Can an Oxford University Mathematician solve a High School Physics Exam? (with
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- Опубліковано 19 гру 2023
- Oxford Mathematician Dr Tom Crawford is challenged by Lewis from @PhysicsOnline to try some questions from an A-level Physics exam. Find the accompanying Maple Learn worksheet with some Physics questions to try for yourself here: learn.maplesoft.com/doc/v9ten...
Part 2 where Lewis tries some Maths exam questions set by Tom is here: • Oxford Mathematician C...
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Find out more about Maple Learn on the Maplesoft UA-cam channel: / @maplesoft
The questions featured in the video are taken from the OCR 2020 A Level Physics exam paper entitled “Modelling Physics”. You can download the exam paper for yourself here: tomrocksmaths.files.wordpress...
The Mark Scheme is here: tomrocksmaths.files.wordpress...
The questions covered in the video are as follows:
1:26 - Q16: Force Diagram
20:47 - Q18: Projectile Motion
49:44 - Multiple choice section: Q1, Q2, Q3, Q4, Q5, Q10, Q13
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.ac.uk/
For more maths content check out Tom's website tomrocksmaths.com/
You can also follow Tom on Facebook, Twitter and Instagram @tomrocksmaths.
/ tomrocksmaths
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Get your Tom Rocks Maths merchandise here:
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With thanks to
Physics Online
Lewis Matheson
St Edmund Hall
University of Oxford
Nicoguaro: en.wikipedia.org/wiki/Stress%...
Find the accompanying Maple Learn worksheet with some Physics questions to try for yourself here: learn.maplesoft.com/doc/v9ten9mzua/physics-online-a-level-worksheet
sir I have a maths problem from probability which sort of made and hv no idea what the correct answer is :
Q. A person throws 4 dice simultaneously. Find the probability that the last four digits of that person's Phone Number is the same as the 4 digit number as given by the 4 dice
Great video! It's really funny he used calculus to manually derive one of the most basic and best known physics formula. :-)
He used calculus to derive v=s/t 😂
He only assumed equality of the gravitational mass and inertial mass at the end of his work showing some deep insight.
Real chads don’t memorize basic formulas but derive on the spot
I mean that's how it's derived in the first place! It's much better to derive it than to memorize it.
@@martinshoosterman granted sure but there’s a degree to which you can intuit what speed means in normal language to just know the equation
I traveled back to my school days when our physics teacher always started a topic by demonstrating mathematically every formula we'd use. Physics is about understanding what's happening and thinking how to use math accordingly, so this collab is something i'd totally watch many times in the future. Thank you very much! both of you.
Your teacher sounds like a cool person.
Your channel is such a great source of exercices for the math teacher I am (here in France).
My high school students will love them.
Thank you !
i studied Physics and i love how Tom uses his math knowledge and common sense to get the results. Seeing the train of thought he uses to find the answers is so interesting... thanks for the video and of course i will check the Physics online channel!
So about the arrow question that took a long time to derive... I loved that! It's really a testament to how your background and the glasses you look at a problem through can hugely influence how you're solving it.
When I teach kinematics to my students, we tackle it from a completely different angle. When reasoning in the physics way, it's quite doable to realize that the answer is just a simple proportionality. Tom did all the thinking steps necessary to get there, and then shot right past it.
*Solution*
When shooting something under an angle, the orthogonal components are independant. Vertically, the only force is gravity downwards. Conversely, the downwards acceleration is constant g. Or a = -g
Because the acceleration is constant, the velocity is going to change linearly. So we can just use the formula for average acceleration.
a = -g = Δv / Δt
So Δt = - Δv / g
I like so much this concept, I wish that it will be part 2 in the future.
So Tom, I haven't always been a huge fan of the channel. Like, I appreciate you, I like your appearances on numberphile, etc... I wanted to reach out to say that I've been LOVING these recent videos, where you interview people and apply math to their expertise. This video, your video with another roof. I dunno, I just wanted to say your recent videos have been so good. I've really enjoyed them.
Hey! Amazing math vids btw. Keep up the good content 👍
I love the math way of doing physics by Dr. Tom! :D
I very much enjoyed this!
This is a helpful trick also for students to ask how to find the reference frame where the momentum is 0 in total before a fullt inelastic collision, the transformation from the frame of the question to the frame where the initial momentum is 0 has to involve the correct velocity to answer the question :).
For Q13, if the % errors are small (which they are) a simple expansion shows (1+/-e1)/[(1+/-e2)(1+/-e3)^2] = 1 +/- e1 -/+ e2 -/+ 2*e3 i..e. the maximum error is just the sum of each % error as Lewis says.
I'm really enjoying this. Thanks! By the way, when I was in school we learned the trig identities (SOHCAHTOA) as "Some People Have Curly Brown Hair Turned Permanently Black" where P is perpendicular, H is hypotenuse= and B is Base.
For Q13 you can take logs and use implicit differentiation to get dE/E = dF/F - de/e - 2dd/d. Change minus signs to plus and you have the equation for uncertainty in E. (uncertainty in F plus uncertainty in e plus 2 times uncertainty in d).
Elaborate pls
@@jameswillows4032E=4F/(e pi d^2). Take logs gives ln(E)=ln(4)+ln(F)-ln(e)-ln(pi)-2ln(d). Differentiate and change minus signs to plus gives a “formula” for the maximum fractional error. dE/E=dF/F+de/e+2dd/d. I don’t know if this is introduced at A level but is what I learnt As an undergraduate student, probably in a lab session.
Elasticity is the room between conservation of momentum and conservation of energy in all its forms. A completely elastic process conserves kinetic energy and momentum, a completely inelastic collision conserves momentum but as much kinetic energy as is possible gets dissipated into other forms of energy. You can get rid of all of it in a sense, but that is sort of frame dependent :) the difference in kinetic energy in one frame can be completely not conserved between before and after, for example two pieces of clay crashing in the frame where their final velocity is 0, there the momenum is 0 all the time but the kinetic energy goes to 0, and for all completely inelastic collisions it is the same, that is there is some frame in which the momentum is 0 before and after and that frame is the frame where the kinetic energy will be 0. For the in between case if you look at the 0 total momentum frame there will still be kinetic energy after the crash. That is a good intuetive picture for velocities i think, also it is identical to lookikg at the momentum piecewise before andnafter for each body and so on.
For physics with conserving potentials just always write down the kinetic and potential energies in the margin without thinking and usually it stops you needingnto think at all.
I loved you both in this interesting video. Thank you.
It is very interesting how all those forces, velocitys, etc. have values. In Germany you very rarely get values to the corresponding variables. I think this helps significantly with understanding the physics behind the questions rather than just answering 1 question with given values.
Are you actually German?
Or if you want to have an intuition for inelastic deformation, basically if you squeeze a ball andnthe force integrated over the inward and outward journey are either equal or you lose some energy because the integral of the force on the way back to normal is smaller. Like a rubber band for example, it has close behavior being stretched and unstretched, but not quite, it takes more energy to draw it out than you get back by releasing it, and this is basically mostly because it is a devilish little heat engine along the way, when you stretch it out it gets hot and so no matter what as long as you didn't also change the temperature in the room to match it will dissipate some heat before you release it again, then it will get colder than the environment and take up some heat, but the process can't be 100 % effichienent, and it is a fun version of carnots heat engine that is also easier to explain, but also because it will vibrate and things like that it loses energy to air currents and sound as well, but that is a recommended example for a twist on teaching thermodynamics and Carnot.
Fantastic video! : )
Thanks!
I would go v=- root of g times the height,
Kinetic energy in the y direction is 1/2mVy^2
Potential energy = - mgh
It is okey to do it as long as the kinetic energy in y and x are independent. Which is not okey for air resistance for example.
To find the height you just use the initial velocity with these three and then set the time derrivative of the y position (with initial conditions) equal to - root g times h, since we found h so easily. Then you just do some algebra to put t on one side and there you go :).
When you integrate over some trajectory with a sum of force that is constant in direction, draw your potential, it will always be a bunch of parallel lines 90degrees of your force, no matter what the trajectory is, the velocity simoly has to be defined with respect to what lines of potential energy you have crossed and what is still "above you" so to speak. When the lines of equal potential are constant in space you never ever habe to worry about forces in different directions mixing andncausing trouble, only if you habe friction or some potential that is funky do you have to worry. And this theorem can be extended to funky potentials that are curved but continuous, meaning no discontinuety in the lines of equal potential, alan they never cross, and the gradient of the potential aka its derrivatives are well defined everywhere. When that is the case, no matter how skrewey you potential is with respect to the trajectory, you will just have to take the difference between the starting point in the potential and whatever other point along the trajectory you are looking at, and the difference in potential energy must equal the difference in kinetic energy. This is sort of a definition of a well defined potential. And all forces with a well defined potential conserves energy, if you had air resistance there is also kind of a trick by looking at the potential and modyfying it along the trajectory such that you have a second potential along andnaround your trajectory that solves some equation based on the combined variables so to speak, but that is a bit trickier as i am sure you know. But basically for air resistance you just have some force depending on velocity, and you can formulate your gravitational potential as such, but with a dissipative air resistance going up and a negative one going down, and you just sum up the potentials and then you are just left with a new potential along your trajectory which is not conservative, but has only one variable which is velocity, and you can integrate from an initial velocity whatever lenght along the trajectory you want in a simplified way ^^. Physics is pretty fun sometimes
Now the velocity version of the gravity potential cannget messy to define, there are simpler ways to do it, but that one is funnier. But yeah for simplified gravitational potentials or even complicated potentials it is very nice to have an intuiton for a general case of a rollercoaster, traveling around in a potential of any sort, where the dependence for velocity is always independently of the path of the rail depending only on the position in the potential, which is the defining feature of a potential that conserves energy.
A small angle cuts across the unit circle, really small angle means cos=1 long leg plus small angle =cos.
For occilators, at some non resonance frequency you can use this intuitive picture.
When your driving frequency is not the resonance frequency, and you have no occilation you start producing a small occilation at approximately the frequency of the driving frequency by force because the reaction force from the occilation being out of resonance is small with respect to the driving frequency, when the occilation grows it will become closer andncloser to its natural frequency, because the reaction force over a cycle of the actual frequency and the driving frequency approaches zero once the occilation becomes large with respect to the driving frequency/force. Roughly speaking that is right for a constant natural frequency occilators with no damping, combining a driving frequency and the natural frequency occilation of an occilator sort of works like damping plus driving and a change in the frequency.
ah i see a copy from google
Tom has a PhD in fluid mechanics from DAMTP at Cambridge. I do think he is acts up a bit on his channel an pretends to know less than he does. He is likely to have a better understanding of Newtonian Mechanics, Waves, Materials, and Thermodynamics than the vast majority of physics teachers.
I think he definitely pretended as though he didn’t know the equation needed for the second problem, but I appreciated the derivation.
Just defined the potential energy as negative because then KE plus PE =0 plus some constant starting energy in kinetic or potential energy. Sums and differences andnfunctions of the kinetic and potental energies are always useful so no need to get fancier than that.
Basically physics is maths plus conventions and thinking about the world, the conventions are sort of a mediator between the simplest math and the kinds of stuff we want to work out. Conventions are choosen either to make applications easy to calculate or calculations easy to apply hehe.
This was Feynman esque.. rock on Tom!!
Nice! Now quantize 4d Yang-Mills non-perturbatively.
A funny thing about arrows is that they have a decently large moment of inertia, and they are originally fired pointing up into the airflow and they are turned gradually to match the airflow corresponding to their trajectory, so even without air resistance in the direction of motion, it has a transverse small linear force associated with it that provides lift, this means it is not described by a parabolic or elliptic trajectory even ignoring air resistance as long as you want to account for the air turning it along the trajectory, which turns out to be a much more difficult question than the standard one :-). Ofc the lift force when air resistance is ignored has basically an infinite effichiency and coefficient of lift, but thats details in a realistic case the transverse force is still there in the total forces of the air acting upon the arrow ofc it is just mixed with the air resistance and therefore it isn't this screwy infinitely effichienent wing. But like a backspinning golf ball, the talfins of an arrow provide a tiny amount of lift, which is cool.
interesting! thanks for sharing
Try JEE Advanced(high school engineering entrance exam in India)questions.I'd suggest 2016 Jee advanced physics/maths or probably even the 2022 paper. It covers a much vaster syllabus and the questions are trickier.You'll enjoy solving some of them.
Best of luck 😊
Please make a video on India's toughest IIT exam questions. I would advise you to give it a try
for a person myself who absolutely horrible at maths. this is good stuff.
Mass 1 plus mass 2 = mass3,
Momentum one plus momentum 2 =momentum 3
V3 =momentum 3 /mass3.
Not seeen the rest yet but work done should also include the friction element.
I was waiting for him to reach the wave and circuits section of the paper
What application do you use to write out all of your work?
There was a small error in the problem with the arrow hitting the target on wheels at 48:00. The final momentum shouldn't be "velocity times m_target", but "velocity times (m_target + m_arrow)". The arrow keeps moving with the target after it is hit, so the arrows mass has to be included in the final momentum.
I'm surprised that Lewis didn't point this out.
Presumably, given that the mass of the arrow is so small compared to that of the target, that extra term can be ignored.
In my country we learned sos cas toa it means s(in) = o/s, c(os) = a/s and t(an) = o/a
I like how you turn a high school physics exam into am introductory classical mechanics exam by using calculus to derive seemingly trivial equations.
Well, calculus is how the equations were derived in the first place
@@cl4655 False. Mathematical formalism for SUVAT are derived way before calculus was created by Leibniz and Newton.
it's a lot simpler and easier to understand when you use calculus@@figulapt3784
The occilator question is kind of worded a bit sloppily, it should say in a steady state. Because that is what it is, with damping an occilator has a steady state with respect to the magnitude of the driving input, ofc if there is no driver it just stops, andnif there is no damping it either breaks or grows to infinite amplitude for it resonances, that last bit is what you would want to have an intuiton for to answythis question easily, just imagine a heavy ball on a spring, if you drive it with no breaks at the right frequency it tends to blow up, if you drive it close to the frequency it will also blow up until the shift in phase sort of cancels out with the driving force over time, you are doing work on the occilation you already created so to speak by creating a displacement. If you add damping then, you are changing from infinity at the ressonance to something finite, and something finite to something finite right next to it, so the graph without damping is a spike.
Just add all of the momentum, find the corresponding velocity, if everything sticks together at the end thats it. Almost so simple that it is funny.
As a first year physics student, my practical teacher would be proud that I did Q13 in 10 secs in my head. Just how many of us were yelling "SUVAT!" during Q2? I'm still on the fence as to whether I prefer maths or physics. At my age really should know.
sin(x) ~ x for small x, so sin(10° π/180°) ≈ π/18 ≈ 1/6. Close enough for an end-of-high-school exam, methinks.
LOVE how he extrapolated the force of gravity by thinking in terms of the moon and earth 😀😀
Can you do the Chinese Gaokao mathematics exam¿ that is really challenging
For trig my teacher taught us a bunch of mnemonics and the one that stuck with me was “Some Old Hippie Caught Another Hippie Tripping On Acid” 😂🤣
You forgot to add the metric units when calculating the work.
PLEEEASEE DO AN ALEVEL FURTHER STATS PAPER
🥰
can you do a video on TMUA?
What app are you using on the ipad?
Now the question is, can a high school math teacher solve an Oxford university exam?
completed the whole of physics maths and FM levels yet never knew you could prove the SUVAT equations using F = ma and integration
That’s first semester physics at uni though. Showing how F = m*a results in SUVAT.
That’s where they come from. All of physics is calculus, well and linear algebra and etc but calculus is the backbone
Watching this whilst currently doing A-level Physics. I attempted each question too 🤣
Please try the jee advanced exam conducted in India as their high school exam
You will thoroughly enjoy it because of its difficulty
You should try jee advanced maths questions paper
How does he not know suvat when u learn that in maths??
Tbh physics is one of most logical subject , visuals are beauty of it ,, like i still remember going to my jee advance exam center with papa ji seating back on motor bike i used to keep my eyes on truck wheels and keep thinking about rotational mechanics going around it , ,during that journey to exam center ,i visualise all my rotational mechanics formula and theory in my mind ,❤ tbh my fav subject is physics always ,same thing happen during my nsep inpho and kvpy exams .
@PhysicsOnline what about the wheels rolling
47:00 I'd argue there's some missed details here.
m*v_in = M*v_out misses that the arrow sticks to the target and now moves with the target, so it should me m*v_in = m*v_out + M*v_out.
The conclusion is the same but solving for v_out is a little different.
v_out = v_in * m/(m+M), not v_out = v_in * m/M
There you go i said the wrong words as well, more like :when the weight vector is equal in magnitude and opposite to the gravity vector the acceleration is 0.
Is half a meter practical for the tow bar? The truck has the towing ball at the back, so one end can attach to that. What's the back end attached to? What radius turn can the truck make, given the short towing bar? How many police cars will pull over the combo?
Tom what is your field of research?
V=D/T >>> 90/68=1.3 seconds. Done, NEXT!
Dr. Tom. Would you admit Lewis into Oxford?
😭 he took the longest way to derive the equation. I kept screaming use your projectile equations! He lost me at take the integral. I would have immediately knew it’s a dead end and go back!
It’s insane newton held the chair for maths whilst being a physicist…..shows how many levels ahead scientist vs mathematicians are
But if you where to calculate artillery shell trajectories for example it would have analogous dynamics, which is why artillery has always been so math heavy, the aerodynamics has always been rather complicated, abd you usually never solve it exactly but approximate with all sorts of tables for moisture and winds and different shells and so on, props to karl swartzchild for having time to also come up with a black hole, maybe he was looking for a place to dispose of all the silly leaders that started ww1.
Holy fuck dude. Calm down and stop spamming the comment section
Yes
Btw if the gentleman was misquoted when i said he called mg the weight then my bad ofc.
I rarely loggin to say anything but what in the world was this guy doing to finally get 1.75 sec. LOLOL. I did it in about 30 sec. Here's how, jeezus you guys. Verticle equation with Time is: (Final Y) = Vsin(theta)t -at^2 , putting in the givens you have: -4 = (25)Sin15t -5t^2, which is just 0 = -5t^2+6.47t +4 , very simple quadratic that gives 1.75s. You can even divide through by -5 giving a super easy equation: 0 = t^2 -1.294t - .8 (come on now)
Hey i want to have a video with you on indias JEE advanced exam
U can try israel bagrot ^^
Omg bro just derived Newton's equation using his own invention(calculus)without even knowing
And it is more like the weight of the earth acting on the car that is acting on the car.
great video! love ur channel, but i gotta say: you have the vibe of oli sykes of bring me the horizon hahaha
I love BMTH
These are fun :). To be a bit nitpicky, it really isn't a big deal, it is about language not about anything else. But you said mg downward was gravity and he said it is the weight. I agree with you, it is the force of gravit, the weight is the force from the ground counteracting gravity, when thw weight equals the force of gravity the acceleration is 0 with respect to the coordinates where the ground is stationary. Weight is the force acting on your feet, gravity is the force acting on your mass. :-) but i don't want to be a buzzkill, it could have just slipped through in passing anyway ^^.
Are you confusing the normal reaction force with weight? Weight acts through the centre of mass of the object and always acts 'downwards', the only forces acting at the feet would be the normal reaction and friction. "Gravity" in normal conversation relates to the force of gravity, however in physics it's important not to mix up the "acceleration due to gravity" with "gravity". g in these examples is an acceleration, weight is the force acting on the object as a result of this acceleration, and the normal reaction force is the force from the ground pushing upwards to counteract the object's weight.
You should try the AP Physics C: Mechanics and Electricity/Magnetism exams
My man used calculus to derive the most basic physic Formula v=s/t😂😂
I find it funny that you talk of not having and needing to go get a calculator when *you're currently using an iPad.*
Oozing self-promotional enthusiasm
Your not an idiot you're very smart
hi!
The Physics teacher looks like Mark Zuckerberg and Bill Gates 😅.
Just a heads up: Alan Becker dropped a new "Animation vs" video i think you would like it or not depending on how you view it
absolute banger
As all good engineers know the sin of a small angle is 0 and the cos of a small angle is 1. You sire haveth revolutionised engineering :'D. I disagree you do know what the sin of 10degrees is, it is sagittarius.
What kind of physics teacher is this, not chiding the 'student' for neglecting to put in units?
I was just impressed by his methods - my students who I have taught have this hammered into them!
Fun Times
Jee Advanced
Maths. Is mainly about manipulations of human-invented formulas and notions of mathematics , whereas physics is about more logical reasoning and understanding of the real physical world
why pi d ^2 rather than pi r^2
The formula given is E = 4F/επd²
If the radius were used instead of the diameter, the formula would be E = F/επr²
Remember r = d/2 so r² = (d/2)² = d²/4
ohh, yeah now i realised it ,i missed it during the solution you are right thanks @@Grizzly01-vr4pn
This was the most dumb yet beautiful things ever
Am I the only one who didn't hear the organ?😅
I love that when the elastic energy formula is given to tom he hears F(x) rather than F*x 😂
Maybe not an a level question though :')
Physics is easybro
i hated a level physics
I hated a level physics too. Now doing astrophysics at uni and I love it. The lecturers are who make me want to go to each lecture, at a level my teachers made me want to drop physics. Glad I didn’t!
@@BasicallyVader i think this is what happened with me as well. i loved maths and still do but my physics instructors sucked so much
bro im from india and this paper is insanely easy any 11th grader in india can attemp these questions .
Bro, India only has a literacy rate of 74%, meaning only 74% can read/write. Also, its total world ranking is around 34th for education and for math some sources suggest rank 50th roughly.
Exactly its meant for 11th grade my friend.