Applied maths q's

Shox

hey could anyone help me with these two question...

in the juggler one i got the initial velocity of the balls but have nt a clue where t go from there..

ZorbaTehZ

I didn't know if it was help or a solution you wanted ...

Solution
1995 Q1

(ii)
u=√6g
v=-√6g
a=-g
s=6
t=x

v=u+at
-√6g = √6g - gx
x = 2√(6/g)

x/6 = t
2√(6/g) x 1/6 = t
1/3.√(6/g) = t
t = √(2/3g)

(iii)
u=√6g
v=?
a=-g
s=?*
t=√(2/3g)

s=ut+0.5at^2
s=√(2/3g).√6g + 0.5.(-g).(√(2/3g))^2
s= ...
s= 5/3m

Same as above, only this time take 2t
s=8/3m

Therefore heights are 3m, 5/3m, 8/3m.

ZorbaTehZ

Solution
1992 Q1

This is a difficult question, if you assume that when it leaves the balloon it has an initial velocity of 0, which it doesn't.

Particle
u=u
v=?
a=-g
s=h
t=9

s=ut+0.5at^2
h=9u-81g/2

Balloon
u=u
v=u
a=0
s=h
t=7.2

1=ut+0.5at^2
h=7.2u
u=h/7.2

Substitute it back in.
h = 9(h/7.2) - 81g/2
7.2h = 9h - 291.6g
18g = h
h = 176.4m

JC 2K3

I'm having trouble with 1991 4(i)

Can someone please just post a solution so I can see where my mistake is? I've a feeling I'm oversimplifying it somehow.

cocoa

lol, ironically, it actually helps with this question if you are in fact able to juggle...

since you didn't give me the answer, i need to find the initial velocity...
s=3 v=0 u=initial velocity a=-g
v^2=u^2+2as
0=u^2+2*-g*3
u^2=6g
u=root(6g)

ok, so we know that the time between any one ball being thrown, and the next being caught, is the same, and luckily (unlike in real life) we are allowed imagine that the balls are thrown instantaneously, i.e., the time between catching and throwing is zero. This isn't really emphasised enough in the question, but they do say "immediately thrown" so we can assume as much.

so then you just need to count it out in your head:
first through t second throw t 3rd throw t 4th t 5th t 6th t 1st catch 7th throw t second catch 8th throw

and so on, but we can see that there is 6t between the first throw and the first catch, so we just need the time of that flight, then we divide by 6 and we have t.
v=u+at
-u=u-g6t
t=2u/6g=2(root(6g))/6g=2(root(1/6g))
which is 0.26 but I doubt you would lose marks for leaving it in surd form.

the last question is a bit tricky or just tedious, depending on your method. the simplest way by far, is to simply find a value for s at t, 2t, 3t, 4t, 5t and 6t (or 0). You can also however guess at some of these and/or guess at the symmetry, but it's best to check your answers.

the answer being: 0m, 1.67m, 2.67m, 3m, 2.67m, 1.67m, 0m

edit: wow, super sniped... ouch.

ron-burgandy

how can I assume that the Particles inital vel. (u) is equal to the balloons initial velocity??

i'm not saying that solution is wrong, just asking a question

ZorbaTehZ

Solution
1994 Q4

ZorbaTehZ

ron-burgandy wrote:

how can I assume that the Particles inital vel. (u) is equal to the balloons initial velocity??

i'm not saying that solution is wrong, just asking a question

Try to visualise it. If you were in the balloon and you held your hand out holding the particle and then let go, it would initially have the same speed as the balloon.
The other way of looking at is, when you drop the particle out of the balloon its initial velocity is zero relative to you. But you are moving, ie your in the balloon itself, so relative to the ground it has the same velocity as the balloon

JC 2K3

Ah, so the fact the large block is moving right and there's no friction means that the particle on top experiences a force, relative to the block, of mg/3 to the left?

ZorbaTehZ

JC 2K3 wrote:

Ah, so the fact the large block is moving right and there's no friction means that the particle on top experiences a force, relative to the block, of mg/3 to the left?

Yup, it seems wrong on paper initially - you would think the accelerations should be in the same direction - but if you visualize it, you can see that its right.

Shox

Thanks Zorba

eZe^

ron-burgandy wrote:

how can I assume that the Particles inital vel. (u) is equal to the balloons initial velocity??

i'm not saying that solution is wrong, just asking a question

If youre standing still on a train thats going at 100mph and you jump out, its not going to be like jumping off a wall the same height, youre going to be travelling as fast as the speed of the train was when you initially jumped out.

Nehpets

wow, I am ****ed for this subject

sd123

i posted this a few days before the LC began but never got a chance to see any replies and i cant find it now!

ok for q 10 does anyone think that 2nd order diff. eqtns could come up this year, have they ever come up? i have the papers as far as 83 and no sign of them.

the other question is, does anyone think that power could come up for q. 10 (b) cos i havent a clue how to do it:o

last, could anyone please explain to me the difference between shortest distance and shortest time in relative vel. qs crossing rivers etc. , and how to do them;)

ZorbaTehZ

sd123 wrote:

i posted this a few days before the LC began but never got a chance to see any replies and i cant find it now!

ok for q 10 does anyone think that 2nd order diff. eqtns could come up this year, have they ever come up? i have the papers as far as 83 and no sign of them.

Technically yes. There are a number of examples in Oliver Murphy's book on how to do them (if you have it), you need to redefine the second deferential. As you said yourself, they've never come up, so I would think that it is very unlikely.

sd123 wrote:

the other question is, does anyone think that power could come up for q. 10 (b) cos i havent a clue how to do it:o

Power has come up at least thrice in the past 20 years. You will be using the formula P=Fv if does come up. Iirc there was one extremely difficult question involving Power in one of the late 1980 papers, so don't do that one try the others.

sd123 wrote:

last, could anyone please explain to me the difference between shortest distance and shortest time in relative vel. qs crossing rivers etc. , and how to do them;)

Shortest time, you ignore the current and direct yourself straight across the river. You will land further downstream, but it's the quickest time. Shortest distance, is that you aim the vector of your swimmer/boat relative to the current in such a way, that your vector of the boat/swimmer ends up being straight across. A number of nice diagrams showing it here: skoool.ie

sd123

thanks Zobrah, nice one:D

Thomas_S_Hunterson

JC 2K3 wrote:

I'm having trouble with 1991 4(i)

Can someone please just post a solution so I can see where my mistake is? I've a feeling I'm oversimplifying it somehow.

That question shouldn't have been asked, it's beyond the scope of the leaving cert course or so I'm told, although quite do-able

JC 2K3

Yeah, they've asked a couple of dodgy questions over the years.

There's been nothing too dodgy since SEC took over, AFAIK, however.

Also, re: second order differentials. Accelleration comes up all the time. Now fair enough, you might just have learned off that a = d^2s/dt^2 = dv/dt = vdv/ds, but you apply the same principle when working out any second order differential problem. It's not all that difficult.

I absolutely love Q10, I think I'm just good at visualising things as equations, and I like the concepts of differentiation and integration in the first place.

ZorbaTehZ

I assume sd123 is referring to Part A when he talks about second order differentials. Something like d2x/dt2 = 3x^2/3, which has got nothing to do with acceleration.

One would never use d2s/dt2 in Part B anyways because you'd have to replace it to work out the equation.

eZe^

A horizontal frictionless plane rotates with constant angular velocity (omega) about a vertical axis which intersects the plane at point C. A particle on the plane travels around C with the same angular velocity as the plane and is held at the same position relative to the plane by tying it to C with a light string of length R.

If the string is shortened by pulling it at a constant speed v though a hole at C, show that, when the length of the string is reduced to r, it will have rotated through the angle ((Omega) (R - r)^3) / vr, relative to its original position on the plane. (which is rotating)



That better not be on the course, it was in my friends applied math notes when he did it last year... I seriously dont even know where to start... Please please someone tell me we cant get that!!!

ZorbaTehZ

lol, I would probably shit myself if we got something like that.

JC 2K3

ZorbaTehZ wrote:

I assume sd123 is referring to Part A when he talks about second order differentials. Something like d2x/dt2 = 3x^2/3, which has got nothing to do with acceleration.

But that's based on the exact same principle as what you do when working out acceleration.
ie:

d2x/dt2 = 3x^2/3
d/dt(dx/dt) = 3x^2/3

let dx/dt = v
d2x/dt2 = dv/dt
dv/dt = dv/dx*dx/dt = vdv/dx

vdv/dx = 3x^2/3

Which is the exact same thing you'd do if you had a = 3s^2/3.

Solve for v to get another differential equation, solve that to complete the question.

eZe^

ZorbaTehZ wrote:

lol, I would probably shit myself if we got something like that.

Rang my friend, it turns out its from a college applied math problem sheet... Thank god.. If I saw that Id just get up and leave.. hahaha

ZorbaTehZ

Huh? If you had d2s/dt2 in a Part B you would just simply let that equal a. If you get it in Part A you need to do that thing you did there.

ZorbaTehZ

JC 2K3 wrote:

But that's based on the exact same principle as what you do when working out acceleration.
ie:

d2x/dt2 = 3x^2/3
d/dt(dx/dt) = 3x^2/3

let dx/dt = v
d2x/dt2 = dv/dt
dv/dt = dv/dx*dx/dt = vdv/dx

vdv/dx = 3x^2/3

Which is the exact same thing you'd do if you had a = 3s^2/3.

Solve for v to get another differential equation, solve that to complete the question.

O wait I see you're doing it some other way. It's much quicker if you just swap out the differentials, and then resolve it.

ZorbaTehZ

On the topic of difficult questions, what is the hardest question you've come across?
Theres been a number of them for me that took ages to solve.
Q5 A 1999
Q10 B 1987
Q6 B 2000
Q7 (ii) 2001

JC 2K3

ZorbaTehZ wrote:

O wait I see you're doing it some other way. It's much quicker if you just swap out the differentials, and then resolve it.

Well, it is rather fast as it is(I never actually have to write out all I wrote there, i was just showing what I did).

I'm not going to change my method now anyways

On the subject of differentiation in app maths, does anyone else use it outside Q10 to solve problems? I've found myself instinctively using it in Q1, 2 and 3 in questions on max and min points having studied it for Maths.

mathew

JC 2K3 wrote:

Well, it is rather fast as it is(I never actually have to write out all I wrote there, i was just showing what I did).

I'm not going to change my method now anyways

On the subject of differentiation in app maths, does anyone else use it outside Q10 to solve problems? I've found myself instinctively using it in Q1, 2 and 3 in questions on max and min points having studied it for Maths.

I find using it on question 1 gives the "wrong" answers. ie, different ones to whats on the marking scheme.
Annoying really, cause the use of the formulae give fundamentally wrong answers...


Best of luck in it everyone. Ill be on tomorrow for analysis!!

JC 2K3

mathew wrote:

I find using it on question 1 gives the "wrong" answers. ie, different ones to whats on the marking scheme.

I don't....

JC 2K3

eZe^ wrote:

A horizontal frictionless plane rotates with constant angular velocity (omega) about a vertical axis which intersects the plane at point C. A particle on the plane travels around C with the same angular velocity as the plane and is held at the same position relative to the plane by tying it to C with a light string of length R.

If the string is shortened by pulling it at a constant speed v though a hole at C, show that, when the length of the string is reduced to r, it will have rotated through the angle ((Omega) (R - r)^3) / vr, relative to its original position on the plane. (which is rotating)

Lol, tricky enough, but I don't think it's undoable. Basically, as you reduce the radius, you increase the angular velocity. Think of the "plane"'s movement as simply the angle the particle on the string would have gone through had the radius remained constant, ie. the angle you're looking for at the end is the angle the string would have rotated through had the radius remained constant minus the angle the string actually rotated through.

Shox

just a quick question, if they say draw a velocity time graph. does it have t be on graph paper???