EDEXCEL mechanics 3/ june 06 pm

What is that ??
I thought this was the correct way. I also found 4.

To those people who got 12.9 as the max angle before the object would topple, did you have to subtract the Centre of mass (part b) from the length of the object? I thought i read that the centre of mass was measured from the base, so it was simply arctan (2a/CoM). Did i read wrong? The max angle i got was around 30 something.

I also got the max angle as around 30...i really hope it was right, i started to worry when i saw people posting that they got 12.9
I really need good UMS from this exam.

Yeah, I got that the max angle was around 30, but I wouldn't trust my answers, I phailz at M3

Ah, well I failed the paper in January, so my answers probably arent' the most likely to be acurate either. Maybe we're missing something.

Quite possibly. Oh well M3 isn't in my conditional offer, so if the worst comes the worst and I've completely failed it, it isn't the end of the world for me

Yeh, i've just redone the question...it was 12.9. The answer of 31.3 is when you use the length of the distance of the cetre of mass from O, not from the base of the cylinder.
Ah well, we may get 2 out of the 4 marks, as we said it wont topple, which is true, and we kind of used the right method.
I need it, FP2 will go horrendously, because the algebra is so random and tricky for me, and i need an A
Ah well, a mirracle could always occur.

Heres hoping they lose our papers!

dude really?
it cant be THAT bad lol
im exepcting around 15-20 marks gone :P

it was 0.016m away from B is that what you got? because you had to take what you found out away from 0.12m

I'm expecting 15-20 marks total!

Alright everyone, I am gonna forget about M3 now and focus on M4. Goodluck for any exams you still got.

One more thing, I have seen the thread for Jun07. Everyone was saying that A would be more than 65/75. So basically, after reading everyones comments I am sure the boundary would not be more than 58/75.

.A.

it cant be 0.016m away from B
think about it...it takes 1.5 seconds to go to A from B so to come back from A to O it wont take longer than 0.75 seconds...
hence from B to A to O it should take no more than 2.25 seconds...
...so the particle is more than 0.12 m away from be...
use x = A sin wt and ull get -0.104 meters
This means that the particle is 0.104 meters from O near A
tada...0.104 + 0.12 =0.224 meters
but then again!!! i might be wrong :P

it shudnt be more than 55/75
coz in jan08 the paper was superbly easy and the boundary was 55

Was I the only one who found that paper very tricky?

As for question 6(c):

$\displaystyle[br]\sqrt{6}t = \sqrt{x^2 + 2x} + c \newline[br]t = 0, x = 0, \Rightarrow c = 0 \newline[br]\to \sqrt{6}t = \sqrt{x^2 + 2x} \newline[br]t = 2, \Rightarrow 2\sqrt{6}t = \sqrt{x^2 + 2x} \newline[br]24 = x^2 + 2x \newline[br]x^2 + 2x - 24 = 0 \newline[br](x + 6)(x - 4) = 0 \newline[br]x > 0 \to x = 4 \newline[br]$

lets get this right before I leave.

A to B it takes 1.5 sec. So A to 0 it takes 0.75 sec. The question said that the ball is directed towards A initally. So it takes 0.75 s to go to A from 0. Then another 0.75 to come back to 0. So far 1.5 s.

The question asked for t=2s. So we are left with 0.5s now.

Its at centre position.

X = asinwt. a=amplitude (0.12) w=2pi/3 t=0.5. This will give you distance away from O in 0.5s. OB=0.12

P = particle position.

PB = OB - PO

= 0.016

.A.

Still, I myself was told by the mods to not put up recent papers, so I don't really want to risk a ban. I will send papers to anyone who PMs me their email address though.