Hi,(Original post by ∞∞∞)
Good revision guide for pat is by dr matthew french there are some for sale on eBay. I used one last year and i passed my pat, it helps. Look on eBay oxford pat guide
How is the book like? Apart from model solutions to past PAT papers, how does it help with the exam?
Thanks in advance!!! 😁
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 21072016 23:39

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 22072016 00:20
(Original post by tangotangopapa2)
Hi,
How is the book like? Apart from model solutions to past PAT papers, how does it help with the exam?
Thanks in advance!!! 😁
Posted from TSR MobilePost rating:1 
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 223
 24072016 09:39
Ok im going to give you guys advice which i wish i was given last year.
So first to establish credibility I'm going to oxford for physics this year ( 95 on last years Pat )
Okay so first thing you must know A level maths ( intergration by parts and substitution is needed ) sometimes )
The maths section is the easiest one ( of the two xD ), you really have to be good at algebraic manipulations and graph sketching
You have to be able to sketch the regular graphs of course ( cubics, quadratics etc ) but also those of the form f(x)/g(x) and harder trigonometric curves ( sin^3 (x) ) for example or (sinx/x)
Coordinates is a must ( with circles )
Oh and every year they give a question on finding the area shaded ( geometry ) that really is free marks, get good at it
Doing some MAT questions might help with this
Else i also did some STEP for the maths ( STEP 1 easy questions are enough but you might want to do the rest to prepare for interviews )
Now the physics part might be a little harder if you have only been exposed to A level Physics type questions. You really need to forget those easy things and try Bpho papers. ( paper 1 round 1 and round 2 should be enough ) Basically you should be able to use A level maths in relevant physics questions. ( for sketching curves for example )
Past Pat papers are also excellent rescources i suggest you do last years paper as soon as possible ( because its the easiest xD ) keep 2010 in a drawer for last.
You should Know A level material ( capacitors, gravitation, shm, waves etc )
A Mechanics problems might also help here ( the harder ones )
90+ should be easy ( full marks is not impossible )
side note : if your the type of kid that likes puzzle and hard problems, all you have to do is to know the basics of A level topics and you're set.Post rating:6 
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 24072016 21:09
This questions might sound a bit silly lol...
question 14 part a and b:
http://www.physics.ox.ac.uk/olympiad...AS_2010_MS.pdf
so basically power emitted from the hole is 24 W because the temperature of the wall is the same as the temperature of the heater?
so at any point inside the heater, power would be 24 W then? is there a formula which connects these other than P= work done/time?
and part b) if there is no radiation escaping, temperature will increase inside? 
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 225
 24072016 21:49
(Original post by lawlieto)
This questions might sound a bit silly lol...
question 14 part a and b:
http://www.physics.ox.ac.uk/olympiad...AS_2010_MS.pdf
so basically power emitted from the hole is 24 W because the temperature of the wall is the same as the temperature of the heater?
so at any point inside the heater, power would be 24 W then? is there a formula which connects these other than P= work done/time?
and part b) if there is no radiation escaping, temperature will increase inside?
so at any point inside the heater, power would be 24 W then?
Sorry, I didn't exactly get the question. Power is the rate of change of energy. At any arbitrary point inside the system the energy received is same as the energy lost so that constant temperature is maintained and hence net power is zero. The total power of the heater is 24 watt, so the power of part of the heater should be a fraction of it.
In the second part of the question, the hole is closed and there is no way to emit energy from that closed system. The energy to the system is constantly added by the heater (Note that this is not against first law of thermodynamics as 'total energy in closed and adiabatic (isolated or no heat exchange) system remains constant' as electric energy is constantly being converted into heat energy and that energy is shown by raise in temperature) the temperature will increase.Last edited by tangotangopapa2; 24072016 at 22:28. 
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 24072016 23:44
(Original post by tangotangopapa2)
The first part of the question: As the temperature of the wall is same as that of the heater, the power (energy) emitted is no longer used to increase the internal energy of the wall. Since the heat radiated from elsewhere is negligible, all energy produced by the heater is radiated via the hole. So the power (energy per time) is 24W.
so at any point inside the heater, power would be 24 W then?
Sorry, I didn't exactly get the question. Power is the rate of change of energy. At any arbitrary point inside the system the energy received is same as the energy lost so that constant temperature is maintained and hence net power is zero. The total power of the heater is 24 watt, so the power of part of the heater should be a fraction of it.
In the second part of the question, the hole is closed and there is no way to emit energy from that closed system. The energy to the system is constantly added by the heater (Note that this is not against first law of thermodynamics as 'total energy in closed and adiabatic (isolated or no heat exchange) system remains constant' as electric energy is constantly being converted into heat energy and that energy is shown by raise in temperature) the temperature will increase.
"any point inside a heater" = an arbitrary point within the furnace (sorry, I meant the furnace, not the heater) ie within the walls. SO let's say, I pick a point randomly 10 cm from the heater within the sphere, would power "emitted" at that point be 0W then as it has a constant T?
I struggle to understand some basic concepts in physics... I found maths and FM easy but I self taught the A level physics starting at the end of January for the June exam series (I'm expecting a low A or a high B) and sometimes I just feel so stupid haha.. 
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 227
 25072016 09:29
(Original post by lawlieto)
Thanks, I'd give you reps but I gave too many reps to you and TSR wouldn't let me give more unless I rep others.
"any point inside a heater" = an arbitrary point within the furnace (sorry, I meant the furnace, not the heater) ie within the walls. SO let's say, I pick a point randomly 10 cm from the heater within the sphere, would power "emitted" at that point be 0W then as it has a constant T?
I struggle to understand some basic concepts in physics... I found maths and FM easy but I self taught the A level physics starting at the end of January for the June exam series (I'm expecting a low A or a high B) and sometimes I just feel so stupid haha..
Power is the rate of change of energy and any point of the furnace other than the hole does not emit any energy. So as to see further details let's try to magnify the situation. Your arbitrary point receives some energy from neighboring point and passes that to next point. This process goes on throughout the furnace except at the heater itself and the hole. The heater always emits energy and hole always discharges energy to the surrounding. So power of heater and hole (during equilibrium i.e. when temperature of the furnace is same as that of the heater) is same.
Now, if you pick a point at 10cm from heater within the sphere. The power emitted at that point would be zero as that does not emit any radiation to the surrounding. (You don't normally calculate power of a point unless that point is the point power source. Otherwise, even if you have a large system emitting a lot of energy say 5000 watt uniformly throughout the surface. If you take a fraction say onefourth of the emitting surface then the power of that surface is total energy emitted from that section per time, which would be 5000/4 watt. If you take a point, the energy emitted about that point would be extremely small compared to the total energy emitted by the power source so is zero.
In the above question, the power through hole was 24 watt because all the energy emitted by the heater was emitted by that hole.
I am extremely sorry if I have complicated the stuff.
Tbh, I never really liked A level physics syllabus. You should be comfortable if you pick up a book like Feynman's lectures in physics (there is pdf version available).
p It was already March when I started selfstudying for A levels (both AS and A2; Physics, Maths and F. Maths, a terrible experience). Having good background knowledge in physics, I chose Mechanics 14. I thought those would be extra easy if I carefully study Mechanics in Physics, which was my big mistake. Mechanics 2,3 and 4 went extra disastrous. I am expecting bottom B or C in Further maths. I feel stupid too.p 
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 228
 25072016 15:10
Hi guys. Ive done a few past year papers and found the maths section fairly ok, not as hard as the physics section. Ive looked at https://oxfordpat.wordpress.com/ as well as http://www.physicsandmathstutor.com/pat/ and have found that the oxfordpat solution seems to use calculus to solve the questions e.g 2012 Q 22
So I was wondering how important is calculus physics for the PAT and whether it is necessary to know it.Could i just do with algebraic physics? Its harder to apply calculus to physics as we have never been thought to do so that way. 
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 229
 25072016 16:14
Hey guys!
Are you using the BPHO round 1/paper 1 papers to practice for the PAT? coz i believe they are of a higher difficulty than the PAT past papers! 
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 230
 25072016 17:07
(Original post by kj5576)
Hey guys!
Are you using the BPHO round 1/paper 1 papers to practice for the PAT? coz i believe they are of a higher difficulty than the PAT past papers!
Posted from TSR Mobile 
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 231
 25072016 17:20
(Original post by tangotangopapa2)
I am thinking of using BPHO round 1 and STEP 1 papers for next few months and only then PAT papers a month before exam. Not sure how it will turn out.
Posted from TSR Mobile 
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 232
 25072016 17:38
(Original post by lyamlim97)
Hi guys. Ive done a few past year papers and found the maths section fairly ok, not as hard as the physics section. Ive looked at https://oxfordpat.wordpress.com/ as well as http://www.physicsandmathstutor.com/pat/ and have found that the oxfordpat solution seems to use calculus to solve the questions e.g 2012 Q 22
So I was wondering how important is calculus physics for the PAT and whether it is necessary to know it.Could i just do with algebraic physics? Its harder to apply calculus to physics as we have never been thought to do so that way.
Posted from TSR Mobile 
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 233
 25072016 18:59
(Original post by kj5576)
That's a great idea! Learning harder stuff over summer and then getting some PAT practice b4 the exam! It's just that bpho papers are harder and maybe involve more physics than the PAT? so maybe we might have to learn more
I feel like I know most of the BPHO topics. It's just that my problem solving skills are lacking which I wish to improve this summer. 
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 234
 25072016 20:15
(Original post by rohan.nuck)
Ok im going to give you guys advice which i wish i was given last year.
So first to establish credibility I'm going to oxford for physics this year ( 95 on last years Pat )
Okay so first thing you must know A level maths ( intergration by parts and substitution is needed ) sometimes )
The maths section is the easiest one ( of the two xD ), you really have to be good at algebraic manipulations and graph sketching
You have to be able to sketch the regular graphs of course ( cubics, quadratics etc ) but also those of the form f(x)/g(x) and harder trigonometric curves ( sin^3 (x) ) for example or (sinx/x)
Coordinates is a must ( with circles )
Oh and every year they give a question on finding the area shaded ( geometry ) that really is free marks, get good at it
Doing some MAT questions might help with this
Else i also did some STEP for the maths ( STEP 1 easy questions are enough but you might want to do the rest to prepare for interviews )
Now the physics part might be a little harder if you have only been exposed to A level Physics type questions. You really need to forget those easy things and try Bpho papers. ( paper 1 round 1 and round 2 should be enough ) Basically you should be able to use A level maths in relevant physics questions. ( for sketching curves for example )
Past Pat papers are also excellent rescources i suggest you do last years paper as soon as possible ( because its the easiest xD ) keep 2010 in a drawer for last.
You should Know A level material ( capacitors, gravitation, shm, waves etc )
A Mechanics problems might also help here ( the harder ones )
90+ should be easy ( full marks is not impossible )
side note : if your the type of kid that likes puzzle and hard problems, all you have to do is to know the basics of A level topics and you're set. 
 Follow
 235
 26072016 20:23
I suggest you do the maths section of last years paper or 2014 and check if u can cope with that ( 1hr ) just so you know if you need much more stuff or just a little bit.
Now for the physics part. Pat is not a regular exam, they want to see how u think rather than what you know. So basically when you study the A level stuff for the pat you should really focus on understanding the concepts and ideas ( some derivations might be helpful here ). You should know the relevant formulas and when they are used/how they are used/and why they are used. you should focus on trying to make the maths in physics 'natural' for you ( forming an energy equation in terms of some unknown) maybe im not that clear here. Let me give you an example : in physics papers( edexcel cie etc) most of the time you have to subsitute values in an equation and thats it but for the pat the situation will be more complex. Basically the calculations are way longer. ( harder mechanics problems are kind of what you should expect )
I cannot remember if waves/youngs double slit are in AS or not but if you have done this then this is a good comparaison: In your exam paper you might be asked to calculate the distance between the grating and screen given the relevant data, but in the pat you may be asked the same question except that this time the rays are not perpendicular to the grating but they are inclined at an angle theta
So i would say understanding the basics and concepts well and being at ease with the maths is what you should focus on. there is really no need for you yo focus on details in notes etc Practice long calculations from bpho or A2 challenge
For the A levels its also a good idea to have an idea of the derivation of formulas. these may come in handy ( especially for s.h.m, waves )
It might also be a good idea to kind of get an idea of the role of differentiation and integration in mechanics and shm 
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 236
 27072016 13:21
I have another question
A tank contains water to a depth of 1.0 m. Water emerges from a small hole in the vertical side of the tank at 20 cm below the surface. Determine:
(i) the speed at which the water emerges from the hole the mark scheme says use 0.5mv^2=mgh, and it uses 0.2 m for height, BUT that would imply that all the water coming from the hole is falling through a height of 0.2 m. if you just drill the hole, water next to that place will starts leaking immediately and that water hasn't fallen through any height previously, and also, as water is leaking, there will be less and less water on top of the water in line with the hole, therefore less pressure will push the water out, and it will be slowing down, so it should be modelled by a differential equation? I'm confused. tangotangopapa2 I need you. Also I would've thought speed would be constant only if the rate of leaking is equal to let's say a rate of resupplying water on top of the container?Last edited by lawlieto; 27072016 at 13:22. 
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 27072016 13:27
(Original post by tangotangopapa2)
You are very welcome.
Power is the rate of change of energy and any point of the furnace other than the hole does not emit any energy. So as to see further details let's try to magnify the situation. Your arbitrary point receives some energy from neighboring point and passes that to next point. This process goes on throughout the furnace except at the heater itself and the hole. The heater always emits energy and hole always discharges energy to the surrounding. So power of heater and hole (during equilibrium i.e. when temperature of the furnace is same as that of the heater) is same.
Now, if you pick a point at 10cm from heater within the sphere. The power emitted at that point would be zero as that does not emit any radiation to the surrounding. (You don't normally calculate power of a point unless that point is the point power source. Otherwise, even if you have a large system emitting a lot of energy say 5000 watt uniformly throughout the surface. If you take a fraction say onefourth of the emitting surface then the power of that surface is total energy emitted from that section per time, which would be 5000/4 watt. If you take a point, the energy emitted about that point would be extremely small compared to the total energy emitted by the power source so is zero.
In the above question, the power through hole was 24 watt because all the energy emitted by the heater was emitted by that hole.
I am extremely sorry if I have complicated the stuff.
Tbh, I never really liked A level physics syllabus. You should be comfortable if you pick up a book like Feynman's lectures in physics (there is pdf version available).
p It was already March when I started selfstudying for A levels (both AS and A2; Physics, Maths and F. Maths, a terrible experience). Having good background knowledge in physics, I chose Mechanics 14. I thought those would be extra easy if I carefully study Mechanics in Physics, which was my big mistake. Mechanics 2,3 and 4 went extra disastrous. I am expecting bottom B or C in Further maths. I feel stupid too.pLast edited by lawlieto; 27072016 at 13:28. 
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 238
 27072016 17:18
(Original post by lawlieto)
I have another question
A tank contains water to a depth of 1.0 m. Water emerges from a small hole in the vertical side of the tank at 20 cm below the surface. Determine:
(i) the speed at which the water emerges from the hole the mark scheme says use 0.5mv^2=mgh, and it uses 0.2 m for height, BUT that would imply that all the water coming from the hole is falling through a height of 0.2 m. if you just drill the hole, water next to that place will starts leaking immediately and that water hasn't fallen through any height previously, and also, as water is leaking, there will be less and less water on top of the water in line with the hole, therefore less pressure will push the water out, and it will be slowing down, so it should be modelled by a differential equation? I'm confused. tangotangopapa2 I need you. Also I would've thought speed would be constant only if the rate of leaking is equal to let's say a rate of resupplying water on top of the container?
Initially at the depth of 20 cm Fluid potential energy inside tank at depth h is (density X g X depth). The fluid kinetic energy outside the hole is (1/2 X density X v^2). [Note that these fluid energies do not have dimension of energy but have dimension of pressure]. Equating these two equations gives [v=sqrt(2 X g X h)]. This velocity is the speed of efflux water when the surface is h height higher than the hole.
Note that, it does not mean that water drop has fallen through 20 cm.
All you need to know is: The velocity to the water is provided by energy due to pressure of water. This energy at depth is numerically equal to the gravitational potential difference between surface and hole. This gives 1/2 m v^2 = mgh which has been used in the mark scheme.
Yes, you are right. As the height of water surface decreases, the water will be slowing down and v is sqrt(2XgXh). As velocity is function of h here and h is decreasing so velocity is also decreasing and it is not constant. You don't need to model the situation by differential equation. But the question is only interested in what happens initially, not throughout the motion.
For some reason, I feel like I have further complicated the stuff. I apologise for that.
In fact, there is a theorem known as Torricelli's theorem relating the speed of fluid flowing out of an opening to the height of fluid above the opening. https://en.wikipedia.org/wiki/Torricelli%27s_law
Torricelli's law states that the speed of efflux, v, of a fluid through a sharpedged hole at the bottom of a tank filled to a depth h is the same as the speed that a body (in this case a drop of water) would acquire in falling freely from a height h. So, . This is obtained my equating 1/2 mv^2 = mgh.
Another way of approaching this problem (which is in fact much simpler) is noting the fact that pressure can be written as energy / volume. As (Pressure = Force / Area = Force X distance / Volume = energy/volume). Pressure = dgh for liquid where d is the density. and Energy / Volume = 1/2 mv^2 / volume and writing (m/volume as d = density) we have P = 1/2 d v^2. Equating dgh = 0.5 d v^2 the result follows. Hope this helps.Last edited by tangotangopapa2; 27072016 at 20:29. 
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 27072016 18:19
(Original post by lawlieto)
I have another question
A tank contains water to a depth of 1.0 m. Water emerges from a small hole in the vertical side of the tank at 20 cm below the surface. Determine:
(i) the speed at which the water emerges from the hole the mark scheme says use 0.5mv^2=mgh, and it uses 0.2 m for height, BUT that would imply that all the water coming from the hole is falling through a height of 0.2 m. if you just drill the hole, water next to that place will starts leaking immediately and that water hasn't fallen through any height previously, and also, as water is leaking, there will be less and less water on top of the water in line with the hole, therefore less pressure will push the water out, and it will be slowing down, so it should be modelled by a differential equation? I'm confused. tangotangopapa2 I need you. Also I would've thought speed would be constant only if the rate of leaking is equal to let's say a rate of resupplying water on top of the container? 
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 240
 27072016 19:53
all the calculus u may need is v=dx/dt a=dv/dt and w= integral(F)ds and probably I=dq/dt
you may get a feel for these maybe in A level f.maths mechanics questions
try doing this : an object of mass M is released from rest from a plane, it experiences a resistive force of kv ( v= velocity). find its speed at time T.
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Updated: December 4, 2016
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