STEP I 2014 solutions

So after the substitution we have

$\alpha u_{n+1} + \beta = -p \left(\alpha u_n + \beta \right)^2 + q \left( \alpha u_n + \beta \right) + r$

Rearrange a little

$\displaystyle u_{n+1} = - \frac p {\alpha} \left(\alpha u_n + \beta \right)^2 + \frac q {\alpha} \left( \alpha u_n + \beta \right) + \frac r {\alpha} - \frac {\beta} {\alpha}$

Then equating coefficients with $u_{n+1} = 4 u_n (1 - u_n)$ we obtain

$- \dfrac p {\alpha} \alpha^2 = - 4$

Might as well do the only question I could get a full solution for

STEP I 2014 question 8

part i)


part ii)


part iii)

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STEP I, Q11. (Interesting question, this!).

For the 2nd part. Label the M1, M2 pulley P, and let the tension it its string be T. Let the tension in the other string be S. For legibility, write A for the acceleration a2.

Again, the acceleration of M is g - T/M and the acceleration of m is g-T/m. But relative to the pulley, the accelerations are (g+A) - T/M and (g+A) -T/m. So (replacing g in the earlier calculation with g+A) we end up with a
new value for the tension $T = 2(g+A)\dfrac{m_1m_2}{m_1+m_2} = 2(g+A)\mu$.

Now, since the pulley is weightless, the net force on it must be zero, so we have 2T = S, so $S = 4(g+A)\mu$. Now since the acceleration of M is A, we know the net force on M is MA.

So we must have Mg - S = MA. So $Mg - 4(g+A)\mu = MA$. So $A(M+4\mu) = g(M-4\mu)$ so $A = g\dfrac{M-4\mu}{M+4\mu}$ as required.

From the first part of the question, we know that the force on the pulley due to the tension in the string is$\displaystyle\ \frac{4mM}{M+m}g$.

Dfranklin, do you think the last part of this solution is sufficient to gain all marks for that part?

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Suppose the other weight in the system is 0. Then the "pulley from the first part" is actually going to be in free fall, and so the tension in that string is 0. (In other words, your expression for the force on the pulley can't be correct for the general 3mass/2pulley scenario).

Bump

Please correct me if I am mistaken, but since the pulley (P1) is light (so we consider it as 'weightless'(?)), surely it will not be in free fall in the case M=0? (Since it will only move if acted on by a force, and it does not have any weight?)

If this is the case, then it would be the force on the pulley due to the string around it which cases it to fall, and so the tension in the string around the pulley (P1) would not be zero (and would still be what we found it to be previously, I believe).

To be honest, I don't really like the answer to part (iii). In fact, I'm not even 100% sure what is meant by much of the wording, which makes deciding whether it's actually right problematic. Of course, that may just be me being dim.

Although it's tempting to try to use (i) and (ii) to answer (iii) (and needing to do so is a fairly typical STEP pattern), I don't think it's useful here.

The first step is still to differentiate $y = (1-\sqrt(x))^2$ to get $\displaystyle \frac{dy}{dx} = \frac{\sqrt{x} - 1}{\sqrt{x}}$ and compare with the gradient of $L_a$ which is $(a-1)/a$. That leads us to consider the tangent to the curve for a particular value of x, say x = b^2 (we can use b^2 instead of b since we know 0 <x ). The tangent has gradient (b-1) / b, and so has the form y = x (b-1)/b + K for some value K. To find K, we note that to lie on C, we must have y = (1-b)^2 when x = b. So (1-b)^2 = b(b-1) + K, so K = 1-b. So the tangent has equation y = x (b-1)/b + (1-b), which is exactly the same as the equation for L_b. So every tangent is indeed of the form L_a for some a, namely, when x = b^2, a = b.

Tbh i think your explanation is practically the same as his and you have used the same reasoning, so cheers

Something else that might help convince you. Suppose (for simplicity), that m1 = m2. As M gets very small, using the formula we derive in the question, we see that M's acceleration tends to -g and so m1 and m2 have acceleration tending g. But if T is the tension in the string over P1, we know that m1 and m2 must have acceleration g-T/m1. It follows that as $M \to 0$ we must have $T \to 0$ also.

I disagree, but whatever...

But if you read them theyre essentially the same?! You just gave more detail at the end, similar to how i did in the exam but the solution was a bit less detailed than exam conditions

Question 9:

Resolving in the standard way we get:

$y = U \sin \theta t - \frac{1}{2} g t^2$

$x =U \cos \theta t - \frac{1}{2}kgt^2$

So we get $T_H$ when $\frac{dy}{dt} = 0$, thus:

$\displaystyle T_H = \frac{U \sin \theta}{g}$

And so by symmetry we get $\displaystyle T_L = \frac{2U \sin \theta}{g}$

Now we define $\displaystyle T = \frac{U \cos \theta}{kg}$. So $T$ is the point at which the $x$ velocity is zero. For the sketches we consider the ratios:

$\displaystyle \frac{T_H}{T} = k \tan \theta$

$\displaystyle \frac{T_L}{T} = 2k \tan \theta$

So when $k \tan \theta < \frac{1}{2}$ we have the particle reaching its maximum height and hitting the ground without the $x$ velocity changing direction.

When $\frac{1}{2} < k \tan \theta < 1$ we have the particle reaching its maximum height then moving back towards $O$ then hitting the ground.

When $k \tan \theta > 1$ we have the particle moving back towards the origin before reaching its maximum height.

Finally when $k \tan \theta = 1$ we have the particle stationary when $t = T = T_H$. Also we have:

$k y = U \cos \theta - \frac{1}{2} k g t^2 = x$

so the trajectory is a straight line.

The last part might be a little unclear, if people can't understand then I'll rewrite it.

STEP 1 Q2

i)
Integrating by parts
$\displaystyle \int \ln x \, \mathrm d x = x \ln x - \int \, \mathrm d x = x \ln x - x + C$

Now substitute $x \mapsto 2-x$ and remember to divide by $-1$.

ii)


Symmetry about the y axis due to $x^2$ term. Intersects x axis at $\pm \sqrt 3$ and $\pm \sqrt 5$. Intersects y axis at $\ln 4$.

iii)
$\displaystyle \int_0^{\sqrt 3} \ln {|x^2 - 4|} \, \mathrm d x = \int_0^{\sqrt{3}} \ln (4 - x^2) = \int_0^{\sqrt{3}} \ln (2 - x) + \ln (2 + x) \, \mathrm d x \\ = 4 \ln (2 + \sqrt {3} ) - 2 \sqrt{3}$

Note that $- \ln(2 - \sqrt{3}) = \ln(2 + \sqrt{3})$.

iv)
$\displaystyle \int_0^2 |\ln|x^2-4|| \, \mathrm d x = \int_0^2 |\ln(4-x^2)| \, \mathrm d x \\ = \int_0^{\sqrt{3}} \ln(4-x^2) \, \mathrm d x - \int_{\sqrt{3}}^2 \ln(4-x^2) \, \mathrm d x = 4 - 4 \sqrt 3 - 8 \ln 2 + 8 \ln (2 + \sqrt 3 )$

Now multiply by $2$ to get $8 - 8 \sqrt 3 - 16 \ln 2 + 16 \ln (2 + \sqrt 3 )$

Hi Tarquin;

My calculation of Q2 part three is 16\sqrt2-8.

I'm wondering whether is it possible that you make a minor mistake?

Mathematica agrees with Tarquin.