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. Mistakes are more likely this year as I have had to compile these solutions more rapidly than usual due to a close family bereavement.
For the 2021 solutions I have included the time taken next to each question. Note that there is wide variability in the time taken to solve each question. It is definitely a good strategy to be well rested for the exam with no cramming the night before! The direction the ENGAA took this year was to include some slightly different questions, testing on less familiar ideas. In order to prepare for any changes in future years I recommend attempting a large number of problems from a wide variety of sources. 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.

Hints

Section 1

Part A

1. Count powers of $5$, $x$ and $y$ in your head then record each in turn on your paper. Time taken 0:29.

2. The density of the gas must have increased because it is now occupying less volume. Apply $PV=$ a constant to work out by what factor the density has increased. Time taken 0:42.

3. Try to find the optimal number of steps to reduce the time to solve the question. Time taken 0:38.

4. We need the voltage on the secondary, for which we need the voltage on the primary which we can calculate from the power and input current. Once this voltage has been found then the output power and hence the efficiency can be calculated. Time taken 1:23.

5. Noting that the curved areas are the same helps to work the answer out more quickly. Time taken 0:33.

6. Working out the spring constant $k$ from Hooke's law enables you to apply $\frac{1}{2}kx^2$. Time taken 1:06.

7. Take an imaginary price for both items, say $£100$ if it helps to visualise the outcome. Alternatively, note that both processes multiply the price by $1.1\times0.9$. Time taken 0:18.

8. Define the voltage of the supply to be $V$ and the resistance of a single lamp to be $R$. Working either with $P=\frac{V^2}{R}$ or $P=I^2R$, work out the power dissipated in each scenario, for example $I=\frac{V}{20R}$ etc. Time taken 4:17.

9. Apply Pythagoras' Theorem to the right-hand triangle then convert from $SQ$ to $RQ$, preferably in your head then apply Pythagoras to the left-hand triangle. Time taken 1:03.

10. Use force $=$ rate of change of momentum then add the resistive force to this to work out the total force. Time taken 1:15.

11. Equate the curve and the line then use the quadratic formula to work out the difference between the roots. Time taken 1:02.

12. Think of the opposing speeds as giving rise to an effective speed of the wave (from the boat's perspective) of those speeds added together. Time taken 0:41.

13. Evaluate the familiar trigonometric values and simplify. Time saving - notice how all the rational terms are different in the solutions meaning you don't have to work out the $\sqrt3$ term. Time taken 1:58.

14. Work out what the potential difference across the bulb must be when the current is $0.25A$. This eliminates all but two of the graphs. Think about what happens to the resistance of the bulb as the current through it increases. Time taken 0:59.

15. Make equations from the information given and solve, noting that the number of red sweets is positive. Time taken 2:13 (numerical slips).

16. Be organised and make a table of $X$, $Y$, $R$ and $S$ at the time of formation and after the years have elapsed then make a fraction as suggested in the question. Time taken 1:22.

17. Convert from the smaller to the larger cuboid in your head and make an equation from 3D Pythagoras. Time taken 2:51 (question misread).

18. Work out how much the temperature of the water has reduced due to the ice melting. Think of there being two waters now; $180g$ at the new temperature and $20g$ at $0^{\circ}C$ and make a weighted average of their temperatures. Time taken 1:21.

19. Note that $x<1$ and place quantities on the numerator or denominator depending on whether they would increase or decrease the cost of the journey. Time taken 1:46.

20. Work out the two times in terms of the length of the rod, $L$, the velocity of the pulse $v$ and $x$. Subtract and solve for $x$. Time taken 1:20.

Part B

21. Note that one of the terms is independent of $x$ and will disappear under differentiation so can be ignored. Time taken 0:27.

22. Work out $F$ from the information given then apply Newton's Second Law. Time taken 0:45.

23. Use the sum to $n$ terms of an arithmetic series, subtract and simplify. Time taken 1:23.

24. Work out how many wavelengths the distance in the diagram corresponds to. Particles will have been drawn away from any rarefactions and towards any compressions. Time taken 1:13.

25. Complete the square for the quadratic and take a square root, not forgetting your $\pm$. Time taken 0:36.

26. Use $P=\frac{V^2}{R}$ to work out the potential difference across the parallel combination and hence its resistance by applying the theory of potential dividers. Time taken 0:57.

27. Draw a diagram of the circle and curve and connect the centre of the circle to the intersections of the chord and the circle. Time taken 0:32.

28. Resolve the force in the cable and apply moments. Time taken 0:51.

29. Combine the logarithms using the laws of logarithms then bring the power down to cancel with the $3$ which should now be on the denominator. Time taken 0:28.

30. Apply $s=ut+\frac{1}{2}at^2$ to both the given distances and subtract to find out the speed at the third drop minus the speed at the first drop then divide by the time this takes. Time taken 0:59.

31. Note that $sin\theta$ and $cos\theta$ are equal (in the relevant range) when $\theta=\frac{\pi}{4}$ which is a little over $0.75$. Time taken 0:59.

32. Apply the Young's modulus formula and Energy stored$=\frac{1}{2}Fe$. Time taken 1:10.

33. The sum to infinity of the first geometric progression is standard. The second geometric progression has a first term which is $r$ times that of the first progression and its common ratio is also $r$ times that of the first progression. Form two equations from this and solve. Time taken 2:12.

34. Apply a straightforward conservation of momentum calculation. Time taken 0:37.

35. Factorise the second equation then make an obvious cancellation to avoid a cubic. Resist the temptation to assume that the later solutions you find won't be equal to any you find earlier in your working. Time taken 0:46.

36. Apply $R=\frac{\rho L}{A}$ and think about the effect the reshaping will have on the quantities in this formula. Time taken 0:52.

37. Rationalise the denominators and look for cancellations. Time taken 2:45.

38. Integrate twice, noting that the object starts at rest. Time taken 1:11.

39. Define $x$ to be the $x$ co-ordinate of the point $P$. Work out the area of the rectangle in terms of $x$ and differentiate. Time taken 1:50.

40. Apply conservation of momentum and energy. Time taken 0:54.



Section 2

1. Ignore the $180^{\circ}$ and pretend that the waves should arrive in phase since this is equivalent. Therefore the path difference must be one wavelength since we require the lowest frequency. Apply $v=f\lambda$. Time taken 0:59.

2. Take the acute angle to be one that is greater than $45^{\circ}$ but is also familiar, ie $60$ and work out $W$ and $F$ in terms of $N$. Time taken 3:13.

3. Reading the question carefully(!), work out the resistance of the wire and apply Ohm's law. Time taken 3:24 (question misread (parallel instead of series)).

4. Make an inequality involving the force down the plane on $M$ and downwards on $m$. Time taken 0:39.

5. This is a clever question. When the collision takes place, the second object is stationary therefore the velocity must halve (not vanish). For the second terminal velocity the weight is greater but the resistance is not therefore we should expect the second terminal velocity to be greater than the first. Time taken 1:18.

6. Apply Power$=$Force$\times$Velocity and Newton's Second Law then set the acceleration equal to zero. Time taken 2:18.

7. Use the information given about the battery to work out its emf and the current drawn then apply $P=I^2R$. Time taken 0:45.

8. The spring constant is given, so calculate the tension in the spring via moments then the extension via Hooke's law then apply Energy stored$=\frac{1}{2}kx^2$. Time taken 2:15.

9. Differentiate twice then substitute in the time and mass, recalling that force is rate of change of momentum. Time taken 0:30.

10. Visualise the new shape of the wavefront after the last part of it has just crossed the boundary, in particular, would you expect the left and rightmost edges of it to be vertical or not?. Time taken 1:50.

11. Apply Kirchhoff's laws, taking care with the signs. Time taken 2:44

12. Think about the effect of increasing the length of something while keeping its mass (and hence volume) constant. The Young's modulus of the sample must remain constant so it may help to apply this formula. Time taken 1:31.

13. Take a time frame such as one second to work out how much mass approaches and how its velocity changes. Time taken 1:16.

14. Work out the ratio of the forces in the two wires via resolving and applying horizontal equilibrium. Since the Young's modulus is a constant, the ratio of strains must equal of the ratio of the stresses. Time taken 1:47.

15. Work out the refractive index of the glass using the information given about the speed of light in it. Then apply Snell's law to obtain sin of the critical angle. This is also the cosine of the marked angle. To get the inequality correct, note that we require the angle to be large (i.e. approaching $90^{\circ}$) for refraction. Time taken 1:26.

16. Apply $s=ut+\frac{1}{2}at^2$ twice. Time taken 1:38.

17. The difference between the two pairs of readings can be used to find the density of the liquid and therefore the first volume of the liquid then the volume of the powder. We have its mass so the density follows. Time taken 2:23.

18. Use SUVAT to work out the vertical velocity when the stone lands and the time taken. Use the kinetic energy to work out its final speed then, along with the vertical component, calculate the horizontal component of the velocity which remains unchanged throughout the flight. Time taken 1:44.

19. In order to work out the energy transferred, we wish to add up the power for all of the time interval but must note that the power is changing over this interval which means a special kind of sum is required. The time taken can be worked out from the information and equation given in the question. Time taken 1:08.

20. The force on the end of the trough depends on how deep we are in it. Use $P=\rho gh$ to give the pressure at a certain depth which you could label as $x$. Considering a small strip of the end of the trough of width $dx$ at depth $x$, apply Force=Pressure$\times$Area to give the force on the strip at this depth. These forces need to be added but they change as the depth changes which (as in question 19) requires a special kind of sum to evaluate the total. Time taken 2:55.



Full solutions

Section 1

Part A

1. Count powers of $5$, $x$ and $y$ in your head then record each in turn on your paper. Time taken 0:29.
For the powers of $5$ we have $1-3+1=-1$, for $x$ we have $1-6+2=-3$ and for $y$ we have $2-3+1=0$.
2. The density of the gas must have increased because it is now occupying less volume. Apply $PV=$ a constant to work out by what factor the density has increased. Time taken 0:42.
The volume has decreased by a factor of $1.2$ so the density must have increased by this factor.
3. Try to find the optimal number of steps to reduce the time to solve the question. Time taken 0:38.
$\frac{q}{3}=\frac{1}{\frac{4}{r}-\frac{p}{2}}=\frac{1}{\frac{8-pr}{2r}}=\frac{2r}{8-pr}$
4. We need the voltage on the secondary, for which we need the voltage on the primary which we can calculate from the power and input current. Once this voltage has been found then the output power and hence the efficiency can be calculated. Time taken 1:23.
$V=\frac{P}{I}=\frac{3000}{12.5}=\frac{24000}{100}=240$
$\frac{N_S}{N_P}=\frac{V_S}{V_P}=\frac{1}{4}$
$V_S=60$
$P=60\times40=2400$
$\frac{24}{30}\times100=80$
5. Noting that the curved areas are the same helps to work the answer out more quickly. Time taken 0:33.
$2\times \frac{\pi x^2}{4}-2\times\frac{\pi y^2}{4}$
6. Working out the spring constant $k$ from Hooke's law enables you to apply $\frac{1}{2}kx^2$. Time taken 1:06.
$k=\frac{5}{0.02}=250$
$\frac{1}{2}\times 250\times(0.02)^2=\frac{2.5\times2}{100}$
7. Take an imaginary price for both items, say $£100$ if it helps to visualise the outcome. Alternatively, note that both processes multiply the price by $1.1\times0.9$. Time taken 0:18.
$£100 \rightarrow £90 \rightarrow £99$ is a reduction
$£100 \rightarrow £110 \rightarrow £99$ is a reduction
8. Define the voltage of the supply to be $V$ and the resistance of a single lamp to be $R$. Working either with $P=\frac{V^2}{R}$ or $P=I^2R$, work out the power dissipated in each scenario, for example $I=\frac{V}{20R}$ etc. Time taken 4:17.
$I=\frac{V}{20R}$
$P=20I^2R=\frac{V^2}{20R}$
similarly, for the $19$ bulb case, $P=\frac{V^2}{19R}$
9. Apply Pythagoras' Theorem to the right-hand triangle then convert from $SQ$ to $RQ$, preferably in your head then apply Pythagoras to the left-hand triangle. Time taken 1:03.
$\sqrt{8^2-4^2}=\sqrt{48}=4\sqrt{3}$
$RQ=3\sqrt3$
$\sqrt{64-27}$
10. Use force $=$ rate of change of momentum then add the resistive force to this to work out the total force. Time taken 1:15.
$F-1.8\times10^7=\frac{dp}{dt}=3\times10^6$

11. Equate the curve and the line then use the quadratic formula to work out the difference between the roots. Time taken 1:02.
$x^2-4x+5=2x+c$
$x^2-6x+5-c=0$
$x=\frac{6\pm\sqrt{36-4(5-c)}}{2}$
The quantity which is subject to the $\pm$ is $\sqrt{4+c}$
$2\sqrt{4+c}=8$
$4+c=16$
12. Think of the opposing speeds as giving rise to an effective speed of the wave (from the boat's perspective) of those speeds added together. Time taken 0:41.
The effective speed of the wave is $11ms^{-1}$.
$11=\frac{\lambda}{8}$
13. Evaluate the familiar trigonometric values and simplify. Time saving - notice how all the rational terms are different in the solutions meaning you don't have to work out the $\sqrt3$ term. Time taken 1:58.
$(\sqrt3-2)^3$
rational terms$=-8-2\times3\times3=-26$
14. Work out what the potential difference across the bulb must be when the current is $0.25A$. This eliminates all but two of the graphs. Think about what happens to the resistance of the bulb as the current through it increases. Time taken 0:59.
$\frac{6}{R+8}=\frac{1}{4}$
$R=16$
The gradient must increase as the current increases due to the effect the extra heat has on the resistance of the filament.
15. Make equations from the information given and solve, noting that the number of red sweets is positive. Time taken 2:13 (numerical slips).
$\frac{r}{9}\times\frac{r-1}{8}=\frac{5}{12}$
$r^2-r-30=0$
$(r-6)(r+5)=0$
$r=6$
$\frac{3}{9}\times\frac{2}{8}$
16. Be organised and make a table of $X$, $Y$, $R$ and $S$ at the time of formation and after the years have elapsed then make a fraction as suggested in the question. Time taken 1:22.
$X,Y,R,S$
$\frac{N}{16},\frac{N}{4},\frac{15N}{16}+N,\frac{3N}{4}+N$
The required fraction is $\frac{\frac{15}{16}+1}{\frac{3}{4}+1}=\frac{31}{28}$
17. Convert from the smaller to the larger cuboid in your head and make an equation from 3D Pythagoras. Time taken 2:51 (question misread).
$16+36+4x^2=77$
$4x^2=25$
18. Work out how much the temperature of the water has reduced due to the ice melting. Think of there being two waters now; $180g$ at the new temperature and $20g$ at $0^{\circ}C$ and make a weighted average of their temperatures. Time taken 1:21.
Find the new temperature of the hot water
$ml=mc\Delta T$
$20\times3000=180\times4\times\Delta T$
$\Delta T=\frac{100}{12}=\frac{50}{6}=8\frac{1}{3}$
Average temperature of the water $=\frac{16\frac{2}{3}\times9+0\times1}{10}=15$
19. Note that $x<1$ and place quantities on the numerator or denominator depending on whether they would increase or decrease the cost of the journey. Time taken 1:46.
The more miles you do the more you pay so place $m$ on the numerator.
We convert to kilometres by dividing by $x$ since $x<1$ and the count needs to be larger in kilometres.
The further you go per litre the less your journey will cost so place $f$ on the denominator.
The higher the cost per litre the more expensive your journey will be so place $p$ on the numerator.
So far we have been working in pence but require pounds so divide by $100$.
20. Work out the two times in terms of the length of the rod, $L$, the velocity of the pulse $v$ and $x$. Subtract and solve for $x$. Time taken 1:20.
$t_1=\frac{2(L-x)}{V}$
$t_2=\frac{2L}{V}$
$x=(\frac{t_2-t_1}{2})V$ but $V=\frac{2L}{t_2}$
Part B

21. Note that one of the terms is independent of $x$ and will disappear under differentiation so can be ignored. Time taken 0:27.
$4x+\frac{1}{4x}$
$4-\frac{1}{4x^2}$
$4-1$
22. Work out $F$ from the information given then apply Newton's Second Law. Time taken 0:45.
$2F=24sin60=\frac{24\sqrt3}{2}$
$F=6\sqrt3$
$\frac{6\sqrt3}{3}\times4$
23. Use the sum to $n$ terms of an arithmetic series, subtract and simplify. Time taken 1:23.
$(\frac{n+1}{2})(2a+nd)$
$(\frac{n-1}{2})(2a+(n-2)d)$
subtracting gives $2a+d(\frac{n^2+n-n^2+3n-2}{2})$
24. Work out how many wavelengths the distance in the diagram corresponds to. Particles will have been drawn away from any rarefactions and towards any compressions. Time taken 1:13.
$\frac{5\lambda}{2}=0.33$
$\lambda=\frac{0.66}{5}$
$f=\frac{5\times330}{0.66}=2500$
25. Complete the square for the quadratic and take a square root, not forgetting your $\pm$. Time taken 0:36.
$((x+2)^2-1)^2=1$
$(x+2)^2=0$ or $2$
26. Use $P=\frac{V^2}{R}$ to work out the potential difference across the parallel combination and hence its resistance by applying the theory of potential dividers. Time taken 0:57.
$\frac{V^2}{R}=9$ so $6V$ is dropped across the $4\Omega$ resistor and $24V$ across the $8\Omega$ resistor.
Therefore the resistance of the parallel combination must be $\frac{1}{4}$ of the $8\Omega$ resistor.
$\frac{1}{\frac{1}{4}+\frac{1}{R}}=2$
$\frac{4R}{R+4}=2$
$4R=2R+8$

27. Draw a diagram of the circle and curve and connect the centre of the circle to the intersections of the chord and the circle. Time taken 0:32.
Sketching the circle, axes, the line $x=1$ and the radii mentioned above gives a sector of the circle and a triangle with coordinates $(1,\pm\sqrt3)$ and $(0,0)$. The area of the sector is $\frac{4\pi}{3}$ and the area of the triangle, via $\frac{1}{2}bh$ is $\frac{1}{2}\times2\sqrt3$.
28. Resolve the force in the cable and apply moments. Time taken 0:51.
The plank is uniform so its weight acts at a distance of $1.5m$ from the pivot. The cable acts at $4m$ from the pivot.
Recalling that $sin60=\frac{\sqrt3}{2}$ we have $4\times\frac{75\sqrt3}{2}=\frac{3W}{2}$
$W=100\sqrt3$
29. Combine the logarithms using the laws of logarithms then bring the power down to cancel with the $3$ which should now be on the denominator. Time taken 0:28.
$\frac{log_{10}(3\times4\times6)^3}{3}$
30. Apply $s=ut+\frac{1}{2}at^2$ to both the given distances and subtract to find out the speed at the third drop minus the speed at the first drop then divide by the time this takes. Time taken 0:59.
$x=ut+\frac{1}{2}at^2$
$y=vt+\frac{1}{2}at^2$
$\frac{y-x}{t}=v-u$
need $\frac{v-u}{2t}$ because $2t$ elapses between the first and third drops.
31. Note that $sin\theta$ and $cos\theta$ are equal (in the relevant range) when $\theta=\frac{\pi}{4}$ which is a little over $0.75$. Time taken 0:59.
We would like to select the answer that gets the furthest up the $y$-axis. Starting at $(\frac{\pi}{4},\frac{1}{\sqrt2})$, we have the choice of going left on the cosine curve or right on the sine curve to get a larger value. However, the $x$ value of the correct answer also needs to be furthest away from $\frac{\pi}{4}$ and since $\frac{\pi}{4}$ is a little over $0.75$, the furthest away we can get among the options given is $0.5$ so we select the cosine curve and the answer is $A$.
32. Apply the Young's modulus formula and Energy stored$=\frac{1}{2}Fe$. Time taken 1:10.
$Y=\frac{Fl}{Ae}=\frac{72\times5\times10^6}{2\times1.8\times5\times10^{-4}}=20\times10^{10}$
$\frac{1}{2}Fe=\frac{36\times10^{-3}}{2}$
33. The sum to infinity of the first geometric progression is standard. The second geometric progression has a first term which is $r$ times that of the first progression and its common ratio is also $r$ times that of the first progression. Form two equations from this and solve. Time taken 2:12.
$\frac{a}{1-r}=\frac{8}{5}$, $5a=8-8r$
$\frac{ar}{1-r^2}=\frac{3}{5}$, $5ar=3-3r^2$
Eliminating $a$, $8r-8r^2=3-3r^2$
$5r^2-8r+3=0$
$(5r-3)(r-1)=0$
$r=\frac{3}{5}$
$a=\frac{8-\frac{24}{5}}{5}=\frac{16}{25}$
34. Apply a straightforward conservation of momentum calculation. Time taken 0:37.
$40\times4=30\times0+10v$
$v=16$
35. Factorise the second equation then make an obvious cancellation to avoid a cubic. Resist the temptation to assume that the later solutions you find won't be equal to any you find earlier in your working. Time taken 0:46.
$(x-4)(x^2-2x-8)=-(x-4)^2$
$x=4$
$x^2-2x-8=-x+4$
$x^2-x-12=0$
$(x-4)(x+3)=0$
$x=-3$ or $4$ but we have $x=4$ from before.
36. Apply $R=\frac{\rho L}{A}$ and think about the effect the reshaping will have on the quantities in this formula. Time taken 0:52.
The length is multiplied by $4$ and the area is quartered and the effect this has on $R=\frac{\rho L}{A}$ is to multiply it by $16$.
37. Rationalise the denominators and look for cancellations. Time taken 2:45.
$\frac{\sqrt3(\sqrt9-\sqrt7)}{2}+\frac{\sqrt3(\sqrt8-\sqrt6)}{2}+\frac{\sqrt3(\sqrt7-\sqrt5)}{2}+...+\frac{\sqrt3(\sqrt3-1)}{2}=\frac{3-1+2\sqrt2-\sqrt2}{2}$
38. Integrate twice, noting that the object starts at rest. Time taken 1:11.
$v=4t-0.18t^2$
$s=2t^2-0.06t^3$
$200-\frac{6000}{100}$
39. Define $x$ to be the $x$ co-ordinate of the point $P$. Work out the area of the rectangle in terms of $x$ and differentiate. Time taken 1:50.
$(15-x)\sqrt x$
$\frac{15}{2\sqrt x}-\frac{3\sqrt x}{2}=0$
$x=5$
Area=$10\sqrt5$
40. Apply conservation of momentum and energy. Time taken 0:54.
The trolleys have velocities $v$ one way and $4v$ the other.
$W=\frac{1}{2}\times4mv^2+\frac{1}{2}m\times16v^2$
$=2mv^2+8mv^2$
$6mv^2=\frac{6W}{10}$

Section 2

1. Ignore the $180^{\circ}$ and pretend that the waves should arrive in phase since this is equivalent. Therefore the path difference must be one wavelength since we require the lowest frequency. Apply $v=f\lambda$. Time taken 0:59.
$\lambda=6$
$v=f\lambda$
$f=\frac{336}{6}$
2. Take the acute angle to be one that is greater than $45^{\circ}$ but is also familiar, ie $60$ and work out $W$ and $F$ in terms of $N$. Time taken 3:13.
$Wcos60=N$, $W=2N$
$Wsin60=F$, $W=\frac{2F}{3}$
$N=\frac{F}{\sqrt3}$, $F=\sqrt3N$
3. Reading the question carefully(!), work out the resistance of the wire and apply Ohm's law. Time taken 3:24 (question misread (parallel instead of series)).
$R=\frac{20}{10^7\times0.1\times10^6}=20$
$I=\frac{V}{70}=\frac{1}{5}$
4. Make an inequality involving the force down the plane on $M$ and downwards on $m$. Time taken 0:39.
$Mgsin\theta>mg$
$M>\frac{m}{\frac{4}{5}}$
5. This is a clever question. When the collision takes place, the second object is stationary therefore the velocity must halve (not vanish). For the second terminal velocity the weight is greater but the resistance is not therefore we should expect the second terminal velocity to be greater than the first. Time taken 1:18.

6. Apply Power$=$Force$\times$Velocity and Newton's Second Law then set the acceleration equal to zero. Time taken 2:18.
$P=Fv$
With no resistance force, $F=\frac{P}{v}=ma$
Bringing in the resistance force, $a=\frac{P}{mv}-kv^2$
For maximum speed, $a=0$ and $P=mkv^3$.
7. Use the information given about the battery to work out its emf and the current drawn then apply $P=I^2R$. Time taken 0:45.
The battery consists of two branches, each with emf $4.5$ in series with a resistance of $0.6$ which is equivalent to a $4.5V$ source in series with a resistance of $0.3$. $I=\frac{V}{R}=\frac{4.5}{1.5}=3$
$P=I^2R=9\times1.2=10.8$
8. The spring constant is given, so calculate the tension in the spring via moments then the extension via Hooke's law then apply Energy stored$=\frac{1}{2}kx^2$. Time taken 2:15.
$2Tsin\theta=mg$
$T=\frac{mg}{2sin\theta}=\frac{8}{4}\times5=10$
$x=\frac{1}{2}$
$\frac{1}{2}\times20\times(\frac{1}{2})^2$
9. Differentiate twice then substitute in the time and mass, recalling that force is rate of change of momentum. Time taken 0:30.
$12\times2\times5\times2$
10. Visualise the new shape of the wavefront after the last part of it has just crossed the boundary, in particular, would you expect the left and rightmost edges of it to be vertical or not?. Time taken 1:50.
The wavefront must stretch out so that its radius of curvature is smaller than before it reached the boundary because the parts of it that reach the boundary first have a headstart on the rest of the wavefront and hence get disproportionately further ahead. For the centre of these arcs to be on the boundary however, the left and right edges of them would have to be vertical but this isn't possible because the velocity in the new medium would have to be infinite to achieve this. Therefore the centre is lower than it was previously but not as low as the boundary.
11. Apply Kirchhoff's laws, taking care with the signs. Time taken 2:44
Working in $mA$ and defining $I_1$ to be the current travelling anti-clockwise in the right-hand loop and $I_2$ to be the current travelling upwards in the middle we have, from the left-hand loop:
$4=10R$ so $R=\frac{2}{5}$
for the outer loop, $2+4=-I_1\times5R$, $I_1=-3$
12. Think about the effect of increasing the length of something while keeping its mass (and hence volume) constant. The Young's modulus of the sample must remain constant so it may help to apply this formula. Time taken 1:31.
$Y=\frac{Fl}{Ae}$
If the length increases and the cross-sectional area decreases then, in order to preserve the Young's modulus, the extension for the same force must increase and furthermore it must increase with a power greater than $1$ since if the length doubles then the cross-sectional area must halve which multiplies the extension by $4$ for the same force.
13. Take a time frame such as one second to work out how much mass approaches and how its velocity changes. Time taken 1:16.
The area halves so due to the incompressible nature of the water, the velocity must double to $1ms^{-1}$.
In one second, a mass of $0.5\times0.004\times1000$ gains a velocity of $0.5ms^{-1}$
Applying $\frac{1}{2}mv^2$ we have $\frac{1}{2}\times2\times(1^2-(\frac{1}{2})^2)=\frac{3}{4}$

14. Work out the ratio of the forces in the two wires via resolving and applying horizontal equilibrium. Since the Young's modulus is a constant, the ratio of strains must be equal the ratio of the stresses. Time taken 1:47.
$\frac{\sqrt3F_Q}{2}=F_P$
$\frac{F_P}{F_Q}=\frac{\sqrt3}{2}$
Since stress is given by $\frac{Force}{Area}$, the ratio of stresses is $2\sqrt3$.
The Young's modulus is a constant, so the ratio of strains is the same as this.
15. Work out the refractive index of the glass using the information given about the speed of light in it. Then apply Snell's law to obtain sin of the critical angle. This is also the cosine of the marked angle. To get the inequality correct, note that we require the angle to be large (i.e. approaching $90^{\circ}$) for refraction. Time taken 1:26.
$n=\frac{3}{2}$
$\frac{3}{2}sin\theta_c=\frac{6}{5}\times sin90$
$sin\theta_c=\frac{4}{5}$
$cos\theta=\frac{4}{5}$ and we require $\theta$ to be larger than this for refraction.
16. Apply $s=ut+\frac{1}{2}at^2$ twice. Time taken 1:38.
For the first leg of the journey $s=100$.
For the second leg we have $20\times10-\frac{1.5\times100}{2}=125$
17. The difference between the two pairs of readings can be used to find the density of the liquid and therefore the first volume of the liquid then the volume of the powder. We have its mass so the density follows. Time taken 2:23.
Working in $g$ and $cm^{-3}$ we have the density of the liquid $=\frac{12.6}{15}$. Time saving - don't evaluate this, keep it as a fraction.
The first addition of the liquid has volume given by its mass over its density $=\frac{6.3\times15}{12.6}=7.5$
Therefore the initial volume of the powder is $2.5$ and the density is $\frac{8.7}{2.5}$ which I decided to multiply numerator and denominator by $4$ to give $\frac{32+2.8}{10}$.
18. Use SUVAT to work out the vertical velocity when the stone lands and the time taken. Use the kinetic energy to work out its final speed then, along with the vertical component, calculate the horizontal component of the velocity which remains unchanged throughout the flight. Time taken 1:44.
Applying $v^2=u^2+2as$ vertically we have $v_v^2=1600$ so the vertical component of velocity upon landing is $40$.
From the kinetic energy we have $0.05\times v^2=125$.
Noticing that the resultant velocity of $50$ and the vertical component of $40$ means that we have a $3$, $4$, $5$ triangle and the horizontal component of velocity is $30$. Observe that the stone gains $40$ in the vertical direction so this process must have taken $4$ seconds to give the answer.
19. In order to work out the energy transferred, we wish to add up the power for all of the time interval but must note that the power is changing over this interval which means a special kind of sum is required. The time taken can be worked out from the information and equation given in the question. Time taken 1:08.
$P=IV=kt^2V$
In order to add up all the energies over time, since they are changing as time goes on we will require an integral. The component fails when $t=\sqrt{\frac{I_F}{k}}$
$E=\int_0^{\sqrt{\frac{I_F}{k}}}kt^2Vdt=\frac{kV}{3}\times(\frac{I_F}{k})^{\frac{3}{2}}$

20. The force on the end of the trough depends on how deep we are in it. Use $P=\rho gh$ to give the pressure at a certain depth which you could label as $x$. Considering a small strip of the end of the trough of width $dx$ at depth $x$, apply Force=Pressure$\times$Area to give the force on the strip at this depth. These forces need to be added but they change as the depth changes which (as in question 19) requires a special kind of sum to evaluate the total. Time taken 2:55.
Defining $x$ to be the depth of the water we have $P=\rho gx$
$F=P\times A$. Since the area changes as $x$ increases, we need to consider a small strip of the end of the trough. It is a rectangle at depth $x$ with height $dx$ (which is small) and width $0.6-x$. The width is the distance of the slanted rectangular face of the trough to the vertical rectangular face. Substituting this into $F=P\times A$ we have:
$F=\rho gx(0.6-x)dx$
In order to work out the total force we require an integral to work out the contributions of all these strips in the limit as $dx \rightarrow 0$.
$F_{total}=\int_0^{0.6}\rho g(\frac{3x}{5}-x^2)dx=\rho g[\frac{3x^2}{10}-\frac{x^3}{3}]_0^{0.6}=10000(\frac{27}{250}-\frac{9}{125})=10000\times\frac{9}{250}$