Mathematics: Post your doubts here!

[salvatore]

I need some serious help in mechanics. In a question regarding the work done = change in total energy, I don't understand how to decide which one from K.E and P.E has a greater value so that I can subtract well. Please pleaseee help me understand the concept.
Here is an example:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_41.pdf
qn no. 5(ii).. how do you decide which one (K.E/P.E) has a greater value?

I'll really reallyyy appreciate any help provided.

Thanks a lot

 

[sitooon]

I need some serious help in mechanics. In a question regarding the work done = change in total energy, I don't understand how to decide which one from K.E and P.E has a greater value so that I can subtract well. Please pleaseee help me understand the concept.
Here is an example:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_41.pdf
qn no. 5(ii).. how do you decide which one (K.E/P.E) has a greater value?

I'll really reallyyy appreciate any help provided.

Thanks a lot

I will explain you th whole concept and try to understand ...
First thing you need to consider , is to know if surface is smooth/ rough ( has friction )by reading question.
then follow the formula wd (workdone ) by driving force = wd by resistive force ( resisting motion ) +- K.E gained/lost +- P.E gained/lost.
if it increses in speed KE is gained ( Add in eqn ) if decreased in speed its decresed KE is lost ( subtract from eqn )..
If it increses in Height ( moves up ) PE is gained ADD in eqn ... and if comes down PE is lost and subtract in eqn
so when something is gained ADD and when lost MINUS as you see i wrote +- in eqn depending whether its lost or gained .
we know wd formula is = F * d

In this question ,,, here its same speed .. mean no gain/loss in KE .. so its not included in equation and there is no Wd by driving force so both are ZERO in eqn
So our eqn is wd (workdone ) by driving force = wd by resistive force ( resisting motion ) + K.E + P.E gained
0 = 0.39 + 0 + mgh

 

[salvatore]

I will explain you th whole concept and try to understand ...
First thing you need to consider , is to know if surface is smooth/ rough ( has friction )by reading question.
then follow the formula wd (workdone ) by driving force = wd by resistive force ( resisting motion ) +- K.E gained/lost +- P.E gained/lost.
if it increses in speed KE is gained ( Add in eqn ) if decreased in speed its decresed KE is lost ( subtract from eqn )..
If it increses in Height ( moves up ) PE is gained ADD in eqn ... and if comes down PE is lost and subtract in eqn
so when something is gained ADD and when lost MINUS as you see i wrote +- in eqn depending whether its lost or gained .
we know wd formula is = F * d

In this question ,,, here its same speed .. mean no gain/loss in KE .. so its not included in equation and there is no Wd by driving force so both are ZERO in eqn
So our eqn is wd (workdone ) by driving force = wd by resistive force ( resisting motion ) + K.E + P.E gained
0 = 0.39 + 0 + mgh

Thanks a lot for the explanation, I really appreciate it.
I understood what you said about adding/subtracting the P.E/K.E values but I still don't get when you have to do (K.E - P.E) or (P.E - K.E).. how do you know which one is greater than the other?
Regarding the question paper that I had linked, the marking scheme actually calculates the K.E lost which opposes your previous statement. If you have any other explanation about why they used W.D = P.E gained - K.E lost + w.d against friction, please help me out.

Thanks

 

[aniroula]

I am stuck in finding the value of Z in binomial distribution.
question
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_6.pdf
markscheme
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_ms_6.pdf
I am having problem in question number 4(i) to calculate the value of Z. please help! Thanks in advance.

 

[Snowysangel]

Can someone pleas e solve this

 

[salvatore]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s08_qp_3.pdf
Please help me with qn no. 7(ii).

I got the partial fractions in (i) as:
1 + (1/2)(1/x+1) + (3/2)(1/x+3)

I'll appreciate any help. Thanks

 

[TheZodiac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf
Please help me with qn no. 7(ii).

I got the partial fractions in (i) as:
1 + (1/2)(1/x+1) + (3/2)(1/x+3)

I'll appreciate any help. Thanks

To make it easier, integrate it separately:
∫1 dx + 1/2∫ 1/(x+1)dx - 3/2∫1/(x+3) dx

x + 1/2ln(x+1) - 3/2ln (x+3)
Put the limits
3 + 1/2 ln4 - 3/2 ln 2
3 + 2/2 ln 2 - 3/2 ln2
3 - 1/2 ln2

 

[salvatore]

To make it easier, integrate it separately:
∫1 dx + 1/2∫ 1/(x+1)dx - 3/2∫1/(x+3) dx

x + 1/2ln(x+1) - 3/2ln (x+3)
Put the limits
3 + 1/2 ln4 - 3/2 ln 2
3 + 2/2 ln 2 - 3/2 ln2
3 - 1/2 ln2

Thanks

 

[sitooon]

Find factors of'
z^4 -16

 

[Igcse stuff]

really need your help guys https://www.xtremepapers.com/community/threads/need-mathematics-p1-s1-help.31538/

 

[snowbrood]

Q7 anyone please it is from mechanics

 

[Igcse stuff]

intergrate ∫(Inx)/(x^1/2) dx within the limits 4 to 1 ....can someone answer this question for me

 

[soumayya]

Find factors of'
z^4 -16

z^4 - 16 = (z^2)^2 - (4)^2
= ( z^2 - 4)(z^2 +4)
= ( z - 2)( z + 2)(z^2 + 4)

 

[KWIKIW]

Hello,

I need some help in the 2004 may june S1 paper question no.5
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s04_qp_6.pdf

How can we know whether to apply permutations or combinations for (a) ?

 

[TheZodiac]

intergrate ∫(Inx)/(x^1/2) dx within the limits 4 to 1 ....can someone answer this question for me

I hope it's readable.

 

[arsiss]

Does anyone know some website from where I can get A-levels revision notes? I'v lost mine and the xtreme papers revision notes are not available

 

[Starlight97]

A stone is projected vertically upward with a speed of 21 m/ s .find the distance travelled by the stone in the first 3 seconds of its motion . Plzzz anyone help me out

 

[arsiss]

A stone is projected vertically upward with a speed of 21 m/ s .find the distance travelled by the stone in the first 3 seconds of its motion . Plzzz anyone help me out

s=ut+1/2at^2
u = 21
a =-9.81
t=3 so t^2=9
so s=21*3-1/2*9.81*9
s= 18.86m

 

[snowbrood]

can anyone solve this question please the answers are 39.5cm and 28.6cm respectively help anyone i am goin to faint

 

[sitooon]

How to draw argand diagrams ?
i know the basic ideas of argand diagram , but how to draw some complex no,'s