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# OCR (non mei) M2 Wednesday 18th May 2016 watch

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1. (Original post by swingrx)
Please sign this petition to lower the grade boundaries!
[

QUOTE=TheBride;64926843]Hitler Reacts
First, starting a petition is retarded. How well candidates did on average will dictate the grade boundaries, starting a petition so "everybody wins" is just whiny.

Secondly, posting a petition for a core 1 paper in an M2 thread, with the intent to trick people into signing a petition for a paper which they didn't even take, is scummy.
2. (Original post by marioman)
This thread is for the M2 paper, not the C1 paper. Ignoring the fact that your petition is silly anyway, I'm sure that most M2 candidates wouldn't have had any trouble scoring a high mark had they sat C1 this morning.
It is in an attempt to trick people into signing the petition. He replies to comments with the petition wanting us to think it is an M2 petition, not to mention signing a petition for grade boundaries is retarded in itself.

Also hi! I remember you from the COMP1 thread last year.
3. (Original post by marioman)
This thread is for the M2 paper, not the C1 paper. Ignoring the fact that your petition is silly anyway, I'm sure that most M2 candidates wouldn't have had any trouble scoring a high mark had they sat C1 this morning.
Can confirm, resat C1 this morning (and found it decent) before getting absolutely rekt by the steaming pile of horse manure that was the M2
4. (Original post by Miningstew)
Using Xsanda’s solutions as a template:

1) Car on a plane:
i) v = 22.5 m/s [3]
ii) P = 35.9 kW [3]

2) Particle up a plane:
i) Wd = 8.43 J [3]
ii) v = 10.8 m/s [4]

3 Arch shape:
i) show that: 13a/3π [5]
ii) T = 8.82 N [3]
iii) θ = 10.9º left of upwards vertical [4] (or 79.1º above the horizontal to the left)

4 Particle around a cone:
i) show that m/6 (2√3 g - 3aω²) [6]
ii) T = 39.6N, V =1.19 m/s [3]

i) show that 5W/24 (1+6x/a) [5]
ii) max x = 1.7a [3]
iii) tan = 39/14 [3]

6 collisions horizontally:
i) a: 5.5 m/s, b: 8 m/s, both in opposite direction to their original motion [8]
ii) e=3/4 [2]

7 projectiles:
i) 18.8m high, 61.1 m AB [4]
ii) 87.3 Ns [4]
iii) 11.9 m [9]
Good work you two. Only just had chance to do the paper and I concur with all of these except 1(ii) should be 35.8kW.

It was in my opinion an unnecessarily hard paper. I would guess some very low grade boundaries and, as mentioned above, there's nothing to stop the marks needed for an A dropping below 50; it's happened a few times before (in D1, FP2).
5. (Original post by Mr M (jr))
Good work you two. Only just had chance to do the paper and I concur with all of these except 1(ii) should be 35.8kW.

It was in my opinion an unnecessarily hard paper. I would guess some very low grade boundaries and, as mentioned above, there's nothing to stop the marks needed for an A dropping below 50; it's happened a few times before (in D1, FP2).
Did you think this one was harder than June 2013?
6. (Original post by marioman)
Did you think this one was harder than June 2013?
I think that overall the paper was harder, not because of difficulty of the questions, but because of time. I just about finished and had no time to check, and a lot of people didn't even finish, its getting rediculous.
7. (Original post by jacobe)
Correct, that's what I am doing. I got 98 in S2 and well...less in M2 so I am pretty sure you can switch em around.
A* criteria correct. Switching gets done automatically, by prioritising:

1. Highest grade possible in maths
2. Sticking to 1, highest grade possible in further maths
3. Sticking to 1 & 2, highest UMS in maths.

Only module restrictions are C's go in maths, FP's go in further maths and if you have, for example, M2 counting towards maths then you must also have M1 counting towards maths.
8. (Original post by marioman)
Did you think this one was harder than June 2013?
My guess is most would find it harder than June 2013. Certainly it was more time (and paper) consuming. In fact I used nearly twice as much paper for this one. The only sort of hard question in June 2013 not in this one was the lamina at an angle with both components of the weight contributing to the moments.
9. I don't suppose anyone could show me a solution for the last part of question 7?
10. For the last part of the last question, I got something like 0.373 something like this..... And I am pretty sure I did make every steps correct, how can you get 11.6?
11. (Original post by himeandme)
For the last part of the last question, I got something like 0.373 something like this..... And I am pretty sure I did make every steps correct, how can you get 11.6?
I had that initially and then I went back and changed it and I can't remember at all what I did
12. Guys can anyone remember question 2?
13. (Original post by Rordy1999)
Guys can anyone remember question 2?
You needed to calculate the WD by the friction in part one.
Then use that and the final velocity to work out the initial velocity.
14. Can someone please help me understand question 4. I know I got part 1 correct, and I got the tension = 39.6N, but I cant remember how to workout or what I got for the velocity. can someone explain the method please
15. (Original post by Mandos)
You needed to calculate the WD by the friction in part one.
Then use that and the final velocity to work out the initial velocity.
Thanks but i need to remember the numbers given in the question
16. (Original post by Mandos)
Can someone please help me understand question 4. I know I got part 1 correct, and I got the tension = 39.6N, but I cant remember how to workout or what I got for the velocity. can someone explain the method please
The greatest possible speed with the particle still touching the cone is when Normal reaction force (R) is equal to 0. So using newtons second law, Tsin30 = m(v^2)/r where m is mass and r is radius. Then just solve for v.
17. How is it that in this paper question 6(i) was worth 8 marks when in the Jan 2010 Question 5, the exact same question was worth 12 marks? How will they distribute the marks out for the question this time?
18. (Original post by EuanF)
For Q3, I got to 20/4+pi which I checked and saw was numerically the same on a calculator, so i just went from 20/4+pi to 13/3pi

Reckon I got the mark?
I'm going to assume you meant 20/(4+pi), which is not the same as 13/(3pi), therefore you will not get the mark.
19. (Original post by SamDavies1998)
I'm going to assume you meant 20/(4+pi), which is not the same as 13/(3pi), therefore you will not get the mark.
No, it is. Did 13/3pi - that and got 0.

or at least i did something similar and got 0
20. (Original post by EuanF)
No, it is. Did 13/3pi - that and got 0.

or at least i did something similar and got 0
Type both into a calculator and you'll see they aren't.

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