion uh what what does that mean well in real life things don't tend to move just either left or right or up and down generally speaking they're going to move in some kind of combination of angles so for example let's say I have a ball over here and I'm going to give it an initial velocity of let's say 10 meters per second and that is an angle of let's say 35 degrees okay so what we can do with that is if I were to extend this I can see that it's going to go in a massive Arc and land somewhere over here now what if I told you just this information we can work out exactly how far away from the ball from the starting position it ends up so I'm going to call this distance s okay so with this we have to use so this is a projectile a projectile is just anything that is moving through the air now we have to consider two Dimensions okay up and down and left to right so what I'm going to do is I'm going to take the vertical stuff first so if I look vertically do I know what the distance is well you can actually say what the distance is because remember this is displacement not distance so the key thing to remember here is that this is the vertical displacement so depending on what we're talking about like where we're ending up so if we're ending up back where we started the same vertical displacement the actual displacement is technically zero because again just think vertically the ball goes up it goes back down overall it hasn't moved right from its original position it ends up where it started so the displacement is zero now here's where things can get a bit tricky if they ask you what's the maximum height reached by the um projectile then your displacement isn't zero it's a maximum value so this is specifically as I'm talking about the end of the arc here now what about the initial velocity well we do have it but it's not 10 right because keep in mind 10 is diagonal there's a horizontal and a vertical component so go back to the vector mechanics video If you're not unsure but if I were to draw it out like this I can draw two components right I have a horizontal component and a vertical one again remember this is 35 degrees and this is 10. okay now if I want the vertical component I'm looking for the opposite side so I have sine 35 equals the thing we want I'm going to call it X over 10. so X is now 10 sine 35 we can work what that is out using a calculator or we can save it for the end so it really it really is up to you so we could just do so again I'm in radians mode so I'm going to quickly change back over so again anytime I use trig I'm constantly thinking degrees radians degrees radians what am I using so for now just for sick of Simplicity I'll keep like this but normally I would actually keep it as is so I'll keep it as 10 sine 35 so that is 5.74 meters per second what about the final velocity well this is where things can get a bit interesting if you are not sure leave it as a question mark because that's totally fine however I know that this object is going to spend half its time going upwards and half its time going downwards so its final velocity is actually the same as its initial velocity but negative but again we don't need to include this in fact we don't really need to include this either I'll show you why now here's what a is a is actually super super easy okay a is always just going to be gravity okay so again I'm taking upwards as positive in this question and to the right is positive since gravity acts downwards it's going to be minus 9.81 and the time is I don't know so we can do here is using we have everything except time so we can use whatever equation we so please and the best one perhaps to use might be you can use S equals u t plus half of 80 squared right so find all the times where s is equal to zero you could use um V squared equals U squared plus 2as to get what um actually that won't give you anything never mind you can also use the fact that a is equal to B minus U over T and rearrange for T So then T is equal to V minus U over a so we have minus 5.75 minus 5.74 over minus 9.81 and what that would do is it gives you the time of flight for the ball so for the projector so this is another thing I want you to keep in mind is the the language that they use what's the maximum height means that you're working out halfway along here so it's the maximum here if they say what's the time of flight or how long is the projectile in the air for you're looking at the time it takes to get to its final position so minus 5.74 minus 5.74 all over minus 9.81 gives me a time of 1.17 seconds so this thing ends up in the air for 1.17 seconds so now well how does that help me I wanted to work out this distance well now I can look at the horizontal component right I'll keep it as worries s is what I'm trying to work out initial velocity instead of 10 sine 35 will be 10 cos 35. to keep in mind a lot of these kind of um questions that we that I'm doing this example specifically it's a very hard example however so in your exam they may give you an easier one but they can can also give you a very hard one there's actually no limit to how difficult they can make it as long as it's one of the skills that you need to know but that's going to be so here's an interesting thing what about the final velocity some people would say zero but we're talking about the horizontal velocity it isn't zero okay and this is the reason why acceleration is zero so whenever we deal with horizontal well whenever we do projectile motion we're making a big assumption we're assuming there is no air resistance or frictional forces which means the acceleration along the horizontal axes is always zero okay now again in other questions they may introduce air resistance but it will never be in a projectile motion question okay they might say what effect it will have that's it the reason for that is there would be air resistance both vertically and horizontally which makes it really tricky to kind of deal with so there is no acceleration for projectile motion questions unless they specifically State otherwise which again I highly doubt so funnily enough my final velocity is actually going to be the same as my initial one and lastly what is the time well the time this is another um key thing will always be the same as the vertical time think about it if it takes 1.17 seconds for it to vertically come back that is the exact same amount of time as it takes to travel horizontally right because the ball has to end up here does it make sense that it would take more or less time for it to go up and then down rather than just across if that's the case it wouldn't end up in the same place right if it took two seconds to go horizontally but one point something seconds to go vertically won't end up in the same place so the two things I want you to remember are this yeah and the fact that these two things are the same vertical and horizontal time is always the same 100 of the time it has to be otherwise I mean you're talking about this object kind of changing into two objects or some weird stuff okay the time is always the same so if you work out the time vertically you can use that in your horizontal calculations and if you're working out horizontally you can work it out in your vertical calculations which is really important because when a is zero we can't all of the equations that have u and v in it become quite strange you just get V equals U which yeah okay I can see that obviously so that's something quite important you need to have a time in these questions and well when you're working out horizontal things so that's why the time is the same okay that's why it's a really useful thing to remember and again remember uh acceleration has to be zero for horizontal so now we can use again we don't want to use equations that have u and v in them because I'll just get U equals V which is obvious so I can use anything I want to work out s um u t plus half a t squared could be fun um so why is it fun well if you look a is zero so this entire back thing's zero so actually what I'm just doing is distance equals speed times time which again is something that we should be 100 percent okay with by now so again we want to use our calculator to do it we have 10 cos 35 close the bracket times uh 1.17 oops 9.58 meters okay again this is a general example this comes up in all exam boards so expect to be able to do this this is a very hard example though okay um but they can't give you weird situations like this question from the AQA paper if you notice this is actually part two because part one involved something different which I'm I can't remember what it is but it's not to project on motion so I've ignored it so we have a golf that like balls that lands in a sand pit and he needs to hit the ball to get out of the sand pit so he wants to go from here up to here okay now for the first part it just shows the vertical component of view is approximately six meters per second and it tells me the horizontal component is 8.8 so if I were to draw this out right I'm trying to find this which I'm going to call Y and they've given me X is 8.8 and we know that this is 35 degrees okay so all I'm trying to work out is y so we can use tan for example could be a good one so 10 35 equals y over 8.8 so then Y is just 8.8 5 and again always drawing diagrams because that helps you immensely immeasurably in um a level physics so 8.8 1035 gives you 6.16 meters per second which is approximately 6 meters per second so when they say approximately what you want to do is make sure you answer rounds to that value so approximately six meters per second means that to one Sig fig um I should get six which I do okay and that's all you really need to care about so even if I got 6.45 it still counts as approximately the ball is traveling horizontally as it reaches X as shown in figure five so it kind of goes up and then it really does level off assume that the weight is the sorry assuming that weight is the only force acting on the ball when it is in the air so what do they mean by this well it just means that the only acceleration we have to deal with is gravitational acceleration that's all they mean by that okay which again is exactly what I said in the previous part so it's you know nothing surprising there it says assume that way you see any uh Force calculate the time for the ball to travel to X well let me think about this we're looking we're going to look vertically why because I have no information horizontally except what speed it's going I don't have the acceleration I've done at the time I don't have anything else well the acceleration is probably zero um in most cases so I'm going to look vertically only do I know what s is nope do I know what U is well U is the vertical component which we worked out here which is 6.16 now what would V be well V when it reaches this top if you look it says it's traveling horizontally which means there is no vertical component to its velocity which means its velocity must end up at zero again you can't assume this unless they give you a hint in the question acceleration well we already know acceleration is again I'm going to take upwards as positive because I've already kind of implied that so it's minus 9.81 time I don't know but that's what I want so I Want U V A and T in an equation so what could we use well we could use a is equal to B minus U over t rearrange that fatigue would give us B minus U over a so effectively we get 0 minus 6.16 and again this is where I said in both of the previous videos don't be afraid of the minus signs because as you can see they're going to cancel out just sub it in see what happens if you get a math error then you know you've done something wrong but don't assume you're going to fail because uh that's not a good way to look at things so technically I could just do 6.16 you know divide it actually should we do zero ones answer I don't know why I look for that on my keyboard over 9.81 I'm using the answer button because that was um more accurate and it says calculate the time you should get 6 as 0. 0.63 which I have so let's do these are all to three significant figures so I'm going to write 0.628 seconds because it's all to three significant figures and then calculate the vertical distance of X above the initial position of the ball so now we're just using our same suvat but for S so we can use anything we want because we have everything but s so the most obvious one would be S equals u t half a t squared so U was 6.16 again always using brackets T is 0.628 plus half minus 9.81 0.628 squared and again I'm going to use the answer button just to save um the most accurate answer we have and that plus 0.5 minus 9.81 and Sarah squared yep that's perfect and we get 1.93 meters perfect so again they've broken it down quite far for you and most of the time they will but again not always so the golfer Returns the ball to its original position in the sandpit he wants the ball to land at X but this time with the smaller horizontal velocity than in figure five sketch a possible trajectory so one thing to note is the overall energy if I talk about kinetic energy that is the overall velocity so it's the vertical and the horizontal kind of combined into one okay so what he's saying is it's going to land in the same position but with a smaller horizontal velocity and if you notice here it's only traveling horizontally there's no vertical component right but at this position they will have the same energy so we're not changing about how hard he hits it we're just changing that there will be less horizontal velocity the way to do that is to make sure that there is Vertical Velocity at this position just go through that again the overall velocity will be the same but we want to make the horizontal component smaller which means you have to make the vertical component bigger so what we can do is all we have to do is make sure instead of Landing so it lands horizontally we want it to kind of dip back down so we could do something like this right and at that point they'll I mean it'll be all Vertical Velocity actually so explain your reasoning well all we want to say in this part is well why did we choose it well we're going to just say exactly what I said so one other thing to note is you could also add like a little thing where this is greater than 35 degrees the idea is that it's a steeper angle so we can explain our reasoning so it's two marks so let's think so a steeper or let's say larger angle and I'll label this one because again remember I want you guys to bullet point your answers okay larger angle at the beginning lowers or increases vertical component which results in smaller horizontal component that's it two marks and again it's exactly what I said but it's more precise so as the marketing open here so the kind of marks that you get all you need to do is increase the angle here and that will and make sure it has some vertical drop at X so you can't go like up but then still Level off here it needs to go above the point here okay perfect next question so in areas of the world where a plane is unable to land three four airdrops can be used to deliver supplies so a plane is traveling at 75 meters per second and at a height of 63 meters so I actually kind of like to draw on the diagram itself so it's traveling horizontally right it's not a rocket it's not going straight up and again the height 63 meters draw the path of the falling package now this is where things can get a bit interesting if we ignore air resistance it's going to fall basically diagonally okay so it's going to fall like that but as it gets to it's going to start off diagonal but as it accelerates down it means that it's going to start curving so it's going to look more like this it's going to be diagonal and then a curve okay and again we don't so the key point about this question is we don't have to end up in the drop zone we just need to show it so at the beginning it's diagonal and then it slows down the reason for this is we are ignoring air resistance which means this velocity will stay the same but right and here's a big part it's accelerating downwards which means the downwards component is increasing okay so at the beginning we just have horizontal then as we start to fall the vertical component begins to increase greater and greater and greater okay which means it's going to accelerate downwards where it's not accelerating at all horizontally in other words going to be a parabola kind of shape and we need to show the time taken for the supplies to reach the ground is about four seconds so again if we look at this vertically why because this is just a dropping question right this is the same as just if I've dropped something we have 63 meters for s the initial Vertical Velocity is zero initial uh sorry the final velocity is we don't know a is 9.81 because again it's gravity and we're taking downwards as positive and we're going to take it to the right is positive as well why because the package is moving to the right and downwards so if we can avoid negatives all the better also keep in mind that s is a vector that is going downwards so if you take output is positive s is negative your velocity is negative your acceleration is negative Etc et cetera Etc and we're working out time so s u a and t so that reminds me of the equation s equals u t plus half a t squared and remember this is actually a question an equation that we always use when we're dropping something because we can do the same thing and now T is just going to be equal to so times both sides by two so we get 2s divide by a all square rooted so we have 2 times 63 1 that gives us our value and we just need to work out what that value is square root over 2 times 63 over 9.81 3.58 seconds which is indeed it rounds to 4 seconds the reason why they always make these show like questions is so if you get this part wrong you can do part three and get full marks so if you notice my value for time is different to this right that's fine if you use this value of four you will get four marks here even though your answer will be different if you use the correct answer so don't worry about that it's called error carried forward they're completely fine with it even if you use your let's say you write um 1.38 right so it's not even four and you decide to use that 1.3 in the next part again our error carried forward they will recalculate the answer using your value and you can get the full marks okay calculate the horizontal distance well remember actually I'm not going to say that just yet I'm going to give you a chance to think about it while I write out the values s is exactly what we want to know U is 75 where did I get that from horizontal as traveling at a speed of 75 meters per second V I'm going to leave that blank a is zero we're going to ignore air resistance so V is that but the important one is that time is the same think about it right we're we are ignoring a lot of stuff when we do Sue that right so when it hits the ground obviously it's going to stop but our equations don't take that into consideration so we have to use our brains on ideologic so what I'm saying is the second this box hits the ground 3.58 seconds after it's dropped is going to stop so my time is going to be the exact same in real life sure it might slide it may do this that the other but if I increase this value for time my equation will just keep going and I can get some really weird values you can get like negative s and things like that so you have to use your logic unfortunately I know it's a very scary thing but the times for horizontal and vertical is always the same to prevent anything weird happening okay so again what happens this box drops it hits the floor and instantly it's going to stop moving left right up or down it's just gonna boom hit okay so that's why the time is going to remain the same and again we can use um UT yeah so again it's just um distances speed times time but if you want to know where that comes from it's just s equals u t plus half a t squared but again a is zero so it's just speed equals you know sorry distance equals speed times time so 75 times 3.58 and again I'm going to use my exact value just um because I can 268.8 oh that's actually that's actually a look right so three significant figures two significant figures so we should really say 270 meters but what I like to do is actually write out like a bit more sometimes the 268.8 meters so I like to write this one and then write the rounded version so I can get marks either way again unless it specifies you know to an appropriate degree of accuracy you don't technically have to always round so you can leave it as that and yeah that's all we have today so hopefully this makes you a bit more confident when it comes to projectile motion it ain't that bad it's okay um but obviously I can completely understand the fear but anyways I'll see you in the next one