Mathematics: Post your doubts here!

[XPFMember]

aoa wr wb..n welcome!

 

[cHeStEr]

Hey again.
But the question seems a lil different.
Find the equations of the tangents of the graph of y=x^2 AT THE POINT AT WHICH IT MEETS THE LINE WITH THE EQUATION y=x+2.
So Here they're asking us to find the point where it meets the line with equation y=x+2
the answer given is the same as yours but it doesn't satisfy the line eqn thing.

Heyyy thanks once again.

 

[Rk1877]

Hi i need help in mechanics(m1) may/june 2004 paper question number 4(ii),cannot understand how to solve,how R and F are found.Also need help in may/june 2003 paper question number 6(v),don't understand the formula and why acceleration*2 is used.Thank You for your help.

 

[abcde]

Hi i need help in mechanics(m1) may/june 2004 paper question number 4(ii),cannot understand how to solve,how R and F are found.Also need help in may/june 2003 paper question number 6(v),don't understand the formula and why acceleration*2 is used.Thank You for your help.

Post the links of the past paper(s) you need help in.

 

[Rk1877]

Post the links of the past paper(s) you need help in.

Question 4(ii)
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s04_qp_4.pdf
Question 6(v)
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s03_qp_4.pdf
Thanks

 

[abcde]

Question 4(ii)
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s04_qp_4.pdf
Question 6(v)
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s03_qp_4.pdf
Thanks

Q4(ii)
sin theta = 0.7/2.5
=> theta = 16.26020*
Contact force, R = 2 cos 16.26020* = 1.92 N.
=> Frictional force = 0.15 x 1.92 = 0.288 N.
=> Work done against resistance = 0.288 x 2.5 = 0.72 N.
Loss in G.P.E = gain in K.E + work done against resistance
So, K.E on ground = (1.4 - 0.72)J = 0.68 J.

Q6)(v) When the block passes through A after rebounding, it continues its motion before finally coming to rest. We need to determine the distance it covers after passing through A and before coming to rest.
Initial speed, u = 1.5 m/s
Final speed, v = 0 m/s
a = - 0.25 m/s^2
Using v^2 = u^2 + 2as, you get s = 4.5 m.
Therefore, total distance = AB + BA + 4.5 = 2(20) + 4.5 = 44.5 m.

 

[Rk1877]

Q4(ii) View attachment 3786
sin theta = 0.7/2.5
=> theta = 16.26020*
Contact force, R = 2 cos 16.26020* = 1.92 N.
=> Frictional force = 0.15 x 1.92 = 0.288 N.
=> Work done against resistance = 0.288 x 2.5 = 0.72 N.
Loss in G.P.E = gain in K.E + work done against resistance
So, K.E on ground = (1.4 - 0.72)J = 0.68 J.

Q6)(v) When the block passes through A after rebounding, it continues its motion before finally coming to rest. We need to determine the distance it covers after passing through A and before coming to rest.
Initial speed, u = 1.5 m/s
Final speed, v = 0 m/s
a = - 0.25 m/s^2
Using v^2 = u^2 + 2as, you get s = 4.5 m.
Therefore, total distance = AB + BA + 4.5 = 2(20) + 4.5 = 44.5 m.

Thanks alot.Just one doubt,how do you know that loss in p.e=gain in k.e+w.d against resistance and why do we use k.e=p.e lost in the 1st part of that question?.Can you explain the reasons and when to use which formula's?.VERY CONFUSED.Thanks a ton for all your help.

 

[abcde]

Thanks alot.Just one doubt,how do you know that loss in p.e=gain in k.e+w.d against resistance and why do we use k.e=p.e lost in the 1st part of that question?.Can you explain the reasons and when to use which formula's?.VERY CONFUSED.Thanks a ton for all your help.

The plane is smooth in the first part. So there's no resistance and no work done against resistance. The presence of friction in the second part indicates that some work will need to be done against resistance by the driving force.
These formulas are used based on the Principle of Conservation of Energy. Since energy is constant, all of the G.P.E must be converted into K.E when there's no friction. In case of friction, some of the G.P.E will be converted to K.E and some will be the work done against friction. U'll know which formula to use by carefully examining each scenario.
Hope you understand.

 

[Rk1877]

The plane is smooth in the first part. So there's no resistance and no work done against resistance. The presence of friction in the second part indicates that some work will need to be done against resistance by the driving force.
These formulas are used based on the Principle of Conservation of Energy. Since energy is constant, all of the G.P.E must be converted into K.E when there's no friction. In case of friction, some of the G.P.E will be converted to K.E and some will be the work done against friction. U'll know which formula to use by carefully examining each scenario.
Hope you understand.

Thanks alot.My teacher could not explain con of energy properly hence the confusion lol.

 

[Rk1877]

Also need help in:
Question 4(ii) and question 6 of may/june 2005.Thanks alot.
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s05_qp_4.pdf

 

[abcde]

Also need help in:
Question 4(ii) and question 6 of may/june 2005.Thanks alot.
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s05_qp_4.pdf

AoA!
If you could do Q4(i), there's little reason for being unable to do (ii). Notwithstanding that, here's an explanation:

As the system is in limiting equilibrium, the tension would equal A's weight i.e. 2N.
As A is about to move upwards, B is about to move rightwards. So the frictional force now will be in the same direction as the tension.
First, analyse the vertical forces.
In equilibrium, R + 1.8 = 3. => R = 1.2 N.
Considering forces horizontally, X = 2 + 2/3 (1.2) = 2.8 N.

Q6) (i) Distance moved = area under graph
=> 1/2 x v x 5 = 10
=> v = 4 m/s.

(ii) V/3 = 2.
=> V = 6 m/s.

(iii) Distance moved = area under graph
1/2 x 6 x (t+ 24.5 - 15) = 34.5
=> t + 9.5 = 11.5
So, t = 2 s.

(iv) time taken = 24.5 - 20 = 4.5
=> a = -6/4.5 = - 4/3
So, the deceleration is 4/3 m/s^2.
Hope this helps.

 

[Rk1877]

AoA!
If you could do Q4(i), there's little reason for being unable to do (ii). Notwithstanding that, here's an explanation:
View attachment 3793
As the system is in limiting equilibrium, the tension would equal A's weight i.e. 2N.
As A is about to move upwards, B is about to move rightwards. So the frictional force now will be in the same direction as the tension.
First, analyse the vertical forces.
In equilibrium, R + 1.8 = 3. => R = 1.2 N.
Considering forces horizontally, X = 2 + 2/3 (1.2) = 2.8 N.

Q6) (i) Distance moved = area under graph
=> 1/2 x v x 5 = 10
=> v = 4 m/s.

(ii) V/3 = 2.
=> V = 6 m/s.

(iii) Distance moved = area under graph
1/2 x 6 x (t+ 24.5 - 15) = 34.5
=> t + 9.5 = 11.5
So, t = 2 s.

(iv) time taken = 24.5 - 20 = 4.5
=> a = -6/4.5 = - 4/3
So, the deceleration is 4/3 m/s^2.
Hope this helps.

Thanks alot.

 

[abcde]

Q5) (i) The force exerted on the string by the pulley is given by 2T cos x, where x is the angle between the two tensions.
=> 2 T cos 45* = 4 (2)^1/2
=> T = 4 N.
(ii) 4 = 0.8 R
=> R = 4/0.8 = 5N
10 m = 5
So, m = 0.5 kg.
(iii) mass of Q = 4/10 = 0.4 kg (since T = weight of Q when T = 4 N).
Now, a mass of 0.1 kg has been attached to Q, making the total 0.5 kg. You can form 2 equations:
0.5(10) - T = 0.5 a (Resultant force = ma at Q)
T - 0.8(5) = 0.5 a (Resultant force = ma at P)
Upon elimination, T = 4.5 N.

 

[CaptainDanger]

http://www.xtremepapers.com/CIE/ind...el/9709 - Mathematics/&file=9709_s06_qp_1.pdf

http://www.xtremepapers.com/CIE/ind...el/9709 - Mathematics/&file=9709_s06_ms_1.pdf

Q10 first part... I am able to do it but after doing the 2nd part of it... How do we do it without getting to the 2nd part?

 

[abcde]

http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9709 - Mathematics/&file=9709_s06_qp_1.pdf

http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9709 - Mathematics/&file=9709_s06_ms_1.pdf

Q10 first part... I am able to do it but after doing the 2nd part of it... How do we do it without getting to the 2nd part?

(i) Derive the equation. dy/dx = 3x^2 - 6x - 9
At min.pt. this is equal to zero.
=> x^2 - 2x - 3 =0
=> x = 3 and x = -1.
At min. pt. x = 3 and y = 0 (apparent from the sketch). Insert these values into the curve's eq to get k. In the second part, you then insert x = -1 into the eq.

 

[Broly]

Hi,everyone I have a question concerning Maths chap Coord geometry

The line AB makes an angle of 120 degrees with the positive direction of the x-axis. Find the equations of the straight lines(root 3,1-root 3)respectively parallel and perpendicular to AB.

The answer is Y time Root 3= x - 3
I hope you guys could understand the question, what i want is the full expl of how to do it. Thanks in advance.

 

[Doctor Nemo]

Hi,everyone I have a question concerning Maths chap Coord geometry

The line AB makes an angle of 120 degrees with the positive direction of the x-axis. Find the equations of the straight lines(root 3,1-root 3)respectively parallel and perpendicular to AB.

The answer is Y time Root 3= x - 3
I hope you guys could understand the question, what i want is the full expl of how to do it. Thanks in advance.

If the line makes an angle of 120 degrees with the positive x-axis then the gradient of the line is equal to the tan (120) which is equal to -sqrt(3). The line perpendicular will have a gradient of 1/sqrt(3). The rest of the question I don´t understand.

 

[zain942010]

Please Help !
November 2003 Maths P3 Question 7 (iii)

 

[abcde]

Please Help !
November 2003 Maths P3 Question 7 (iii)

Please post the link for the paper. Always.

 

[Khan_971]

Q10 Part (b) http://www.mediafire.com/download.php?krh8zc8hr8xm25h
Jazak Allah