Oxford University Mathematician takes Oxford PHYSICS Admissions Test (PAT) - with
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- เผยแพร่เมื่อ 9 ก.ย. 2024
- University of Oxford Mathematician Dr Tom Crawford is challenged by Zhelyo from @zhelyo_physics to try some questions from the Oxford Physics Admissions Test (PAT). The exam is taken by students applying for undergraduate Physics at the University of Oxford.
Part 2 where Zhelyo tries some questions from the Oxford Maths Admissions Test (MAT) set by Tom is here: • Oxford University Math...
The questions featured in the video are taken from the 2018 Oxford Physics Admissions Test. You can download the exam paper for yourself here: tomrocksmaths....
The questions covered in the video are as follows:
Q1-10: multiple choice
Q18: gravitational force
Produced by Dr Tom Crawford at the University of Oxford. Tom is Public Engagement Lead at the Oxford University Department of Continuing Education: www.conted.ox....
For more maths content check out Tom's website tomrocksmaths....
You can also follow Tom on Facebook, Twitter and Instagram @tomrocksmaths.
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With thanks to
Zhelyo @zhelyo_physics
St Edmund Hall
University of Oxford Department for Continuing Education
Freesound: Rich Craft Chime
Freesound: Suicdxsaturday
Watch part 2 where Zhelyo tries some questions from the Oxford Maths Admissions Test (MAT) here: th-cam.com/video/Q6SCglxjRFQ/w-d-xo.html
22:09 I think the answer is A. This is my method.
I need to first define some terms
1. A point that is held fixed with zero displacement will be called a Node and will be denoted as N.
2. The point of maximum displacement (i.e the peak of the wave) will be called an Anti-Node denoted as A.
Assuming a sine wave Nodes are formed where the wave is zero. Ie 0, π, 2π,.... antinodes are formed at π/2, 3π/2,....
Hence, it follows that after a Node, there must be an Anti-Node and after an Anti-Node there must be a Node, and hence the successions are N,A,N,A,N,.... Where we could have started with Antinodes first and then Node and then Anti-Node and so on i.e A,N,A,N,....
Consider, the sine wave in [0,2π], 0 is a Node and so is 2π, therefore a wavelength is the distance between 0 and 2π, hence half a wavelength will be the distance between 0 and π, i.e between two successive Nodes. Let's denote half a wavelength as (1/2)λ.
With all this our definition now let's tackle the problem.
Since, the ends of the string are fixed with zero displacement, this implies that the ends must be nodes N. Since the center of the string is also fixed, it must also be a Node N, this the possibilities we can have is
N (for first end), N(for middle) and N(for second end)
But since we must find at least an Anti-Node between successive Node, the only possibility are: (note I wrote the nodes at the middle and center in parentheses so that it would be clear to understand)
1) (N)A(N)A(N)
2) (N)ANA(N)ANA(N)
3) (N)ANANA(N)ANANA(N)
4) (N)ANANANA(N)ANANANA(N)
And so on (notice the pattern).
Since the length of two successive Nodes separated by an Anti-Node is (1/2)λ
The lengths of each pattern is therefore
1) 2•(1/2)λ
2) 4•(1/2)λ
3) 6•(1/2)λ
4) 8•(1/2)λ
And so on...
This lengths must be equal to the length of the string L since they are formed by the string, thus
L=2k(1/2)λ
Setting 2k=m
L=m(1/2)λ
Therefore λ=2L/m option A.
By this we see that m is restricted to be an even positive integer. For this to hold true.
Now I realize I didn't quite look at option B correctly. You are right....
We just divide 2 by 2 and we get
L=kλ
λ=L/k
In the case of the question instead of k we use m so option b is correct. You are right.
You said at the end that you hadn't stopped smiling. I honestly feel like you are always smiling in your videos! It's really cool to see someone so excited and in love with the subject!
FIRST TIME LOOKING AT A PERSON WHO STRUGGLES IN PHYSICS BUT WHO IS ACTUALLY AN MATHEMATICIAN
Why is bro screaming?!
They don't do real world problems very often they play with numbers and abstract things
Hey! I am 50 years out of college but I love the content. You Gentlemen are enthusiastic and you have fun. I love math and science. Thank you for making my day. When you love what you do, you never work a day in your life. Hope to see a lot more. ENJOY😊
Check out part 2 tomorrow on this link: th-cam.com/video/Q6SCglxjRFQ/w-d-xo.htmlsi=-hWmW1UCNXoLYXZt. : )
Looking forward to seeing the edit on this! So much fun to film!
Hidden in the solution of the gravitational hanging spheres problem is the approximation tan(theta) = sin(theta). This is acceptable as the angle will be small.
Well that is not acceptable for a mathematician😅
I immediately thought sin(θ) = θ for small θ. Its used in deriving the equation for simple harmonic motion.
@@blackhole3407 It very much is for a dynamicist or analyst. But you'd have to be very careful about the way you go about it. The reason you can do that is because you're expanding sin 𝜽 and tan 𝜽 in a Taylor series. It's a very standard thing to do to expand to first order, i.e linearize. Linearizing sin 𝜽 and tan 𝜽 gives you sin 𝜽 ≈ tan 𝜽 ≈ 𝜽.
But in this question, we aren't expanding to first order, but to second order. This is because we are keeping terms on the order of 𝛅x^2, which means that our expansion needs to be to second order throughout. Therefore we should expand sin 𝜽 and tan 𝜽 to powers of 𝜽^2. A physicist who's not a theorist (they deal with perturbations all day) would very likely gloss over this detail and expand one side in 𝛅x to the first order, and the other side of the equation in 𝛅x to second order, which is wrong. HOWEVER in this particular case they luck out, because sin 𝜽 = tan 𝜽 to the second order as well, since both are odd functions. A mathematician or a theorist would note this detail, and still get the same answer as the naive (and conceptually incorrect) first order expansion.
There's definitely a very big difference in philosophy between physics and math profs for making test questions, probably comes from the differences in philosophy and mindset of physicists and mathematicians in general
Interesting, what do you think the difference is? I think a lot of physics questions in entrance exams and olympiads tend to be inspired from famous experiments, current events or just a random phenomenon an examiner saw and thought was amazing! : )
You do have to consider the tension from the rope because there is no "horizontal component of the gravity". The horizontal component that is balanced with the gravitational force between the two spheres is the horizontal component of the tension. The vertical component of the tension is equal to mg, so the horizontal component is mg tan θ. Here we then use the assumption that δx
So you mean that at 44:45 the equation ("F horizontal= sin(..)..." ist wrong cause he is mixing up the horizontal portion with the "tangential" portion ? (Which is approx. the same after δx
Great work Zhelyo (and Tom)! Looking forward to doing some of these with our students when we get back to work!! Have a good summer!
Thanks Andy! Some excellent questions throughout the papers, have a great summer!
You can think of force as equivalent with the gradient of a potential. If the gradient is zero, there is no force, if the gradient is not zero there is a force. When the ball is hanging at an angle, it is hanging straight down the gradient of the combined potential. Now we know that the ball will not hang around in any other location, and so it must be the only place where the potential and mechancal force on the rope are parallel, equal and opposite.
Zphysics carried me through Physics 🙏
fantastic to hear, thanks for the comment!
Hey Tom. I am from South Africa and a huge fan. Would love to see you sit a South African high school exam.
I agree, it would be pretty interesting. Hopefully Tom agrees to it.
Sum the potentials and take the gradient, which position along the balls freedom in position does the ball hang parallel to the gradient of the potential.
For 6 I was really confused why both of you didn't chose A until I realized I had read the question wrong and the string was fixed in the middle. For 7 I used calculus to derived one of the kinematic equations and used that because I couldn't remember it but the energy approach is much simpler.
2 GOATs one video
For question 4, I would have just used the mean of the binomial distribution np. If p is the probability of turning left, the np = 3/3 = 1. So the mean number of turns left is 1, which means 2 turns right. Or you can say p is the probability of turning right, so 6/3 = 2, 2 turns right 1 left as well. And that is C.
I guessed that it was C with other reason
By the Definition, two thirds will Go right and one third left, that means the most likely combination is going twice right and once left (Sorry for my english)
came at a perfect time! i was actually planning on checking out some of the pat questions to get used to them and see if my level is worth applying :)
definitely worth it and you will also learn a great deal of amazing physics along the way! I have a long How I'd prepare for the PAT Video on my channel you might find useful if you do decide to apply. Good luck!
Great video!
I don't think question 2 5:26 is correct. The trajectory 2 is a valid trajectory. But using the scale values, we can see two other trajectories are ruled out:
The ellipse, 1, should have it's closest distance along its major axis at approximately (-3, -3.5), which is at a distance of about 4.61 from the star. But there are two closer points at about (-2, 3) and (2, -3), each about 3.6 from the star.
The parabola, 4, should have its closest point to the star along its axis of symmetry at (0, -2) at a distance of 2. But two points at about (-1.1, -1.1) and (1.1, -1.1) are each only about 1.55 from the focus.
It's hard to tell with the hyperbola 5, but it looks like its closest point is on its axis of symmetry and is therefore a valid trajectory.
But, using the provided scales, trajectories 1, 3, and 4 are all invalid trajectories. The correct answer isn't an option.
39:00 literally the exact same reaction I get from my maths friend when I tell him something like this 😂
That's a pretty complicated set of units you put on screen for voltage but I can see that the units work out. I just think of it as energy per unit charge or Joules per coulomb. If I forget that, which I did for the third question, I just recall the integral definition of voltage and the units of N/C for an electric field to work it out in my head. If you forget the units of electric field but remember Coulomb's law and F=qE (Lorentz force law without magnetic field) then that's also easy to quickly work out in your head. The relationship between electric field and voltage is similar to the relationship between force and potential energy (or the amount of work required to travel somewhere from a chosen reference point, often infinitely far away for a system with only central forces). The relationship between electric field and force is similar to force and acceleration with charge being the bridge between them, instead of mass.
La combinaison la plus probable est donc d’aller à droite deux fois et une fois à gauche, avec une probabilité de 12/27❤❤❤
Solution :
Pour résoudre cet exercice, nous devons trouver la valeur maximale de ( a ) telle que la somme des quantités de thé ne dépasse pas 3 fois la quantité de la première tasse.
La quantité totale de thé bue est une série géométrique :
S=1+a+a2+a3+…
Donc a< ou égal à 2/3
❤❤❤❤
Je fais la licence 1 en informatique mais je trouve tes exos sympa
i sat the MAT 2024 this paper is quite the challenge 💀
The best duo ever ❤😂
Hi Tom, another great video! Just out of interest, what iPad app do you use for note taking?
I accidentally missed your reply, and so deleted my post because I was not sure of my computations. My apologies. The post was about Question 18. I argued that a more systematic use of the fact that delta x
Cool, I also have degree maths but work with ai assistant system (for analysing any code base (using the ai system to devolop itself)) on company having lots of physics and engineering workers maybe physics is not all out anyway (for me)!
The Einstein graphic tee 😭
for question 7, wouldnt it be better wording to find the minimum force required to stop before hitting the cat? Because answer D is double the force of answer A so it would stop the car even earlier. Or am i missing something?
No you're absolutely right.
The question is badly worded: it just asks what force needs to be applied to the car to not hit the cat. If you apply the force A the car will stop just before hitting the cat. If you apply D the car will reverse direction and accelerate to speed u in that direction. It won't stop, but since it is moving in the opposite direction, it still won't hit the cat.
i might be a genius, i did q 1
The PAT is similar to the MAT but the eigenfunctions are always non degenerate.
could you please solve the maths section of jee advanced paper from india
Letss gooooooooo❤🎉
Yo, can anyone tell me where I can find this vans shirt? That thing is fire
I ordered it on ASOS :)
@@TomRocksMaths thank you :) keep making those wonderful videos!
Reupload? Could have sworn I’ve seen this a few weeks ago
I feel your hayfever pain
To be fair, this is for undergrads, just go gain entry. A mathematician will do just fine. What would be more interesting, is having someone who is an undergrad trying to do this again, or someone trying to get in.
how do you prep for this test? any books?
excellent question! You may find my video on this useful: th-cam.com/video/LXwEv6PKpqg/w-d-xo.htmlsi=bEPQJM8o3i20F9BN , I also have a playlist here: th-cam.com/play/PLSygKZqfTjPDvik2yQTKYOLe7tCSGkOpB.html&si=CnL7e9ngr6aCJbF3 , there is now also a practice test on the Oxford website with a lot of helpful resources. Good luck!
@@zhelyo_physics thank you so much
Podriamos decir que son dos vectores que en si forman la base canonica de las ciencias....
Just from the PAT, looks like it is much easier to get into Oxford Physics undergrad than an IIT or Tsinghua.
The input parameters in physics are taken from an experiment, that is why L is the length of the string.
Where is this amazing Einstein shirt from?
Please attempt jee advanced paper of India. That's really a tough paper specially the mathematics part of the paper considerd the toughest.
The first one definitely was the easiest one.
Why is the level of maths here lower than GCSE 😭😭😭
this exam killed me man
Please try Jee Adv
17:23 don’t judge me
the guy in the black shirt is sooo hot omg whatever u say beautiful
You look like Fundy
Physics is nothing but applied maths in the realm of “physics”.
the line between Applied Maths and Physics has always been very blurry to me. I guess in Applied Maths often the collective behaviour of a large system of particles is considered (e.g. fluid dynamics) but also somehow the same applied to ions (magnetohydrodynamics) is physics? Hm :) It's all fun to think about.
@@zhelyo_physics I meant physics is mathematical modeling of physical phenomena, and yes statistical mechanics is one of the tools.
@@iteerrex8166 It isn't entirely - treating physics just as maths is actually problematic in some circumstances - there are many conceptual ideas that step outside pure maths.
The assumptions you make and the way you build some models - especially particle physics doesn't really work if you just do maths.
@@andrewgilday4549 Of course, whatever the topic and field of study, after a qualitative understanding have taken place, mathematics quantifies it. I hope that is clear, English is not native language.
It is applied math until you are asked to explain some random equation
the questions are basic stuff for a smart indian student. she need not be even upto jee level. a good smart well prepared cbse exam kid would answer these quite easily. having said that i like the fact that oxford is testing the math skills before the student enters. what indian students will perhaps lack is laboratoty access to conduct experiments. in india laboratory skills are almost non existant.
Are you in a over-niceties contest or is it just the new hyper politically correct world? Seriously guys, it's cloying.
Jee aspirants 💀bc itna easy