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OCR Physics A - G484: The Newtonian World - 18/06/12

can someone explain how to work out 4b(iii) and 5b

http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61507_pp_11_jan_gce_g484.pdf

(edited 11 years ago)

Reply 181

sarah-xx

Original post by D556mm

Does anyone have any idea of how high the grade boundaries could be?

Well I hope that they won't be as high as they were in January. Needed 48 for an A! Not sure what it will be this time, really depends on how difficult the paper will be...

Reply 182

jahani08

Original post by D556mm

Does anyone have any idea of how high the grade boundaries could be?

not sure about the other grades but an a is usually around 48/60. there are usually about 4/5 marks between each grade boundary

good luck :smile:

Reply 183

D556mm

Original post by sarah-xx

Well I hope that they won't be as high as they were in January. Needed 48 for an A! Not sure what it will be this time, really depends on how difficult the paper will be...

Original post by jahani08

not sure about the other grades but an a is usually around 48/60. there are usually about 4/5 marks between each grade boundary

good luck :smile:

Thanks guys :biggrin:

just worried because of the extra time allowed :P also, this is my first time doing this paper as my teacher was an A-hole and didn't let most of the class take it because a Physics resit was within a couple of days of this exam and he wanted us to concentrate more on that. But yeah, lets not hope it's like the Jan 12 boundaries :biggrin:

Reply 184

Beresford George

Original post by jahani08

not sure about the other grades but an a is usually around 48/60. there are usually about 4/5 marks between each grade boundary

good luck :smile:

January was 3 marks between each grade. Hopefully they wont be that stupid this time!!!!!

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Reply 185

Jukeboxing

8

Original post by lollizard123

can someone explain how to work out 4b(iii) and 5b

http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61507_pp_11_jan_gce_g484.pdf

Remember the equation x = Acos(2*pi*ft)

Plug in the values that you know.

d = 2.5cos(2 \times \pi \times 2.22 \times 10^{-5} \times t) + 15.5

The reason I put the + 15.5 is because its asking for depth of the water, which has minimum at 13m and maximum at 18m.

If we omitted the 15.5 it would just give us the displacement of the water rather than the depth of the water.

15.5 is the equilibrium position 18-13/2 =2.5, 13+2.5 = 15.5

(edited 11 years ago)

Reply 186

lollizard123

Original post by Jukeboxing

Remember the equation x = Acos(2*pi*ft)

Plug in the values that you know.

d = 2.5cos(2 \times \pi \times 2.22 \times 10^{-5} \times t) + 15

The reason I put the + 15 is because its asking for depth of the water, which has minimum at 15m and maximum at 18m.

If we omitted the 15 it would just give us the displacement of the water rather than the depth of the water.

thank you!

Reply 187

ambitiousdan

can someone explain how they proved that g is directly proportional to r on question 2(c)(iii). I don't understand what they're trying to imply in the mark scheme. http://pdf.ocr.org.uk/download/assess_mat/ocr_7976_sam_gce_unit_g484.pdf?

Reply 188

jibjohn

Hello, I always get stuck on "derive the equation
T²=((4π²/(GM))r³
from first principles"
I think I'm almost there, but i don't do maths so I'm not entirely confident, thanks in advance!

Reply 189

Shahman123

OP

Anyone have any model answers for

Describe simple examples of free oscillations.

Describe with graphical illustrations the changes in displacement, velocity and acceleration of simple harmonic motion.

Describe the effects of damping on an oscillating system.

Describe solids, liquids and gases in terms of the spacing, ordering and motion of atoms.

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Reply 190

mack94

3

Original post by lollizard123

thank you!

On the mark scheme it says 15.5? I really dont know where they get the .5 from!
If you understand the question can you explain it to me please :smile:

Reply 191

Jukeboxing

8

Original post by jibjohn

Hello, I always get stuck on "derive the equation
T²=((4π²/(GM))r³
from first principles"
I think I'm almost there, but i don't do maths so I'm not entirely confident, thanks in advance!

v = \frac{2\pi r}{T} F = \frac{mv^2}{r} F = \frac{GMm}{r^2}

\frac{mv^2}{r}

= \frac{GMm}{r^2}

Cancel out m and r, so we get

v^2 = \frac{GM}{r}

Replace v

\frac{4\pi^2r^2}{T^2}

= \frac{GM}{r}

[br]{4\pi^2r^2} \times \frac{r}{GM} = T^2[br]

[br][br]T^2 = \frac{4\pi^2r^3}{GM}[br][br]

Reply 192

sarah-xx

Original post by jibjohn

Hello, I always get stuck on "derive the equation
T²=((4π²/(GM))r³
from first principles"
I think I'm almost there, but i don't do maths so I'm not entirely confident, thanks in advance!

F=GMm/r^2

F=mv^2/r

equal them to eachother and let v^2 be the subject.

Sub v=2*pi*r/T into v^2 (remember that everything gets squared when you do this, e.g 4*pi^2*r^2/T^2)

& rearange to make T^2 the subject

hope that doesn't confuse you too much

Jukeboxing's post is much clearer as you can see...

(edited 11 years ago)

Reply 193

Jukeboxing

8

Original post by lollizard123

thank you!

Original post by mack94

On the mark scheme it says 15.5? I really dont know where they get the .5 from!
If you understand the question can you explain it to me please :smile:

Sorry it is 15.5m, not 15m. The equilibrium point is at 15.5m

I misread the question.

(edited 11 years ago)

Reply 194

Picture~Perfect

19

Good luck everyone :smile:

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Reply 195

Oliv

Original post by mack94

On the mark scheme it says 15.5? I really dont know where they get the .5 from!
If you understand the question can you explain it to me please :smile:

The minimum height is 13m, the maximum height is 18m so the equilibrium position is halfway between this at a depth of 15.5m

Reply 196

h_jackson

Original post by jibjohn

Hello, I always get stuck on "derive the equation
T²=((4π²/(GM))r³
from first principles"
I think I'm almost there, but i don't do maths so I'm not entirely confident, thanks in advance!

Start with mv²/r=GMm/r² as they both use Force

But v=2'pi'r/T so that goes to

GM/r²= (2'pi'r/T)²

Then 4'pi'²r³/GM= T²

Also this proves keplers T² is proportional to R³ as (4'pi'²/GM) is a constant
Hope this helps :biggrin:

Reply 197

jibjohn

Thanks Jukeboxing! (mustn't forget h_jackson and sarah-xx)
One more thing my cgp guide say's that it's ok to define newtons law of gravitation with
F=-GMm/r²
then saying what each letter means, is this really good enough?

(edited 11 years ago)

Reply 198

Jukeboxing

8

Original post by jibjohn

Thanks Jukeboxing! (mustn't forget h_jackson and sarah-xx)
One more thing my cgp guide say's that it's ok to define newtons law of gravitation with
F=-GMm/r²
then saying what each letter means, is this really good enough?

That should be fine.

I prefer to define it in words.

Edit: I just realised you said you would define each symbol

(edited 11 years ago)

Reply 199

Lindizya

Original post by lollizard123

can someone explain how to work out 4b(iii) and 5b

http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61507_pp_11_jan_gce_g484.pdf

question 5 b
you know that 1/2mv^2=3/2kT
Implying the kinetic of hydrogen is equal to the kinetic energy of oxygen.
so basically 1/2mv^2=1/2mv^2 put values given into equation then solve by making the v the subject of the formula to find what the mean speed is.

OCR Physics A - G484: The Newtonian World - 18/06/12

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