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# Unofficial M1 Mark Scheme?

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1. Definitely No Arsey?
2. (Original post by samma1)
hey, can you just remind me what question 2 was? i cant remember....
A lift accelerating with a mass inside. Had to find tension for part A and Normal reaction of the mass for part B
3. Im thinking i got low 60s, hopefully an A
4. was μ meant to be given to 2sf or is 3sf ok?
5. If anyone knows the marks for each question could you please post them

thanks
6. (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

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?
7. can anyone remember question 8
8. (Original post by VA123)
was μ meant to be given to 2sf or is 3sf ok?
I give it to 2 sf because I used the value for g as 9.8 in my calculations when finding the weights of the mass which is also to 2 sf
9. 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.
10. So on the last question i completely missed out the fact it said 'and direction' so i didn't say southwest or 225degs or anything but I did just out of chance draw the arrow in the right direction on the diagram that they provided. Does this get you any marks or no?
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:
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
Q1 10 marks
2 is 6
3 is 7
4 is 12
7 is 11
8 is 12
12. (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
I said North East for 8b get the marks ?
13. (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
Yo I think the very last question was not a bearing. So it would be 45 degrees
14. (Original post by kprime2)
I'll be posting model answers tomorrow (edexcel)
Thanks kprime2. Good man.
15. (Original post by marshmellow:))
I give it to 2 sf because I used the value for g as 9.8 in my calculations when finding the weights of the mass which is also to 2 sf
Ah ok, hopefully they'll allow it!
16. I gave μ=0.7, will this lack of significant figures make me lose marks?
17. could someone show the working for 5 bc i got 1.07 for μ and obviously im wrong but i have no idea what i did lmao ;;A;;
18. I completely guessed 7a) F=2.5i and forgot to write j component. Showed no working. How many marks could i get??
19. (Original post by Jackhawkins21)
My teacher reckons it will be
A=60
A*=66
70=100ums
Its an AS module, you cant get an A*?
20. Guys,
How did u do q 2b?

It said find the force on the pan exerted by the brick, not the other way around.
i got the answer to be 13.95 (or 14). Some other people also got this.

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