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Leaving Cert Applied Maths 2017 Aftermath

Mldj

Im sorry but did anyone else think that was the worst applied maths exam ever?
Like I went in there today hoping for a H1 but after doing that paper I'll honestly be lucky to get a H4. I'm actually shaking. I spend four classes a week for the last two years working my arse off to get a good grade in this subject and for it to end like that is such a disappointment. I feel like I might as well have not done the subject at all and concentrated on the others. What tf was that wedge question? Wtf was that spacecraft thing? Sorry for the rant but im so pisse off

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carefulnowted

I feel the exact same as you. I'm trying to forget about it because we're FINISHED at last, but I know what you mean.

I had a lot of trouble with Q10, the last part of Q4, and the first part of Q3 (not that I got everything else right, but those were the questions that stumped me)

VG31

That must have been the hardest exam in years! What were you meant to do in Q1B? I though it might be something to do with angular velocity but I had no idea how to get it. The wedge question was very confusing as well.

I was hoping for a H3 but I probably got a H6. I feel like all the time I spent doing applied maths after school was a waste since I reckon I didn't get below a H4 in any other subject.

judeboy101

Remember its graded on a curve. Loads of ppl had a bad experience the year the "tennis court" q3 paper and the adjusted the entire marking scheme to fit the curve. If you were always a H1 you will be after this, correctors conferences can work miracles

Red F Warrior

Just think, if ye didn't work your ass off for those two years, ye may have failed it so ye didn't waste your time. It sounds like it was a crap paper but it is an example of how things don't always go to plan. It may end up being marked very easily if it was really tough. Forget about it and enjoy your summer.

Liordi

I'm very very annoyed with that exam.

I thought that 1B, 2B, 3A, the ending of 4A and the phrasing of 5A&B were very very difficult.

japester

As was mentioned earlier, the exam will definitely be marked to fit a curve so don't fret too much. I will post up my answers to the exam over the next few days for anyone that might be interested in comparing. But the key thing now is to try to put it to the back of your mind and enjoy yourselves fully now the shackles are off

Liordi

japester wrote: »

As was mentioned earlier, the exam will definitely be marked to fit a curve so don't fret too much. I will post up my answers to the exam over the next few days for anyone that might be interested in comparing. But the key thing now is to try to put it to the back of your mind and enjoy yourselves fully now the shackles are off

Did you think it was considerably harder than previous years?

lostatsea

It was a badly thought out paper.
Question 4 (b) mentions the slope beta which is the angle. I felt part (ii) should have been asked before part (i) and the hint given caused confusion.
In relative velocity you were asked to find the true velocity and then it was no longer required as you reverted to the relative velocity. Then they talked about being visible for 12 minutes - but 12 mins from what. I presume they meant 6 mins either side of the closest approach.
It was full of algebraic results.
Part b of differential equations may have required some knowledge of gravity from Physics.
It is hard to know how the SEC is pitching this subject.

Liordi

Was 1b very random too? I've done exam questions from like 20 years and I've never seen anything like it.

Also, what was 3A (ii), 'find how long it takes them to reach p' even though they started at P?

lostatsea

I agree Liordi. I talked to a number of teachers tonight and they felt it was an unsatisfactory paper. Whoever is writing the paper is loose with language which is a weakness when dealing with mathematical subjects. I'll go through the paper tomorrow and make further comments.

japester

Liordi wrote: »

Did you think it was considerably harder than previous years?

Hi Liordi, I did my LC more than 25 years ago and each year I torture myself by doing the Applied Maths and Maths Higher Level papers to see how I'd get on. I do them at my leisure though, which is an important difference, so I only did 3 questions from the AM paper yesterday evening. There's no way I'd be sharp enough to complete either Maths or AM papers at all within the given time frames (personally I'd like to see both Maths papers and the AM paper given 3 hours each rather than 2.5 hours!! And I can't get over the fact that there are absolutely no choices at all on the LC Maths paper, there were good choices back when I did my LC).

Back to your question, Q1 wasn't too bad from my own perspective, Q2 I thought was fine also and Q3 (a) was tricky in that you needed to use Sy = -h as it was dropping from a cliff. Also there was definitely a mistake in Q3 where it referred to "reaching P" (it is being thrown from P) rather than "the ground" - this would certainly have thrown some people and I'd imagine that Q will be marked a little easier as a result.

As I say, I'll put up some of my own answers when I've got a few more questions done, maybe later today.

Mldj

I went back over the paper (something I promised myself I wouldnt do lol). I thought questions 1 and 2 were fine, I thought 3a was hard. I tried to solve it using the sec^2X = 1 + Tan^2X method but TanX=0 kept appearing as an answer for me which just didnt make sense. Question 4a was fine. I set up every single equation for the wedge and still couldnt prove what they wanted...I resolved the forces for the particle horizontally and vertically, and I resolved the forces on the wedge itself horizontally and vertically. Nothing! I do physics and I found 10b very dofficult to understand. Also, for question 5a could anyone tell me if it is okay to have your answers in terms of m?

jogd1234

Mldj wrote: »

I went back over the paper (something I promised myself I wouldnt do lol). I thought questions 1 and 2 were fine, I thought 3a was hard. I tried to solve it using the sec^2X = 1 + Tan^2X method but TanX=0 kept appearing as an answer for me which just didnt make sense. Question 4a was fine. I set up every single equation for the wedge and still couldnt prove what they wanted...I resolved the forces for the particle horizontally and vertically, and I resolved the forces on the wedge itself horizontally and vertically. Nothing! I do physics and I found 10b very dofficult to understand. Also, for question 5a could anyone tell me if it is okay to have your answers in terms of m?

Yes I have no idea how to get the kpcosx form they were looking for in 4. In 5a AFAIK it came out as a number, not in terms of m.

Overall I think q1 was mostly okay (not sure though), 2 was fine, 3 was fine, the wedge part on 4 was very tough and I stupidly didn't even attempt the bit about the reaction on the particle in q4. After that q5 was fine, and I thought 10 was good even though I didn't check back over it & everyone's complaining. Q4 will be my downfall, still hopeful I did enough to get the H1. Glad to be done anyway 😀

japester

Here are some of my own answers for the first 5 questions:

Q1 (a) k = 2
(b) (i) sqrt(2gr) m/s (ii) d = r/u metres (iii) r = 5 metres

Q2 (a) |Vw| = 4.07 m/s, direction is 79.38 degrees North of East
(b) (i) Vq = 15sqrt(3)/2 i - 15/2 j km/h
(ii) Theta = 35.71 degrees
(iii) Distance from P to Q at this stage = 10.57km

Q3 (a) (i) The 2 possible angles of projection are zero degrees and 71.565 degrees

(ii) if theta = 0 degrees, time it takes to reach the ground is sqrt(2h/g) seconds

if theta = 71.565 degrees, time taken is sqrt(20h/g) seconds

(b) (i) worked out fine by doing the necessary calculations for determining time when Sy was zero and then subbing time value into the formula for Sx

(ii) worked out fine also by noting that if the particle strikes the plane at right angles then Vx = 0 upon impact when Sy = 0. Subbing the time value at the point of impact (already determined) in to the formula for Vx allows the sine of the angle of projection to be determined as 1/sqrt(3), from which the cosine can be determined as sqrt(2/3). Then using the range formula already determined sub in for Sin(theta) and Cos(theta) to get the desired result.

Q3 was a fairly long question though and you'd be seriously pushed to get it done in 25 minutes even with all your wits about you!

Q4 (a) (i) T = 8mg/5 N (ii) B has risen 0.0136m at this point (iii) R = 6mg/5 N

(b) (i) diagram showing the forces involved on the particle and wedge

(ii) I had to cheat here because I was getting nowhere fast and had made a hames of the equations of motion!! Anyway, once I had determined the correct equations of motion involved I was able to use them to determine k as 0.2. Even with the equations of motion, this was not a trivial task by any means - I did it as follows:

From the equation involving forces parallel to the direction of motion for the particle I determined that p = gSinB/(1-k(CosBCosB))

The normal reaction on the wedge from the particle was given by
R = 4mkp/tanB

I then subbed these into the equation involving forces perpendicular to the direction of motion for the particle to give me

mgCosB - 4mkgSinB/(tanB(1-k(CosBCosB))) = mkgSinBCosBSinB/(1-k(CosBCosB))

after dividing across by mgCosB I then get

1 - 4k/(1-k(CosBCosB)) = kSinBSinB/(1-k(CosBCosB))

multiplying across by (1-k(CosBCosB)) I then get

1-k(CosBCosB) - 4k = kSinBSinB so

1-k(CosBCosB + SinBSinB) - 4k = 0

1-k(1) - 4k = 0 so 5k = 1 and k = 0.2 ..... very hard work!!

(iii) Already have p = gSinB/(1-k(CosBCosB)) from earlier so I sub in for k and do some mathematical manipulation to give me

p = gSinB/[0.8 + 0.2 - 0.2CosBCosB] = gSinB/[0.8 + 0.2(1-CosBCosB)]

= gSinB/[0.8 +0.2SinBSinB]

Multiplying above and below by 5 I then get

p = 5gSinB/(4 + SinBSinB) = 49SinB/(4 + SinBSinB)

It was another very long question though and you'd need to be really on your game to be able to manipulate the equations correctly to work out k.

Q5

(a) (i) v = 1.714 m/s (ii) e = 0.857
(b) (i) Speed of P before collison = 70v/(27CosA) m/s
(ii) I calculated the angle B, the angle P moves off at after the collision, relative to the line joining the spheres centres, as 80.173 degrees. So I reckon the angle through which the motion of P is deflected is therefore 30+ 90 + (90-80.173) = 129.83 degrees.

I found Q5 doable (apart from the slips I made) but again you'd have to be on your game to get it done in 25 minutes!

I will try to tackle a few more questions tomorrow and put up my answers to them

reason vs religion

Well done as always, Japester. Inclined to agree with you that it wasn't as hard a paper as it at first might seem. Q6a was needlessly convoluted, though, and Q10b was somewhat unusual. But, then, it is Applied Maths, and those sorts of uncertainties should be expected.

Just a quick quibble with your answers. I've only worked through Q1, and I think b (ii) should have been R=gr/μ, giving an answer of r=5 for (iii).

japester

Thanks a lot Reason, I still get a kick out of having a go at these questions but I think I am getting slower every year with them! I just had a look through my solution to Q1 (b) (ii) there and you could very well be right but I can't see where I might be going wrong with it. For me, from Q to R the frictional force is coming into play and this is what brings the baggage to a halt in the end. So, the way I am seeing it, is that the Frictional Force Fr = -(mew)R = -(mew)mg and this is equivalent to ma, making the acceleration a = -(mew)g, effectively a decelerating force making it negative.

I am then plugging this result into my kinematics formula

v*v = u*u + 2as to give me 0 = sqrt(2gr)*sqrt(2gr) +2(-mew)g(d)

giving me 0 = 2gr - 2(mew)gd and so I get d = r/mew

Definitely let me know if I have made a mistake here as I am always looking to improve!

reason vs religion

Sorry, japester, you're absolutely right. I mistakenly used the mass rather than weight as the resultant force. Bit of a rookie error. I still think you may be wrong with the third part, though. Forgivingly, because not having g in my acceleration means the g doesn't cancel from my answer for R in (ii) the answer both ways is r=5.

japester

For anyone who might be interested here are my answers to Q6 and the first part of Q7. I tried all kinds of things to get Q7(b) out but no joy I'm afraid. If anyone has a solution to this part I would love to see it.

Q6(a) 1:sqrt(2)
(b)

(i) I worked this out eventually by using the Law of Conservation of Energy to get an expression for v*v and then copping that SinB = (2a-h)/3a to give me an expression for h in terms of SinB. I then subbed in for h in the v*v equation to give me the desired result.

(ii) The equation for T = 3mgSinB - mg. Therefore, as far as I can see anyway, the max value for T here is the value where B is a max but angle B is limited by the size of angle theta. Know that Cos(theta) = 2/3 and this is also the max value of SinB here so max tension is T = mg N. The minimum tension allowed in this case, given that the string is meant to remain taut at all times, must be zero N since if it falls below this the string is slack. If the value of angle B is low enough then it will turn out that 3mgSinB < mg and then the string becomes slack.

Q7

(a)

(i) Reaction at Q = 61.2 N and reaction at P is 16.97 N

(ii) As I mentioned earlier, I tried many things but had no luck coming up with the desired result here

japester

reason vs religion wrote: »

Sorry, japester, you're absolutely right. I mistakenly used the mass rather than weight as the resultant force. Bit of a rookie error. I still think you may be wrong with the third part, though. Forgivingly, because not having g in my acceleration means the g doesn't cancel from my answer for R in (ii) the answer both ways is r=5.

Hi Reason, for the next part I see the speed at point Q is sqrt(2gr) m/s and we are told the speed when the baggage was half-way along the length d is 7 m/s so I am then using the formula

v*v = u*u+2as to get a value for r

This becomes 7*7 = sqrt(2gr)*sqrt(2gr) + 2(-mew/g)*(r/[2*mew])

so 49 = 2gr-r/g which gives me r = 2.513m

I'm wondering if you used d instead of d/2 in your value for distance for the formula, but I still don't think that would have given you an answer of 5. Your answer looks to be around twice mine.

reason vs religion

japester wrote: »

Hi Reason, for the next part I see the speed at point Q is sqrt(2gr) m/s and we are told the speed when the baggage was half-way along the length d is 7 m/s so I am then using the formula

v*v = u*u+2as to get a value for r

This becomes 7*7 = sqrt(2gr)*sqrt(2gr) + 2(-mew/g)*(r/[2*mew])

so 49 = 2gr-r/g which gives me r = 2.513m

I'm wondering if you used d instead of d/2 in your value for distance for the formula, but I still don't think that would have given you an answer of 5. Your answer looks to be around twice mine.

In your second line, you sub in a=-μ/g, when it should be a=-μg.

I'll have a look at that elusive Q7!

reason vs religion

This is 7b. What threw you?

japester

reason vs religion wrote: »

In your second line, you sub in a=-μ/g, when it should be a=-μg.

I'll have a look at that elusive Q7!

Thanks very much for that Reason, isn't it crazy that I made the mistake originally but then never copped it either when posting up my solution

You are on the ball, I'll take a peek at Q7 now.

ray giraffe

5(b) I'm getting speed=10v/sqrt(1+99(sinα)^2) and angle of deflection is 50.17 degrees

6(b) I'm getting maximum tension is 2mg since the string will swing back to vertical

7(a) (Not sure on this one) I'm getting reaction at Q is 4.583 at angle 40.89 to the horizontal and reaction at P is 6.928 at angle 60 to horizontal

japester

reason vs religion wrote: »

This is 7b. What threw you?

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

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.

japester

ray giraffe wrote: »

5(b) I'm getting speed=10v/sqrt(1+99(sinα)^2) and angle of deflection is 50.17 degrees

6(b) I'm getting maximum tension is 2mg since the string will swing back to vertical

7(a) (Not sure on this one) I'm getting reaction at Q is 4.583 at angle 40.89 to the horizontal and reaction at P is 6.928 at angle 60 to horizontal

Hi Ray,
I could easily have made a mistake in my own calculations but I've had a look over Q5 (b) (i) and I still get the same value for speed of 20v/27CosA m/s - maybe Reason or another reader of the thread can tell us what they got.

On Q6(b) you are dead right to say that the theoretical max tension is 2mg once the particle swings back to the vertical (making B = 90 degrees at that point). I may have been reading into the question too much as I was taking it that the angle B had to be limited to some value between 0 and (90-theta), meaning for me the maximum tension would be when B is exactly 90-theta and that's how I was coming out with Tmax = mg N rather than your value of 2mg N.

With Q7 (a) I have

forces up = forces down => Rpeg = 18g

looking at the forces acting on the rod QR and resolving these vertically I get

Rpeg = 6g +RqSinA + RrSinA = 6g + 2RSinA = 18g => RSinA = 6g

(I am taking it that by symmetry Rr = Rq = R)

looking at the forces acting on the rod QP and resolving these vertically I get

RSinA = 6g + Rpdown => 6g = 6g + Rpdown so Rpdown must be zero meaning that the reaction at P must only have a horizontal component

Now taking the moments of all forces on the rod QP about P gives:

6g(0.15Sin30) + RCosA(0.3Sin60) = RSinA(0.3Cos60)

As a result of this I end up with angle A = 73.897 degrees, thus making Rq = R = 61.2 N.

Knowing that the reaction at P has only a horizontal component means that Rphorizontal = RrCosA = RCosA = 16.97 N

Again, my calculations could easily have gone awry here so any feedback from other readers of the thread is most welcome

ray giraffe

My 5(b)(i) attached

(Turns out it's wrong! as japester noted!)

japester

ray giraffe wrote: »

My 5(b)(i) attached

Hi Ray, I had a good look at your solution there and I may well be wrong, but I think there is an error where you are looking at the 3m object having a speed of v after the collision, the question states it is the 7m object that goes off with this speed.

Having read your solution I did spot another silly error in my own one though, which related to the NEL formula (I had relative velocity before collision as the numerator by accident

). Having worked with my own solution following this correction I end up with kind of similar results as follows:

u = 70v/27CosA m/s

I now get the angle the 3m object goes off at relative to the positive sense of the x-axis to be 85.05 degrees which (in my head at least!!) means the angle the object was deflected through was 30+90+4.95 = 124.95 degrees. As always, I stand corrected on these answers

ray giraffe

Thank you Japester!

I'm getting the 3m mass after collision to have velocity (ucos(30)/10)i+(usin(30))j

Same as yours?

Also question 7(a) has weight of each bar is 6 newtons, not mass of 6kg as you have.

I don't fully understand statics. Why do we not consider a tension in the rods PQ and PR?

japester

ray giraffe wrote: »

Thank you Japester!

I'm getting the 3m mass after collision to have velocity (ucos(30)/10)i+(usin(30))j

Same as yours?

Also question 7(a) has weight of each bar is 6 newtons, not mass of 6kg as you have.

I don't fully understand statics. Why do we not consider a tension in the rods PQ and PR?

Yep, your velocity for the 3m mass after the collision matches my own

Thanks also for letting me know about my slip in Q7 (a), can you imagine all the marks I'd be losing for these mistakes!!

ray giraffe

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