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OCR MEI AS Mathematics M1 10/06/2013

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Original post by jelmes96
Ok it's really annoting me, people thinking that 7.65s (my answer) was too long. I'm sorry, but let's think about it.

It's already going at 8ms^-1. There's an acceleration against motion, at -gSin15. so, it's got to go at 0ms^-1, which would take about 3. odd seconds. Then, it's got to travel downwards, at 13m. From the first 3 seconds, it's going to go slightly further than 13m. So to you, who put 7.67, WELL DONE! :biggrin: :P


thanks for that :biggrin: my rounding was a little off i think but yeah 7.67 ish 7.65 is probably exactly right
Reply 161
Avatar for jelmes96
jelmes96
3
S = 321 N
T = 80.4 N

if angle alpha is 5.71, Tan5.71 = roughly 0.1 = a / 30. Therefore a= 30 x 0.1 = 3m. Voila!
Reply 162
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Axion
15
Original post by JG1027
did you do pythagoras to find T and do tan to find the angle and - it from 90 degrees?


yeah pythag for 2 but sine rule to show everything else

cant remember though
Reply 163
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jelmes96
3
Original post by jg1027
did you do pythagoras to find t and do tan to find the angle and - it from 90 degrees?


sqrt (250^2 + 25^2) = 251 n
Reply 164
Avatar for Axion
Axion
15
Original post by jelmes96
S = 321 N
T = 80.4 N

if angle alpha is 5.71, Tan5.71 = roughly 0.1 = a / 30. Therefore a= 30 x 0.1 = 3m. Voila!


yeah i got those two forces.

Howeverm I did sine rule to find it was 3m
Reply 165
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JG1027
Original post by jelmes96
sqrt (250^2 + 25^2) = 251 n
so basically we found c ? :smile:
Reply 166
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jelmes96
3
Original post by JG1027
so basically we found c ? :smile:


Pretty much. That's why they wanted us to draw a triangle of forces.

For the first question, did people get 16g as the weight and reaction force?
I have a paper I can post, just to confirm, am I allowed to?
Reply 168
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jelmes96
3
Original post by stefanocattaneo
I have a paper I can post, just to confirm, am I allowed to?


Yes you can. MEI M1 is only taken at 9.00 UK time.
Going by TSR guidelines: "Exams from the examination boards AQA, OCR, CCEA, WJEC and CIE (other than orals and practicals) are fine to talk about straight after the exams. In any situations where exams from these boards are taken internationally, the exam board is satisfied that appropriate controls are already in place."

Here is the paper :smile: If someone could get going on an unofficial Mark Scheme that would be fab x

Since its too big for TSR, you can download it here: https://dl.dropboxusercontent.com/u/248326/M1%20June%202013.pdf
Reply 170
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jaetpa
1
Original post by stefanocattaneo
Going by TSR guidelines: "Exams from the examination boards AQA, OCR, CCEA, WJEC and CIE (other than orals and practicals) are fine to talk about straight after the exams. In any situations where exams from these boards are taken internationally, the exam board is satisfied that appropriate controls are already in place."

Here is the paper :smile: If someone could get going on an unofficial Mark Scheme that would be fab x

Since its too big for TSR, you can download it here: https://dl.dropboxusercontent.com/u/248326/M1%20June%202013.pdf


Cheers for this :smile: on another note has anyone got the D1 paper, haven't seen that one surface yet
Reply 171
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jelmes96
3
I'm on it. It will take about 45 minutes
Reply 172
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Axion
15
Original post by jelmes96
Pretty much. That's why they wanted us to draw a triangle of forces.

For the first question, did people get 16g as the weight and reaction force?


Up : 9g and 7g

Down 16g

You had to write the numerical value for each (i.e. substitute g for 9.8)
what do we reckon grade boundaries will be like??
Original post by rachel1508
what do we reckon grade boundaries will be like??


Judging by the general reaction to the paper, I reckon somewhere along the lines of June 2006:

A - 55
B - 47
C - 40


Posted from TSR Mobile
Reply 175
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qwerty1D
Original post by jelmes96
S = 321 N
T = 80.4 N

if angle alpha is 5.71, Tan5.71 = roughly 0.1 = a / 30. Therefore a= 30 x 0.1 = 3m. Voila!

i put t = 80.66 obviosuly due to rounding etc do u think this will be correct
Reply 176
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jelmes96
3
Give me 5 minutes to upload
Original post by jelmes96
Give me 5 minutes to upload


Cheers mate :smile:


Posted from TSR Mobile
Original post by jelmes96
Give me 5 minutes to upload


thanks :smile:
Reply 179
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jelmes96
3
Original post by Axion
Up : 9g and 7g

Down 16g

You had to write the numerical value for each (i.e. substitute g for 9.8)


Well, g is a constant, as 9.8, so having put 16g, won't matter. 16g is a number, it's just neat :biggrin: On the papers they use g as a constant, so why can't we? Anyway, I've had some problems, but it's uploading now.

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