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AQA Mechanics 1B 24 May 2012 Unofficial Markscheme

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Original post by karchun
+1

This was posted from The Student Room's Android App on my GT-N7000


+1?
Reply 21
Original post by 1platinum
Can't you do v=u+at?


you do you use v = u+at, and then you equate i = -j values from the equation to work out that time is 20. Then you substitute back into equation to get v = i -j and then to find speed you do root(1^2 + 1^2) which gives you root2 or 1.41
Reply 22
Original post by darkshadow1111
No, this is not trial and error. You dont just put as many answers as you want and then get the marks.


oh noooo.... ewbsfbgssjk im an idiot

edit: oh wait no i remembered i didnt put several answers but i put different working outs so are my method marks invalidated too>??
(edited 11 years ago)
Original post by louisjevans
Use v=u +at, sub in u and a

i component equals negative j component because its SE.

So t = 20
Therefore v= i - j

Speed = root (1^2 + 1^2)
= 1.41





This was posted from The Student Room's iPhone/iPad App

I got t=40 for some reason.
Reply 24
Whats the grade boundaries usually for mechanics?
Reply 25
Original post by louisjevans
Use v=u +at, sub in u and a

i component equals negative j component because its SE.

So t = 20

Therefore v= i - j

Speed = root (1^2 + 1^2)
= 1.41





This was posted from The Student Room's iPhone/iPad App


I equated i = -j, then i factored out the t and i'm pretty sure i got t = 40 seconds. Then speed as 5.86 or something....
Reply 26
can someone remember question 6? like values or exact question, I jsut want to make sure I got c) right, because I forgot what i wrote for it. Would appreciate if someone would upload the question paper.
Original post by tom472
I equated i = -j, then i factored out the t and i'm pretty sure i got t = 40 seconds. Then speed as 5.86 or something....


I think my friend got that too, but we think he used the position equation rather than velocity?
Original post by tom472
I equated i = -j, then i factored out the t and i'm pretty sure i got t = 40 seconds. Then speed as 5.86 or something....


Exactly what i done and got. Hopefully were right. My teacher is going to do the exam in 2 hours so i will post the right answers.
Original post by Miyata
you do you use v = u+at, and then you equate i = -j values from the equation to work out that time is 20. Then you substitute back into equation to get v = i -j and then to find speed you do root(1^2 + 1^2) which gives you root2 or 1.41


I got 5.88 though?
Reply 30
Original post by tom472
I equated i = -j, then i factored out the t and i'm pretty sure i got t = 40 seconds. Then speed as 5.86 or something....


Did you make new equation v = u + at? You might have used equation from the previous question which was for r.
Reply 31
I didn't know how to find t :frown: But i knew how to do the next 6marker but i needed t in order to do it FML
Original post by darkshadow1111
I got t=40 for some reason.


-1-0.1t = -(3-0.2t)
0.1t = 2
T = 20




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Reply 33
Original post by louisjevans
-1-0.1t = -(3-0.2t)
0.1t = 2
T = 20




This was posted from The Student Room's iPhone/iPad App


OMG i Can't believe i didn't realise that! :angry:
Reply 34
I think there was a question before it where you used v=u+at to find a time when its travelling east of the origin. I used s=ut+1/2at^2 for the last one though, i may be wrong gah. i thought it was an ok paper overall.
Original post by darkshadow1111
Exactly what i done and got. Hopefully were right. My teacher is going to do the exam in 2 hours so i will post the right answers.


That would be delightful :smile:


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Reply 36
Original post by tom472
I think there was a question before it where you used v=u+at to find a time when its travelling east of the origin. I used s=ut+1/2at^2 for the last one though, i may be wrong gah. i thought it was an ok paper overall.


Last one asked you to find Speed. To find speed you find v, and then square root i and j component and root it. The first part asked for s though. and more precisely it is r, not s but w/e.
(edited 11 years ago)
For part 1(b) isn't the bearing 022 degrees rather than 338 degrees?
Original post by darkshadow1111
Werent those two minuses.
Plus 7c i got 5.38


Same i got 5.38 for the speed when it is travelling south-east
Original post by 1platinum
For part 1(b) isn't the bearing 022 degrees rather than 338 degrees?


22 degrees anticlockwise from north

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