G481 Mechanics 20 May 2014 Unoffical mark scheme

It depends on the exact wording of the question. I thought it said "Estimate the mass.....explain your answer" in which case you would need to state 16.25kg.

Do you think the grade boundaries might be lower this year as there were no January modules aswell as the fact it was hard???

Wahaaay, full UMS for you

Yeah, it said estimate a value

x

I think A will be around 40-43 bc this was MUCH harder than June 13 and a lot more people complained.
It looks like I got 42-47. This was honestly the worst paper I've ever done.

I hope it's like 50 for full ums lol



I think so too, and lack of Jan exams

F = kx

Oh bloody hell

I tried doing:

F = ma

1/2 F x = strain energy
F = (s.e)(2)/x

(s.e)(2)/x = (3/9.81)a

And it went wrong


I'm saying min 50, max 55

I'm i a similar situation to you, ballsed up the last question, but that's because I ran out of it niece, messed up the change in energy ap it and just did have time for the last bit, so I'm saying max about 55 but I think I absolutely nailed the rest of it, Exceptt the 1 mark part at the end of the moments question

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

For the mass of the stupid dinasaur I put down, "around 16kg" would that be fine??

F = kx

Oh bloody hell

I tried doing:

F = ma

1/2 F x = strain energy
F = (s.e)(2)/x

(s.e)(2)/x = (3/9.81)a

And it went wrong


I'm saying min 50, max 55