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* Applied Maths * predictions / discussion / aftermath (1 thread only please)
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@BL1993, I've just been tinkering around with Q1 (b) (iii) again (a sucker for punishment!) and my intuition tells me that the answer you're getting of 2t cannot be right because the time for the original journey is actually t+t1+t2 = 2t and the time for the journey with the speed limit must intuitively be longer than 2t. I am taking the distance to be covered as 3vt/2 (based on the original journey) and then I note that the time for the acceleration part of the second journey takes (2/3)t1 (this is the same t1 as in the first journey) and the time for the deceleration for the second journey takes (2/3)t2 (again, this t2 refers to the time for deceleration in the first journey - intuitively, these times are lower than the original times, as expected, because it will take less time for the body to get to the lower speed given the same acceleration and likewise for the deceleration.
So then I form an equation for the distances of the 2 journeys to get the time T, where the car is traveling with constant velocity in the 2nd journey. So, 3vt/2 = 2vt1/9+2vt2/9+(2v/3)T, giving me a value for T of (23/12)t (I am leaving t1+t2 = t here). Now, I can find the total time for the second journey by adding this to the accceleration and deceleration times to get (23/12)t + (2/3)t = (31/12)t. So the time for the second journey is a little over 2 and a half times the time taken for the original journey.
I could still be wrong about this final answer but I've a very strong feeling that the time must be greater than 2t, based on intuition alone.
If I'm wrong, I'd be very interested to see anyone elses solution.
@ everyone
I've just done Q2 and the answers I'm getting match with those of a few people. They are:
(a) -i - 8j , shortest distance = 49.61m (b) 28.96deg<=theta<=60deg0 -
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@medman101, I'm taking it that from (ii) t1+t2 = t. Therefore, for the first journey, the total time spent traveling is t1 + t + t2 = t + (t1+t2) = t + t = 2t. Hope this helps to see where I'm coming from 0
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I've just done Q3 and Q4. The answers I'm getting seem to be consistent with a number of people. They are
Q3(a) (i) 50 (ii) 71.56 deg (b) (i) 56.31 deg (ii) 0.7071 (that was a long question to do 3 hours would be much better for this exam - one slip and you're done for!!)
Q4(a) 2.1528m/s (b) 6 simultaneous equations :eek: the tensions are 73N, 24N and 21.9N0 -
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randylonghorn wrote: »Over 6 years!
And none of them are dissecting an applied maths exam the Saturday night after my LC finished, I'll guarantee you that!
Boards wasnt invented the saturday the day after my applied marys exam. Great subject. Still have murphy's book and look over it from time to time. Long living cert though0 -
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@geographic, I did plenty of lying in my own day too when my answers didn't exactly "fit" with what it should have been - and the amount of time I spent trying to work out answers that would be given in the question in the guise of various formulae!!! Especially trigonometry related ones where pg 9 of the log tables would be needed :eek: The exam would really need to be 5 hours long in my case as I'm a fairly slow thinker and the applied maths questions need some thought in many cases!!!
@godtabh, I'm in the same boat as you, except I sold on my Applied Maths book unfortunately. That lovely chocolate brown cover and those quotations!! It was by pure chance that I came across an old applied maths set of leaving cert papers lately and I got hooked!! I did my exam back in 1990 but didn't do very well I'm afraid so I said to myself, why did you only get a D in applied maths when you got an A in honours maths and a B in honours physics? - I guess I considered it my 7th subject at the time and between a very nice, but too easygoing, teacher (who in fairness did know his stuff whenever I had the mind to pay attention!) and the fact that I was just too lazy to knuckle down to study properly for it and the fact that I always seemed to mess up every question and the fact that I used to spend way too much time trying to get quoted anwsers, it just didn't happen for me I've been going through past papers lately and couldn't wait for this years exam to be put online so that I could try it out. I'm like a dog with a bone now :P I understand where Randy is coming from though, and believe me, I certainly enjoyed myself after I had finished that applied maths exam back in 1990 
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Not too many people seem to have done Q8 (moments of inertia) but for anyone who might have, here are the answers I am getting
(a) standard proof
(b)
(i) 5.886 rad/s
(ii) 1.74s, the length of the equivalent simple pendulum is 75.15cm
This is a relatively short topic and its surprising more people don't attempt it as part (a) is almost always a standard proof (there are only 5 of them in total to remember) and (b) usually involves reasonably straightforward calculations involving PCE and formulae for compound and simple pendulums. Of course, I can't really talk as I never went near this chapter when I was doing the LC
The look of the proofs were enough to send a shiver down my spine :eek: 0 -
Not too many people seem to have done Q8 (moments of inertia) but for anyone who might have, here are the answers I am getting
(a) standard proof
(b)
(i) 5.886 rad/s
(ii) 1.74s, the length of the equivalent simple pendulum is 75.15cm
This is a relatively short topic and its surprising more people don't attempt it as part (a) is almost always a standard proof (there are only 5 of them in total to remember) and (b) usually involves reasonably straightforward calculations involving PCE and formulae for compound and simple pendulums. Of course, I can't really talk as I never went near this chapter when I was doing the LC The look of the proofs were enough to send a shiver down my spine :eek:
Yeh usually a nice, quick question, got full marks in it in my mocks. Think those answers look familiar, can't fully remember though didn't take any down..:p I knew I'd spend too much time checking every single one if I did...0 -
For anyone who attempted the SHM/Circular motion question (Q6) - it was a very doable question to be fair - the following are the answers I got:
(a)
(i) In order to prove that the body is undergoing SHM you must prove that the acceleration of the body from a fixed point O is proportional to the displacement of the body from this same fixed point and in the opposite direction to the displacement. So, if the displacement is x, we need to prove that acceleration = -kx to prove it is performing SHM.
Here x = asin(wt + e). We differentiate this twice wrt time to get the acceleration. Doing this I end up with acceleration = -sqr(w)asin(wt+e) = -sqr(w)x. We are told that w is a constant so acceleration is proportional to x and the sign tells us it is acting in the opposite direction to the displacement
(ii) w = 4 rad/s, a = 2.6m, e = 0.3947 rad
(b) w = 7 rad/s
Only one more question to do now 0 -
I think I may be the only one left here I've just completed the paper now by doing Q7 (statics) and I'm getting the following answers:
(a) (i) To show that the extension is 7cm is a very quick calculation. We are told that the angle is arctan(3/4) and we know that the cosine of the angle can also be got by dividing 32 by the (new) length of the string. The new length of the string must be 40cm (since 32/40 = cos(alpha) = 4/5). And so the extension must be 40-33 = 7cm
(ii) I'm not 100% sure if I am doing this part right. I am taking it that the extension in the string is caused by the tension force along that string. I equate the tensions and the weight to get T = 196N and then I use Hooke's Law to find the string constant. This gives me a value of 134.4, maybe someone could confirm that for me if they have it done.
(b) just spotted a mistake I made in this part - I'll post my solution to it as soon as I've figured it out 0 -
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@geographic - thanks for the support, its needed in this question!!! Wow, that part (b) is a proper humdinger!! That’s one that I’d have spent way too much time on in the exam back in the day trying to prove it – and still not getting it right most likely Statics was never my strong point!!
I started by drawing the force diagram and then equating forces up and down. I then took moments about point B and equated the clockwise with the anticlockwise. This was the part I spent loads of time on – I couldn’t figure what the perpendicular distance of the weight line of force from point B was. Eventually (after a long time!!) I saw that the angle around B is (180 – 2theta) and so the angle I used for finding the perpendicular for the weight force was (90-2theta). So that moment was WCos(90-2theta) = WSin(2theta) = 2WSin(theta)Cos(theta) (you really must know your trig well). Then I had 3 equations to work with:
TSin(theta) = R
W – mew(R) = TCos(theta) and
T = WCos(theta)
Eliminating T and R from the second equation I ended up with 1/Cos(theta) – mew(Sin(theta)) = Cos(theta). Mutiplying across by Cos(theta) and rearranging and noting that 1-sqr(Cos(theta)) = sqr(Sin(theta)) gave me
–mew(Sin(theta))Cos(theta) = sqr(Sin(theta))
Isolating mew gave me mew = -tan(theta). tan(theta) can be negative, in which case mew is positive and so mew > tan(theta). Likewise tan(theta) can be positive, in which case mew is negative. I'm chancing my arm here but I'm going to say that mew must always be positive and so we can ignore this "absurd" possibility. Therefore it must always be the case that mew >= tan(theta) [note that I may have taken the friction force in the wrong direction to begin with, if I had taken the other direction I would end up with mew = tan(theta) and again, because mew must always be positive mew >= tan(theta).
I could easily have made an error here though so if anyone sees any error with it, you might let me know.
Thats it until LC 2012 I guess!! Hope ye all have a great summer break - I'll check back every so often just to see if anyone has posted solutions to the questions, especially this one, as it especially tricky.
:pac:0 -
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Thanks for the feedback geographic. I've been checking out that last part again and I believe the answer I should have got was mew = tan(theta) (I took the direction of F to be wrong in the original case).
This corresponds to the general definition of mew, which is the ratio of the frictional force F to the normal reaction R i.e. F/R. If you think of it in terms of a diagram, then theta would be the angle of friction (the angle between the resultant of F and R and R itself), so in the traditional diagram illustrating mew, F, R and the angle of friction (theta), tan(theta) would actually be equal to F/R. Now (assuming the simple case of a book on a rough table, where mew > 0) as long as the force acting on the book remains below the value of the limiting friction force (the max amount of frictional force that can be called into play) then the book will not move (it is in equilibrium). When the value of the force acting on the book equals the frictional force F, then the limiting frictional force has been reached and the book is just at the point of moving (in limiting equilibrium). And finally, when the force acting on the book exceeds the limiting frictional force then the book moves. Remembering that tan(theta) = F/R then, if the frictional force called into play is less than the limiting frictional force, tan(theta) is smaller (as F is smaller) and mew will be > tan(theta) in this case. When tan(theta) = Limiting Frictional Force/R then mew = tan(theta) and the body is at the point of limiting equilibrium.
So, after all that, to answer the question, once we have that mew = tan(theta) [the limiting frictional force case], we can automatically say that mew >= tan(theta) for all cases where the system is in equilibrium.
One other piece of useless (?) information on coefficient of friction - it is actually possible for the coefficient of friction to be > 1. It just means that the frictional force exceeds the normal reaction force for the 2 surfaces in contact. Apparently, the coefficient of friction between cast-iron and copper is 1.05 and and the coefficient of friction between rubber and some other materials can be easily between 1 and 2!! I'm still learning all these years later 0
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