STEP I, II, III 2002 Solutions

For Q4 in III, I know there is an algebraic method to show what they want but is it cool if I say let x = a, and y=b.

Then I'd draw the cubics $y=a^3+a^2$ and $y=b^3-b^2$ and show that for every corresponding value on the curves, the x value on the a curve is always less than the x value on the b curve?

Edit: I can show the second part really easily from the curves as well.

Question 5, I'll pick up where Dadeymi left off:

$(b-1)\left(ab^2 + (a-1)b +(a-1)\right)=0$

Now, $a_1 \not= a \implies k \not= \frac{1}{1-a} \implies b \not= 1$

So $\left(ab^2 + (a-1)b +(a-1)\right)=0$

$\implies \left( b + \frac{a-1}{2a} \right)^2 = \frac{1-a}{a} - \frac{a^2 - 2a +1}{4a^2} = \frac{-5a^2 + 6a - 1}{4a^2}$

$\implies b = \frac{1-a}{2a} \pm \frac{\sqrt{-5a^2 +6a - 1}}{2a}$

$\implies k = \frac{1}{2a} \pm \frac{\sqrt{-5a^2 +6a - 1}}{2a(1-a)}$

You have a sign error.

Probably not terribly useful quoting someone who last posted to TSR in 2011...

STEP I 2002 Question 7

$\displaystyle I = \int_0^a \frac{\cos x}{\sin x + \cos x} \, dx$ and $\displaystyle J = \int_0^a \frac{\sin x}{\sin x + \cos x} \, dx$ with $0 \leq a < \frac{3\pi}{4}$

(Note that in the interval $0 \leq x < \frac{3\pi}{4}$, $\sin x + \cos x \not= 0$)





$\displaystyle (I+J)+(I-J) = a + \ln (\sin a + \cos a) \Rightarrow \boxed{2I = a + \ln (\sin a + \cos a)}$

Question 10, STEP I, 2002

Using NLR with elasyic collisions, speed of seperation = speed of approach so collision with the end of the cylinder dont change the speed. This gives $v_n - u = v_{n-1} + u \Leftrightarrow v_n = v_{n-1} +2u$, which is an arithmetic progression with initial value v and common difference 2u so $v_n = v +2un$ as required.

The diference in the distances from the piston and end of cylinder is the distance the piston moves in that time so $d_n-d_{n+1} = ut_n$

$t_n$ is the time take to reach the end of the cylinder + the time taken to reach the piston afterwards.
Giving $t_n = \dfrac{d_n}{v_n}+\dfrac{d_{n+1}}{v_n} = \dfrac{d_n +d_{n+1}}{v_n}$

Using our previous result
$d_n-d_{n+1} = ut_n = u\dfrac{d_n +d_{n+1}}{v_n}$
$\Leftrightarrow (v+2un)(d_n-d_{n+1}) = u(d_n +d_{n+1})$
$\Leftrightarrow (v+u(2n+1))d_{n+1} = (v+u(2n-1))d_n$
$\Leftrightarrow d_{n+1} = \dfrac{v+u(2n-1)}{v+u(2n+1)}d_n$
Which was to be shown.

$d_n= \dfrac{v+u(2n-3)}{v+u(2n-1)}d_{n-1} =\dfrac{v+u(2n-3)}{v+u(2n-1)} \times \dfrac{v+u(2n-5)}{v+u(2n-3)} \times \cdots \times \dfrac{v+u}{v+3u} d_1$
$\Rightarrow d_n = \dfrac{v+u}{v+u(2n-1)}d_1$

I the case $u = v$ we have $d_n = \dfrac{d_1}{n}$

Giving $ut_n = d_n - d_{n+1} = \dfrac{d_1}{n}-\dfrac{d_1}{n+1} = \dfrac{d_1}{n(n+1)}$

Obviously I can see that the result follows from the fact that the COR=1, however the result then seems to conflict with the idea that the KE of the system remains constant (if the speed of the particle is increasing). Could any kind soul help me to understand this? thanks

STEP III, Question 1

...



As a tends to infinty the volume tends to $2\pi$

STEP III, Question 1

a)



b)

II/7: (Scary scary vectors.)

For the last part, λ=2/3, which is not allowed.