Firstly, I would like to caution students regarding the use of solutions when preparing for exams. They are certainly a double-edged sword and since you are planning to invest an enormous amount of time, effort and money into getting a degree, it would be a good idea to use solutions properly from now on. In all cases you should only consult a solution once you think your answer is correct or if you have spent a significant amount of time trying to reach an answer. Coming up with your own conclusion after several minutes of independent thought is far more worthwhile than learning directly from the solution itself. You should also make sure you are critical of all solutions presented to you. Not only will this help your understanding, but solutions may have errors. Therefore if you find that you can't reconcile a solution with yours then take the time to get in touch with the author to discuss the discrepancy. For example, please do contact me if you have spotted a mistake below or would like further guidance on any of the questions.
I will firstly set out a series of hints. Try to do the questions from these alone and if you are still having difficulty then scroll down to the full solutions. The full solutions are set out in 'time-saving' form; I recommend that you don't write much more than this when you sit the exam.

Hints

Section 2

1. Work out the frequency using the time given in the question and apply $c=f\lambda$

2. Consider conservation of energy for the speed. For the time, make a thought experiment where you make the slope on the left more shallow and the slope on the right more steep.

3. Make a conservation of momentum calculation in the rest frame of the space probe. If you like you can boost to the frame of the stationary observer but this is not necessary.

4. Work out how much energy is lost to friction over what distance and apply Work done $=$ Force $\times$ distance.

5. Use the vertical motion to find the time taken and apply this to the horizontal motion.

6. Use a potential divider to work out the resistance of the wire and apply $R=\frac{\rho L}{A}$.

7. Define a tension in the string, make two equations for the forces acting on the blocks and solve.

8. Define the internal resistance to be $r$ and use $V=IR$ for the whole circuit.

9. Find the centre of mass of the rod using a weighted average technique.

10. Relate the Young's modulus and the Power via the Force applied.

11. The phase relationship between the waves at the buildings can be used to work out what fraction of a wavelength they are apart. Then apply $c=f\lambda$.

12. The force, $F$ at terminal velocity is $mg$. At $v_0$ it is half of this. Make two equations and solve for the terminal velocity.

13. Define $e$ as the extension of one spring with $100g$ hanging from it. Work out what the extension of the combined spring is in terms of this to find $e$ then apply Hooke's law.

14. Apply Snell's law using the dimensions of the block to find $\theta_2$.

15. Apply conservation of momentum and conservation of energy but with an appropriately placed $\frac{3}{4}$.

16. Apply SUVAT to both particles while taking care of signs and initial positions, equate positions and solve for $t$.

17. We don't know the force at $P$ so take moments about there. Use familiar triangles to work out the distances from the pivot.

18. Apply Kirchhoff's laws to the circuit.

19. The pressure arises in two different ways. The first is from the fact that the rock is arresting the motion of the water so we have a rate of change of momentum. There is also a height of water sitting on the rock which creates its own pressure. Calculate each of these and add them.

20. Form an inequality for $R$ using a potential divider then take logarithms of both sides. The choice of base should be clear from a quick glance at the answers.



Full solutions

Section 2

1. Work out the frequency using the time given in the question and apply $c=f\lambda$
$T=0.2\times4=0.8$
$f=\frac{1}{T}=\frac{5}{4}$
Using $c=f\lambda$, $\lambda=\frac{60\times4}{5}=48cm$
2. Consider conservation of energy for the speed. For the time, make a thought experiment where you make the slope on the left more shallow and the slope on the right more steep.
The blocks gain kinetic energy through losing gravitational potential energy ($mgh$). Since $g$ and $h$ are the same for each block they must have the same speed. For the time, imagine making the right hand slope steeper and the left one shallower. Clearly it will take longer for the block to fall along the shallower slope.
3. Make a conservation of momentum calculation in the rest frame of the space probe. If you like you can boost to the frame of the stationary observer but this is not necessary.
Imagine that the space probe is travelling at a speed $u$. Furthermore, for the purposes of this question it is acceptable to assume arbitrarily that the mass of the probe before the explosion is $2m$ and the mass of each of the parts after the explosion is $m$. We can also take the direction of the new parts to be the same as the original probe. If this mattered to the outcome of the question then we would need to be given more information.
First we run alongside the space probe so that we are travelling at the same speed as it. From our perspective the momentum of the probe is zero. When it splits into two parts they will be travelling at equal and opposite speeds $v$. At this point you could deduce that the kinetic energy has increased but if you want to be sure, boost back to the frame of the earth which will entail adding $u$ to all velocities. Before the explosion we have $KE=\frac{1}{2}2mu^2$ and after we have $KE=\frac{1}{2}m[(v-u)^2+(v+u)^2]$ which is $mv^2$ larger than the kinetic energy before the explosion.
4. Work out how much energy is lost to friction over what distance and apply work done $=$ force $\times$ distance.
The work done against gravity is $mgh=150J$. The remaining $60J$ is lost against friction. Recognising a $3$, $4$, $5$ triangle we have that the frictional force is $\frac{60}{5}=12N$ by applying work done=force $\times$ distance.
5. Use the vertical motion to find the time taken and apply this to the horizontal motion.
Applying SUVAT vertically, we have $4=5t^2$ so $t=\frac{2}{\sqrt5}$.
Now using distance = speed $\times$ time we have $vt=\frac{6\sqrt5}{5}$ giving $v=3$
6. Use a potential divider to work out the resistance of the wire and apply $R=\frac{\rho L}{A}$.
By inspection the resistance of the wire is $200\Omega$. To see this, observe that $1000\Omega$ receives $1V$ therefore the remaining $0.2V$ must drop across the wire so its resistance is one fifth of the resistor.
Applying $R=\frac{\rho L}{A}$ we have $200=\frac{\rho\times4}{0.02\times10^{-6}}$. At this point look at the first two significant figures of all the answers and the order of magnitude rather than calculate $\rho$.
7. Define a tension in the string, make two equations for the forces acting on the blocks and solve.
$6g-T=6a$
$T-4gsin30-15=4a$
$60-20-15=10a$
8. Define the internal resistance to be $r$ and use $V=IR$ for the whole circuit.
The parallel combination of two $10\Omega$ resistors is equivalent to one resistor of $5\Omega$
$\frac{20}{r+8}=2$
Spot the answer by inspection from here.
9. Find the centre of mass of the rod using a weighted average technique.
Multiply the masses by their positions, sum the results then divide by the total mass.
$\frac{0.4\times10+0.6\times30+1\times50}{2}=\frac{4+18+50}{2}$
10. Relate the Young's modulus and the Power via the Force applied. Don't be thrown by the four cables; this merely increases the cross-sectional area by a factor of $4$.
$Y=\frac{Fl}{Ae}$
$P=Fv=\frac{YAe}{l}\times0.2=2\times10^{11}\times4\times2\times10^{-4}\times0.0025\times0.2$
From here, mentally combine the $4$ and $0.0025$ to start with a $1$. Therefore the answer starts with an $8$. It is straightforward to distinguish between the two answers which start with $8$.
11. The phase relationship between the waves at the buildings can be used to work out what fraction of a wavelength they are apart. Then apply $c=f\lambda$.
A phase difference of $\frac{\pi}{3}$ is equivalent to one sixth of a wavelength so $\lambda=6000m$. Using $f=\frac{1}{T}$, apply $c=f\lambda$.
12. The force, $F$ at terminal velocity is $mg$. At $v_0$ it is half of this. Make two equations and solve for the terminal velocity.
$mg=kv_T^n$
$\frac{mg}{2}=kv_0^n$
$v_T^n=2v_0^n$
13. Define $e$ as the extension of one spring with $100g$ hanging from it. Work out what the extension of the combined spring is in terms of this to find $e$ then apply Hooke's law.
The upper spring extends $2e$ whereas the lower spring extends $e$. The unstretched double spring is $24cm$ so $e=2cm$. Apply Hooke's law to either spring to determine $k$.
14. Apply Snell's law using the dimensions of the block to find $\theta_2$.
By Pythagoras' Theorem the hypotenuse of the right-angled triangle is $2.5\sqrt5$.
$sin60=nsin\theta_2$ where $\theta_2$ is the angle of refraction.
$n=\frac{\sqrt3}{2}\times\frac{2.5\sqrt5}{2.5}=\frac{\sqrt{15}}{2}$
15. Apply conservation of momentum and conservation of energy but with an appropriately placed $\frac{3}{4}$.
Let the speeds pf the $3kg$ and $1kg$ spheres be $u$ and $v$ respectively.
Kinetic energy before $=\frac{1}{2}\times3\times2^2+\frac{1}{2}\times1\times6^2=24$
Kinetic energy after $=18=\frac{1}{2}\times3\times u^2+\frac{1}{2}\times1\times v^2$
Conservation of momentum: $3u=v$
$18=(\frac{1}{6}+\frac{1}{2})v^2$
$v^2=27$
16. Apply SUVAT to both particles while taking care of signs and initial positions, equate positions and solve for $t$.
$3t^2=60-14t+t^2$
$2t^2+14t-60=0$
$t^2+7t-30=0$
$(t+10)(t-3)=0$
17. We don't know the force at $P$ so take moments about there. Use familiar triangles to work out the distances from the pivot.
Noting that $PO=2PS$, triangle $PSO$ is half an equilateral triangle and $PS=2\sqrt3$.
Define the base of the sphere to be $B$. Triangles $OPB$ and $PSO$ are congruent ($SSS$) so $SPB=60^{\circ}$.
Let $F$ be the force between the disc and the rod and apply moments:
$2\sqrt3F=2mgcos60=mg$ so $F=\frac{5g}{2\sqrt3}$.
18. Apply Kirchhoff's laws to the circuit.
Let $I_1$, $I_2$ and $I_3$ be the currents in the three vertical sections from left to right respectively. Also cancel all zeros on the end of the voltages and resistances.
From the left loop, counter-clockwise:
$2=4I_2-4I_1$
From the right loop, clockwise:
$1=4I_2-4I_3$
From the outer loop, counter-clockwise:
$1=4I_3-4I_1$
Applying the current law ($I_1+I_2+I_3=0$) to the third equation:
$1=4I_3+4I_2+4I_3$
Substitute for $I_2$ from the second equation:
$1=4I_3+1+4I_3+4I_3$
$I_3=0$, $I_2=0.25A$
19. The pressure arises in two different ways. The first is from the fact that the rock is arresting the motion of the water so we have a rate of change of momentum. There is also a height of water sitting on the rock which creates its own pressure. Calculate each of these and add them.
From the falling water (which reaches the rock at $30ms^{-1}$ from SUVAT), force = rate of change of momentum. Take the time frame to be $1s$ in which time $40kg$ loses $30ms^{-1}$ so:
$F=40\times30=1200N$
Using pressure = $\frac{force}{area}$ we have $P=\frac{1200}{2}=600Pa$
From the sitting water,
$P=\rho gh=1000\times10\times0.05=500Pa$
20. Form an inequality for $R$ using a potential divider then take logarithms of both sides. The choice of base should be clear from a quick glance at the answers.
For the potential difference across $X$ to be greater than $2V$, we require that the resistance of the variable resistor must be less than $1000\Omega$ (potential divider).
$R<1000$
$R_0 b^{-\mu T}<1000$
$log_b R_0 - \mu T < log_b 1000$

$T>\frac{1}{\mu}(log_b R_0-log_b 1000)$