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EDEXCEL mechanics 3/ june 06 pm

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Reply 20

mattbennsion

aster100

what was the integral?

yep that was the integral
(x+1)[(x^2 + 2x)^(-0.5)]

Reply 21

bballer4life

MaLvI

i found that it wouldnt topple
In the limiting case angle comes out to be 12.9 which is greater than 12 so toppling does not occur.

oh yea, your right. i work out the anlgle's the 12 and 12.9, but i must of been brain dead to get it the wrong way round.
ah well

Reply 22

AdvancedWave

I worked out the position of the center of mass on the slope and found that it would topple.

Reply 23

aster100

mattbennsion

yep that was the integral
(x+1)[(x^2 + 2x)^(-0.5)]

(x^2 + 2x)^1/2 ?

Reply 24

.A.

OMG, we just needed to substitute U = x^2 + 2X.

NOOOOOOOOOOOOOOOOOOOOOOO

.A.

Reply 25

mattbennsion

aster100

(x^2 + 2x)^1/2 ?

ye thats what i ended up with

Reply 26

chidona

I did the toppling question a slightly different way, I just used the angle they gave and then worked out (hard to describe, but you may follow) the distance along the base where the vertical would pass through. If it would topple it would have to be greater than the base radius (2a), but it turned out to be 1.8 ... a so it didn't topple.

The last part of the last question... ugh. I knew what to do, I just went braindead at the integral - I made a silly error when rooting it and subsequently stuffed it up. Oh well. Method marks! =D

The rest of the paper was pretty good, I thought. May have made some silly mistakes here and there, but I was fairly sure I was systematic throughout the thing. A LOT of energy consideration came up, more than usual.

The only proof question I wasn't able to derive was the one with the circle and the two particles coalescing - I needed rt (ag/8) and I always came out with rt(ag/2) . Hmm, only 3 marks, and I've most likely got a method mark from it.

Reply 27

Candescence

Well the integral is easy, i just couldn't figure out how to get to it -_-

Reply 28

The Sherminator

bballer4life

oh yea, your right. i work out the anlgle's the 12 and 12.9, but i must of been brain dead to get it the wrong way round.
ah well

Shiz. I did that wrong :frown:

I said it wouldn't topple, but my angle was weird....
I can see this paper falling apart for me.

What was the method you used to work out the angle?
I just found tan x = 6a / x bar
??

Reply 29

chidona

The Sherminator

Shiz. I did that wrong :frown:

I said it wouldn't topple, but my angle was weird....
I can see this paper falling apart for me.

What was the method you used to work out the angle?
I just found tan x = 6a / x bar
??

6a? The base radius of the cylinder was 2a - that's how it was placed on the slop. I have a feeling you assumed it was on the larger plane face of the hemisphere, which is the wrong way up.

Reply 30

The Sherminator

chidona

6a? The base radius of the cylinder was 2a - that's how it was placed on the slop. I have a feeling you assumed it was on the larger plane face of the hemisphere, which is the wrong way up.

Oh damn. Thats why the angle was funny!!!!!!! Argh, what an easy question to throw marks on.
Hopefully method marks :frown:

Reply 31

.A.

SHM motion

X away from O was 1.1....

so 0.07........away from B.

Vertical circular motion question.

First was prove.

The angle in last part was 23 i think

.A.

Reply 32

Clefairy_Clefairy

Yeah last question, could not do at all and wasted at least half an hour pondering over it!
Also that w^2 question caught me out for a bit....
Only 6 questions for the whole paper, but they were so not the standard ones!!

Reply 33

ahezhara

For the vertical circular motion question, I got something like 20.4 degrees.
I had 1.6 cm away from B...
My w was 2PI/3, a was 0.12m, maximum v was 2PI/25... 0.016...m away from B...

Reply 34

luckyu

for question 3 the horizontal circular motion i never included the reaction of the table i forgot but i got w^2 and what did u guys get for part b to that question.. 2a/5g something like that??

Reply 35

luckyu

i think i gt pi/25 not 2pi/25

Reply 36

alex_hk90

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

As for question 6(c):

Unparseable latex formula:

\displaystyle[br]v^2 = 6(1 - \frac{1}{(x + 1)^2}) = 6(\frac{x^2 + 2x}{(x + 1)^2}) \newline[br]v = \frac{dx}{dt} = \sqrt{6}\frac{\sqrt{x^2 + 2x}}{x + 1} \newline[br]\int{\sqrt{6}}dt = \int{\frac{x + 1}{\sqrt{x^2 + 2x}}dx \newline[br]

\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} = \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]

Reply 37

kcfch

i really dont remember which question is was but one of them we had to find the tension, does anyone recall getting a T = 3/2 mg? im convinced i did it wrong :frown:

oh and i think the second part of that question was a find w (omega) in terms of r and h or something if that helps jog memories

Reply 38

.A.

1.6 cm

The time period was 3s it was initially moving towards A away from O.

The question said find the distance away from B in 2 s. So in 1.5 sec it goes from O to A and then comes back to O.

Then you use X=asinwt. t=5, a=1.2 (i think) This gives you distance away from O. Then you subtract this from amplitude to get distance away from B.

.A.

Reply 39

luckyu

i got for the circular motion cos = 15/16 does any one agree?