Mathematics: Post your doubts here!

[Ahmed Aqdam]

View attachment 52399
just have a bit of confusion ...how was c found when acceleration was integrated? nd why did we use 2.5s as time to find the total distance??

The equation will be equated to v found i part (i) at t=0 as the speed of P will be same at A.
Time taken from O to A is 0.5s so after A it will be 2.5s.

 

[Haya Ahmed]

http://onlineexamhelp.com/wp-content/uploads/2014/02/9709_w13_qp_43.pdf

q5 (ii)

 

[ashcull14]

http://onlineexamhelp.com/wp-content/uploads/2014/02/9709_w13_qp_43.pdf

q5 (ii)

total time = 552s
total distance 12000m
graph forms a trapezium .....through area of trapezium i cudnt get the corect ans so i used the area of rectangle - area of two triangles (T1 and T3)
AREA OF RECTANGLE= total time * velocity = 552 *V
Distance of T1 (triangle T1) = (s=ut +1/2at^2) = 0 + (0.5 *0.3 *(V/0.3)^2) = 5/3 V
Distance of second triangle T3= (s=ut - 1/2 at^2) = V^2 - 1/2 V^2 = 1/2 V^2 (negative sign with a as it is decelerating)
now subtract distances obtained above (triangle T1 and T3) from 552V and equate it with 12000
552 V- (5/3 V +1/2 V^2) = 12000
552 V- 13/6 V^2 =12000 (LCM =6)
3312 V - 13 V^2 =72000
13 V^2 -3312 +72000 (ANS)

 

[qurratul ain yafu]

hi i m giving igcse... can u solve my doubts.... plz

 

[ashcull14]

hi i m giving igcse... can u solve my doubts.... plz

https://www.xtremepapers.com/community/threads/mathematics-post-your-doubts-here.2565/

 

[Wolfgangs]

Can someone please help me with this question?

 

[ashcull14]

How is the acceleration 6 ??? can someone help me wid the application of the formula above?

 

[ashcull14]

Can someone please help me with this question?

which paper is it?

 

[TheJDOG]

Can someone please help me with this question?

Hey, I have solved this question before it's not hard, all you have to do is to find everything in terms of r and pie and you have to equate the area of shade to half the area of the circle and yeah pretty much basic algebra, I'll do this for you tomorrow really need to sleep

 

[xxxvip]

http://onlineexamhelp.com/wp-content/uploads/2014/02/9709_w13_qp_31.pdf
pls help Q2, Q4.. i cant get it

 

[forest822]

Hello. Can anyone explain my doubts? Thank You.

1. A lorry of mass 15000kg climbs from bottom to the top of a straight hill, of length 1440m, at a constant speed of 15ms-1. The top of the hill is 16m above the level of the bottom of the hill. The resistance to motion is constant and equal to 1800N. Find the work done by the driving force. (ANS: 4.99x10^6)

Why my answer is wrong according to the working below. F-1800=0 F=1800N W=1800x1440=2.592x10^6

2. OCT/NOV 2013 9709/41 Questions No.2 http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_w13_qp_4/ ( ANS: 80N)

30(0.6)+40(0.8)-F=0 F=50N
5/8 ( Normal Reaction )= 50
N.R= 80

80+30(3/5)=40(4/5)+W
W= 66N?

 

[ashcull14]

Hello. Can anyone explain my doubts? Thank You.

1. A lorry of mass 15000kg climbs from bottom to the top of a straight hill, of length 1440m, at a constant speed of 15ms-1. The top of the hill is 16m above the level of the bottom of the hill. The resistance to motion is constant and equal to 1800N. Find the work done by the driving force. (ANS: 4.99x10^6)

Why my answer is wrong according to the working below. F-1800=0 F=1800N W=1800x1440=2.592x10^6

2. OCT/NOV 2013 9709/41 Questions No.2 http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_w13_qp_4/ ( ANS: 80N)

30(0.6)+40(0.8)-F=0 F=50N
5/8 ( Normal Reaction )= 50
N.R= 80

80+30(3/5)=40(4/5)+W
W= 66N?

QUESTION 1
at an inclined plane u should keep some factors in mind
especially the energy losses and gain
usually in order to find the WD BY DRIVING FORCE in such cases the following formula is used:-
WD BY DRIVING FORCE= WD by Kinetic energy + WD BY FRICTION + Gain in gpe
since the lorry has a constant speed the kinetic energy here will be zero...since its due to change in motion
the formula u used here is Friction * distance thats = WD BY FRICTION HERE...
for WD by driving force u need to calculate the energy changes first

(AS KE =0J)
WD BY DRIVING FORCE= Gain in GPE + WD by Friction
gAIN IN GPE = mgh = 15000*10*16 = 2.4*10^7 J
WD BY FRICTION = 18000* 1440 = 2.59*10^7 J
thus WD BY DRIVING FORCE =( 2.4*10^7 J) + (2.59*10^7 J) = 4.99x10^6 J

 

[Topperman]

Hey guys...
I'm going to write my AS 9709 Mathematics Paper 06 (Statistics) in about one month from now in the June 2015 series.
I am stuck on normal distribution...when using continuity correction, when do we round up and when do we round down?
Like for example :
October/November 2006 Paper 06:
Q.7
A manufacturer makes two sizes of elastic bands: large and small. 40% of the bands produced are
large bands and 60% are small bands.
An office pack contains 150 elastic bands.
(iii) Using a suitable approximation, calculate the probability that the number of small bands in the
office pack is between 88 and 97 inclusive. [6]
so I round 97 to 97.5 or to 96.5 and 88 to 88.5 or 87.5?
Any help will be much appreciated.
Thanks.

 

[forest822]

QUESTION 1
at an inclined plane u should keep some factors in mind
especially the energy losses and gain
usually in order to find the WD BY DRIVING FORCE in such cases the following formula is used:-
WD BY DRIVING FORCE= WD by Kinetic energy + WD BY FRICTION + Gain in gpe
since the lorry has a constant speed the kinetic energy here will be zero...since its due to change in motion
the formula u used here is Friction * distance thats = WD BY FRICTION HERE...
for WD by driving force u need to calculate the energy changes first

(AS KE =0J)
WD BY DRIVING FORCE= Gain in GPE + WD by Friction
gAIN IN GPE = mgh = 15000*10*16 = 2.4*10^7 J
WD BY FRICTION = 18000* 1440 = 2.59*10^7 J
thus WD BY DRIVING FORCE =( 2.4*10^7 J) + (2.59*10^7 J) = 4.99x10^6 J

Thank you for your explantions. May I know why WD BY DRIVING FORCE= WD by Kinetic energy + WD BY FRICTION + Gain in gpe? I can't imagine it.

 

[qwertypoiu]

View attachment 52453View attachment 52454
How is the acceleration 6 ??? can someone help me wid the application of the formula above?

I suggest you don't bother with that formula. Whenever there is a pulley problem, apply Newton's Second Law to both particles, in terms of a and T, then solve the simultaneous equations by adding them:
resultant force = mass*acceleration
For B:
6-T=0.6a
For A:
T-2=0.2a
Adding the equations together:
4 = 0.8a
a=5

 

[ashcull14]

I suggest you don't bother with that formula. Whenever there is a pulley problem, apply Newton's Second Law to both particles, in terms of a and T, then solve the simultaneous equations by adding them:
resultant force = mass*acceleration
For B:
6-T=0.6a
For A:
T-2=0.2a
Adding the equations together:
4 = 0.8a
a=5

View attachment 52489 View attachment 52489

i did the same but the acceleration is 6 m/s^2

 

[qwertypoiu]

i did the same but the acceleration is 6 m/s^2

The mark scheme you attached says 5??

 

[asadalam]

Hey guys...
I'm going to write my AS 9709 Mathematics Paper 06 (Statistics) in about one month from now in the June 2015 series.
I am stuck on normal distribution...when using continuity correction, when do we round up and when do we round down?
Like for example :
October/November 2006 Paper 06:
Q.7
A manufacturer makes two sizes of elastic bands: large and small. 40% of the bands produced are
large bands and 60% are small bands.
An office pack contains 150 elastic bands.
(iii) Using a suitable approximation, calculate the probability that the number of small bands in the
office pack is between 88 and 97 inclusive. [6]
so I round 97 to 97.5 or to 96.5 and 88 to 88.5 or 87.5?
Any help will be much appreciated.
Thanks.

Use CC when data is non continuous like no of people,it can be 34,33 but nothing in between,and dont use it in continuous like height it can be anything in decimals between 180 cm and 181 cm.

 

[ashcull14]

The mark scheme you attached says 5??

im so DUMB so sorry nd thnks a lot

 

[qwertypoiu]

im so DUMB so sorry nd thnks a lot

Haha happens sometimes
And you're welcome