OCR MEI AS Mathematics M1 10/06/2013

p.s. sorry about my bad handwriting. I was rushing.

Reply 181

Axion

15

Original post by jelmes96

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.

Bro, no joke. It said magnitude and in all past papers they require the exact number. Not trying to take the mick or anything :/

Reply 182

Original post by Axion

Bro, no joke. It said magnitude and in all past papers they require the exact number. Not trying to take the mick or anything :/

My teacher said it will be fine, and if it isn't, oh well. It's just one mark xD

Reply 183

Well, here it is guys...

http://db.tt/STbgP2wY

Reply 184

Axion

15

Original post by jelmes96

My teacher said it will be fine, and if it isn't, oh well. It's just one mark xD

lol why does everyone always try and cite a teacher to back up their claims :wink:

. Not saying you are lying, but in general when 80% of them are lying :P

True :P, i've lost lots of one marks all over the paper!

Thanks for the paper btw bro! you may have lost two marks on question 1 thoguh as you didn't separate the normal reaction force from below and above into 7g and 9p respectively

Repped you :smile:

(edited 10 years ago)

Reply 185

For question 4, xi = yj (i.e 2i = 2j) if the resultant is 045 degrees right? So (16-t^2) = (31-8t) which gives 0=t^2-8t+15. Well if you solve this quadratic, you get to positive values, however the answer is looking for one? Have I done something wrong, or is the question worded incorrectly?

Couldn't do any of 7 or 8 after part iii.

LOL

*cries*

Also, and this honestly isn't an excuse for my sub-C grade, our teacher for Mechanics has been absolutely awful, and everyone in my class who left that exam was either crying, shouting or just like YOLO. If there is a substantial difference with our class' results and everyone else, will something be done? We're considering making a formal complaint. We weren't taught 3D vectors, tension or F=ma stuff whatsoever.

According the the unofficial mark scheme, I got less than 50%, that's for sure.

(edited 10 years ago)

Reply 186

Original post by jelmes96

Well, here it is guys...

http://db.tt/STbgP2wY

Reply 187

JG1027

wasnt one of the forces for part 5 -10newtons?

Reply 188

Axion

15

Original post by ryanm95

For question 4, xi = yj (i.e 2i = 2j) if the resultant is 045 degrees right? So (16-t^2) = (31-8t) which gives 0=t^2-8t+15. Well if you solve this quadratic, you get to positive values, however the answer is looking for one? Have I done something wrong, or is the question worded incorrectly?

Couldn't do any of 7 or 8 after part iii.

LOL

*cries*

you are correct, but if you sub the values 3 and 5 back in, you get one set of positive is and js with 3 and one set of negatives is and js with 5

hence you take 3 as the answer

Reply 189

Original post by Axion

you are correct, but if you sub the values 3 and 5 back in, you get one set of positive is and js with 3 and one set of negatives is and js with 5

hence you take 3 as the answer

Well s**t

Reply 190

Original post by JG1027

wasnt one of the forces for part 5 -10newtons?

I got 10 and 50, 10 if it's going right, 50 if it's going left. The acceleration didn't specify a direction, so either could be positive I thought.

Reply 191

stefanocattaneo

Original post by jelmes96

Well, here it is guys...

http://db.tt/STbgP2wY

Looks good to me, though I think you need the angle in the triangle.

44/72 for me :frown:

Posted from TSR Mobile

(edited 10 years ago)

Reply 192