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# Unofficial M1 Mark Scheme?

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1. for question 2 part b i got 15.7N because i did F=2(0.5)+1.5(9.8), how many marks is that
2. (Original post by E1Ephan7)
I completely guessed 7a) F=2.5i and forgot to write j component. Showed no working. How many marks could i get??
Zero
3. Any working out for the 0.74 question
4. Marks:
Question 1 A -3, B-3, C-4
Question 2 A-3, B-3
Question 3 A-7
Question 4 A-4, B-8
Question 5 A-10
Question 6 A-7
Question 7 A-7, B-4
Question 8 A-8, B-4
5. I got 0.65 for the coefficient of friction wtfff i thought i aced it
6. how many marks was 1c worth?
I got 0.65 for the coefficient of friction wtfff i thought i aced it
And I got 0.873
8. How many marks would I lose for 3sfgs on the coefficient of friction for q5??? i.e 0.727
9. (Original post by VA123)
was μ meant to be given to 2sf or is 3sf ok?
3is fine, i gave it to 3 as well
10. (Original post by student bang)
And I got 0.873
Rip. At least were not the only ones
11. (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:
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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:
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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

Woo think I only lost about 5 marks,
12. (Original post by student bang)
Good luck in life with this attitude
Good luck in life with that name.

13. What do people estimate the raw mark to be for 90 or 100 UMS
14. (Original post by Verish)
Its an AS module, you cant get an A*?
Probably means 90 UMS.
15. Anyone has the actual questions?
16. (Original post by geniuses)
Dear Failures,

I have just realised i have achieved 100% in this exam.

Good luck next year!
Same, but I think its a bit rude to call people failures. I'm sure loads of people did really well and tried their best.
17. (Original post by Ilyas_99)
Same, but I think its a bit rude to call people failures. I'm sure loads of people did really well and tried their best.
You and me Bro, we will rule the 7th dimension.
18. All you had to do is 2Tcos(45), as the forces where at right angles to each other.
(Original post by vvlp)
For the magnitude of the force on the pulley I didn't do root T^2 + T^2 because i got confused when it said work out the resultant magnitude and it messed me up. I thought the resultant force was the root of ((T-friction)^2 + (weight - T)^2)

Was this stupid of me? I thought 11.47^2 with an angle of 45 from the horizontal was too simple for 4 marks.
(Original post by vvlp)
For the magnitude of the force on the pulley I didn't do root T^2 + T^2 because i got confused when it said work out the resultant magnitude and it messed me up. I thought the resultant force was the root of ((T-friction)^2 + (weight - T)^2)

Was this stupid of me? I thought 11.47^2 with an angle of 45 from the horizontal was too simple for 4 marks.
19. (Original post by geniuses)
You and me Bro, we will rule the 7th dimension.
I'm a girl

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