Leaving Cert Applied Maths 2017 Aftermath

japester

ray giraffe wrote: »

(ucos(30)/10)i+(usin(30))j is at an angle 80.17 degrees to the i direction yes?

Hi Ray, another slip on my part I'm afraid This time I forgot to multiply by 0.5 in an equation! So with that sorted, I also am getting the 3m mass going off at the angle you state above. Is this the answer to the question directly or is "the angle through which the direction of motion of P is deflected as a result of the collision" 30 + 90 + (90-80.17) = 129.83 degrees instead? Many thanks for your contributions.

ray giraffe

I'm getting the angle of deflection to be 80.17-30=50.17 as both angles are measured anticlockwise from the vector i.

japester

ray giraffe wrote: »

I'm getting the angle of deflection to be 80.17-30=50.17 as both angles are measured anticlockwise from the vector i.

That sounds very sensible to me Ray .... my brain needs this exercise Did you work through Q8 yet? Again I could be wrong on it but I'm getting 21ma^2 for the MoI for the lamina, I'm able to show the value in (b) (ii) fine also using moments about A and I'm ending up with a very large value of 86.73 degrees for theta which I am very suspicious of

ray giraffe

The mass of the material that would be in the hole is m

I'm getting
[(1/2)(4m)(4a)^2+(4m)(4a)^2]-[(1/2)(m)(2a)^2+(m)(5a)^2]=69ma^2

I'm ending up with exactly 120 degrees for theta!

japester

ray giraffe wrote: »

The mass of the material that would be in the hole is m

I'm getting
[(1/2)(4m)(4a)^2+(4m)(4a)^2]-[(1/2)(m)(2a)^2+(m)(5a)^2]=69ma^2

I'm ending up with exactly 120 degrees for theta!

Thanks a million once more Ray, I've made a few more slips along the way I had the exact same expression as you for the MoI above except for some unknown reason I used 2a rather than 4a for the parallel axes part for the disk as a whole, which accounts for my MoI value.

Beyond this I also made a serious blunder in reading the question, where I read sqrt(11g/23a) as a linear velocity rather than an angular one!!

Having fixed that and working with my original equations for total KE lost = total PE gained, I do indeed come out with the same value of theta = 120 degrees as you. I guess this value is perfectly valid and just means that the lamina gets close to performing a full revolution given the initial angular velocity applied but just stops short of it by 60 degrees.

Many thanks again for all your contributions with these, they really do help me bigtime

japester

For Q9 I ended up with the following answers:

(a) volume of iron = 19.476 cm^3

(b)

(i) depth of mercury = 5.735 cm

(ii) volume of oil required = 2265 cm^3

As always I'm open to correction on any and all of these

japester

Last but not least, my answers to Q10 are as follows:

(a)

(i) v = 2.5(1-e^-10t) m/s

(ii) t = 0.2302 seconds and distance traveled = 0.35 metres

(b)

(i) proved this by just noting that at the surface of the earth, where x = R, the acceleration is only due to gravity, g => at that point F = ma = mg = k/(R*R) and so k = mg(R^2)

(ii) I reckon I may have gone wrong somewhere on this part so all help would be appreciated. I started off with a = gR*R/(x*x) and then used a = v.(dv/dx) to determine v in terms of x using integration. However, I end up with

v = sqrt[2gR(1/5-R/x)] for v

the problem is that when the spacecraft strikes the earth x = R and so 1/5 - R/R becomes -4/5 and I end up with a negative value to get the square root of!! No doubt another slip on my part but any help is appreciated.

Quinoalike

japester wrote: »

Last but not least, my answers to Q10 are as follows:

(a)

(i) v = 2.5(1-e^-10t) m/s

(ii) t = 0.2302 seconds and distance traveled = 0.35 metres

(b)

(i) proved this by just noting that at the surface of the earth, where x = R, the acceleration is only due to gravity, g => at that point F = ma = mg = k/(R*R) and so k = mg(R^2)

(ii) I reckon I may have gone wrong somewhere on this part so all help would be appreciated. I started off with a = gR*R/(x*x) and then used a = v.(dv/dx) to determine v in terms of x using integration. However, I end up with

v = sqrt[2gR(1/5-R/x)] for v

the problem is that when the spacecraft strikes the earth x = R and so 1/5 - R/R becomes -4/5 and I end up with a negative value to get the square root of!! No doubt another slip on my part but any help is appreciated.

You're answers to Q8 didn't post properly for me ,any chance you could post again ? Also Q5 also worked out 20/27cosa for myself. Thanks!

japester

Quinoalike wrote: »

You're answers to Q8 didn't post properly for me ,any chance you could post again ? Also Q5 also worked out 20/27cosa for myself. Thanks!

Hi Quinoalike, for Q8 I got the following answers (many thanks to Ray Giraffe for setting me straight on these!!)

(a) text-book proof

(b)

(i) Moment of Inertia = 69(a^2)m (Ray showed how this is worked out in an earlier post)

(ii) I was able to prove that the distance of the centre of gravity from A is 11a/3 by effectively turning the lamina on its side and then taking moments about the point A. Basically, if x is the distance of the CoG from A then

3mx = 4m(4a)-m(5a) since 3m is the mass of the lamina with the hole in it and the sum of the moments is the moment of the sum about A.

From here you'll see that x = 11a/3 as required

(iii) In this part I used the Law of conservation of energy to basically say the kinetic energy lost = potential energy gained as the lamina rises. So

0.5I(w^2) = 3mgh

I = 69(a^2)m and h = (11a/3)(1-CosT), while w = sqrt(11g/23a)

so it turns out that 1.5 = 1 - CosT giving a value for angle T of 120 degrees (thanks again to Ray for setting me straight on this one!!)

hope this helps you out

Also, if anyone could check out my answers for Q9 and Q10 I would greatly appreciate any feedback, I'm pretty sure I messed up somewhere in Q10 (b)

ray giraffe

All Japester's answers perfect except for:

Last part of 9 should be 996.6 cm^3
[a member of https://groups.google.com/forum/#!forum/appliedmaths also got this answer]

Last part of 10, remember the acceleration is the direction of decreasing x, so a=-gR^2/x^2

japester

ray giraffe wrote: »

All Japester's answers perfect except for:

Last part of 9 should be 996.6 cm^3
[a member of https://groups.google.com/forum/#!forum/appliedmaths also got this answer]

Last part of 10, remember the acceleration is the direction of decreasing x, so a=-gR^2/x^2

Thanks a million Ray, I knew I slipped up somewhere in 10 (b) Is there any chance you could outline the solution for the last part of 9b so that I can see where I was going wrong? Unfortunately I can't see the solution in the link provided unless I join the group, which I prefer not to right now.

japester

Hi Ray, you can ignore my earlier request for a solution to Q9 (b) as I managed to spot my mistake - once I had determined the depth, d, of the oil I was mistakely using the formula for the volume of the oil displaced by the cylinder to calculate the volume of oil Once I used the correct formula

pi*d*(Rtank*Rtank-Rcyl*Rcyl)

I ended up with 996.7 cm^3 which tallies with your own and the one in the google group.

So that's another AM paper put to bed for another year. I'm always interested to see how I'll fare out each year and how much I can remember 12 months later.

Overall I think it was a fair enough paper with not too many surprises for those who had studied well for it and had done plenty of past exam papers.

Now just have to wait another 12 months for the next one!! How sad am I

ray giraffe

The general opinion is that it was a hard paper. A large proportion of students struggled to finish 6 questions in the 2hrs30mins.

Glad you enjoyed working through it!

japester

ray giraffe wrote: »

The general opinion is that it was a hard paper. A large proportion of students struggled to finish 6 questions in the 2hrs30mins.

Glad you enjoyed working through it!

Thanks again Ray for all your help there. It's quite a difference when you're in your 40s and able to take each question at your lesiure though and hum and haw about them at length!! If I was doing that exam for real, I would really struggle to complete 4 questions in the time given as I just think too much about the questions before I am comfortable putting pen to paper. I would really love to see that exam made 3 hours long for the sake of students but I guess that is extremely unlikely, especially given the generally high marks achieved by students for the paper.

For those who took the exam for real though, I am confident that the curve will apply and those who generally achieved high marks all along will still do well out the paper. Time to kick back and relax now though, and put it far far from your minds

Mldj

I read through all the answers here and im feeling much more at peace with the paper. The only question I made a COMPLETE balls of was question 5 which is usually quite safe. Ive decided to stop dissecting the paper now, time to let it go😂

reason vs religion

japester wrote: »

Well I was the one making rookie errors now You see I ended up with

2N/Sin(theta) = 2W for my forces up = forces down equation...

Another mistaken division!

...and also, when taking moments about B I ended up with

NxSin(theta) = WaSin(theta) [x for me was the distance from peg to B]

so I only have myself to blame for such basic mistakes but sincere thanks to you for posting up your solution there, I had tried so many combinations you wouldn't believe it! When you think of it, wasn't Q7 (b) relatively short? Still and all, I guess you can understand why many students won't tackle the statics question though.

Yeah, that x business got a bit messy!


I'm not great at judging the standard, in truth. I am somewhat sceptical, though, of the accounts of students fresh out of the exam. And I think teachers can sometimes have their judgement clouded by those accounts. I was startled by the opinion of Noel Cunningham, the respected teacher, who posted here a year or two ago decrying the standard of that year's exam as difficult, when it seemed unequivocally manageable, even gentle, to me. As you say, the marking process involves manipulating students into a predetermined distribution; provided the exam poses the same difficulty to all students, that process means one should do just as well as with an easy paper. A difficult paper would only become problematic if it stressed some students, causing them to underperform, or if the questions were significantly unusual such that a degree of luck became involved in whether students came upon the right answering strategy, or if it prevented some students from penetrating questions. I don't think this year's paper is in danger of any of those.

You ask about Q7, though. I'm not sure. It's definitely shorter than ones from previous years, many of which have been among the most challenging on their respective papers. But I think that, while short, it's a distillation of most of the concepts from that question. A comprehensive Q7; definitely not easy.

See you next year!

tina1040

Were results better than expected?

carefulnowted

For me, yeah definitely. I managed a H2 when I was just hoping not to fail after that exam.

judeboy101

carefulnowted wrote: »

For me, yeah definitely. I managed a H2 when I was just hoping not to fail after that exam.

The A1's halved according to stats

Mldj

I got a H1. I started a rant post about the exam and then ended up getting a H1. Like I was actually shocked I did not expect that at all😂

Liordi

I ended up with a H3. Slightly disappointed as I didn't use it for points so it feels like it was pointless having it as an 8th subject.