Mathematics: Post your doubts here!

[fishfish]

Hey thanks alot! I get it now!

I'm just gonna explain the whole part..
For this part, we have:
u = 0 (initial velocity)
s = 0.6 m
v = ? (thats what we're looking for)
a = 5.846 (from part i)

lets use the equation v^2 = v^2 + 2as
v^2 = 0 + (2 x 5 x 0.6)
v = 2.65 m/s

For A:
There is no force acting on it, so forward force = 0
The only force is the horizontal component of the weight; mgsin(theta)
From F = ma,
0 - mgsin(alpha) = ma
-gsin(alpha) = a
a = -10 x 16/65
a = (-32/13) m/s^2

Now lets find the maximum distance travelled by A after B hits the ground (v = 0)
The final speed of B = initial speed of A

v^2 = u^2 + 2as
0 = (2.65)^2 + (2 x -32/13 x s)
s = 1.425m

So the distance of A from P will be:
2.5 - (0.6 + 1.425)

distance = 0.475m

Hope this helped

hanks

 

[hamza.k143]

The p1 paper was out before the paper

 

[Suchal Riaz]

thanks for the animation ...

the smaller ball stops moving as the bigger ball hits the ground?

yeah okay.. u can tell me tomorrow after chem paper... if that's fine with u?

ok no problem. at that i had a kind of brain freeze. tomorrow inshallah i will. if u want i can solve it now.

 

[kitkat <3 :P]

The p1 paper was out before the paper

wth hell is this

 

[GCE As and a level]

The p1 paper was out before the paper

man this is impossible cuz this site ( http://dynamic-marketing.blogspot.com/ )downloaded the paper (http://3.bp.blogspot.com/-B9kCXfhwbrU/U2uK3cLPunI/AAAAAAAACdg/JAE9R36GnSg/s1600/photo 3.JPG) more than 24 hrs after the exam and the paper with the background is the same.....so just one of the guys took a screen shot and told u that the paper was out before
......Thts what i think

wth hell is this

 

[Snowysangel]

ok no problem. at that i had a kind of brain freeze. tomorrow inshallah i will. if u want i can solve it now.

But if the smaller ball was hitting the floor, the bigger ball would move some more when the other ball hits the ground due to intertia right?

 

[cool Asviva]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_41.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_ms_41.pdf
question # 7ii
anyone please explane how come constant term 50? show calculation. u=30 m/s right?

 

[aniroula]

How to change 1/(N)(1800-N) into partial fraction. I know the format is A/N +B/(1800-N) but the answer is different. answer in markscheme is A=1/2 and B=1/2
please help.

 

[kitkat <3 :P]

How to change 1/(N)(1800-N) into partial fraction. I know the format is A/N +B/(1800-N) but the answer is different. answer in markscheme is A=1/2 and B=1/2
please help.

M1??

 

[aniroula]

M1??

nope S1

 

[RahatMT]

(b) In a car magazine, 25% of the pages are used for selling second-hand cars,
62 1/2 % of the remaining pages are used for features,
and the other 36 pages are used for reviews.
Work out the total number of pages in the magazine.

can somebody show me the working along with an explanation...

 

[kitkat <3 :P]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_ms_41.pdf
question # 7ii
anyone please explane how come constant term 50? show calculation. u=30 m/s right?

look we will put the value of v which is 30 and t which is 10 in the equation so we will get the constant
a=-0.4t
v=-0.2t^2 +k
30=-0.2(10)^2 +K
30=-20+K
k=50

 

[kitkat <3 :P]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_41.pdf
Q3 please :/

 

[princeali97]

Is any1 giving M1 here? I have M1 so if you are here and looking for help then pls you are not alone

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_41.pdf
Q3 please :/

I guess the problem is in the third part. An easy way to do this question is to consider the movement of the signal from its highest point, where its u = 0. Now we know that total time is underroot 17s. Thus when conisdering only the returing time, it would become underroot 17/2s

Now use S = 1/2 (10) (underroot 17/2)square
S = 5 (17/4)'
S = 21.25m

Thus its height above the cliff was 21.25 Now just use the third equation of motion with v = 0 and you will get the answer

 

[kitkat <3 :P]

I guess the problem is in the third part. An easy way to do this question is to consider the movement of the signal from its highest point, where its u = 0. Now we know that total time is underroot 17s. Thus when conisdering only the returing time, it would become underroot 17/2s

Now use S = 1/2 (10) (underroot 17/2)square
S = 5 (17/4)'
S = 21.25m

Thus its height above the cliff was 21.25 Now just use the third equation of motion with v = 0 and you will get the answer

thanks and yeah it was 3rd part i didnt mention

 

[princeali97]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_41.pdf
Q3 please :/

i) v=o (max.height) , s=45(40+5) , g= -10(moving upwards)
v^2=u^2+2gh
putting the values u easily get u=30m/s
(ii) There are two ways of solving this one
.1. u find the time for which the signal is 40 m above the ground:
S=ut+1/2gt^2
40=30t-5t^2
t=2,4 ----> so length of time for which it is above 40 m (i.e top of cliff) =4-2=2m
2.U can calculate the time taken for the signal to travel from its max height (i.e 45m) to top of cliff (i.e 40m)
S=5 , u=0(max.height) ,g=10(moving downwards)
5=(0)t+5t^2
t=1.Now,this value of t is the time taken for the signal to move from its max.height to top of cliff.So,since u need to find the time for which signal is above the cliff's top u multiply by 2 becoz the time taken to travel from 40m(cliff's top) to max,height will be same.
(iii)First calculate the height of signal above the cliff's top.
U=0 (max height).T= root 17 / 2 (because root 17 is the time for which the signal goes upto max.height and then from max.height to cliff's top.So dividing it by 2 gives time taken by signal to move from max.height to cliff's top)
So s=1/2gt^2
S=0.5(10)9(root 17/2)^2
S=21.25.
So u get the max height of signal above the ground i.e 40 + 21.25=61.25m,v=0(max.height),g=-10(going upwards)
V^2=u^2+2gh
u=35m/s. Sorry for making this explanation so long.But i just tried myself for making it as easy as possible.

 

[kitkat <3 :P]

t

i) v=o (max.height) , s=45(40+5) , g= -10(moving upwards)
v^2=u^2+2gh
putting the values u easily get u=30m/s
(ii) There are two ways of solving this one
.1. u find the time for which the signal is 40 m above the ground:
S=ut+1/2gt^2
40=30t-5t^2
t=2,4 ----> so length of time for which it is above 40 m (i.e top of cliff) =4-2=2m
2.U can calculate the time taken for the signal to travel from its max height (i.e 45m) to top of cliff (i.e 40m)
S=5 , u=0(max.height) ,g=10(moving downwards)
5=(0)t+5t^2
t=1.Now,this value of t is the time taken for the signal to move from its max.height to top of cliff.So,since u need to find the time for which signal is above the cliff's top u multiply by 2 becoz the time taken to travel from 40m(cliff's top) to max,height will be same.
(iii)First calculate the height of signal above the cliff's top.
U=0 (max height).T= root 17 / 2 (because root 17 is the time for which the signal goes upto max.height and then from max.height to cliff's top.So dividing it by 2 gives time taken by signal to move from max.height to cliff's top)
So s=1/2gt^2
S=0.5(10)9(root 17/2)^2
S=21.25.
So u get the max height of signal above the ground i.e 40 + 21.25=61.25m,v=0(max.height),g=-10(going upwards)
V^2=u^2+2gh
u=35m/s. Sorry for making this explanation so long.But i just tried myself for making it as easy as possible.

Thanks alot but i was stuck in the 3rd part but thanks alot

 

[Snowysangel]

I guess the problem is in the third part. An easy way to do this question is to consider the movement of the signal from its highest point, where its u = 0. Now we know that total time is underroot 17s. Thus when conisdering only the returing time, it would become underroot 17/2s

Now use S = 1/2 (10) (underroot 17/2)square
S = 5 (17/4)'
S = 21.25m

Thus its height above the cliff was 21.25 Now just use the third equation of motion with v = 0 and you will get the answer

Why is the formula s=1/2 at2 when u is not zero? What gave you taken u as then?

 

[kitkat <3 :P]

Why is the formula s=1/2 at2 when u is not zero? What gave you taken u as then?

u is zero ._.