Applied maths solutions

Im looking for a set of AM solutions that can show me how the answer was obtained. Do they exist? I just gave it a quick google and nothing came up. Im pretty desperate at this point

help meh plz! :(

daveirl

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Wow thats impressive.could you post up 2001 q. 1, 2, 3, 5, 10.


tnx.

daveirl

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Where can i get the marking schemes?

daveirl

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Tnx alot dave.


If youre not using the solutions nemore could I take them off your hands?

daveirl

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It would be a bit long winded if I tried to ask you all the questions I want to ask:)


Could you maybe photocopy them and I'd pay you back for it?

daveirl

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Tallow Lad

Any Chance you still got sum o dem solutions

MathsManiac

daveirl wrote:

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Actually, it's the state exams commission website that you need: www.examinations.ie
Click on "Examination Material Archive" and you can get the marking schemes, which include the solutions. They go back as far as 2001.

Also, Oliver Murphy, a teacher in Belvedere College and the Institute of Education, has a booklet containing papers and answers (not full solutions) of over 20 years. He sells it for €10, which he gives to charity. Details here:
http://www.discoveringmaths.com/students/appliedmaths/

Good luck with the subject; it's great!

spiderblade

would u by any chance have the 1994 solutions now

OhMSGlive

Does anyone have any PASS App. Maths solutions from 99 backwards? Much obliged if you do.

bonez025

Does anyone have the 95 paper? Its not on examinations.ie .
Please, I've done that far back.

pinkballetdance

I have the solutions(not for Q 6 or 8 though) but I woouldnt be able to get them up until the probably the saturday before the exams here are the answers though from oliver murphys book.

1. a i 5u b i √(6g) ii √(6g) /3g iii 5/3,8/3,3 metres
2 a i 64-6t, 62-8t ii 4, 13.6 b i p=0.6 ; q=0.45 ii 357.14 seconds
3 a i 14.7 m and 9 m ii 7.056m iii 12.6m/s b i 40.9* (*=> degrees!)
ii tan L = -Vy/Vx <0 => obtuse
4 i g/9 ii g/9 iii g/11 iv g/22
5 a i v1 = u(1-3e) ; v2= U(1+2e) ii e >1/3 => v1<0 and v2>0
9. a 0.5 litres b i 2/3 ii 0.098 newtons iii 128 cm3
10. i y= √(8tan-1 x +1)

Q1 i 6g under root ii 6g under root , 3g not under root . ie (root 6g) divided by 3g
Q 10 all under square root and tan-1 =arc tan

Hope this helps!

Dicksboro_man

Hey guys,
didnt wanna bump an old thread but can anyone do 2005 q4B(i) pleeease?

MathsManiac

The solutions are in the marking scheme on the SEC website:
http://www.examinations.ie/archive/markingschemes/2005/LC020ALP1EV.pdf

Dicksboro_man

yeah, i checked them out before posting but didnt get how they got the answer
our teacher has all the solutions from over the years up on moodle but this question is missing... I consider myself a pretty good app.maths student, got an a1 in the mock and everything but i just dont get what they're at in that q. :rolleyes:

MathsManiac

The first thing that happens is that the 3 kg mass falls freely under gravity for 1.5 metres (because the string is 4 m long and the peg is 2.5 m above the table). The first three lines of the solution deal with that bit, telling you what the speed of the the 3 kg mass is at the instant before the string goes taut.

Then the string goes taut, so some of the momentum of the falling mass gets transferred (via the string) to the stationary mass. The principle of conservation of momentum is used to find the common speed of the two masses at the instant after the string goes taut. That's the final two lines of the solution to (b)(i).

Dicksboro_man

MathsManiac wrote: »

The first thing that happens is that the 3 kg mass falls freely under gravity for 1.5 metres (because the string is 4 m long and the peg is 2.5 m above the table). The first three lines of the solution deals with that bit, telling you what the speed of the the 3 kg mass is at the instant before the string goes taut.

Then the string goes taut, so some of the momentum of the falling mass gets transferred (via the string) to the stationary mass. The principle of conservation of momentum is used to find the common speed of the two masses at the instant after the string goes taut. That's the final two lines of the solution to (b)(i).

thanks a million buddy, i see whats happening now