Paper 1 Derivations A Level Physics

Added: 2026-05-21

[00:00:00]Let's go over some paper one derivations for Alevel physics. Just a little note that this is not an exhaustive list and lots of different derivations may appear at the exams as show questions even if they're not listed in the specification.

[00:00:14]We're going to start off with the kinetic energy derivation. Now, imagine that I am accelerating an object on the constant acceleration. And let's say that we're going from position one to position two. And we're going to just apply a force over a distance d. There will be some work done on the object which will be equal to the change of kinetic energy. The work done will be given by the force multiplied by the distance which is going to be also equal to delta E. Now the force will just be mass time acceleration time d which is given by delta E. And what about this distance D? Well, I can probably use the suvat uh equations and I'll be able to use perhaps let's say that we start off at a speed 0 and we arrive here at a speed v. I could probably use v ^ 2 is = u ^2 + 2 a s. Now my distance or displacement s will now just be d. So I'm just going to change the symbol. And my initial speed was zero. So I can remove this term. So I can just rearrange for the displacement d which will just be v ^ 2 / 2 a. And now I can just plug this into here. So what do I get? I get that m a * v ^ 2 / 2 a is equal to my change of energy. But look at this. The acceleration cancels out and I'm left with that my change of energy is just equal to mv^2 / 2. And this is actually the acquired kinetic energy of movement. The kinetic energy.

[00:01:54]Okay, we've got the kinetic energy derived. Let's also derive the gravitational potential energy. If we have an object which is moving between a couple of positions because there's a gravitational force or the weight that's acting on it, the work done is just going to be given by the force multiplied by the distance traveled. And let's say that uh let's say that we need to travel some sort of a distance d or shall we just call it a height. Well, we can immediately see how the force is just mg and the distance traveled is just h. In other words, our work done is equal to mgh, which is once again the change of energy if we've if we have fallen or if we've moved that distance in a gravitational field. And this means that GPE is just delta E which is just mgh. Moving on to power. We also need to be able to derive this equation that power is force time velocity. Well, if we have work done which is force multiplied by distance and power in general is the rate of work done. So we can say that this here will be force * distance / t d / t. This here is just the speed. So therefore, the power is force time velocity. Please note that if the force is at an angle, we need the components in that direction. So if that's the case, the power just becomes F v cosine of theta. We also need to be able to show that F is equal to MA is just a special case of the net force being the rate of change of momentum. So the change of momentum is just delta brackets mv over delta t. But now if the mass is constant there will not be any change meaning that we can take it outside of the delta. So what we can write is that f is m delta v over delta t. And this over here is now the acceleration.

[00:04:02]Meaning that f is just m a.

[00:04:07]Moving on to thermal physics, we need to know the derivation of a half m c^² or the rms speed and that being equal to 3 kt. The easiest way is to start off with pv is equal to n kt. And also another equation which we given in the formula booklet for OCR which is PV is equal to 1 over3 N M C².

[00:04:34]Okay, we can set those two equal to one another. So we can get N KT is 1 over 3 N M C^ squ. Notice how the ends cancel out. And what we can do is move the three over. So we get 3 KT is equal to M C^² and then we just divide by 2. So we get 3 KT is a half MC^ 2. The important thing to note is that this here is the kinetic energy of an individual molecule. Well the average or mean kinetic energy. Okay. Next one. T^ squ is proportional to R cubed in gravity.

[00:05:14]Well, the way we derived this is by looking at the net gravitational force which we can set that to be the centripedal force. So, gm m over r² is equal to m omega^ 2 multiply this by r. Okay, so let's cancel out these. And then what we're going to say is that g m is equal to omega^ 2 r cubed. Now omega is 2<unk>i / t. So let's bring this into here. So g m will be 2<unk>i / t^ 2 r cubed.

[00:06:03]G M will be 4<unk>i^² over t^² r cubed.

[00:06:12]That means that t² will be 4<unk>² over gm r cubed.

[00:06:22]Okay. Escape velocity. So the total energy needs to be equal to zero. So the total energy is the kinetic energy.

[00:06:32]Let's call it E K + GP has to be equal to zero. In other words, a half MV² minus G M / R is= 0. This cancels out. A half V ^2 is G M / R. V is equal to square root of 2 GM / R. Okay, let's do the speed of a satellite. And there's a couple of different equations you can go. The gravitational force is GM m / r 2. Set that equal to mv^ 2 / r for a circular orbit.

[00:07:15]And then we get that v ^2 is gm / r. In other words, v is equal to the square root of gm / r. If you given just the radius and the time period, we can also say that V is distance over time.

[00:07:34]The distance is just the circumference.

[00:07:36]So that's going to be 2 pi r / the time period. And those two are equivalent depending on what you're given. The total energy of a satellite is equal to half of the potential energy. Well, how do we know this? because a half mv² plus well minus minus g m over r. This here is just the potential energy which is negative and if we add them up we get the total potential energy v it was we've just derived that to be square root of gm / r. So that means that v² will be given by gm / r.

[00:08:20]We can just plug this into here. So we get a half m gm / r takeway gm / r is equal to total. Okay. So this is minus gmm / r take away half of it. So we're going to be left with minus gm over 2r is equal to e total. And this over here is half the potential energy which is equal to the total energy. Show that the age of the universe is given by 1 / Hubble's constant. We're going to start off with the fact that speed is equal to Hubble's constant multiplied by distance. But additionally speed is also distance over the amount of time. Then we just need to equate the two and then we get that h multiplied by d is just given by distance over time. Now we can cancel out the distance and what we get is that hub's constant is 1 / time i.e 1 over the age of the universe. Up next we have the parse derivation. So this will occur when we have an angle of one arcsec.

[00:09:42]Now the one arcsec is actually 1 over 3,600th of a degree. And in this earth sun and then nearby star right hand triangle we can use the tangent. So the tangent of this angle here which is the parallax is going to be the opposite which is 1 au astronomical unit divided by the distance d in par divided by the distance d which will just be given by one par. And we can see what that value of one parseek is in meters because the tangent of P is just one au divided by 1 parse.

[00:10:30]So one parseek is 1 AU / the tangent of P. So 1 au is just 1.5 * 10 11 / by the tangent of 1 over 3,600°.

[00:10:55]Uh now this here should give us around 3.09 * 10 ^ of 16 m which is roughly the value of a pass. I think it's given as 3.1 * 10 ^ of 16 in your formula booklet. So let's pick a point on the screen that will end up being a maximum.

[00:11:17]So let's just pick this point. The path difference or the path from one of these slits, let's go from the middle of the slit is going to be approximately this one over here. On the other hand, the path to the other slit will be just a little bit bigger. Let's just draw that over here. We can find the path difference, which is simply going to be just this distance over here, which will end up having a 90° triangle like so. So the key bit about this proof is that our distance to the screen D has to be significantly bigger than the slit separation S. I've only drawn this relatively big simply so that we can visualize this a little bit better.

[00:12:18]Let's just say that this here is the central maximum and this here is let's say the first maxima that we can see after the central maxima. Okay. So the power of difference in this case should be exactly equal to the wavelength and the two rays in a way are going to differ by this distance over here. Now let's pick one of the angles. So I'm going to call this angle theta. I can say that the sign of the angle sine of theta which will be equal to the opposite divided by the hypotenuse. Now my opposite in this case is simply the wavelength which is the path difference for the first fringe divide that by the hypotenuse which in this case is just the slit spacing because this triangle is actually really stretched out in the limit that d significantly bigger than the slit separation. Then we can say that sine of theta is approximately equal to theta. So therefore we can say that theta is approximately equal to lambda / s. Now what we want to do as well is have a look at the two points across here. So this here was the central maximum. Let's call that 0.1 and 2. and the distance between them is actually W. Let's so let's just be a little bit more precise. The we can actually invoke them into another triangle because this distance D here will have a 90° to um to the screen and meaning that those two triangles will actually be just similar. So if I was to make this triangle like this, I mean this one right over here. Well, because of that similarity, this angle here is also theta. So by looking at this second triangle we can say that tan of theta which is approximately equal to theta is then going to be equal to the opposite / the adjacent which is w / d.

[00:14:55]So this means that lambda / s is equal to w / d in the limit that the distance d significantly larger than the slit separation. Just rearranging for w we get that w is going to be equal to lambda d / s. And let's go over the derivation of d sin theta is equal to n lambda. So imagine I have a diffraction grating and this here are just two slits and the distance between them is just d.

[00:15:30]If p is the first order defraction uh maxima in other words n is equal to one and right over here we're going to have the central maximum n is equal to zero. So the first step in this derivation is to consider the path difference. So those two paths are going to differ by this distance here. If the angle to the first fringe is theta, the yellow triangle right over here is typically assumed to be similar to this one over here, which I believe is actually a tiny approximation, but this is typically true because the slit separation is significantly smaller than the distance to the actual screen. So we can say that the blue triangle essentially is similar to this yellow uh triangle. And if those triangles are similar then this angle here will just be theta. And now we have this triangle in which we're going to have 90° here d and then theta. Well because s of theta is equal to the opposite / the hypotenuse. And our opposite in this case is actually just the path difference. So in this case sine of theta will just be equal to the path difference over the hypotenuse which is just d meaning that our path difference will just be equal to d multiplied by sin theta. But because the waves arrive in phase at P for constructive interference, the path difference is just going to be equal to N lambda. This is just the condition for constructive interference. And there we have it that n lambda is equal to d sin of theta. We also need to be able to derive that kinetic energy is P ^2 over 2M where P is the momentum. We're going to start off with our regular formula for kinetic energy which is just a half m v^2. Then I'm going to multiply the top and the bottom by m.

[00:17:54]And look at that. What we're left with is a half m 2 v^ 2 over m. m^2 v^² is just momentum squared. So this here is just going to be P ^2 over 2M.

[00:18:08]Let's derive the centripedal acceleration equation. Just a little note that there's multiple different ways of deriving it. And I'm just going to do one of them, but if you use a different way such as similar triangles or something else, this may be totally valid. So let's assume that the object is rotating say in this direction. The velocity will be tangential at both of those positions. So that means that right over here the velocity will be this way. The velocity will have the same magnitude but a different direction in the previous position. Something like this.

[00:18:44]I'm going to call this vector v1. Then I'm going to call this vector v2. Their magnitudes will be the same but their directions will be different. So I'm going to call that magnitude just to be equal to v which will be important later on. Now it's time to construct our vector triangle. Well, what we want to do is find our change of velocity, which is delta V. This is typically defined as V2 takeway V1. So my V2 is horizontal in this direction.

[00:19:16]Now V1 will look just like V_sub_1, but it will have the opposite direction. And I'm going to arrange those vectors tip to tail. This one here is negative v1 and this one here is v2. The vector just over here, let's call that delta v.

[00:19:38]Notice how it's pointing directly towards the center of rotation as we expect. Now, because they're magnitudes are the same, I am just going to call these round v2 and minus v1. I'm just going to say that they have a magnitude of v. Okay. So what are we actually looking for in this problem? We're aiming to find the acceleration which is given by delta v over delta t. The best way to tackle this is to use the small angle approximation. Let's call this angle theta. The angle swept the angle which has been swept in a time t. Well, we can say that sine of theta is going to be given by delta v over v. It doesn't really matter whether we've chosen v_sub1 or v2 because they're going to be equal. And by the small angle approximation sine of theta is about equal to theta.

[00:20:34]So therefore we can say that theta is going to be given by delta v over v. Of course we can make delta v the subject.

[00:20:44]So I can say that delta v will be theta * v. Wait a minute. But a was given by delta v over deltat t.

[00:20:58]In other words, this here will be given by theta * v / delta t. We're almost there. The final thing I'm going to consider is that v is given by omega r, which is going to be essentially given by delta theta. In this case, we just called it theta / delta t * r. So if we want to eliminate theta from this equation, we can just make that the subject and say that theta be given by v deltat t / r. Let's plug this back into this equation. So this is going to be v deltat t * v over r deltat t. Notice how the delta t's cancel and what we're left with is that a is given by v ^ 2 / r. Let's start off with resistance in series and parallel for OCR paper 2. So let's say that we have a total voltage V2 Vub1 and V2 resistance R1 R2 and total resistance in the circuit R total because V total will be given by V1 + V2. This is Kirkov's second law. The sum of the emfs is equal to the sum of the PDS in a closed loop. Then if we had a current I in the circuit. I total will be I R1 + I R2. These here will just cancel out leaving us R2 is R total is R1 + R2. On the other hand, if we had a parallel circuit and let's say that the current here split into I1 and I2 and let's say the PD was V, meaning that there's no internal resistance in the circuit. So, it's V across all the branches. And also, let's call this one R1 and R2. We can say that I total will be I1 + I2. But I total is just V / R.

[00:23:01]So v / r total will be given by v / r1 + v / r2. These here cancel out. 1 / r total is 1 / r1 + 1 / r2. We can also do a couple of easier ones. So since electrical power is vi and power is electrical energy divided by time work done or electrical energy is given by power multiplied by time which is going to be given by vi.

[00:23:35]Additionally, since power is vi and because v is equal to i, we can say that power is given by i 2 * r. If we substitute the i. So i being v / r into the power equation, we can also get the other equation that p is v * v / r. In other words, v ^2 / r. Okay, folks. I hope this video has been very, very useful to you. If you're revising for paper one, you also really need to check out this video right over here. That will help you get the best exam result.