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.

For the 2019 solutions I have set out the work to give an insight into how to do the problems more quickly. I have only included what I wrote to show how writing less can help you complete more questions. I have also included the time taken next to each question in Section 1. I didn't time my Section 2 but it took around 50 minutes. Note that there is wide variability in the time taken to solve each question, ranging from 12 seconds to over 5 minutes. Some questions which took a long time were as a result of a numerical slip early in the question leading to the wrong quadratic. It is definitely a good strategy to be well rested for the exam with no cramming the night before! 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 time-saving solutions.

Hints

Section 1

Part A

1. Do most of the question by inspection. Time taken 0:20.

2. Read off the graph to obtain the potential difference across the diode then subtract this from the voltage of the battery and apply Ohm's law. Time taken 0:54.

3. Rearrange without writing all steps. Time taken 0:35.

4. Note that in a vacuum all EM waves have the same speed. Work out how the ratio of the frequencies compares to the ratio of the wavelengths. Time taken 0:36.

5. This can be done by inspection using the percentage multiplier method. Time taken 0:12.

6. Use the pressure of a column of liquid in combination with atmospheric pressure. Time taken 1:04.

7. Form two equations from the information given and solve. Time taken 0:43.

8. Work out how the current changes from primary to secondary and use this to calculate the power dissipated. Time taken 5:23 (initially I misread the question and came back to this at the end. I erroneously thought that the $1500\Omega$ was for both cables, not just the secondary).

9. Factorise and cancel first. Time taken 0:36.

10. Use prior knowledge or common sense to work out what affects the rate of heat transfer and write very little. Time taken 0:31.

11. Recall that volume scale factor is the cube of the linear scale factor. Time taken 0:23.

12. Write the number of alpha decays next to the right hand column and use a process of elimination to identify the answer. Time taken 1:13.

13. Use Pythagoras to form a quadratic and solve. Time taken 2:42 (I made some slips which led to the wrong quadratic).

14. From the numbers given you can work out the change in velocity and hence the acceleration then the force. Time taken 0:41.

15. Using the information given make a quadratic equation and solve. Time taken 3:44 (More numerical slips).

16. Know or work out the resistance of the parallel combination then deduce the voltage dropped over each $\frac{R}{3}$. Time taken 0:45.

17. Use familiar triangles to make quick progress. Time taken 0:52.

18. Use mass=density$\times$volume and be on the lookout for cancellations. Time taken 1:45.

19. Work out the slant height of the pyramid then the height of the triangles. Accept this question may take longer than others. Time taken 2:45.

20. Start each sample off with N nuclei. Make a table of the amount present after 3 and 6 years. Time taken 0:53.

Part B

21. Make a rough sketch and apply the area of a trapezium. Time taken 0:37.

22. Convert gravitational potential energy to the energy consumed by the power station in one second then multiply by $60$. Time taken 2:16.

23. Differentiate, use the information given to give two equations then solve. Time taken 0:39.

24. Apply conservation of momentum and Impulse$=$Force$\times$time. Time taken 0:55.

25. Rationalise the denominator in one step and write as little as possible. Time taken 1:18.

26. Apply Ohm's law and $R=\frac{\rho L}{A}$. Time taken 0:39.

27. Use trigonometric identities to form a quadratic in $cosx$ and use this to work out $sinx$ and hence $tanx$. Time taken 1:48.

28. Take moments about the hinge. Time taken 0:37.

29. Use familiar triangles and know that the area of an equilateral triangle of side $s$ is $\frac{\sqrt3s^2}{4}$ (via $\frac{1}{2}absinC$, not $\frac{1}{2}bh$). Time taken 2:17 (more unforced errors).

30. Work out the path difference and what fraction of a wavelength this represents and hence the phase. Time taken 0:43.

31. Write as a quadratic in $5^x$. Time taken 1:02.

32. Don't be put off by this taking place on the moon - it just means there is a different value of $g$. Time taken 0:35.

33. Know that if it suits your purposes you can split an integral from $a$ to $b$ to two integrals from $a$ to $c$ and $c$ to $b$ with $a < c < b $. Apply this to get two simultaneous equations in two integrals (instead of three) and solve. Time taken 1:11.

34. The parallel combination must have a steeper line since it has less resistance. Work out this resistance and form an equation for the unknown resistance and solve. Time taken 1:07.

35. Find the coefficient of $t^3$ then integrate Time taken 0:57.

36. Find the extension via the Young's modulus then apply $EPE=\frac{1}{2}kx^2$. Time taken 1:26.

37. Solve for $tan\alpha$ without much writing then for $\beta$ by inspection. Time taken 0:47.

38. Note that the angles given and sought are between the ray and the boundary, not the ray and the normal. This can be overcome by replacing $sin$ with $cos$ in Snell's law. Time taken 2:16.

39. Spot a root then factorise into a linear term/quadratic by inspection. Time taken 0:59.

40. Work out the acceleration from the graph and make a resultant force calculation. Time taken 0:35.

Total time 49:21

Section 2

1. Apply Snell's law twice.

2. Write the resistances on the diagrams given by inspection and interpret using $P=\frac{V^2}{R}$.

3. Consider the forces on the mass on the plane.

4. Try to apply the principles of a potential divider by inspection.

5. Use $c=f\lambda$ and the information given taking care to correctly calculate the amplitude of the resultant wave.

6. Apply Snell's law and $v=f\lambda$. The angle is given between the light and the boundary (not between the light and the normal) which can be remedied by replacing $sin$ with $cos$ in Snell's law.

7. Use $Pressure=\frac{Force}{Area}$ and include the contribution of the atmosphere.

8. Use the pressure exerted by a column of liquid in conjuction with $PV=$ a constant.

9. Convert Kinetic and Gravitational Potential Energy to that lost by friction.

10. Use vertical equilibrium to find the weight of the beam then apply moments about its centre of mass with an unknown for the distance between the centre of mass and the pivot.

11. Apply SUVAT while not writing to much to get started.

12. Use the information about the spring to find the maximum force applied and hence the time taken for the energy to be imparted to the spring.

13. Assign $V$ as the voltage of a cell and $R$ as the resistance of a resistor. Work out the power before and after in terms of these.

14. Ignore what is irrelevant and focus only on the forces on the caravan.

15. Write the information given about the resistances on the diagram. Think of the upper and lower branches as two separate potential dividers to work out what will happen to the reading on the voltmeter.

16. Spot that there must be energy lost to air resistance otherwise the graph would be a straight line. This has direct implications for statements 2 and 3. For statement 1 observe the gradient of the graph.

17. Apply SUVAT to the landing to find how $h$ depends on $v$. Then apply it to the topmost point of the flight to find $v$ then solve for $h$.

18. We don't know anything about the force at the hinge so apply moments about that point.

19. Perform a conservation of momentum calculation to work out the energy lost.

20. Find out the refractive index of medium X then work out whether medium Y is more or less optically dense than X and apply Snell's law.

Time taken roughly 50 minutes

Full solutions

Section 1

Part A

1. Do most of the question by inspection. Time taken 0:20.

$7+3-$

$4\sqrt{21}$

2. Read off the graph to obtain the potential difference across the diode then subtract this from the voltage of the battery and apply Ohm's law. Time taken 0:54.

On a sketch of a series circuit - $6$ near the cell and $4.8$ near the resistor.

$I=\frac{V}{R}=\frac{4.8}{8\times10^{-3}}$

3. Rearrange without writing all steps. Time taken 0:35.

$\sqrt{\frac{3-y}{4}}=1-\frac{x}{2}$

$2(1-\sqrt{\frac{3-y}{4}})$

4. Note that in a vacuum all EM waves have the same speed. Work out how the ratio of the frequencies compares to the ratio of the wavelengths. Time taken 0:36.

A vertical line next to the first four options. $c=f\lambda$.

5. This can be done by inspection using the percentage multiplier method. Time taken 0:12.

$F\alpha v^2$

6. Use the pressure of a column of liquid in combination with atmospheric pressure. Time taken 1:04.

$P=\rho gh=4\times10^5Pa$

$h=\frac{4\times10^5}{10000}$

7. Form two equations from the information given and solve. Time taken 0:43.

$6=4p+2q$

$-4=16p+4q$

$3=2p+q$

$-2=8p+2q$

$4p=-8$

$p=-2$, $q=7$

8. Work out how the current changes from primary to secondary and use this to calculate the power dissipated. Time taken 5:23 (initially I misread the question and came back to this at the end. I erroneously thought that the $1500\Omega$ was for both cables, not just the secondary).

$\frac{V_s}{V_p}=\frac{N_s}{N_p}$

$4A$, $\frac{1}{5}A$

$P=I^2R=60$

9. Factorise and cancel first. Time taken 0:36.

$4-\frac{(3x+1)}{x(3x+1)(x-1)}$

$\frac{4x(x-1)-1}{x(x-1)}$

10. Use prior knowledge or common sense to work out what affects the rate of heat transfer and write very little. Time taken 0:31.

First I put a mark next to all rows with $40^{\circ}C$, since this gives the lowest temperature difference. Then I marked rows which already had a mark and also had the highest length of bar ($0.80m$). Finally of the rows with two marks I marked the row with the smallest $d$. The answer was the row with three marks.

11. Recall that volume scale factor is the cube of the linear scale factor. Time taken 0:23.

$192\times(\frac{3}{4})^3$

$3^4$

12. Write the number of alpha decays next to the right hand column and use a process of elimination to identify the answer. Time taken 1:13.

By inspection I could tell that the first three rows were not correct because there are seven decays. To the right of the third row I wrote $5, Z-10, 2$ (that's alpha decays, followed by the atomic number after that many decays followed by the number of beta decays required to make up the seven decays) and to the right of the fourth row $6, Z-12, 1$ and to the right of the fifth row $4, Z-8, 3$. From this I could spot the solution which worked.

13. Use Pythagoras to form a quadratic and solve. Time taken 2:42 (I made some slips which led to the wrong quadratic).

$x^2+8x+16+4x^2+8x+4=9x^2$

$4x^2-16x-20=0$

$x^2-4x-5$

$(x+1)(x-5)$

$\frac{1}{2}\times9\times12$

14. From the numbers given you can work out the change in velocity and hence the acceleration then the force. Time taken 0:41.

$\frac{1}{2}mv^2, 2v^2, 32, 200$

$v:4\rightarrow10$

$a=2$

15. Using the information given make a quadratic equation and solve. Time taken 3:44 (More numerical slips).

$(3x+2)(8-2x)=22$

$(3x+2)(4-x)=11$

$8-3x^2+10x=11$

$3x^2-10x+3=0$

$(3x-1)(x-3)$

16. Know or work out the resistance of the parallel combination then deduce the voltage dropped over each $\frac{R}{3}$. Time taken 0:45.

On the diagram above the resistors from left to right: $R$, $\frac{2R}{3}$ for the parallel combination, $R$. I then realised that $1V$ was dropped across every $\frac{R}{3}$ so $8V$ are dropped in total.

17. Use familiar triangles to make quick progress. Time taken 0:52.

First I drew a $3,4,5$ triangle then wrote $\times\frac{4}{5}$ then $4\times\frac{4}{5}\times 2$. This is because I realised that $QRS$ consists of two smaller $3,4,5$ triangles, having been enlarged by scale factor $\frac{4}{5}$. Half of the line QS corresponds to a "$4$" side of a $3,4,5$ triangle so the length is $4\times\frac{4}{5}\times 2$

18. Use mass=density$\times$volume and be on the lookout for cancellations. Time taken 1:45.

$m=\rho V$

$=100\times(0.22\times 0.22-\pi \times(0.07)^2)\times1$

$=100(0.15\times0.22)$

$=15\times0.22$

19. Work out the slant height of the pyramid then the height of the triangles. Accept this question may take longer than others. Time taken 2:45.

Volume$=48h$

$\frac{12}{\sqrt2}$ (from familiar triangles this is the distance from the centre of the base to a corner)

A triangle of base $\frac{12}{\sqrt2}$, height $h$ and hypotenuse $s$ ($s$ is the slant height).

$s=\sqrt{h^2+72}$

A triangle of base $6$, height $x$ and hypotenuse $\sqrt{h^2+72}$ (this is half one of the triangular faces).

Area$=144+4\times\frac{1}{2}\times12\times x$

$=144+24\sqrt{h^2+36}=48h$

$6+\sqrt{h^2+36}=2h$

$h^2+36=4h^2-24h+36$

$3h^2=24h$, $h=8$

20. Start each sample off with $N$ nuclei. Make a table of the amount present after 3 and 6 years. Time taken 0:53.

$3: \frac{N}{2}, \frac{3N}{2}$ where the fractions are numbers of $X$ and $Y$ respectively.

$6: \frac{N}{4}, \frac{7N}{4}$

Part B

21. Make a rough sketch and apply the area of a trapezium. Time taken 0:37.

A sketch of the lines in the question

$\frac{7}{2}\times6$

22. Convert gravitational potential energy to the energy consumed by the power station in one second then multiply by $60$. Time taken 2:16.

$mgh$

$\rho Vgh=2000\times10^6$

$V_{60}=frac{2000\times10^6\times60}{1000\times10\times150}$

I then cancelled as far as possible, glanced at the answers then focused on the most significant number of my answer and the number of zeros.

23. Differentiate, use the information given to give two equations then solve. Time taken 0:39.

$3x^2+2px+q=0$

$3-2p+q=0$

$27+6p+q=0$

$8p=-24$

24. Apply conservation of momentum and Impulse$=$Force$\times$time. Time taken 0:55.

I drew a diagram showing the velocities of the objects before the collision.

Impulse$=500\times10$

$F=25000$

I then looked back at the Impulse equation to realise that car Q had lost $5ms^{-1}$.

25. Rationalise the denominator in one step and write as little as possible. Time taken 1:18.

$\frac{(1+\sqrt2)^3}{-1}$

$3\sqrt2+2\sqrt2$

26. Apply Ohm's law and $R=\frac{\rho L}{A}$. Time taken 0:39.

I drew a rough cylinder with length $0.5$ and cross-sectional area $4\times 10^{-7}$.

$V=IR$

$=\frac{I\rho L}{A}=\frac{2\times1.6}{4}$

27. Use trigonometric identities to form a quadratic in $cosx$ and use this to work out $sinx$ and hence $tanx$. Time taken 1:48.

$7cosx+\frac{sin^2x}{cosx}=5$

$7cos^2x-5cosx+1-cos^2x=0$

$6cos^2x-5cosx+1=0$

$(2y-1)(3y-1)$

$cosx=\frac{1}{2}$ or $\frac{1}{3}$

$sinx=\frac{\sqrt3}{2}$ or $\frac{2sqrt2}{3}$

$tanx=\sqrt3$

28. Take moments about the hinge. Time taken 0:37.

$\frac{0.4\sqrt3\times14g}{2}=F\times 0.7$

$40\sqrt3$

29. Use familiar triangles and know that the area of an equilateral triangle of side $s$ is $\frac{\sqrt3s^2}{4}$ (via $\frac{1}{2}absinC$, not $\frac{1}{2}bh$). Time taken 2:17 (more unforced errors).

I drew a triangle from the centre of the circle to the midpoint of a side to a corner with $4$ as the base, $\frac{4}{\sqrt3}$ as the height (recognising this as half an equilateral triangle) and hypotenuse $\frac{8}{\sqrt3}$.

$\frac{64\pi}{3}-16\sqrt3$ using Area=$\frac{\sqrt3s^2}{4}$ for the area of an equilateral triangle of side $s$.

30. Work out the path difference and what fraction of a wavelength this represents and hence the phase. Time taken 0:43.

$10cm$ path difference

$c=f\lambda$

$\lambda=\frac{c}{f}=\frac{4}{5}=80cm$

31. Write as a quadratic in $5^x$. Time taken 1:02.

$15\times y^2-y-2=0$

$(3y+1)(5y-2)$

32. Don't be put off by this taking place on the moon - it just means there is a different value of $g$. Time taken 0:35.

$v=\sqrt{2gh}$

$80=\sqrt{2\times1.6\times h}$

$\frac{6400}{2\times1.6}$

$\frac{32000}{16}$

33. Know that if it suits your purposes you can split an integral from $a$ to $b$ to two integrals from $a$ to $c$ and $c$ to $b$ with $a < c < b $. Apply this to get two simultaneous equations in two integrals (instead of three) and solve. Time taken 1:11.

Next to the second equation I wrote $=4\int_{-2}^{2}f(x)dx-\int_2^4f(x)dx$.

Next to the target I wrote $=4-\int_{-2}^22f(x)dx$ (i.e. just rearranging the first equation).

I then wrote $4+2=8$, $3=1$ but I can't remember why I wrote this so suggest it might be clearer to work out $\int_{-2}^2f(x)$ to be $\frac{11}{6}$ by substituting my second equation into my first equation then substituting this into the first equation given to yield the answer as $4-2\frac{11}{6}$.

34. The parallel combination must have a steeper line since it has less resistance. Work out this resistance and form an equation for the unknown resistance and solve. Time taken 1:07.

$R=\frac{V}{I}$

$\frac{4}{20\times 10^{-3}}=200\Omega$

I then drew a sketch of a $300\Omega$ resistor in parallel with an unknown $R$.

$\frac{1}{\frac{1}{R}+\frac{1}{300}}=200$

$\frac{300R}{300+R}=200$

$60000+200R$

35. Find the coefficient of $t^3$ then integrate Time taken 0:57.

$\frac{7!}{4!3!}=\frac{7\times6\times5}{6}$

$\frac{(2t)^33^4\times35}{4}$

$2\times3^4\times35$

$70\times81$

36. Find the extension via the Young's modulus then apply $EPE=\frac{1}{2}kx^2$. Time taken 1:26.

$Y=\frac{Fl}{Ae}$

$e=\frac{40\times2.4}{2\times10^{-6}\times1.2\times10^{11}}=\frac{20\times2}{10^5}$

$E=\frac{20\times40}{10^5}$

37. Solve for $tan\alpha$ without much writing then for $\beta$ by inspection. Time taken 0:47.

$14tan\alpha=14$

$\alpha=45$

$\beta=30$

38. Note that the angles given and sought are between the ray and the boundary, not the ray and the normal. This can be overcome by replacing $sin$ with $cos$ in Snell's law. Time taken 2:16.

$ncos\alpha=1$

$cos\alpha=ncos\beta$

$cos\beta=\frac{1}{n^2}$

39. Spot a root then factorise into a linear term/quadratic by inspection. Time taken 0:59.

$-1-2+7$

$(x+1)(x^2-3x-4)$

$(x+1)(x+1)(x-4)$

40. Work out the acceleration from the graph and make a resultant force calculation. Time taken 0:35.

$a=\frac{2}{5}$

$m(g-a)$

$80(10-\frac{2}{5})$

Total time 49:21

Section 2

1. Apply Snell's law twice.

$sin60=nsin45$

$n=\frac{\sqrt3\sqrt2}{2}$

$sin45=\frac{\sqrt6}{2}siny$

$\sqrt{\frac{2}{6}}$

2. Write the resistances on the diagrams given by inspection and interpret using $P=\frac{V^2}{R}$.

On the diagram: $2R$, $\frac{3R}{2}$, $\frac{2R}{3}$

Using $P=\frac{V^2}{R}$ note that high resistance means low power.

3. Consider the forces on the mass on the plane.

$\frac{2g}{2}+5$

4. Try to apply the principles of a potential divider by inspection.

A vertical line next to options B and C.

5. Use $c=f\lambda$ and the information given taking care to correctly calculate the amplitude of the resultant wave.

$c=f\lambda$

$3.2=f\times0.08$

$T=0.025$

2400 cycles

$2400\times4\times1.5\times2$ (amplitudes combine)

6. Apply Snell's law and $v=f\lambda$. The angle is given between the light and the boundary (not between the light and the normal) which can be remedied by replacing $sin$ with $cos$ in Snell's law.

A diagram showing a ray at $25^{\circ}$ to the normal. (This turned out to be a slight detour - I should have used $cos65$).

$sin25=\frac{1}{n}=cos65$

$\frac{\lambda_{air}}{n}=\frac{v_{air}}{nf}$

7. Use $Pressure=\frac{Force}{Area}$ and include the contribution of the atmosphere.

$F=mg$

$=\frac{1}{3}\times140\times Area\times\rho\times g$

$700\times140\times g + 100000$

8. Use the pressure exerted by a column of liquid in conjuction with $PV=$ a constant.

$x\rho g+P$ is pressure

PV$=$constant

$4\pi r^3\alpha\frac{1}{x\rho g+P}$

$\frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3}=\frac{x\rho g+P}{\rho gh+P}$

9. Convert Kinetic and Gravitational Potential Energy to that lost by friction.

A diagram of a slope

$\frac{1}{2}\times2\times64$

$64J$ against friction

GPE: $2\times10\times h=40$

$\frac{104}{4}$

10. Use vertical equilibrium to find the weight of the beam then apply moments about its centre of mass with an unknown for the distance between the centre of mass and the pivot.

A diagram of the beam and the pivot, taking the centre of mass of the beam to be a distance $y$ to the left of the pivot.

$0.6X+800=X$

around pivot

$2000y=3\times0.6\times2000$

11. Apply SUVAT while not writing to much to get started.

$s=ut+\frac{1}{2}t^2$

$s_3-s_2=12.2$

$s_4-s_3=14.4$

$12.2=u+\frac{a}{2}(3^2-2^2)$

$14.4=u+\frac{a}{2}(4^2-3^2)$

I then crossed out the brackets above and replaced them with $5$ and $7$ respectively (because $3^2-2^2=5$ and $4^2-3^2=7$)

$\frac{12.2}{5}-\frac{u}{5}=\frac{14.4}{7}-\frac{u}{7}$

$12.2\times7-7u=14.4\times5-5u$

$u=\frac{12.2\times7-14.4\times5}{2}=6.1\times7-7.2\times5$

12. Use the information about the spring to find the maximum force applied and hence the time taken for the energy to be imparted to the spring.

$0.25=\frac{1}{2}kx^2$

$x^2=\frac{1}{100}$

$x=0.1m$

$F=5N$

therefore takes $25s$

$\frac{0.25}{25}$

13. Assign $V$ as the voltage of a cell and $R$ as the resistance of a resistor. Work out the power before and after in terms of these.

$I=\frac{2V}{3R}$

$I^2R=\frac{4V^2}{9R}=P$

$I=\frac{V}{2R}$

$I^2R=\frac{V^2}{4R}$

14. Ignore what is irrelevant and focus only on the forces on the caravan.

A diagram of a connected $M$ and $m$ with towbar of length $l$ and resistive forces as detailed in the question.

$E=\frac{Fl}{Ae}$

$F-D_2=Ma$

$\frac{EAe}{l}-D_2=Ma$

15. Write the information given about the resistances on the diagram. Think of the upper and lower branches as two separate potential dividers to work out what will happen to the reading on the voltmeter.

I did this question largely by inspection. Any change which reduces the resistance of P or S or increases the resistance of Q or R will act to reduce the voltage drop across P and reduce the voltage drop across S which will increase the potential difference across the voltmeter which stands at $\frac{V}{3}$ before the change.

16. Spot that there must be energy lost to air resistance otherwise the graph would be a straight line. This has direct implications for statements 2 and 3. For statement 1 observe the gradient of the graph.

For statement 1, observe that the gradient is strictly decreasing. For 2, note that the velocity is generally decreasing so later journeys over the same distance will take longer later in the experiment. For 3, use the fact that energy is being dissipated continually so later stages will be less energetic than earlier ones.

17. Apply SUVAT to the landing to find how $h$ depends on $v$. Then apply it to the topmost point of the flight to find $v$ then solve for $h$.

A graph of a negative parabola cut short a little way after the maximum.

$h=\frac{v}{2}t-\frac{gt^2}{2}$

$h=\frac{v}{2}-\frac{g}{2}$

at top, vertically $\frac{v}{2}=0.6g$ (this equation comes from the fact that the vertical component of the velocity takes $0.6s$ to become reduced to zero by gravity).

18. We don't know anything about the force at the hinge so apply moments about that point.

I marked on the diagram the weight of the drawbridge and the angle between the cable and the ramp as $90-\frac{\theta}{2}$. In my mind I called the length of the drawbridge $l$.

moments

$\frac{L}{2}Wsin\theta=lTsin(90-\frac{\theta}{2})$

$T=\frac{Wsin\theta}{2cos(\frac{\theta}{2})}=Wsin(\frac{\theta}{2})$ (from here the largest $W$ occurs for the largest $\theta$)

19. Perform a conservation of momentum calculation to work out the energy lost.

A rough diagram of the particles before and after the collision.

$v=\sqrt{\frac{2E}{m}}$

$\sqrt{2Em}=(m+M)u$

lost$=\frac{1}{2}(m+M)u^2-E$

$=\frac{2Em}{2(m+M)}-E=E(\frac{m}{m+M}-1)$

20. Find out the refractive index of medium X then work out whether medium Y is more or less optically dense than X and apply Snell's law.

A diagram of a light ray in medium X approaching the boundary at $45^{\circ}$ to the normal.

$n=\sqrt2$

A diagram of a light ray in medium Y approaching the boundary at $60^{\circ}$ to the normal.

$n=\frac{2\sqrt3}{3}$

A vertical line next to the first four options and a diagram of medium X below medium Y.

$\sqrt2sin\theta_c=\frac{2\sqrt3}{3}$

$sin\theta_c=\sqrt{\frac{2}{3}}$

$\frac{1}{\sqrt2}<\sqrt{\frac{2}{3}}<\frac{\sqrt3}{2}$

Time taken roughly 50 minutes

For the 2019 solutions I have set out the work to give an insight into how to do the problems more quickly. I have only included what I wrote to show how writing less can help you complete more questions. I have also included the time taken next to each question in Section 1. I didn't time my Section 2 but it took around 50 minutes. Note that there is wide variability in the time taken to solve each question, ranging from 12 seconds to over 5 minutes. Some questions which took a long time were as a result of a numerical slip early in the question leading to the wrong quadratic. It is definitely a good strategy to be well rested for the exam with no cramming the night before! 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 time-saving solutions.

Hints

Section 1

Part A

1. Do most of the question by inspection. Time taken 0:20.

2. Read off the graph to obtain the potential difference across the diode then subtract this from the voltage of the battery and apply Ohm's law. Time taken 0:54.

3. Rearrange without writing all steps. Time taken 0:35.

4. Note that in a vacuum all EM waves have the same speed. Work out how the ratio of the frequencies compares to the ratio of the wavelengths. Time taken 0:36.

5. This can be done by inspection using the percentage multiplier method. Time taken 0:12.

6. Use the pressure of a column of liquid in combination with atmospheric pressure. Time taken 1:04.

7. Form two equations from the information given and solve. Time taken 0:43.

8. Work out how the current changes from primary to secondary and use this to calculate the power dissipated. Time taken 5:23 (initially I misread the question and came back to this at the end. I erroneously thought that the $1500\Omega$ was for both cables, not just the secondary).

9. Factorise and cancel first. Time taken 0:36.

10. Use prior knowledge or common sense to work out what affects the rate of heat transfer and write very little. Time taken 0:31.

11. Recall that volume scale factor is the cube of the linear scale factor. Time taken 0:23.

12. Write the number of alpha decays next to the right hand column and use a process of elimination to identify the answer. Time taken 1:13.

13. Use Pythagoras to form a quadratic and solve. Time taken 2:42 (I made some slips which led to the wrong quadratic).

14. From the numbers given you can work out the change in velocity and hence the acceleration then the force. Time taken 0:41.

15. Using the information given make a quadratic equation and solve. Time taken 3:44 (More numerical slips).

16. Know or work out the resistance of the parallel combination then deduce the voltage dropped over each $\frac{R}{3}$. Time taken 0:45.

17. Use familiar triangles to make quick progress. Time taken 0:52.

18. Use mass=density$\times$volume and be on the lookout for cancellations. Time taken 1:45.

19. Work out the slant height of the pyramid then the height of the triangles. Accept this question may take longer than others. Time taken 2:45.

20. Start each sample off with N nuclei. Make a table of the amount present after 3 and 6 years. Time taken 0:53.

Part B

21. Make a rough sketch and apply the area of a trapezium. Time taken 0:37.

22. Convert gravitational potential energy to the energy consumed by the power station in one second then multiply by $60$. Time taken 2:16.

23. Differentiate, use the information given to give two equations then solve. Time taken 0:39.

24. Apply conservation of momentum and Impulse$=$Force$\times$time. Time taken 0:55.

25. Rationalise the denominator in one step and write as little as possible. Time taken 1:18.

26. Apply Ohm's law and $R=\frac{\rho L}{A}$. Time taken 0:39.

27. Use trigonometric identities to form a quadratic in $cosx$ and use this to work out $sinx$ and hence $tanx$. Time taken 1:48.

28. Take moments about the hinge. Time taken 0:37.

29. Use familiar triangles and know that the area of an equilateral triangle of side $s$ is $\frac{\sqrt3s^2}{4}$ (via $\frac{1}{2}absinC$, not $\frac{1}{2}bh$). Time taken 2:17 (more unforced errors).

30. Work out the path difference and what fraction of a wavelength this represents and hence the phase. Time taken 0:43.

31. Write as a quadratic in $5^x$. Time taken 1:02.

32. Don't be put off by this taking place on the moon - it just means there is a different value of $g$. Time taken 0:35.

33. Know that if it suits your purposes you can split an integral from $a$ to $b$ to two integrals from $a$ to $c$ and $c$ to $b$ with $a < c < b $. Apply this to get two simultaneous equations in two integrals (instead of three) and solve. Time taken 1:11.

34. The parallel combination must have a steeper line since it has less resistance. Work out this resistance and form an equation for the unknown resistance and solve. Time taken 1:07.

35. Find the coefficient of $t^3$ then integrate Time taken 0:57.

36. Find the extension via the Young's modulus then apply $EPE=\frac{1}{2}kx^2$. Time taken 1:26.

37. Solve for $tan\alpha$ without much writing then for $\beta$ by inspection. Time taken 0:47.

38. Note that the angles given and sought are between the ray and the boundary, not the ray and the normal. This can be overcome by replacing $sin$ with $cos$ in Snell's law. Time taken 2:16.

39. Spot a root then factorise into a linear term/quadratic by inspection. Time taken 0:59.

40. Work out the acceleration from the graph and make a resultant force calculation. Time taken 0:35.

Total time 49:21

Section 2

1. Apply Snell's law twice.

2. Write the resistances on the diagrams given by inspection and interpret using $P=\frac{V^2}{R}$.

3. Consider the forces on the mass on the plane.

4. Try to apply the principles of a potential divider by inspection.

5. Use $c=f\lambda$ and the information given taking care to correctly calculate the amplitude of the resultant wave.

6. Apply Snell's law and $v=f\lambda$. The angle is given between the light and the boundary (not between the light and the normal) which can be remedied by replacing $sin$ with $cos$ in Snell's law.

7. Use $Pressure=\frac{Force}{Area}$ and include the contribution of the atmosphere.

8. Use the pressure exerted by a column of liquid in conjuction with $PV=$ a constant.

9. Convert Kinetic and Gravitational Potential Energy to that lost by friction.

10. Use vertical equilibrium to find the weight of the beam then apply moments about its centre of mass with an unknown for the distance between the centre of mass and the pivot.

11. Apply SUVAT while not writing to much to get started.

12. Use the information about the spring to find the maximum force applied and hence the time taken for the energy to be imparted to the spring.

13. Assign $V$ as the voltage of a cell and $R$ as the resistance of a resistor. Work out the power before and after in terms of these.

14. Ignore what is irrelevant and focus only on the forces on the caravan.

15. Write the information given about the resistances on the diagram. Think of the upper and lower branches as two separate potential dividers to work out what will happen to the reading on the voltmeter.

16. Spot that there must be energy lost to air resistance otherwise the graph would be a straight line. This has direct implications for statements 2 and 3. For statement 1 observe the gradient of the graph.

17. Apply SUVAT to the landing to find how $h$ depends on $v$. Then apply it to the topmost point of the flight to find $v$ then solve for $h$.

18. We don't know anything about the force at the hinge so apply moments about that point.

19. Perform a conservation of momentum calculation to work out the energy lost.

20. Find out the refractive index of medium X then work out whether medium Y is more or less optically dense than X and apply Snell's law.

Time taken roughly 50 minutes

Full solutions

Section 1

Part A

1. Do most of the question by inspection. Time taken 0:20.

$7+3-$

$4\sqrt{21}$

2. Read off the graph to obtain the potential difference across the diode then subtract this from the voltage of the battery and apply Ohm's law. Time taken 0:54.

On a sketch of a series circuit - $6$ near the cell and $4.8$ near the resistor.

$I=\frac{V}{R}=\frac{4.8}{8\times10^{-3}}$

3. Rearrange without writing all steps. Time taken 0:35.

$\sqrt{\frac{3-y}{4}}=1-\frac{x}{2}$

$2(1-\sqrt{\frac{3-y}{4}})$

4. Note that in a vacuum all EM waves have the same speed. Work out how the ratio of the frequencies compares to the ratio of the wavelengths. Time taken 0:36.

A vertical line next to the first four options. $c=f\lambda$.

5. This can be done by inspection using the percentage multiplier method. Time taken 0:12.

$F\alpha v^2$

6. Use the pressure of a column of liquid in combination with atmospheric pressure. Time taken 1:04.

$P=\rho gh=4\times10^5Pa$

$h=\frac{4\times10^5}{10000}$

7. Form two equations from the information given and solve. Time taken 0:43.

$6=4p+2q$

$-4=16p+4q$

$3=2p+q$

$-2=8p+2q$

$4p=-8$

$p=-2$, $q=7$

8. Work out how the current changes from primary to secondary and use this to calculate the power dissipated. Time taken 5:23 (initially I misread the question and came back to this at the end. I erroneously thought that the $1500\Omega$ was for both cables, not just the secondary).

$\frac{V_s}{V_p}=\frac{N_s}{N_p}$

$4A$, $\frac{1}{5}A$

$P=I^2R=60$

9. Factorise and cancel first. Time taken 0:36.

$4-\frac{(3x+1)}{x(3x+1)(x-1)}$

$\frac{4x(x-1)-1}{x(x-1)}$

10. Use prior knowledge or common sense to work out what affects the rate of heat transfer and write very little. Time taken 0:31.

First I put a mark next to all rows with $40^{\circ}C$, since this gives the lowest temperature difference. Then I marked rows which already had a mark and also had the highest length of bar ($0.80m$). Finally of the rows with two marks I marked the row with the smallest $d$. The answer was the row with three marks.

11. Recall that volume scale factor is the cube of the linear scale factor. Time taken 0:23.

$192\times(\frac{3}{4})^3$

$3^4$

12. Write the number of alpha decays next to the right hand column and use a process of elimination to identify the answer. Time taken 1:13.

By inspection I could tell that the first three rows were not correct because there are seven decays. To the right of the third row I wrote $5, Z-10, 2$ (that's alpha decays, followed by the atomic number after that many decays followed by the number of beta decays required to make up the seven decays) and to the right of the fourth row $6, Z-12, 1$ and to the right of the fifth row $4, Z-8, 3$. From this I could spot the solution which worked.

13. Use Pythagoras to form a quadratic and solve. Time taken 2:42 (I made some slips which led to the wrong quadratic).

$x^2+8x+16+4x^2+8x+4=9x^2$

$4x^2-16x-20=0$

$x^2-4x-5$

$(x+1)(x-5)$

$\frac{1}{2}\times9\times12$

14. From the numbers given you can work out the change in velocity and hence the acceleration then the force. Time taken 0:41.

$\frac{1}{2}mv^2, 2v^2, 32, 200$

$v:4\rightarrow10$

$a=2$

15. Using the information given make a quadratic equation and solve. Time taken 3:44 (More numerical slips).

$(3x+2)(8-2x)=22$

$(3x+2)(4-x)=11$

$8-3x^2+10x=11$

$3x^2-10x+3=0$

$(3x-1)(x-3)$

16. Know or work out the resistance of the parallel combination then deduce the voltage dropped over each $\frac{R}{3}$. Time taken 0:45.

On the diagram above the resistors from left to right: $R$, $\frac{2R}{3}$ for the parallel combination, $R$. I then realised that $1V$ was dropped across every $\frac{R}{3}$ so $8V$ are dropped in total.

17. Use familiar triangles to make quick progress. Time taken 0:52.

First I drew a $3,4,5$ triangle then wrote $\times\frac{4}{5}$ then $4\times\frac{4}{5}\times 2$. This is because I realised that $QRS$ consists of two smaller $3,4,5$ triangles, having been enlarged by scale factor $\frac{4}{5}$. Half of the line QS corresponds to a "$4$" side of a $3,4,5$ triangle so the length is $4\times\frac{4}{5}\times 2$

18. Use mass=density$\times$volume and be on the lookout for cancellations. Time taken 1:45.

$m=\rho V$

$=100\times(0.22\times 0.22-\pi \times(0.07)^2)\times1$

$=100(0.15\times0.22)$

$=15\times0.22$

19. Work out the slant height of the pyramid then the height of the triangles. Accept this question may take longer than others. Time taken 2:45.

Volume$=48h$

$\frac{12}{\sqrt2}$ (from familiar triangles this is the distance from the centre of the base to a corner)

A triangle of base $\frac{12}{\sqrt2}$, height $h$ and hypotenuse $s$ ($s$ is the slant height).

$s=\sqrt{h^2+72}$

A triangle of base $6$, height $x$ and hypotenuse $\sqrt{h^2+72}$ (this is half one of the triangular faces).

Area$=144+4\times\frac{1}{2}\times12\times x$

$=144+24\sqrt{h^2+36}=48h$

$6+\sqrt{h^2+36}=2h$

$h^2+36=4h^2-24h+36$

$3h^2=24h$, $h=8$

20. Start each sample off with $N$ nuclei. Make a table of the amount present after 3 and 6 years. Time taken 0:53.

$3: \frac{N}{2}, \frac{3N}{2}$ where the fractions are numbers of $X$ and $Y$ respectively.

$6: \frac{N}{4}, \frac{7N}{4}$

Part B

21. Make a rough sketch and apply the area of a trapezium. Time taken 0:37.

A sketch of the lines in the question

$\frac{7}{2}\times6$

22. Convert gravitational potential energy to the energy consumed by the power station in one second then multiply by $60$. Time taken 2:16.

$mgh$

$\rho Vgh=2000\times10^6$

$V_{60}=frac{2000\times10^6\times60}{1000\times10\times150}$

I then cancelled as far as possible, glanced at the answers then focused on the most significant number of my answer and the number of zeros.

23. Differentiate, use the information given to give two equations then solve. Time taken 0:39.

$3x^2+2px+q=0$

$3-2p+q=0$

$27+6p+q=0$

$8p=-24$

24. Apply conservation of momentum and Impulse$=$Force$\times$time. Time taken 0:55.

I drew a diagram showing the velocities of the objects before the collision.

Impulse$=500\times10$

$F=25000$

I then looked back at the Impulse equation to realise that car Q had lost $5ms^{-1}$.

25. Rationalise the denominator in one step and write as little as possible. Time taken 1:18.

$\frac{(1+\sqrt2)^3}{-1}$

$3\sqrt2+2\sqrt2$

26. Apply Ohm's law and $R=\frac{\rho L}{A}$. Time taken 0:39.

I drew a rough cylinder with length $0.5$ and cross-sectional area $4\times 10^{-7}$.

$V=IR$

$=\frac{I\rho L}{A}=\frac{2\times1.6}{4}$

27. Use trigonometric identities to form a quadratic in $cosx$ and use this to work out $sinx$ and hence $tanx$. Time taken 1:48.

$7cosx+\frac{sin^2x}{cosx}=5$

$7cos^2x-5cosx+1-cos^2x=0$

$6cos^2x-5cosx+1=0$

$(2y-1)(3y-1)$

$cosx=\frac{1}{2}$ or $\frac{1}{3}$

$sinx=\frac{\sqrt3}{2}$ or $\frac{2sqrt2}{3}$

$tanx=\sqrt3$

28. Take moments about the hinge. Time taken 0:37.

$\frac{0.4\sqrt3\times14g}{2}=F\times 0.7$

$40\sqrt3$

29. Use familiar triangles and know that the area of an equilateral triangle of side $s$ is $\frac{\sqrt3s^2}{4}$ (via $\frac{1}{2}absinC$, not $\frac{1}{2}bh$). Time taken 2:17 (more unforced errors).

I drew a triangle from the centre of the circle to the midpoint of a side to a corner with $4$ as the base, $\frac{4}{\sqrt3}$ as the height (recognising this as half an equilateral triangle) and hypotenuse $\frac{8}{\sqrt3}$.

$\frac{64\pi}{3}-16\sqrt3$ using Area=$\frac{\sqrt3s^2}{4}$ for the area of an equilateral triangle of side $s$.

30. Work out the path difference and what fraction of a wavelength this represents and hence the phase. Time taken 0:43.

$10cm$ path difference

$c=f\lambda$

$\lambda=\frac{c}{f}=\frac{4}{5}=80cm$

31. Write as a quadratic in $5^x$. Time taken 1:02.

$15\times y^2-y-2=0$

$(3y+1)(5y-2)$

32. Don't be put off by this taking place on the moon - it just means there is a different value of $g$. Time taken 0:35.

$v=\sqrt{2gh}$

$80=\sqrt{2\times1.6\times h}$

$\frac{6400}{2\times1.6}$

$\frac{32000}{16}$

33. Know that if it suits your purposes you can split an integral from $a$ to $b$ to two integrals from $a$ to $c$ and $c$ to $b$ with $a < c < b $. Apply this to get two simultaneous equations in two integrals (instead of three) and solve. Time taken 1:11.

Next to the second equation I wrote $=4\int_{-2}^{2}f(x)dx-\int_2^4f(x)dx$.

Next to the target I wrote $=4-\int_{-2}^22f(x)dx$ (i.e. just rearranging the first equation).

I then wrote $4+2=8$, $3=1$ but I can't remember why I wrote this so suggest it might be clearer to work out $\int_{-2}^2f(x)$ to be $\frac{11}{6}$ by substituting my second equation into my first equation then substituting this into the first equation given to yield the answer as $4-2\frac{11}{6}$.

34. The parallel combination must have a steeper line since it has less resistance. Work out this resistance and form an equation for the unknown resistance and solve. Time taken 1:07.

$R=\frac{V}{I}$

$\frac{4}{20\times 10^{-3}}=200\Omega$

I then drew a sketch of a $300\Omega$ resistor in parallel with an unknown $R$.

$\frac{1}{\frac{1}{R}+\frac{1}{300}}=200$

$\frac{300R}{300+R}=200$

$60000+200R$

35. Find the coefficient of $t^3$ then integrate Time taken 0:57.

$\frac{7!}{4!3!}=\frac{7\times6\times5}{6}$

$\frac{(2t)^33^4\times35}{4}$

$2\times3^4\times35$

$70\times81$

36. Find the extension via the Young's modulus then apply $EPE=\frac{1}{2}kx^2$. Time taken 1:26.

$Y=\frac{Fl}{Ae}$

$e=\frac{40\times2.4}{2\times10^{-6}\times1.2\times10^{11}}=\frac{20\times2}{10^5}$

$E=\frac{20\times40}{10^5}$

37. Solve for $tan\alpha$ without much writing then for $\beta$ by inspection. Time taken 0:47.

$14tan\alpha=14$

$\alpha=45$

$\beta=30$

38. Note that the angles given and sought are between the ray and the boundary, not the ray and the normal. This can be overcome by replacing $sin$ with $cos$ in Snell's law. Time taken 2:16.

$ncos\alpha=1$

$cos\alpha=ncos\beta$

$cos\beta=\frac{1}{n^2}$

39. Spot a root then factorise into a linear term/quadratic by inspection. Time taken 0:59.

$-1-2+7$

$(x+1)(x^2-3x-4)$

$(x+1)(x+1)(x-4)$

40. Work out the acceleration from the graph and make a resultant force calculation. Time taken 0:35.

$a=\frac{2}{5}$

$m(g-a)$

$80(10-\frac{2}{5})$

Total time 49:21

Section 2

1. Apply Snell's law twice.

$sin60=nsin45$

$n=\frac{\sqrt3\sqrt2}{2}$

$sin45=\frac{\sqrt6}{2}siny$

$\sqrt{\frac{2}{6}}$

2. Write the resistances on the diagrams given by inspection and interpret using $P=\frac{V^2}{R}$.

On the diagram: $2R$, $\frac{3R}{2}$, $\frac{2R}{3}$

Using $P=\frac{V^2}{R}$ note that high resistance means low power.

3. Consider the forces on the mass on the plane.

$\frac{2g}{2}+5$

4. Try to apply the principles of a potential divider by inspection.

A vertical line next to options B and C.

5. Use $c=f\lambda$ and the information given taking care to correctly calculate the amplitude of the resultant wave.

$c=f\lambda$

$3.2=f\times0.08$

$T=0.025$

2400 cycles

$2400\times4\times1.5\times2$ (amplitudes combine)

6. Apply Snell's law and $v=f\lambda$. The angle is given between the light and the boundary (not between the light and the normal) which can be remedied by replacing $sin$ with $cos$ in Snell's law.

A diagram showing a ray at $25^{\circ}$ to the normal. (This turned out to be a slight detour - I should have used $cos65$).

$sin25=\frac{1}{n}=cos65$

$\frac{\lambda_{air}}{n}=\frac{v_{air}}{nf}$

7. Use $Pressure=\frac{Force}{Area}$ and include the contribution of the atmosphere.

$F=mg$

$=\frac{1}{3}\times140\times Area\times\rho\times g$

$700\times140\times g + 100000$

8. Use the pressure exerted by a column of liquid in conjuction with $PV=$ a constant.

$x\rho g+P$ is pressure

PV$=$constant

$4\pi r^3\alpha\frac{1}{x\rho g+P}$

$\frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3}=\frac{x\rho g+P}{\rho gh+P}$

9. Convert Kinetic and Gravitational Potential Energy to that lost by friction.

A diagram of a slope

$\frac{1}{2}\times2\times64$

$64J$ against friction

GPE: $2\times10\times h=40$

$\frac{104}{4}$

10. Use vertical equilibrium to find the weight of the beam then apply moments about its centre of mass with an unknown for the distance between the centre of mass and the pivot.

A diagram of the beam and the pivot, taking the centre of mass of the beam to be a distance $y$ to the left of the pivot.

$0.6X+800=X$

around pivot

$2000y=3\times0.6\times2000$

11. Apply SUVAT while not writing to much to get started.

$s=ut+\frac{1}{2}t^2$

$s_3-s_2=12.2$

$s_4-s_3=14.4$

$12.2=u+\frac{a}{2}(3^2-2^2)$

$14.4=u+\frac{a}{2}(4^2-3^2)$

I then crossed out the brackets above and replaced them with $5$ and $7$ respectively (because $3^2-2^2=5$ and $4^2-3^2=7$)

$\frac{12.2}{5}-\frac{u}{5}=\frac{14.4}{7}-\frac{u}{7}$

$12.2\times7-7u=14.4\times5-5u$

$u=\frac{12.2\times7-14.4\times5}{2}=6.1\times7-7.2\times5$

12. Use the information about the spring to find the maximum force applied and hence the time taken for the energy to be imparted to the spring.

$0.25=\frac{1}{2}kx^2$

$x^2=\frac{1}{100}$

$x=0.1m$

$F=5N$

therefore takes $25s$

$\frac{0.25}{25}$

13. Assign $V$ as the voltage of a cell and $R$ as the resistance of a resistor. Work out the power before and after in terms of these.

$I=\frac{2V}{3R}$

$I^2R=\frac{4V^2}{9R}=P$

$I=\frac{V}{2R}$

$I^2R=\frac{V^2}{4R}$

14. Ignore what is irrelevant and focus only on the forces on the caravan.

A diagram of a connected $M$ and $m$ with towbar of length $l$ and resistive forces as detailed in the question.

$E=\frac{Fl}{Ae}$

$F-D_2=Ma$

$\frac{EAe}{l}-D_2=Ma$

15. Write the information given about the resistances on the diagram. Think of the upper and lower branches as two separate potential dividers to work out what will happen to the reading on the voltmeter.

I did this question largely by inspection. Any change which reduces the resistance of P or S or increases the resistance of Q or R will act to reduce the voltage drop across P and reduce the voltage drop across S which will increase the potential difference across the voltmeter which stands at $\frac{V}{3}$ before the change.

16. Spot that there must be energy lost to air resistance otherwise the graph would be a straight line. This has direct implications for statements 2 and 3. For statement 1 observe the gradient of the graph.

For statement 1, observe that the gradient is strictly decreasing. For 2, note that the velocity is generally decreasing so later journeys over the same distance will take longer later in the experiment. For 3, use the fact that energy is being dissipated continually so later stages will be less energetic than earlier ones.

17. Apply SUVAT to the landing to find how $h$ depends on $v$. Then apply it to the topmost point of the flight to find $v$ then solve for $h$.

A graph of a negative parabola cut short a little way after the maximum.

$h=\frac{v}{2}t-\frac{gt^2}{2}$

$h=\frac{v}{2}-\frac{g}{2}$

at top, vertically $\frac{v}{2}=0.6g$ (this equation comes from the fact that the vertical component of the velocity takes $0.6s$ to become reduced to zero by gravity).

18. We don't know anything about the force at the hinge so apply moments about that point.

I marked on the diagram the weight of the drawbridge and the angle between the cable and the ramp as $90-\frac{\theta}{2}$. In my mind I called the length of the drawbridge $l$.

moments

$\frac{L}{2}Wsin\theta=lTsin(90-\frac{\theta}{2})$

$T=\frac{Wsin\theta}{2cos(\frac{\theta}{2})}=Wsin(\frac{\theta}{2})$ (from here the largest $W$ occurs for the largest $\theta$)

19. Perform a conservation of momentum calculation to work out the energy lost.

A rough diagram of the particles before and after the collision.

$v=\sqrt{\frac{2E}{m}}$

$\sqrt{2Em}=(m+M)u$

lost$=\frac{1}{2}(m+M)u^2-E$

$=\frac{2Em}{2(m+M)}-E=E(\frac{m}{m+M}-1)$

20. Find out the refractive index of medium X then work out whether medium Y is more or less optically dense than X and apply Snell's law.

A diagram of a light ray in medium X approaching the boundary at $45^{\circ}$ to the normal.

$n=\sqrt2$

A diagram of a light ray in medium Y approaching the boundary at $60^{\circ}$ to the normal.

$n=\frac{2\sqrt3}{3}$

A vertical line next to the first four options and a diagram of medium X below medium Y.

$\sqrt2sin\theta_c=\frac{2\sqrt3}{3}$

$sin\theta_c=\sqrt{\frac{2}{3}}$

$\frac{1}{\sqrt2}<\sqrt{\frac{2}{3}}<\frac{\sqrt3}{2}$

Time taken roughly 50 minutes