Edexcel AS/A2 Mathematics M2 - 17th June 2016 - Official Thread Watch

alvosm
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#81
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are there any proofs we should learn?
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ImJared
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(Original post by Louisb19)
Can someone help with M2 2015 2b please.

I don't understand what the mark scheme is saying.
The second part of this question requires the fact that the lamina is in equilibrium and therefore the moment about any point is equal to zero. I took the point O as my origin and took moments about it. Remember that moments are distance multiplied by perpendicular force.

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alvosm
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and what c3 do we need to know specifically? bc i did c3 last year
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apocolyptic
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Name:  Screen Shot 2016-06-15 at 9.58.09 PM.png
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Size:  41.2 KBPart b please... I cant solve it and Im not able to get what the markingscheme is saying
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ImJared
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(Original post by apocolyptic)
Name:  Screen Shot 2016-06-15 at 9.58.09 PM.png
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Size:  41.2 KBPart b please... I cant solve it and Im not able to get what the markingscheme is saying
Again, it is easiest to do this one using moments. Find the distance from O to the new centre of mass, then take a moment around the new centre of mass to find k. Another thing to note is that the new centre of mass is always going to be between the old centre and the additional masses. This means, for some questions where the masses are equal, you can simply find the distance between the old centre and the added mass to find the new centre of mass.

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yareelit
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(Original post by Inges)
Sure & yeah me too I got D for M1 :/ eek!
I got an E in M1 and am resitting
We have a special bond

Good luck bro
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aprilth
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Can anyone help me? I don't seem to quite understand why they used the horizontal initial speed instead of calculating the horizontal final speed...
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ImJared
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(Original post by aprilth)
Can anyone help me? I don't seem to quite understand why they used the horizontal initial speed instead of calculating the horizontal final speed...
The horizontal final speed is the same as the horizontal initial speed because there are no forces acting horizontally. The only velocity which changes is the vertical velocity which is affected by gravity. If you put the horizontal velocity in to the equation v = u + at you can see that, as a is equal to zero, v = u.

I just rushed it and here are the answers I got:

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josephinemar25
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Size:  128.6 KB Hi there, I am really struggling with (b) and (c) of this question. If anyone could help me that would be great
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izofficiallyme
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Does anyone know where I can actually find the June 2015 paper?
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Maetras
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(Original post by ImJared)
Again, it is easiest to do this one using moments. Find the distance from O to the new centre of mass, then take a moment around the new centre of mass to find k. Another thing to note is that the new centre of mass is always going to be between the old centre and the additional masses. This means, for some questions where the masses are equal, you can simply find the distance between the old centre and the added mass to find the new centre of mass.

Hi could you explain why the new centre of mass can't lie on the segment OA?
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Maetras
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Hello I don't understand how 0.9a=0.5(a+acostheta) in this question. Could someone explain it to me please?
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Maetras
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(Original post by izofficiallyme)
Does anyone know where I can actually find the June 2015 paper?
I posted it as an attachment on page 4 I think
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alevelstresss
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Is anyone just not worried for this exam? I worked out i only need like 30% to still get an A*
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OldMathsAccount
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Does anyone know why the original 2013 paper was withdrawn? Just curious.
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OldMathsAccount
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(Original post by alevelstresss)
Is anyone just not worried for this exam? I worked out i only need like 30% to still get an A*
I'm with you. After M3 it's a pretty straightforward unit, just some tricky methods I need to go over tomorrow. I don't think it'll matter too much in terms of ums for me.. knock on wood.
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ImJared
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(Original post by target21859)
Hi could you explain why the new centre of mass can't lie on the segment OA?
I assume you mean the M2 2015 paper; it's because if you think about it as two shapes put together, the new centre of mass will be between the two centre of masses. You work out the centre of mass in the first part of the question and you know that an extra mass was added to one of the points therefore the centre of mass will shift towards the new mass added (i.e. it will be somewhere between the two points depending on the ratios). Hope that helps?
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Maetras
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(Original post by ImJared)
I assume you mean the M2 2015 paper; it's because if you think about it as two shapes put together, the new centre of mass will be between the two centre of masses. You work out the centre of mass in the first part of the question and you know that an extra mass was added to one of the points therefore the centre of mass will shift towards the new mass added (i.e. it will be somewhere between the two points depending on the ratios). Hope that helps?
I understand that the centre of mass is shifted but I don't understand why it can't be much closer to the original centre of mass in this case somewhere along OA. Oh also I'm talking about question 4.
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Nirm
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For M2 Jan 2013 Question 3. I do not understand why the R(b) component is multiplied by sin(a). Can someone give me a break down with triangles in M2 with components or when it is sin(x)/cos(x) times by whatever? I thought I had it figured but I guess not, I always get them wrong when in moments or slopes. I was watching examsolutions (https://www.youtube.com/watch?v=FDo1ix2Kjuc) but he just stated it and didn't go much into detail.
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ImJared
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(Original post by target21859)
I understand that the centre of mass is shifted but I don't understand why it can't be much closer to the original centre of mass in this case somewhere along OA. Oh also I'm talking about question 4.
Ah sorry, didn't realise you meant that question. The A you are talking about is actually just a little angle diagram I created, there is no point A on the actual diagram. As you rightly said, the centre is shifted somewhere along that line but as you are given the angle, you can work out the new centre of mass. We are mostly interested in what the distances away from the centre of mass are rather than the location of the new centre itself.
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