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

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?

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😂