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Further Mathematics: Post your doubts here!

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Hannahang

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Hi Everyone
I got stuck with this question from further maths papers, november 2003 Q3 paper 2 about circular motion. Please help...thxs very much

I cannot draw the diagram here but I will type out the Qs....

A particle P of mass m is connected to a fixed point O by a light inextensible string OP of length r and is moving in a vertical circle centre O. At its lowest point, P has speed U. When the string makes an angle alpha with the downward vertical it encounters a small fixed peg Qwhere OQ = 1/2r. The string proceeds to wrap itself around peg so that P begins to move in a vertical circle with centre Q (see diagram) . Given that the particle describes a complete circle about Q, show that U^2 > equal to gr (7/2 - cos alpha)

I got stuck on how to solve this question..please help...thxs
 
H

Hannahang

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Hi everyone,

I got stuck with this question from CIE further maths past year papers, november 2004 Q11 paper 2
Please help..I cannot draw the diagram here but I will type the Qs...

This is the link to the diagram from extreme papers

http://www.google.com.bn/url?sa=t&r...w4HgBA&usg=AFQjCNHS-H5M_GXizeXYhyFjyFN5FXEg9A

A ring Pof mass m kg is threaded on a smooth semicircle wire with centre C and radius a m, fixed in a horizontal plane. A light elastic string connects P to D, a fixed point on the axis of the semicircle through a small distance from its equilibrium position E and is released from rest at time t = 0. In the subsequent motion, angle PCD = theta and beta is the angle which PD makes with the tangent at P. Given that theta is small enough for theta square and higher powers to be neglected, show that

(i) PD = am
(ii) the tension TN in the string is constant
(iii) Beta = pie/2 - 2theta

I have solved the first 2 parts....The part (iii) and last 2 parts are the ones I got stuck with....

Write down an approximate equation of motion of P on the wire.

The period of the motion of P is approximately 2 s. Find the value of T in terms of m and a and find the solution of the approximate equation of motion for which theta > 0 when t = 0 and 0.5 s after release , d theta/ dt = -0.03 rad per second..

Thxs
 
H

Hannahang

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Hi everyone, I got stuck with this question from CIE past year paper november 2003 Q9 paper 2 Please help! Thxs

The line L1 passes through the point A with position vector i - j - 2k and is parallel to the vector 3i - 4j - 2k. The line L2 passes through the point ( 1 + 5 cos t ) i - (1 + 5 sin t )j - 14k, where 0< inclusive t < 2 pie. and is parallel to the vector 15i + 8j -3k. The point P and Q are on L1 and L2 respectively, and PQ is perpendicular to both L1 and L2.

Find the length of PQ in terms of t.

I got this part. The ans is PQ = (20cos t + 15 sin t -144) / 13 .

This is the part I got stuck with ...Hence show that the lines L1 and L2 intersect and find the maximum length of PQ as t varies...

The proving part I am not sure and the max length...It should be 13.

Thxs
 
H

Hannahang

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Hi Everyone, I got stuck on this question on circular motion from november 2007 Q4 paper 2....Please help.... I found this topic very confusing

A smooth hemisphere of radius a has its circular face fixed to the surface of a horizontal table. The centre of the face is O. A particle P of mass m is placed at the highest point of the hemisphere and is given a horizontal velocity u. The particle does not immediately lose contact with the surface of the hemisphere. Assuming all resistances to motion are neglected, find the set of possible values of u.

Show that P loses contact with the surface before OP becomes horizontal.

I checked the examiner report but I don't really understand why the upward centripetal force is less than the weight.....is not centripetal force always act toward the centre of the circle?
Please help..thxs
 
H

Hannahang

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Hi Everyone...Please help me with this question from november 2002 Paper 2 Q4 on circular motion...It is quite difficult....pls pls pls help me...thxs...

The Qs starts with A toboggan travels along the path ABC shown in the diagram....

Thxs very much
 
S

Sakib

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Is there anyone sitting for the Further Maths exam this session?? I am! Wish me luck guys!
Hannahang, all your queries are from paper 2 while my exam tomorrow is on paper 1, so sorry can't answer them now.
Only the O/N 2003 question is from paper 1 (you made a typo and wrote that p-2 as well). Here goes that solution-
i) u could already do it, so no further explanation needed.
ii) if the 2 lines intersected, the shortest distance between them would be ZERO (for obvious reasons). So, taking PQ=0, yields (20cos t +15sin t) = 144. Now that's impossible, so PQ never intersects. For maximum distance, differentiating the distance with respect to time and then taking that expression = 0 yields the desired angle and then plugging that angle in the expression of length yields the desired length.
iii) Angle between the 2 planes is effectively the angle between the 2 lines L1 and L2 (Hope this picture is clear to you)

Hope that helps! :D I'll try to post the answer to the other queries of paper 2 after P-1 is over tomorrow!
Once again, Wish me luck everyone, I badly need it!
 
H

Hannahang

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Hi Sakib....thxs so much for your help...Infinity of good luck for your FMTH paper 1 exam tomorrow!!!
 
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Hannahang

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And To everyone who is sitting for FMTH Paper 1 tomorrow....All the best!!!!!!! U guys can do it....
 
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Hannahang

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Hi Everyone...I know u all are sitting for the FMTH paper 1 exam tomorrow.....I have been pratising paper 1 and I got stuck on this question from D.E, november 2004 Q12 Paper 1...I managed to solve the part (i) but I don't understand why the answers in part (ii) got R...I could prove e^(-3/2t) sin (2t + theta) but u are supposed to prove Re^(-3/2t)sin (2t + theta)...I don't know where the R comes from....If anyone knows why....could u pls explain....thxs very much
 
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Sakib

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C.F. is e^(-1.5t) (Acos 2t + Bsin 2t) where A and B are dependent on the initial conditions of y. So R=sqr. root(A^2+B^2)
 
S

Sakib

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To everyone who has given the Further math exam (9231/12) How was it??? I gave probably the most terrific exam!! Really happy!! What about you guys??
 
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Sakib

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Guys, I need help!
What is the degrees of freedom for chi-squared test?? (during the goodness of fit tests)
CIE themselves are confused!!
In O/N 2012 they took it as (n-2) (in the mark scheme) but in M/J 2013 they took it as (n-1)!
What should I take??
 
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Hannahang

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Hi Sakib the degrees of freedom of chi-squared depends on the number of restrictions..

If u calculated ur poisson mean from your data, then you will have two restrictions, one is sum of E = sum of O and the other the mean estimated from the data....

Hope that helps...
 
H

Hannahang

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If u r free Sakib or anyone, could you guys help me to solve the Qs that I asked on pg 23....thxs very much
 
Relon

Relon

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Guys can any one tell me, if am going to take both A-level mathematics and A-level further mathematics, both of them will be counted or just one ?
 
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