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# Unofficial M1 Mark Scheme?

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1. (Original post by Jacob2215)
How many marks would I lose for 3sfgs on the coefficient of friction for q5??? i.e 0.727
none
2. 45 degrees isn't a direction, just a size, a bit like a scalar
(Original post by SirRaza97)
Yo I think the very last question was not a bearing. So it would be 45 degrees
3. (Original post by Michal-sweaty-g)
45 degrees isn't a direction, just a size, a bit like a scalar
Well 45 degress to the vertical/horizontal axis. That is a direction. I just drew an arrow in the diagram and put 45 degrees.
4. (Original post by mk_98)
I said force acts at 45 degrees below horizontal. Would that be enough?
no because you need to include the direction, which can be shown in a diagram or say pointing to south west direction, so i think you lost a mark for that
5. (Original post by SirRaza97)
Well 45 degress to the vertical/horizontal axis. That is a direction. I just drew an arrow in the diagram and put 45 degrees.
"45 degrees from the horizontal table" can be south west or north east, i think the examiners want to see "south west" or a diagram with an arrow
6. (Original post by suhaylpatel786)
...
Agreed.
7. (Original post by KloppOClock)
i thought it was an okay paper, maybe 64 for an A?

what do you guys think?

also, does anyone remember the mark distribution of the questions?
I think it was along the lines of ;

1a)
b)
c)

2a) 3marks
b) 3marks

3) 8marks

4a) 4marks
b) 8marks

5) 10marks

6)

7a) 7marks
b) 4marks

8) 8marks
b) 4marks

8. (Original post by ChilliLemon)
For question 2) (b)

I wrote the reaction force is equal to 15.45N.
Buth then i wrote the force exerted on the steel pan is 15.45 - 1.5g, so i took away the weight of the brick, and got 0.75N.

Will i stil get full marks because i wrote dwon that R=15.45N, though this was not my final answer?
no i am sure that you wont get full mark for that, because you havent rounded up to 2 or 3 sf.
9. For the one where you had to calculate friction (mu), I put 0.727 to 3 s.f. (i think). Is that right guys?
10. (Original post by thelegend99)
For the one where you had to calculate friction (mu), I put 0.727 to 3 s.f. (i think). Is that right guys?
yes you are correct.
11. As far as I can tell I got it all right except the very last q in which I put southeast as the direction instead of southwest.. 73/75?
13. Please can someone post full working for the moments question?
Im trying to work our where I messed up as I got d=2 and M=90
Thank you
14. 1a) find bearing 3
b) equations 3
c) position vector 4

2a) tension in string 3
b) reaction on thing inside the pan 3

3) particle question, 7

4a) sketch speed time graph, 4
b) solve for T, 8

5) Find coefficient of friction,10

6) moments, 7

7a) find F2, 7
b) find speed, 4

8) find T, 8
b) resultant on pulley, 4
15. (Original post by geniuses)
You and me Bro, we will rule the 7th dimension.
yeah you can use those 2 inchers to have a sword fight over allegedly getting 100% ;D
16. (Original post by KloppOClock)
Message me if anything is wrong.
Also, if anyone remembers how many marks each question was worth, please comment so I can add them on too.

Question 1:
Spoiler:
Show
1(a) 104, (90+14)

1(b) p = (400+15t)i + (20t)j
q = (20t)i + (800-5t)j

1(c) The 'j' vectors for both are equal, as it is due west.
800-5t = 20t
t = 32
therefore q = 640i + 640j
Question 2:
Spoiler:
Show
2(a) T = 20.6N
2(b) 15.45 N => 15.5 N (3sf)
Question 3:
Spoiler:
Show
3 Ns
Question 4:
Spoiler:
Show
4(a)
4(b) Area under graph = 975
Area for slower car for first 25 seconds = 750
975-750 = 225
1/2 * b * 30 =225
b = 15
total time = 15+25 = 40
so area under faster car = 975 = 1/2 * (40)(T+40)
T = 8.75 s
Question 5: (10 Marks)
Spoiler:
Show
μ=0.73 to 2 significant figures
Question 6: (7 Marks?)
Spoiler:
Show
For 1st situation, R(T) = 0
For 2nd situation, R(S) = 0

M(S) => 0.5M = 30d-15
M(T) => M = 60 - 15d
Therefore,
d = 1.2m
M = 42kg
Question 7:
Spoiler:
Show
7(a) F2 = 2.5i + 2.5j (this required simultaneous equations)
(b) V = 12i +5j, thus speed = 13 ms-1
Question 8:
Spoiler:
Show
8(a) Fmax = 0.3g
acceleration = 0.6g
Tension = 11.76N = 11.8 N (3sf)

8(b) RF = √(11.76)^2 + (11.76)^2 = 16.6
It has a bearing of 225
At 8, Did you have to write down the bearing? All I did was work out the Tension and the second part I used 2Tcos(45)
17. What do u think the boundary will be for an A?

I think 59-60
18. (Original post by examhater44)
What do u think the boundary will be for an A?

I think 59-60
yeah between 59-62 for an A, was easier than last year in my opinion but the second vectors question might have caused a few problems. Maybe 72/73 for 100 UMS
19. Can some one remember what question 2 was especially the working out for 2b?
20. (Original post by KloppOClock)
i thought it was an okay paper, maybe 64 for an A?

what do you guys think?

also, does anyone remember the mark distribution of the questions?
I would like to think it would be lower than you have predicted based on the last 3 years' papers where an A was about 59.

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