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# AQA Maths Mechanics 2B June 2012 Unofficial Mark Scheme

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1. For the two tensions I got 7.84N (or around there, can't remember) at A and 11.76N at B.
2. (Original post by Signed Solution)
For the two tensions I got 7.84N (or around there, can't remember) at A and 11.76N at B.
That sounds like what I got.
3. 0.0668 Correct. The rebound velocity was v = ((96-11gu)^0.5)/2

0.5mv^2 - 5,5mgu - 120e^2 /2l =0 (e=0.5)
this leads to 96-11gu /20= 2.2gu +3
u=36/55g

which is, 0.0668 ( you need to do it to 3sf)
4. (Original post by Signed Solution)
For the two tensions I got 7.84N (or around there, can't remember) at A and 11.76N at B.
got same
5. Dropped 5 marks-ish. Knowing how high the boundaries will be, I doubt it'll be an A*
6. (Original post by Iepnauy)
Dropped 5 marks-ish. Knowing how high the boundaries will be, I doubt it'll be an A*
If you only dropped five marks, then you'll definitely get above 90% on this paper. (:
7. The highest i have ever seen for an a* is 69 so you should be fine
(Original post by Iepnauy)
Dropped 5 marks-ish. Knowing how high the boundaries will be, I doubt it'll be an A*
8. Can anyone tell me what they got for part b of 6.. swinging child tenison in string one...?
9. (Original post by jph12)
Question one: Ski jumper
1a)29792
b)23088.8
ci)52880.8
cii)37.3

Tick

Question 2:
2ai)a=12t+8e-4t
aii)7.08
b)28.3
c)2t2+0.5e-4t+8t-0.5

Tick
Question 3: Centre of mass
3ai)4.8
aii)7.2
aii)A=7.84N, B=11.76N

Tick

Question 4: Trig motion vectors
4a) Didn't complete. ): Hard question! I mentioned that r would never equal zero, don't know if that will get me marks.
b)-12sin3ti-12cos3tj
c)-36cos3ti+36sinstj
d)k=-9
e)Toward centre of the circle? North west?

4a - I showed that the modulus of r is always 4 and hence it must be going in a circle. I then showed that at t=0 r = 4i and at t=60 r = -4i and thus if radius is 4 then centre must be at origin. Anyone to back me up ?
4e - If something is moving in a circle about the origin then surely the acceleration must be towards the center ? Where are people getting NW from ?

Question 5: Horizontal circular motion
5a)7
b)2.1
c)0.27

5c - w = 2pi/t so t = 2pi/7 and so t= 0.898 sec, no ?
Question 6: Swinging child
6a)1.51
b)236.7

Tick

Question 7: Differential equation
7a)-1.96(v-5)
b)v=2e-1.96t+5
Tick

Question 8: Elasticity question
8a)9.80
bi)sqroot(96-11gu)
bii)mew=0.334

8b - 0.668.
I will edit according to the general concensus. Some right answers for question eight, please, and the ones I couldn't do. (:

(Original post by vedderfan94)
Nice job man, I've got these:

3b) Tension at A = 18.4N, Tension at B = 1.2N

4a) I found the magnitude and showed that r always equals 4, therefore it moves in a circle. Not sure if that's right.

8a) 9.80
bi) sqroot(96-11gu)
bii) Think I got this wrong. Correct answer seems to be 0.067 or around that :/
Yea I got all your answers except for 3b I got TA=7.84 N and TB= 11.76N
(Original post by SwimGood)
Also, I think I wrote North West for the direction on the vectors question
But if it's in circular motion then isn't the acceleration to the center ? Where did you get NW ?
(Original post by Signed Solution)
I think in the first question there shouldn't be any newtons.

For question 8, I remember getting root(96-11gmew) or something like that and ended up with mew=0.0668.
I got that value for mew

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Updated: June 21, 2012
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