G481 Mechanics 20 May 2014 Unoffical mark scheme

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

Scroll to see replies

Reply 1

BrokenS0ulz

For the weight of the cable car I used the cosine rule and drew a vector triangle. Is this sufficient to score marks even if final answer is wrong? Also is rounding 0.249m to 0.25 m going to cost me a mark?
thanks for posting this :smile:

(edited 9 years ago)

Reply 2

L'Evil Fish

Original post by teachercol

OCR G481 Mechanics Tues 20th May

Can I ask. The very last q

Where did 5 come from?

Reply 3

teachercol

Original post by BrokenS0ulz

This approach is fine.
2sf answer to 0.25 is fine

Reply 4

teachercol

Original post by L'Evil Fish

Can I ask. The very last q

Where did 5 come from?

Its the upward force on the mass due to the spring ( = kx)

Reply 5

Red Fox

50 marks I reckon I got, think full ums will be lower than 55 but you know it better than me :tongue:

(edited 9 years ago)

Reply 6

randlemcmurphy

For 4)a)I) I said the horizontal component was independent of vertical component, where acceleration occurs, therefore there is not component of acceleration for horizontal component as cos90=0, so there is no change in velocity and it stays constant.
What do you make of this answer?

EDIT: In your opinion was this harder than G841 June 2013?

(edited 9 years ago)

Reply 7

L'Evil Fish

Original post by teachercol

Its the upward force on the mass due to the spring ( = kx)

F = kx

Oh bloody hell :lol:

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

Original post by Red Fox

50 marks I reckon I got, think full ums will be lower than 55 but you know it better than me :tongue:

I'm saying min 50, max 55

Reply 8

Kate20

I think I scored around 49-52 depending on if I remember my answers correctly! Thanks so much for posting this mark scheme!

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Reply 9

Chung224

52/60 fuk yeahhhhhhhhh

Reply 10

eddie1221

So would you say this exam was easier than last years? I thought this years seemed harder, especially how some of the questions were set out, so I thought the grade boundaries would have lowered abit :frown:

Reply 11

Username13

Thanks, but do you really think the grade boundaries will be higher this year??

I'm thinking I got about 43 which was last years grade boundary for an A.. considering most people that I've seen found this harder..

Reply 12

Awdrgyjo

For the moments question - I used sin 40, because sin 90 = 1 so usually it's not in the equation when it is perpendicular but cos 90 equals 0? Sorry just confused why it's cos not sin. Thankyou! :smile:

Reply 13

ChrisCareyy

For the last part of the last question when calculating the acceleration I kinda messed up, I did:

F=ma

where m = 3.0/9.81 = 0.30
&
where F = kx => 50 x 0.1 = 5

so if F=ma, then a=F/m

so a = 5/0.30 = 16.7ms^-2

How many marks do you think I would have gained (if any) considering I did not take the resultant force into account.

Reply 14

lewis123456

Thanks, some strangely worded questions, e.g the what happens to the work done?, which were confusing. Think I wrote the weight of the cable car as x10^5 not ^4 for some unknown reason aswell!

Reply 15

Hoppo7

Do you know the exact wording of question 7)b)I) and 7)b)II)

Reply 16

Tonkin

Not too bad, I think I got 47 give or take 3 marks. Thanks for posting the unofficial mark scheme. IMO I thought this paper was easier than last years.