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

Does anyone have the grade boundaries for past M3 papers that they could post? Thanks :smile:

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

Reply 41

connerzandmrslushy

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 :smile:

All Grade Boundaries Jan2005-Jan2011.pdf349.7KB

Reply 42

aurao2003

3

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)
So, AD = 2l +x+y
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.

Reply 43

aurao2003

3

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)
So, AD = 2l +x+y
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.

Reply 44

gestation

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)
So, AD = 2l +x+y
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
so AD is 3.5l

sorry if this is horribly wrong btw... :smile:

(edited 11 years ago)

Reply 45

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!

Reply 46

bencwbrown

11

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

Reply 47

SkyStorm

Original post by browb003

Hope this helps

That helps a lot, thanks very much. :biggrin:

Reply 48

thethinker

1

I'm so screwed for this exam...

Reply 49

No_matter_how_many

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!

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

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