Oxford PAT 2016

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Same extent. The cut-off point for interview is the same, I believe.

Might be easier to get in though. Last year i think half of the interviewed students were admitted for engineering. So if you get a decent score you should be fine

Reply 481

ASadTeddyBear

The PAT syllabus includes electrostatic forces and magnetism: electromagnets; motors and generators; thermionic emission and energy of accelerated electron beams.
I've been wondering how much I should know about these topics.
For electrostatic forces does this include electric fields + Coulomb's law + EM induction (flux linkage or alternating current generator or transformers), or is it something else?
For magnetism does this include to motor effect/flux density/left hand rule/charges in a magnetic field or something else?

The syllabus does not go into depth about this at all, and I don't want to spend time learning these topics if I should be learning something different.

I'd really appreciate some help - thanks in advance!

Reply 482

PhysicsMathsTut

Original post by ASadTeddyBear

Hi, I suggest you know the basics. Keep in mind that based on previous years, there are many more questions on electricity/circuits. For physics, mechanics is the big part. Learn all A-level stuff (including A2) plus levers and pulleys. Also remember to read about the solar system and eclipses, which is not normally taught in A-level.

I’ve also seen many questions about what’s included in maths and whether further maths topics comes up. Here’s a mapping of the maths topics you need to know onto the Edexcel textbook chapters. Further maths is not required. Sketching curves can be tricky, as most students don’t go beyond simple transformations in A-level. Make sure you practice that.

Algebra (C1 ch1,2; C2 ch1; C3 ch1; C4 ch1)
Binomial expansion (C2 ch5; C4 ch3)
Coordinate geometry (C1 ch5; C2 ch4)
Differentiation (C1 ch7; C2 ch9
Inequalities (C1 ch3)
Simultaneous equations (C1 ch3)
Integration (C1 ch8; C2 ch11; C4 ch6.1,6.2,6.4)
Exponentials & logarithms (C2 ch3; C3 ch3)
Probability (S1 ch5)
Series (C1 ch6; C2 ch7)
Sketching curves (C1 ch4, C3 ch5)
Trigonometry (C2 ch8,10)

I realised that my solutions haven’t been accessible on the website for a while. I apologise for that, I forgot to update the old google drive links. They are working here now.

(edited 7 years ago)

Reply 483

rohan.nuck

Original post by PhysicsMathsTut

Seriously look up solar system and eclipses. I didn't study that and flunked one whole interview because i had no idea what they were talking about

Reply 484

student1856

Original post by rohan.nuck

Seriously look up solar system and eclipses. I didn't study that and flunked one whole interview because i had no idea what they were talking about

Hi, thanks very much for your help on the other question. I just had one more thing I was unsure of if you don't mind helping. It's question 11)c on this paper. I just don't understand how they got an answer of n+1/2+1 for 313rps.
http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_AS_%202011_MS.pdf I'd be very grateful if you could tell me what you thought. I think it has something to do with extinction which is mentioned in part a but I'm not entirely sure what this means in terms of light. Thanks again.

(edited 7 years ago)

Reply 485

rohan.nuck

Original post by student1856

I'm sorry I typed a long thing to explain it to you but then it got deleted by accident. I'll write it again tomorrow xD

Reply 486

student1856

Original post by rohan.nuck

I'm sorry I typed a long thing to explain it to you but then it got deleted by accident. I'll write it again tomorrow xD

No worries, there's no rush. :smile:

Reply 487

rohan.nuck

Original post by student1856

No worries, there's no rush. :smile:

Okay. So i think you have no understood the experiment fully. I'll try and explain it as clearly as possible. So Basically we are firing a laser/light at the top of the rotating cog. Obviously, the light will only go through if the cog is in a position where the laser/light is between two teeth. the light then goes through to the mirror and is reflected back to the cog.Again this reflected light will only go through if the cog is positioned in such a way that the reflected light is between two teeth. This light goes through to the screen and is observed. Now, the situation that the light is exactly between two teeth is highly unlikely. The light will most of the time be partially obstructed by a tooth. But sometimes, it can happen that the light is fully obstructed by a tooth and thus the reflected light does not go through to the screen and is 'extinguished'. When will this happen ? lets consider the simplest situation. light goes through to teeth initially ( call them T1 and T2). As the light goes through to the mirror and is reflected to the cog, the cog would have rotated. The smallest amount it could have rotated so that the light is extinguished is the case when T1 or T2 has obstructed the reflected light. Either way it has rotated by 1/2 a tooth. From there it can easily be seen that for the light to be extinguished the cog must rotate through n+1/2 teeth in the time the light goes from the cog and back to the cog. At 283 rpm n+1/2 teeth have passed in that time ( and thus light is extinguished). Imagine gradually increasing the rpm of the cog. The tooth that was previously blocking the light (T3) will gradually let some light trough and at some point all the light will go through and the tooth adjacent to T3 will start partially blocking the light. When that tooth completely blocks the light, the light will be extinguished again. The question says at 313 rpm the light is extinguished again. So since now T3 has also passed, 1 more tooth has passed which gives n+1/2+1

Reply 488

PhoenixHellRider

Hi, so I dont really understand some of the proofs such as the differentiation/integration ones for sin, e, cos, etc. (especially as we haven't been taught them yet) so should I just memorise them

Reply 489

rohan.nuck

Original post by PhoenixHellRider

Hi, so I dont really understand some of the proofs such as the differentiation/integration ones for sin, e, cos, etc. (especially as we haven't been taught them yet) so should I just memorise them

Memorise them, you really will not need to know the derivation. As long as u understand the principle ( ie you can differentiate x^2 from first principles you should be fine.

Reply 490

lawlieto

Question 3:
http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_Paper1_2007_QP.pdf

when it says "one ten thousandth of the surface area of the sphere" what sphere is it talking about? I guess Earth but when it is calculating the surface area of the sphere it uses the distance between the earth and the satellite as the radius in 4pi r^2.
Why is that distance used as the radius of the sphere? :s-smilie:

here's the markscheme: http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_Paper1_2007_MS.pdf

Thanks :biggrin:

Reply 491

username2452153

Original post by lawlieto

Question 3:
http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_Paper1_2007_QP.pdf

when it says "one ten thousandth of the surface area of the sphere" what sphere is it talking about?

Edit: It is talking about the imaginary sphere centered at the source. Exactly same kind of sphere you would be drawing when finding radially outward electric field due to point charge.

I guess Earth but when it is calculating the surface area of the sphere it uses the distance between the earth and the satellite as the radius in 4pi r^2.
Why is that distance used as the radius of the sphere? :s-smilie:

here's the markscheme: http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_Paper1_2007_MS.pdf

Thanks :biggrin:

Normally, a point source would radiate energy equally in all direction in the space. But the beam from the transmitter is focused directly towards the earth. Unfortunately, the beam is not in straight line (this would have made life lot easier). The beam diverges with angular diameter of 2 degree. This means the beam does not cover the whole sphere but just the one with angular diameter of 2 degree. To make calculation lot simpler, the question further gives an approximation that 2 degree angular diameter corresponds to one ten thousandth of the total surface area of sphere centering the source. At the distance (telescope-earth distance), the beam only falls within the area, which is one tenth of the surface area of sphere centered at the source. So, the corresponding area is found out by first drawing imaginary sphere of 10km centering at the transmitter and finding out the area which corresponds to 2 degree angular diameter using the information in the question. (Imagine this similar to drawing a gaussian sphere).

Then a further assumption has been made that every point within this area receives the same amount of energy per second.

Hope this helps. :smile:

(edited 7 years ago)

Reply 492

username2452153

Original post by lawlieto

here's the markscheme: http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_Paper1_2007_MS.pdf

Thanks :biggrin:

Original post by tangotangopapa2

Normally, a point source would radiate energy equally in all direction in the space. But the beam from the transmitter is focused directly towards the earth. Unfortunately, the beam is not in straight line (this would have made life lot easier). The beam diverges with angular diameter of 2 degree. This means the beam does not cover the whole sphere but just the one with angular diameter of 2 degree. To make calculation lot simpler, the question further gives an approximation that 2 degree angular diameter corresponds to one ten thousandth of the total surface area of sphere centering the source. At the distance (telescope-earth distance), the beam only falls within the area, which is one tenth of the surface area of sphere centered at the source. So, the corresponding area is found out by first drawing imaginary sphere of 10km centering at the transmitter and finding out the area which corresponds to 2 degree angular diameter using the information in the question. (Imagine this similar to drawing a gausian sphere).

Then a further assumption has been made that every point within this area receives the same amount of energy per second.

Hope this helps. :smile:

The following is the figure to supplement the explanation. Sorry for making it look messy. (I am terrible at drawing :tongue:

)

Reply 493

rohan.nuck

Original post by tangotangopapa2

The following is the figure to supplement the explanation. Sorry for making it look messy. (I am terrible at drawing :tongue:

)

Can u tell what u used to make this picture ? or is it just paint ? XD

Reply 494

username2452153

Original post by rohan.nuck

Can u tell what u used to make this picture ? or is it just paint ? XD

Of course Paint!!! :biggrin:

Reply 495

rohan.nuck

Original post by tangotangopapa2

Of course Paint!!! :biggrin:

How on earth can u write on paint. U must be an alien.

Reply 496

username2452153

Original post by rohan.nuck

How on earth can u write on paint. U must be an alien.

I write on it using a graphic tablet. XD

Reply 497

RPFeyn

Original post by PhysicsMathsTut

I do WJEC Maths instead of Edexcel so can anyone point out to me what this Edexcel textbook is? I've done a quick google search and found some PDF's but I'm not sure if its the right one! Can anyone help?