ULTIMATE GUIDE!!! Further Mechanics Simplified

Added: 2026-05-23

[00:00:01]Hey guys, today we're going to be doing further mechanics. We're going to just do the first four chapters. Linear motion under variable force I feel is simply pure mathematics, but you just need to do use a bit of conceptual understanding.

[00:00:14]As for momentum, that chapter is quite repetitive in the way they make the questions. So, I think that's fairly easily analyzable.

[00:00:25]And it's quite easy to solve momentum and linear motion under variable force questions. But as for the other chapters, I feel that questions they have a broad variety of questions that they test on. And so, I feel demonstrating understanding on the questions they already have, especially the niche ones in each of them, would give you the highest level of understanding and therefore could improve your performance to the max.

[00:00:51]So, let's start off with projectile motion, okay? Let's start this question.

[00:00:54]By differentiating the equation of the trajectory or otherwise, find the coordinates of A.

[00:00:59]Okay. So, this is the question right here. I've taken a look at it.

[00:01:03]And it is projected from a point O horizontal plane. So, the very first thing I would do whenever I see a question like this is to draw the horizontal plane, the full plane just like this.

[00:01:14]Okay.

[00:01:16]Next up, I would draw my projectile trajectory, okay? So, it starts off with this velocity, okay?

[00:01:24]Now, we have the velocity 25.

[00:01:28]And it is at an angle theta, okay? I I like to label it as theta. I'm not going to actually find out the theta values, okay?

[00:01:37]Um and yes, it moves at a velocity of 25 m/s.

[00:01:42]Now, I would draw dotted lines like this.

[00:01:47]And yeah, okay. That may not be the most accurate projectile motion trajectory.

[00:01:53]Uh but yes, with a pencil, I would draw this trajectory right here.

[00:01:58]And this just helps me analyze how exactly the projectile should be moving, okay?

[00:02:02]Now, we're looking at point A in which the direction of motion of P makes a 45° angle with the downward vertical. So, looking at a 45° downward vertical. So, immediately, okay, when you're looking over here, we can see that the velocity actually changes. Okay? Because as you go higher and higher, your velocity your vertical velocity is maxed out right here.

[00:02:25]As you go higher and higher, you tend to lose the vertical velocity.

[00:02:30]It becomes zero here. And then it starts to decrease like this.

[00:02:35]Decrease in um it would be increasing in magnitude, uh but decreasing when you look at in terms of like the number line.

[00:02:43]Okay? So, this is how the velocity would be changing. And therefore, my resultant velocity uh is going to keep changing as well.

[00:02:49]So, it could be like this. Then it can be like this. Then it could be perfectly horizontal. Then we're going to look something like maybe something like this.

[00:02:59]And like this.

[00:03:01]So, we're looking at one stage.

[00:03:05]Okay? So, we're looking at one stage of this entire trajectory where the angle between the horizontal, let's say like we're taking this as an example.

[00:03:14]The angle between the horizontal and the velocity is 45°.

[00:03:19]That's what the question is trying to ask us.

[00:03:22]Now, if I'm going to just magnify this section over here.

[00:03:25]Okay?

[00:03:26]We're basically if this was the horizontal, we're looking at the point A in which the velocity is V like this.

[00:03:34]And it makes an angle of 45°.

[00:03:38]This is what the question is asking.

[00:03:39]Now, we already know that the horizontal velocity stays constant. So, we can label the horizontal velocity vector as UX.

[00:03:47]Uh vertical velocity in fact does change, so it's VY.

[00:03:50]Now, 45°, well, now you got to think Vy by Ux equals to opposite over adjacent when you look at this triangle. So, it's tan 45 degrees.

[00:04:06]Okay? Now, Vy is Vy. Ux can be written as Vx. Now, why what is the meaning of Vx and Ux in the context of my explanation? Ux represents the initial velocity in the x direction.

[00:04:20]So, in this case, Ux equals to 25 times cos theta.

[00:04:26]Okay? Now, cos theta can easily be found if tan theta is 4 by 3, cos theta is going to be uh 3 by 5. So, Ux equals to Um yeah, so Ux equals Ux equals to 15.

[00:04:39]Now, Uy would just be the initial velocity in the y direction, which is 25 sin theta.

[00:04:45]Okay? And this would be 20.

[00:04:49]Now, we know vertical velocity varies.

[00:04:51]Horizontal velocity stays the same. So, when we talking about Vx, which is the final horizontal velocity, it is always same as Ux and therefore, Vy by Ux is the same as is the same as Vy over Vx, which is equal to tan 45 degrees.

[00:05:09]Okay? Now, do note this is not going to be looking at tan 45 degrees because we're doing above below the horizontal, we're supposed to use -45, okay?

[00:05:19]So, -45 degrees.

[00:05:21]So, what happens when you do tan -45 degrees? Well, that means that Vy by Vx equals to -1, okay?

[00:05:29]Now, what is Vy? In fact, what is V? V equals to dx over dt.

[00:05:37]Okay? This is what happens when you're doing with the horizontal displacement.

[00:05:40]Okay? This is if horizontal displacement is x. If my horizontal displacement was y, as you can you could probably understand this linear motion of a variable force.

[00:05:49]Uh but yeah, in the sense of Y then V equals to dy over dt.

[00:05:55]So when we do Vy over Vx we're basically doing dy over dt divided by dx over dt.

[00:06:04]Okay? And now dt's cancel out in the numerator and denominator and we get dy over dx which is equal to minus one.

[00:06:11]This is exactly why they are saying to differentiate the equation of the trajectory because then when you differentiate it you can equate the solution to minus one and then we can solve for the answer.

[00:06:23]Alternatively, we know that Vy equals to U sin theta minus gt.

[00:06:31]Um then we could use this formula and then uh plug in the values and that's a bit more, you know, tedious. I would recommend you use differentiation. Now, equation of the trajectory is Y equals to X tan theta minus gx squared by 2u squared secant squared theta.

[00:06:54]Differentiating this we get dy over dx equals to tan theta minus now when you do all this uh you would get we would get gx by u squared secant squared theta, okay?

[00:07:17]Now, we could just plug in the values right here. So yeah, we could There are a number of ways you could do this.

[00:07:24]We could just plug in the values right here instead of simplifying them. So tan theta equals to 4 by 3, cos theta is 3 by 5 so secant squared must be 5 by 3 uh the whole squared and yeah, we could do all this and our result our result would be 4 by 3 minus I'm going to input this uh the uh in my calculator now.

[00:07:48]Okay, and yeah, just Mhm.

[00:08:00]Okay.

[00:08:04]Yeah, okay. So, I'm getting uh Yeah. So, I'm getting 4 by 3 minus 2 by 45 x.

[00:08:19]Uh yeah, I'm just checking it out again.

[00:08:23]Yeah.

[00:08:27]Uh okay, got it. And this is equal to uh dy by dx, which is -1.

[00:08:31]Now, if I solve for x, I'm going to get x equals to 105 by 2.

[00:08:38]And by plugging this value of x into your equation of trajectory to get y using this equation, you would see that y equals to 35 by 4.

[00:08:50]Okay, so these are the two answers to solving that five-marker question. And it's really that simple if you just understand how it works. Now, when it strikes a smooth barrier, it rebounds and lands on a horizontal plane inclined at 45° to the horizontal. Find the speed of P immediately before it collides with the barrier.

[00:09:07]Okay, fine.

[00:09:09]Um Okay, since dy by dx equals to -1, which is equal to vy over vx, we know that vy equals to -vx.

[00:09:22]We also know that u cos theta gives you ux, which is equal to vx, which is equal to 25 * 3 by 5, which is 15.

[00:09:35]Okay.

[00:09:36]And so, Vy must be -15.

[00:09:40]So, that means V equals to well, if your horizontal is 15 and your vertical is 15 in terms of magnitude, your resultant would just be well, using Pythagoras theorem, a squared plus b squared. So, V squared is equal to 15 squared plus -15 squared.

[00:09:58]It doesn't matter if you put the minus sign uh sign. And then you get 450.

[00:10:02]And then V equals to the root of 450.

[00:10:06]And that gives you 15 root 2.

[00:10:10]It's really that simple. Okay? That gives you the final speed immediately before it collides with the barrier.

[00:10:16]Now, the next question uses coefficient of restitution, which I guess does link to momentum a bit, but not too much actually.

[00:10:25]So, we know that coefficient of um restitution um it multiplies your velocity with the factor with the you know, with the coefficient of restitution. So, if my velocity is five and the coefficient of restitution is 1/5, my new velocity becomes one.

[00:10:43]Provided that the velo- velocity is directly colliding, perpendicularly colliding to the barrier. Okay? So, that's why we look in momentum based questions. If you collide it like this, you'll see that the vertical velocity always remains same because it's parallel.

[00:11:01]But, the horizontal velocity is the only thing that gets reduced.

[00:11:05]And why does it get reduced? Because the coefficient of restitution.

[00:11:08]Okay? Yeah, only the velocities that act perpendicularly to it or the component of velocity that acts perpendicular to it gets reduced by the factor of the coefficient of restitution.

[00:11:19]In our case, okay? It's making an angle 45 degrees with the downward vertical, 45 degrees.

[00:11:30]Okay?

[00:11:31]So, it's it's let's say that that's 45°.

[00:11:34]Okay, at this very instant, I'm just going to, you know, extend it a bit more for better visibility.

[00:11:40]And it collides with the barrier like this.

[00:11:46]Okay.

[00:11:48]And this is inclined at 45° as well.

[00:11:52]This instantly tells me that this must be 90.

[00:11:55]And if it's perpendicular, and the velocity at um this stage was 15 root two, I can comfortably say that my full velocity gets reduced uh by the coefficient of restitution. Why? Because it is colliding perpendicular to the barrier.

[00:12:10]Therefore, my new velocity is going to be 15 root two by nine or five root two by three.

[00:12:19]Okay. Now, after collision, it's going to come like this.

[00:12:23]You can see that the change in height is going to be 35 by four.

[00:12:28]But, we need to understand that's going down, so we use negative 35 by four.

[00:12:33]And now you can use the equation of the trajectory, okay? Um y equals to x tan theta minus gx squared by 2u squared secant squared theta, okay? Uh you can plug in your values for uh tan theta, secant squared theta, u, g, and you'd get a quadratic in x. As uh like just for the checkpoint in case you want to see if you're doing it right, uh you should be getting this quadratic over here.

[00:13:06]Um hm.

[00:13:09]Okay, if you got this quadratic by, you know, inputting the values correctly, um yeah, yeah, you're obviously in the right track, okay? And solving for this quadratic yields you, well, >> [clears throat] >> yeah, uh solving for this quadratic will yield you one uh main solution, which is 5 by 2.

[00:13:28]And this is going to give you that this Um, alternatively, of course, we can use the uh, suvat way, where you can use this.

[00:13:40]Uh, oops, it seems like there's a glitch there.

[00:13:43]Yeah.

[00:13:44]Yes, alternatively, you could just use suvat. We could, you know, use the quadratic in t. Uh, and then you when you use the quadratic in t, you get t equals to 3/2.

[00:13:54]And then you could use x equals to ut cos theta.

[00:13:58]Okay, and uh, yeah, when you do this, uh, you'll ultimately get the same result of 5/2.

[00:14:05]So, that's projectile motion for you.

[00:14:07]And that's the most That's the hardest question I found in the past few years.

[00:14:11]And yeah, I do hope that this explanation does enhance your understanding of further mathematics in the terms of the projectile motion.

[00:14:19]Okay, now we're going to equilibrium of rigid bodies. This is probably the miscellaneous questions that you may receive. Sometimes, well, I've seen questions where you get a rect- a square like this.

[00:14:31]Right? And it's tied to rope. I think we've seen this question before.

[00:14:35]Um, yeah. It's these kind of questions that may, you know, freak you out during exams, because you're normally, uh, you know, um, used to seeing shape-based questions like you got a cone over here, and maybe a cylinder is cut off.

[00:14:50]Now, find center of mass. Well, we know how to do that. We just need to use the volume, need to find the mass, use x mean, y mean. It's all very, uh, planned out. But this is not planned out. This is, in fact, one of the more confusing ones that I'm going to make it extremely easy to follow.

[00:15:05]Okay, before even reading the question, okay, uh, maybe a brief skim, you need to understand what's going on, okay?

[00:15:13]Immediately when you see it, you can see that there's a plane. It's inclined.

[00:15:17]There's a sphere. It's about to go down, obviously. And there's a string to fall. That depends on coefficient of friction.

[00:15:26]Now, it says that the ring of weight W.

[00:15:28]The moment I read weight W, I'm going to go to the center and just mark W.

[00:15:35]Okay, this is very important. Radius A, that's already marked, center O, done.

[00:15:41]Is at rest on a rough This This is translating to equilibrium.

[00:15:46]Okay, this is just equilibrium.

[00:15:49]It means that all the stuff has to balance out. And that is inclined at horizontal at an angle alpha.

[00:15:55]Uh okay. Well, okay. Now, it says that the plane of the ring is perpendicular to the inclined surface and parallel to this. So, if I extend this this dotted line all the way, you'd see that it is parallel to the plane.

[00:16:09]Okay?

[00:16:10]Using the, you know, if this PQ is also parallel, this angle is also alpha. And if this angle is alpha, that means this angle is also alpha.

[00:16:20]That's already a few values That's already a rough understanding of what we're dealing with right here. And that's how, you know, labeling the diagram makes it so much more easier.

[00:16:29]I think I can write a better alpha though, so Yeah.

[00:16:38]Yeah, let's call that alpha.

[00:16:41]Right. Okay. So, uh this is uh alpha, the angle.

[00:16:46]Now, the point P on the circumference of the ring is such that all of this parallel to the surface. Yes, we know this.

[00:16:53]Uh light inextensible string is attached to PQ. Yep, yep, yep. Okay. This is important as well. System is in limiting equilibrium, meaning it's in the very, you know, edge of it.

[00:17:03]And the coefficient of friction between the edge and the surface is mu.

[00:17:07]Okay. So, immediately, when just by looking at this, okay?

[00:17:10]I don't think anybody could, you know, convince you that when you're having a when you're having this system over here, the ball is about to go upwards, okay? Everybody knows the ball is not going to go upwards. Right? The ball is not going to go upwards. No matter how many like how many strings have you attach, the ball isn't going upwards. Instead, it's going to be stopped from going downwards. So, we say that the direction of motion is this way.

[00:17:36]Okay? If the direction of motion is this way, that means friction must act this way.

[00:17:42]Okay? And that gives me a few other values that I can add to my diagram. Let's erase the unnecessary ones.

[00:17:53]Okay. So, we know that the friction, which always acts in contact with the two surfaces, is going to act this way.

[00:18:00]This is friction. R is always perpendicular.

[00:18:04]And yeah, it's always going to be perpendicular.

[00:18:08]This is R.

[00:18:10]And this is going to be friction, F.

[00:18:14]Okay. So, we already have a couple of values.

[00:18:17]We also know that the W, this weight, is also at an angle alpha with the center.

[00:18:24]Okay? Yeah. So, we have a lot of values right here. Now, what are we going to do with them? You may ask. Well, okay.

[00:18:30]Now, there are three main rules that you need to understand such that a system can be in equilibrium.

[00:18:36]Firstly, the sum of the forces acting on the x-axis must be equal to zero.

[00:18:43]Okay?

[00:18:45]Sum of Y equals zero as well. That means the Y directional forces.

[00:18:49]And sum of the torques must also be zero.

[00:18:52]Torque is just um well, you can consider it to be moment in this case, okay? So, let's let's start off with whatever we can do right now.

[00:19:01]So, when I say X, I usually mean parallel to the plane. And when I say Y, I mean uh perpendicular to the plane.

[00:19:09]So, looking at parallel to the plane, which is sum X equals zero.

[00:19:13]We can resolve this as R this.

[00:19:17]I think that's the best way to draw it, okay? Now, looking at all the forces that act parallel to the plane, okay? We have a frictional force F.

[00:19:25]Okay?

[00:19:26]We have, well, the tension in the string is obviously going to be T.

[00:19:32]So, we have a T cos alpha acting parallel to the plane.

[00:19:37]So, F plus T cos alpha.

[00:19:44]Our force R is only perpendicular to the parallel, so it has no horizontal component when you're when you're considering the horizontal to be the plane.

[00:19:53]And our weight, okay, since this is an angle alpha when you magnify this, let's magnify this.

[00:19:58]Okay? This is O.

[00:20:01]This is what's perpendicular to the plane. Let's consider this the plane.

[00:20:04]This is W.

[00:20:07]This angle is also alpha.

[00:20:10]Which means that the horizontal component, I mean, the parallel component to this is going to be W sin alpha.

[00:20:20]Okay? This is important. Now, we can resolve vertically as well if needed. Um let's resolve vertically, or I would say perpendicular to the plane. Okay? Yeah.

[00:20:32]So, perpendicular to the plane, we have, well, R.

[00:20:36]R is upwards. Do we have any forces? We can't consider F in our case because it is parallel to the plane.

[00:20:42]Uh we have W acting downwards and T is also acting downwards. So, R equals to W cos alpha plus T sin alpha.

[00:20:55]And as for the final one, the sum of all the torques. This is a very interesting one, okay?

[00:21:01]Uh let me just erase this. Okay, yeah.

[00:21:05]Sum the torques. This is where, you know, might make the mistake on and it's valid. Okay? Ideally, you could choose any point that you want to. You could choose this point, you could choose this point, this point. Doesn't matter. It's entirely your choice. Um >> [clears throat] >> For circles, it is preferable that you use the center, but in most cases you use um the point of contact, okay, in any system.

[00:21:31]Um I'm going to be using, you know, moments about O because it's far simpler to do it this way.

[00:21:36]Uh but yes, of course, you could just use this one as well. Okay? Let's start off with O, okay? When you're looking at O, there are, well, a couple of forces acting on it.

[00:21:46]Now, weight is acting directly at O, meaning that the moment is zero. Why?

[00:21:51]Because FD equals to moment, okay?

[00:21:55]Now, D equals to zero at the when you're in the force is acting on that point.

[00:22:00]So, W produces no moment.

[00:22:03]R produces no moment, either, because R is acting directly in line with the axis.

[00:22:08]Okay? F, however, is acting perpendicular.

[00:22:12]Uh and well, P is also acting perpendicular to it.

[00:22:16]Now, what's the separation between O and F?

[00:22:20]This is E. A, sorry, this is A.

[00:22:23]And F is acting perpendicular, directly perpendicular to the axis of O.

[00:22:30]This means that my anticlockwise moment, which is this way, is FA.

[00:22:37]And this must be equal to the clockwise moment of this.

[00:22:41]Now, P is not directly acting perpendicular to the axis of O.

[00:22:47]In fact, T cos alpha doesn't, but T sin alpha, which is this one, does act perpendicular. So, we say this is going to be T sin alpha A times A.

[00:23:02]And then we can, you know, cut off A on both sides. We get F equals to T sin alpha.

[00:23:09]And this would be the third solution.

[00:23:12]Okay? Uh these are the first two.

[00:23:15]And yes, now that we have this, we can use our values. So, if tan alpha is equal to 1 by 2, sin alpha equals to 1 by root 5, and cos alpha equals to 2 by root 5.

[00:23:35]How do I, you know, get these values mentally? Uh it's very simple. This is the opposite. This is the adjacent. So, doing opposite squared plus adjacent squared gives you five, and then rooting that gives you the hypotenuse, which is the result of the denominators for both sin and cos. Now, this is opposite, O, so it must be over here for sin. And two is adjacent, which must be in the numerator for cos. So, yeah, that's that's just a quick way to find all the trigonometric ratios. Now, once you get all of these values, you can just plug, you know, we have F equals to T sin alpha. We can plug that in right into this equation.

[00:24:10]We can use our, you know, trig values, and you should be getting your answer as T equals to 1 by 3 W. Okay?

[00:24:19]So, yeah, it's relatively quite simple.

[00:24:22]Okay? Now, if you were to find the value of mu, okay?

[00:24:25]Uh this is where they, you know, test a little bit of that AS uh level mechanics in regular 9709.

[00:24:32]Well, we know F equals to mu R.

[00:24:34]We know T sin alpha equals to F.

[00:24:37]And I think you can see where I'm getting to with this. We know we can we can synthesize a value for this, okay?

[00:24:44]Uh in terms of W.

[00:24:46]We could just say that R equals to W cos alpha, which is 2 root 5 W plus T sin alpha. T is equal to 1 by 3 W. Sine alpha would be well, 1 by root 5, so it's going to be 1 by 3 root 5 W.

[00:25:10]And yeah, that is what you'd be getting R as. And you know, if you simplify everything, that's going to be well 7 3 root 5 W. Yep, this is what you're going to get.

[00:25:28]And then F equals to mu R, so F equals to T sine alpha. Um T is 1 by 3 W.

[00:25:34]Um hm.

[00:25:37]And yeah, we just did this before and sine alpha is 1 by root 5.

[00:25:42]And yeah, now that we do this, we get our answer for F as 1 by 3 root 5 W.

[00:25:52]Now F equals to mu R, so that means that F by R equals to mu.

[00:25:56]And so this is F.

[00:25:58]We divide it by R.

[00:26:01]And you can see that the numerator is the only thing that doesn't get canceled out. And that gives you 1 by 7.

[00:26:07]And that's just a really quick way of doing it. So the key steps is to resolve perpendicular and parallel to the plane or you know, the horizontal, however the question demands it. Find the torque, which is the tricky part, with respect to any point you'd want, and then resolve using your three simultaneous equations.

[00:26:26]Yeah, so that marks the end of that equilibrium of rigid bodies. This one's also really simple, okay? Now, the main reason I picked this one is cuz uh there are This is actually very common base question. It's really easy to solve.

[00:26:39]Except I just want to, you know, show you trick that I learned in this.

[00:26:43]Okay, so normally we would, you know, split the two triangles or I would initially.

[00:26:49]I went ahead and split the two triangles like this.

[00:26:53]And then I use the shape area method, right? So, we use the shape.

[00:26:58]We formulate the area.

[00:27:00]We give the mass, although that's that's not used a lot these days.

[00:27:04]So, yeah, we get the mass and then we get the center of mass.

[00:27:09]Which would be X mean, Y mean.

[00:27:11]And then we would use the whole mass one times X one mean and then yeah, we would find X mean and Y mean, blah, blah, blah, and so on.

[00:27:20]But for triangles, there's a there's a cool trick that I found.

[00:27:25]Um so, looking at this one, okay? We note down the coordinates with respect to point O.

[00:27:31]This would be KA, 0.

[00:27:34]This one would be KA plus H.

[00:27:39]Okay?

[00:27:40]{comma} And yeah, now to find the vertical distance, well, we know this is 2A.

[00:27:46]45 degrees, so tan 45 degrees equals to H by this. So, this must also be H because only if your opposite and adjacent is one, the ratio is one, do you get this 45 degree angle?

[00:27:57]Okay? So, basically what I'm trying to say is tan 45 equals to H by H. It has to be, which is equal to one.

[00:28:05]And this is H, that means that this is 2A minus H.

[00:28:09]Um okay, yeah. As a result, this this coordinate is going to be H.

[00:28:15]And as for this coordinate, it's going to be well, proper 2A.

[00:28:19]No, no, no. The X coordinate is KA.

[00:28:22]Y coordinate is 2A.

[00:28:25]Okay, with that being said, now there's a really neat formula you could use.

[00:28:29]X mean equals to X one plus X two plus X three divided by three.

[00:28:37]And yeah, you could do the same thing for Y mean as well.

[00:28:45]Yeah, okay. Now, this is the simple part.

[00:28:48]Um X mean would be KA + KA + H + KA. So, X mean equals to 3 KA + H / 3, which is basically the answer they were looking for in your question.

[00:29:02]Okay. Now, they're also finding, you know, for Y mean. And yeah, Y mean equals to Well, we could do the same thing. 2A + H + 0. This is 0. Uh we get 1 / 3 2A + H.

[00:29:16]Yep. So, that that that marks the end of it.

[00:29:19]Um the remaining questions are quite, you know, neat and they're on niche. I mean, we know how to solve these questions, but uh repetitive part practice. Main reason I chose this question to go through is just the for the triangle method that I didn't come across before and would probably save you some time.

[00:29:35]Okay. With that being said, we're done with uh this question.

[00:29:42]Okay. Now, we're going to circular motion. A circular motion very, very simple.

[00:29:47]Um once again, this is a miscellaneous question.

[00:29:50]And I'm going to show you how to solve these kind of questions, okay? So, we don't see a lot of hemispherical shell kind of questions, okay? So, what does that even mean? Think of it as a bowl, okay?

[00:30:00]Um as one of my comments earlier, uh you know, Yeah, as the comments earlier uh told me, it's best to understand that this is just an empty sphere.

[00:30:13]And that is true, and I think that it's really good depiction of like a bowl.

[00:30:18]Okay. So, we have this bowl over here.

[00:30:20]We take the center, just like that.

[00:30:23]Now, we have to depict what they're showing, okay? So, the particle of mass M is moving in a horizontal circle with angle omega one angular speed omega one and smooth inner surface of this, and angle between upward vertical and normal reaction is theta one. Well, this Okay, a lot a lot of words. Um I can already tell that the angle Okay, now it's saying angular speed is increased.

[00:30:46]Okay, so angular the angle between the vertical and normal must also be increased.

[00:30:51]Um that's proved by just doing tan inverse of these two. We could just find out theta one theta two.

[00:30:57]Uh but yes, now I'm going to just draw the vertical.

[00:31:02]And now I'm going to arrange for two motions, okay? Uh Okay, yeah.

[00:31:10]Yep, now we've got to arrange for two motions. Let's look at the first one right here.

[00:31:18]Yep, that's the one. Okay.

[00:31:20]Uh for the first one, well, if you're on a hemispherical shell, you're obviously going to get a certain contact force.

[00:31:27]Okay, so let's call this R1.

[00:31:31]Okay? And this contact force must be, you know, aligned towards the center.

[00:31:35]Okay, because it always does. It never does vertically. It's always perpendicular to the surface, and it's when you're perpendicular to a circle, it points towards directly to the center.

[00:31:45]Um this will, you know, rotate with this kind of a circle form.

[00:31:53]You'd get a circle like this, I think.

[00:31:55]Okay, it's it's my best depiction of a circle that I could do.

[00:31:58]Um yeah, this is the kind of circle it would rotate around. So, it's going to go like this and this and this.

[00:32:04]You could just probably put like a ping-pong ball in a bowl and see what happens when you rotate it, but keeping the same uh vertical height.

[00:32:12]Now, this is theta one.

[00:32:16]Okay, brilliant. Now, let's move on to the next one.

[00:32:20]Okay, now we've done it in such a way that now we have a particle here, okay? The same particle, by the way. I'm just going to use a different color to depict this.

[00:32:33]And well, it's going to be like with so And now the normal reaction force is changing, okay? It's not going to remain constant. It's going to change.

[00:32:41]This is R2.

[00:32:43]And this time, it makes a circle like this.

[00:32:48]Okay? So, it moves on like this.

[00:32:51]Okay, this is the kind of circle that it's going to make. So, I think you can understand what I'm trying to say, okay?

[00:32:55]And so, it's it's still in the same level. It's just moving like this and this and this, like that. Okay? That's what the initial one is doing. Except, this one is at a higher angle.

[00:33:06]This is at theta two.

[00:33:09]Okay? Now, this is moving with speed omega one.

[00:33:14]And the first one, sorry, this is moving at omega two.

[00:33:18]Omega two, and this one's moving at omega one.

[00:33:23]Fine. So, if you were to find the ratio, um first thing you need to understand is that the mass is the same.

[00:33:29]And always acts downwards.

[00:33:33]So, resolving all the forces, let's resolve vertically.

[00:33:37]We'd get, well, R2. Now, we know this is theta uh this is theta two.

[00:33:43]And this is theta one. So, let's resolve for the first case. I'm going to use the green.

[00:33:48]Um R1 cos theta one. This would represent the vertical component of R.

[00:33:56]And this should be matching mg.

[00:34:00]This is equal to R1.

[00:34:02]So, R1 equals to mg divided by cos theta one.

[00:34:08]Now, cos theta one is going to be equal to 4/5. I as I said before, you could use the earlier technique to find trigonometric ratios.

[00:34:19]This is going to be 3/5. And this time, they swap in this case.

[00:34:24]4/5.

[00:34:26]Cos theta equals to 3/5.

[00:34:30]Okay, so um the mass is m.

[00:34:33]G is yeah, so we would just do this.

[00:34:38]And we would get mg divided by cos theta one. This is theta one.

[00:34:43]two two and then this is going to be 5 by 4 mg.

[00:34:49]And when you're looking at the you know, when you're resolving horizontally you'd see that the only horizontal component acting it'd be it's going to be R.

[00:35:01]R1 in this case. So it's going to be R1 sin theta one because theta one is your angle over here. If it was here then it'd be cos theta.

[00:35:10]And this is equal to mR omega squared.

[00:35:14]Okay, now when you have [snorts] mR omega squared you would get 5 by 4 mg times now sin theta one is 3 by 5.

[00:35:24]Okay, and this is equal to m times R. Now the radius is R like this, right?

[00:35:31]So the radius will also be R here.

[00:35:34]The radius will also be R over here.

[00:35:37]Okay, so the radius of the circle would just be well, this is theta one as well. So this is going to be R sin theta.

[00:35:44]So R times sin theta which is 3 by 5.

[00:35:47]times omega one squared.

[00:35:50]Cut 3 by 5 on both sides, cut m on both sides.

[00:35:54]We would get what else do we get?

[00:35:58]Can we simplify this any further?

[00:36:03]Okay, well we get 5 by 4 g equals to R omega one squared.

[00:36:12]Okay, we could do the similar method for R2 as well.

[00:36:15]We would get R2 equals to 5 by 3 mg.

[00:36:20]And we could you know proceed and we would get R omega two and the whole square equals to 5 by 3 Okay.

[00:36:31]So, we have r omega 2 squared equals to 5/3 mg. We have r omega 1 squared equals to 5/4 mg. Now, we need to find omega 1 by omega 2.

[00:36:42]So, if I did this Okay, I could technically do it.

[00:36:49]If I squared this though, that means you could just still cut the r off and we get omega 1 by omega 2 the whole square and this is equal to 5/4 mg. Sorry, 5/4 g / 5/3 g.

[00:37:06]G's cancel out. Five is here. Five is here.

[00:37:09]This is three. And so, omega 1 by omega 2 the whole square equals to 3/4.

[00:37:17]And when you root both sides you'd get your answer for omega 1 by omega 2 which is equal to root 3 by 2.

[00:37:25]And that's the basics of the circular motion questions of hemispherical circles.

[00:37:29]Let's proceed with this one. Okay, so this is obviously one of the most intimidating questions you could possibly get in the exam day. Okay, it's like a double particle system. How do you do this? Well, I'm here to teach you not to solve this question but how to solve questions like these. Okay.

[00:37:46]Now, immediately the first thing that you should understand is free body diagrams. Okay, where you resolve the forces. Looking at P over here. Okay, let's let's look at P.

[00:37:55]This is a particle P.

[00:37:58]Let's just look at the forces that's acting on it. Okay.

[00:38:01]There's a tension in the string OP.

[00:38:03]Let's label this Do we have any labels that we can call it?

[00:38:08]Mhm.

[00:38:10]No. No. No. No any labels at all. Okay.

[00:38:13]Let's call this tension A. Okay, TA.

[00:38:17]Okay, and this is going to be tension B.

[00:38:25]>> This is also tension B.

[00:38:28]Now, the first question that may come into your mind is how did I know the direction of TB and TA? It's simple.

[00:38:34]Whenever you have a string, okay?

[00:38:37]Whenever you have a string attached to two points A and B, the tension always acts towards the center or rather it always acts away from the point, okay?

[00:38:47]Um whether tension is positive or negative entirely matters in the context. But in this case, okay?

[00:38:54]If you're talking about A moving this way, the tension is going to be negative.

[00:38:58]Negative tension because the resultant force is always going to be this side.

[00:39:02]If we talk about B moving this way, well, pulling the string this way would actually cause it to have positive tension.

[00:39:09]Okay, but in the in the in the end, both of them have the same magnitude all throughout the string. Always has the same magnitude, okay? So, TA over here is the same as TA over here, TB here, TB here. Same same.

[00:39:21]But if they're attached differently like this, then the tensions can vary.

[00:39:24]Um on to the question itself.

[00:39:28]Looking at this now, I'm I'm just going to, you know, clear the unnecessary info.

[00:39:33]Yeah.

[00:39:34]Let's use red.

[00:39:35]Okay, yeah. So, looking at the system for B, let's look at B.

[00:39:40]Okay, there's a force TA acting like this.

[00:39:46]And there's a force TA like this.

[00:39:48]Um that's an angle theta, but we're not going to care about we're not going to care about that.

[00:39:54]Um okay, and there is also another force.

[00:39:59]Second, yeah.

[00:40:01]Yeah, there's there's a another tension acting let's let's put it a bit more of a deviated angle for better visibility.

[00:40:08]This TB and this is going to be mg.

[00:40:12]Okay, well.

[00:40:14]Uh yeah, now Now that's settled, we can proceed.

[00:40:17]Uh wait, what just happened?

[00:40:20]Yeah, sorry.

[00:40:21]Okay, yeah. With that being said, we can proceed now. I'm just going to use the pen.

[00:40:26]Okay, this is angle beta.

[00:40:29]And this is going to be angle alpha.

[00:40:32]Let's resolve the forces for this particle P right here.

[00:40:35]Uh when you look at it, the upward force the only upward force that's acting for P would be called TA cos alpha equals to mg plus TB cos beta.

[00:40:55]Okay, this is very, very important.

[00:40:58]Okay, it's it's extremely important to learn how to resolve forces like this.

[00:41:03]Okay?

[00:41:03]Now, as for, you know, resolving the forces horizontally, well, the forces horizontally would be this.

[00:41:12]So, that's TA sin alpha minus TB sin beta. Why did I say minus sin beta?

[00:41:22]Because the horizontal component B is acting towards the right as opposed to the TA which is acting to the left. So, this should be equal to zero normally, but because it's moving in a horizontal circle, we're going to say it's going to be mr omega squared.

[00:41:37]And this shows you how to simplify equations.

[00:41:40]It makes understanding a whole lot easier now that we've drawn the diagram.

[00:41:45]Now, it doesn't look so intimidating and that's the beauty of mechanics. You know, everything appears complex initially, but once you know how to break down the question, um it's it becomes very, very simple.

[00:41:56]Uh okay, wait 1 second.

[00:42:01]Yeah.

[00:42:04]I'm just going to It seems that there was an error initially. Yeah, so let me just write down the equations.

[00:42:11]So, for P, we have uh TA cos alpha equals to TB cos beta plus mg.

[00:42:28]And we also have TA sin alpha minus TB sin beta equals to MR omega squared.

[00:42:44]Okay. Well, we have that. Now, we can using the similar method, we can solve for uh Q. Again, looking at Q, we have This is Q.

[00:42:55]Now, let's look at the forces that acting on Q.

[00:42:59]Downward force mg.

[00:43:02]There's a top left force of the tension.

[00:43:05]Tension's acting this way, right? So, yeah.

[00:43:08]Uh that's TB. Any other forces that we have for for Q?

[00:43:13]No, there's nothing else. Well, okay.

[00:43:15]This is at an angle beta.

[00:43:18]Okay. Well, now I can already say that um TB cos beta would be equal to mg.

[00:43:27]And TB sin beta equals to MR omega squared.

[00:43:36]Okay. Um R would be varying, of course.

[00:43:39]Um but this is just to, you know, just giving you a simpler understanding of what's going on over here.

[00:43:44]Now, I could just solve the question over here, but the main purpose of this video is to understand how to how to solve a question, okay?

[00:43:53]You would need an ideal starting point for this. Now that you know all these equations, you could just plug in values and so and you can find the answer for every uh question that we have over here.

[00:44:04]Okay. So, for example, okay, let's start off with the first one. Find the tension in string OP.

[00:44:09]Well, we know that mass is 0.05 and we don't we also know that T2 cos beta equals to well, mg where m is the mass over here.

[00:44:20]Okay, because why do I say T2 cos beta?

[00:44:23]We need to use this over here.

[00:44:26]Okay, now we don't put 2mg because it's the mass of Q.

[00:44:30]Okay, so T2 cos beta equals to the mass of Q, which is 0.04.

[00:44:38]And times G, okay. Now, we put this in in this equation over here and we get >> [cough] [laughter] >> we get 0.04g plus 0.05g because this is the mass of uh P.

[00:44:56]And this gives us 0.09g.

[00:45:00]Uh cos alpha, well, we know that sin alpha is 0.8 by 1.

[00:45:06]Sin alpha equals to 4 by 5 basically.

[00:45:10]And so cos alpha equals to 3 by 5.

[00:45:16]Or this is just going to be 0.6 and 1, okay. So, yeah, this is and this is all equal to T TA.

[00:45:25]No, so this is this is TB, sorry.

[00:45:28]Yeah, and this is all equal to TA cos alpha.

[00:45:34]Which is 0.09g. So, 0.09 divided by 5 divided by 3 by 5 will give you 0.15g.

[00:45:46]Yeah, and this is equal to the tension in A. Using this value, you could plug it in this. We know the values of alpha.

[00:45:53]We can do this. We already have certain values and we can solve for the rest of the equations.

[00:45:58]So, yeah, that's how easy circular motion questions can be provided that you understand the concept. Okay? Once again, this video is purely based off concept building, not based on how to solve uh like how to solve every question based on this. It's to understand how to get to that starting point, and then using your brain to solve the equations. That's the easy part.

[00:46:20]How you derive the equations is the hard one, and that's exactly what this is doing.

[00:46:24]Uh at least I hope to do.

[00:46:26]Okay, now that we have another circular motion question. And the reason I have a lot of circular motions is cuz these tend to be the hardest questions there.

[00:46:33]Uh but yes, and this one combines momentum as well.

[00:46:37]So, let's start off with the easy one because we kind of need this one.

[00:46:41]Um for any time you have a circular motion question, just understand that the energy always remains the same. So, EI equals to equals to EF, where I represents initial energy, F represents the final part.

[00:46:53]Okay?

[00:46:54]So, if I take this point and this point or this point, they all have the same amount of energy.

[00:46:59]Their velocities and heights may change, but it's all the same. Okay?

[00:47:03]So, you could think of it if you have if you've taken A-level physics, think of it as harmonic motion. Okay?

[00:47:09]This would represent um if Yeah, okay.

[00:47:14]Let me just draw like this. Yeah, okay.

[00:47:17]So, if it is, you know, oscillating from here all the way to here, and the midpoint is right here.

[00:47:24]Okay? Over here, kinetic energy zero, here it's max, here it's zero. Okay? So, this is what the KE graph would look like.

[00:47:32]If you drew for potential energy though, you're going to have max at the end points, and you have zero over here. All right?

[00:47:40]But in the end, the energy remains a constant like this.

[00:47:44]So, the the picture I'm trying to paint right here is that the energy is always the same, and it's either that the kinetic energy increases when GP decreases, or it could be the other way around. So, that's yeah, based on how the question is.

[00:48:01]And yes, let's find the initial energy in this case.

[00:48:05]Um 1/2 mv squared directly gives you your kinetic energy.

[00:48:09]This would be 17 by 2 m g a.

[00:48:13]Looking at the height, this is going to be a cos 60.

[00:48:17]Uh cos 60° is 1 by 2.

[00:48:22]So, this height is 1 by 2a. This height is a. And I always consider the GPE to be the point, right?

[00:48:30]Like right at the bottom. If you look at this point also, it is valid, but it's preferable that you look at the bottom.

[00:48:35]And so that you don't have to deal with negative GPEs or anything. Okay, we can keep it positive.

[00:48:41]Uh yeah, so the GPE would be well, it's going to be m g * h, which is 3 by 2a.

[00:48:50]Which is equal to 3 by 2 m g a. So, total energy is this plus this, which gives you 10 m g a.

[00:48:58]And at this lowest point, there is zero kinetic I mean, there is zero potential energy. So, 10 m g a equals to 1/2 mv squared.

[00:49:08]Uh m cancels out. Sorry, m cancels out here. Uh times it two, 20 g a equals to v squared. And so, v equals to root 20 g a.

[00:49:19]Which is equal to uh 2 root 5 g a.

[00:49:24]Okay, if you just want to, you know, get it in terms of thirds.

[00:49:29]>> [sighs and gasps] >> Okay, now as for collisions.

[00:49:32]Okay, once again, I told you that energy is never, you know, truly lost or reconverted. Now, when when you deal with collisions, it does however change, okay?

[00:49:43]Um as a result of a collision between p and q, so q is also here.

[00:49:49]P goes all the way up over here. Okay, now this is where tension resolution comes in, okay?

[00:49:55]At Let's look at the tension over here.

[00:49:57]Let's say that the string was here for a second. The tension acts T, this acts mg.

[00:50:03]So, when you resolve the forces, you get T minus mg, and this is your centripetal force.

[00:50:07]So, you can say mv squared by r or mv squared by a.

[00:50:11]Okay? But, when you look at the top point, though, mg acts in this direction.

[00:50:16]T also acts in this direction. So, T plus mg equals to mv squared over a.

[00:50:22]Now, the tension in the string is different. So, this will be T1, this will be T2.

[00:50:26]And yeah, the tensions are different, velocities are different.

[00:50:30]Um velocity at the top is always the lowest because this has the highest GPE, and therefore the since the total energy is the same, KE must be lowest here.

[00:50:38]Uh but, that's not relevant to the question. But, what I'm trying to say is how you solve resolve forces. So, when you look in this point over here as well, you'd see that mg is acting in the same level as the horizontal.

[00:50:51]Okay? And since they're both underneath the horizontal, this angle of theta over here is the same as this angle of theta. And this is in line with mg cos theta.

[00:51:01]Which is why some questions have T plus mg cos theta.

[00:51:07]Okay? And when when it acts over here at these points, you have Well, if this is theta, well, that means this is also theta.

[00:51:16]And this is mg, and that means this is theta.

[00:51:20]And therefore, you get Well, T here, and you get mg cos theta acting the other way. And that's how you get questions like T minus mg cos theta.

[00:51:29]Where you have to resolve the forces.

[00:51:31]I'm just showing you how we resolve the forces here.

[00:51:34]Looking at the top point, we have T plus mg equals to mv squared by a.

[00:51:43]Now, uh yeah, okay.

[00:51:49]Now it says that it moves along a circular path, becomes slack. Okay, when you say that it becomes slack, it means that the tension over here is zero.

[00:51:58]So that means mg must be equal to mv squared by a.

[00:52:03]So that means well, we can cancel out m's. So we get v squared equals to ga. Okay. Uh we can leave it at this because we're leading with energies. And whenever we use energies, kinetic energy is always half mv squared.

[00:52:17]Uh but for the sake of it, we could write v equals to root ga.

[00:52:20]Okay, but this is what we're going to be using more.

[00:52:23]Now the kinetic energy is well, v squared equals to ga.

[00:52:27]Potential energy will be well, 2 mga. Why? Because the height is totally 2a.

[00:52:36]And so the total energy ei equals to uh half mv squared, which is half times mass times ga.

[00:52:46]That's your kinetic energy plus your gravitational potential energy, gpe.

[00:52:51]And that's going to be 5/2 mga.

[00:52:54]And remember, right after the collision, p is still in the same horizontal level as q, and therefore it's all kinetic energy.

[00:53:06]I think this is where, you know, it becomes, you know, easy. So m m cancelled, two two cancelled, and v squared equals to 5ga.

[00:53:14]And so v equals to root 5ga.

[00:53:17]And that just goes to show how easy questions can be, you know, assuming that you understand the concepts behind vertical circles as well.

[00:53:26]Now, this is also This is more of a momentum-based question, but whenever you see perfectly elastic, now normally, whenever you deal with collisions, the kinetic energy is not always the same, okay? It is going to vary.

[00:53:38]Um so if your initial energy was 9 J, it's only going to decrease. It's not going to increase, decrease. It may be 8 joules perhaps.

[00:53:45]Okay, that 1 joule can be lost by sound or heat or whatever. Okay? But if the collision is perfectly elastic, that means 9 joules remains 9 joules.

[00:53:55]And yeah, the initial kinetic energy, well, when you looked over here when you looked over here, our V squared was 280.

[00:54:04]Or whenever, we knew that the initial energy was 10 mgA.

[00:54:08]You could look at it at this point come back energy PE or you could, you know, just use root 20 ag.

[00:54:15]Okay, and then you could use 1/2 mv squared to get the kinetic energy. So, the total energy is 10 mgA.

[00:54:21]We can find the total we we know the total energy after the collision uh due to due to P, it's going to be 1/2 times mass times Well, now V squared is uh 5 gA.

[00:54:36]It could be 5 by 2 gA.

[00:54:39]And this is going to be adding with 1/2 mv squared where this time V represents the speed of Q after the collision.

[00:54:46]Okay, this is VQ.

[00:54:48]And yeah, this is the first way and since momentum is conserved, we could use the momentum uh I'm just again and yeah, we could use momentum conservation laws and we'd find out that K equals to 3. Once again, like I said, not to show you how to do every question and just to show you how to do the questions enough so that you understand how it's done.

[00:55:10]Uh okay, lastly, we have Hooke's law. Okay, these questions are a bit tricky at times and this is what I found in the October November series.

[00:55:19]Um hopefully it'd be challenging enough to understand.

[00:55:22]Okay, so light elastic string of natural length A, modulus of elasticity 5 mg fixed point O. So, I start off with drawing this.

[00:55:32]Okay, I always start off with this, no matter what. Now, it's very important you read string or spring. Why? Because in a string you have like this. Okay, you extend it then it's going you can extend a bit more.

[00:55:44]You can extend more. But if you contract it, it's going to look something like this.

[00:55:49]Okay, there's zero tension in the string. But if you do with springs, Okay, if you extend it, well, it will still have a tension.

[00:55:57]If you contract it, it's still going to have a tension. This would be compression.

[00:56:03]This would be expansion.

[00:56:06]Okay, uh or you know, extension. Yeah, extension, sorry.

[00:56:11]Uh but yeah, that's how it works.

[00:56:14]And yeah, this this this goes to show that you you need to pay high attention to the question.

[00:56:21]Uh because they it can trick you up.

[00:56:23]Okay, now let's start off with diagram. So, initially it's supposed to be looking like this. Okay?

[00:56:30]Natural length. This is A. Okay.

[00:56:34]Um seems to be simple so far.

[00:56:37]Now it says that these two masses are hanged in equilibrium, M and 4M. So, before I do anything, I just draw the next one.

[00:56:46]Okay.

[00:56:48]And I'm going to label this as X, my extension.

[00:56:53]Okay, it's going to be an extension because the mass would cause it to um you know, uh extend.

[00:56:59]So, immediately I'm thinking, okay, well, at this point there must be a force acting downwards, which is the weight. But there must also be a force equalizing it, which is the tension.

[00:57:08]I know my weight is going to be M plus 4M, G, which is 5mg.

[00:57:16]And I know that my tension is the same as lambda X by L.

[00:57:20]Lambda being modulus of elasticity.

[00:57:24]X being the extension of the string from the natural length.

[00:57:28]Oh yeah, extension of spring, and then divided by natural length A.

[00:57:31]Since the tension and the weight must be equal at equilibrium, that means that 5 mgx / A = 5 mg.

[00:57:42]And simplifying or factorizing out 5 mg in both sides, we get x = A.

[00:57:49]Yeah, so I did all that work to get that x = A.

[00:57:53]This itself should gain you at least a mark or two.

[00:57:56]Um yeah, proceeding from this just let me erase everything.

[00:58:04]Okay, yeah, this is where I make the updates to my diagram, okay? And at sometimes, if I want the examiner to see that I've done my calculation properly, I would write down x and I will cross it out, write A, okay? Just to understand it a bit better.

[00:58:18]Then, we're releasing the particle, okay?

[00:58:22]I'm drawing it as a rectangle, just for no reason at all.

[00:58:26]Um yeah, when you're releasing particle Q away from it, okay? So, you're taking away Q, so the particle now attached has only a mass mg.

[00:58:34]Okay?

[00:58:34]And the force acting upwards of it was 5 mg.

[00:58:39]Okay, that was at the time, but we don't need to deal with this right now.

[00:58:42]Uh particle Q is detached in the string, P is at rest, blah blah blah blah. Okay, find the thing.

[00:58:49]So, at this very stage, okay? At the very lowest point, let's consider this lowest point as the baseline.

[00:58:56]Why What is a baseline? This is simply going to be where GPE equals to zero.

[00:59:02]Okay? At this baseline, there are three factors to consider, okay? When dealing with an energy equation, always remember there are three energies you need to be concerned of.

[00:59:14]Kinetic energy, gravitational potential energy, and elastic potential energy.

[00:59:21]Okay, this These are the three extremely important ones. Now, in circular motion, you don't get a lot of elastic uh you know energy used. So, you get this is very negligible. But, in Hooke's law this is this is the pinnacle of Hooke's law based question. So, this is expected to see.

[00:59:36]This is the full energy equation that you are expected to see. At times, well I've not seen many questions with it, but at times there's also work done by friction, which is going to be new r times the distance it travels. But, this is very optional, you know, you don't see a lot of this anyways.

[00:59:54]But, yeah, this is what you should be concerned of. Now, I like to use EI equals to EF. So, energy over here is EI, energy over where when 7 root 5 AG, let's take this as this point over here.

[01:00:07]This can be EF, okay? I know that these two must be equal.

[01:00:14]So, looking at this stage, it is saying that it is Q is detached from string, and so please release from the rest. Okay, this is key. This means that my initial energy has my kinetic energy is zero.

[01:00:25]Okay? My kinetic energy is zero.

[01:00:28]I took the baseline as this point over here. So, my GPE is also zero.

[01:00:33]And as for my elastic potential energy though, my extension is A.

[01:00:38]So, that tells me that 5 mg A the whole square divided by 2 A.

[01:00:47]Natural length. Now, I'm not going to simplify this. I'll tell you exactly why soon enough.

[01:00:52]Um but, in exam you obviously it is worth simplifying. In this case though, it's convenient not to. Now, in this stage we have 1/2 mv squared.

[01:01:01]And that's going to be 1/2 times mass times 7 by 5 AG.

[01:01:05]Mhm, v squared is going to be, you know, 7 by 5 AG. That's what I'm saying. So, this should be 7 by 10 mgA.

[01:01:14]Okay.

[01:01:15]mgh, how would we solve that? We don't know the height, do we? Okay, well Yeah, okay. So, we don't know the height. Then that's not going to be much of a problem either.

[01:01:29]Um let's take a height. Let's take a completely made up height.

[01:01:33]Let's call uh this H.

[01:01:39]Okay?

[01:01:40]So, my extension would be H minus A.

[01:01:45]And this height would be 2 A minus H.

[01:01:51]That being said, now I have my GPA GPE as well. So, this is going to be mg times 2 A minus H plus 5 mg H minus A the whole square divided by 2 A.

[01:02:08]And I know that this is equal to 5 mg A whole square by 2 A.

[01:02:17]Now, why I didn't simplify this is because it's quite convenient to bring this to this side and do it. Uh completely unnecessary, although it's your it's very optional.

[01:02:27]And yeah, we could just simplify each side now. 10 mgA plus 2 mgA. Yeah, this is the part where I said it's really simple, you know, you don't need uh further assistance as long as you know how to do the question.

[01:02:39]And this is equal to 7 27 by 10 mgA minus mgH which is equal to 5 mg by 2 A times A square.

[01:02:55]Okay, let's let's bracket. A square minus H minus A the whole square.

[01:03:02]Okay, now I'm going to erase I'm going to erase some of this to make space.

[01:03:13]Okay.

[01:03:18]Let's Yeah.

[01:03:22]Okay, yeah. So, proceeding from this, uh now it's I think the simple parts.

[01:03:29]Uh let's use a red pen.

[01:03:32]Okay, we can cancel out m g's on both sides. m g m g cancelled.

[01:03:37]We get 27 by 10 a minus h equals to um 5 by 2 a times now factorize a Simplifying this would be a squared minus h squared minus 2 a h plus a squared.

[01:03:59]Okay, now I'm going to multiply 2 a on both sides, so I get 27 by 5 a squared minus 2 a h equals to 5 times Well, a squared and a squared gets cancelled out here. So, I'm getting 2 a h minus h squared.

[01:04:19]And so, 27 by a squared minus 2 a h equals to 10 a h minus h squared. I can multiply 5 on both sides. Uh wait, no, this is 5.

[01:04:32]This is 5 and this is a squared. Uh yeah, I can multiply 5 on both sides. I would get 27 a squared minus 10 a h equals to 50 a h minus h squared. I can write a quadratic. I'd get this.

[01:04:50]27 a squared and this is equal to zero.

[01:04:54]Now, solving for your values for h h should be equal to 9 by 5 a or 3 by 5 a.

[01:05:04]What do we pick? Well, we know that h is clearly going to be an extension.

[01:05:09]And as a and as a result And as a result, it must be greater than A. H must be greater than A.

[01:05:17]3 by 5 A is lower than A. 9 by 5 A is greater than A. So, H must be 9 by 5 A.

[01:05:23]Hence, you have your answer.

[01:05:26]That's it. And that just, you know, goes to show how simple questions can be when they're analyzed deeply in this context.

[01:05:34]Now, that is the end of this um crash course, I would say.

[01:05:39]This was basically the video I've been planning to make for some time cuz I wanted to collect data on the most miscellaneous questions that you could get so that your practice for the final exam may be, well, um crystal clear, I guess.

[01:05:54]And your concepts can be, you know, perfected. Uh I do apologize for making this video quite late, and it is it was quite delayed due to work. Uh but, yeah.

[01:06:04]>> [clears throat] >> I'd say that this video should be able to provide with you uh enough uh enough understanding to tackle the exam. I do hope it goes well for those who are writing the exam tomorrow. And to them, I wish them all the best. Um you guys got this. And yeah, just stay tuned for more videos like this.

[01:06:23]Uh see you soon next time, and goodbye.