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Original post by L'Evil Fish
No:tongue:

And for good reason. If your school use it as a mock, you want it to be accurate


Do you know what full UMS is for both of these papers generally. 52 g484 =90ums?


Original post by Makashima
I hope I am not too late! :P

500 individuals boxes! NO, too effort!
2 marks usually mean worth 2 minutes or so...thus the method must be quick/easy

Instead of counting individual box
Do this:
Therefore each boxes = 500 x 0.2x10^-3 = 0.1N.s

so here there is 16 complete boxes
what about the uncompleted boxes? well using your own judgement, how much complete boxes can be made from those pieces?

For me I say maybe 4-6 boxes but Im going to go with 4...
4+16 = 20 boxes

20 ( 500 x 0.2 x10^-3) = 2Ns



MS: 2.2 Ns {allow 2.0 to 2.4}


This question xD

I worked it out counting the squares but you could have also used the impulse value given in the question later and work ed backwards to get I=2.3Ns.
Original post by EmmaBxoxo
Do you know what full UMS is for both of these papers generally. 52 g484 =90ums?




This question xD

I worked it out counting the squares but you could have also used the impulse value given in the question later and work ed backwards to get I=2.3Ns.


Urmmm... Probably 53+ yeah

G485 not sure, haven't even looked at a paper
Original post by L'Evil Fish
Urmmm... Probably 53+ yeah

G485 not sure, haven't even looked at a paper


Phy is soooo much more interesting than chem.

That's pretty good to be fair, is it true that it was 39 for an A last year!

About 60% lol
Original post by Damien_Dalgaard
Phy is soooo much more interesting than chem.

That's pretty good to be fair, is it true that it was 39 for an A last year!

About 60% lol


No way lol

Physics is quite **** this year
Chem is harder

I think so yeah!
Reply 104
Original post by EmmaBxoxo
This question xD

I worked it out counting the squares but you could have also used the impulse value given in the question later and work ed backwards to get I=2.3Ns.


I put 3 points on the graph into my calculator to find a quadratic equation of the graph and then integrated for a laugh. It came out as an exact number like 2.2, so clearly the people who wrote the exam were doing some proper maths!

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Original post by Elcor
I put 3 points on the graph into my calculator to find a quadratic equation of the graph and then integrated for a laugh. It came out as an exact number like 2.2, so clearly the people who wrote the exam were doing some proper maths!

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Woah what calc do you have
Original post by L'Evil Fish
No way lol

Physics is quite **** this year
Chem is harder

I think so yeah!


Chem 5 is basically extended maths.

chem 4 is boring af mayn.

Papers are getting harder though especially last year, also they are so heavy on application and less recall.

p.s. ocr paper writers have an unhealthy obsession over skeletal formulae srs.
Reply 107
Original post by L'Evil Fish
Woah what calc do you have


The usual Casio fx-991ES PLUS, I just know a trick with the simultaneous equation solving mode. :wink:
Original post by Elcor
The usual Casio fx-991ES PLUS, I just know a trick with the simultaneous equation solving mode. :wink:


Oh that makes sense

Solve them simultaneously

Get two roots out

And Bob is your uncle using the sum and product of them

Quite clever fp

I think that's what you meant anyway

Actually that makes no sense

How do you do iy?
(edited 9 years ago)
Original post by L'Evil Fish
Oh that makes sense

Solve them simultaneously

Get two roots out

And Bob is your uncle using the sum and product of them

Quite clever fp

I think that's what you meant anyway

Actually that makes no sense

How do you do iy?


For the two physics modules how much have you got left to learn or have you covered everything?
Reply 110
Original post by L'Evil Fish
Oh that makes sense

Solve them simultaneously

Get two roots out

And Bob is your uncle using the sum and product of them

Quite clever fp

I think that's what you meant anyway

Actually that makes no sense

How do you do iy?


Well you know at every point y=ax^2+bx+c

Plug in 3 coordinates and set the coefficient of c to 1 because it's a constant. It will then tell you a, b and c (but I think it calls them X, Y and Z). It's the second setting in the EQN mode.


Btw you'll never have to use that it's just fun

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Original post by Damien_Dalgaard
For the two physics modules how much have you got left to learn or have you covered everything?

We have all the magnetic stuff and everything after capacitors

And then some radioactivity and all of medical
Original post by Elcor
Well you know at every point y=ax^2+bx+c

Plug in 3 coordinates and set the coefficient of c to 1 because it's a constant. It will then tell you a, b and c (but I think it calls them X, Y and Z). It's the second setting in the EQN mode.


Btw you'll never have to use that it's just fun

Posted from TSR Mobile


So say it was

y = 3x² + 7x + 2

You plug in

3 Co ordinates eg

1 1 1 12
4 2 1 64
0 0 1 2

And it'll give you a B and c? That's cool

Yeah ik, but it's nice to know how to get that up
Reply 112
Original post by L'Evil Fish
We have all the magnetic stuff and everything after capacitors

And then some radioactivity and all of medical


So say it was

y = 3x² + 7x + 2

You plug in

3 Co ordinates eg

1 1 1 12
4 2 1 64
0 0 1 2

And it'll give you a B and c? That's cool

Yeah ik, but it's nice to know how to get that up


Exactly

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It's not working
Reply 114
Original post by L'Evil Fish
It's not working


2nd y-coordinate should be 28.
Original post by Elcor
2nd y-coordinate should be 28.


Lmao don't ask what was going through my mind

So cool
Original post by Ki Yung Na
Do you guys know what equations, if any - aren't given in the formula booklet that we should know for either unit?


For G485 you don't get the equation for Faraday's law of magnetic induction, you only ge the statement 'induced emf = - rate of change of magnetic flux linkage'
Can't think of any for G484
Original post by sarcastic-sal
For G485 you don't get the equation for Faraday's law of magnetic induction, you only ge the statement 'induced emf = - rate of change of magnetic flux linkage'
Can't think of any for G484


Do we get a flux linkage equation? If so that's not too bad. I think we might be given pretty much all of them.

Regarding definitions; are you using the textbook? I was told by my friends that many of the textbook definitions don't get the marks for a lot of definition questions.
Original post by Ki Yung Na
Do we get a flux linkage equation? If so that's not too bad. I think we might be given pretty much all of them.

Regarding definitions; are you using the textbook? I was told by my friends that many of the textbook definitions don't get the marks for a lot of definition questions.


The latter is true, I just learn them from markschemes now

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Original post by Ki Yung Na
Do we get a flux linkage equation? If so that's not too bad. I think we might be given pretty much all of them.

Regarding definitions; are you using the textbook? I was told by my friends that many of the textbook definitions don't get the marks for a lot of definition questions.


You get φ = BAcosθ so as long as you know that flux linkage is just an 'extension' of magnetic flux (i.e flux linkage is φN) then it's fine. I think they deliberately didn't include all of the extra equations (which are just the above applied to a coil or when a wire/coil is moving) because magnetic fields/electromagnetism is the hardest topic and it kind of shows you have a solid understanding if you can cope without being guided by the data book.

I'm pretty much using the specification and past paper markschemes to learn definitions, and if they've never came up and if they're mentioned in a vague way on the spec I have three textbooks (plus another non-OCR specific one) and I try to learn a definition which is quoted by two of the three.

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