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# The Edexcel M3 (14/06/12 - AM) Revision Thread

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

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1. Re: The Edexcel M3 (14/06/12 - AM) Revision Thread
Does anyone have the grade boundaries for past M3 papers that they could post? Thanks

Started doing past papers yesterday, they don't seem as bad as I was expecting tbh, though I am still using notes to help me. :P
2. Re: The Edexcel M3 (14/06/12 - AM) Revision Thread
I have attached the file for all the garde boundaries for the maths modules untill 2011

I think they're still pretty much the same
Attached Files
3. All Grade Boundaries Jan2005-Jan2011.pdf (341.5 KB, 595 views)
4. Re: The Edexcel M3 (14/06/12 - AM) Revision Thread
Hi Guys
Sorry for being late. Got caught up in other you know what!!!
Can anyone help with this question?
A light elastic string AB has natural length l and modulus of elasticity 2mg. Another light elastic string CD has natural length l and modulus of elasticity 4mg. The strings are joined at their ends B and C and the end A is attached to a fixed point. A particle of mass m is hung from the end D and is at rest in equilibrium. Find the length AD.

this is what I did

Let Tension AB is T2 and tension CD be T1. since system is at rest
T2-T1 = mg (Equation 1)
AD = l +x +y +l (Where x and y are the extensions of AB and CD respectively)
Using Hookes law
T2 = 2mgx/l

Similarly, T1 = 4mgy/l

substituting the above into equation 1
2x-4y =l
This is as far as I can go. I wasn't given the distance between the the fixed point and AD. Am I missing something? Kindly comment.
5. Re: The Edexcel M3 (14/06/12 - AM) Revision Thread
Hi Guys
Sorry for being late. Got caught up in other you know what!!!
Can anyone help with this question?
A light elastic string AB has natural length l and modulus of elasticity 2mg. Another light elastic string CD has natural length l and modulus of elasticity 4mg. The strings are joined at their ends B and C and the end A is attached to a fixed point. A particle of mass m is hung from the end D and is at rest in equilibrium. Find the length AD.

this is what I did

Let Tension AB be T2 and tension CD be T1. since system is at rest
T2-T1 = mg (Equation 1)
AD = l +x +y +l (Where x and y are the extensions of AB and CD respectively)
Using Hookes law
T2 = 2mgx/l

Similarly, T1 = 4mgy/l

substituting the above into equation 1
2x-4y =l
This is as far as I can go. I wasn't given the distance between the the fixed point and AD. Am I missing something? Kindly comment.
6. Re: The Edexcel M3 (14/06/12 - AM) Revision Thread
(Original post by aurao2003)
Hi Guys
Sorry for being late. Got caught up in other you know what!!!
Can anyone help with this question?
A light elastic string AB has natural length l and modulus of elasticity 2mg. Another light elastic string CD has natural length l and modulus of elasticity 4mg. The strings are joined at their ends B and C and the end A is attached to a fixed point. A particle of mass m is hung from the end D and is at rest in equilibrium. Find the length AD.

this is what I did

Let Tension AB be T2 and tension CD be T1. since system is at rest
T2-T1 = mg (Equation 1)
AD = l +x +y +l (Where x and y are the extensions of AB and CD respectively)
Using Hookes law
T2 = 2mgx/l

Similarly, T1 = 4mgy/l

substituting the above into equation 1
2x-4y =l
This is as far as I can go. I wasn't given the distance between the the fixed point and AD. Am I missing something? Kindly comment.

hey there. i have an idea how to tackle this problem but i may be wayyy off. In physics when you are taught about hookes law you use F= k*x where k is a constant and x is the extension. And when string/springs are in series and you want to model them as one big string/spring you do:

1/k(overall spring constant) = 1/k(1st spring) + 1/k(2nd spring)
i hope that layout makes sense.. (its basically the reciprocal of the spring constants added together is equal to the reciprocal of the overall spring constant)

Relating this to the tension equation we use in m3. T= lambda*x/l comparing this to F= k*x. K= lambda/l. hence using the earlier stated rule...
l/lamda = l/4mg + l/2mg rearranging that we get lambda is 4mg/3 .
Therefore overall we can say mg = (4mg*x)/6l solve giving x=1.5*l

sorry if this is horribly wrong btw...
Last edited by gestation; 12-06-2012 at 03:18.
7. Re: The Edexcel M3 (14/06/12 - AM) Revision Thread
I think I am probably missing the obvious, but does anyone know how to work out question 2b from the january 2011 paper. I follow what they have done in the mark scheme, but I don't see how OG=(lambda/(7+lambda))*2r. Can somebody please put me out of my misery!
8. Re: The Edexcel M3 (14/06/12 - AM) Revision Thread
(Original post by SkyStorm)
I think I am probably missing the obvious, but does anyone know how to work out question 2b from the january 2011 paper. I follow what they have done in the mark scheme, but I don't see how OG=(lambda/(7+lambda))*2r. Can somebody please put me out of my misery!
Hope this helps

9. Re: The Edexcel M3 (14/06/12 - AM) Revision Thread
(Original post by browb003)
Hope this helps

That helps a lot, thanks very much.
10. Re: The Edexcel M3 (14/06/12 - AM) Revision Thread
I'm so screwed for this exam...
11. Re: The Edexcel M3 (14/06/12 - AM) Revision Thread
I think the missed fact is that the tension in the system is constant, as it is with all string/spring systems at rest. Using this and equating the two tensions, the answer came out easily. Hope this helps and is right!

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Last updated: June 14, 2012
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