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# AQA Mechanics 1B 24 May 2012 Unofficial Markscheme Watch

1. (Original post by karchun)
+1

This was posted from The Student Room's Android App on my GT-N7000
+1?
2. (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
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>??
4. (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.
5. Whats the grade boundaries usually for mechanics?
6. (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....
7. 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.
8. (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?
9. (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.
10. (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?
11. (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.
12. I didn't know how to find t But i knew how to do the next 6marker but i needed t in order to do it FML
I got t=40 for some reason.
-1-0.1t = -(3-0.2t)
0.1t = 2
T = 20

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14. (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!
15. 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.
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

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17. (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.
18. For part 1(b) isn't the bearing 022 degrees rather than 338 degrees?
Werent those two minuses.
Plus 7c i got 5.38
Same i got 5.38 for the speed when it is travelling south-east
20. (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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