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# AQA Mechanics 1B 24 May 2012 Unofficial Markscheme Tweet

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1. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by SGodfrey)
I made a massive mistake for the one where you had to find the speed when the object was travelling south east, I did all the working for when its travelling North east. I also put the range of time when y is above 5 metres as an inequality. How many marks do you estimate I will have lost for these mistakes?
Provided you followed through correctly on the South East one, probably 2 or 3 (was it of 6?).
For the other, provided you solved the correct quadratic and so found the two values of t when s=5 then probably all but 1 or 2 of the marks.
2. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by Bord3r)
Provided you followed through correctly on the South East one, probably 2 or 3 (was it of 6?).
For the other, provided you solved the correct quadratic and so found the two values of t when s=5 then probably all but 1 or 2 of the marks.
Yeah it was out of 6 Ah not too bad then thankyou :P
3. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
For 7c i got t=40 and the speed to be 5.83 or square root of 34... Did anyone else get this???
4. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
What was question 8c? because whatever it was i seem to remember getting 39.2-19.6 = 19.6 and thinking hmm thats wrong but left it.
how many marks was it worth and how do you get the correct answer of forty something?
5. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by Khan_ah)
For 7c i got t=40 and the speed to be 5.83 or square root of 34... Did anyone else get this???
Yeah loads of people did, including me. But apparently its wrong
6. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by Khan_ah)
For 7c i got t=40 and the speed to be 5.83 or square root of 34... Did anyone else get this???
I got that but it's wrong because we used the equation from part (a) which was actually s=ut+0.5at^2.
7. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by jph12)
I saved all my answers on my graphical calculator. I do not remember the questions, and will edit this thread according to the general feedback of the answers. Some of these may be wrong, again, these are simply my answers.
Updated version, 19:03 24th May 2012. I have included some very general "questions", so people can jog their memories to the question. A small summary follows the x), and the answer is the bit after the colon (. Again, any errors, please let me know so I can edit them. Question 6c) changed according to the general consensus.

1 - Ship on water:
a)Resultant Velocity: 5.39
b)Bearing: 338

2 - Momentum
2)Mass: 0.5

3 - Car braking to a halt
ai)Acceleration of car the car, using uvast, I think: -2 (therefore, deceleration).
3aii)Can't remember: 10
3aiii) Without any other forces except braking force - 2800N
3b) With an air resistance of 200N - 2600

4 - Particle in equilibrium. Force of 20N going at an angle of alpha, and in the opposite direction, a force of 10N.
a)Find alpha: 60
b)Weight: 17.3
c)Mass: 1.77

5 - Block and particle system.
a)When smooth, acceleration: 5.88
bi)When rough, tension: 122.4
bii)NR(?): 117.6
biii)Coefficient of friction: 0.735
c)Assumptions made: No air resistance, block is uniform in mass.

6 - Some stupid girl with a sledge, who has most probably cost a lot of people a few marks, for forgetting the Tcos30.
a)Diagram, I think, of the forces acting on the sledge when modelled as a particle.
b)Find NR in terms of T: NR = 78.4 - 0.5T
c)Find tension when acceleration is 0.05: 23.5

7 - Vector question. Unsure of the following questions. i and j vectors.
a)Unsure. Maybe express t in terms of s?: s = (-i+3j)+0.5(0.1i-0.2j)t^2
b)Again, unsure: 30
c)Find the speed when going SE. Needed to form an equation of v=u+at, not s = (-i+3j)+0.5(0.1i-0.2j)t^2: 1.41

8 - Projectile. She fires at an angle of alpha with speed 22.4.
a)Find alpha: 61 degrees
b)Find maximum height: 19.6
c)Find horizontal distance(?): 43.4
d)This is wrong, however, partially right, need confirmation of a full answer: 3.69
e)Hmm, again, can't remember, can anyone?: 10.8
8e was the smallest possible speed. Also, can someone PLEASE explain part 8c) with 43.4m i got 19.6 so can someone explain the answer and tell me how many marks it was ? pleaseeeeee
8. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by EAP)
8e was the smallest possible speed. Also, can someone PLEASE explain part 8c) with 43.4m i got 19.6 so can someone explain the answer and tell me how many marks it was ? pleaseeeeee
Using vectors and calculus
Horizontal
Vertical

a=[ 0 ]
[-9.8]
v=[22.4cos0] 0=theta
[-9.8t+22.4sin0]
s=[22.4cos0t]
[-4.9t^2 +22.4sin0t]
they said that t=2 half way there at its highest point so t=4 when the particle reaches B
given sin0=0.875
sin-1(0.875)=61 degrees i think cant remember exactly
therefore, horizontal displacement which is the distance AB=
22.4 x cos(61) - which was root15/8 x 4
=43.4
it was 3 marks
19.6 was the one before when you worked out the height at the highest point (or when vertical velocity is 0.
9. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
To clarify I put two modelling assumptions as:
. block modeled as particle
. air resistance negligible
but I put a third one down for ****s and giggles. It said on the front assume g=9.8ms^-2, however the third assumption I made was that it was carried out on Earth (my assumption was not that g=9.8).
Obviously g, being a constant, will be 9.8, but if the experiment was carried out on the moon, the answers would have been different.
10. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
Can anyone remember how many marks any of the questions were worth? That could be added to the unofficial mark scheme?
11. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by AndyHoughton)
Using vectors and calculus
Horizontal
Vertical

a=[ 0 ]
[-9.8]
v=[22.4cos0] 0=theta
[-9.8t+22.4sin0]
s=[22.4cos0t]
[-4.9t^2 +22.4sin0t]
they said that t=2 half way there at its highest point so t=4 when the particle reaches B
given sin0=0.875
sin-1(0.875)=61 degrees i think cant remember exactly
therefore, horizontal displacement which is the distance AB=
22.4 x cos(61) - which was root15/8 x 4
=43.4
it was 3 marks
19.6 was the one before when you worked out the height at the highest point (or when vertical velocity is 0.
thanks god knows what i did because i got 19.6m twice haha oops. I have definitely lost 9 marks in this exam, could i still get an a if i lost no other marks from stupid mistakes? i kinda need a high a to pull me up but that doesnt seem likely :/
12. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
For 8(a): Am I not right that the question was asking you to prove that sin(theta)=0.875, not finding the angle?
13. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by EAP)
thanks god knows what i did because i got 19.6m twice haha oops. I have definitely lost 9 marks in this exam, could i still get an a if i lost no other marks from stupid mistakes? i kinda need a high a to pull me up but that doesnt seem likely :/
I think it had 5 parts (the last question)
a - 3 marks show that, probably got that!
b -either 3 or 4 which was 19.6 (you got that im guessing)
c -3 marks you didnt get, might get 1 mark?
d-5 marks i think, the quadratic was -4.9t^2+19.6t-5=0 and minus the big value of t
e 10.8, sub in value into initial velocity thing 2 marks
depends what you lost if you got the rest, out of all my friends only me and someone else haven't actually made any mistakes! so i thought the grade boundaries would be really high but you might be in luck, just keep revising and you'll be fine, past papers work best
14. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by 1platinum)
I got that but it's wrong because we used the equation from part (a) which was actually s=ut+0.5at^2.
How many marks do you wel get for that question? those of us who did that?
15. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by 1platinum)
I got 5.88 for 7(c) and for 8(d) 3.45.
Definitely got 3.45 for 8(d) but got sqrt(2) or (1.41) for 7(c)
16. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by AndyHoughton)
I think it had 5 parts (the last question)
a - 3 marks show that, probably got that!
b -either 3 or 4 which was 19.6 (you got that im guessing)
c -3 marks you didnt get, might get 1 mark?
d-5 marks i think, the quadratic was -4.9t^2+19.6t-5=0 and minus the big value of t
e 10.8, sub in value into initial velocity thing 2 marks
depends what you lost if you got the rest, out of all my friends only me and someone else haven't actually made any mistakes! so i thought the grade boundaries would be really high but you might be in luck, just keep revising and you'll be fine, past papers work best
think i got a,b,d and e right. but i got the big six marker wrong earlier and part c wrong here. other than that i think i got all the other answers right but i may have used the wrong methods haha oops im gonna say i probs got about 64/65 out of 75. Reckon that will get me an a ?
17. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
How many marks do you wel get for that question? those of us who did that?
I think we will still get three or four marks because essentially we used the correct method.

(Original post by alan_9016)
Definitely got 3.45 for 8(d) but got sqrt(2) or (1.41) for 7(c)
For 8(d) you took the higher value away from the lower value to get 3.45?
18. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by 1platinum)
For 8(d) you took the higher value away from the lower value to get 3.45?
Yes I did
19. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
(Original post by 1platinum)
I think we will still get three or four marks because essentially we used the correct method.

For 8(d) you took the higher value away from the lower value to get 3.45?
3.46
20. Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
3.46
3.45 if you used the exact values instead of the rounded up versions
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