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# Unofficial M1 Mark Scheme?

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1. for question 2 part b, i got 5.15N because the question was about the force on the scale pan, so i thought it would be:R-0.5g = 0.5 X 0.5which would give you 5.15 im not sure though im slightly confused :/
2. can i express answers in terms of g because i don't border to multiply it.
3. does anyone remember what the question 7 part b was? ive done part a and need to upload part b
4. (Original post by Jackhawkins21)
My teacher reckons it will be
A=60
A*=66
70=100ums
I think thats quite low, considering the paper was a lot easier than last year, maybe 62-63 for an A?
5. (Original post by KloppOClock)
does anyone remember what the question 7 part b was? ive done part a and need to upload part b
resultant force on a pulley
6. (Original post by lai812matthew)
can i express answers in terms of g because i don't border to multiply it.
yes
7. Guys how many marks do you think I would have lost for the following:

1c. I found t=32 but got something like 640i 320j as the final answer

7a. I got the value of 3/2 for lambda but again got the wrong answer overall due to a careless error
8. (Original post by lai812matthew)
resultant force on a pulley
that was question 8, im on about the details of the question about 7b where you had to work out velocity of a vector at a time
9. How did u get 225 for bearing, isnt it 45 as tension was acting east and north so resultant would be in north east which is 45 as bearing.

Posted from TSR Mobile
10. (Original post by Major-fury)
yeah that's fine you'll get full marks but could you help me with question 2 part b please? stating the method ty.
Thanks

For 2 part b what you had to do - it was worded very funny but you had to work out the R force from the scale pan upwards which was,
Resolving upwards : R-1.5g = 1.5a
R= 1.5g + 1.5a
= 1.5*9.8 + 1.5 * 0.5
= 14.7 + 0.75
= 15.45N
= 15.5N - 3 sig fig.
11. (Original post by Assmaster2)
Thanks

For 2 part b what you had to do - it was worded very funny but you had to work out the R force from the scale pan upwards which was,
Resolving upwards : R-1.5g = 1.5a
R= 1.5g + 1.5a
= 1.5*9.8 + 1.5 * 0.5
= 14.7 + 0.75
= 15.45N
= 15.5N - 3 sig fig.
yeh i see what i did wrong by accident fk ... should ahve changed the mass
12. (Original post by KloppOClock)
that was question 8, im on about the details of the question about 7b where you had to work out velocity of a vector at a time
Yes, here it is (according to what you have labelled as Q7 on the first page):

If the initial velocity is 3i - 22j and the acceleration is 3i + 9j, find the speed when t = 3s.

For the section where we are asked the velocities of P and Q as a function of time, I remember it was something like this:
P moves with velocity 15i + 20j and Q moves with velocity 20i-5j. At t = 0, P has position vector 400i and Q has position vector 800i. Find an expression for the position vectors of P and Q at time t.
13. (Original post by Maim56)
How many marks was 8b

Posted from TSR Mobile
4 marks
Yes, here it is:

If the initial velocity is 3i - 22j and the acceleration is 3i + 9j, find the speed when t = 3s.
thanks
15. (Original post by neek101)
for question 2 part b, i got 5.15N because the question was about the force on the scale pan, so i thought it would be:R-0.5g = 0.5 X 0.5which would give you 5.15 im not sure though im slightly confused :/
You had to use the mass of the brick, not the steel thing. Or maybe you just read the masses the wrong way around?
16. did everything correct for the moments except i supidly did 6-d not 4-d....how many marks will i have lost you think?.....also anyone got the final question....I cant remember my answer
17. (Original post by suhaylpatel786)

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

2(a) T = 20.6N
2(b) 15.45 N => 15.5 N (3sf)

3(a) Acceleration = -g/8
velocity after rebound = 3.5
thus Impulse = 3 Ns

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

5) μ = 0.73

6) 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

7(a) F2 = 2.5i + 2.5j (this required simultaneous equations)
(b) V = 12i +5j, thus speed = 13 ms-1

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
Hello, for question 3 I put 1.225 for acceleration but I was aware that it's actually a deceleration of -1.225 and I did get the correct answer. Do you think I'll lose one or two marks for that? All the other answers are the same as urs.
18. For question 1c I equated the i's instead of the j's! stupid mistake I know, but if I did it right do you guys reckon I'd get any marks?
19. If u get below an E (40ums) does that mean u get automatically 0 UMS (U) and it counts as if u've never done the exam?
20. (Original post by KloppOClock)
IMPORTANT NOTE: After speaking with some maths teachers, they have said that they believe the grade boundaries will be relatively low due to the simultaneous vector question and question three!

*So people stop asking, when using 'g' 2 or 3 significant figures is fine.

Question 1: (10 marks)
Spoiler:
Show
1(a) 104, (3 marks)

1(b) p = (400+15t)i + (20t)j (3 marks)
q = (20t)i + (800-5t)j

1(c) Q = 640i + 640j (4 marks)

Question 2: (6 marks)
Spoiler:
Show
2(a) T = 20.6N (3 marks)
2(b) 15N or 15.5N (3 marks)
Question 3: (7 marks)
Spoiler:
Show
3 Ns (7 marks)

Question 4: (12 marks)
Spoiler:
Show
4(a) (4 marks)

4(b) T = 8.75 s (8 marks)

Question 5: (10 marks)
Spoiler:
Show
μ=0.73 or μ=0.727 (10 marks)

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 (4 marks)
M = 42kg (3 marks)
Question 7: (11 marks)
Spoiler:
Show
7(a) F2 = 2.5i + 2.5j (this required simultaneous equations) (7 marks)
(b) V = 12i +5j, thus speed = 13 ms-1 (4 marks)
Question 8: (12 marks)
Spoiler:
Show
8(a) Fmax = 0.3g
acceleration = 0.6g
Tension = 11.76N = 11.8 N (3sf) (8 marks)

8(b) RF = √(11.76)^2 + (11.76)^2 = 16.6
It has a bearing of 225
or
45 degree angle between horizontal plane and vertical rope
or
draw a diagram and annotate angles (4 marks)
Model Answers: (click spoiler to show images).
Lower than 59 last year? I think I got 61ish, do you think that's enough to get me an A?

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