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2016 Unofficial M1 Mark Scheme? 2016

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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 :/
(edited 7 years ago)
can i express answers in terms of g because i don't border to multiply it.
does anyone remember what the question 7 part b was? ive done part a and need to upload part b
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?
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
Original post by lai812matthew
can i express answers in terms of g because i don't border to multiply it.


yes
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
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
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
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 :smile:

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.
Original post by Assmaster2
Thanks :smile:

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
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.
(edited 7 years ago)
Original post by Maim56
How many marks was 8b

Posted from TSR Mobile


4 marks
Original post by hadtosignup
Yes, here it is:

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


thanks
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?
(edited 7 years ago)
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
Original post by suhaylpatel786
These are my answers:

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.
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?
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?
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! :banana2:

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

Question 1: (10 marks)

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Question 2: (6 marks)

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Question 3: (7 marks)

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Question 4: (12 marks)

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Question 5: (10 marks)

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Question 6: (7 marks)

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Question 7: (11 marks)

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Question 8: (12 marks)

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Model Answers: (click spoiler to show images)

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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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