A LEVEL MATHS: 2D Vectors Exam Practice LESSON 24

Another one of the A level maths tuition lessons. So today, we're coming up towards the end of AS mechanics, we're going to do two dimensional vectors. So two dimensional vectors, you have two notations that you can use for them, which you would have seen before, you have the column notation. And you also have something called ij notation, which looks something like this, where i and j are something called unit vectors, which basically means they just give you direction. Now, it's important to note that if on a normal kind of x y axis, like this, i and j just represent x and y respectively. So j is always pointing straight up, and i is always pointing to the right. So that's pretty much what we're going to use for today. And we're going to use our other mechanics knowledge as well. So things like when forces are in equilibrium, for example, resultant forces, etc, that's all going to come up. We also need to know how to work out the magnitude of a vector. So for example, if I want to find the magnitude of this vector here, what you do is, so I'll just put it in brackets, is you basically choose Pythagoras's theorem on it. So you do one squared plus two squared, which is square root of one plus four, which is root five. And the reason why this is important is if you remember, in mechanics, we talk a lot about a lot of vectors, for example, velocity, acceleration, displacement. Now, if you have velocity, for example, the magnitude of velocity will get rid of the direction, and you'll just have the speed left over. So remember, with velocity, acceleration, and displacement, they're all going to actually be given as vectors like this in this topic anyway, so in ij notation or column notation. So if they ask you for the speed, you wouldn't write something like this, you would instead find the magnitude. And again, that works for all of them. So you have displacement, that would give you just how far you've traveled, so just the distance. In terms of acceleration, you can have an acceleration vector. And when you find the magnitude, you're just left with regular acceleration. So just basically the magnitude, the size of it. So when you take the magnitude, it means you get rid of the direction. And the only thing you have left is basically how big the vector is, just the size. So without further ado, let's just get into some questions in order to kind of apply this knowledge. So this is one of the most common questions they can give you. They'll tell you that a bunch of forces are acting on a particle and it's in equilibrium. Or they'll say that there is a resultant force. But the point of this question is for you to essentially find out two unknowns. Now looking at this, you might think it's a bit impossible if you've never done it before. But I assure you, regardless of whether or not they're in equilibrium, or they are equal to some kind of resultant vector, you do the same thing. First thing we're going to do is I'm going to write this in column notation. You don't have to do this, but I'm going to write them in column notation. And I'm going to add them all together because if forces are in equilibrium, the resultant should be zero. So if I just have a look at these vectors, they should all add up to give me zero. Now the reason why this isn't impossible, the reason why this can work, is I now have two equations. If I look at the top part here, I have 3. So I'm just looking at the top sections here. So these values here. I have 3 minus 2 plus 4q should equal zero. So then you have 1 plus 4q equals zero. 4q equals negative one, and therefore q is equal to negative one over four. And the reason for this is the ij components we spoke about before are completely independent of each other. Because the i component tells me how far left and right I'm going, and the j component tells me how far up and down I'm going. So you can treat them as completely separate things. So we've done that for the first one here, so now we're going to do the same thing for the bottom section. So these values here, again we can treat them all completely separate from each other, equals zero. So we have 2p plus 5 plus 5 equals zero. So we have 2p plus 10 equals zero. 2p equals minus 10, which means p is equal to minus 5. So I hope that makes some amount of sense. So that is probably the most common question that they can ask you. Again, sometimes they'll make it actually equal to some kind of resultant vector. So it maybe could have been like 3 and 4, in which case all you would do is replace the zeros with 3 and 4. But it's the exact same thing. Okay, moving on to the next exam question that we have. They can also ask you to work out angles in terms of vectors. So particle p of mass 0.8 kilograms moves under the action of a force F newtons. The acceleration is minus 3i plus 5j. Find the angle between the acceleration and the vector i. So if you remember, if I just draw this out again, the i always acts to the right and j always acts upwards. Now here we have, I'm actually going to have to draw the negative axes as well, we have negative 3i plus 5j. So it's going 3 to the left and 5 up. So we have a vector that goes something like this. Negative 3i plus 5j. And we're trying to look for this angle here. Now all you have to do here is, we're actually looking for the reflex angle here, but we can't really work that out directly. So what we're going to do is work out this other angle. So what if I just, that's the overall angle we're working out. But for now, what we're going to work out is this angle here, which I'm going to call phi. So from the question, we have negative 3. So this bottom length here is 3 and this top length is 5. So all we need to do is use trigonometry. So the angles we have, both sides, is opposite and adjacent. So we're going to use tan. So tan phi equals opposite 5 over 3. And phi is just equal to inverse tan. So if we use our calculator quickly, make sure we're in degrees mode, which I'm not. Then all we do is do inverse tan 5 over 3. 59 degrees. And again, because we're actually looking for the angle between the acceleration and the vector i, we're actually going to add 180 degrees on. And that will be our final answer. So actually, let's write it in black, that makes more sense. 239 degrees. So hopefully that's okay. Again, that is only two marks, so it's meant to be relatively tricky. So for part b, it says find the magnitude of f. So we don't know what f is yet. However, if you remember Newton's second law, f is just mass times acceleration. So if we have a look at what we have, we have a mass 0.8. And we have an acceleration, it's minus 3i plus 5j. So if we expand that, 3 times 0.8 is going to be negative 2.4i. And that's going to be 4j. So that is what my vector is for force, because remember force is a vector. If you ever have a scalar multiplied by a vector, it always gives you a vector. However, it says find the magnitude. So this is where we're actually going to do what we said earlier. If you want, you can actually make this a vector by putting a little arrow on its head. And we get square root 2.4 squared plus 4 squared. So we have square root 2.4 squared plus 4 squared. And that will give us our answer of 4.66 Newtons. Do three significant figures. Okay, so for question three, a particle of p of mass 2kg moves with a constant acceleration under the action of a force. The initial velocity is minus 2i plus 6j, and after four seconds, it's 7i plus 2j. Find three significant figures, the magnitude of the acceleration. So when we did SUVAT, I mentioned that everything except time was a vector. So just because they've given us vectors in the question doesn't mean that we can't actually use our SUVAT equations. So s, we have no idea. The initial velocity is minus 2i plus 6j. Our final velocity is 7i plus 2j. Our acceleration is what we're actually looking for, and the time is equal to four seconds. So we want an equation that links v, u, a, and t. So that is going to be v equals u plus at, and we can work it out from there. So our value for acceleration is going to be v minus u divided by t. So let's put that on. So we have 7i plus 2j, and remember because we're doing negative u, we need to flip the signs of all of this. So minus minus 2i is going to become plus 2i, and then minus 6j, and that's all over the time, which is four. So we have 7i plus 2i is 9i, and we have 2j minus 6j is going to be minus 4j over 4. Isn't acceleration negative 9.8? No, that's only in free fall. So if you drop something, then it's going to be negative 9.8 for the acceleration. That's only when you're dropping something. In this case, we don't know if it's being dropped or what's happening to it. So in other words, we have 9 over 4i minus j. So that's my acceleration. However, it asks me to three significant figures, what's the magnitude of the acceleration? How are you meant to know? Basically, if it ever says that something has been dropped, or it's in free fall, then you can assume that acceleration is 9.8. But if it doesn't state that it's being dropped, or thrown or something like that, then you can't assume it. So that's how you're supposed to know. So you can only kind of say that if in the question, it says that it's dropped, it's thrown, or it's under free fall or something like that. So we're going to have square root 9 over 4 squared minus 1 squared. Yeah, it said constant acceleration, but it didn't say that it's moving under free fall. It's just saying that it's some constant acceleration, which doesn't have to be 9.8. We have 2.46 to three significant figures. Part b, it's asking us to work out the angle between the force and the vector i. So in this question, you can actually get away with just finding out the angle for this vector, and it'll be the exact same for the force. But that might be a bit confusing. So I'm just going to write F equals ma, let's work out what the force vector is. That equals the mass, which was 2 kilograms, 9 over 4 i minus j. And again, you could have just taken the angle between this vector here, but that might have been a bit confusing to some of you. So what I'm going to do instead, let's do it like the slightly longer way. Okay, and we're doing it between the vector of the force, so the vector of the force and the vector i. So if I draw my grid, i is going to be this way. And for the force, it's 9 over 2 to the right, and it's negative 2 down. So it's actually going to be a force that looks something like this. So now we have 9 over 2 along the bottom, and we have 2 down here. And we're going to work out this angle theta, which is the exact angle that they actually want us to work out. So in this case, we have tan theta equals opposite over adjacent, we have 2 over 9 over 2, which means theta equals inverse tan 2 over 9 over 2, the exact same thing as usual, 2 over and then 4.5. And to three significant figures, the next one. So question four, it says the resultant of two forces is i minus 14j. It says given that f1 is 2pi minus 4qj, and f2 is 3qi plus 4pj, find the values of p and q. So we're going to do the exact same thing again, I'm going to turn them into column vectors. So f1 is going to look like this, 2p and minus 4q. And if it's a resultant, that means you're going to be adding the forces together. So that'd be 3q or p equals one and minus 14. Now the problem here is that in each kind of column, so if I write these out as separate equations, now we have 2p plus 3q equals one. And then we also have minus 4q plus 4p equals minus 14. So we have two unknowns, but we also have two equations. So what we can do is actually use simultaneous equations to work out the answer. So what I would probably do is if I look at equation two, I'm going to divide everything by two, and we're left with minus 2q plus 2p equals minus seven. And if I look at equation one, there's a 2p there. So I'm going to use substitution for this method, although you could quite easily do elimination as well. So if I rewrite this over here, we have 2p minus 2q equals minus seven. So we could just subtract the two equations. 2p minus 2p is zero, 3q minus minus 2q is 5q, one minus minus seven is eight. So that means q is equal to 8 over 5. So that's one unknown found. And for the other one, we're just going to substitute this back into the original equation in order to figure out which one it is. Which equation would be easiest? I'm just going to substitute it into this one, as long as they're both kind of bad options. We have minus two, 8 over 5 plus 2p equals minus seven. So we have minus 16 over 5 plus 2p equals minus seven. So I'm going to add that to both sides. We have 2p equals so minus 7 plus 16 over 5 would be 19 over 5. Good. So it'd be minus 19 over 5. And then divide both sides by two to get p is equal to minus 19 over 10. Perfect. Okay, so any questions about that? So in that question, instead of giving us kind of an equilibrium, they said instead that they're equal to some resultant vector. So we wrote that out. And then we wrote our two separate equations. Remember, we can look at the top kind of row and the bottom row completely separately, which can help us out with solving the equations. But in this case, we actually had a simultaneous equation. So hopefully that's fairly easy. Let's go to the next one then. So particle p moves with a constant velocity of 4i minus j. And it says find the speed. So as we wrote a bit earlier, speed is just equal to the magnitude of velocity. So when you take the magnitude, you essentially just remove the direction from a vector. So you can change between vectors and scalars. Well, in this case, just from a vector to a scalar. So you can just get rid of the kind of direction part of it. So we're going to do is take the magnitude of that, which is 4 squared plus 1 squared, which is 16 plus 1. So in other words, it's square root 17 meters per second. And then lastly, part b, it says find the direction of motion of p. And it wants it as a bearing. So when I drew my grid before, when it had the x y axis, that's completely fine. But the y axis is also the north line, because it's straight up. So we have 4 and minus 1. So it's going to go 4 to the right and 1 down. It's going to end up somewhere like this for our So again, we're working out a angle, and we're just going to add 90 degrees to it. So that is 1 and this is 4. So we're working out this angle here. So theta is equal to, once again, inverse tan of opposite over adjacent, so 1 over 4, which is 14 degrees, which means the actual bearing now, you have to add on the 90 degrees here, 104 degrees. Hopefully not too bad so far. I think it's a fairly quick three marks that you can get. But this whole question was five marks in total. And I think that I was pretty fast. Overall. Perfect. So question six, we have a particle p moves with constant acceleration 3i minus 4j metres per second squared. At time t equals zero, p has speed of u metres per second. At time t equals three, p has velocity of minus 5i plus 2j. So it says find the value of u. So technically speaking, you can just work out the magnitude. But we can also do it using normal suvat. So s is equal to don't know, don't care. u is equal to the u vector. It's basically what we want to work out. We can call it question mark as well. b is equal to minus 5i plus 2j. a is equal to 3i minus 4j. And lastly, t is equal to three seconds. So we have v, a and t and we want u. So we're going to use v equals u plus at, which means u is equal to v minus at, which in this case is going to be minus 5i plus 2j minus 3 times 3i minus 4j. So we have minus 5i minus 9i, which would be minus 14i. Then we have 2j plus 12j, which is going to be 14j. So again, it's minus three times minus four makes it plus 12, two plus 12 is 14. Now, it does actually ask for the speed, right, kind of stealthily, it doesn't ask for the vector. So in terms of what the actual speed is for you, it's going to be the magnitude of the vector. So this whole thing here is actually the actually it's not the magnitude, it's the vector. So we're just going to do square root 14 squared plus 14 squared. And that will be our final answer, which is 19.79898989. So it'd just be 19.8 metres per second to three significant figures. And again, we usually round to three significant figures. So that is why we've got that. Okay, that's all good. Okay, with this one, it's a bit more interesting. So it says the resultant of two forces F1 and F2 is parallel to ij. So if you remember what parallel vectors means, it means that they are the same. However, they've been multiplied by some constant, which I'm just going to call k, because constant starts with a k, of course. So we're just going to essentially add these two vectors together and get this. So just as a kind of forewarning, the resultant means we're going to add these two things together. So we have 3i minus 2j plus pi plus 2pj. And again, you could have done this as a kind of column vector. So just as a column vector, it would look like this. 3 minus 2 plus p 2p equals k k. So we have two equations now we have 3 plus p equals k. And we also have minus 2 plus 2p equals k. So all we need to do now is sub in for k. So we can equate these two equations here. And that will give us so if we bring them together, we have 3 plus p equals minus 2 plus 2p. So I'm going to bring the p's to this side and the numbers to this side, that will give me 5 equals p, and I'm done. Okay, hopefully that makes a fair amount of sense. Again, it's not too bad once you get into it.