Mathematics: Post your doubts here!

[Suchal Riaz]

Thank you, kind sir. Wish you the best of luck with your exams, assuming you are sitting one.

comment on you username: "NO , I procrastinate"

 

[MiniSacBall]

QP: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_43.pdf
Q7 pii
MS: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_ms_43.pdf

Q7 pii, I know that when velocity is maximum acceleration is zero but idk how to solve this part, I am getting t=0 , when solving a=0.
Help Me.

 

[Suchal Riaz]

QP: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_43.pdf
Q7 pii
MS: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_43.pdf

Q7 pii, I know that when velocity is maximum acceleration is zero but idk how to solve this part, I am getting t=0 , when solving a=0.
Help Me.

this is how to factorise in cases like these. you can substitute this value in V to find max velocity.

 

[AbbbbY]

Anyone here who wants the solutions to this years Math P1?

 

[Snowysangel]

The top of a cliff is 40 metres above the level of the sea. A man in a boat, close to the bottom of the cliff, is in difficulty and fires a distress signal vertically upwards from sea level.

This signal is above the level of the top of the cliff for (root) 17
seconds

Find the speed of projection of the signal.

Someone pls help thx!!!

Is this m1?

 

[Suchal Riaz]

View attachment 41973
plz explain iii and iv

do you understand it after video or should i explain it?

 

[daredevil]

do you understand it after video or should i explain it?

can u explain it?

the video didnt work... and my internet connection is too slow ryt now to load it on proxy or anything... but thankss anyways for that!

 

[Suchal Riaz]

can u explain it?

the video didnt work... and my internet connection is too slow ryt now to load it on proxy or anything... but thankss anyways for that!

this is like an animtion to help you see what happens.

*lighter ball stops at maximum point (instanenously at rest) then moves downwards. it's acceleration after heaver ball hits the ground is 10 m/s²

 

[fishfish]

hi i really need help im this question

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_42.pdf

Q7, the second part of the part ii

 

[abruzzi]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf
Pleaseee help me with qn no. 7(i). From the marking scheme, why is K.E = 0.5 x Vapproach - Vreturn? Change in K.e is usually (final speed - initial speed), so why have they done the other way round here?
I'll be really grateful for your help
Suchal Riaz or anyone else

I'm having the same problem.. somebody please explain this

 

[salvatore]

hi i really need help im this question

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_42.pdf

Q7, the second part of the part ii

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

 

[Tushar968]

hey how do u do q7 on may june 2012 paper 4\
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_41.pdf

 

[Suchal Riaz]

daredevil please remind me to explain that part to you. now i have to go. sorry for being busy and delaying your doubt for long time.

 

[daredevil]

daredevil please remind me to explain that part to you. now i have to go. sorry for being busy and delaying your doubt for long time.

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?

 

[daredevil]

daredevil please remind me to explain that part to you. now i have to go. sorry for being busy and delaying your doubt for long time.

ohh it moves for a maximum height... until its velocity becomes zero and then falls back to the 'black line' where it comes to rest?

 

[mohammad hurani]

please explain step by step

 

[usama321]

please explain step by step

First part

V = 0 + .3*T1
T1=V/.3
T3 =552-T2
0 = V - 1(T3)
putting in T3
0 = V - 1(552-T2)
T2 = 552 - V

Total distance travelled = 1/2 (sum of parallel sides) * height and equal it to 12000 (total journey was 12 km)
1/2 (552 + (T2-T1) ) * V = 12000
1/2 (552 + 552-V - V/.3) *V = 12000
(1104 -13/3V) *v = 24000
3312V - 13V^2 = 72000
The rest is pretty easy

 

[Rutzaba]

best of luck guys

 

[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