The Student Room Group

AQA A2 Mathematics MM2B Mechanics 2 - Monday 22nd June 2015 [Exam Discussion Thread]

Scroll to see replies

My teacher wen through all the papers and said the hardest one she's seen is the Jan 13 paper. It was a normal paper for the majority of the questions, but the last two questions required serious thinking


Posted from TSR Mobile
Reply 81
Original post by bitofageek
My teacher wen through all the papers and said the hardest one she's seen is the Jan 13 paper. It was a normal paper for the majority of the questions, but the last two questions required serious thinking


Posted from TSR Mobile


Ah, I've only done up to Jan 2010. Do you happen to know what topics they were on?


Posted from TSR Mobile
Original post by CD223
Ah, I've only done up to Jan 2010. Do you happen to know what topics they were on?


Posted from TSR Mobile


Yeah hookes law was one. The other was a mix of moments and vertical circular motion


Posted from TSR Mobile
Reply 83
Original post by bitofageek
Yeah hookes law was one. The other was a mix of moments and vertical circular motion


Posted from TSR Mobile


Ah crap, I've not even finished Hooke's Law. It's funny how such a simple concept can lead to such horrid complex questions.


Posted from TSR Mobile
Original post by CD223
Without giving too much away, what made it hard? Was it just not as guided as previous papers in breaking down step by step what to do for each question?


Posted from TSR Mobile


The majority was fine, the first 5 questions were pretty much standard M2. The last 3 questions came at stuff from a different angle - I can totally understand where the answer has come from looking at the mark scheme, but in the limited time available I just didn't 'click' as to how to do it.
Reply 85
Original post by sarcastic-sal
The majority was fine, the first 5 questions were pretty much standard M2. The last 3 questions came at stuff from a different angle - I can totally understand where the answer has come from looking at the mark scheme, but in the limited time available I just didn't 'click' as to how to do it.


That worries me. My M2 mock is 8th May. Is there any good way I can prepare?


Posted from TSR Mobile
Original post by CD223
That worries me. My M2 mock is 8th May. Is there any good way I can prepare?


Posted from TSR Mobile


I think doing at least one of the more recent papers beforehand (don't go crazy and do all of them though, save them for closer to the real thing). I'd also go through the paper with a highlighter and circle/draw attention to anything important beforehand (I didn't do this and I think it might have helped).
Reply 87
Original post by sarcastic-sal
I think doing at least one of the more recent papers beforehand (don't go crazy and do all of them though, save them for closer to the real thing). I'd also go through the paper with a highlighter and circle/draw attention to anything important beforehand (I didn't do this and I think it might have helped).


Thank you for the tips! :smile: I'll try and recent paper. Maybe June 2013?


Posted from TSR Mobile
Reply 88
Can someone help me with this question? (Part B). I've formed a quadratic but it's not the same as the one in the question as its in terms of x, the extension.

ImageUploadedByStudent Room1430332532.654778.jpg


Posted from TSR Mobile
Reply 89
Original post by CD223
Can someone help me with this question? (Part B). I've formed a quadratic but it's not the same as the one in the question as its in terms of x, the extension.

ImageUploadedByStudent Room1430332532.654778.jpg


Posted from TSR Mobile


Alrighty I'll try and help you out :biggrin:

So for part bi the statement " the maximum length of the string in the subsequent motion is L" basically means that L is the sum of the extension and natural length. So L=e+2 L=e+2 where ee is the extension (I think you've got it as x) which means e=2L e=2-L . Ok great we've got the extension in terms of LL now so that means we can use it in the EPE formula in place of the ee.

EPE=560(e)24[br]e=L2[br]EPE=560(L2)24[br]EPE=140(L2)2[br]EPE=140(L24L+4) EPE = \dfrac {560(e)^2}{4}[br]e=L-2[br]EPE = \dfrac {560(L-2)^2}{4}[br]EPE = 140(L-2)^2[br]EPE = 140(L^2 - 4L + 4)

Using conservation of energy the Initial GPE = EPE + KE + Final GPE . If we consider the maximum height to be L then equation becomes Initial GPE = EPE (the KE disappears because the final velocity is zero at the maximum length of the string)

Ok Initial GPE = EPE
20gL=140(L24L+4) 20gL=140(L^2 - 4L + 4)

and bish bash bosh you should get 5L227L+20=0 5L^2-27L+20=0
Reply 90
Original post by Shadez
Alrighty I'll try and help you out :biggrin:

So for part bi the statement " the maximum length of the string in the subsequent motion is L" basically means that L is the sum of the extension and natural length. So L=e+2 L=e+2 where ee is the extension (I think you've got it as x) which means e=2L e=2-L . Ok great we've got the extension in terms of LL now so that means we can use it in the EPE formula in place of the ee.

EPE=560(e)24[br]e=L2[br]EPE=560(L2)24[br]EPE=140(L2)2[br]EPE=140(L24L+4) EPE = \dfrac {560(e)^2}{4}[br]e=L-2[br]EPE = \dfrac {560(L-2)^2}{4}[br]EPE = 140(L-2)^2[br]EPE = 140(L^2 - 4L + 4)

Using conservation of energy the Initial GPE = EPE + KE + Final GPE . If we consider the maximum height to be L then equation becomes Initial GPE = EPE (the KE disappears because the final velocity is zero at the maximum length of the string)

Ok Initial GPE = EPE
20gL=140(L24L+4) 20gL=140(L^2 - 4L + 4)

and bish bash bosh you should get 5L227L+20=0 5L^2-27L+20=0


Thank you!

How are you finding M2 revision? Do you do any other modules?


Posted from TSR Mobile
Reply 91
How is everyone finding this unit?
Reply 92
Original post by CD223
Thank you!

How are you finding M2 revision? Do you do any other modules?


Posted from TSR Mobile


Tricky question since I've not done much recently :biggrin:. But it's definitely the hardest module out of the ones I'm doing (FP1, M1 and M2). Actually come to think of it, it's the hardest out of all the modules I've done (D1-2 and C1-4).
You've posted some good questions which keeps me on my toes. Keep them coming :woo:
Reply 93
Original post by Tiwa
How is everyone finding this unit?


Harder than C3 and C4 to get the knack of to be honest! You?


Posted from TSR Mobile
Reply 94
Original post by Shadez
Tricky question since I've not done much recently :biggrin:. But it's definitely the hardest module out of the ones I'm doing (FP1, M1 and M2). Actually come to think of it, it's the hardest out of all the modules I've done (D1-2 and C1-4).
You've posted some good questions which keeps me on my toes. Keep them coming :woo:


I totally agree! Harder than C3 and C4 I think!

Thanks for your help - if I see another tricky one I'll post it here :wink:


Posted from TSR Mobile
Reply 95
Original post by CD223
Harder than C3 and C4 to get the knack of to be honest! You?


Posted from TSR Mobile

Agreed.
Reply 96
Original post by Tiwa
Agreed.


Have you done many papers? I feel I've done quite a few "older" ones but still can't get it right!


Posted from TSR Mobile
Reply 97
Original post by CD223
Have you done many papers? I feel I've done quite a few "older" ones but still can't get it right!


Posted from TSR Mobile


About 5?

btw, could you help me with c part 2 of this question?
Reply 98
Original post by Tiwa
About 5?

btw, could you help me with c part 2 of this question?


Attachment not found

ImageUploadedByStudent Room1430417077.990619.jpg


Posted from TSR Mobile
Reply 99
Original post by CD223
Attachment not found

ImageUploadedByStudent Room1430417077.990619.jpg


Posted from TSR Mobile


Thanks so much! But I don't understand where the angle 70 comes from...

Quick Reply

Latest

Trending

Trending