STEP 2005 Solutions Thread Watch

nuodai
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#41
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#41
(Original post by GHOSH-5)
Not sure about the bit I've boxed.  \cos \theta = 1 \iff \theta = 2n\pi for some integer n.
Indeed, but the (2n-1)\pi terms (and the others) correspond to \sin 2t = 0.
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SimonM
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#42
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STEP II, Question 5

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Tangents from the same point are equal.

Therefore we can write the side lengths as a = x+y, b = y+z, c = z+a

The radius of the incircle will be z (since there is a right angle at A, so tangents from a will be //el to the radius).

Therefore 2r = 2z = c+b -a as required.

Since ABC is right angled at A, the radius of the circumcircle will be a/2.

Therefore:

\displaystyle R = \frac{\frac{1}{2}bc -\left(\frac{c+b-a}{2}\right)^2 \pi}{\left ( \frac{a}{2} \right)^2 \pi}

So

\displaystyle \pi R = 4\left ( \frac{1}{2} \frac{bc}{a^2} - \frac{1}{4}\left ( \frac{c+b}{a} - 1 \right)^2 \pi \right)

\displaystyle \pi R = 2\frac{bc}{a^2} - (q-1)^2 \pi

\left ( \frac{b+c}{a} \right)^2 = \frac{b^2+c^2}{a^2} + \frac{2bc}{a^2}

But since b^2+c^2 = a^2 (Pythagoras' theorem)

q^2 = 1 + \frac{2bc}{a^2}

So we have

\displaystyle \pi R = q^2-1 - (q-1)^2 \pi
\displaystyle \pi R = q^2 - 1 - q^2 \pi + 2q \pi - \pi
\displaystyle \pi R = q^2(1-\pi) + 2\pi q -(\pi+1)

Considering this as a quadratic in q

(2\pi)^2 \geq 4(\pi-1)(\pi + 1 +\pi R)

Therefore

\displaystyle \frac{\pi^2}{\pi-1} \geq \pi(R+1) + 1
\displaystyle \frac{\pi}{\pi-1} - \frac{1}{\pi} \geq R+1

So R \ge \frac{1}{\pi-1} - \frac{1}{\pi} = \frac{1}{\pi(\pi-1)}
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SimonM
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#43
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STEP III, Question 8

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|a-c|^2 = (a-c)(a-c)^* = aa^* + cc^* - ac^* -ca^*

By Pythagoras' theorem

|a|^2 + |a-c|^2 = |c|^2 \Leftrightarrow aa^* + aa^* + cc^* - ac^* -ca^* = cc^* \Leftrightarrow 2aa^* = ac^* + ca^*

Since b \mapsto \frac{1}{b^*} is self inverse, we need only show one direction.

|a-c|^2 = |ab-c|^2 \Leftrightarrow  aa^* + cc^* - ac^* -ca^* = aa^*bb^* + cc^* -abc^* -ca^*b^*

Since OA is tangent, substituting from before

cc^* - aa^* = aa^*bb^* + cc^* -abc^* -ca^*b^*

\Leftrightarrow

-aa^* = aa^*bb^* -abc^* -ca^*b^*

\Leftrightarrow

-\frac{aa^*}{bb^*} = aa^* -\frac{ac^*}{b^*} -\frac{ca^*}{b}

\Leftrightarrow

-aa^* = \frac{aa^*}{bb^*}-\frac{ac^*}{b^*} -\frac{ca^*}{b}

\Leftrightarrow

cc^*-aa^* = \frac{aa^*}{bb^*}+cc^*-\frac{ac^*}{b^*} -\frac{ca^*}{b}

\Leftrightarrow

|c-a|^2 = |\frac{a}{b^*} - c|^2 as required.

By hypothesis, |ab-c|^2 = |\frac{a}{b^*} - c|^2

Taking the difference and expanding

\displaystyle 0 = aa^* \left ( bb^* - \frac{1}{bb^*}\right ) - ca^* \left ( b^* - \frac{1}{b} \right) - c^*a \left ( b - \frac{1}{b^*} \right )

Dividing by bb^* -1 (hence bb^* \not = 1)

\displaystyle 0 = \frac{aa^*(bb^*+1)}{bb^*} - \frac{ca^*}{b} - \frac{c^*a}{b^*}

\displaystyle 0 = aa^* + |\frac{a}{b^*} - c|^2 - cc^* \Leftrightarrow cc^*-aa^* = |\frac{a}{b^*} - c|^2 = |c-a|^2 and we're done
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Oh I Really Don't Care
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#44
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#44
lulz

I just got negged for my solution to STEP III q.1.

Apparently my graphs are not up to the exacting standards that TSR requires.
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Glutamic Acid
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#45
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#45
II/10:

:)
Equating the vertical displacements, remembering that the anti-missile missile travels a time t-T seconds:
100 \cdot \dfrac{3}{5}t - \dfrac{1}{2} \cdot 10t^2 = 200 \cdot \dfrac{4}{5} (t-T) - \dfrac{1}{2} \cdot 10 (t-T)^2 \implies 60t - 5t^2 = 160(t-T) - 5(t-T)^2

The horizontal distances will add to 180: 80t + 120(t-T) = 180, and solving these simultaneously gives T = 1.
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Sk1lLz
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#46
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STEP I 2005 Question 10:

First part
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From A colliding with B:

Let  v_A represent the velocity of A after colliding with A.

1. NEL:  e = \frac{v - v_A}{u}

2. CLM:  au = av_A + bv

 \Rightarrow v_A = \frac{au - bv}{a}

Substitute into 1.:

3.  v = \frac{au(e + 1)}{a + b}

From B colliding with C:

Let \omega_B represent the velocity of B after colliding with C.

4. NEL:  e = \frac{\omega - \omega_B}{v}

5. CLM:  bv = b\omega_B + c\omega

\Rightarrow \omega_B = \frac{bv - c\omega}{b}

Substitute into 4.:

6. \omega = \frac{bv(e + 1)}{(b + c)}

Now, substitute 3. into 6.:  \omega = \frac{abu(e + 1)^2}{(a + b)(b + c)}

i)
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 \frac {a}{b} = \frac {b}{c} \Rightarrow b^2 = ac

 \frac {b}{c} = e \Rightarrow b = ce

 \frac {a}{b} = e \Rightarrow a = be

 ab = c^2e^3

Use 6.: \omega = \frac{c^2e^3u(e + 1)^2}{c^2e^3 + c^2e^2 + c^2e + c^2e^2}

 \omega = ue^2

For v, use 3.:  v= \frac{beu(e + 1)}{b(e + 1)}

 v = ue

ii)
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\frac{b}{a} = \frac{c}{b} \Rightarrow b^2 = ac

\frac{c}{b} = e \Rightarrow c = be

\frac{b}{a} = e \Rightarrow b = ae

bc = a^2e^3

Use 6.:  \omega = \frac{a^2ue(e + 1)^2}{a^2e + a^2e^3 + a^2e^2 + a^2e^2}

 \omega = u

Use 3.:  v = \frac{au(e + 1)}{a(e + 1)}

 v = u

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∏ Squared Over 6
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#47
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#47
(Original post by SimonM)
STEP III, Question 3

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f(x) = x^2+px+q
g(x) = x^2+rx+s

f(g(x)) = (x^2+rx+s)^2+p(x^2+rx+s)+q = q + p s + s^2 + (p r  + 2 r s )x + (p + r^2 + 2 s )x^2 + 

 2 r x^3 + x^4

Comparing coefficients, we have

a = 2r
b = p+2s+r^2
c = pr+2rs
d = q+ps+s^2

Therefore if we can find suitable values for p,q,r,s we have:

c = \left ( b - \left ( \frac{a}{2} \right )^2 \right ) \left ( \frac{a}{2} \right )

If a,b,c satisfy those equations, then we can solve simultaneously to find p,r,s and take q to be whatever value we need.

Expanding

(x^2+vx+w)^2-k = x^4+2vx^3+(v^2+2w)x^2+2vwx+w^2-k

This clearly satisfies our equation, so we're done.

f=x^4 - 4x^3 + 10x^2 - 12x + 4

(10-2^2)(-2) =-12 so it satisfies our equation.

Therefore:

f=(x^2-2x+3)^2-5=0

Therefore we have

x^2-2x+3 = \pm  \sqrt{5}

Therefore \displaystyle x \in \left \{ 1- \sqrt{\sqrt{5}-2},   1+ \sqrt{\sqrt{5}-2}, 1- i\sqrt{\sqrt{5}+2},   1+ i\sqrt{\sqrt{5}+2} \right \}
In particular:
(Original post by SimonM)
(x^2+vx+w)^2-k = x^4+2vx^3+(v^2+2w)x^2+2vwx+w^2-k

This clearly satisfies our equation, so we're done.
I have just done this question and for the part above (in the same way as the first part) I got simultaneous equations and found the condition on a, b and c, which turned out to be the same as in the first part. I'm just wondering how you could tell that it "clearly satisfies our equation". What have I missed?

Thanks
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SimonM
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#48
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#48
((v^2+2w)-(2v/2)^2)(v/2) = 2vw
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∏ Squared Over 6
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#49
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#49
(Original post by SimonM)
((v^2+2w)-(2v/2)^2)(v/2) = 2vw
ok, thanks, that's pretty obvious really :rolleyes:
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The Bigtime
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#50
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#50
(Original post by Unbounded)
STEP I, Question 3, 2005
(i)
 \dfrac{x}{x-a} + \dfrac{x}{x-b} = 1

 \iff x(x-b) + x(x-a) = (x-a)(x-b)

 \iff x^2 -bx +x^2 -ax = x^2 -bx -ax +ab

 \iff x^2 -ab = 0

 x = \pm \sqrt{ab}

So there are two distinct real roots if, and only if, a and b are both positive or both negative. Q.E.D.
(ii)

 \dfrac{x}{x-a} + \dfrac{x}{x-b} = 1+c

 \iff x(x-b)+x(x-a) = (1+c)(x^2-(a+b)x+ab)

 \iff x^2 -ab = cx^2 -c(a+b)x +abc

 \iff x^2(1-c) -x(ca+cb) -ab(1+c) = 0

There is exactly one real solution when the discriminant of this quadratic in x is equal to zero.

 \therefore c^2(a+b)^2 +4ab(1-c)(1+c) = 0

 \iff c^2 (a+b)^2 - 4abc^2 + 4ab = 0

 \iff c^2(4ab-(a+b)^2) = 4ab

 \iff c^2 = \dfrac{4ab}{4ab -(a+b)^2}

 = 1 + \dfrac{(a+b)^2}{4ab-a^2-2ab-b^2} = 1 - \left(\dfrac{a+b}{a-b} \right)^2 \ \ \ \square

Noting that c^2 is nonzero and real,  c^2 > 0

But also that  \left(\dfrac{a+b}{a-b} \right)^2 \geq 0 \iff 1- \left(\dfrac{a+b}{a-b} \right)^2 \leq 1

 \iff 0 < c^2 \leq 1 \ \ \ \square
Good solution, but how do we know c^2>0? Surely c^2=0 is a possibility?
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The Bigtime
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#51
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#51
(Original post by Unbounded)
I recall they told us that c is non-zero.
They told us c does not equal one, so c could equal zero. I believe I have a good argument why c cannot equal zero, but you might like to look for it yourself. (As your solutions show you're a cut above me at this STEP lark, so I'm sure you'll find it, but i'm fairly sure your solution is incomplete).
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matt2k8
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#52
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#52
(Original post by Unbounded)
I recall they told us that c is non-zero.
It doesn't, you need to exclude the possibility that c = 0 seperately
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miml
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#53
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#53
It's actually quite a nice contradiction..
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The Bigtime
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#54
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(Original post by Unbounded)
Ah apologies:

Suppose that c = 0, then the equation for (ii) becomes exactly the same as (i). Hence we require a and b to both be positive or negative. Then the condition of exactly one root, which is  c^2 = \frac{-4ab}{(a-b)^2} no longer holds, since the LHS is zero whereas the RHS is negative - contradiction. Hence c is non-zero.

Cheers for pointing this out - I'll edit the solution post.
No problem, I still have one small qualm though. In the case c=0, if a or b=0, then surely there is only one solution, x=0?

Also out of curiosity, how many marks would you (or someone else) estimate your original solution would have got out of 20? I ask because its the kind of overlook I make all the time.
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The Bigtime
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#55
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(Original post by Unbounded)
We in fact are told a and b are non-zero.
I'd be surprised to lose more than 3 marks.
So we are told that, my turn to overlook what the question tells us :o:
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matt2k8
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#56
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#56
When I did the question I used the argument that if c = 0, then we must have ab =0, which is impossible as a and b are both non-zero. I also think it would've been around 2-3 marks.
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The Bigtime
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#57
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#57
I also think that the solution to the final part of I question 7 can be simplified further by expanding the sine function.
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Farhan.Hanif93
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#58
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(Original post by Sk1lLz)
STEP I 2005 Question 10:

First part
Spoiler:
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From A colliding with B:

Let  v_A represent the velocity of A after colliding with A.

1. NEL:  e = \frac{v - v_A}{u}

2. CLM:  au = av_A + bv

 \Rightarrow v_A = \frac{au - bv}{a}

Substitute into 1.:

3.  v = \frac{au(e + 1)}{a + b}

From B colliding with C:

Let \omega_B represent the velocity of B after colliding with C.

4. NEL:  e = \frac{\omega - \omega_B}{v}

5. CLM:  bv = b\omega_B + c\omega

\Rightarrow \omega_B = \frac{bv - c\omega}{b}

Substitute into 4.:

6. \omega = \frac{bv(e + 1)}{(b + c)}

Now, substitute 3. into 6.:  \omega = \frac{abu(e + 1)^2}{(a + b)(b + c)}

i)
Spoiler:
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 \frac {a}{b} = \frac {b}{c} \Rightarrow b^2 = ac

 \frac {b}{c} = e \Rightarrow b = ce

 \frac {a}{b} = e \Rightarrow a = be

 ab = c^2e^3

Use 6.: \omega = \frac{c^2e^3u(e + 1)^2}{c^2e^3 + c^2e^2 + c^2e + c^2e^2}

 \omega = ue^2

For v, use 3.:  v= \frac{beu(e + 1)}{b(e + 1)}

 v = ue

ii)
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\frac{b}{a} = \frac{c}{b} \Rightarrow b^2 = ac

\frac{c}{b} = e \Rightarrow c = be

\frac{b}{a} = e \Rightarrow b = ae

bc = a^2e^3

Use 6.:  \omega = \frac{a^2ue(e + 1)^2}{a^2e + a^2e^3 + a^2e^2 + a^2e^2}

 \omega = u

Use 3.:  v = \frac{au(e + 1)}{a(e + 1)}

 v = u

I don't agree with this. I think you've misread the question. It asks you to find the final velocities of each of the three particles but haven't you just found the final velocity of C and the initial velocity of B following the collision with A?
From your solution, the final velocities of A and B are v_A and w_B but you didn't solve for them. I'll post the remainder of the solution after college.
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Farhan.Hanif93
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#59
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#59
(Original post by SimonM)
...
STEP II, Q9

(i)
Diagram

Consider the particles individually when they are at the point of slipping upwards.

Considering A:
Resolving perpendicular to the slope: R_m - mg\cos \frac{\pi}{6} = 0 \implies R_m = \frac{\sqrt3}{2}mg

Resolving parallel to the slope: T-\frac{1}{2}mg - \frac{\sqrt3}{6}R_m =0 \implies T=\frac{3}{4}mg by subbing in R_m from above and rearranging.

Considering B:
Resolving perpendicular to the slope: R_{2m}- 2mg\cos \frac{\pi}{6} + P \sin \theta =0 \implies R_{2m} = \sqrt 3mg - P\sin \theta

Resolving parallel to the slope: P\cos \theta - mg - \frac{\sqrt 3}{3}R_{2m} - T =0 \implies \frac{11}{4}mg = P(\cos \theta + \frac{\sqrt 3}{3}\sin \theta ) by subbing in R_{2m} from above and rearranging.

Expressing \frac{\sqrt 3}{3}\sin \theta +\cos \theta} in the form R\sin (\theta + \alpha) gives it to be equivalent to \frac{2}{\sqrt3}\sin (\theta + \tan^{-1} \frac{3}{\sqrt{3}}).

Note that this implies that P= \frac{11 mg \sqrt {3}}{8 \sin \left( \theta + \tan ^{-1} \frac{3}{\sqrt 3}\right)}, which is minimised when \sin\left(\theta + \tan^{-1} \frac{3}{\sqrt{3}}\right) is maximised i.e. when \theta + \tan^{-1} \frac{3}{\sqrt{3}} = \frac{\pi}{2}.

Therefore \boxed{P_{min} = \frac{11\sqrt3}{8}mg} and P must act at an angle of \boxed{\frac{2\pi}{3}-\tan ^{-1} \frac{3}{\sqrt 3}} to the horizontal i.e. parallel to the slope.

(ii)
If the particles are not sliding down the slope then they are either on the point of slipping down it or moving up it. We've considered the latter in part (i) so we need to find the smallest value of P for which the system is at the point of slipping downwards. If you were to draw a diagram for this, it would be the same as the one for part (i) except the frictional forces will be acting in the opposite direction.

Considering A:
Resolving perpendicular to the slope: R_m - mg\cos \frac{\pi}{3} = 0 \implies R_m = \frac{\sqrt3}{2}mg

Resolving parallel to the slope: T + \frac{\sqrt3}{6}R_m -\frac{1}{2}mg =0 \implies T=\frac{1}{4}mg

Considering B:
Resolving perpendicular to the slope: R_{2m} - \sqrt 3mg + P\sin \theta =0  \implies R_{2m}=\sqrt 3mg-P\sin \theta

Resolving parallel to the slope: P\cos \theta + \frac{\sqrt3}{3}R_{2m} - mg -T=0 \implies P=\dfrac{mg}{4(\cos \theta - \frac{\sqrt 3}{3}\sin \theta)}

Note that cos \theta - \frac{\sqrt 3}{3}\sin \theta \equiv \frac{2}{\sqrt 3}\cos \left(\theta + \tan ^{-1}\frac{\sqrt3}{3}\right).

Therefore \boxed{P_{min} = \frac{\sqrt 3}{8}mg} in this case. This is smaller than the answer to part (i), therefore is the correct answer.
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Farhan.Hanif93
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#60
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(Original post by SimonM)
...
That was a long one!
STEP II, Q11
First part
Diagram

Where F is the tension in the string, A is the object on the slope and B is the hanging object. (I chose not to use the conventional T because this letter is assigned to something else later on)

Considering B:
Resolving downwards: m_2g-F=m_2a \implies F=m_2(g-a) (\alpha)

Considering A
Resolving perpendicular to the slope: R-m_1g\cos (\arctan \frac{3}{4}) =0 \implies R=m_1g\cos (\arctan \frac{3}{4})

Resolving parallel to the slope: F-m_1g\sin (\arctan \frac{3}{4}) - \mu R = m_1a (\beta)

Note that:
Useful triangle

Using SOHCAHTOA on this triangle yields that:
\sin (\arctan \frac{3}{4})= \frac{3}{5}
\cos (\arctan \frac{3}{4}) = \frac{4}{5}

Therefore using these results and subbing (\alpha) into (\beta):
(\beta) : m_1a + m_2a =m_2g - m_1g \implies \boxed{a=\frac{m_2-m_1}{m_2+m_1}g} as required.

Second part
Considering the velocity, v_1, of A at the point where the string breaks:

u=0, t=T, a=\frac{m_2-m_1}{m_2+m_1}g, v=v_1

Therefore, using v=u+at:

v_1=\frac{m_2-m_1}{m_2+m_1}gT.

In this case, A can be modelled as follows:
Diagram

Where a_2 is the acceleration of A after the string breaks.

Resolving perpendicular to the slope: R-\frac{4}{5}m_1g=0 \implies R=\frac{4}{5}m_1g

Resolving down the slope: \frac{3}{5}m_1g + \mu R = m_1a_2 \implies a_2 = g

Considering the motion of A between the time where the string breaks and the particle reaches it's maximum height (let this time be denoted by t_1):

u=v_1, a=-g, v=0, t=t_1

Therefore using v=u+at:

t_1=\frac{m_2-m_1}{m_2+m_1}T

Note that the time taken (after release) to reach maximum height is given by:

T+t_1 = \boxed{\dfrac{2m_2T}{m_2+m_1}}

Third part
Let the maximum displacement of the particle be denoted by d; let the distance between the initial position and the position at which the string breaks be d_1 and let the distance between the string's breaking position and the maximum height of motion be d_2 such that d=d_1+d_2

Considering upwards motion for the distance d_1:

u=0, a=\frac{m_2-m_1}{m_2+m_1}g, t=T, s=d_1

Therefore, using s=ut +\frac{1}{2}at^2:

d_1 = \dfrac{(m_2-m_1)gT^2}{2(m_2+m_1)}

Considering upwards motion from d_1 to d_2:

u=v_1, a=-g, v=0 s=d_2

Therefore, using v^2=u^2+2as:

d_2 = \left(\dfrac{m_2-m_1}{\sqrt2(m_2+m_1)} \right)^2gT^2

\implies d = d_1+d_2 = \dfrac{(m_2-m_1)m_2}{(m_2+m_1)^2}gT^2

Note that, by resolving forces of the object when it is at it's maximum height, the acceleration down the slope is \frac{g}{5} ms^{-2}.

Considering motion down the slope from max height to initial position:

u=0, a=\frac{g}{5}, t=T+t_1, s=d

Therefore, using s=ut+\frac{1}{2}at^2:

\implies \dfrac{(m_2-m_1)m_2}{(m_2+m_1)^2}gT^2 = \dfrac{2m^2_2gT^2}{5(m_2+m_1)^2}

\implies \boxed{m_1:m_2 = 3:5}
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