Edexcel IAL Physics Unit 1 Predictions For May 2026

Added: 2026-05-11

[00:00:02]Hello everyone, welcome to the IIL unit one predictions for May 2026.

[00:00:09]And even though I have said this is the guide to grade A, this video alone is definitely not the guide to grade A.

[00:00:18]There are several several several other um resources involved uh that I'll be sharing with you all. um if you follow all of those then you might have a chance at grade A. I repeat this video alone is not going to ensure um you know the best grade for you.

[00:00:40]What I've included in this um description is the most frequently asked questions and answers of unit one.

[00:00:51]And when you download this, make sure to add the free definition sheet in the card. Get it for free. Link is in the description. And let's get started.

[00:01:05]Like I said, this is not the only resource that that you should depend on for your preparation for your exam tomorrow.

[00:01:15]I have made a full syllabus revision video for unit one. I will highly highly highly recommend you to go to that one as well.

[00:01:25]Now, so this is you can say fresh course of sort of preparations and this is what you can get uh started with right but make sure this is not the only thing uh that you end your preparation with. So there are three sections of unit one uh in the broader scheme of things. One is mechanics, second is fluid mechanics and three is solid materials. What I have done in this video is compiled a list of um what I feel is uh the most important topics.

[00:02:04]I wouldn't say that is most likely. Of course, the most important ones are the most likely to appear in the exam tomorrow. But this is what I just felt myself. Okay, this is just my personal preference. I have no access to the script uh prior to the exams.

[00:02:23]So, let's first look at the sections of mechanics.

[00:02:29]So very very very likely you'll get a projectile motion question and to answer this question uh confidently you you should have a very clear idea about the three scenarios in projectile motion.

[00:02:47]So the moment you see uh the diagram in your exam paper, it should instantly pop up in your head which specific scenario are you dealing with because every different scenario has a different feature and aspect to it. So the first scenario is where an object is launched and lands at the same height from which it is launched. Meaning the total vertical displacement is zero. That is scenario one. and throughout the whole um motion the acceleration is - 9.81.

[00:03:22]In scenario two, an object is launched from a height but horizontally. Since it's launched horizontally, it has no initial vertical motion and so no initial vertical velocity. But since it's moving in only one direction, the acceleration throughout the motion is just 9.81.

[00:03:44]Scenario three is the most interesting one. It's where an object is launched at an angle from a height. And you'll see the projectile lands below the point it is launched from. Meaning the displacement as in final displacement is negative because it is below the point of origin. So another thing I need to discuss with all of you is you should have a very clear understanding of the origin because in recent papers I'm saying there are a lot of questions where you need to identify where the origin is to draw a graph or sketch a graph properly and do calculations with it.

[00:04:23]And there is a whole uh video on projectile motion in my channel. You can refer to that also. I have discussed some very important topics in the full syllabus division video as well.

[00:04:35]Next up, what I feel is most likely to appear in the exam tomorrow is a question with law of conservation of momentum.

[00:04:44]So in most questions, most papers you'll see uh the law of conservation of momentum is applied in collisions but explosions uh are also there in your syllabus. Right? So when you apply total momentum before collision equals to total momentum after collision the same thing applies to explosions. Total momentum before explosion is equal to total momentum after explosion. For example uh in this scenario this uh object before it explodes. U has a mass but doesn't have a velocity. It is at rest. So its momentum is zero. And after explosion the fragments move in opposite directions to conserve momentum. And if you add up these momenta momenta momentum being a vector quantity since they're in opposite directions their momenta cancel each other out. Okay. So making the total momentum zero again. So this is how you apply law of conservation of momentum in explosions as well. And if the question comes with collision, there might be uh at the se there might be a section at the end where they ask you whether it's an elastic or an inelastic collision.

[00:05:56]Right? So when it comes to elastic or inelastic collisions, what you have to remember is you have to find out the total kindinetic energy before and after. If the total kinetic energy before is equal to the total kinetic energy after then and only then you can say that this is an elastic collision.

[00:06:17]then uh very likely there will be a moment as in principle of moments question. So let's look at a very very very simple scenario here right. So this is a beam and the beam is in equilibrium because the total clockwise moment is equal to the total anticlockwise moment.

[00:06:38]So in questions like this they can directly ask you what do you mean by about equilibrium right? So uh these questions are clearly answered in the definition sheet uh equil definition of equilibrium definition of center of gravity right all of these are there in the definition sheet also in the fact sheet but what I want to show you in this video is how they can elevate a concept just look at a simple IGCAC level scenario like this right so what they can do using this is they can apply this in a specific context So it is hung from this position and the dish and the rice is producing a clockwise moment and the weight acting from the center of gravity of the beam is producing an anticlockwise moment.

[00:07:27]And since this is in equilibrium again the same concept applies total clockwise moment equals to total anticlockwise moment. Now issue is somewhere else.

[00:07:39]When it comes to unit one, there are a lot of concepts that include and incorporate vectors in them and moments is one of them. So look at this scenario.

[00:07:53]In this scenario, um if I take moments about this point because the question asked me to do that what you do is look at the tension, right? So this tension is not creating the clockwise moment. It is basically the horizontal component of this tension. So the horizontal component there is no angle here. So without angle is sin theta. That makes it t sin 44.

[00:08:25]And whenever you um encounter questions with mments, always remember you have to consider uh the force and the perpendicular distance from the force to the pivot. perpendicular distance. So if I'm dealing with this force, I have to consider the perpendicular distance from this force to the pivot. This is the distance. So t sin 44 * 1.5 is basically the clockwise moment. and 25 * 1.8 because force and perpendicular distance 25 * 1.8 gives you the anticlockwise moment about this point.

[00:09:13]So you can find out t from here. You can actually pause the video and find out t.

[00:09:19]Okay, let me give you the answer. And I'm not going to show you how to do this because this is not a video for solution sol uh solving papers or solution of papers. This is a video um to show what you need to revise. Right? So I'll pause the video and I'll write down the answer. You do the calculation and see if your answer matches with mine.

[00:09:44]So I hope you have done it. The value of t is 43.2.

[00:09:50]2 Newton.

[00:09:55]Another trend I'm noticing in the recent papers is they're bringing in concepts from M1.

[00:10:01]So this used to be a very common concept in unit one as well, but it's been a while they haven't introduced this question in the recent papers. So I thought I might just include this in the revision video. So what happens is the larger mass m accelerates downward and the smaller mass m accelerates upward.

[00:10:25]There is tension in the rope and they both have the same acceleration.

[00:10:31]So what happens is since it accelerates downward the weight mg is greater than this tension T and since this accelerates upward its weight small mg is smaller than the tension.

[00:10:50]So for this for this uh the larger mass since mg is greater mg minus t equals m a and for the smaller one p minus mg because this tension is bringing this up right so t has to be greater minus mg equals to since this is the small m small m a and a is the same for both because they they're connected so they'll have the same acceleration.

[00:11:31]There were several questions like this but it's been a while. Uh Pearson has stopped asking these questions but this is still in M1 so I just thought I should include this in the revision video and I'll tell you why.

[00:11:44]Look at this one. This is a classic mechanics question but this has become a staple question for physics unit one as well. So what happens is um a box or a block slides down an inclined plane.

[00:12:00]Right? So when the block slides down an inclined plane if there is friction the whole scenario changes. And let's first assume there is no friction. Okay? If there is no friction then the only force acting on it along the plane is mg sin theta. And guess what this mg sin theta is the resultant force. And you know the equation for resultant force that is m a. So mg sin theta equals to ma. From here you can find out the acceleration because the m's get canceled out. But if you bring in friction then the whole dynamic changes then mg sin theta as in the component of weight down the plane because this is the only force that is in the direction of the motion minus this is the other force in the opposite direction of the motion friction let's call this f_sub_r right and now this is the resultant force that is ma So I hope you have figured out the two scenarios with friction and without friction.

[00:13:16]So there is a 50/50 chance of a scale vector diagram appearing in the exam tomorrow.

[00:13:23]So because in the in all the recent papers they have asked uh scale vector diagram question.

[00:13:31]So there is a high probability there is a minimum probability actually.

[00:13:37]So whenever it comes to scale vector diagrams always remember the vectors must be arranged tip to tail. Okay, always tip to tail. And for this specific question I want to show you a variant. Okay. So you can draw a vector parallelogram as well because there are uh two ways of drawing this. Let me show you both instead of just saying let me just show you. So a sailing boat is moving through the water. The force of the wind on the boat is 350 Nton towards the north. The force of water on the boat is 410° at an angle of 120° from the north. All right. Determine the magnitude and direction of the resultant force. So I need to find the resultant of these two forces. Fine.

[00:14:20]So always remember vectors must be arranged tip to tail. This is 350.

[00:14:29]And I place the tail of 410 at the tip of 350 like this.

[00:14:39]This is 410.

[00:14:42]And guess what?

[00:14:45]The resultant cannot form a complete loop. If it formed a complete loop, it would appear as though the body is in equilibrium. Right? So when a body is in equilibrium then and only then all the vectors form a complete loop and since there is a resultant the resultant one must be in the opposite direction. So these two in the same direction this is the resultant F.

[00:15:14]So like I said you can do this two ways.

[00:15:18]So for the first scenario what I did was I placed the tip uh I placed the tail of 410 on the tip of 350. I can just do the reverse. What I can do is I can bring in the 410 and at the tip of 410 I can place 350 tip of 410. Click and place 350 like this.

[00:16:02]This is actually let me take out 410 from inside so that it looks less clumsy.

[00:16:09]This is 410 and this is 350 and guess what? This is the resultant F and it cannot form a complete loop. This will be in the opposite direction to the other two.

[00:16:24]And if you join these two, what you'll end up with is a vector parallelogram instead of a vector triangle. If you join these two like this, you'll end up with a vector parallelogram.

[00:16:43]So you have two options. And of course, you have to do the scaling.

[00:16:50]And it's imperative that you mention the scale you have chosen for drawing the scale vector diagram.

[00:16:59]So the next section is about Newton's laws of motion. And I'm a bit surprised to see that how Pearson has completely forgotten about Newton's first law. Okay, it's also referred to as the law of inertia. So all of these scenarios are associated with Newton's first law.

[00:17:20]Let's first look at the first scenario.

[00:17:23]Uh here you have a card cup and a coin.

[00:17:26]When you flick on the card, the card flies away, but the coin remains where it is and drops directly below into the cup. So how this works is the force only acts on the card. The card does not exert any force on the coin because friction between card and coin is negligible. After the card flies away, now there is a resultant force on the coin vertically downward. So it accelerates vertically downward. And every time you bring in acceleration, you refer to Newton's second law. So you can say that this is a combination of Newton's first law and second law. How first law? Because when the force acts on the card, only the card moves away.

[00:18:11]There is no force on the coin. So the coin remains horizontally at rest.

[00:18:15]according to Newton's first law. Same thing goes with tricks like this. So you must have seen in TV series or videos, viral videos where the guy pulls down the tablecloth from under the cutlery, right? And the plates and everything remain where they are. So this is again a classic example of Newton's first law.

[00:18:39]because the force acts only on the clo.

[00:18:43]The clo exerts no force on the plate because friction between plate and uh clo is negligible.

[00:18:53]So this is not a very common concept nowadays. What is common is this everybody?

[00:19:03]Right? So here you can see a swimmer exerting a force on the wall.

[00:19:10]to move forward. So as her feet exerts a force on the wall, according to Newton's third law, the wall exerts an equal and opposite force on the feet forward.

[00:19:23]There is a resultant force on her body or her feet forward causing her to accelerate. You know what? The swimmer will only accelerate as long as she's in contact with the wall because there is a resulting force only as long as she's in contact with the wall.

[00:19:39]So this is a combination of Newton's second law and third law. This is way more likely.

[00:19:47]Let's move on to the next section.

[00:19:50]Okay. So now we are talking this is energy conservation. So I strongly believe there will be some questions either in section A or section B uh regarding conservation of energy or energy conservation. So here this is the very simple scenario. Um a block slides down an inclined plane. Okay.

[00:20:18]If the surface is not smooth when the block starts off it has it is starts it starts at a height. So it has gravitational potential energy. In an ideal world this Egraph would have been transferred to the kinetic energy at the bottom. But this is not an ideal world. What happens is there is friction and as the block slides down there is work done against friction. So if you subtract the work done against friction as inverive forces what you end up with after subtracting the work done is the value in the form of kinetic energy. So this is the classic example and actually you know what you can go in reverse. There was actually a question where uh it was about a disc right? So a disc was launched from the bottom and it rose to the top and the exact same thing at the bottom it had kinetic energy. So this kinetic energy was not fully transformed transferred to gravitational potential energy. On the way up it had to do some work against friction. So going up it's kinetic energy minus work against friction which is equal to the gravitational potential energy.

[00:21:42]So you can go both ways going down or going up. Going down you start off with gravitational potential energy. Going up you start off within energy.

[00:21:57]Next. Okay. So this concept has become a staple uh unit one concept in recent papers. It started from May 22 with the sorry Jan 22 from the pinball machine question. So the scenario is this box or block initially is pressed against this spring. Okay.

[00:22:26]So when the spring is compressed, it has elastic potential energy.

[00:22:36]Okay.

[00:22:38]When it decompresses, this elastic potential energy is transferred to two forms. Since while it decompresses, it goes up. Right? So this elastic potential energy is transferred to the gravitational potential energy as well as the kinetic energy. But in a lot of questions they ignore this part. So a lot of times what they do is this elastic potential energy is directly transferred to the kindinetic energy over here when the spring decompresses.

[00:23:15]So this small change in e graph is at times just ignored.

[00:23:20]So it's better not to ignore depending on the context of the question.

[00:23:24]And finally in the third scenario you have a car moving with a constant speed. If it has a constant speed it has some kinetic energy. When the brakes are applied, hear me out. When the brakes are applied, the kinetic energy drops to zero because the speed drops down to zero. So where does the kinetic energy go? all of the kinetic energy uh is dissipated due to work done against uh resistive forces.

[00:23:58]Even more appropriate would be work done by the breaking force. So if you want to write down an equation for this specific scenario, it should be kinetic energy is equal to what? If you break this down and if the mass and the velocity is given it's half mv² = to you just recall the equation for work done is f * s and this f is actually the breaking force.

[00:24:31]So for the next question, I deliberately chose this because you need to understand some very specific situations and previously I also told you uh you have to have a very strong idea about this concept called origin. Right? So let's look at this question.

[00:24:51]The displacement time graph for a sky diver between jumping from the aeroplane and opening this parachute is shown. So you see the displacement starts from zero. So here what they have considered is um the point of jumping has been considered to be the origin and since you are landing below the point of origin all the displacements are negative. Okay. And uh how do I put this? A simpler uh graph for this would have been which you already know would have been like this.

[00:25:28]So this is displacement time like this, right? And the corresponding velocity time graph would have been the most common one that you already know. Velocity time graph the velocity becomes constant as in the sky averages terminal velocity. Here it's a it becomes a straight line. So the gradient of straight line is constant. Gradient of displacement time graph gives velocity. So the velocity becomes constant. Right? But what they have done here is they have given a downward sloping line.

[00:26:26]Meaning all the um the gradient at any given point is negative. Downward sloping lines always have a negative gradient. And guess what? The gradient of displacement time graph gives me velocity. So all the velocities will be in the negative axis.

[00:26:47]And the graph that they expected you to draw is not in the positive axis because a downward sloping line always has a negative gradient. And this is the graph that was expected from you for this paper up to 20 20 something. So I have to extend this line and go all the way up to 20 something. Yeah, something like this.

[00:27:14]So these are all new stuff. not previously asked ever. So you need to be wary of these concepts as well. Always look for the origin. And another example that I already discussed in my full syllabus revision video is from uh June 23 where uh diver jumps from a platform.

[00:27:35]That's where the water surface was considered to be the origin. Any point can be considered to be the origin. by the way.

[00:27:44]Let's move on to the next section that is fluids.

[00:27:48]So I just have a gut feeling that this paper might have something to do with when an object floats or whether an object will float or not. So in this scenario since this wooden block is floating of course the forces on it are balanced. What are the two forces? One is of course its weight.

[00:28:12]The other one is the up thrust.

[00:28:16]Right? But every time you come across up thrust, you just recall the definition of up thrust. It is the weight of fluid displaced. Okay? Up thrust is the weight of fluid displaced. And you can see how much fluid has this block displaced.

[00:28:34]It is equivalent to the volume of the block that is underwater.

[00:28:40]So this is the volume of the block that is underwater and this is the volume that will correspond to the thrust.

[00:28:54]Then this is a very very very straightforward concept. This might appear in your exam regarding Stokes law. Sometimes they ask you whether Stoke's law is applied or not or whether Stoke's law is applicable. So in questions like this all you have to do is find out Stoke's law force using F= 6 pi at RP and then separately find out the drag force using if it's a falling ball then always remember to draw a diagram question like this. If it's a falling spear, the forces acting on it would be its weight upward will be up thrust and drag. And if it is falling with terminal velocity, all the upward forces must be equal to all the downward forces. So in this situation, weight should be equal to up thrust plus drag.

[00:29:49]And from here you find out drag separately and you just compare if the stoke's law force is equal to drag force. If they are then and only then you say stoke law is applied. And the other most common question is what are the conditions for stoke's law? So the conditions are the speed must be low fluid flow around it must be laminar and it must be small.

[00:30:19]And also remember in unit one every time you come across a question associated with temperature you directly link it with viscosity because temperature is always associated with viscosity.

[00:30:37]Now let's move on to uh solid material section.

[00:30:41]um in the most recent paper that is Jan 26 you had to find out the combined spring constant so this is a very good time to just review the whole concept so let's say if you have just one spring you of spring constant K and if you add another identical spring with it in series what happens is the spring constant halves and conversely if you add another spring with the one in parallel. The spring constant is doubled.

[00:31:18]And I can refer to you um the question from October 2020.

[00:31:26]Question number question 16 C. Actually this is the question. So the springs were identical to the spring used in the original investigation.

[00:31:37]Explain why the spring constant for this arrangement of the springs would half.

[00:31:41]Like I just said, if you connect them in series, the spring constant halves. The reason is same force acts through both springs.

[00:31:51]Each spring has the same extension but the extension of the combination is doubled.

[00:32:01]The force remains the same. So f= kx f remains the same but since you are connecting both in series and each spring extends by the same amount the extension of the combination is doubled.

[00:32:18]Guess what? K will be halfed. The exact reverse thing happens when you connect them in parallel.

[00:32:29]Also remember how energy changes. Okay, this can be a MCQ question or in a section B question as well. Okay, how the energy stored in the spring changes when you connect them in series and when you connect them in parallel compared to when you connect them in parallel. Let's move to the next one. Okay, so this used to be a very common question. It's been a while they haven't asked this. So I just had a gut feeling this might appear in the exam.

[00:32:59]If it doesn't, no worries. I'll show things around this that will be definitely beneficial for your exam tomorrow. So you have a force extension weight extension as in force extension graph. How do you find out the yan modulus from the force extension graph?

[00:33:18]Right? So if you recall the equation for yan modulus so this breakdown will be really really really important this out.

[00:33:25]So young modulus is basically stress by strain.

[00:33:30]Do one more breakdown. Stress is force per unit area divided by strain is extension by original length. So young modulus equals F by A time this will get reversed. L by d x. Now you do a flip as in position swap.

[00:33:53]Bring d x over here and a over here. So yan modulus equals f by d x * l by a.

[00:34:07]Guess what? Force by extension is basically the gradient of the force extension graph. So what you do is find out the gradient of this graph and multiply the gradient with L by A as in length divided by cross-sectional area and you'll get the young modulus.

[00:34:32]So this is a variant and a simpler way to directly find out the young modulus is from a stress strain graph.

[00:34:41]But there is an exception. If your stress strain graph is curved, you have to find out the gradient from the straight section initial straight section from the origin. So you have to draw a tangent over here and keep on moving this as long as it is aligned with your graph like this. Yeah. And then you find out the gradient of this line.

[00:35:12]The gradient of the stress strain graph directly gives you the yang modulus. All right. So now that we have covered what does the gradient of the graph give us. Let's first find out let's now find out what does the area under the stress rate graph give us. So let's just assume uh up to the proportionality limit. Okay. So the area is uh if it's a straight line it should be the area of a triangle. So area of a triangle is half base into height as in half into stress into strain.

[00:35:52]So area would be half stress is force per unit area. Strain is extension by original length.

[00:36:05]Now I want you all to separate this in your head. What is half FX and what is AL? So if the sample is in the form of a wire which is basically a cylinder. So this is the cross-sectional area and this is the length and area time cross-sectional sorry cross-sectional area time length gives you the volume equals a * l and recall what half fx gives you. So area is half fs gives you the energy stored divided by a l gives you the volume.

[00:36:56]So energy per unit volume is quantity we call energy density. So the area under stress strain graph gives you energy density. And if there is a reverse question where they ask you how can you calculate the energy what you do is calculate the area and multiply it with the volume of the sample.

[00:37:20]One thing I forgot to mention when it comes to springs doesn't have to be parallel or series. There are three uh equations that you have to remember for the energy stored or the work done in stretching a spring. One is e = to/ fx the one we just saw. The other one is a = half k x² and the third one is a = f² by k. So this one is not very common but I have seen several questions uh that you have to use only this one or half f² by k my bad/ f² by k to calculate the energy stored.

[00:38:07]And now just some final suggestions from me. Uh do not not reply rely only on the predictions. Uh use this video as a stepping stone. Use this video as a starting point. Make sure this is not what you end your revision with. You just start your revision with this and you just go all in. Review the fact sheet. Very important. I've given the link in the description. Memorize the definition sheet. I can promise you you'll get some definitions from the definition sheet. And my personal request to all of you is do not start from section A. Always start with section B because there is a time constraint. If you start with section A, you might lose a lot of time solving unimportant questions and if you even if you get all of these right, you only end up with 10 marks. Whereas section B contains most of the marks.

[00:39:00]Do not forget the units in your answers and give units for the interim calculations.

[00:39:09]Maintain the SFU as a significant figure. Um, never ever give your answers to less than two significant figures.

[00:39:18]And if it's a show that question, the rule is give your answer to at least one significant figure higher than the show that value.

[00:39:25]Review the full syllabus revision video that will be really helpful. And review the full paper solutions. I have some full paper. For example, one of the most recent ones would be May 25. Okay, that's available on my channel. just see how I encountered and approach every single question in a whole syllabus, whole uh paper solution video and solve some papers yourself. Just looking at the videos is not going to cut it. I repeat, please make sure you solve some papers, right? This is really, really, really important. And good luck with your exam tomorrow. Let me know how the exam goes and if this video helps. I'll be waiting for you in the comment section. Thanks for watching.