Applied Maths Aftermath

JC 2K3

DtotheK wrote:

YESSSS!!!

and i think his thing is the same as u^2(cosA)^2.. or else it's close ish. i just simplified it in the case minute when i notices the sin^2 and cos^ =1 stuff

There was no need for all the hassle, there's no Ek lost in the j direction, so it's just:
(1/2)4(ucosA)^2 - (1/2)8(ucosA/2)^2 = 2(ucosA)^2 - (ucosA)^2 = (ucosA)^2

Jonathan

Has anybody got answers to the inertia question? i feel im the only one that did it.

cocoa

contested Qs in the answer list: Q2(a)(ii) and Q5(b)(i)
2(a)(ii) draw the line of A's relative velocity, it happens to have slope 4/3
you know that at one point on the line, it is 8km south west of B
find the closest distance it comes to B. (line y=4x/3 and point (4root2,4root2))
this distance happens to be 4root2. you now have a triangle of side 8km, where the boat starts, 4root2km, where the boat is closest and 4root2 km, between the two. this triangle is repeated, so the total distance spent inside 8km of B is 8root2 km
divide this by the vel got in part i and you have root(2)/5 hours
which is 17 minutes.

5(b)(i) we agree the first velocity is usinA (and therefore upward) so all of the first sphere's horizontal velocity is imparted. with e=1/2 this velocity is also halved. therefore B's velocity is ucosA/2

lemur option

yes john you are the only one that did it.

JC 2K3

lemur option wrote:

anyone care to show an answer for 10 (b) a worked out answer I believe I got quite close to JC 2K3's answer works down to 4.41 something.

It works out as 44.12, it's correct, but one hell of a messy equation.

Thinking about it now, I probably got most of the marks in 10.

Shox

I did Q,1,2,3,4,5,10, and got exact same asnwers as you Zorba.. every one. SO i hope all yours are right!!!

Gangsta

JC 2K3 wrote:

It WAS my favourite question.

Seriously, IIRC my equation at the end of part (b)(i) was:

v = root(3200 - 3200e^-15/16)

....

dude that's right I got it and a few other's get it. You then just find the limit as x approaches infinity. Soft paper overall, especially considering how much of a b!tch it can be.

YES IT'S FCUKING OVER!!!!!

PS. for 5 (B)ii i got 90-A too, I'm pretty sure that's right. I believe they're opposite angles in a right-angled triangle seeing as the particle's i component becomes 0.

Jonathan

Q10 Bi (a bit vague... not in mood to write it up properly)

Re arrange the acceleration so you have one fraction (3200-v^2)/(3200)
let a = v dv/ds
re arrange and integrate (u need to use a substitution) both sides using limits from v to 0 and 1500 to 0

the ans i got was 44.12 ms^-1

Q10bii

The max speed occurs when the acceleration is 0 as any greater acceleration will mean the car's speed will decrease. just set the acceleration=0 and work out v

ZorbaTehZ

@Cocoa
This is what I did for Q2. Where am I going wrong?
Line has slope 4/3, angle is 53.13. Angle between line connecting both boats, is 45 to the horizontal. Therefore 8.13 degrees at the corner of the triangle. You've got an isosceles triangle, I used Sine Formula and got a distance of 15.83. Distance over velocity ie 15.83/40 = 0.39575, which is 23.745 minutes.

As for the other question, 0.5uCosa == ucosA/2.

EDIT: Unless by ucosA/2, you mean half angle of alpha?

Roonels

just wondering wat the equations for the conncected particles were? thanks....its really bugging me cause i knew how to do it but...messed it up:mad: oh well no more leaving cert!:D

cocoa

i misread the cosA thing, my mistake. I'm still looking at your method for the boat Q

ZorbaTehZ

Shox wrote:

I did Q,1,2,3,4,5,10, and got exact same asnwers as you Zorba.. every one. SO i hope all yours are right!!!

lol, I hope so too.

cocoa

argh, i made a terrible mistake, rushed through the formula. the closest distance is actually 4root(2)/5. Use Pythagoras rule to find the side we actually want: root(288/5) and multiply by two to get the distance we want... and it's 15.18
well that settles it.. I was wrong...

ZorbaTehZ

cocoa wrote:

(snip) ... it happens to have slope 4/3
you know that at one point on the line, it is 8km south west of B
find the closest distance it comes to B. (line y=4x/3 and point (4root2,4root2))
this distance happens to be 4root2 ...

Now I didn't check out the rest but my initial reaction would be that 4root2 (5.65) is quiet big ...

JC 2K3

Feck it, for 2(b) I did everything right, but added 5 and 3cos30 rather than subtracting them, why I dont know(just mixed up a sign I guess, did it more with equations rather than visualising it). But meh, my method was right.

ZorbaTehZ

cocoa wrote:

argh, i made a terrible mistake, rushed through the formula. the closest distance is actually 4root(2)/5. Use Pythagoras rule to find the side we actually want: root(288/5) and multiply by two to get the distance we want... and it's 15.18
well that settles it.. I was wrong...

kk, nevermind the post above.

JC 2K3

Roonels wrote:

just wondering wat the equations for the conncected particles were? thanks....its really bugging me cause i knew how to do it but...messed it up:mad: oh well no more leaving cert!:D

depends what way direction you let the accelerations go.

If you let the 6 kg mass go down and the 4 kg one go up, the equations are:

4a = T - 4g
m((a-b)/2) = mg - 2T
6a = 6g - T

I made the mistake of letting the acceleration of B equal (a+b)/2

E92

I'm happy with it overall. DELIGHTED to be finished. Hopefully never have to do the Leaving Cert ever again.
Q10 a beauty. Q2 a beauty. Q4 mostly good,but I think I made an error in b part ii, and I didn't get out part iii either.
Q3 was a balls, oddly enough. Q1 a) was simple, 1 b) was anything but, never got that one out either. Lastly q5 wasnt TOO bad but wasnt great either. Overall, a B grade of some sort, possibly a B2, well hopefully anyway.

7woohoo7

Q9. (a) What was the length of the column ? I got 27.2cm

rjt

7woohoo7 wrote:

Q9. (a) What was the length of the column ? I got 27.2cm

Yeah, this is what I got.

Feddd

In the forces question, initially I had 6kg and 4kg going up and B going down. I know it said later that thats not the case but considering thats a later part it shouldn't have any impact on the first part? Got a slighty different answer to others -48gm/(5M-48) but because its the standard way of working out that question and having A+B/2 then do I still get the marks?

And for Q2 a part I got 7 mins. Prolly made a mistake but cba workng it out now. I looks like i never doubles it but I'm sure I did.

In Q1 B I felt that there wasn't a chance in helll it was gonna come out for me but said i would do the various eqns anyway, and then it suddenely came out! Was cool the way it didn't look like it was coming out and then suddenely does. ^^


Q1: easy a difficult b. Same answers as Zor.

Q2. Easy a and very very easy b. Must have made a mistake on the a though as I am getting 7 mins. =/

Q3. Very very nice.

Q4. a part was nice, b a bit of a bitch.

Q5. Abstract in the way it was deflected to that angle. Said (90-A) but said +0u just in case!

Q10. Depending on how you handled the e, theres a possibility your calculater wouldn't be able to handle it. Got 35.39 m/s for part (i) and same as Zor for (ii).

7woohoo7

I got 27.2cm doing it one way and 28cm doing it another way (didn't get to study it much!). The 27.2 cm way seemed more correct. Really hoping its right.

ZorbaTehZ

Feddd wrote:

In the forces question, initially I had 6kg and 4kg going up and B going down. I know it said later that thats not the case but considering thats a later part it shouldn't have any impact on the first part? Got a slighty different answer to others -48gm/(5M-48) but because its the standard way of working out that question and having A+B/2 then do I still get the marks?

That's exactly the method I used, and would use for any type of question like that - tbh I thought that was the only proper way too. JC 2K3 used a different method above which I initially thought must be wrong but I haven't properly looked at it, but I suppose there must be other ways to consider the acceleration.

Gangsta

ZorbaTehZ wrote:

That's exactly the method I used, and would use for any type of question like that - tbh I thought that was the only proper way too. JC 2K3 used a different method above which I initially thought must be wrong but I haven't properly looked at it, but I suppose there must be other ways to consider the acceleration.

Ok I'm just really curious.....ZorbaTehz what school are you in?

JC 2K3

ZorbaTehZ wrote:

That's exactly the method I used, and would use for any type of question like that - tbh I thought that was the only proper way too. JC 2K3 used a different method above which I initially thought must be wrong but I haven't properly looked at it, but I suppose there must be other ways to consider the acceleration.

You don't know the mass of B, so you don't know what direction the accelerations are going to be in.

It's either both go up and the acceleration is ((a+b)/2), both go down and it's ((-a-b)/2) or one goes up and the other down and it's ((a-b)/2).

lemur option

JC2K3, I see where you are coming from. There is reason to put (a-b)/2 in certain circumstances but in this case it doesn't matter. as long as your equations are set up so that your "a" and "b" are opposite to the direction of the pulley's accelleration (a+b)/2 then this formula will be ok for all scenarios.

setting a and b down and (a+b)/2 up

if both a and b are down a and b are positive(down), so (a+b)/2 (up)
if both are up a and b are negative, so (a+b)/2 (up)=> -k/2 (up)= k/2 (down)
if a is up (-) and b is down (+), (a+b)/2 (up) =>(b-a)/2 (up or down depending on the difference)

so the best bet is to put (a and b), and (a+b)/2 in opposite directions because it works in all cases

JC 2K3

Yup.

But who cares now?

cianc1

Jeez, all you people with your correct answers, am I the only one here who knows they won't get an A in this terrible paper??

JC 2K3

Well if I do get an A it'll be by the skin of my teeth and due to attempt marks and my Irish bonus.