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# OCR M3 (non MEI) Wednesday 8th June watch

1. Can't remember which questions were which answers but I remember a couple I put down, no idea if they're right or not:
e>5/9 for spheres, CoF=0.495 for R=73/50W question, Amplitude of SHM on 7(iv)=0.675 and centre of oscillation where OP is 0.7, (extension=0.1)
U<(5ag)^(1/2) for vertical circle,
Max KE=1.04J of P between two strings. Can't remember any other questions, think I was 1.43 in Q1, (Downwards for 1(ii))
2. Well that was quite a tough paper, thankfully only just managed to finish it (meaning silly mistakes will likely cut me down), managed to get R=73/50 N however unsure on my answer for set of values for U on the circular motion question, I got U>2root(ag) however this is the initial speed of the particle and therefore not likely to be correct. also got time = 0.70..(s) for the time on last question. Anyone get similar answers?
3. I dropped about 10-12 marks I think. Mainly on the SHM question. I understand all the theory and always make mistakes inevitably when I do the big SHM question. I'll read the mark scheme and instantly know where I went wrong and I know the maths and everything behind the right answer, I just always make mistakes on the SHM question, and did once again today.
4. (Original post by Simonium1010)
I dropped about 10-12 marks I think. Mainly on the SHM question. I understand all the theory and always make mistakes inevitably when I do the big SHM question. I'll read the mark scheme and instantly know where I went wrong and I know the maths and everything behind the right answer, I just always make mistakes on the SHM question, and did once again today.
That's what's annoying, knowing all of the theory but timing/stress causes silly mistakes
5. (Original post by JoeyTr)
Can't remember which questions were which answers but I remember a couple I put down, no idea if they're right or not:
e>5/9 for spheres, CoF=0.495 for R=73/50W question, Amplitude of SHM on 7(iv)=0.675 and centre of oscillation where OP is 0.7, (extension=0.1)
U<(5ag)^(1/2) for vertical circle,
Max KE=1.04J of P between two strings. Can't remember any other questions, think I was 1.43 in Q1, (Downwards for 1(ii))
For Q1 i got I=0.0989N at 111 degrees to the original direction of motion.
6. (Original post by tobibo)
Well that was quite a tough paper, thankfully only just managed to finish it (meaning silly mistakes will likely cut me down), managed to get R=73/50 N however unsure on my answer for set of values for U on the circular motion question, I got U>2root(ag) however this is the initial speed of the particle and therefore not likely to be correct. also got time = 0.70..(s) for the time on last question. Anyone get similar answers?
Everyone I know got different answers for the final question, I got t=0.682 which I know is wrong. Also pretty sure I got u>7(a)^1/2 ...

I really gotta hope that it's like and 61 for an A* and 55 for 80 UMS.
7. (Original post by JoeyTr)
Can't remember which questions were which answers but I remember a couple I put down, no idea if they're right or not:
e>5/9 for spheres, CoF=0.495 for R=73/50W question, Amplitude of SHM on 7(iv)=0.675 and centre of oscillation where OP is 0.7, (extension=0.1)
U<(5ag)^(1/2) for vertical circle,
Max KE=1.04J of P between two strings. Can't remember any other questions, think I was 1.43 in Q1, (Downwards for 1(ii))
That max KE one, thank god, I only realised with time running out that KEmax when a=0 therefore T's cancel
8. (Original post by Nathanburns01)
For Q1 i got I=0.0989N at 111 degrees to the original direction of motion.
I got that
9. (Original post by Nathanburns01)
For Q1 i got I=0.0989N at 111 degrees to the original direction of motion.
Pretty sure it was 68.3 degrees to the original direction, 180 minus that is 111.7, but that's the angle to the opposite direction.
Had to be less than 90 cos the horizontal component of its velocity increased.
10. Did anyone get v=(15/2)sin(2t)+4?
11. well that was atrocious
12. Who didn't finish? I flopped that time wise
13. (Original post by cam5morrison)
well that was atrocious
Hate to be cynical but that's reassuring. As reassuring as one can find a comment after that atrocity...
If it helps, you're just joining the vast majority in thinking that. As long as 56/72 is an A i'm good.
14. (Original post by XY123)
Who didn't finish? I flopped that time wise
I did, but started the last 7-marker with 3 minutes so didn't get to go back and check the answer that I knew was wrong, and also meant that I rushed the last question and got the time and the whole question wrong.
15. (Original post by JoeyTr)
Good solid paper I thought, I think many people would've struggled with the R=73/50W and the last question. Didn't get the last one fully, only got the time of Q falling and the amplitude of their motion, but that was it before running out of time.
I got R=73/50W, what did you get for coefficient of friction? I got like 0.49 not sure if thats right.

Last one I has some problems with too, I got 0.7m below O for centre of motion.
16. (Original post by Miracle-)
I got R=73/50W, what did you get for coefficient of friction? I got like 0.49 not sure if thats right.

Last one I has some problems with too, I got 0.7m below O for centre of motion.
Think I got 36/73 for that, messed up amplitude for question 7 though, divided by l not 2l

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17. so for the last question, I balanced forces correctly and got acceleration = -392x, so w2=(392) - (think that's right), and then I got the suvat bit correct where you work out that the speed after they coalesce is 2.1ms-1, but after that, I tried to conserve energy and did so wrongly, getting an incorrect amplitude of 0.105-0.025= ~0.8, but then for the final step here's my working:

x=asin(wt), so x=asin(root(392)t)

but even though my value of a is wrong, surely the time at which the particle reaches its highest height is when x=a (it's at its amplitude and so x/a is just 1, meaning sin(wt)=1 and then wt= pi, 5(pi)/2, etc..., and the first case is when it's at the lowest point, so the highest point is (root(392))t = 5(pi)/2, and t= ~0.4, and adding on the 2/7 which is the initial time that Q falls for, I got T=0.682

However, as I right this I think maybe I should have considered x=-a too, meaning wt=3(pi)/2 would be the next solution, and so in hindsight I think the answer should have been 3(pi)/2 divided by root(392), = 0.238, + 2/7 = 0.524 ??

If that IS the answer, I'd imagine I would just lose 1 or 2 method marks, as well as the answer mark, and 4/7 is alright considering I did fine on questions 1, 2, 3, and 6.
18. (Original post by Nathanburns01)
Did anyone get v=(15/2)sin(2t)+4?
Yeah, I can't remember the average speed though
19. (Original post by drandy76)
Think I got 36/73 for that, messed up amplitude for question 7 though, divided by l not 2l

Posted from TSR Mobile
Yes same I got 36/73 thats 0.493.

Why would you divide by 2l for the amplitude?
20. (Original post by JoeyTr)
Yeah, I can't remember the average speed though
I got 8.77

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