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# OCR MEI AS Mathematics M1 10/06/2013 Watch

1. what did everyone get for the last bit when you had to find the time
2. (Original post by JG1027)
what did everyone get for the last bit when you had to find the time
7.67 seconds...
3. (Original post by JG1027)
what did everyone get for the last bit when you had to find the time
I defo did that question wrong. I got 12. Something seconds lol...
If the rope broke does that mean F=0?????
4. (Original post by johnny2spoons)
7.67 seconds...
What did you all get for the first question

I had 9g up, 16g down and a further 7g up
5. the girl couldn't get the 60 degree angle because it required a force greater than her weight so she would be lifted off the floor LOL
6. If I can remember yeah f=0 which gives you the new acceleration so you put in equation to find t, I think it was a quadratic. I got 7.66 or thereabouts
7. (Original post by Axion)
What did you all get for the first question

I had 9g up, 16g down and a further 7g up
yeah same here
8. How did everyone resolve the last question?
The sleigh weighed 40kg n the rope breaks at 8ms-1 n she is 13.0m up.
The rope breaks n your suppose to work out the time....
I used s= ut + 0.5at2
N used the quadratic formula....but got 12.summat :/
9. What do you think the grade boundaries will be like ?
10. (Original post by Jojo_555)
How did everyone resolve the last question?
The sleigh weighed 40kg n the rope breaks at 8ms-1 n she is 13.0m up.
The rope breaks n your suppose to work out the time....
I used s= ut + 0.5at2
N used the quadratic formula....but got 12.summat :/
F=ma to start where f=-40gsin15 and f=40a so -40gsin15=40a and then a=-2.5 something use xuvat

x=-13
u=8
a=-2.5 summat
t=t
then x=ut + 1/2at^2 sub in values and use quadratic formula.... about 7.67s i think...
11. (Original post by johnny2spoons)
F=ma to start where f=-40gsin15 and f=40a so -40gsin15=40a and then a=-2.5 something use xuvat

x=-13
u=8
a=-2.5 summat
t=t
then x=ut + 1/2at^2 sub in values and use quadratic formula.... about 7.67s i think...
so basically was the only force acting the component of weight down slope ?
12. (Original post by Axion)
What did you all get for the first question

I had 9g up, 16g down and a further 7g up
I got 7g down, 16g up and a further 9g down?

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13. the first question confused me i thought i had to draw extra boxes the wording should have been better
14. (Original post by JG1027)
so basically was the only force acting the component of weight down slope ?
yes
15. (Original post by johnny2spoons)
F=ma to start where f=-40gsin15 and f=40a so -40gsin15=40a and then a=-2.5 something use xuvat

x=-13
u=8
a=-2.5 summat
t=t
then x=ut + 1/2at^2 sub in values and use quadratic formula.... about 7.67s i think...
Like holy sh*t. I did all of that but I stupidly put x=13 :/
Cuz I was thinking of the equation S= So + ut + 0.5t2
My S being 0 cuz she goes back to her original position and So being 13
16. (Original post by johnny2spoons)
yes
i did that but i used v=u+at to find the time would that be wrong? because v=0 u=8 a=2.5 and t=? so rearrange and i got -3.2s so 3.2 s
17. Wait... i got like 1.44s on the last question

views on that?
18. Does anyone remember what they got for that vector equation thing?
Finding out the vector a?
I remember getting summat like (0.4, 14, -35) or something :/
19. (Original post by stefanocattaneo)
I got 7g down, 16g up and a further 9g down?

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0.o wrong way round bro...

btw I wrote the actual values rather than just writing 7g
20. (Original post by Jojo_555)
Does anyone remember what they got for that vector equation thing?
Finding out the vector a?
I remember getting summat like (0.4, 14, -35) or something :/
i got 7/0.4= 17.5

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