1987 Specimen STEP solutions thread Watch

Dirac Spinor
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(Original post by brianeverit)
There is a solutions thread for 1987 with ALL questions answered.
this is the specimen paper rather than the actual one.
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brianeverit
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(Original post by ben-smith)
this is the specimen paper rather than the actual one.
Apologies, I didn't read carefully enough.
Incidentall;y, several of these questions are included in Siklos's A.P.I.M. as follows
Paper I /6 APIM No.5
Paper II/13 APIM No.9, II/2 APIM No. 16 II/16 APIM No. 23
Paper III/13 APIM No. 24 III/14 APIM No. 26 III/11 APIM No. 31
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brianeverit
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Specimen.I.16
\int_0^{\infty}f(t)dt=1\mathrm{\ so\ }\int_0^1 \frac{1}{2}dt+\int_1^{\infty}ke^  {-2t}dt=1
i.e. [\frac{t}{2}]_0^1+k[-\frac{1}{2}e^{-2t}]_1^{\infty}=1\Rightarrow \frac{1}{2}+\frac{k}{2}e^{-2}=1 \Rightarrow k=e^2
E[T]=\int_0^1 \frac{t}{2}dt+e^2\int_1^{\infty}  te^{-2t}dt =[\frac{t^2}{4}]_0^1+e^2[-\frac{t}{2}e^{-2t}]_1^{\infty}+e^2\int_1^{\infty}\f  rac{1}{2}e^{-2t}dt=\frac{1}{4}+\freac{1}{2}+e  ^2[-\frac{1}{4}e^{-2t}=\frac{1}{4}+ \frac{1}{2}+ \frac{1}{4}=1
F(t)= \int_0^t f(x)dx= \int_0^t \frac{1}{2}dx= \frac{t}{2} \mathrm{\ for\ }0\leq t<1
And  F(t)=\frac{1}{2}+\int_1^te^{2-2x}dx=1-\frac{1}{2}e^{2-2t} \mathrm{\ for\ }t>1
If  P(n) is the probability of spending n 10 pence pieces then
 P(1)=F(0.5)=\frac{1}{4},\ P(2)=F(1)-F(0.5)=\frac{1}{4}
 P(3)=F(1.5)-F(1)=1-\frac{1}{2}e^{-1}-\frac{1}{2}=\frac{1}{2}(1-\frac{1}{e})
And in general  P(n)=F(n/2)-F((n-1)/2)=\frac{1}{2}[(1-\frac{1}{e^n/2}-\frac{1}{e^{(n-1)/2}}]
E(C)=10 \sum_{n=0}^\infty nP(n)=10\left[\frac{1}{4}+2\times\frac{1}{4}+ 3 \times \frac{1}{2}(1- \frac{1}{e})+4 \times \frac{1}{2}(\frac{1}{e}- \frac{1}{e^2})+5 \times \frac{1}{2}(\frac{1}{e^2}- \frac{1}{e^3})+...\right]
=10\left[\frac{1}{4}+\frac{1}{2}+\frac{3}  {2}+\frac{1}{2}\left(e^{-1}+e^{-2}+e^{-3}+...\right)\right]=10\left[\frac{9}{4}+\frac{1}{2}\left( \frac{e^{-1}}{1-e^{-1}}\right)\right]
=22\frac{1}{2}+\frac{5}{2}\times  \frac{1}{e-1}=22\frac{1}{2}+\frac{5}{e-1} as required
This result assumes that all calls are multiples of 10 minutes and of course, most will not be exactly. Hence it will differ from average length of call times 20p
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brianeverit
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(Original post by ben-smith)
STEP I Q2


PB=\sqrt{x^2+h^2},PC=\sqrt{x^2+(  d+h)^2}
Cosine rule:
cos\theta=\dfrac{d^2-x^2-h^2-x^2-(h+d)^2}{-2\sqrt{(x^2+h^2)(x^2+(d+h)^2)}}

\Rightarrow Tan\theta=\frac{\sqrt{1-cos^2\theta}}{cos\theta}

=\dfrac{\sqrt{(x^2+h^2)(x^2+(d+h  )^2)-(x^2+hd+x^2)^2}}{x^2+hd+h^2}}

=\dfrac{\sqrt{x^4+x^2h^2+2x^2hd+  x^2d^2+h^2x^2+h^4+2h^3d+h^2d^2-x^4-2x^2hd-2x^2h^2-h^2d^2-2h^3d-h^4}}{x^2+hd+h^2}

=\dfrac{xd}{x^2+hd+h^2}
Note that max[\theta] \Rightarrow max[Tan\theta]
so we require an x such that \frac{d}{dx}[Tan\theta]=0. (obviously will be a max because min occurs a x=infinity).
\frac{d}{dx}[Tan\theta]=\frac{d(h^2+hd-x^2)}{(x^2+hd+h^2)^2}

\Rightarrow x^2=hd+h^2 \Rightarrow x=\sqrt{h(h+d)}
tan theta is found more easily by noting that theta=angle APC minus angle APB and using tangent addition theorem.
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brianeverit
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(Original post by abra-cad-abra)
how did you know that that first angle was pi/7. i just guessed it as its a 7sided shape so makes sense and looking at the answer i was working towards it seemed right
Produce EQ and use exterior angle of triangle theorem.
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brianeverit
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(Original post by ben-smith)
STEP I Q7
 z=1+e^{2i\alpha}=1+cos2\alpha+is  in2\alpha

\Rightarrow |z|=\sqrt{(1+cos2\alpha)^2+(sin2  \alpha)^2}=\sqrt{2+2cos\alpha}=2  cos\alpha

arg(z)=arctan(\dfrac{sin2\alpha}  {1+cos2 \alpha})=arctan(Tan(2\alpha/2))= \alpha

\displaystyle \Sigma^{n}_{r=0} \displaystyle \binom{n}{r} sin(2r+1)=\Im[\displaystyle \Sigma \displaystyle \binom{n}{r} e^{i(2r+1)\alpha}]=\Im[e^{i\alpha} \displaystyle \Sigma \displaystyle \binom{n}{r} e^{i2r\alpha}]

=\Im[e^{i\alpha}(1+e^i\alpha)^n]=\Im[e^{i\alpha}(2cos\alpha e^{i\alpha})^n]

=2^n cos^n\alpha\Im[e^{i(n+1)\alpha}]=2^n cos^n\alpha sin(n+1)\alpha
I don't understand your summation.
Isn't \displaystyle \Sigma^n_{r=0} \displaystyle \binom{n}{r}e^{2ir\alpha}=(1+e^{  2i\alpha})^n \mathrm{\ not\ }(1+e^{i\alpha})^n
though there seems to be a compensating error in the next line and I agree with the final solution.
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brianeverit
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(Original post by ben-smith)
STEP I Q7
 z=1+e^{2i\alpha}=1+cos2\alpha+is  in2\alpha

\Rightarrow |z|=\sqrt{(1+cos2\alpha)^2+(sin2  \alpha)^2}=\sqrt{2+2cos\alpha}=2  cos\alpha

arg(z)=arctan(\dfrac{sin2\alpha}  {1+cos2 \alpha})=arctan(Tan(2\alpha/2))= \alpha

\displaystyle \Sigma^{n}_{r=0} \displaystyle \binom{n}{r} sin(2r+1)=\Im[\displaystyle \Sigma \displaystyle \binom{n}{r} e^{i(2r+1)\alpha}]=\Im[e^{i\alpha} \displaystyle \Sigma \displaystyle \binom{n}{r} e^{i2r\alpha}]

=\Im[e^{i\alpha}(1+e^i\alpha)^n]=\Im[e^{i\alpha}(2cos\alpha e^{i\alpha})^n]

=2^n cos^n\alpha\Im[e^{i(n+1)\alpha}]=2^n cos^n\alpha sin(n+1)\alpha
I don't understand your summation.
Isn't \displaystyle \Sigma^n_{r=0} \displaystyle \binom{n}{r}e^{2ir\alpha}=(1+e^{  2i\alpha})^n \mathrm{\ not\ }(1+e^{i\alpha})^n
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brianeverit
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(Original post by bogstandardname)
Another solution for an uncompleted question, STEP I Question 8 is attached
You cannot have the variable in your integral as one of the limits.
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brianeverit
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(Original post by Farhan.Hanif93)
STEP I - Q11
Let the cross section of the cylinder have centre O and radius a. Let the contact force exerted by the ramp on the cylinder have magnitude R. Note that there is no reaction force exerted by the horizontal surface as the cylinder is on the point of losing contact with it. Friction must be limiting on the vertical surface. On the point of rolling:

\Re (\mathrm{up} \mathrm{the} \mathrm{ramp}): \dfrac{P}{\sqrt 2} - \dfrac{\mu P}{\sqrt 2} - \dfrac{Mg}{\sqrt 2} - \nu R =0 (1)

\Re (\mathrm{perpendicular} \mathrm{to} \mathrm{ramp}): R - \dfrac{Mg}{\sqrt 2} - \dfrac{P}{\sqrt 2} - \dfrac{\mu P}{\sqrt 2} = 0 (2)

Taking moments about O: a\nu R = a\mu P (3)

Subbing (3) into (1): P(1-\mu -\sqrt 2 \mu) - Mg=0 \implies \boxed{P=\dfrac{Mg}{1-\mu(1+\sqrt 2)}}, as required.

Note that P>0 \implies 1-\mu (1+\sqrt 2) > 0 \implies \boxed{\mu < \dfrac{1}{\sqrt 2 + 1} = \sqrt 2 -1}, as required.

Furthermore, in order for the cylinder to be rolling up the slope, we require the clockwise moment to be greater than or equal to the anti-clockwise moment.

By (2): R = \dfrac{Mg(2- \sqrt 2 \mu)}{\sqrt 2(1-\mu (1+\sqrt 2))}

We require \nu R \geq \mu P

\implies \dfrac{\nu (2-\sqrt 2\mu)}{\sqrt 2(1-\mu (1+\sqrt 2))} \geq \dfrac{\mu}{1-\mu (1+\sqrt 2)}

\implies  \nu (\sqrt 2 - \mu)  \geq  \mu

By the earlier result, \sqrt 2 - \mu > 1 so it's safe to divide through by it to obtain the required result \boxed{\nu \geq \dfrac{\mu}{\sqrt 2 - \mu}}.
You cannot assume that the friction force on the incline is  \nu Rsince friction is not limiting there.
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brianeverit
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(Original post by bogstandardname)
Another solution for an uncompleted question, STEP I Question 8 is attached
Why bother using an integrating factor when the differential equation is variable separable?
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Yung_ramanujan
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(Original post by brianeverit)
Specimen.I.16
\int_0^{\infty}f(t)dt=1\mathrm{\ so\ }\int_0^1 \frac{1}{2}dt+\int_1^{\infty}ke^  {-2t}dt=1
i.e. [\frac{t}{2}]_0^1+k[-\frac{1}{2}e^{-2t}]_1^{\infty}=1\Rightarrow \frac{1}{2}+\frac{k}{2}e^{-2}=1 \Rightarrow k=e^2
E[T]=\int_0^1 \frac{t}{2}dt+e^2\int_1^{\infty}  te^{-2t}dt =[\frac{t^2}{4}]_0^1+e^2[-\frac{t}{2}e^{-2t}]_1^{\infty}+e^2\int_1^{\infty}\f  rac{1}{2}e^{-2t}dt=\frac{1}{4}+\freac{1}{2}+e  ^2[-\frac{1}{4}e^{-2t}=\frac{1}{4}+ \frac{1}{2}+ \frac{1}{4}=1
F(t)= \int_0^t f(x)dx= \int_0^t \frac{1}{2}dx= \frac{t}{2} \mathrm{\ for\ }0\leq t<1
And  F(t)=\frac{1}{2}+\int_1^te^{2-2x}dx=1-\frac{1}{2}e^{2-2t} \mathrm{\ for\ }t>1
If  P(n) is the probability of spending n 10 pence pieces then
 P(1)=F(0.5)=\frac{1}{4},\ P(2)=F(1)-F(0.5)=\frac{1}{4}
 P(3)=F(1.5)-F(1)=1-\frac{1}{2}e^{-1}-\frac{1}{2}=\frac{1}{2}(1-\frac{1}{e})
And in general  P(n)=F(n/2)-F((n-1)/2)=\frac{1}{2}[(1-\frac{1}{e^n/2}-\frac{1}{e^{(n-1)/2}}]
E(C)=10 \sum_{n=0}^\infty nP(n)=10\left[\frac{1}{4}+2\times\frac{1}{4}+ 3 \times \frac{1}{2}(1- \frac{1}{e})+4 \times \frac{1}{2}(\frac{1}{e}- \frac{1}{e^2})+5 \times \frac{1}{2}(\frac{1}{e^2}- \frac{1}{e^3})+...\right]
=10\left[\frac{1}{4}+\frac{1}{2}+\frac{3}  {2}+\frac{1}{2}\left(e^{-1}+e^{-2}+e^{-3}+...\right)\right]=10\left[\frac{9}{4}+\frac{1}{2}\left( \frac{e^{-1}}{1-e^{-1}}\right)\right]
=22\frac{1}{2}+\frac{5}{2}\times  \frac{1}{e-1}=22\frac{1}{2}+\frac{5}{e-1} as required
This result assumes that all calls are multiples of 10 minutes and of course, most will not be exactly. Hence it will differ from average length of call times 20p
Why is P(1) =1/4 . You have to pay 10p at beginning of the call. shouldnt it p(2)=1/4 and p(1)=1 ?
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Yung_ramanujan
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For Q15 I got E(C)=0 by symmetry and Var (C)= 8.5

Then for the context defining new random variable K as 36 independent random variables from C. So E(K)=0 and Var(K)=306.

Then would i use the normal distribution definition of outlier to determine if she should believe it to be a mistake?

(Original post by brianeverit)
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brianeverit
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(Original post by Yung_ramanujan)
Why is P(1) =1/4 . You have to pay 10p at beginning of the call. shouldnt it p(2)=1/4 and p(1)=1 ?
P(1) is the probability that the call lasts no more than half a minute.
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brianeverit
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Specimen Paper I. Questions 10 and 15.
These seem to be the only two without solutions so here are my attempts.
Attached files
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brianeverit
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Paper II
Question 1
 \frac{3x^2+2(a+b)x+ab}{x^3+(a+b)  x^2+abx}=\frac{3x^2+2(a+b)x+ab}{  x(x+a)(x+b)}=\frac{A}{x}+\frac{B  }{x+a}+\frac{C}{x+b}
[\impliesA(x+a)(x+b)+Bx(x+b)+Cx(x  +a)=3x^2+2(a+b)x+ab
x=0 \implies abA=ab \implies A=1
x=-a \implies -aB(b-a)=3a^2-2a^2-2ab+ab \implies B=\frac{a^2-ab}{a^2-ab}=1
Comparing coefficients of x^2,\ A+B+C=3\implies C=1
So \frac{3x^2+2(a+b)x+ab}{x^3+(a+b)  x^2+abx}=\frac{1}{x}+\frac{1}{x+  a}+\frac{1}{x+b} There are two special cases to consider, a=b \mathrm{\ and\ }a+b=0
a=b \implies \mathrm{f}(x)=\frac{3x^2+4ax+a^2  }{x^3+2ax^2+a^2x}=\frac{(3x+a)(x  +a)}{x(x+a)^2}=\frac{1}{x}+\frac  {2}{x+a} (Cover up rule)
a+b=0 \implies b=-a \implies \mathrm{f}(x)=\frac{3x^2-a^2}{x^3-a^2x}=\frac{3x^2-a^2}{x(x+a)(x-a)} =\frac{1}{x}+\frac{1}{x+a}+\frac  {1}{x-a}
If f(x)=1 \mathrm{\ with\ }x,a,b, \mathrm{\ all\ positive\ integers\ then}
Case 1. a\neqb, without loss of generality we may assume a<b \mathrm{\ so\ }x+b>x+a>x
Hence, \frac{1}{x}>\frac{1}{x+a}>\frac{  1}{x+b} \implies \frac{1}{x}>\frac{1}{3} \implies x<3 \mathrm{\ i.e.\ }x=2 since we cannot have x=1
So  \frac{1}{2+a}+\frac{1}{2+b}= \frac{1}{2} with  a,b integers, so a=1,b=4
Case 2. a=b \implies \frac{1}{x}+\frac{2}{x+a}=1
Again we must have x=2 \mathrm{\ so\ } \frac{2}{2+a}=\frac{1}{2} \implies a=2
So the only possible positive integer solutions are x=2,a=1,b=4 \mathrm{\ and\ } x=2,a=b=2
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brianeverit
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Question 2
 \frac{y-z}{b-c}=\frac{y-z}{(y^2-xz)-(z^2-xy)}=\frac{y-z}{(y^2-z^2)+x(y-z)}=\frac{1}{x+y+z}
Similarly, by symmetry, \frac{z-x}{c-a}=\frac{x-y}{a-b}=\frtac{1}{x+y+z}
(y-z)^2+(z-x)^2+(x-y)^2=\frac{(b-c)^2+(c-a)^2+(a-b)^2}{(x+y+z)^2}
\implies (x+y+z)^2=\frac{2(a^2+b^2+c^2)-2(ab+bc+ca)}[2(x^2+y^2+z^2)-2(xy+yz+zx)}=\frac{a^2+b^2+c^2-bc-ca-ab}{(x^2-yz)+(y^2-zx)+(z^2-xy)}=\frac{a^2+b^2+c^2-bc-ca-ab}{a+b+c}=\Delta^2
Hence, x+y+z=\Delta as required.
y-z=\frac{b-c}{\Delta}\implies y=z+\frac{b-c}{\delta} and similarly x=z+\frac{a-c}{\delta}
Substituting in x+y+z=\delta gives 3z+\frac{a-c}{\delta}+\frac{b-c}{\delta}=\delta \implies z=\frac{\Delta^2-(a-c)-(b-c)}{3\Delta}=\frac{\frac{a^2+b^2  +c^2-bc-ca-ab}{a+b+c}-a-b+2c}{3\Delta}
=\frac{(a^2+b^2+c^2-bc-ca-ab)-(a+b+c)(a+b-2c)}{3\Delta(a+b+c)}=\frac{3c^2-3ab}{3\Delta(a+b+c)}=\frac{c^2-ab}{\Delta(a+b+c)}
And, by symmetry, x=\frac{a^2-bc}{\Delta(a+b+c)} \mathrm{\ and\ }y=\frac{b^2-ca}{\Delta(a+b+c)}
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brianeverit
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Question 3
To prove (\cos\theta+i\sin\theta)^n=\cosn  \theta+i\sin n\theta
Statement is obviously true for n=1 so assume true for n=k
So (\cos\theta+i\sin\theta)^k=\cos k\theta+i \sin k\theta hence
 (\cos \theta+i\sin \theta)^{k+1}=(\cos k\theta+i\sin k\theta)(\cos\theta+i \sin\theta)
=(\cos k\theta \cos \theta-\sin k\theta \sin\theta)+i(\cos k\theta \sin\theta+\sin k\thewta \cos \theta)
=\cos(k+1)\theta+i\sin(k+1)\thet  a so true for n=k\implies \mathrm{\ true\ for\ }n=k+1 and hence true for all integer values of n by induction.
Let z=\cos\theta+i \sin\theta then x^3\cos3y+2x^2 \cos2y+2x\cosy=\mathrm{\ Re}((xz)^3+2(xz)^2+2xz) \mathrm{\ and\ }x^3\sin3y+2x^2\sin2y+2x\siny= \mathrm{\ Im}((xz)^3+2(xz)^2+2xz)
\mathrm{Hence\ }((xz)^3+2(xz)^2+2xz=-1\implies ((xz)^3+2(xz)^2+2xz+1=0 \implies (xz+1)(x^2z^2-xz+1)=0 \implies xz=1 or \frac{1}{2}(1\pm\sqrt3)
Clearly we must have |x|=1 \implies x=\pm1,z=-1 \mathrm{\ or\ }\frac{1}{2}\pm\frac{\sqrt3}{2}
x=1,z=-1\implies y=(2k+1)\pi
x=1,z=\frac{1}{2}\pm\frac{\sqrt3  }{2} \implies y=(2k\pi\pm\frac{1}{3})\pi
x=-1,z=1\implies y=(2k+1)\pi
x=-1,z=-(\frac{1}{2}\pm\frac{\sqrt3}{2} \implies y=(2k\pm\frac{2}{3})\pi
\mathrm{So\ solutions\ are\ } x=1,y=(2k+1)\pi, x=1,y=(2k\pm\frac{1}{3})\pi,
 x=-1,y=2k\pi,\ x=-1,y=(2k\pm\frac{2}{3})\pi.
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brianeverit
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Paper II
Question 10
10.(i) If A is a toad then he is telling the truth and he is a frog - contradiction, so he is lying and hence must be a frog, and since he said B was a frog then this is also a lie and so B is a toad.
(ii) If C is a frog then his statement is a lie and would therefore be a contradiction. So C must be a toad and since he is telling the truth, D must be a frog.
(iii) If E is a toad then both G and H are toads but they cannot both be telling the truth. So E is a frog and G and H are not both toads which means that G is telling a lie and hence is a frog whilst H is telling the truth and hence is a toad.
(iv) I, J and K cannot all be frogs as otherwise I would be telling the truth. So I is a frog
Either J or K must be a toad. If K is a toad and J a frog or J a toad and K a frog then J is telling the truth, hence J must be the toad and K a frog
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brianeverit
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Question 11
Let A,B denote the particles projected from A and B re3spectively. Since they meet when both are at their greatest height then their maximum heights must be equal.
Max height for A is \frac{V^2\sin^245^o}{2g} \mathrm{\ and\ for\ B\ }\frac{U^2\sin^230^o}{2g}
So 

]\frac{V^2\sin^245^o}{2g} =\frac{U^2\sin^230^o}{2g}\implie  s\frac{V^2}{4g}=\frac{U^2}{8g} \implies U^2=2V^2 as required.
Time to reach maximum height is given by \frac{v\sin\theta}{g} so A has traveled a distance
\frac{V^2\sin45^o \cos45^o}{g}=\frac{V^2}{2g} \mathrm{\ and\ B\ has\ traveled\ }\frac{U^2\sin30^o\cos30^o}{g}= \frac{U^2\sqrt3}{4g}= \frac{V^2\sqrt3}{2g} hence
a=\frac{V^2}{2g}+\frac{V^2\sqrt}  {2g}=\frac{V^2}{2g}(1+\sqrt3) as required.
If particles coalesce on impact then, by conservation of momentum, velocity of combined particle after impact in directionAB, V' \mathrm{\ is\ given\ by\ }2mV'=mV\cos45^o-mU\cos30^o\implies V'=\frac{V}{2\sqrt2}(1-\sqrt3).
So particle hits ground with vertical velocity \frac{V}{\sqrt2} and horizontal velocity   \frac{V}{2\sqrt2}(\sqrt3-1) so angle to the horizontal is
\phi=\tan^{-1}\Left( \frac{V}{\sqrt2} \times \frac{2\sqrt2}{V(\sqrt3-L1)}\Right)=\tan^{-1}\frac{2}{\sqrt3-1}=\tan^{-1}(\sqrt3+1) as required.
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brianeverit
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Question 14
MI of disc}= \frac{Ma^2}{2} and force acting on it is  -Mak(\omega_1-\omega_2) at a distance of  \frac{a}{2} from the axis
So  \frac{Ma^2}{2} \frac{d\omega_1}{dt}=-Mak(\omega_1-\omega_2)\frac{a}{2}\implies \frac{d\omega_1}{dt}=-k(\omega_1-\omega_2)
Similarly, the particle has MI \frac{1}{2}  \left(\frac{a}{2}\right)^2 and is acted on by a force Mak(\omega_1-\omega_2)
So \frac{Ma^2}{8}\frac{d\omega_2}{d  t}=Mak(\omega_1-\omega_2)\implies\frac{d\omega_2  }{dt}=4k(\omega_1-\omega_2)
Hence \frac{d\omega_2}{dt}=-4\frac{d\omega_1}{dt}\implies \omega_2=4\omega_1+c
\omega_1=\Omega \mathrm{\ and\ }\omega_2=0 \mathrm{\ at\ }t=0\implies c=4\Omega
So  4\omega_1+\omega_2=4\Omega as required.
\frac{d\omega_1}{dt}=-k(\omega_1-\omega_2)=-k\omega_1+4k(\Omega-\omega_1)=4k\Omega-5\omega_1
Separating the variables and integrating we have \int\frac{d\omega_1}{5\omega_1-4\Omega}=-\int kdt
\implies \frac{1}{5}\ln(5\omega_1-4\Omega)=-kt+A
\omega_1=\Omega \mathrm{\ at\ }t=0 \implies A=\frac{1}{5}\ln\Omega \mathrm{\ so\ }\ln\frac{5\omega_1-4\Omega}{\Omega}=-5kt

 \implies 5\omega_1-4\Omega=\Omega e^{-5kt} \implies \omega_1=\frac{\Omega}{5}(4+e^{-5kt}
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University open days

  • Arts University Bournemouth
    Art and Design Foundation Diploma Further education
    Sat, 25 May '19
  • SOAS University of London
    Postgraduate Open Day Postgraduate
    Wed, 29 May '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Thu, 30 May '19

How did your AQA GCSE Physics Paper 1 go?

Loved the paper - Feeling positive (350)
30.54%
The paper was reasonable (460)
40.14%
Not feeling great about that exam... (197)
17.19%
It was TERRIBLE (139)
12.13%

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