Applied maths: Help needed

Mutant_Fruit

1997 Q.1. (a). Can't do it. i need a solution.

1997 Q2. The easiest possible relative velocity question, and i cna't do it. Its ****ED!

subway_ie

I've got a solution for question 1 someplace... along with about 5000 other sheets of paper, could take awhile to find it.

General method of doing that type of relative velocity question:
Vw = velocity of wind
Vm = velocity of man
Vw = xi + yj
Vm = 0i + 4j

wVm = xi + (y - 4)j

Slope of wVm = j/i = (y - 4)/x = tan(angle)

Blah blah blah, do that for the second condition. You'll probably end up multiplying top and bottom of an exquation by x or y and end up with a quadratic.

Mutant_Fruit

exatly what i did, but i can't get the fecker out. Its driving me INSANE!

subway_ie

Well, I'll give the relative velocity thing a try... but here's my solution for Question 1: not too sure about its reliability though... I have a habit of "fixing" my answers so that they look right.

Part (iii) was on another sheet:

fd = 30u^2

Sq = 0
=> 3ut -.5ft^2 = 0
=> 6u = ft^2
=> 6u = f( d/5u )
=> fd = 30u^2

Mutant_Fruit

hrmm... we wonders where you got s1 + s2 + s3 = 31....

subway_ie

So here's what I got for question 2:

Vw = ai + bj
Vm = 4j
Vw/m = Vw - Vm = (AI +BJ) - 4J = AI + (B-5)J

Speed = modulus (Ww/m) = root (a^2 + (b - 4)^2) = 3


Second condition:

Vm = -3i
Vwm = (a+3)i + bj

Modulus(Vw/m) = root((a+3)^2 + b^2) = 4

Ok, so you use the two equations in bold, solve for a and b, then test both values in the original equations to see if they hold. A = -3, B = 4

=> -3 i+ 4j is a solution
=> -3i + 4j is a possible velocity

For the second part of the question, you add root((a+3)^2 + b^2) = 4 and root (a^2 + (b - 4)^2) = 3 and get a in terms of b. With that value, you sub it back into root (a^2 + (b - 4)^2) = 3 to get a b value. You'll end up with a quadratic, solving for b will give you two possible solutions, one of them being the one from part (i), the second being 0.84i + 1.12j


I did alot of this in my head... so not guaranteed to be 100% right. Bit of a rush job.

subway_ie

Originally posted by Mutant_Fruit
hrmm... we wonders where you got s1 + s2 + s3 = 31....

It should be s1 + s2 + s3 = 30... dunno where I got 31 from. Be advised though; don't put too much value on my solutions... more often than not they're either completely wrong, or only half wrong and "fixed". I'm fairly confident that I've mastered the fine art of applied maths deception.

*Wonders why my applied maths solution is called "Jes-receipt-06"* :dunno:

Mutant_Fruit

sorry, i can't follow the solution at all.

I must be missing something rediculously simple... where are you getting the distances from?

EDIT: and i got the right magnitudes etc in the relative velocity. I'm just stupid and can't solve em

subway_ie

Originally posted by Mutant_Fruit
sorry, i can't follow the solution at all.

I must be missing something rediculously simple... where are you getting the distances from?

Ok, well it says that the distance travelled while accelerating is 6m. That's the area of section A, the first triangle of the velocity-time graph. It says that the decelleration is twice that of the acceleration => the distance travelled while decelerating is .5 that while accelerating = 3m. The total distance travelled is 30m - that means 6m + 3m + xm = 30. X being the distance travelled at the constant speed. Then you can show that x = 30 - 9 = 21m.

Then using the fact that the area under a velocity-time graph is equal to the distance travelled, you can solve for t1, t2, t3 for each potion of the journey. Since you know that t1 + t2 + t3 = 6, you just multiply accross by "v" and you get 6v = 38 (that's when you make the mistake about 31m instead of 30m.... but it's the same method). The rest is fairly easy to follow I think?

Mutant_Fruit

**slowly gets ready to explode**

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I realised that (t3) = (0.5)(t1) but never for the life of me did i make the connection about the distances travelled. Lol, i must be feckin stupid today.

subway_ie

Originally posted by Mutant_Fruit
I realised that (t3) = (0.5)(t1) but never for the life of me did i make the connection about the distances travelled. Lol, i must be feckin stupid today.

Happens to me all the time, especially in Q1 and 2. Just make sure you keep reading them again and again before you start doing them. At least now you'll know to read the question thoroughly on friday? Either that, or you'll keep thinking about how you couldn't do one of the easiest linear velocity questions ever, and be consumed by self-hate, forcing you to leave the exam after 30 mins and breakdown, in a blubbering mess of tears and self-disgust.

Mutant_Fruit

my plan, spend at most 2 minutes thinking before i move on.

Ther is not a second to waste in an applied maths exam. 150 mins just ain't enough.

The Shol'va

Originally posted by Mutant_Fruit
**slowly gets ready to explode**

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I realised that (t3) = (0.5)(t1) but never for the life of me did i make the connection about the distances travelled. Lol, i must be feckin stupid today.

It's not just today... :P

Roll on Friday! :P

subway_ie

I've been trying to do this question for about 30 mins (it's a back up question... don't really know how to do the second parts of them, but the theorem proof is always an easy 20 marks). Anyway, I had a look at the solutions, and it says that:

Ipq = (4/3)ml^2 = Ipr

Okay, so where are they getting the 4/3 from? Should it not be 1/3?
[EDIT] Just realised that it's not about an axis through its centre, it's about one end :rolleyes: [/EDIT]

Iqr = 1/3ml^2 + m{(root3)l}^2

I don't have a clue where that part is coming from. The perpindicular distance from P to qr is root3... so I'm guessing that's where it came from - but why do they add the m{(root3)l}^2 part and take it from an axis about its centre???

Mutant_Fruit

The M(I) of the rod is 1/3ml^2 PLUS the distance to the axis squared (parallel axis therom).

The perpendicular distance is l(root3) (the angle is 60degrees, so 2lSin60....)

Thats wher it comes from.

EDIT: The reason behind it is that once you move PQ using the parallel axis therom, you end up with the axis through P going through the centre of the rod, so thats why its (1/3 ml^2) PLUS [l(root3)]^2

subway_ie

Thanks for that. I don't normally do this question, so all I really know are the theorems. Need to brush up on the part b's.

Mutant_Fruit

2001 Q7.

I need a worked solution for this that actually makes sense. I cant follow the marking scheme answer.

subway_ie

Can't help you out with that... never do statics, hydrostatics or SHM. There's actually a fairly good maths/physics/general science forum called (very imaginatively) Physics Forums. Just scan in the question, and ask if they can give you any help. Tell them what you've done so far to try and do it and they should give you some pointers. There's one guy there - HallsOfIvy - that generally gives the best solutions. Usually it only takes a few hours - a day at max - to get a reply.

Mutant_Fruit

checking it out now, thanks for the tip.

subway_ie

They don't seem to be very helpful today... I think HallsOfIvy is the only one who actually gives straightforward, clear advice. I've lost all motivation now though. Can't even bring myself to opening the applied book. Just want to get in there, do the exam and get it over with right now.

DS

Yeah big time. I'm getting very bored with the leaving cert now. It doesn't feel the slightest bit important anymore. Even though I'll be well ****ing pissed off if applied doesn't go to plan. I'm just dying to get drunk. Had a pint with the economics lads who finished today and it went straight to my head. My liver's really been living the life for the past month. Poor bastard doesn't have a clue what's coming.

subway_ie

Ok, tomorrows plan:

Do 2004 Mock Paper, revise:
Question 1 - V-type motion, along with collision/passing out.
Question 2 - Apparent wind velocities, intercept course, longest time in view.
Question 3 - "unusual" formulas, like the perpindicular distance above the horizontal, bouncing down a slope, etc.
Question 4 - Wedge questions, including wedges with attached pulleys.
Question 5 - Have a look at the more unusal ones, like 1992 and 1990. All the rest are usually the same thing over and over.
Question 8 - Rigid Body Roatation proofs and formulas.
Question 10 - Looking over part (b)s, especially ones involving power, force equations, etc.

If I get through all that between tomorrow and a few hours on Friday, then I should be fairly set. The only thing that'll feck me up is Friday morning... trying to do some actual work without worrying about the exam too much. Up until now, they've all either been in the morning or I've had two a day. Evening exams are never good.

Mutant_Fruit

I spent 20 minutes today looking at Q6 on 2002, and i just couldn't do it.

It was then that i realised i'm not doing the SHM question.... lol. Thats how tuned out from reality i am dur to the stress of this whole feckin LC.

DS

2004 Mock Paper

Cheers for that. Woah, JPEGs, Murphy may be a genius but he's no webmaster. My printer's going to have some fun with this.

To be honest I'm not concerned about revision now. It's all down to the paper, where the tricks are, and how long it takes me to find them. The best thing you can do is go in with a clear head. I'd say don't do anything Friday morning. Have a sleep in. It's not like other subjects where you might pick up marks for reading certain things before the exam, there's nothing to remember. It's the one subject where cramming is pointless, assuming you're not a chancer and you'll have already encountered all the 3/4 variations they can ask in each question. It's just a case of tackling the problems on the day.

Mutant_Fruit

there's nothing to remember.

Erm, there are PLENTY of tricks to cram. If a man just catches a bus, what are the two conditions neccessary to do the question?

If a projectile lands horizontally, whats the condition for that.

Prove the moment of inertia for a disc of mass m.

Theres plenty of stuff to know.

Morf3h

Originally posted by Mutant_Fruit
Erm, there are PLENTY of tricks to cram. If a man just catches a bus, what are the two conditions neccessary to do the question?

If a projectile lands horizontally, whats the condition for that.

Prove the moment of inertia for a disc of mass m.

Theres plenty of stuff to know.

WELL...........?????

Tell me Tell me!!

subway_ie

Originally posted by Morf3h
WELL...........?????

Tell me Tell me!!

Just read through your book... if you're using the brown "Fundamental Applied Mathematics" by Oliver Murphy, then you just need to look through all the examples and learn the stuff on the last page of every chapter. Plus the proofs of the moments of inertia for the various bodies. That's about it really - doesn't sound like much but there is *a lot* to learn.

BangBeater

2003 Collisions Part (b) on the IMPULSE

How is that done?!

Checked the marking scheme and it says m(o) - m(v2)?!! Huh?!

By the way - Cheers for the Mock exam! Was looking for something I hadn't done already!
I'll owe ya a pint at oxegen if u're goin!!

DS

Impulse is change in momentum -> mv - mu

subway_ie

Originally posted by BangBeater

By the way - Cheers for the Mock exam! Was looking for something I hadn't done already!

The mock is fairly good - some parts are a bit unusual. There's a mistake in the solutions though, to question 3 - he does the solution to the particled landing perpinduclar to the plane, when the question asks for horizontal. It's nice to get something fresh before the exam though.

Mutant_Fruit

Q2:
Case1:
Vathlete = i - 2j
Vwa = -ai + aj
Vw = (1-a)i + (a-2)j (i.e. Vwa + Va = Vw -Va + Va)

Case2:
Va = 4i
Vwa = bj
Vw = 4i + bj


But Vw = Vw
Therefore i=i and j=j
therefore: (1-a)=4 AND (a-2)=b

Solving there i get Vw= 4i - 5j. Not the solution he got... whats wrong!

EDIT: i did it three times...
EDIT2: His solution is wrong! HAH, i'm not stupid after all. He thinks the wind is coming from the north east!, but its actually coming from south-east (i.e. going north WEST) Therefore he should have set (a-1) = MINUS(b+2) i believe.

My solution is right.