* Applied Maths * predictions / discussion / aftermath (1 thread only please)

ride-the-spiral · 24-06-2011 4:49pm

mr_brightside wrote: »

Does anyone know is there such thing as attempt marks, or is it the five or nothing? Is it in any way like the maths where one stupid slip only costs you a mark if all the actual maths bits are right?

Though Q6 was unreal, seemed a bit too good to be honest.. any one get omega = 7 ? And then 10 a i think i made a balls of, should it have worked out really evenly or was it sort of awkward?

Yeup, the applied maths marking scheme just doesn't emphasise it as much as the normal maths MS.

And to Brian, sorry but I can't actually remember and didn't write my answers down, all i remember was they weren't very "nice".

KoolAidRelic · 24-06-2011 4:51pm

A slip is -1, and a blunder is -3 - just like normal maths.

I got omega as 7 as well.

mr_brightside · 24-06-2011 4:53pm

Also in the linear motion, the shortest time i got was (11t/6), which doesn't make sense cause that'd be shorter than the original journey which was at a greater speed... anyone get it out?

And just one more (i swear), alpha in q3(a)... 75. something anyone?

Athos902 · 24-06-2011 4:55pm

I thought it was quite nice overall.

Did anybody else get e=-0.75 for Q5 (b) (ii)?

Geog ariphic · 24-06-2011 4:57pm

mr_brightside wrote: »

.. any one get omega = 7 ? And then 10 a i think i made a balls of, should it have worked out really evenly or was it sort of awkward?

Got c=7, i think it was all logs.

anna_k · 24-06-2011 5:42pm

Athos902 wrote: »

I thought it was quite nice overall.

Did anybody else get e=-0.75 for Q5 (b) (ii)?

i'm probably completely wrong here but surely e must be positive?
(btw, i can't remember what i got for e exactly...some nice positive fraction)

john1741 · 24-06-2011 5:48pm

Athos902 wrote: »

I thought it was quite nice overall.

Did anybody else get e=-0.75 for Q5 (b) (ii)?

I got +0.75, take the absolute value because 0<e<1

BL1993 · 24-06-2011 6:23pm

I can confirm that these are the correct answers for question 10:

Q10) a) y=66.231
b) i) 44.629m
ii) 0.336 ms-1

japester · 24-06-2011 6:38pm

I could be wrong but I've just done Q10 there and I get the same results as you for (a) and (b)(i) but for b(ii) I am getting 16.249m/s which would seem about right given that u = 40m/s and v = 0m/s (in the case of a uniform deceleration, we would expect the average speed to be 20m/s). Your value (I think!!) is way too low.

Geog ariphic · 24-06-2011 6:39pm

@BL1993, that last one looks VERY familiar - if i was a gambling man i'd say that was what i had down for average velocity. Did u just divide max distance by total time aswell? xD

BL1993 · 24-06-2011 6:48pm

@japster: did you intergrate again using dv/dt and not vdv/dx? Because you will be getting a tan inverse and not a log for the speed side. Well it may appear to be low but the maths doesn't lie. If the time it takes to reach 0 is big then the average speed is going to be low.

@Geo: Yes, it's the only way it can be done

electrictrad · 24-06-2011 6:53pm

Athos902 wrote: »

Did anybody else get e=-0.75 for Q5 (b) (ii)?

anna_k wrote: »

i'm probably completely wrong here but surely e must be positive?
(btw, i can't remember what i got for e exactly...some nice positive fraction)

Got a nice positive fraction too. . .7/8. . .I'm probably wrong though. . .

BL1993 wrote: »

I can confirm that these are the correct answers for question 10:

Q10) a) y=66.231
b) i) 44.629m
ii) 0.336 ms-1

I got a) y = 2e^3.5
b) i) 608m approx
ii) around 4.3 m/s

. . .so I hope to God you're not the SEC. . .

mr_brightside wrote: »

Also in the linear motion, the shortest time i got was (11t/6), which doesn't make sense cause that'd be shorter than the original journey which was at a greater speed... anyone get it out?

And just one more (i swear), alpha in q3(a)... 75. something anyone?

I got shortest time as 31t/12, and x=50 for Q3a) i). . .how did people do ii)?. . .I said sy/sx must be tan 45 (as i=j for my part i) and worked ut out that way. . .probably wrong though

japester · 24-06-2011 7:00pm

@ BL1993, I've just checked my calculation there for Q10 (b) (ii) and I do see that I made a slip! I did the integration alright to find the time for the total deceleration and in my corrrect (??) version I am now getting 2.318s for the time and therefore 19.253m/s for the average speed. I completely appreciate what you are saying about the fact that it could take a long time for the deceleration and therefore average speed could be very low.

Geog ariphic · 24-06-2011 7:07pm

electrictrad wrote: »

Got a nice positive fraction too. . .7/8. . .I'm probably wrong though. . .

ok i lied earlier, i DO remember getting 7/8 now, for e. Allow me to go compltely wild and say whoot. *Whoot*

Bte ppl are giving their answers as 200lnsomething and e to powers, and they usually have it like that in marking schemes, but you can just work it all out to decimals and get ur marks right?
I know correctors are usually fair but i don't trust anything to do with correcting applied maths. Some of it is so off the wall.

japester · 24-06-2011 7:20pm

@BL1993 I've just spotted where our answers differ in Q10(b)(ii), when you are getting arctan(40/80), you are getting the angle as 26.56 degrees. However, I am converting this angle to radians to get 0.4636 radians. This accounts for the different results we are getting. Any past questions that I have come across involving arctan appear to expect the angle to be expressed in radians - its a dirty little twist in the question really though

Geog ariphic · 24-06-2011 7:25pm

They cannot hold you for not giving your angle in particular units when they do not specify and you just go with whats normal for angles, SURELY?

BL1993 · 24-06-2011 7:33pm

@electricrad: For part (a) your answer is the same as mine if you fully evaluate it. However, I don't believe that your second part is right. I'm pretty sure that my first part is right for (b) but now that I read japster's post, I believe that he is in fact right in saying that the angle must be converted to radians. Thanks japster

@Geo: Well when dealing with tan or sine inverse, I always had to use radians to solve problems so i don't see why it wouldn't be the case here. Well spotted japster.

Geog ariphic · 24-06-2011 7:43pm

You always 'had to' use radians to solve? why?
Alright, whenever there is an arcsine or arctan formula to solve it always equals constant times pi. But just beacuse you are finding the angle, if u have say tan(theta) = 8, does not mean you have to give the answer in radians.
Does anyone kno, have they actually penalised this before?

BL1993 · 24-06-2011 7:49pm

Because it is the S.I unit of angular measure

Gyaradose · 24-06-2011 8:15pm

ANy idea on what the hell question 3 (a) was? if you rounded the values of sine and/or tan or even the initial speed the quadratic cannot be done, i.e. b(squared) - 4ac is negative, but using the true value for the initial speed b(squared) - 4ac =0

japester · 24-06-2011 8:23pm

Not many people attempt Q9 but for anyone that did, part (a) was very reasonable (U-tube) and part(b) took a bit of thinking about but still fairly doable. The answers I got for this Q are (a) 1.44cm3 and (b) 3/7. Just be interested to know if they tally with anyone who tried it out.

Gyaradose · 24-06-2011 8:30pm

japester wrote: »

Not many people attempt Q9 but for anyone that did, part (a) was very reasonable (U-tube) and part(b) took a bit of thinking about but still fairly doable. The answers I got for this Q are (a) 1.44cm3 and (b) 3/7. Just be interested to know if they tally with anyone who tried it out.

I got the same for part a but my part b was dreadfull so I cant confirm that

anna_k · 24-06-2011 9:19pm

japester wrote: »

Not many people attempt Q9 but for anyone that did, part (a) was very reasonable (U-tube) and part(b) took a bit of thinking about but still fairly doable. The answers I got for this Q are (a) 1.44cm3 and (b) 3/7. Just be interested to know if they tally with anyone who tried it out.

I got 3/7 for 9b) too! yay!!...part a) on the other hand...it was the last question i did ( i did b before a for some reason) so i was running out of time...god only knows what stupid answer i put down..but i'm in no doubt that it was wrong...and u-tubes are so easy :mad:

japester · 24-06-2011 10:00pm

I've just done Q1 there and the answers I'm coming out with seem to match with those of redredwine earlier. They are (a) 0.875 s (b) (i) speed-time graph (ii) t1+t2 = t and (iii) 5t/3. I'm not 100% sure of the answer I'm getting to (iii) however, because I am getting an equation for s that uses the acceleration and deceleration from part (ii) (as instructed) to get the expressions for these parts of the journey in terms of t but then I am saying that the distance traveled for the "flat" part of the journey is (2/3)vt which is not the case surely (the t in this part is not the same "t" from part (ii) i.e. it will have to be greater in part (iii) to compensate for the fact that the distances covered in the acceleration and deceleration phases are now lower and the total distance covered must be a constant. Seriously confused with this part

Maybe someone can shed some light on it - I know I came across one like it somewhere before but I probably made a mess of it too then!!

Geog ariphic · 24-06-2011 10:12pm

japester wrote: »

(iii) however, because I am getting an equation for s that uses the acceleration and deceleration from part (ii) (as instructed) to get the expressions for these parts of the journey in terms of t but then I am saying that the distance traveled for the "flat" part of the journey is (2/3)vt which is not the case surely (the t in this part is not the same "t" from part (ii) i.e. it will have to be greater in part

Yes ur right you cannot use the same t values. i called them t1, t2, t in the first situation, ta, tb, tc in the second (ta+tb+tc=T).
However i never used the acceleration, so ended up with T(+/-)tb=t (i forget whether it was plus or minus)

so maybe u can make something with that, im not sure.

BL1993 · 24-06-2011 11:13pm

I am a million percent sure(did question out 3 times) that the answer to q.1 part 3 is 2t. let T1, T and T2 equal the times that the body accelerates, moves at constant velocity and decelerates respectively. The acceleration is equal to the slope of the graph if you drew it out. I.e, a=v/t1 which is equal to 2v/3T1`and you rearranged it until you get T1=2t1/3. Do the same for the deceleration. Then you find the distance it travels in case 1 and 2 and let them equal, mess around with the equation until you get T=2t. I cba arsed typing it out but I hope you guys see the point. Remember to bear in mind that t1+t2 = t

japester · 25-06-2011 12:56am

@ BL1993 - I see what you're doing alright and I was getting the acceleration and deceleration times in a similar manner, with reference to the earlier situation. I'm getting the acceleration time for the new situation as (2/3)t1 and similarly for the deceleration, it will be (2/3)t2. I'm happy with this much of it and the distance can be determined for these parts. As I said earlier, getting the time that the body moves with constant velocity takes a fair bit of effort and I'm basically saying that the distance covered must be the same in both cases. I'm using the speed-time graph in the original situation to get the area and then doing likewise for the second situation. I am letting the time spent traveling with constant speed in the second situation be T and this must be > t from the first situation in order that the body covers the distance (as the speed limit is lower). In my equations here I am getting terms in t1, t2 and T, which I can then solve for T, provided that I leave (t1+t2) be equal to t. And this is where I have my serious doubts

I believe that the expression t = t1 + t2 is true only provided that the average speed is 3v/4, as it was in the original case. In the second case, the average speed cannot be 3v/4 (it must be lower) and therefore how can I let t = (t1+t2) in my equations? I'm probably making some basic error here but I can't figure this one! For what its worth, if I do actually assume that t = (t1 + t2) is valid, then I am getting a value of T = (23/12)t for the constant speed in the second situation and then I end up with 3t/2 for my answer. Its a tricky little bugger

japester · 25-06-2011 1:03am

@ geographic, thanks for the advice on that question - as I mentioned in my last post, its a tricky little one and I think that its possibly simpler than I am giving it credit for

I can't imagine that it could be worth more than 10 marks, given that (i) and (ii) take a bit of time themselves and therefore I suspect that the solution is more straightforward than I think!

Geog ariphic · 25-06-2011 1:13am

You're right there i'd say, however, and i may be reading this wrong, but u seem to have confused time notation from the first scenario and the second?

And yes i did get t1+t2=t in the first part, but if u are calling any parts of the time in ur second scenario t,t1 or t2 (which u cant since they were all used and all three times are now different), then no it would not be the same.

BL1993 · 25-06-2011 10:24am

Times in case one: t1, t, t2
Times in case 2: T1, T, T2

I probably shouldn't have sued capitals and should have called them x, y and z :P sorry about that.

japester · 25-06-2011 11:57am

@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<=60deg

MedMan101 · 25-06-2011 12:13pm

japester wrote: »

@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.

but you got t1+t+t2 wrong! its equal to t not 2t! and then you simply sub in 2v/3 for v in the original equation to get 2t in part (iii)

BL1993 · 25-06-2011 12:15pm

Oh damn, 2t is the time i get for the time it travels at constant velocity the second time >.<. Let me do it out again. :L

BL1993 · 25-06-2011 12:26pm

I did the question out again and I get the same result as you japster. 31t/12 is correct.

japester · 25-06-2011 12:45pm

@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

japester · 25-06-2011 5:22pm

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 21.9N

Geog ariphic · 25-06-2011 5:57pm

japester wrote: »

@ 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<=60deg

I remember getting everything there except 60 degrees, and if you've confirmed this with others then i assumed i rushed it an got it wrong.
Thanks

randylonghorn · 25-06-2011 7:26pm

Lord's sake, lads, would ye forget about it and go do something fun! :P

MedMan101 · 25-06-2011 8:30pm

randylonghorn wrote: »

Lord's sake, lads, would ye forget about it and go do something fun! :P

Randylonghorn: Posts: 15,269.....says you..!:p

randylonghorn · 25-06-2011 8:33pm

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!

Geog ariphic · 25-06-2011 9:14pm

Now Randy, calm yourself.
Noone forced you to come here. Some of us are interested in doing mathemtical and/or engineering or just science courses next year, and some of us just can't leave a problem alone until we've solved it satisfactorily.
Let us have our little chat. :P

japester · 25-06-2011 11:19pm

I've just done Q5 and I got the following answers:

(a)

(i) To prove this, I used the coefficient of restitution formula (NEL) and the PCM formula to get 2 equations. Using the NEL formula I end up with Vq = 2ue + Vp and using the PCM formula I end up with Vq = -(1/3)u - (2/3)Vp. I equate these to end up with Vp = (-u/5)(1+6e). We can be certain that the sphere P will rebound provided that the value of Vp is negative (it was going in the positive direction to begin with along +ve x axis). Vp can only be negative if (1+6e) > 0. As the range of allowable values of e are 0<e<1 then it must be the case that 1+6e is always positive and so we can conclude that the sphere P rebounds for all value of e.

(ii) To prove this I follow a similar path to get equations involving Vp using NEL and PCM. Then I end up equating these to get an expression for Vq in terms of e. This time it is Vq = (-4/5)(1-4e). For sphere Q to rebound this must turn out to be positive (as it was travelling in the -x direction before impact). So (1-4e) must turn out to be negative. This can only happen if e>0.25. So the range of values of e for which Q will rebound is 0.25<e<1

(b)

(i) I am ending up with k = (1-e)/2 here

(ii) e = 0.75

Geog ariphic · 25-06-2011 11:30pm

i had the same answer for k.

For the first part, Vp, i had that or something quite similar, but it was positive! I just lied and said it was the opposite direction, but obviously it affected everything else and i couldnt get a real value of e. (i.e. I believe i obtained 14.43... As ONE of its possible value)

godtabh · 25-06-2011 11:32pm

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 though

randylonghorn · 25-06-2011 11:44pm

Geog ariphic wrote: »

Now Randy, calm yourself.
Noone forced you to come here. Some of us are interested in doing mathemtical and/or engineering or just science courses next year, and some of us just can't leave a problem alone until we've solved it satisfactorily.
Let us have our little chat. :P

Fair enough, we all know you can't leave it alone, geog!! >_>

japester · 25-06-2011 11:54pm

@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

japester · 26-06-2011 12:56am

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:

luciemc · 26-06-2011 2:21am

japester wrote: »

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...

japester · 26-06-2011 1:26pm

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

japester · 26-06-2011 6:06pm

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

* Applied Maths * predictions / discussion / aftermath (1 thread only please)

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