Thanks for your very kind comments. My videos are intended to cover the ground for AS and A2 level of the Edexcel, AQA and OCR syllabus. Alas, I wasn't able to distinguish between the material needed for AS and that needed for A2. However, all the AS material ought to be there.
I realise the formulae look pretty horrid. But as I tried to explain in the video, in an exam you would usually be given information about several of the unknowns and you would be asked to calculate the remaining unknown.
soh cah toa is the funniest thing ive seen since the right hand rule, when i lifted my eyes off a test in magnetism and saw the whole class starring into their hands and making different shapes with it. once again, great video, thank you very much!
II am not socially aware of the syllabus for me to give you any meaningful advice. I try to cover material for a range of different exam boards. I hope that my videos would be helpful in final revision and I wish you all the best in your exams.
ive been with khanacademy for awhile now but you make it even easier. you dont go off topic and you dont leave us wondering why something works the way it does. and you dont spend too much time explaining something. thanks please continue videos for as long as possible you help a lot of people
i agree. although i respect khan academy, i'm never able to learn from his videos. it seems like he stumbles and stutters a lot, and i just have a hard time looking at his harsh graphical aids.
I never did physics at A level because i wasn't interested enough at the time, but here I am learning it now, so thank you very much for making it more compelling and understandable.
The basic answer is that you must have the enough energy in the form of kinetic energy in a collision between two particles to account for at least the rest of mass energy of the created particles. Any surplus kinetic energy of the incoming particles will contribute to the kinetic energy of the creative particles. Much will depend on whether the incoming particles annihilate or whether there is simply additional particle creation.
Thank you very much! You're the best physics course videos out there, you are clear with your explanations, don't go off tangent and actually tell us _how_ something works so we can have a chance at understanding the logic behind something, instead of just "here's the name for this new concept, now use it". Would gladly donate for your work!
The point is that two particles travelling in opposite directions can, in something like the large hadron collider, each be travelling at a speed very close to c. This makes the energy of the collision very high and much higher than can be obtained by moving particles hitting a stationary target.
Indeed. I am no snooker player but I understand that if the cue ball is aimed other than at the centre of the other ball (ie slightly to the side) then the effect I have drawn would arise.
Well in practice very few elastic collisions take place because this suggests that kinetic energy is completely conserved, whereas any kind of collision is likely to give rise to sound, heat, light energy. But you can have elastic collisions at particle physics level.
Thanks a lot, this helped me a freaking lot. i have my exams tom and this I was confused with but now i understand this ! thanx for posting this , u saved my ass
i just can't without commenting on this blessed video u see i 'm watching this video in 2020 (i mean after sooo many yrs) and still it's effective i srsly can't understand a shit when our physics sir teaches and this video............just saved my ASS!!!!!!!
I recommend taking a look into Feynmann diagrams and the equation 'E=MC2' which covers the behaviour of young sub-atomic particles such as when particles decay more partcles arrive and how energy and particle matter can interact relative too motion (Usually C)
19:11 If I saw those two formulas in GCSE, I would have straight up gotten a heart attack. It's mad to think how much your knowledge can improve in less than a year.
Hello! Great video, but I wanted to ask when you're calculating the momentum of the y-dimension, isn't the negative direction already inside the velocity, rather than out of it? I thought that the formula could be always written with pluses, and you took care of the minuses when calculating the velocity. Thank you so much for sharing your wealth of knowledge!
Correct, you can write it with pluses if you are consistent with your angles. In the video alpha (~40°) and beta (~30°) are measured to the same axis but in different directions. This error is undone by later manually adding or subtracting by hand. If you measure all angles in the same direction (e.g. alpha=40° and beta=-30° or 330°), you convey the direction-information to the formula where sin(beta) now has the same magnitude but negative sign. Hence you can just add all terms, no manual decisions necessary.
parv bakshi Never seen such a good teacher. He teached in kings college in the past so no wonder for his competence level and such an entousiasm !! let s make a collect for him . He deserves that largely !!
Intuitively, in the first example, why there's an angle between the two particles while the first particle travels straight to the second particle that is static. Isn't the consequence of the collision is also straight? Or it's an 'assumption' to have this consequence that there's an angle between the particles?
give this man a nobel peace prize for his contribution to A level physics !!! . by the way sir Can you show us how to solve the unknowns using simulataneous equations
Hello Doc. Another efficient video on a classical subject. So thanks again ! There is anyway something to be precised. In your first example (one snnoker ball at velocity v hitting another ball at rest), if the hitting ball is hitting the rest ball just in its center, the two balls will go in the same direction (alpha=beta=0). To get the situation you described in the scheme, the hitting ball should hit the second ball at the different point than the center of its section. Do you agree ?
@12:07 Initially, there is energy only in x component (y=0 and x=mv1), AND after collision, the sum of both x components ARE EQUAL to total initial energy; wouldn’t that leave no energy left for y-components..? As a snooker fan, I would strongly presume, that Ronnie would scorn these kind of developements, with a long session of nail biting, and eventually skip the whole tournament.
Ello mates, my name is Dr. Physics, and I like to sip tea with my mum while eating British crumpets.(insert accent while reading). Just kidding Dr. I love your videos, thanks for the big help.
I love you sir. I'm going to do my AS level exams this May/June and I'm currently studying the A level portions in my school, and this has helped me a lot. However, do you have any videos that are specifically meant for the AS level curriculum?
Concerning the two balls colliding along the x axis( Horizontally) I would have thought that the outcome would be horizontal: Or in snooker jargon the fist ball would "follow through"so that both would end up in the pocket.
Sir, You are amazing. Anyways I am retaking my AS level physics exam in a week. Paper 2 is in 8th November. I am retaking as I got this horrendous 53% grade this May. For paper 1and2 I am thinking that i would go through your videos of the chapters included in AS and your videos of exam questions and then solve recent papers 2010-2012 which includes 18 papers for practice. Is this enough to aim for an 'A'? Or do i need further knowledge. If so sir, then please advise me what i should do next.
sir,thank u very much for these videos,it helped a lot,but i wanna say that plz add videos for experimental analysis of A-level physics.i m quite weak in experimental co-ordinations.
Why not use the convention of u= initial velocity? That's the only part of the video that confused me in this case, I found the rest extremely helpful for revising for my test.
100 000 views and only 800 likes i think that this man didnt put this videos because he wanted to have a lot of likes and be famous but if he is glad to spend his free time to make this VERY helpful videos we should give him likes and help his chanell grow and maybe he can even take some money from yt. Like its the less think we can do because we are able to look VERY helpful videos for free.
Hi, this may be a silly question but I am confused as to why the angles for the direction of the "ball" with respect to the x-axis and then with the same ball with respect to the y-axis are the same. For example, in the simplified example at 9:13, why is x=V2cosA and y=V2cosA. Why is it not x=V2cosA and y=V2sin(90-A) as (90-A) would be the angle from your drawn line to the y-axis. Hope this makes sense.
My question is that when Considering particle collisions how does the Energy Needed to create new particles relate to the KE, for example a particle Aimed at a target, and a particle Colliding with with another particle moving in the opposite direction.
When in the y-direction both balls have ZERO momentum before the collision (as in the video), how can they have an angle bigger than ZERO after the collision (as in the video too!)???
Sjaak van Dijk I was trying to represent the situation in, say, snooker where the cue ball is hit with the cue such that it hits the other ball slightly off centre. In those circumstances the two balls will move off in different directions.
So. Four post-impact unknowns and (for elastic collisions) three equations. How would we predict the complete post-impact calculation? That is, to calculate all four unknowns.
I think I have just figured out my own question. Is it because of the change from cos to sine? If you were to use the angle between the direction of the ball with respect to the y-axis would the equations be x=V2cosA and y=V2cos(90-A)? Which would be slightly more work and therefore pointless? So it is easier to work out y=V2sinA over y=V2cos(90-A). Am I making any sense at all?
i notice that you dont have a video for SI base unit and vectors. i know its a really basic part of physics but i still wish there's a video on it. physics have never been my strong point so every little bit counts i guess?
I just had one question, light is simply described as an energy packet of photons. We already know that light has momentum. However, momentum is the product of mass & velocity. But it is said that photons have no mass, rather simply, light particles have NO mass. Then how can it have momentum? :/
This is related to Einstein's famous equation E = mc² In special relativity energy is also E² = (pc)² + (m₀c²)² Since the rest mass m₀ of the photon is zero this reduces to E = pc Putting the two together: E = mc² = (mc)c = pc, one can see that p = mc which is a product of mass and velocity. But this is now a "relativistic mass" for the photon multiplied with its light speed c. Energy is related to mass by E = mc², so the energy of the photon gives it a relativistic mass equivalent and hence also a momentum. In quantum mechanics momentum is Planck's constant h divided by wavelength λ, thus p = h/λ From before: E = pc => p = E/c, so E/c = h/λ The energy of the photon is then related to its wavelength by E = hc/λ The relativistic mass equivalent for the photon can then be calculated with its energy or wavelength or frequency: m = E/c² = hf/c² = h/λc
physics.stackexchange.com/questions/2229/if-photons-have-no-mass-how-can-they-have-momentum if you are interested, have a look at compton effect wich is photon hiting electron but be familiar with special relativity before ;)
The thing is that particles that light are massless but their energy "weights" so you know that the Energy of a foton is given by hf, where h is the planck costant and f is the foton's frequency, so you take E = mc^2 and arrenge the formula so it gives you m = E / c^2 = hf/c^2, but we know by waves that c = λf so f = c/λ so you go back to our mass formula and you got: m = hf / c^2 = (hc/λ)/c^2 = h/cλ and since h is a really small number and c is a big one, the mass of a foton is very small so m ≈ 0 and if so p ≈ 0
I have a question, I am an aspiring physicist and I wish to learn everything I can about all branches of physics and I want to know, in what is the proper order to learn in your video playlists, from start to finish? Like which ones should I begin with and which ones should I conclude? I imagine I should start with classical mechanics and the final should be nuclear physics but what about between that? Thank you professor.
+lrobertiii My videos are organised into playlists and in each playlists the videos are in the order I recommend for watching. As to the order of playlists, I would start with GCSE, then A Level, then any order you like for the uni level material except I would watch the Quantum Mechanics playlist before you watch the particle physics playlist.
At 4:47 when he writes the incorrect equation, is that just an improper way to write the equation, or even if you plug the numbers in it wouldn’t work?
Where does the formula for momentum p=mv come from? Is there any derivation or logic to prove this relation? And what is the physical meaning of momentum?
so,if we given that the collision is elastic ,do we take the velocities as they are or do we have to resolve their components before we use kinetic energy formula ?
