STEP Maths I, II, III 1993 Solutions

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
TSR launches Learn Together! - Our new subscription to help improve your learning 16-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
Search Thread
Advanced Search
Sign in to Reply
  1. nota bene's Avatar
    221 13-06-2009  01:16
    • TSR Idol
    • Posts: 9,970
    Re: STEP Maths I, II, III 1993 Solutions
    (Original post by generalebriety)
    STEP III Q4:
    (iii)

    \displaystyle S = \sum_{r=2}^\infty \frac{r2^{r-2}}{3^{r-1}} = \frac{1}{3}\sum_{r=2}^\infty r\bigg( \frac{2}{3} \bigg)^{r-2}.\\ \text{Let } u_n = \sum_{r=n}^\infty \bigg( \frac{2}{3} \bigg)^{r-2} = 3\bigg( \frac{2}{3} \bigg)^{n-2}.\\ \text{Then } S = \frac{1}{3}(2u_2+u_3+u_4+u_5+\do ts). \\ u_2 = 3\\ u_2+u_3+u_4+\dots = \frac{3}{1 - \frac{2}{3}} = 9.\\ \therefore S = \frac{1}{3}(3+9) = 4.
    Alternative way:

    \frac{1}{2}\displaystyle\sum_{r= 2}^{\infty}r(\frac{2}{3})^{r-1}
    Let's study S=\displaystyle\sum_{r=2}^{\inft y}r(\frac{2}{3})^{r-1}=2\frac{2}{3}+3(\frac{2}{3})^2 +4(\frac{2}{3})^3+...
    Note that \frac{2}{3}S=2\frac{2}{3})^2+3(\ frac{2}{3})^3+4(\frac{2}{3})^4+. ..
    Now S-(2/3)S=(1/3)S=\frac{2}{3}+\frac{2}{3}+(\frac{2 }{3})^2+(\frac{2}{3})^3+...
    This is almost a GP, so add (1/3) to this and we have \frac{1}{3}+\frac{1}{3}S=\displa ystyle\sum_{r=0}^{\infty}(\frac{ 2}{3})^r=\frac{1}{1-\frac{2}{3}}=3

    Thus (1/3)S=3-(1/3)=8/3 i.e. S=8 and we're looking for (1/2)S, which is 4.
  2. Glutamic Acid's Avatar
    222 13-06-2009  01:45
    • TSR Legend
    • Location: E.I.R.E. / S.E. / Cam
    Re: STEP Maths I, II, III 1993 Solutions
    I'll post my alternatives for question 4 too.

    (i) Note that \tanh^{-1}x = x + \dfrac{x^3}{3} + \dfrac{x^5}{5} + ... \implies \dfrac{\tanh^{-1}x}{x} = 1 + \dfrac{x^2}{3} + \dfrac{x^4}{5} + ...

    Letting x = 1/2 and using the logarithmic form of arctanh(x) gives 2 \times \dfrac{1}{2} \ln \left( \dfrac{1 + \frac{1}{2}}{1 - \frac{1}{2}} \right) = \ln 3

    (iii) \displaystyle \sum_{r=0}^{\infty} x^r = (1-x)^{-1} \Rightarrow \sum_{r=0}^{\infty} rx^{r-1} = (1-x)^{-2} -- the former is the sum of a geometric series and the latter is its derivative. Let x = 2/3 and we get
    \displaystyle \frac{1}{2} \sum_{r=0}^{\infty} r \left( \frac{2}{3} \right)^{r-1} = \frac{9}{2}. To find the sum from r = 2 to infinity we need to subtract the terms for r = 1 and 0 which gives 9/2 - 1/2 = 4.
  3. Glutamic Acid's Avatar
    223 13-06-2009  14:24
    • TSR Legend
    • Location: E.I.R.E. / S.E. / Cam
    Re: STEP Maths I, II, III 1993 Solutions
    III/14

    The particle will undergo circular motion in a vertical circle. Let the velocity when it starts/rebounds be v. At a height h, it gains mgh in potential energy. At this height, let it have velocity x. Change in kinetic = Change in potential
    So \dfrac{1}{2}mv^2 - \dfrac{1}{2}mx^2 = mgh \implies x^2 = v^2 - 2gh
    The radial acceleration will equate to the tension in the string plus the weight acting radially inwards, so \dfrac{mx^2}{c} = T + mg \sin \theta, but \sin \theta is clearly h/c. Combining and rearranging these two equations gives T = \dfrac{m}{c}(v^2 - 3gh) so when the string is slack v^2 - 3gh = 0. Clearly for v^2 = 64^2 \cdot 6gc, h will not lie inside the circle. So we keep halving this value (as the coefficient of restitution is 1/2) until v = \dfrac{1}{2} \sqrt{6gc} giving h = \dfrac{c}{2}; the first valid value of h as it lies inside the semicircle. This happens on the way from A to B.

    Now it gets messy. As x^2 = v^2 - 2gh, x = \sqrt{\dfrac{gc}{2}}. The vertical angle will be \cos \theta = \dfrac{\sqrt{3}}{2}, so using our kinematic equation one wishes to solve -\dfrac{c}{2} = \dfrac{1}{2} \sqrt{\dfrac{3gc}{2}} - \dfrac{1}{2}gt^2 for the larger value of t, giving t = \dfrac{(\sqrt{\frac{11}{2}} + \sqrt{\frac{3}{2}}) \sqrt{gc}}{2g}, and multiplying this by the horizontal component (\sqrt{\dfrac{gc}{8}}) gives a distance travelled of \dfrac{c}{4}(\sqrt{3} + \sqrt{11}). By geometry, the distance from A when the string is slack is c - \sqrt{3}{2}c giving the distance from A when it particle rebounds as (4 - \sqrt{3} + \sqrt{11}) \dfrac{c}{4} \approx 1.4c, so it's closer to B. The algebra for the second part was horrible (though the concept simple), and I wonder if I've gone wrong.
  4. Mathletics's Avatar
    224 15-06-2009  09:35
    • Adored and Respected Member
    • Posts: 477
    Re: STEP Maths I, II, III 1993 Solutions
    (Original post by Speleo)
    I/4

    I like pencil vv

    N.B. these are rough solutions, I'd write much more explanation of what was happening in the real exam.
    Hey could you explain why, for the second integral, you used limits 2\pi and \pi instead of 2\pi and 0 as the question asks?
  5. jj193's Avatar
    225 24-02-2010  20:25
    • Peer Of The TSR Realm
    • Location: Manchester
    Re: STEP Maths I, II, III 1993 Solutions
    STEP I Question 8 - gen's solution is STEP III Q8.





    edit
    Attached Thumbnails
    Click image for larger version.  Name: STEP I '93 Q8 a.jpg  Views: 22  Size: 311.4 KB  ID: 81705   Click image for larger version.  Name: STEP I '93 Q8 b.jpg  Views: 17  Size: 197.3 KB  ID: 81706  
    Last edited by jj193; 28-02-2010 at 15:34.
  6. jj193's Avatar
    226 16-05-2010  19:05
    • Peer Of The TSR Realm
    • Location: Manchester
    Re: STEP Maths I, II, III 1993 Solutions
    I just did STEP II Q7, got the 'infinite descent' part but I don't really understand infinite descent (yet I know/think it is this). Why should it be that if I can keep dividing a,b,c by 5 - forever - that a,b,c are 0. I geuss 0/5=0 etc. and zero is still in the intergers. I've just convinced myself it's ok, I still don't like it.

    It seems like a huge jump.

    Would this be acceptable as an ending:
    If an interger can be divided infinitely many times, then the interger must be zero because if it is not it must be infinite which isn't an interger.
  7. miml's Avatar
    227 16-05-2010  19:14
    • Overlord in Training
    • Location: Warwickshire
    • Posts: 3,103
    Re: STEP Maths I, II, III 1993 Solutions
    I think a neater way (although I don't see anything wrong with your method) is to say,

    Assume a is non zero, and an integer
    Then there exists some a that contains infinitely many factors of 5 (this bit is crucial. 0 is definitely in the integers, and has infinitely many factors)
    No such integer exists
    By contradiction a=0

    Similarly for b,c
    Therefore only solutions when a=b=c=0
  8. themaths's Avatar
    228 07-05-2011  14:33
    • New Member
    • Posts: 15
    Re: STEP Maths I, II, III 1993 Solutions
    (Original post by Rabite)
    I think I did Q2 in the Further Maths A.
    But it seems quite easy so I've probably made a mistake. It's still red on the front page, so if no one else has typed it out already, I'll do so~

    [edit] Here it is anyway.
    \int \cos{mx}\cos{nx} dx = ½ \int \cos (m+n)x + \cos (m-n)x dx
    By the product/sum formulae that no one remembers.
    = ½[\frac{1}{m+n} \sin(m+n)x + \frac{1}{m-n} \sin(m-n)x]_0 ^{2\pi}

    But if m=±n, one of the fractions explodes. So in that case the question is:

    \int \cos^2 {mx} dx

    =½ \int 1+ cos{2mx}dx

    = ½[x+\frac{1}{2m} \sin{2mx}]_0 ^{2\pi}

    =\pi

    If m=n=0, the integral turns to 2\pi.

    As for the second bit.
    Let x = sinh²t

    dx = 2sinhtcosht dt

    I = \int \sqrt{\frac{x+1}{x}}dx

    = \int \sqrt{\frac{\sinh^2 t +1}{\sinh^2 t}}(2\sinh t\cosh t)dt

    = \int 2 \frac{\cosh t}{\sinh t} (\sinh t \ cosh t) dt

    = \int 2 \cosh^2 t dt

    = \int 1+ \cosh 2t dt = t + ½ \sinh 2t Ignoring the +c for now

    = t + \sinh t \cosh t = \sinh ^{-1}(\sqrt{x}) + \sqrt(x(x+x)) +c
    Which you can rewrite using the log form of arsinh.
    Alternatively for the second integral:
    Let x=tan^2 {z}
    dx/dz = 2tan {z} sec^2 {z}

    =\int \sqrt{sec^2{z}/tan^2{z}}.2sec^2{z}tan{z}dz
    =\int 2sec^3{z}dz

    Let:
    I= \int sec^3{z}dz
    I= \int sec{z}(1+tan^2{z})dz
    I= \int sec{z}dz + \int sec{z}tan^2{z}dz

    Now consider the differential of
    sec{z}tan{z}
    By the product rule
    =sec{z}tan^2{z} + sec^3{z}

    Thus
    \int sec{z}tan^2{z}dz +\int sec^3{z}dz = sec{z}tan{z}

    So we now have two simultaneous equation where we can cancel the ugly integral:
    (1) \int sec^3{z}dz = \int sec{z}dz + \int sec{z}tan^2{z}dz
    (2) \int sec^3{z}dz = sec{z}tan{z} - \int sec{z}tan^2{z}dz

    Add (1) and (2)
    2\int sec^3{z}dz = \int sec{z}dz + sec{z}tan{z}
    2\int sec^3{z}dz = \ln(sec{z}+tan{z}) + sec{z}tan{z}

    x=tan^2{z}
    x+1=sec^2{z}
    \int \sqrt{1+\frac{1}{x}}dx = \ln(\sqrt{x} + \sqrt{x+1}) + \sqrt{x(x+1)} + C

    The method with the hyperbolic functions is much nicer and removes a horrific sec cubed term but i thought i would have a go regardless using good old trg functions and i believe this is a good method to do it if you have no hyperbolic background.
    I wish my STEP exam this year will have a question like this
  9. ben-smith's Avatar
    229 28-05-2011  17:22
    • Overlord in Training
    • Location: Hilbert Space
    • Posts: 2,366
    Re: STEP Maths I, II, III 1993 Solutions
    (Original post by SimonM)
    (Updated as far as #213) SimonM - 11.05.2009
    ...
    STEP II Q14
    Consider the general collision between a particle of mass m and a fixed surface of coefficient of restitution a:
    speed before = v \Rightarrow speed after= av
    From which it follows that the kinetic energy after such collisions is:
    \frac{1}{2}m(av)^2=a^2\frac{1}{2 }mv^2=a^2(original K.E.)

    Applying this to the problem:
    let E_n be the kinetic energy of the particle just before the nth impact with the ceiling.
    By conservation of energy:
    E_1=\frac{1}{2}m(\sqrt{2kgh})^2-mgh=mgh(k-1) \therefore E_2=a^4mgh(k-1)+a^2mgh-mgh \therefore E_3=a^8mgh(k-1)+a^6mgh-a^4mgh+a^2mgh-mgh
    and, more generally:
    E_{n+1}=a^4E_n+a^2mgh-mgh
    Conjecture(\gamma):
    E_n=mgh(a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^2+1})
    letting n=1:
    E_1=mgh(a^{4-4}(k-1)+\frac{a^{4-4}-1}{a^2+1})=mgh(k-1) and \gamma therefore holds for n=1.
    Similarly, let n=2:
    E_2=mgh(a^{4}(k-1)+\frac{a^{4}-1}{a^2+1})=mgh(a^{4}(k-1)+a^2-1)=mgha^4(k-1)+a^2mgh-mgh and so\gamma holds for n=2 as well.
    Now let's assume the result:
    E_x=mgh(a^{4x-4}(k-1)+\frac{a^{4x-4}-1}{a^2+1}).
    Using this result, let n=x+1:
    E_{x+1}=a^4E_x+a^2mgh-mgh=a^4mgh(a^{4x-4}(k-1)+\frac{a^{4x-4}-1}{a^2+1})+a^2mgh-mgh =mgh(a^{4x}(k-1)+\frac{a^{4x}-a^4+(a^2-1)(a^2+1)}{a^2+1})=mgh(a^{4x}(k-1)+\frac{a^{4x}-1}{a^2+1})
    Therefore, by mathematical induction, \gamma (the required result) holds \forall n\in \mathbb{Z^+}

    Next part:
    the maximum number of times the ball can hit the ceiling is the n that satisfies:
    E_n=mgh(a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^2+1})=0
    so:
    a^{4n-4}(k-1)=-\frac{a^{4n-4}-1}{a^2+1} \Rightarrow a^{4(n-1)}(k-1)(a^2+1)=1-a^{4n-4} \Rightarrow a^{4(n-1)}=\frac{1}{(k-1)(a^2+1)+1}
    So, taking logs of both sides:
    4(n-1)ln(a)=-ln((k-1)(a^2+1)+1) \Rightarrow n-1=-\frac{ln(a^2(k-1)+k)}{4lna} \Rightarrow n=1-\frac{ln(a^2(k-1)+k)}{4lna}
    As required.
    Last edited by ben-smith; 28-05-2011 at 17:46.
  10. brianeverit's Avatar
    230 10-06-2011  16:38
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1993 Solutions
    STEP number 9
    \text{For first vat, chemical }A\text{ flows in at a rate of }a \text{ litres/sec}
    \text{If }V\text{ is the volume of}A\text{ in vat at time }T\text{ then proportion of }A\text{ is }\dfrac{V}{K} \text{ so it flows out at }-b\dfrac{V}{K} \text{ litres/sec}
    \text{Hence, }\dfrac{dV}{dt}=a-\dfrac{bV}{K}=\dfrac{aK-bV}{K}
    \text{Solving this equation we have }\int\dfrac{dV}{aK-bV}=\int\dfrac{dt}{K}\Rightarrow-\dfrac{1}{b}\ln(aK-bV)=\dfrac{1}{K}-\dfrac{1}{b}\lnaK \text{ since }V=0 \text{ at }t=0
    \text{i.e. }\ln\left( \dfrac{aK-bV}{aK}\right)=-\dfrac{bt}{K}\RightarrowaK-bV=aK\text{exp}\left(-\dfrac{bt}{K}\right) \text{ or }V=\dfrac{aK}{b}\left(1-\text{e}^{-\frac{bt}{K}}\right)\text{ as required}
    \text{Considering now the second vat, }A\text{ flows in at a rate of }\dfrac{bV}{K} \text{ litres/sec and out at a rate of }\dfrac{cV}{L}
    \text{where }V\text{ is now the volume of }A\text{ in the second vat, so }\dfrac{dV}{dt}=\dfrac{bV}{K}-\dfrac{cV}{L}=\dfrac{(bL-cK)}{KL}
    \text{so }\int\left(\dfrac{dV}{(bL-cK)V}}\right)=\int\dfrac{dt}{KL} \Rightarrow -\dfrac{1}{c}\ln{(bL-cK)V}=\dfrac{t}{KL}-\dfrac{1}{c}\ln{bL} \text{ since }V=0\text{ at }t=0
    \text{i.e. }\ln\left(\dfrac{(bL-cK)V}{bL}\right)=-\dfrac{ct}{KL}\Rightarrow (bL-cK)V=bL\text{e}^{-\frac{ct}{KL}}\text{ or }V=\dfrac{bt}{bL-cK}\left(1-\text{e}^{-\frac{ct}{KL}}\right)
  11. brianeverit's Avatar
    231 12-06-2011  10:59
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1993 Solutions
    Can anyone help with STEP III Number 11. I cannot get the given answer for the direction of the initial axis of rotation.
  12. SimonM's Avatar
    232 12-06-2011  11:16
    • PS Helper
    • TSR Idol
    • Posts: 9,190
    Re: STEP Maths I, II, III 1993 Solutions
    STEP III Question 11

    Spoiler:
    Show
    It is an easy calculation to see that the COM is the midpoint of BC, 1/4 of the length "away from it".

    So the angle of dangle which it makes hanging from A is: \arctan 2/3

    From B it is \arctan 1/2

    So the change in angle for B is \frac{\pi}{2} - \arctan 2/3 + \arctan 1/2 = \arctan3/2 + \arctan 1/2 = \arctan \frac{3/2+1/2}{1-3/4} = \arctan 8


    I can't get the given answer either
    Last edited by SimonM; 12-06-2011 at 11:23.
  13. DFranklin's Avatar
    233 12-06-2011  11:38
    • TSR Royalty
    • Location: London
    • Posts: 18,041
    Re: STEP Maths I, II, III 1993 Solutions
    (Original post by brianeverit)
    ..
    STEP 1993 Q11: Note - done without a good diagram - might be a minor direction mistake somewhere.


    Choose a coord frame s.t. C = (0,0), D = (1,0), A = (0, 1), B = (1, 1).
    Then AB has CM (0.5, 1), BC has CM (1, 0.5) and CD has CM (0.5, 0).
    It follows that the CM of the entire wire is [ (0.5, 1) + (1, 0.5) + (0.5, 0)] / 3 = (2/3, 1/2).

    If we hang from A, then BC lies below the horizontal and makes an angle arctan((1/2)/(2/3)) = arctan(3/4) with the horizontal.
    If we hang from B, then BC lies below the horizontal and makes an angle arctan((1/2)/(1/3)) = arctan(3/2) with the horizontal (going the other way).

    The tan of the angle between these is

    tan(arctan(3/4)+arctan(3/2)) = \frac{3/4+3/2}{1-(3/4)(3/2)} = \frac{9/4}{-1/8} = -18.

    Since this is negative, it's the oblique angle between the lines; the acute angle will be arctan(18) as required.
    Last edited by DFranklin; 12-06-2011 at 11:41.
  14. SimonM's Avatar
    234 12-06-2011  11:42
    • PS Helper
    • TSR Idol
    • Posts: 9,190
    Re: STEP Maths I, II, III 1993 Solutions
    Ah, my diagram was crap.
  15. ben-smith's Avatar
    235 12-06-2011  12:21
    • Overlord in Training
    • Location: Hilbert Space
    • Posts: 2,366
    Re: STEP Maths I, II, III 1993 Solutions
    guys, you were confusing me. I think you have mixed up STEP III with STEP I. The papers have been mislabelled.
    Post rating:
    1
  16. DFranklin's Avatar
    236 12-06-2011  12:35
    • TSR Royalty
    • Location: London
    • Posts: 18,041
    Re: STEP Maths I, II, III 1993 Solutions
    (Original post by ben-smith)
    guys, you were confusing me. I think you have mixed up STEP III with STEP I. The papers have been mislabelled.
    Thanks. I knew about the mislabelling, but thought (incorrectly) it was 1992 and earlier.

    (Original post by Simon M)
    ..
    Care to try again? Not sure I'll have time.
  17. DFranklin's Avatar
    237 12-06-2011  13:04
    • TSR Royalty
    • Location: London
    • Posts: 18,041
    Re: STEP Maths I, II, III 1993 Solutions
    (Original post by brianeverit)
    Can anyone help with STEP III Number 11. I cannot get the given answer for the direction of the initial axis of rotation.
    OK, for the actual correct question you asked about...

    I think you can just take moments about each axis. It's easy to make mistakes, but after 3 attempts, I got their answer using this method.

    I think you can also define OA = i, OB = j, OC = k and then simply find \sum {\bf r} \wedge {\bf F} where r is the position of a point on the line of force and F is the force itself. (Note that this tallies with the intuitive result that forces through O are irrelevant).
  18. ben-smith's Avatar
    238 12-06-2011  13:15
    • Overlord in Training
    • Location: Hilbert Space
    • Posts: 2,366
    Re: STEP Maths I, II, III 1993 Solutions
    (Original post by DFranklin)
    OK, for the actual correct question you asked about...

    I think you can just take moments about each axis. It's easy to make mistakes, but after 3 attempts, I got their answer using this method.

    I think you can also define OA = i, OB = j, OC = k and then simply find \sum {\bf r} \wedge {\bf F} where r is the position of a point on the line of force and F is the force itself. (Note that this tallies with the intuitive result that forces through O are irrelevant).
    It's reassuring that you had to try 3 times because I keep getting an annoyingly similar answer. I keep getting (1,2,1) as opposed to (1,1,2). Did you get that? If so, where did you go wrong?
    EDIT: can I just swap the 2 and the 1 around by just redefining the axes or is that cheating?
    Last edited by ben-smith; 12-06-2011 at 13:16.
  19. DFranklin's Avatar
    239 12-06-2011  13:26
    • TSR Royalty
    • Location: London
    • Posts: 18,041
    Re: STEP Maths I, II, III 1993 Solutions
    (Original post by ben-smith)
    It's reassuring that you had to try 3 times because I keep getting an annoyingly similar answer. I keep getting (1,2,1) as opposed to (1,1,2). Did you get that? If so, where did you go wrong?
    EDIT: can I just swap the 2 and the 1 around by just redefining the axes or is that cheating?
    You can't swap the axes because you'd also swap the forces, and nothing would change. What I kept getting was (1, -1, 2) - sign errors.

    As I calculate it, taking moments anticlockwise about OB you have: -1 (from the force O'A) and that's it.

    Taking moments anticlockwise about OC you have: -1 (from AC'), -2 (from C'B) and 1 (from O'A) for a total of -2.

    Taking moments anticlockwise about OA you have -1 (from A'C).

    I think.

    So I get (-1, -1, -2) which is obviously the same axis as (1,1,2).
  20. brianeverit's Avatar
    240 13-06-2011  18:31
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1993 Solutions
    1993 STEP III number 13

    \text{When }P \text{ has descended a distance }s \text{, let speed of system be }v \text{ then}
    \text{k.e. of disc }=\dfrac{1}{2} \times \dfrac{4ma^2}{2} \times \dfrac{v^2}{a^2}=mv^2
    \text{k.e. of chain }=\dfrac{1}{2}(4+\pi)mv^2
    \text{k.e. of particle }= \dfrac{1}{2}mv^2 \text{ so gain of k.e. Is }\dfrac{1}{2}(\pi +7)mv^2
    \text{p.e. of system relative to initial position }=\dfrac{(2a-x)xmg}{2a}-\dfrac{(2a+x)xmg}{2a}-xmg=-\dfrac{2(x^2+ax)mg}{2a}
    \text{so by conservation of energy, }\dfrac{)x^2+ax)mg}{a}=\dfrac{1} {2}(\pi+7)mv^2
    \Rightarrow (\pi)+7av^2=2g(ax+x^2) \text{ as required}
    \text{Now consider the portion of the chain below Q as a variable mass }M \text{including mass P}
    \text{when Q has fallen a distance }2a \text{, total weight of chainand mass P is }(3m+2m)g=5mg
    \text{By Newton's second law }Mg-T=\dfrac{d}{dt}(Mv)= \dotM v+M\dotv \text{ where }T \text{ is the tension at Q}
    \dot{M}= \dfrac{v}{a}m \text{ and }v^2= \dfrac{2g(ax+x^2)}{(\pi+7)a} \Rightarrow 2v \dot{v}= \dfrac{2g(a+2x)v}{(\pi+7)a}
    \Rightarrow \dot{v}= \dfrac{g(a+2x)}{(\pi+7)}a= \dfrac{5g}{\pi+7} \text{ when }x=2a

    \text{also }v^2=\dfrac{12g}{\pi+7} \text{ when }x=2a
    \text{hence, }T=5mg-\dfrac{v^2m}{a}-5m \times \dfrac{5g}{\pi+7}=\left(\dfrac{5 \pi+35}{\pi+7}- \dfrac{25}{\pi+7}-\dfrac{12}{\pi+7}\right) mg= \dfrac{5\pi-2}{\pi+7} \text{ as required}
Sign in to Reply
Search Thread
Advanced Search
Share this discussion:  
Useful resources

Articles:

Study Help rules and posting guidelinesStudy Help Member of the Month nominationsGuide to MathematicsMathematics resourcesMathematics websitesMathematics revision notes in the TSR wikiCambridge Assessment admissions tests (including STEP)How to use LaTeXCareers in Mathematics

Useful Dates:

Quick Link:

Unanswered Maths Threads

Groups associated with this forum:

View associated groups
Article updates
Moderators

We have a brilliant team of more than 60 volunteers looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is moderated by:

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22

Registered Office: International House, Queens Road, Brighton, BN1 3XE

Shortcuts

Get Started

TSR Group

Using TSR

Info

Connect with TSR

Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.