**Applied Maths 2012 Before/After**

waiting4

hey cud anyone tell me how you determine which limits to put on top nd which limits to put on the bottom of a deferential eqation? I'm always mixing them up!

Pepperr

I'm pretty sure it doesn't matter which side you put the limits on, as long as you pair them up right, eg if it says y=1 when x=0, then the one and the zero should be on the same side of the integral symbol thing!

RedTexan

paddyhes wrote: »

But V1 is the velocity of the first sphere after the initial impact of the two spheres?

What i'm thinking is:
That if sphere one is knocked backwards it's already a minus.. so the minus infront of V1 will make it so that its heading towards the rebounded sphere 2.

But if sphere one isn't knocked backwards the minus will make it so its heading away (which would make sense method-wise).

But seeing as we don't know the direction of Sphere 1 after impact (because we dont know e) should we not investigate both cases?

I know there's something stupidly simple i'm overlooking but i can't get it to click.

It's because if the sphere continues to roll forward after the collision there is definitely going to be a collision so if the sphere changes direction (indicated by the minus) and is rolling away from the wall, the only way there will be a second collision is if the speed of speed of the 2nd sphere (eV2) is greater than that of the first. Basically if V1 was positive you would be guaranteed a second collision so there is no need to investigate that case.

paddyhes

RedTexan wrote: »

It's because if the sphere continues to roll forward after the collision there is definitely going to be a collision so if the sphere changes direction (indicated by the minus) and is rolling away from the wall, the only way there will be a second collision is if the speed of speed of the 2nd sphere (eV2) is greater than that of the first. Basically if V1 was positive you would be guaranteed a second collision so there is no need to investigate that case.

I know that should make sense but if e is > 3/13 then V1 will already be a minus? So adding the minus will then cover the case of the guaranteed 2nd collision?
If that makes sense

bren2001

paddyhes wrote: »

I know that should make sense but if e is > 3/13 then V1 will already be a minus? So adding the minus will then cover the case of the guaranteed 2nd collision?
If that makes sense

Heres the math behind it:

Using newtons law:

v1 - v2 = -e(u1 - u2)

v2 and u2 is the speed of the wall.

v1 = -eu1

-e [3u/16 (1 + e)] < v1
OR

e [3u/16 (1 + e)] > -v1
ev2 > -v1

paddyhes

bren2001 wrote: »

Heres the math behind it:

Using newtons law:

v1 - v2 = -e(u1 - u2)

v2 and u2 is the speed of the wall.

v1 = -eu1

-e [3u/16 (1 + e)] < v1
OR

e [3u/16 (1 + e)] > -v1
ev2 > -v1

In those workings are you not taking v1 to be the speed of the second sphere before the impact with the wall?

RedTexan

paddyhes wrote: »

In those workings are you not taking v1 to be the speed of the second sphere before the impact with the wall?

V1 isn't the speed of the second sphere before impact with the wall, V1 is the speed of the first sphere after the first collision, V2 is speed of the second sphere after that collision and then eV2 is the speed of the second sphere after colliding with the wall!

paddyhes

RedTexan wrote: »

V1 isn't the speed of the second sphere before impact with the wall, V1 is the speed of the first sphere after the first collision, V2 is speed of the second sphere after that collision and then eV2 is the speed of the second sphere after colliding with the wall!

I know thats why in his workings where he says

"V1-V2= -e(u1-u2)

where v2 is speed of the wall after and speed of wall before"

Is he not working with an equation that doesn't involve the speed of the first sphere after first impact (v1).
He's just labelled what was 'ev2' as 'v1' and his equation is basically:

ev2> -ev2

Sorry i just cant get it to f'n click.

bren2001

paddyhes wrote: »

I know thats why in his workings where he says

"V1-V2= -e(u1-u2)

where v2 is speed of the wall after and speed of wall before"

Is he not working with an equation that doesn't involve the speed of the first sphere after first impact (v1).
He's just labelled what was 'ev2' as 'v1' and his equation is basically:

ev2> -ev2

Sorry i just cant get it to f'n click.

I should have used different letters alright but no I have no essentially said ev2 > -ev". All the equation really shows is how to calculate the speed of the ball after impact with the wall, explaining this - sign is remarkably harder then I thought it would be, theres nothing difficult to it really.

Cathalog

Finally figured out what was wrong with my oblique collisions questions. Silly mathematical errors!

Question: Am I the only one with a stupid calculator that gives [root(2)]/2 instead of just 1/root(2) for sin(45) and cos(45) ?

It's so annoying because I never realise that it can be simplified further and it can cause mistakes...
So, is there away to let the calculator know that I don't mind a root as a denominator? Or should I get into the habit of using the log tables for sin/cos/tan (30/60/90)?

Cheers.

paddyhes

bren2001 wrote: »

I should have used different letters alright but no I have no essentially said ev2 > -ev". All the equation really shows is how to calculate the speed of the ball after impact with the wall, explaining this - sign is remarkably harder then I thought it would be, theres nothing difficult to it really.

I know i should get it but i can't.
I over-thought it to the point where it doesn't make sense anymore and now its really frustrating me.
Oh well.

jos360

waiting4 wrote: »

hey cud anyone tell me how you determine which limits to put on top nd which limits to put on the bottom of a deferential eqation? I'm always mixing them up!

It shouldn't effect you'r answer because switching limits (swapping the top and bottom) is the same as multiplying across by -1 (changing signs of both sides of the equation).

What is important, is that the limits on each side of the equation correspond.
for example, if you're told that a particle starts from rest, and then travels for 5 seconds and you are give some equation for acceleration, you should always rewrite the info like this:

1) when t=0, v=0 (at rest- no velocity and time is 0)
2) when t=5, v=? (travels for 5 seconds and we need to find it's velocity)

The there will be some equation that you manipulate a bit to get something like:

S(....)dv=S(...)dt 'S' is the integration symbol

Now you put in your limits:
We can choose what t value to use for our upper limit, aslong as we use the corresponding v value for the upper limit on the other side of the equation.

It is however good maths practice to always use the lower value for time as the lower time limit, but it won't effect the result.

So just to recap.
Using t=0 and v=0 as upper limits and t=5 and v=? as lower is fine
Using t=5 and v=? as upper limits and t=0 and v=0 as lower is fine
Using t=0 and v=? as upper limits and t=5 and v=0 as lower is NOT OK because v doesn't = ? when t=0 and v doesn't =0 when t=5.

Same goes for any variables (forces, half life...)

bren2001

paddyhes wrote: »

I know i should get it but i can't.
I over-thought it to the point where it doesn't make sense anymore and now its really frustrating me.
Oh well.

Your best bet is to leave it, its a difficult enough A part. Move on and try other things. Getting your head around that will only help IF it comes up. I'd be surprised if it was that difficult as an A part again.

Cathalog wrote: »

Finally figured out what was wrong with my oblique collisions questions. Silly mathematical errors!

Question: Am I the only one with a stupid calculator that gives [root(2)]/2 instead of just 1/root(2) for sin(45) and cos(45) ?

It's so annoying because I never realise that it can be simplified further and it can cause mistakes...
So, is there away to let the calculator know that I don't mind a root as a denominator? Or should I get into the habit of using the log tables for sin/cos/tan (30/60/90)?

Cheers.

All casio (from fx-83es) give the answer in that form. Mathematically it is better to have the "surd" or square root on the top line. It also makes manipulations easier (in theory).

bren2001

jos360 wrote: »

It shouldn't effect you'r answer because switching limits (swapping the top and bottom) is the same as multiplying across by -1 (changing signs of both sides of the equation).

What is important, is that the limits on each side of the equation correspond.
for example, if you're told that a particle starts from rest, and then travels for 5 seconds and you are give some equation for acceleration, you should always rewrite the info like this:

1) when t=0, v=0 (at rest- no velocity and time is 0)
2) when t=5, v=? (travels for 5 seconds and we need to find it's velocity)

The there will be some equation that you manipulate a bit to get something like:

S(....)dv=S(...)dt 'S' is the integration symbol

Now you put in your limits:
We can choose what t value to use for our upper limit, aslong as we use the corresponding v value for the upper limit on the other side of the equation.

It is however good maths practice to always use the lower value for time as the lower time limit, but it won't effect the result.

So just to recap.
Using t=0 and v=0 as upper limits and t=5 and v=? as lower is fine
Using t=5 and v=? as upper limits and t=0 and v=0 as lower is fine
Using t=0 and v=? as upper limits and t=5 and v=0 as lower is NOT OK because v doesn't = ? when t=0 and v doesn't =0 when t=5.

Same goes for any variables (forces, half life...)

Technically speaking these are not LIMITS as leaving cert students understand them to be. They are conditions upon when something is true, there is no need to put them in when integrating, just leave a + C at the end of the integration, fill in the specific solutions to calculate C and the equation is solved.

Cathalog

bren2001 wrote: »

All casio (from fx-83es) give the answer in that form. Mathematically it is better to have the "surd" or square root on the top line. It also makes manipulations easier (in theory).

You see, the thing that worries me most is that the marking schemes have the surd underneath the line.

BeanbagBallbag

Cathalog wrote: »

Finally figured out what was wrong with my oblique collisions questions. Silly mathematical errors!

Question: Am I the only one with a stupid calculator that gives [root(2)]/2 instead of just 1/root(2) for sin(45) and cos(45) ?

It's so annoying because I never realise that it can be simplified further and it can cause mistakes...
So, is there away to let the calculator know that I don't mind a root as a denominator? Or should I get into the habit of using the log tables for sin/cos/tan (30/60/90)?

Cheers.

You can learn to do it yourself if you practice a few. For your example, you just multiply by root2/root2 (which equals 1 so it's fine), and you'll see how it cancels. Once you get used to it you will just recognize them, like (2root3)/3 = 2/root3 and so on.

bren2001

Cathalog wrote: »

You see, the thing that worries me most is that the marking schemes have the surd underneath the line.

Dont worry about stuff like that. As stated above, its the same answer:

root(2)/2

= root(2)/[root(2) * root(2)]

= 1 / root(2)

You wont got docked any marks for giving your answer in the form displayed on the Casio (infact mathematically the marking scheme is wrong, you cant have a surd on a bottom line for many mathematical reasons but we just ignore them for the time being).

Its the same way if you give the answer as 1/2 or 0.5, would you got docked marks? No!

nbar12

bren2001 wrote: »

Dont worry about stuff like that. As stated above, its the same answer:

root(2)/2

= root(2)/[root(2) * root(2)]

= 1 / root(2)

You wont got docked any marks for giving your answer in the form displayed on the Casio (infact mathematically the marking scheme is wrong, you cant have a surd on a bottom line for many mathematical reasons but we just ignore them for the time being).

Its the same way if you give the answer as 1/2 or 0.5, would you got docked marks? No!

I'm actually terrified for tomorrow! I still have to learn Question 10! Ugh I think I might just take the pass paper!

ganon

paddyhes wrote: »

I know i should get it but i can't.
I over-thought it to the point where it doesn't make sense anymore and now its really frustrating me.
Oh well.

The signs and everything are quite confusing, but it does make sense if you get your head around it

English first:

• After rebounding from the wall, sphere q is definitely moving left as it's velocity is definitely negative
• Sphere P may be moving left or right as we don't know if it's positive or negative
• If P is moving right (Q is moving left), the velocity of q will be less than the velocity of p (as minus is always less than plus -<+ ) and a collision is guaranteed
• If P is moving left, they will only collide if the speed of Q is greater than the speed of P. But if they're both negative vectors, that means if the velocity of Q is less than the velocity of P (e.g. -5<-4)

Now equations:
Taking that the velocity of Q after it collides with the wall is -ev2, and that the velocity of P is v1, we want to say Q is less than P
-ev2 < v1
Multiply across by minus one and change the direction of the inequality
ev2 > v1
and now you have what the marking scheme starts with, they skip that vital first line and that's what makes it so confusing!

Hope that helps

paddyhes

ganon wrote: »

The signs and everything are quite confusing, but it does make sense if you get your head around it

English first:

• After rebounding from the wall, sphere q is definitely moving left as it's velocity is definitely negative
• Sphere P may be moving left or right as we don't know if it's positive or negative
• If P is moving right (Q is moving left), the velocity of q will be less than the velocity of p (as minus is always less than plus -<+ ) and a collision is guaranteed
• If P is moving left, they will only collide if the speed of Q is greater than the speed of P. But if they're both negative vectors, that means if the velocity of Q is less than the velocity of P (e.g. -5<-4)

Now equations:
Taking that the velocity of Q after it collides with the wall is -ev2, and that the velocity of P is v1, we want to say Q is less than P
-ev2 < v1
Multiply across by minus one and change the direction of the inequality
ev2 > v1
and now you have what the marking scheme starts with, they skip that vital first line and that's what makes it so confusing!

Hope that helps

THANK YOU.
Jesus christ thank you.
That was driving me crazy!

Incompetent

Tomorrow's gonna be a disaster lads

Failed my mock and haven't done work in it since. Shoulda done economics instead

paddyhes

Incompetent wrote: »

Tomorrow's gonna be a disaster lads

Failed my mock and haven't done work in it since. Shoulda done economics instead

You'll do grand just practice the rest of today.
I'm getting worried about it too now.
Need an A1
B2 in the mocks so hopefully

aarond280

worked it ou that if tyou get full marks in 5 questions and do a part a you can get exactly an a1

Mani09

Lad I'm only going for a C so what are the best questions to get easy marks ?

jos360

bren2001 wrote: »

Technically speaking these are not LIMITS as leaving cert students understand them to be. They are conditions upon when something is true, there is no need to put them in when integrating, just leave a + C at the end of the integration, fill in the specific solutions to calculate C and the equation is solved.

Well 'technically' one of them is a limit. If either of the values for t or v go below 0 in linear motion integrals, the given equation doesn't apply if the body was originally at rest. You're right however that t=5 may not be a limit because the equation holds for greater values of t up until a terminal velocity is reached. But I never specifed in the question, and it's my question, I get to make it up, so I can make it one if I want Muhahaha!!

A body is initially at rest. It is then dropped from a point and reaches terminal velocity after 5 seconds. Find the effect of wind resistance on the body as it falls.

BOOM! there both limits!

Also, it's quicker to use them as limits anyway than to waste paper solving for C, especially since the lower 'values' (aka limits) are usually 0.

What a productive use of pre exam study time...

paddyhes

Mani09 wrote: »

Lad I'm only going for a C so what are the best questions to get easy marks ?

Q1, 4, 10 are usually nice then all the 'a' questions.

jos360

Mani09 wrote: »

Lad I'm only going for a C so what are the best questions to get easy marks ?

Question 8 part a gives 20 marks for a proof every year.
There are 3 proofs, that are all very similar and only about 5 lines long.

Not bad, learn 15 lines for a guaranteed 20 mark fallback?
Also, you can usually bullsh!t out the attempt marks in part b for another 12 marks.

aarond280

well lads doing an impacts and collisions question its 5(b) 2004 looking up the marking scheme (http://www.examinations.ie/archive/markingschemes/2004/LC020ALP1EV.pdf) and can seem to work out why they let (v1)^2 + (usin30')^2 = (v2)^2 and suggestions

paddyhes

aarond280 wrote: »

well lads doing an impacts and collisions question its 5(b) 2004 looking up the marking scheme (http://www.examinations.ie/archive/markingschemes/2004/LC020ALP1EV.pdf) and can seem to work out why they let (v1)^2 + (usin30')^2 = (v2)^2 and suggestions

The magnitude speeds are equal after the collision.

A's magnitude speed = root((usin(30))^2+(((ucos30)(1-e))/2)^2)

B's magnitude speed = root(((ucos(30))(1+e))/2)^2)

Let equal and square both sides to eliminate roots.

Hope that helps.

aarond280

paddyhes wrote: »

The magnitude speeds are equal after the collision.

A's magnitude speed = root((usin(30))^2+(((ucos30)(1-e))/2)^2)

B's magnitude speed = root(((ucos(30))(1-e))/2)^2)

Let equal and square both sides to eliminate roots.

Hope that helps.

Thanks it did, that what happens when you rush into the question and dont read it properly