So, MM1B AQA Jan 09 pm guys, HOW'D IT GO?

why did u get for d = 25 m ? Shouldn't it be 45 m?

Oh that could be it. Damn yeah I messed up the d question instead.
I set s (in suvat) equal to d/2 because that is how far they will have each moved when they were equal. Sounds logical but is it wrong then?

Ohhh no sorry the Force of the child was acting down so yeah

R = Weight + 40 sin 50

why did u get for d = 25 m ? Shouldn't it be 45 m?

I don't think so (sorry ) because I think when you Suvat you only consider 1 of them, not both :/

arhhhhhhh F*** i didnt times my answer by 2 DAMN!!!

If you are referring to me, I don't think the general concensus is me being right.....

nope, i think 0.245 is correct, because both particles start moving at the same time so it must be d/2 for each particle lol

i misread my calculator for 40cos30 to 36.64 hence mu is rong too

ahhh i feel so stupid for reading it off rong, hope to get the ecf marks tho

How odd! I'm sure we don't have to do coursework... although I hope to God I haven't done the same paper as you all, as can't seem to remember any of the questions you're talking about!

Ok... the last question of the paper you're talking about here... was it a collisions questions? Where both particles were deaccelerating at a constant rate of 0.4? For the last part of the question we were asked to find the original distance between the two particles... is this the same paper?

Also we had a 9 mark question on a parcle going down a ramp into a van. Is this the same paper?

aww that's not stupid thats just a silly error. you wouldnt be human if you didn't make them!
yeah, me too

Yeah, I did only consider one of them. So say for the 11kg particle:
s = d/2
u=0
v=unknown
a=0.98 or something
t=0.5

s=ut + (0.5)a(t^2)
d/2 = 0 + 0.5(0.98)(0.5^2)
d=(0.98)(0.5^2)
d=0.245

??

We could work out mu now, I guess.
I remember it was 40N and angle of 50 degrees, mass was 8kg sooooooo

Friction = 40cos50 = 34.6N
8g = R + 40sin30
R = 8*9.8 - 40sin30 = 58.4N

Friction = muR
mu = 34.6/58.4 = 0.592

NOOOOOOO I got it wrong ;(

Edit: Oh, angle was 50 degrees?
I'm just confused now, where have I gone wrong in my working? Because R was indeed given as 98.4N
34.6/98.4 = 0.352 :O

Here's my answers. Ah still cant believe I got u wrong

1) 9 ms^-1

2)

a) v = 0 at t = 0, 30, and 50s
b) Distance = 75 m
c) Total Distance = 115 m
d) Distance = 35 m

3)

a) a = Proved a = 6.30 ms^-2
b) a = 5
c) Part (b) includes frictionthats why the value is less.

4)

a) Peg has no friction (smooth).
b) String is light and inextensible.
c) a = 0.98 ms^-2
d) i) v = 0.49 ms^-1
ii) s = 0.245 m

5)

a) Draw free body diagram includes 4 forces.
b) 98.4 N
c) 34.6 N : well if I got mu wrong then I must have got this wrong too. Damn.
d) u = 0.352 : wrong apparently

6)

a) ( 9i + 12j )
b) 15 N
c) 6kg
d) i) r = ( 0.75i + 1j ) t^2
ii) d = 5 m ( Maybe thats as well wrong anyone got different answer?)

7)

a) Proved the resultant force.
b) I got 292 degrees

8)

a) max h = 23.5 m
b) Show the time taken, proved t = 4.28 s
c) Cant remember. Did the horizontal velocity and the vertical velocity then found the magnitude. Hope thats right!

Anyone got same? different? answers?

i would love an unofficial mark scheme