G481 Mechanics 20 May 2014 Unoffical mark scheme
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Original post by Red Fox
I kinda messed up, hope my UMS is 85 at least. What do you think you got?
Same sentiments as you
53 ish
Original post by L'Evil Fish
Same sentiments as you
53 ish
53 ish
50 is the absolute minimum I got, if mark scheme isn't extremely picky then maybe up to 2/3 more marks. To be fair, it could have gone worse
Original post by BrokenS0ulz
For the mass of the dinosaur I said it would be about 8 times smaller however me being stupid I said its mass will be "around 20kg" I had calculated it to be 16 kg but I thought this sounded too accurate T_T
haha same! I thought 16 was too accurate, like wth. I did put 16 down though. Damn that question was pants whoever thought it was a good idea is a legit moron. For real, there was a lot of stuff on the paper that were either worded weirdly, or was just plain queer. I hope they lower the grade boundaries (for everyones sake, of course...
)Original post by sagar448
For the mass of the stupid dinasaur I put down, "around 16kg" would that be fine??
Your lucky you wrote that, i just wrote "1/8th of the mass" -_-
Original post by teachercol
Most of my A2 students resat this paper. I'd expect them to get 50+ on it.
If the same happens elsewhere they are going to pull the boundaries up.
Overall I'd say it was harder than last year.
Maybe I'm just being cautious about the boundaries.
48 for an A used to be the norm and that's what the examiners are aiming at.
If the same happens elsewhere they are going to pull the boundaries up.
Overall I'd say it was harder than last year.
Maybe I'm just being cautious about the boundaries.
48 for an A used to be the norm and that's what the examiners are aiming at.
I was the only one resitting it from my school. Most other people are resitting G482 though.
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(edited 9 years ago)
Think I got a minimum of 49/60, think it'll be better than the D I got last year haha. Any idea how UMS marks work past an A? What would 49/60 give if an A is 45?
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Original post by I Persia I
Think I got a minimum of 49/60, think it'll be better than the D I got last year haha. Any idea how UMS marks work past an A? What would 49/60 give if an A is 45?
Posted from TSR Mobile
Posted from TSR Mobile
It depends on what a B was. If B was 41, A was 45 then 90% would be 49 and 100% would be 53.
Original post by sabre2th1
haha same! I thought 16 was too accurate, like wth. I did put 16 down though. Damn that question was pants whoever thought it was a good idea is a legit moron. For real, there was a lot of stuff on the paper that were either worded weirdly, or was just plain queer. I hope they lower the grade boundaries (for everyones sake, of course... )
)I actually did get 16.125kg but who would leave the answer like that -_- It just seems so stupid that we are estimating the mass of a dinosaur to that many significant figures :/ Examiners don't seem like nice people and I don't think they'll want to push the boundary for an A any lower than ~42
Original post by Sam596
Damn. I just realised I never calculated the CSA of the cable. I thought it gave the CSA in the question, not the radius...
Ohno
Original post by Sam596
Damn. I just realised I never calculated the CSA of the cable. I thought it gave the CSA in the question, not the radius...
If it makes you feel any better I got the KE question wrong on the first page, I kept the mass in kg but only multiplied the speed by 1000 and forgot to divide by 3600. I put in the equation and will probably get 1-2 mark for ECF we'll see.
I thought that question about the ball bouncing on the wall and the time being the same to be a bit of a bad question.
I got the last part of the last question wrong too, everything else 100% correct. I fluked the moments just plugged in tonnes of random combinations until I got something which seemed to resemble the right answer which i got as 0.25
Original post by teachercol
OCR G481 Mechanics Tues 20th May
My predicted boundaries
100% 55
A 45
B 40
C 35
D 30
E 25
Still got all to play for with the EWP paper. Get revising!
Col
My predicted boundaries
100% 55
A 45
B 40
C 35
D 30
E 25
Still got all to play for with the EWP paper. Get revising!
Col
Are these estimates at the highest you think they could be, or the lowest?
Original post by Kalu.m
Your lucky you wrote that, i just wrote "1/8th of the mass" -_-
Well if the examiner is generous then he/she may give you a mark. I mean there are a lot of "Alternatives" usually on the mark scheme.
what do you mean for 3ii ?
Original post by mgmf46
On the penultimate part of 6, I used the cosine rule to find the angle between the force and direction of motion and then used P=Fvcostheta. It gave the same answer, but would this method be allowed?
Can't see why not. It's a valid method.
After going through that I think the lowest I could have got is 50/51 and the highest is 56, I know there are 4 that I have definitely lost a minimum of 1 mark on. The last one was really annoying, I had the right logic, I just forgot about the 3N :/
Original post by teachercol
OCR G481 Mechanics Tues 20th May
Usual disclaimers. This is just my mark scheme. It is in no sense official. It may contain errors and typos – but I’d be pretty confident its pretty close to the actual mark scheme.
Overall I thought this paper was on the hard side. There weren’t many easy definition marks and plenty of places where folk will go wrong.
My students estimates varied from 30 to 57 out of 60. I’d guess that the A grade will be 44/45 ish and the E grade 25 ish. I don’t know what you’re going to get.
I’m not allowed to post a scan of the paper. Forbidden by OCR. Don’t ask.
Ok so here we go (NB 1.0 E7 means 1.0 x ten to the power of 7)
Q1 a) Velocity = rate of change of displacement (with respect to time ) (1)
b) i) 70 km / h = 70 x 1000/3600 m/s = 19.44 m/s
Ek = 1/ mv^2 = ½ x 130 x 19.44^2
= 2.46E4 J (3)
ii) Volume is 8 x smaller (2 x 2 x 2)
Density is same
So mass is 8 x smaller = 130/8 = (16.25 kg) (2)
TOTAL: 6
Easy question to kick off. Unit conversion Ok. Last part may confuse.
2 Moments with a force at an angle. Expecting a lot of blanks here.
a) Weight x d = T cos 40 x 0.75
So d = 5.1 cos40 x 0.75 / (1.2 x 9,81) = 0.249m (3)
b) Resultant force horizontally must be zero. Tension has a component to
Left so force at support must have a component to right and cant be vertical. (1)
TOTAL: 4
3 a) i) brittle / elastic / obeys Hooke’s law (2)
ii) steeper line that flattens out. (2)
b) i) YM = stress / strain
so strain = stress / YM
= 1.8E7 / 2.0E11 = 9.0E-5 (2)
ii) stress = F / A
so F = stress x A = 1.8E7 x π x (2.6 x 10-2) ^2 = 3.82E4 N (2)
iii) Weight = 2Tsin12 (resolve forces vertically) = 1.59E4N (3)
TOTAL 11
Last bit caused the usual problems
Some messed up the area calc
4 a) i) No horizontal forces act so no horiz acc. (1)
ii) Line that starts horiz and reaches ground nearer to wall.
Takes same time to fall
- because vertical acc is same and vert distance is same (3)
b) Drop ball through a height of one metre and measure time. Repeat and average.
s = ut + 1/2 gt^2 u = 0 so g = 2h/t^2 (3)
c) i) Constant acc down slope (negative)
Slows down (at steady rate)
Reaches max height v=0 at t=1.5s
Rolls back at increasing speed to bottom (at constant acc) (3)
ii) Max height = area under graph (or use s= 1/2 (u +v) t )
= 1/2 x 1.5 x 4.0 = 3.0m (3m will lose a sig fig mark) (2)
TOTAL 12
a) is the tricky bit and many will get aii wrong
5 a) The switch from table tennis ball to tennis ball threw a lot! Poor proof reading.
i) Acc = acc of free fall or 9.81
Only force acting is weight / drag = 0 when v=0 (2)
ii) AT max velocity drag = weight (1)
iii) Gold ball has higher terminal vel.
because mass bigger, wieght is bigger so needs more drag so higher v (3)
b) i) T 25m/s drag = 2000N (from graph - did you spot kN on axis?)
so resultant F = 3200 - 2000 = 1200N
so a = F/m = 1200/8000 = 0.15 ms-2 (3)
ii) Either look up max speed when drag = 3200N = 33 ms-1
or look drag at 40ms-1 drag = 5000N which is bigger than driving force
so would slow down (1)
c) Seat belt allows driver to slow down over a longer time
so acc = dvdt is less
so Force = ma is less.
Using a wide seat belt gives lower pressure (=F/A) on ribs so less damage (3)
TOTAL 13
I think the drag graph will throw many. Tricky.
6a) WD = force x distance moved in direction of force. (1)
b) Some debate about what "what happens to the WD " means.
I just wrote "it becomes heat" due to friction
Some wrote WD increases at a uniform rate - which may be OK (1)
c) 1W = 1J/1s (1)
d) Rate of doing work = gain in gravpe per unit time = mgh/t
h = 60m (pythagoras)
so rate = 5200 x 9.81 x 60 / (1.5 x 60) = 3.4E4 Js-2 (3)
Efficiciency = Pout/ Pin x 100% = 3.4E4/170E3 x100 = 20% (1)
TOTAL7
b is confusing and d is tricky.
7 If you confused length and extension you got wiped out here.
a) k = F/x = 3.0 / (8.0 - 2.0)E-2 = 50 Nm-1 (1)
b) New extension = 10cm
Energy = 1/2 kx^2
so change in energy = 1/2 x 50 x 10E-2 ^2 - 1/2 x 50 x 6.0E-2 ^2 = 0.16J (3)
Object has a force of 5.0N up and 3.0N down so resultant F = 2.0N up
mass = 3.0/9.81 = 0.306kg
so acc = F/m = 2.0/0.306 = 6.54 ms-2
Most got this wrong. SHould have drawn the forces acting
TOTAL 7
Tricky paper.
My predicted boundaries
100% 55
A 45
B 40
C 35
D 30
E 25
Still got all to play for with the EWP paper. Get revising!
Col
Usual disclaimers. This is just my mark scheme. It is in no sense official. It may contain errors and typos – but I’d be pretty confident its pretty close to the actual mark scheme.
Overall I thought this paper was on the hard side. There weren’t many easy definition marks and plenty of places where folk will go wrong.
My students estimates varied from 30 to 57 out of 60. I’d guess that the A grade will be 44/45 ish and the E grade 25 ish. I don’t know what you’re going to get.
I’m not allowed to post a scan of the paper. Forbidden by OCR. Don’t ask.
Ok so here we go (NB 1.0 E7 means 1.0 x ten to the power of 7)
Q1 a) Velocity = rate of change of displacement (with respect to time ) (1)
b) i) 70 km / h = 70 x 1000/3600 m/s = 19.44 m/s
Ek = 1/ mv^2 = ½ x 130 x 19.44^2
= 2.46E4 J (3)
ii) Volume is 8 x smaller (2 x 2 x 2)
Density is same
So mass is 8 x smaller = 130/8 = (16.25 kg) (2)
TOTAL: 6
Easy question to kick off. Unit conversion Ok. Last part may confuse.
2 Moments with a force at an angle. Expecting a lot of blanks here.
a) Weight x d = T cos 40 x 0.75
So d = 5.1 cos40 x 0.75 / (1.2 x 9,81) = 0.249m (3)
b) Resultant force horizontally must be zero. Tension has a component to
Left so force at support must have a component to right and cant be vertical. (1)
TOTAL: 4
3 a) i) brittle / elastic / obeys Hooke’s law (2)
ii) steeper line that flattens out. (2)
b) i) YM = stress / strain
so strain = stress / YM
= 1.8E7 / 2.0E11 = 9.0E-5 (2)
ii) stress = F / A
so F = stress x A = 1.8E7 x π x (2.6 x 10-2) ^2 = 3.82E4 N (2)
iii) Weight = 2Tsin12 (resolve forces vertically) = 1.59E4N (3)
TOTAL 11
Last bit caused the usual problems
Some messed up the area calc
4 a) i) No horizontal forces act so no horiz acc. (1)
ii) Line that starts horiz and reaches ground nearer to wall.
Takes same time to fall
- because vertical acc is same and vert distance is same (3)
b) Drop ball through a height of one metre and measure time. Repeat and average.
s = ut + 1/2 gt^2 u = 0 so g = 2h/t^2 (3)
c) i) Constant acc down slope (negative)
Slows down (at steady rate)
Reaches max height v=0 at t=1.5s
Rolls back at increasing speed to bottom (at constant acc) (3)
ii) Max height = area under graph (or use s= 1/2 (u +v) t )
= 1/2 x 1.5 x 4.0 = 3.0m (3m will lose a sig fig mark) (2)
TOTAL 12
a) is the tricky bit and many will get aii wrong
5 a) The switch from table tennis ball to tennis ball threw a lot! Poor proof reading.
i) Acc = acc of free fall or 9.81
Only force acting is weight / drag = 0 when v=0 (2)
ii) AT max velocity drag = weight (1)
iii) Gold ball has higher terminal vel.
because mass bigger, wieght is bigger so needs more drag so higher v (3)
b) i) T 25m/s drag = 2000N (from graph - did you spot kN on axis?)
so resultant F = 3200 - 2000 = 1200N
so a = F/m = 1200/8000 = 0.15 ms-2 (3)
ii) Either look up max speed when drag = 3200N = 33 ms-1
or look drag at 40ms-1 drag = 5000N which is bigger than driving force
so would slow down (1)
c) Seat belt allows driver to slow down over a longer time
so acc = dvdt is less
so Force = ma is less.
Using a wide seat belt gives lower pressure (=F/A) on ribs so less damage (3)
TOTAL 13
I think the drag graph will throw many. Tricky.
6a) WD = force x distance moved in direction of force. (1)
b) Some debate about what "what happens to the WD " means.
I just wrote "it becomes heat" due to friction
Some wrote WD increases at a uniform rate - which may be OK (1)
c) 1W = 1J/1s (1)
d) Rate of doing work = gain in gravpe per unit time = mgh/t
h = 60m (pythagoras)
so rate = 5200 x 9.81 x 60 / (1.5 x 60) = 3.4E4 Js-2 (3)
Efficiciency = Pout/ Pin x 100% = 3.4E4/170E3 x100 = 20% (1)
TOTAL7
b is confusing and d is tricky.
7 If you confused length and extension you got wiped out here.
a) k = F/x = 3.0 / (8.0 - 2.0)E-2 = 50 Nm-1 (1)
b) New extension = 10cm
Energy = 1/2 kx^2
so change in energy = 1/2 x 50 x 10E-2 ^2 - 1/2 x 50 x 6.0E-2 ^2 = 0.16J (3)
Object has a force of 5.0N up and 3.0N down so resultant F = 2.0N up
mass = 3.0/9.81 = 0.306kg
so acc = F/m = 2.0/0.306 = 6.54 ms-2
Most got this wrong. SHould have drawn the forces acting
TOTAL 7
Tricky paper.
My predicted boundaries
100% 55
A 45
B 40
C 35
D 30
E 25
Still got all to play for with the EWP paper. Get revising!
Col
Very petty question : but on the 6d section where you had to work out efficiency, did it say 'percentage efficiency' in the question? Because I think I stupidly put 20 without the '%'
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