A LEVEL MECHANICS 1: THE ULTIMATE SUMMARY - Revision tips to pass A level Mechanics 9709/4

[00:00:00]Hello everyone. This is teacher Simon here uh taking you through A level mechanics and in this one I'm just going to give you a brief summary. Okay, a summary of what you're supposed to be knowing before you go for your paper.

[00:00:15]Now we have it's like five topics I call them like five. Okay. The first one is chyatics.

[00:00:24]So chyatics is uh split into two. We have where we are having the constant acceleration and where there is constant acceleration then we can use the three equations of motion. Use uh we call them the suvt.

[00:00:45]Okay. The suvat formula.

[00:00:50]Okay. The first one is uh-huh in fact let me put it here. The first one is v is equal to u + 80. The second one is s is equal to u t then plus a half a t squar and then v ^ 2 is = u ^ 2 + 2 a s.

[00:01:22]Remember this formula are given in the list of formula. Okay, they are given in your mf19. Now where do we apply this in uh straight line graphs? We can also use them in vertical projections used in uh motion motion or in fact let's say used in horizontal horizontal motion.

[00:02:01]Okay. Then we also have vertical motion.

[00:02:08]Vertical motion.

[00:02:12]And actually in vertical motion where you project a particle and then it goes up like that then it comes down. So at this point here where it is turning off from then v is going to be equal to zero. So suppose they tell you maybe to find the time if let's say uh this particle here is projected at a given speed okay up there and then it covers this greatest speed I mean the greatest height sometimes we say that greatest height there now you should know that this one is equal to what to zero so if they want us to find the h we can use the v ^ 2 is equal to u ^2 + 2 A uh 2 A S because here it is zero and if this one is given then A here by the way A has to be equal to -10 okay -10 that is if you are moving up okay when you're moving down then the A is what is positive so in other words we can say a is equal to what to g so if you're moving up then it is going to be what a negative now since we are going up here then you would put the negative here.

[00:03:24]Does that make sense? And then from there you will be in position to get the edge. Then sometimes they can also bring questions where they say find the time uh at which this particle is at least uh let's say 10 m is at least 10 m and you know that at least is greater than or equal. Okay. So that means this is the one that you're going to have. This particle can go up here.

[00:03:52]Okay. And then it can also come down here. So when they say when it is at least 10 m above the ground, we are looking at here.

[00:04:05]Okay. So when it will be coming down here, it can also be at 10 m above the ground. Now if this are the 10 m, now from from here you can use your s which is equal to UT then plus a half a t. You see this?

[00:04:22]So you have your S. Let's say it is the 10 m. Then the time taken here is the one that you're looking for. But you have your U. So you can put that UT.

[00:04:33]Then now our G, I mean our A is going to be negative because you're going up.

[00:04:39]Does that make sense? So you put that G here, then T ^ 2. So you are going to get a quadratic equation here which you solve. It will give you T1 and then T2.

[00:04:50]So the t1 that is the time taken for this particle to reach this first 10.

[00:04:55]Are we there? Then the t2 will be from here to there. So this one can be a bigger value to this one. So when they say the time taken for this particle to be at least 10 m above the ground then we shall get the t2 minus the t1. Okay.

[00:05:13]All right. So that is that. Allow me take off. Uh okay. I'm going to leave to leave them there. I think it's okay. So, we can have the T1 and then the T2. So, T1 can be here. Then your T2 can be there. Now, Uh-huh. Then another one that we can have they can say they can talk about the speed. Okay. They say that when the speed is let's say at least also maybe 10 let's say 10 m/s.

[00:05:52]Okay. Now if the speed here is 10 that means of course here uh we can have what let's say it is like 30 m/s and then this one is 10. Remember as it is going up it is coming to rest somewhere here. So the speed keeps on reducing also. Uh-huh. So you can use this as the initial one here. Then the final is here. Then you can find the time here taken. You see that? So we can use v is equal to u + 80. The final is zero. The initial going to take is this one here. At that point they are saying at least or at most whatever. So you put that uh speed there. Then since we are going up that one will be a negative 10 then t maybe I'm just giving an example.

[00:06:39]So the time taken here our t is going to be equal to one. Now that is the time taken to move from here to above here and it is going to be the same time taken to move from this maximum position down here here. So that means I will get that then times 2 the the time I've gotten here I multiply it by two because the motion here is symmetrical.

[00:07:06]Does that make sense? Okay. Then uh what else can we do next? I think that is that for this. Let me just take away this. Now let us proceed. We can also have constant.

[00:07:23]Okay, this is constant acceleration. Now we also have variable acceleration. I don't know how I repeated this. Now for variable acceleration you should know that we don't use these three formula okay we instead use integration and differentiation so we begin with displacement displacement then we go to velocity then we go to acceleration acceleration another thing that I haven't talked about here is when they talk about deceleration you should know that that is negative acceleration that is negative acceleration and you should not write deceleration with a negative. So if acceleration is v minus u out of t and then we put a negative here usually the final the final is zero. So the final is zero. We have maybe something like this and this line is like that. So uh this is the zero and then the v somewhere there. So we are subtracting a big value from a small one. You get a negative then the negative here will make everything negative I mean positive. So that's why deceleration has a positive value but it is negative acceleration. Okay. Uhhuh.

[00:08:55]Then from displacement we know that displace velocity is the rate of change of displacement and also acceleration is the rate of change of velocity. So that means going down this way we are differentiating.

[00:09:15]So we differentiate.

[00:09:19]Uh-huh. Then going up here you know the reverse of differentiation is what? Uh integration. So we integrate this will be integration.

[00:09:33]So uh acceleration is equal to dv dt.

[00:09:39]That means v will be equal to the uh integration of the acceleration with respect to t. Then if uh velocity is equal to d s dt that means s will be equal to the integral of v dt. You see that? Uhhuh. Then another thing is that the total area the total distance traveled total distance traveled is equal to the total area under the graph. total area under the what did I write here? Under the graph if it's a trapezium then you do what?

[00:10:32]A you find the area of that trapezium.

[00:10:35]Okay. Uh if we are having curves we are using the integration here that is for variable acceleration.

[00:10:42]Okay. Now look at this one here.

[00:10:46]here when it comes to the total distance especially here where we are having variable acceleration always find the times because if you let's say sketch and there is a part that goes beyond the t- axis okay so let's say this is t1 maybe this is t2 and maybe this one here belongs to t3 so when you are finding the and this is velocity in meters/s Okay, then this is time in seconds. So if you are finding the distance traveled by the particle, you don't just integrate from zero then to t3 because there is this part that is below the t ais. Its value is going to be a negative when you integrate. So that means it is going to reduce these ones which are positive and then you get a wrong answer. So what are you therefore supposed to do? You have to find those times when the velocity is what? zero here. Then after finding those times when you're finding your uh displacement you're going to say integrate let's say from 0 to t1 then plus the integral from now t1 to t2 and then from then t2 to t3. Are we there?

[00:12:03]Uh-huh. But then you put them in the modular signs. You may not know where or which part is below the t- axis, which one is above the t- axis, but you put those integrations in modular signs because the modulus turns what is negative positive. Is that clear? All right. So that is chyatics. Next we have forces and resultants. So we can have a particle here. The particle is here.

[00:12:34]Okay. And then this particle here, let me try to draw something here. We have the horizontal.

[00:12:45]Oh yes. Then we can also have the vertical here.

[00:12:50]So Uh-huh. Now after having this one here, we might have the forces that are here. This one looks weird. I can have a force there. I can have a force here. I can have maybe a force here whatever.

[00:13:07]Okay. Then in fact let me put uh these ones here dotted these axis.

[00:13:18]So those are the forces that are acting on the particle here. Okay. Uhhuh. That is X and then this is Y. So we have a force let's say P and then it is at this angle theta. Then we have this force here here maybe Q and it is at an angle of beta. And then we can also have this force here R. And then it is at an angle maybe of alpha here. It can be here. It can be here. Whatever maybe let me put it here.

[00:13:54]Now this is beta. So alpha there. So you are supposed to resolve these forces here horizontally and then vertically.

[00:14:04]Okay. The ones that are already along the xaxis and the y-axis those ones we don't resolve them. Okay. Now when you are closing you are putting this one I call it closing because you are getting this hypotenuse and then you are putting it on the adjacent and using so that is what cosine are we there? So what does this mean? This way I'll have P.

[00:14:30]Mhm. Let me use another color. I think green is okay. I can have P cosine of theta. Then opening it simply implies if I bring this one here putting this one here it is going to be parallel. So it's like I'm dealing with the hypotenuse and then the opposite and that is sign. Are we there? So because it is s then when I bring this one up here it will be p then sin of the theta. So opening is s then closing is cosine. Now how about this q here close it. When you close it it is going to give us mhm then cos beta. It is closing this angle.

[00:15:17]And then when you open it up here we are going to have Q then sine of beta. Are we there? Uhhuh. Then this one here r you close this angle here then here you will have r then cos of alpha. If the alpha were here then closing this one here would be r then cosine of alpha.

[00:15:45]Are we together? Uhhuh. So putting this one now we are opening. So opening it here will be a sign. So I'll have arrow then sine of what? Sign of alpha like that. So usually you have seen that in my videos I just put a cross and then I put these components here. Okay. So if they say that the forces are in equilibrium for forces, let me change the color. For forces in equilibrium, Uh-huh. What do we know? The resultant in the x is equal to zero and the resultant in the y is also equal to zero. But then you can also be told to find the resultant of the forces. So the resultant of the forces here is going to be equal to let me just write it as as a vector because usually uh it forms an angle. It has a direction. So arrow x and then arrow y.

[00:16:56]Now this one here the resultant of the forces if I'm finding let's say its magnitude this RF f here is going to be equal to the square root because it can be in between these two here and then they forming an angle of 90° you see that so I can uh find the what I'm going to use what we call the Pythagoras theorem so you will have the r x² then plus the r I squ. Okay. And then the direction of that of that uh resultant of the forces is going to be that what uh the the angle okay it forms with the positive xaxis. Now let us look at something here when it comes to the direction. Suppose the resultant of the forces in the vertical they face up and then the resultant of the forces uh in in fact let me start it from here like this. Then we also have this one here.

[00:18:03]So the resultant in the y is going there. The resultant in the x is going there. That means the resultant the whole resultant will be here which is rf. So this one can form an angle of theta with the positive real axis. Okay.

[00:18:20]So we shall say that when we are finding this uh this one here of course that angle that angle is going to be the octon of the r y out of the rx. Okay.

[00:18:38]So let let me try to shift this one here.

[00:18:43]a little bit.

[00:18:46]Yeah, I think it can be there. Uh-huh.

[00:18:50]So, we can have our theta is going to be equal to the act of r y out of the r x.

[00:19:00]Okay. Uh-huh. So we shall say that when we are giving the direction that the resultant of the forces has the magnitude which you have found here in the direction you give that angle then you say above the positive x-axis. Okay.

[00:19:16]Then we can also have this one. The resultant in the y is going up there but the resultant in the x is going this side here. That means the resultant here uh I think I need to use what to use a ruler also.

[00:19:38]So the resultant is going to be here.

[00:19:42]Uh-huh. Now this is the theta. This is our rx and then this is our ray y. Now that means our angle here. Okay. If you can find this one, remember I can put this one here. That is how they can form the what? If I just shift this one here and then I put it here. That is how they can form the the right angle triangle. Okay. So we shall have the RF f here.

[00:20:15]Mhm. So our angle here theta will be equal to the act of of course this is the second quadrant when we go to trigonometry this one will be negative because this one is even going this side can have a negative but you take just the magnitudes of these forces okay in the arx. So that means the ones that are going this side okay are bigger than the ones that are going this other side. So you get this one is big then minus this one which is small. Are we there? Okay.

[00:20:46]Then this one will be the actton of that ray y over the arrow x. So you say that in the direction maybe it is above the negative x-axis. Are we there? We can also have the resultant where in the yaxis the resultant is down. Okay is going down but then the resultant in the x is going this other side. You see that? So what does this mean? that the resultant of all the forces then will be here. So this is our resultant of the forces RF f this is r y okay okay this is r x so the angle is here usually we talk about the ang the direction we talk about the deviation from the x-axis so if by the way you are given a force here and you are giving the direction here you can say that uh it is this angle or below if there is a force here let's say 60 ntons then you say below the 60 Newton force or below the negative x-axis are we there uhhuh again here you don't include the negatives okay you just take these ones as positives the rx and then the r y then uh let me shift this one also a little bit there We can also have the last scenario where we have the x going this other side and then the y going this other side. So that means the resultant of the forces is in the fourth quadrant here. Okay. This is r x this is r y and this is our resultant of the forces and the angle we have to take is that one there. Okay. So again uh what shall we have here? We shall say that our theta is again the actton of that r of y out of r of x. Now after this they can also tell you they can give you a system like this. Then they say that find the magnitude and the direction of a force that can be added to these ones to make the system to be uh in equilibrium. So what do you do? That means when you have these forces here by having these forces here the system is not at equilibrium.

[00:23:26]So you have to therefore find the resultant of these forces. So suppose the resultant of these forces is maybe here. I'm just giving an example. Maybe here the resultant and it forms this angle here. Let me use maybe gamma. It forms this angle here. Gamma with a positive x-axis. Are we there? Uh-huh.

[00:23:52]Then that means I will just put it opposite to this other side because this is the result that it will be acting this way.

[00:24:02]But the force that I will put there to make this system be in equibrium. It has to be looking this other side in the opposite direction. Okay. And the angle has to be this one here. Okay. Below the negative what? uh below the negative xaxis that is if it is here okay you just put it in the opposite direction does that make sense okay so this is about the forces and resultants then the next one is Newton's laws of motion you know the first one is that a particle continues in its state of rest or in uniform motion unless acted upon by an external force Okay. The second one is uh if a force acts on a particle then the the particle uh moves in the direction of the applied what? Force. So that's why we have F is equal to M A. This is the most important here. Okay. Uh F is equal to M A. And then the third one is that action and reaction are equal but opposite. So if I have let's say a block here which is sitting here on the plane it exerts its weight which acts downward. So mg but then this is the action there has to be a reaction which is the normal reaction that is the error. So they are equal but opposite. Are we there? Now in Newton's laws of motion we can have connected particles. We can have connected particles. Now in connected particles we have pulley. We have on the horizontal plane we have on an inclined plane. Okay there are very many. So maybe scenarios which we can have when it comes to pulley.

[00:26:05]Okay. To pulley I can have this one here. Then we have a particle here and then this one here they have their masses and then uh there you should know that if there is a string that is connecting the two then there has to be tension in that string and the tension is always moving away from the particles then this one has maybe m1 g this one here will have m_sub_2 g if they don't have the same one mass. So if this one is heavier than this other one then we shall have this one going up and this one will go down you see and then here you are going to use the F is equal to M A. So I'm just giving an example. Suppose this one was going up, okay? And then this one was coming down. Then you would say uh using this one here, it will be T which is going up. By the way, let me put the T here. Same tension in here as here. So T then minus this M2G. Then this one will be equal to the M_sub_2 A the acceleration. Okay. Then this other one here since it is going down that means the weight exceeds the tension. So we shall have m1 g then minus the t which will be equal to m1 a. Now to solve this one simultaneously you can just add you can see that the tension here will disappear and then from here you can get the acceleration.

[00:27:43]Are we there? So when you get the acceleration in there, you can then substitute in any of these two equations to find the tension. Are we there?

[00:27:55]Uhhuh. So that is one. We can have very many pulley. We can have them on the table. This one can be here. You have this. Then it comes to that. There's another one which is down here. You see that? Mhm. There is also tension here.

[00:28:12]There's tension here. Then the mg here whatever you see then we can also have again on the table here this one is connected this is a pulley then this one here there is this and then this one is connected to this other side okay uh it is connect this is a pulley here and then it goes through then there is another particle this other side you see so we can have also something like that so that means since these are two strings then we are having two different tensions. So maybe this is tension two.

[00:28:48]You see this tension two then this one will be tension one.

[00:28:53]Always the tensions are moving away from the particle. Master that. So this one is going to be T1. Okay. T1 like that.

[00:29:04]But they also have their weights here and there. You can connect them just like I've told you here. And then you do what? You solve. Then we can also have them on an inclined plane. I can have this one here. You see that? Then uh maybe this one goes here. There is a pulley here. And then this one is at a certain what angle. Then this one there is another one hanging in here. Okay. Uh-huh. Like that. So this one they can even give you the distance. this one uh is from the ground. Let's say it is X in here. And then they say when this system is released from rest, this one goes up to the ground. And then they tell you maybe when it reaches the ground, it doesn't bounce. Okay? Or it doesn't rebound.

[00:29:59]Then from there uh they tell you to find let's say the maximum distance this one travels. So when this one travels down here, it covers a distance of X. Even this one will also travel a distance of X up here. Are we there? But then when this one reaches here, this particle here, when this one doesn't rebound, this particle here can move a further distance and then it comes here. And in most cases, they tell you that this one doesn't reach the pulley. So there is a further distance of Y. Okay, maybe here.

[00:30:36]Now by the time this one covers a smaller distance here the string here is slack. Okay. So that means there is no tension in the string.

[00:30:48]Okay. So when the tension is I mean when the string is slack when the string is slack then the tension in the string is what is zero. So in such a case how do you find the total distance here? First of all, this distance can be given. Okay?

[00:31:09]Uhhuh. You know it. But then you find the speed with which the system was moving. Okay? By the time this one came to this position here, you are going to use the first of all uh you have your t in here and the acceleration you have gotten them using this kind of approach.

[00:31:29]Okay. Now you can use this v² is equal to u ^2 + 2 a s you find the speed here remember the system is released from rest maybe this is 0 square then plus the 2 * the a which you have gotten by calculating using this method here and then times the distance that is here you see that now that will be the speed with which this particle is uh with which it moves here and also that is the same speed this one will hit the flow. Does that make sense? Okay. Then after that continue by finding by the time this one comes here its final speed here is going to be equal to zero. Okay. So this speed we have gotten here is now the initial speed for the final movement. Are we there?

[00:32:23]Then I can say again we substitute in here. But since the string is slack, you have to get the new acceleration new acceleration. Okay, you use of course in here for an inclined plane there has to be mg sin theta. Then the mg cos theta is here the vertical component of the weight because the weight is here. There is this angle here which is the same as this one here. How does that come about?

[00:32:52]I have always shown it here like we are having this particle here on the horizontal plane. This is the weight.

[00:33:01]Now when I try to tilt this one a little bit. Okay. Even the weight first of all there was this one here. This one the way it is here. Now if you tilt to a certain angle even that weight is going to tilt to the same angle. Now close this one here. We have looked at resolving of forces. This one will be mg cos theta. You see that? Then this one will be mg when you open it up. Mg sin theta. But then here there is that.

[00:33:34]Okay. Uh-huh. Then what else? So uh I can say now this one here this particle is going up. Now there is no tension in the string. So I will say zero. Remember we are using f is equal to m a 0 then minus the mg sin theta mg sin theta then is equal to the m a automatically the m will go now our a is supposed to be a negative [snorts] why is it a negative because the particle is decelerating it is coming to rest you see that so this acceleration you're getting here you are now going to use it for the final movement and again you say v ^2 is equal to u ^2 + 2 a s so The final is zero. The initial is now this one you got here for the final movement.

[00:34:23]Then the acceleration here is the one you have gotten here. And then you can be in position to find that s that final speed. I mean that final distance which you are going to add to this one here as your y. Okay. And then you get the final. Then you get that. You see?

[00:34:43]Uhhuh. Then also on an inclined plane we can have a particle here. Uh if there is friction let me talk about the friction here. So if I have my particle here we can have of course this is an angle theta maybe.

[00:35:01]Now this is mg sin theta.

[00:35:06]If this one is a rough plane and you are applying let's say a force here. Okay, that force you are applying here is going to be resisted by the frictional force. So if the particle is going up, that means the frictional force is going to be acting downwards. Are we there? If the particle is going down, the frictional force is going to be acting upwards. Then they can also ask for what we call the maximum maybe force that you can apply here to prevent the particle from uh sliding downwards.

[00:35:45]In fact, they said the minimum the minimum force you can apply onto this particle to prevent it from sliding down. So it is just like leaning against it. Okay? So you are preventing it from sliding down.

[00:36:00]Okay. And now you consider it as if it has slid down. That means the the frictional force will be acting upwards.

[00:36:10]Okay. In other words, when do you get P maximum and then P minimum. So the P here if it is going up this way with the frictional force we have P plus the Mur are we there? then is equal to let's say this is the mg sin theta now p can also be this other side so when I take this one this other side I get this and then I subtract this I will get a smaller value of p that means this one will be giving us the minimum okay p minimum then will be equal to the mg sin theta then minus the mur okay But then P maximum will be equal to the mg sin theta then plus the mu r. Okay the maximum if you are just applying a force here to prevent the particle from sliding down the plane you are just applying the minimum force. So meaning the uh frictional force will be on the same side as the P. Okay. or it will be uh acting upwards. Are we there? But then the maximum force that you will apply is the one that is going to make the particle to be on the verge of now moving up the plane. You see that? Just stand somewhere and maybe you hold let's say a bag or something very heavy on the wall. Okay? Against the wall. So you can actually just bend on it. You are not applying any force there. So that is the minimum you're putting there. However, if you hold it with your hands, okay?

[00:38:04]Then you try to push it a little bit up, you will face resistance. That means the frictional force is acting downwards. So in other words, you are preventing now you are trying to take the particle up.

[00:38:17]Does that make sense? All right. Uhhuh.

[00:38:21][snorts] So I think that is that when it comes to Newton's laws of motion.

[00:38:26]Yeah, I think that is that. Next we have work power and energy. Okay, for energy here we are having two of them.

[00:38:37]The energy we have kinetic kinetic energy which is the energy possessed by the body in motion. it is a half mv² and then we have gravitational potential energy.

[00:38:59]Okay.

[00:39:01]So that gravitational potential energy is the energy possessed by the body.

[00:39:06]Okay. Relative to its position from the ground. Okay. There is that vertical height. That's what I'm talking about.

[00:39:14]So it is mg h. Okay. And then we also have what we call mechanical mechanical energy. Mechanical energy is the sum of kinetic energy and then potential energy. Okay. All right. Then after that we have what we call power.

[00:39:43]Yes. Uh then work. Okay. Let's go to work. Work is equal to force times what? The distance. Force times the distance.

[00:39:57]Uh-huh. Then uh lastly we have what we call power. Power is the rate of doing work. Rate of doing work.

[00:40:15]So it is going to be the force times the distance that is the work not.

[00:40:22]So then we divide by the time anything to do with rate we are dividing by the time right? Distance out of time you know that that is speed. So that means we can have force here then times the velocity or the speed will give us the power. Okay. Now we have what we call work energy principle. Suppose I have my particle here. Okay. I subject it to a given force. Let's say this is the driving force. This particle here moves through and then comes to this position here. Here it starts with maybe the speed U. It reaches here with the speed V. These two are different. Are we there? But in moving here there is this ground here is uh is rough. So there is frictional force also.

[00:41:19]There's frictional force. Now if there is frictional force here first of all I've applied a force. So I'm doing work because this work here I'm doing I mean the force that I'm applying is traveling through a certain distance here. And we say that force times distance is what? Work. So we always say work done by this driving force. Okay. The work energy principle actually states that uh let me write the principle here. The work energy principle states that the total work done on any system is equal to the total change in the energy. Okay. Total work done on a system is equal to the total total change in energy.

[00:42:18]Total change in energy. If you have been following me, I have always written this that work done by the driving force minus work done against resistance is equal to the change in the mechanical energy.

[00:42:37]Okay, that is the same thing that we mean here.

[00:42:42]Change in mechanical energy. So as you are taking this one forward, you are doing work going forward. But there is also work that we are doing against this resistance force. So that's why we say the net work done to take this particle forward is going to be the work done by your driving force minus the work done against the resistance. Then this one will produce the total change in the energy like this is kinetic energy which is going to be changing here. Okay.

[00:43:12]However, we can also be taking the particle up here. Okay. So when I have this one here, let's say the particle is here and then it comes to this other side here. If it has u here, the speed, okay, m/s, then it has V here, m/s.

[00:43:32]By the time it reaches that upper position there, okay, and it has that speed, that means it has kinetic energy.

[00:43:42]Are we there? And then uh because of this generation of the vertical height then it will have also the potential energy. And then here down here taking this as the reference point there is no potential energy because there is no vertical height. So here we are just having the kinetic energy. Uh most times I put I that is for the initial then this one is for the final not.

[00:44:12]So when we are finding the change in mechanical energy we get the final then minus the initial. But then suppose this particle here okay was moving. We can by the way have a distance in there and then this is theta. So we can say sine of theta is equal to h / d. And then that h is going to be d sin what d sin theta. Okay. Uhhuh. And I was saying that if this one is moving at a constant speed that means there there won't be the change in kinetic energy because the kindinetic energy you're having here is the same as the one here. So when you subtract them that one disappears. So the only change is going to be in gravitational potential energy. Are we there? Uhhuh. Then when it comes to the energy method okay we understand here there is the mg sin theta.

[00:45:09]Do not include it amongst the resistance forces. Okay? Do not put it there. It is scattered for in the mgh because when you write here mgh, this h already has the d sin theta there. You see that? So it is already catered for. Do not use it. It is the the work done against the resistance. It is the frictional force. The resistance is the frictional force. Okay. And that is that and by the way in such cases you can also use the the forces you have F is equal to M A. So you can use the driving force here then minus the let's say mg sin theta that is where you're going to use the mg sin theta and maybe another uh resistance force here. Okay.

[00:46:00]Then uh if the acceleration is there if it is traveling at constant speed acceleration with B 0. Okay. Then after that what else are you going to have? So you will get the acceleration in here. Then after getting the acceleration you can now use the v ^2 is equal to u ^2 + 2 a s. So using this one now uh what are you looking for? You can maybe get the uh you can get the you you have already gotten the acceleration. So you can maybe get the distance traveled by this one up here. Okay. Uh like that or you can get the speed. In most cases they are asking for the speed. So that is that for work and energy principle.

[00:46:50]It is this one here. You just put this one in your head. work done by the driving force and also maybe another thing that I need to talk about here.

[00:46:58]Suppose you are given power and then the time taken. You can find the work done by the driving force. Remember power is equal to work done by the driving force then divide by the time taken. So when I cross multiply these two work done by the driving force then can be equal to power time the time. Are we there? All right. So that is that. Then lastly, momentum. Momentum is the product of mass and velocity. Okay. Momentum.

[00:47:29]Momentum is that the spelling momentum is equal to mass time velocity. So since velocity is a vector quantity that means even momentum is a vector quantity. And since it is dependent on mass, the heavier the particle, then the more momentum it will have. Are we there? It will have a big value. Now, what is the principle of conservation of momentum? We say that momentum before collision is equal to momentum after collision. Momentum before collision.

[00:48:12]Here we are looking at particles colliding. Okay. Is equal to momentum after collision.

[00:48:22]Uhhuh. Then let me take away this one here. Now look at you.

[00:48:29]I'm going to put after collision. So we can have like maybe m a u aa plus mb u b then is equal to the final m a mhm u that will be v final. So the final speed v a then plus mb vb something like that but this is when they collide and they don't stick together. Okay.

[00:49:00]In other words, in collision, for collision, we have two types.

[00:49:08]Collisions. So, we have elastic collisions.

[00:49:14]Elastic and then inelastic.

[00:49:19]In elastic collisions, the bodies do not stick together. Bodies don't.

[00:49:28]Coalis. Actually, that is the uh common statement. They use coal. Bodies do not.

[00:49:35]Here bodies.

[00:49:38]Okay. Bodies coales or they stick together after what?

[00:49:44]Collision. So they become one body. So this one here, what I've done here is for elastic elastic collisions.

[00:49:58]Then we can also have inelastic.

[00:50:03]So for inner elastic we are going to have m a u aa then plus mb u b then this one will be equal to m a because it has become one particle we shall add the masses together then uh that particle will be moving with the same what velocity okay that is the principle of conservation of momentum.

[00:50:32]Uhhuh. Then what else do we have in here? You have to be very careful. They can bring for you particles when they are moving in a straight line. Now if this particle here is lighter than this one. Okay, let me just do like this. This one is big. So this one is smaller than this one. That means when they collide, okay, this particle here can move forward. There are two scenarios. It can move forward or it can hit.

[00:51:09]Okay. And then jacks backwards. So it can also move backwards. You see that?

[00:51:16]So long as this one is smaller or weaker. I think lighter, not weaker. Lighter than this other one. It is knocking. Okay. Uh-huh. But so that means when you are substituting in here, master the direction. So if your first direction for this was going this way and then uh after collision now this one is coming this way. So you will put a negative on that because remember velocity we are using velocities by the way in here. We don't use speeds we use velocities.

[00:51:50]Okay? And velocity is a vector quantity.

[00:51:52]Therefore momentum is also a vector quantity. Does that make sense? Okay.

[00:51:58]Now, uh we can also have projections where they can project this particle.

[00:52:06]There is another one maybe coming from here.

[00:52:10]They project this one with a certain speed U. But these two have gotten a distance in between them. Okay. Now when this one is projected and it comes here even this one is projected downwards or it is released from rest they will come here and maybe collide. Okay. So as this one is coming down here let's say this one is going to be s of a then this one is s of b. So when you add these two their expressions here will give you the total distance in between the two.

[00:52:44]However, okay, however, there is a possibility that this one, these two can collide there or this one can uh can be projected. It goes up but then it turns and then this other one finds it when it is going down and it collides with it.

[00:53:02]Okay. So when we are finding or when we are using the principle of conservation of momentum, we have to be very careful here.

[00:53:10]You have to when they are going to collide they are not colliding with the speeds they were projected with here.

[00:53:17]Okay. So you need to find those speeds before the collision. Okay. They can tell you maybe they collide after 1 second. So if they collide after 1 second you use this one. SB first say SA is equal to Mhm. There is maybe this one was coming from it was just dropped.

[00:53:40]So we have s is equal to u plus a t² you'll get zero then plus a half since it is going down this is vertical motion acceleration will be the other the other one due to gravity so it is facing down that means this one will be a positive one are we thereh let's say that is t mhm then that one you can get actually it is squared you can have 5t squ then sb here it is projected with a certain speed here not So u then t uhhuh then plus a half but then this one will be negative what 10 then t ^ 2. So you get some u t here u can be any value they can give it to you then - 5t ^ 2. Now when they meet you will get this expression here then plus this expression you equate to this distance in between them. After that you will be in position to get the value of t there the time taken for them to collide. Okay.

[00:54:43]And but if they give you the time taken for them to collide you can use v is equal to u plus a t for you to find the speed at which they each of them like collides with the other. So if this one is dropped from here the initial speed is zero and then we have it is going down. So that is 10 maybe times the time is one. So that means by the time they collide it is uh they collide when it has a speed of that you see. And then this other one here you have the s of b is equal to u which is given there maybe it is projected at 20 m/s. Then after 1 uh minute, okay, we are also substituting in here.

[00:55:28]V is equal to 20. Uh then after 1 second because this one is going up, this will be -10. You see that? So you will also get some speed here. Now that is going to be the initial those are going to be the initial speeds you're going to use in here. Okay? Before they collide. You don't use the initial ones here. when they when they did what when the particles are projected from those positions. Does that make sense? Okay.

[00:56:02]So, I think that is that I think so. So, that is that if you find it very helpful, kindly give us some thumbs up, right? and then let me know in the comment section uh maybe where you haven't gotten or if it has anywhere made any sense to you. Success in your exams. May God bless