Mathematics: Post your doubts here!

[iKhaled]

Thank you so much..
I know the integration part, I was stuck in making ysq the subject.. :/
and Jazak Allah..

no problem!

 

[GorgeousEyes]

Specimen M1 no.4 ii , why can't i just use 1/2mvsquare=mgh ?

 

[knowitall10]

Same year ques no 4 ii.. :/
iKhaled knowitall10

Farru lol.. u seriously wanna test my math huh? ok.. gimme a while to gather my thoughts

 

[knowitall10]

Same year ques no 4 ii.. :/
iKhaled knowitall10

I'm sorry Farru i gotta go study chem.. my maths exam is all the way on the 13th of June..i'm very sorry i can't help u
Sorry again...ugh.. i feel s bad... *sigh*....

 

[50sum]

the link from the rule "Final Energy = Initial Energy + Driving force - Resistance force" ? or the one "Driving force - Resisting force = mass x acceleration"?

so you get the final and initial energy using "K.E. [1/2 mass x velocity(squared) ] + P.E. (mass x gravity [10] x height)" and substituting the mass, height and velocity as given at each point.

if given a driving force or resistant force you put them in the equation "F.E. = I.E. + DF - RF"

What if it's just a particle(with no driving force) going upwards or downwards a rough slope? Can you explain the conditions for both motions. For example sometimes while going down a rough slope the speed increases, but in some cases it also decrease and comes to rest, like in w_12 qp 41, question 6

 

[iKhaled]

Same year ques no 4 ii.. :/
iKhaled knowitall10

integrate ∫1/2(cos 2x - cos 4x) dx
1/2∫cos 2x - cos 4x dx
= 1/2[-sin2x/2 + sin4x/4]
= -1/4 sin 2x + 1/8 sin 4x

now sub the values of x and u will get the answer given

 

[Pie-man]

But I thought we use Lami's rule when we have coplanar forces.
And how can you find angle CP2X first since we have 7.3/sin90 = 5.5/sinAP1X = W/sinCP2X
you did sinCP2X= (5.5xsin90)/7.3 right ? But this should be for angle AP1X. because 5.5 is its opposite force.

I'd be grateful if you showed me your workings possibly with a diagram. Thanks a lot

all i know is if you have 3 different forces on an object you can use lami
i'm not exactly sure why i got CP2X it just made more sense to me and after getting all the other angles i was sure it was right but it will make more sense if you turn it into a triangle (well for most people i know)

 

[Farru]

I'm sorry Farru i gotta go study chem.. my maths exam is all the way on the 13th of June..i'm very sorry i can't help u
Sorry again...ugh.. i feel s bad... *sigh*....

its okay buddy..
my math paper on 20th may.. :/ Best of luck btw.

 

[Farru]

integrate ∫1/2(cos 2x - cos 4x) dx
1/2∫cos 2x - cos 4x dx
= 1/2[-sin2x/2 + sin4x/4]
= -1/4 sin 2x + 1/8 sin 4x

now sub the values of x and u will get the answer given

OMG!!! it was too easy.. :$ :$
btw Thanks bro... iKhaled

 

[knowitall10]

its okay buddy..
my math paper on 20th may.. :/ Best of luck btw.

Thanks! Jazaki Allahu Khair
Best of luck to u too!!!

 

[Pie-man]

What if it's just a particle(with no driving force) going upwards or downwards a rough slope? Can you explain the conditions for both motions. For example sometimes while going down a rough slope the speed increases, but in some cases it also decrease and comes to rest, like in w_12 qp 41, question 6

well in 6 i) you use the F.E. = I.E. as there is no driving force (he didn't mention one) nor resisting force (as it is a smooth plane)

in ii) i'm thinking of getting the I.E. of C (which get you the work done against friction) and F.E. at the mid point where the V is will be the only unknown

if like you say the speed increases while going down a rough slope then there is a driving force that is greater than the resisting force unlike the question above if there is a driving force then he shall give you its value, or the value of the resistance and ask you to obtain the driving force.

 

[Silent Hunter]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_43.pdf

Pie-man

or any one? its Q6 (ii)

Thanks

 

[Pie-man]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_43.pdf

Pie-man

or any one? its Q6 (ii)

Thanks

i did this like a few pages back :

so here is what i did (the correct one):

1- i got the equation for Sp "1/2 x a x (t+2)(squared)"

2- started to get rid of the brackets "1/2 x 1.75 x (t(squared) + 4t +4)
= 0.875(t(squared) + 4t + 4)
= 0.875t(squared) + 3.5t + 3.5

3- got an expression for Sq "1/2 at(squared)"

4- subtracted them from each other to get the distance in-between to be equal to 4.9

0.875t(squared) +3.5t +3.5 - 1/2 x 1.75 x t(squared)
= the "0.875t(squared) is crossed with the "1/2 x 1.75 x t(squared) leaving the equation to be "4.9=3.5t + 3.5"
3.5t=4.9-3.5
t=1.4/3.5 = 0.4

 

[Silent Hunter]

i did this like a few pages back :

so here is what i did (the correct one):

1- i got the equation for Sp "1/2 x a x (t+2)(squared)"

2- started to get rid of the brackets "1/2 x 1.75 x (t(squared) + 4t +4)
= 0.875(t(squared) + 4t + 4)
= 0.875t(squared) + 3.5t + 3.5

3- got an expression for Sq "1/2 at(squared)"

4- subtracted them from each other to get the distance in-between to be equal to 4.9

0.875t(squared) +3.5t +3.5 - 1/2 x 1.75 x t(squared)
= the "0.875t(squared) is crossed with the "1/2 x 1.75 x t(squared) leaving the equation to be "4.9=3.5t + 3.5"
3.5t=4.9-3.5
t=1.4/3.5 = 0.4

thanks .... but why do we minus distance of P from that of Q ? why not the other way round?

 

[GorgeousEyes]

Nov01 no7 ii please.

 

[Pie-man]

thanks .... but why do we minus distance of P from that of Q ? why not the other way round?

yeah that i have no idea why i did that same mistake the first time i did it .. i was actually waiting to see if you would tell me :/

 

[Silent Hunter]

yeah that i have no idea why i did that same mistake the first time i did it .. i was actually waiting to see if you would tell me :/

me too also cant get it ... maybe another m1 candidate can help us out :\

 

[haha101]

thanks .... but why do we minus distance of P from that of Q ? why not the other way round?

because P is further away and has covered more distance than q . it it were the other way around then we would get a negative distance which is impossible ( I am talking about distance not displacement )

BTW i did it another way

 

[Esme]

thanks .... but why do we minus distance of P from that of Q ? why not the other way round?

I took time taken by P to be 't' and time taken by Q to be (t-2). Then i formed the equations for the distance travelled by both the particles and subtracted distance of Q from distance of P. I got t=2.4. So the time taken by Q is t-2= 2.4-2= 0.4

 

[Esme]

because P is further away and has covered more distance than q . it it were the other way around then we would get a negative distance which is impossible ( I am talking about distance not displacement )

BTW i did it another way

You're right but they subtracted distance of P from distance of Q.. it has to be Sp-Sq=4.9