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 1

Part A

1. Spot a simplification then expand the numerator and denominator whilst writing the minimum of working.

2. Read the question carefully and work out the area in squares, converting to distance afterwards.

3. Take everything to the left hand side and factorise noting that we have a positive quadratic.

4. Note that one of the statements is clearly incorrect and the other two are linked.

5. Rearrange, preferably by inspection.

6. Read the question carefully. The energy is lost via two separate means.

7. Form two simultaneous equations and solve by inspection if possible.

8. Assign a letter for the distance of the source from the midpoint and work out the two transit times in terms of this.

9. Find $P$ as a function of $Q$ and $Q$ as a function of $R$ to find $P$ as a function of $R$.

10. Work through the equations in the answer in order to verify if they are correct.

11. Spot the factorisation of the numerator and simplify with the minimum of writing.

12. Equate the power of the motor to the power required to lift the mass noting that the mass moves at a constant speed.

13. Rewrite each side as a power of 2 and equate the exponents.

14. Concentrate on the right-hand column (the left column's entries are all possible). Recall that a beta decay will increase the number of protons in the nucleus.

15. Create a two-way table

16. Quickly spot that the mass of the atom and the nucleus are being taken as the same in this question. Therefore we require a ratio of volumes.

17. Write the information given in terms of $n$ and solve.

18. Starting from $c=f\lambda$ obtain $f$ and $\lambda$ from the graphs and substitute in.

19. Draw a clear diagram taking care to correctly interpret 'from' in this context.

20. Current is the rate of flow of charge, so work out how much current is flowing and for how long.

21. Draw a rough diagram of the clock and work out how many degrees the minute hand has turned through. You can apply angles around a point because the other angles are easy to work out from your knowledge of the clock face.

22. Use conservation of momentum with the number of passenger carriages as an unknown and solve.

23. Form an equation in $x$ and solve.

24. Work out the quantities in statements 1 and 2. Your recollection of Newton's third law should reveal the accuracy of statement 3.

25. Write the radius of the circle in terms of $x$ using Pythagoras' theorem and subtract the area of the square from the area of the semi-circle.

26. Let there be $N$ nuclei each of $X$ and $Y$. Work out how many of each have decayed, add the results and divide by the initial number.

27. Working efficiently, find the area of the front and multiply by the length and the density.

28. Work out the factor by which the kinetic energy has changed from the information given.

Part B

29. Divide the numerator and denominator by $\sqrt3$ to save time then expand and simplify without writing too many steps.

30. Work out the moment as a result of the horizontal section of the crane first by inspection then use this in an equation balancing moments for the second scenario.

31. Find and list some solutions for each equation and read off how many are in the desired range.

32. $g$ is a constant so eliminate any curved graphs. The direction of the initial velocity and the fact that this direction must change reveal the only possible graph.

33. Factorise fully and solve for the contents of the bracket being zero.

34. Determine what is useful information in the question. There is no need to read all of the answers.

35. Work out the radius of the circle from the length of the arc. Use this to find the area of the sector and the triangle.

36. Briefly note the format of the answers then equate clockwise and anti-clockwise moments for the bar.

37. Write as a single logarithm using the laws of indices.

38. Identify the correct suvat equation quickly and try to avoid any unnecessary calculations.

39. Differentiate once, set equal to zero and substitute values for $x$ or differentiate twice and use a process of elimination.

40. Focus on what is important - the required tension in the string must be equal to the weight of the supported object regardless of what holds the other end of the rope.

41. Use $\frac{1}{2}absinC$ and maximise the following expression efficiently.

42. Work out the initial GPE and the final KE and evaluate the discrepancy

43. Only work out the required coefficient and note the large differences between the answers.

44. Read the question carefully, try to work out the initial downward velocity efficiently and apply suvat.

45. Apply the sum to infinity of a geometric series formula, rearrange and solve the resulting equation.

46. Work out the area efficiently and equate this to $mgh$ while keeping an eye on the answers to help avoid any unnecessary calculations.

47. Work out the first four terms of the sequence, spot a solution (the sign of which is hinted at in the question) to the cubic and perform the addition without substituting into the cubic.

48. Work out the maximum frictional force from the first scenario then apply it to find the resultant.

49. Factorise (preferably mostly by inspection) and interpret the inequality taking care to notice the repeated root.

50. Recall that frictional forces increase to match any opposing forces up to a maximum (which has not been achieved in this scenario).

51. Change $x$ accordingly, preferably by inspection.

52. Use Force=Rate of change of momentum taking care with the signs of the velocities.

53. Set the product of the gradients to $-1$ and either solve for $p$ or substitute likely candidates for $p$ from the answers.

54. Use a suvat equation to relate the speed and height and use a process of elimination.

Section 2

1a. $g$ is the acceleration and hence the gradient of the graph. It is changing which eliminates all but three of the possibilities. From here, substituting values for $t$ reveals the correct answer.

1b. Integrate the expression for velocity to obtain an expression for distance.

1c. Use an energy approach to ascertain how the peaks should change over time.

1d. Substitute in the units for mass and acceleration.

1e. Rearrange for $X$ and substitute units in.

2a. Only one of the devices obeys Ohm's law.

2b. As filament lamps draw more current they heat up which gradually increases their resistance.

2c. Draw a circuit diagram and find the values of each of the currents through the two components.

2d. Use early knowledge of series circuits.

2e. For this fairly high voltage, it should be possible to deduce the voltage across W then the voltage across the resistor followed by the current for the circuit. You then have all the information you need to use $P=IV$ for W.

3a. Hooke's law is obeyed for straight lines on this graph. Strain is extension divided by original length.

3b. Work out the area under the graph (I did this by counting large boxes and multiplying by the area of one large box).

3c. Halving the length of the rope halves its spring constant and its extension. Alternatively, you now have half the rope which is being stretched to the same limit as the whole rope.

3d. Two ropes means you have twice the energy stored.

4a. Coherence is a necessary condition for interference from a double slit. The path difference determines what type of interference. An integer number of wavelengths will give rise to constructive interference.

4b. The introduction of the material introduces an extra path difference for one slit only. In order for e.g. a given maximum to recover from this change the light must now be travelling at an angle to what it was previously.

4c. Work out the wavelength and deduce whether interference will occur and what distance there is between maxima at the receiver.

Full solutions

Section 1

Part A

1. Spot a simplification then expand the numerator and denominator whilst writing the minimum of working.

Divide each bracket by $\sqrt3$

$\frac{(2+1)^2}{(2-1)^2}=9$

2. Read the question carefully and work out the area in squares, converting to distance afterwards.

We are only concerned with the deceleration. There are $10$ squares under the deceleration part of the graph and each square is worth $5\times10=50m$ giving $500m$

3. Take everything to the left hand side and factorise noting that we have a positive quadratic.

$2x^2+x-15\geq 0$

$(2x-5)(x+3)\geq 0$

$x\leq -3$ or $x\geq \frac{5}{2}$

4. Note that one of the statements is clearly incorrect and the other two are linked.

If something expands or contracts then its mass is still conserved. Heating expands the water which decreases its density which is the reason why it floats above the cooler water.

5. Rearrange, preferably by inspection.

As an interim step you could write

$\frac{x}{2}-1=\pm\sqrt{\frac{y+5}{3}}$ and spot the answer from there. I don't recommend using too many steps.

6. Read the question carefully. The energy is lost via two separate means.

The motor is $75\%$ efficient therefore $7kJ$ are lost in this way. It takes $mgh=12kJ$ to lift the car but the motor delivers $21kJ$ so another $9kJ$ are lost on top of this. Therefore the total lost is $16kJ$.

7. Form two simultaneous equations and solve by inspection if possible.

$2x+5y=P$

$3x+2y=Q$

$y=\frac{3P-2Q}{11}$

8. Assign a letter for the distance of the source from the midpoint and work out the two transit times in terms of this.

The time for the longer transit is half the distance between the detectors plus $z$ divided by $c$ and the time for the shorter transit is the same but with minus $z$. $4\times10^{-10}$ is the difference between them.

$\frac{1.5+z}{c}-\frac{1.5-z}{c}=4\times10^{-10}$

$z=\frac{c\times4\times10^{-10}}{2}=6cm$

9. Find $P$ as a function of $Q$ and $Q$ as a function of $R$ to find $P$ as a function of $R$.

$P\alpha Q^2$

$P=kQ^2$

$2=16k$

$k=\frac{1}{8}$

$P=\frac{Q^2}{8}$

$Q=\frac{c}{R}$

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

$c=10$

$Q=\frac{10}{R}$

$P=\frac{100}{8R^2}=\frac{25}{2R^2}$

10. Work through the equations in the answer in order to verify if they are correct.

$w+y$ can't be $240$ because of the presence of the unknown number of neutrons on the right hand side.

From the mass numbers $240=w+y+z$ revealing the answer $B$.

11. Spot the factorisation of the numerator and simplify with the minimum of writing. It is crucial to be able to spot the difference of two squares in these exams.

$2-\frac{x^2(3x+2)(3x-2)}{x^3(2-3x)}=2+\frac{3x+2}{x}=5+\frac{2}{x}$

12. Equate the power of the motor to the power required to lift the mass noting that the mass moves at a constant speed.

$P=\frac{mgh}{t}=mgv$

Also, the power delivered is $P=\frac{4IV}{5}$ because the system is $80\%$ efficient.

$\frac{200\times6}{5}=\frac{4\times12I}{5}$

$50\times6=12I$

$I=25A$

13. Rewrite each side as a power of 2 and equate the exponents.

$2^{3+2x+2x-3x}=2^{\frac{5}{2}}$

$3+x=\frac{5}{2}$

$x=-0.5$

14. Concentrate on the right-hand column (the left column's entries are all possible). Recall that a beta decay will increase the number of protons in the nucleus.

Since beta decay increases the atomic number by $1$ it is not possible for the atomic number to drop before the alpha decay which reveals answer $A$ as the impossible stage.

15. Create a two-way table

Adding the entries for 'German' we have $2X+3Y-35$.

16. Quickly spot that the mass of the atom and the nucleus are being taken as the same in this question. Therefore we require a ratio of volumes.

The ratio is $\frac{M}{V_a}\times\frac{V_n}{M}=\frac{r_n^3}{r_a^3}$

which gives the answer $A$.

17. Write the information given in terms of $n$ and solve.

$\frac{360}{n+3}=\frac{360}{n}-4$

$360n=360(n+3)-4n(n+3)$

$3\times360=4n(n+3)$

$270=n(n+3)$

At this stage I recommend looking at the answers and using trial and improvement rather than expanding and re-factorising, yielding $n=15$.

18. Starting from $c=f\lambda$ obtain $f$ and $\lambda$ from the graphs and substitute in.

$f=\frac{1}{Period}=\frac{1}{\frac{2}{3}(t_2-t_1)}$

$\lambda=2(x_2-x_1)$.

Combining these gives

$c=\frac{2(x_2-x_1)}{\frac{2}{3}(t_2-t_1)}$ or after simplification, $E$.

19. Draw a clear diagram taking care to correctly interpret 'from' in this context.

By alternate angles, $LRC=40^{\circ}$ and since the triangle is isosceles so $CLR=40^{\circ}$. From angles in a triangle, $RCL=100^{\circ}$ and so $NCL=80^{\circ}$ from angles on a straight line.

20. Current is the rate of flow of charge, so work out how much current is flowing and for how long.

$Q=It=\frac{Pt}{V}=\frac{150\times20\times60}{12}=15000C$

21. Draw a rough diagram of the clock and work out how many degrees the minute hand has turned through. You can apply angles around a point because the other angles are easy to work out from your knowledge of the clock face.

The angles between the $8$ and the $12$ and the $12$ and the $4$ are both $120^{\circ}$. Each hour the hour hand moves $30^{\circ}$ so in $40$ minutes it will have moved $20^{\circ}$. All that is left out of $360^{\circ}$ is the angle between the hands which is $100^{\circ}$.

22. Use conservation of momentum with the number of passenger carriages as an unknown and solve.

$2(390+210)=5(140+10X)$

$2\times600=5(140+10X)$

$240=140+10X$

$X=10$

23. Form an equation in $x$ and solve.

$\frac{x}{4+x}\times\frac{x-1}{3+x}=\frac{1}{3}$

$12+7x+x^2=3x^2-3x$

$2x^2-10x-12=0$

$(x-6)(x+1)=0$

$x=6$

24. Work out the quantities in statements 1 and 2. Your recollection of Newton's third law should reveal the accuracy of statement 3.

$KE=900J$

Rate of loss of GPE$=\frac{mgh}{t}=mgv=72\times5\times10=3600$

Newton's third law relates to two bodies and their equal and opposite forces. This is not the case for statement 3 because air resistance is between the parachutist and the air molecules whereas the gravity is between the parachutist and the earth. Therefore the only correct statement is 2.

25. Write the radius of the circle in terms of $x$ using Pythagoras' theorem and subtract the area of the square from the area of the semi-circle.

$r^2=x^2+(\frac{x}{2})^2=\frac{5x^2}{4}$

Shaded area=$\frac{\pi r^2}{2}-x^2=\frac{5\pi x^2}{8}-x^2=\frac{5\pi x^2-8x^2}{8}$

26. Let there be $N$ nuclei each of $X$ and $Y$. Work out how many of each have decayed, add the results and divide by the initial number.

After six hours, there are $\frac{N}{4}$ nuclei of $X$ and $\frac{N}{8}$ nuclei of $Y$. This means there are $\frac{3N}{4}+\frac{7N}{8}$ nuclei of $Z$ out of the original $2N$. Therefore the fraction is $\frac{\frac{6N}{8}+\frac{7N}{8}}{2N}=\frac{13}{16}$.

27. Working efficiently, find the area of the front and multiply by the length and the density.

There are two time-saving tips here. Firstly, evaluate $5^2-4^2$ as $3^2=9$ using Pythagorean triples. Secondly, the answers are fairly widely spread so you don't need to work out the final multiple of $\pi$ exactly.

$M=D\times V=8\times\pi(5^2-4^2)\times16=8\pi\times9\times16=9\pi\times128\approx1000\pi$

28. Work out the factor by which the kinetic energy has changed from the information given.

The KE has been multiplied by $\frac{4}{5}$ and $1.5^2$ which is a factor of $\frac{4}{5}\times(\frac{3}{2})^2=\frac{9}{5}=1.8$.

Part B

29. Divide the numerator and denominator by $\sqrt3$ to save time then expand and simplify without writing too many steps.

$1-(\frac{\sqrt3+1}{2\sqrt3-2})^2=1-(\frac{1}{2}(\frac{\sqrt3+1}{\sqrt3-1}))^2=1-(\frac{1}{2}(\frac{4+2\sqrt3}{2}))^2=1-(1+\frac{\sqrt3}{2})^2=-\frac{3}{4}-\sqrt3$

30. Work out the moment as a result of the horizontal section of the crane first by inspection then use this in an equation balancing moments for the second scenario.

The moment of the horizontal section is $1600\times10$.

$1600\times10+15\times400=2000(10+x)$ where $x$ is the distance the counterweight moves. Cancel two zeros and expand.

$160+60=200+20x$ so $x=1$. It should be clear that the counterweight moves to the right.

31. Find and list some solutions for each equation and read off how many are in the desired range.

$cos2x=\frac{1}{2}$ gives $x=30^{\circ}, 150^{\circ}, 210^{\circ}$ etc.

$sinx=-\frac{1}{2}$ gives $x=210^{\circ}$ etc.

These agree for the first time for $x=210^{\circ}$. There are two values of $x$ below this in the range which give solutions, so three in total.

32. $g$ is a constant so eliminate any curved graphs. The direction of the initial velocity and the fact that this direction must change reveal the only possible graph.

The velocity starts positive and ends negative, changing only via a straight line, so the answer is $A$.

33. Factorise fully and solve for the contents of the bracket being zero.

$3\times(3^x)^2-6\times(3^x)=3\times3^x(3^x-2)$ so $x=log_{3}2$.

34. Determine what is useful information in the question. There is no need to read all of the answers.

The aircraft is travelling at a constant velocity so there is no resultant force. Save time by marking $C$ and moving on.

35. Work out the radius of the circle from the length of the arc. Use this to find the area of the sector and the triangle.

$22\pi=2\pi r\times\frac{11}{12}$

$r=12mm$

$Area=\frac{11}{12}\pi\times12^2+\frac{1}{2}\times 30^2\times sin30=132\pi+\frac{30^2}{2^2}=132\pi+225$

36. Briefly note the format of the answers then equate clockwise and anti-clockwise moments for the bar.

$60g\times2=Fsin60\times4$. Keep $sin60$ intact.

$30g=Fsin60$

$F=\frac{300}{sin60}$

37. Write as a single logarithm using the laws of indices.

$log_2(\frac{(2^{3p})^3}{(\frac{1}{2})^{4q}})=log_2(2^{9p+4q})$

38. Identify the correct suvat equation quickly and try to avoid any unnecessary calculations.

$v^2=u^2+2as=40^2-40\times14.4=40^2-4\times144=40^2-24^2$

Save time at this stage by noting that this is a multiple of a Pythagorean triple $(3,4,5\times 8 $gives$ 24,32,40$ so the answer is $32$.

39. Differentiate once, set equal to zero and substitute values of $x$ or differentiate twice and use a process of elimination.

Method 1 - differentiating once

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

$12+4p+q=0$

$48+8p+q=0$

$4p=-36$ so $p=-9$ and substituting in gives $q=24$

Method 2 - differentiating twice

$\frac{d^2y}{dx^2}=6x+2p$ which is negative when $x=2$ only for $p=-9$ which gives the answer $D$.

40. Focus on what is important - the required tension in the string must be equal to the weight of the supported object regardless of what holds the other end of the rope.

Ignore the left hand mass and the rough surface. The required tension is $1g$ or $10N$.

41. Use $\frac{1}{2}absinC$ and maximise the following expression efficiently.

$\frac{1}{2}absinC=\frac{(8-3x)4x\sqrt{3}}{4}$

You can proceed by differentiating but it is quicker to note that this is a negative quadratic therefore the maximum will be for $x$ is the mean of the roots. Therefore $x=\frac{4}{3}$ and $Area=\frac{(8-4)\times16\sqrt{3}}{3\times4}=\frac{16\sqrt3}{3}$.

42. Work out the initial GPE and the final KE and evaluate the discrepancy

$GPE=mgh=4$

$KE=\frac{1}{2}mv^2=\frac{0.1\times8^2}{2}=3.2$ so the difference is $0.8J$

43. Only work out the required coefficient and note the large differences between the answers.

$\frac{dy}{dx}=18(2+3x)^5$. The coefficient is $18\times^5C_3\times27\times4=270\times72$. Note that the answers are widely spread so you don't need to evaluate this exactly to get $E$.

44. Read the question carefully, try to work out the initial downward velocity efficiently and apply suvat.

Since there is no air resistance, the initial downward velocity can be taken to be $13ms^{-1}$ because no energy is lost so the stone must fall with the same speed it started with.

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

$6=13t+5t^2$

$0=5t^2+13t-6=(5t-2)(t+3)$ therefore $t=\frac{2}{5}$. Remember that you may be able to speed up your quadratic factorisation by looking at the answers.

45. Apply the sum to infinity of a geometric series formula, rearrange and solve the resulting equation.

$\frac{1}{1-\frac{1}{2}sin2x}=\frac{4}{3}$

$3=4-2sin2x$

$sin2x=\frac{1}{2}$

$x=\frac{13}{12}\pi$ or $\frac{17}{12}\pi$

46. Work out the area efficiently and equate this to $mgh$ while keeping an eye on the answers to help avoid any unnecessary calculations.

$Area=\frac{192\times\frac{2}{5}}{2}=\frac{192}{5}$

You can save time here by not evaluating this exactly and just choosing 38.4 from the answers instead.

$38.4=mgh$

$h=\frac{3.84}{0.024}=\frac{3840}{24}=160$, once again by looking at the answers.

47. Work out the first four terms of the sequence, spot a solution (the sign of which is hinted at in the question) to the cubic and perform the addition without substituting into the cubic.

The first four terms are $2, 2p+3, 2p^2+3p+3, 2p^3+3p^2+3p+3$

From the question, $2p^3+3p^2+3p+3=-7$

This can only be negative if $p$ is negative and furthermore, $p$ is an integer. This is a time-pressured exam, so $p$ is unlikely to be a large negative integer. I started (and ended) my search with $p=-2$. The sum of the first four terms is $2-1+8-6+3-7=-1$. Remember that you have already evaluated the cubic as -7, so don't substitute $p=-2$ into it for a second time.

48. Work out the maximum frictional force from the first scenario then apply it to find the resultant.

The maximum frictional force is $\mu mgcos20$. This is equal to $mgsin20$ because the book slides down at a constant speed. Therefore $\mu=tan20$. The resultant force is $mgsin25-\mu mgcos25=mg(sin25-cos25tan20)$ giving the answer $H$.

49. Factorise (preferably mostly by inspection) and interpret the inequality taking care to notice the repeated root.

$\frac{(x-1)(x^2-5x+4)}{x}>0$

$\frac{(x-1)^2(x-4)}{x}>0$

This is the case for $x$ is large and positive and for $x$ large and negative. The curve cuts the $x$-axis at $x=0$, touches at $x=-1$ and cuts at $x=4$ and so is negative or $0$ for $0\leq x \leq 4$ and positive otherwise yielding the answer $A$.

50. Recall that frictional forces increase to match any opposing forces up to a maximum (which has not been achieved in this scenario).

The suitcase is not in limiting equilibrium so the frictional force up the plane must equal the force due to gravity resolved down the plane so the force is $mgsin\theta$.

51. Change $x$ accordingly, preferably by inspection.

The first transformation means that $x$ should be multiplied by $2$. For the translation, $x$ should be replaced by $x+\frac{\pi}{4}$ so $sinx$ has become $sin(2x+\frac{\pi}{2})$.

52. Use Force=Rate of change of momentum taking care with the signs of the velocities.

The momentum change is $mv+mu$ because the velocity changes direction. This takes a time $t_2-t_1$ so the force is $\frac{mv+mu}{t_2-t_1}$ which rearranges to give the answer $B$.

53. Set the product of the gradients to $-1$ and either solve for $p$ or substitute likely candidates for $p$ from the answers.

$(2p^2-p)(p-2)=-1$

$p(2p-1)(p-2)=-1$

Approach 1 - substitute in likely values of $p$

When $p=2$, we have $0$. For $p=1.75$ we have $-1.75\times2.5\times0.25=-\frac{35}{32}$. For $p=1.5$ we have $-1.5\times2\times0.5=-1.5$ so the answer must be $p\approx 1.75$.

Approach 2 - expand and solve the resulting cubic

$2p^3-5p^2+2p+1=0$

By inspection, $p=1$ is a solution. It is a good idea to be able to factorise by inspection:

$(p-1)(2p^2-3p-1)=0$ and solving the quadratic gives $p=\frac{3\pm\sqrt{17}}{2}\approx\frac{3\pm4}{2}$ which gives $p\approx1.75$.

54. Use a suvat equation to relate the speed and height and use a process of elimination.

The flight takes $0.8$ seconds so the upward part takes $0.4$ seconds and starts with velocity $v$.

$s=ut+\frac{1}{2}at^2=0.4v-5\times(0.4)^2=0.4(v-5\times0.4)=0.4\times(v-2)$

The only answer which this works for is $A$.

Section 2

1a. $g$ is the acceleration and hence the gradient of the graph. It is changing which eliminates all but three of the possibilities. From here, substituting values for $t$ reveals the correct answer.

$a=-0.4t$

By integration, $v=-0.2t^2+c$ but $c=0$ because the ball starts at rest.

$v=-0.2t^2$

Substituting in $t=0.6$ and $t=1$ gives $0.072$ and $0.2$ respectively which gives the answer $B$.

1b. Integrate the expression for velocity to get an expression for distance.

$s=-\frac{0.2t^3}{3}+d$ where $d$ is the constant of integration/the height at $t=0$. Let $s=0$

$15d=t^3$

1c. Use an energy approach to ascertain how the peaks should change over time.

The GPE at the top of each bounce is $mgh$ but $g$ is increasing with time. Since energy is conserved and no other forces act on the ball, there cannot be any increase in GPE. Since $g$ is increasing with time this must mean that $h$ decreases over time, so the only possible graph is $P$. As an exercise you might like to try to evaluate the time interval between the first bounce and the next maximum as $(\sqrt2-1)\times^3\sqrt{15d}$ or the height of this maximum as $d(4\sqrt2-5)$, both of which are in agreement with graph $P$ which I believe may be drawn to scale which is a nice touch.

1d. Substitute in the units for mass and acceleration.

This simply reveals the answer as $C$.

1e. Rearrange for $X$ and substitute units in.

The units of $x$ are $\frac{kgms^{-2}}{kgm^{-3}m^2s^{-2}m^2}$. The numerator and denominator cancel entirely so $X$ has no units.

2a. Only one of the devices obeys Ohm's law.

$V=IR$ predicts a straight line for all currents, so device $Y$ is the resistor.

2b. As filament lamps draw more current they heat up which gradually increases their resistance.

This means that we would expect the line to curve over gradually since we gain less and less current for the same increase in voltage. Therefore the filament lamp is device $X$.

2c. Draw a circuit diagram and find the values of each of the currents through the two components.

The circuit should be a cell connected to a parallel combination of a lamp and a resistor. Label the current in the lamp as $I_l$ and in the resistor as $I_r$.

$I_r=\frac{V}{R}=\frac{6}{8}=0.75$

$I_l=1.5$ from the graph knowing that it has a voltage of $6V$ across it.

$I_{total}=I_l+I_r=2.25A$

2d. Use early knowledge of series circuits.

Components in series receive the same current (there is nowhere else for it to go but through each component).

2e. For this fairly high voltage, it should be possible to deduce the voltage across W then the voltage across the resistor followed by the current for the circuit. You then have all the information you need to use $P=IV$ for W.

From the graph, the only current for which the voltages across W and Y sum to $6V$ is $0.5A$ and in this case the voltage across W is $2V$. Using $P=IV=0.5\times2$ we have that the power dissipated is roughly $1W$.

3a. Hooke's law is obeyed for straight lines on this graph. Strain is extension divided by original length.

This means that Hooke's law is obeyed up to P. The fracture is at Q where the extension is $0.05m$. Strain=$\frac{extension}{original length}=\frac{0.05}{0.5}=0.1$ giving the answer $D$.

3b. Work out the area under the graph (I did this by counting large boxes and multiplying by the area of one large box).

There are roughly $6$ large boxes under the graph. Each large box represents $10\times0.01=0.1J$ so the total energy stored is $0.6J$.

3c. Halving the length of the rope halves its spring constant and its extension. Alternatively, you now have half the rope which is being stretched to the same limit as the whole rope.

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

The spring constant has halved but so has the extension which results in $U$ halving. Therefore $\frac{U}{2}=\frac{1}{2}mv_{max}^2$ giving the answer $B$. Alternatively, you can think of half the rope being able to store half the energy and work from there.

3d. Two ropes means you have twice the energy stored.

$U$ has now doubled which means that $v_{max}$ has increased by a factor of $\sqrt2$.

4a. Coherence is a necessary condition for interference from a double slit. The path difference determines what type of interference. An integer number of wavelengths will give rise to constructive interference.

If the path difference is an integer number of wavelengths, the waves will arrive in phase and constructively interfere. If the path difference is an integer number of wavelengths plus half a wavelength then the waves will arrive out of phase and destructively interfere. The maxima near the centre of the pattern will be brightest because a smaller proportion of the light is diffracted as the angle it is diffracted through increases.

4b. The introduction of the material introduces an extra path difference for one slit only. In order for e.g. a given maximum to recover from this change the light must now be travelling at an angle to what it was previously.

The time taken to traverse the material is $\frac{2\times300\times10^{-9}}{c}=\frac{2\times3\times10^{-7}}{3\times10^{8}}$. Without the material the time would have been $\frac{3\times10^{-7}}{3\times10^{8}}$. The difference between these is $\frac{3\times10^{-7}}{3\times10^{8}}$ so the extra path difference introduced is $300nm$. This means that the points which were maxima are now minima and vice versa so the pattern shifts.

4c. Work out the wavelength and deduce whether interference will occur and what distance there is between maxima at the receiver.

$c=f\lambda$

$3\times10^8=600\times10^6\times\lambda$ so $\lambda=\frac{1}{2}m$

Since the antennas are only $0.5m$ apart, diffraction effects will occur and could easily lead to a destructive interference at the receiver, so the answer is $E$.

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 1

Part A

1. Spot a simplification then expand the numerator and denominator whilst writing the minimum of working.

2. Read the question carefully and work out the area in squares, converting to distance afterwards.

3. Take everything to the left hand side and factorise noting that we have a positive quadratic.

4. Note that one of the statements is clearly incorrect and the other two are linked.

5. Rearrange, preferably by inspection.

6. Read the question carefully. The energy is lost via two separate means.

7. Form two simultaneous equations and solve by inspection if possible.

8. Assign a letter for the distance of the source from the midpoint and work out the two transit times in terms of this.

9. Find $P$ as a function of $Q$ and $Q$ as a function of $R$ to find $P$ as a function of $R$.

10. Work through the equations in the answer in order to verify if they are correct.

11. Spot the factorisation of the numerator and simplify with the minimum of writing.

12. Equate the power of the motor to the power required to lift the mass noting that the mass moves at a constant speed.

13. Rewrite each side as a power of 2 and equate the exponents.

14. Concentrate on the right-hand column (the left column's entries are all possible). Recall that a beta decay will increase the number of protons in the nucleus.

15. Create a two-way table

16. Quickly spot that the mass of the atom and the nucleus are being taken as the same in this question. Therefore we require a ratio of volumes.

17. Write the information given in terms of $n$ and solve.

18. Starting from $c=f\lambda$ obtain $f$ and $\lambda$ from the graphs and substitute in.

19. Draw a clear diagram taking care to correctly interpret 'from' in this context.

20. Current is the rate of flow of charge, so work out how much current is flowing and for how long.

21. Draw a rough diagram of the clock and work out how many degrees the minute hand has turned through. You can apply angles around a point because the other angles are easy to work out from your knowledge of the clock face.

22. Use conservation of momentum with the number of passenger carriages as an unknown and solve.

23. Form an equation in $x$ and solve.

24. Work out the quantities in statements 1 and 2. Your recollection of Newton's third law should reveal the accuracy of statement 3.

25. Write the radius of the circle in terms of $x$ using Pythagoras' theorem and subtract the area of the square from the area of the semi-circle.

26. Let there be $N$ nuclei each of $X$ and $Y$. Work out how many of each have decayed, add the results and divide by the initial number.

27. Working efficiently, find the area of the front and multiply by the length and the density.

28. Work out the factor by which the kinetic energy has changed from the information given.

Part B

29. Divide the numerator and denominator by $\sqrt3$ to save time then expand and simplify without writing too many steps.

30. Work out the moment as a result of the horizontal section of the crane first by inspection then use this in an equation balancing moments for the second scenario.

31. Find and list some solutions for each equation and read off how many are in the desired range.

32. $g$ is a constant so eliminate any curved graphs. The direction of the initial velocity and the fact that this direction must change reveal the only possible graph.

33. Factorise fully and solve for the contents of the bracket being zero.

34. Determine what is useful information in the question. There is no need to read all of the answers.

35. Work out the radius of the circle from the length of the arc. Use this to find the area of the sector and the triangle.

36. Briefly note the format of the answers then equate clockwise and anti-clockwise moments for the bar.

37. Write as a single logarithm using the laws of indices.

38. Identify the correct suvat equation quickly and try to avoid any unnecessary calculations.

39. Differentiate once, set equal to zero and substitute values for $x$ or differentiate twice and use a process of elimination.

40. Focus on what is important - the required tension in the string must be equal to the weight of the supported object regardless of what holds the other end of the rope.

41. Use $\frac{1}{2}absinC$ and maximise the following expression efficiently.

42. Work out the initial GPE and the final KE and evaluate the discrepancy

43. Only work out the required coefficient and note the large differences between the answers.

44. Read the question carefully, try to work out the initial downward velocity efficiently and apply suvat.

45. Apply the sum to infinity of a geometric series formula, rearrange and solve the resulting equation.

46. Work out the area efficiently and equate this to $mgh$ while keeping an eye on the answers to help avoid any unnecessary calculations.

47. Work out the first four terms of the sequence, spot a solution (the sign of which is hinted at in the question) to the cubic and perform the addition without substituting into the cubic.

48. Work out the maximum frictional force from the first scenario then apply it to find the resultant.

49. Factorise (preferably mostly by inspection) and interpret the inequality taking care to notice the repeated root.

50. Recall that frictional forces increase to match any opposing forces up to a maximum (which has not been achieved in this scenario).

51. Change $x$ accordingly, preferably by inspection.

52. Use Force=Rate of change of momentum taking care with the signs of the velocities.

53. Set the product of the gradients to $-1$ and either solve for $p$ or substitute likely candidates for $p$ from the answers.

54. Use a suvat equation to relate the speed and height and use a process of elimination.

Section 2

1a. $g$ is the acceleration and hence the gradient of the graph. It is changing which eliminates all but three of the possibilities. From here, substituting values for $t$ reveals the correct answer.

1b. Integrate the expression for velocity to obtain an expression for distance.

1c. Use an energy approach to ascertain how the peaks should change over time.

1d. Substitute in the units for mass and acceleration.

1e. Rearrange for $X$ and substitute units in.

2a. Only one of the devices obeys Ohm's law.

2b. As filament lamps draw more current they heat up which gradually increases their resistance.

2c. Draw a circuit diagram and find the values of each of the currents through the two components.

2d. Use early knowledge of series circuits.

2e. For this fairly high voltage, it should be possible to deduce the voltage across W then the voltage across the resistor followed by the current for the circuit. You then have all the information you need to use $P=IV$ for W.

3a. Hooke's law is obeyed for straight lines on this graph. Strain is extension divided by original length.

3b. Work out the area under the graph (I did this by counting large boxes and multiplying by the area of one large box).

3c. Halving the length of the rope halves its spring constant and its extension. Alternatively, you now have half the rope which is being stretched to the same limit as the whole rope.

3d. Two ropes means you have twice the energy stored.

4a. Coherence is a necessary condition for interference from a double slit. The path difference determines what type of interference. An integer number of wavelengths will give rise to constructive interference.

4b. The introduction of the material introduces an extra path difference for one slit only. In order for e.g. a given maximum to recover from this change the light must now be travelling at an angle to what it was previously.

4c. Work out the wavelength and deduce whether interference will occur and what distance there is between maxima at the receiver.

Full solutions

Section 1

Part A

1. Spot a simplification then expand the numerator and denominator whilst writing the minimum of working.

Divide each bracket by $\sqrt3$

$\frac{(2+1)^2}{(2-1)^2}=9$

2. Read the question carefully and work out the area in squares, converting to distance afterwards.

We are only concerned with the deceleration. There are $10$ squares under the deceleration part of the graph and each square is worth $5\times10=50m$ giving $500m$

3. Take everything to the left hand side and factorise noting that we have a positive quadratic.

$2x^2+x-15\geq 0$

$(2x-5)(x+3)\geq 0$

$x\leq -3$ or $x\geq \frac{5}{2}$

4. Note that one of the statements is clearly incorrect and the other two are linked.

If something expands or contracts then its mass is still conserved. Heating expands the water which decreases its density which is the reason why it floats above the cooler water.

5. Rearrange, preferably by inspection.

As an interim step you could write

$\frac{x}{2}-1=\pm\sqrt{\frac{y+5}{3}}$ and spot the answer from there. I don't recommend using too many steps.

6. Read the question carefully. The energy is lost via two separate means.

The motor is $75\%$ efficient therefore $7kJ$ are lost in this way. It takes $mgh=12kJ$ to lift the car but the motor delivers $21kJ$ so another $9kJ$ are lost on top of this. Therefore the total lost is $16kJ$.

7. Form two simultaneous equations and solve by inspection if possible.

$2x+5y=P$

$3x+2y=Q$

$y=\frac{3P-2Q}{11}$

8. Assign a letter for the distance of the source from the midpoint and work out the two transit times in terms of this.

The time for the longer transit is half the distance between the detectors plus $z$ divided by $c$ and the time for the shorter transit is the same but with minus $z$. $4\times10^{-10}$ is the difference between them.

$\frac{1.5+z}{c}-\frac{1.5-z}{c}=4\times10^{-10}$

$z=\frac{c\times4\times10^{-10}}{2}=6cm$

9. Find $P$ as a function of $Q$ and $Q$ as a function of $R$ to find $P$ as a function of $R$.

$P\alpha Q^2$

$P=kQ^2$

$2=16k$

$k=\frac{1}{8}$

$P=\frac{Q^2}{8}$

$Q=\frac{c}{R}$

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

$c=10$

$Q=\frac{10}{R}$

$P=\frac{100}{8R^2}=\frac{25}{2R^2}$

10. Work through the equations in the answer in order to verify if they are correct.

$w+y$ can't be $240$ because of the presence of the unknown number of neutrons on the right hand side.

From the mass numbers $240=w+y+z$ revealing the answer $B$.

11. Spot the factorisation of the numerator and simplify with the minimum of writing. It is crucial to be able to spot the difference of two squares in these exams.

$2-\frac{x^2(3x+2)(3x-2)}{x^3(2-3x)}=2+\frac{3x+2}{x}=5+\frac{2}{x}$

12. Equate the power of the motor to the power required to lift the mass noting that the mass moves at a constant speed.

$P=\frac{mgh}{t}=mgv$

Also, the power delivered is $P=\frac{4IV}{5}$ because the system is $80\%$ efficient.

$\frac{200\times6}{5}=\frac{4\times12I}{5}$

$50\times6=12I$

$I=25A$

13. Rewrite each side as a power of 2 and equate the exponents.

$2^{3+2x+2x-3x}=2^{\frac{5}{2}}$

$3+x=\frac{5}{2}$

$x=-0.5$

14. Concentrate on the right-hand column (the left column's entries are all possible). Recall that a beta decay will increase the number of protons in the nucleus.

Since beta decay increases the atomic number by $1$ it is not possible for the atomic number to drop before the alpha decay which reveals answer $A$ as the impossible stage.

15. Create a two-way table

Adding the entries for 'German' we have $2X+3Y-35$.

16. Quickly spot that the mass of the atom and the nucleus are being taken as the same in this question. Therefore we require a ratio of volumes.

The ratio is $\frac{M}{V_a}\times\frac{V_n}{M}=\frac{r_n^3}{r_a^3}$

which gives the answer $A$.

17. Write the information given in terms of $n$ and solve.

$\frac{360}{n+3}=\frac{360}{n}-4$

$360n=360(n+3)-4n(n+3)$

$3\times360=4n(n+3)$

$270=n(n+3)$

At this stage I recommend looking at the answers and using trial and improvement rather than expanding and re-factorising, yielding $n=15$.

18. Starting from $c=f\lambda$ obtain $f$ and $\lambda$ from the graphs and substitute in.

$f=\frac{1}{Period}=\frac{1}{\frac{2}{3}(t_2-t_1)}$

$\lambda=2(x_2-x_1)$.

Combining these gives

$c=\frac{2(x_2-x_1)}{\frac{2}{3}(t_2-t_1)}$ or after simplification, $E$.

19. Draw a clear diagram taking care to correctly interpret 'from' in this context.

By alternate angles, $LRC=40^{\circ}$ and since the triangle is isosceles so $CLR=40^{\circ}$. From angles in a triangle, $RCL=100^{\circ}$ and so $NCL=80^{\circ}$ from angles on a straight line.

20. Current is the rate of flow of charge, so work out how much current is flowing and for how long.

$Q=It=\frac{Pt}{V}=\frac{150\times20\times60}{12}=15000C$

21. Draw a rough diagram of the clock and work out how many degrees the minute hand has turned through. You can apply angles around a point because the other angles are easy to work out from your knowledge of the clock face.

The angles between the $8$ and the $12$ and the $12$ and the $4$ are both $120^{\circ}$. Each hour the hour hand moves $30^{\circ}$ so in $40$ minutes it will have moved $20^{\circ}$. All that is left out of $360^{\circ}$ is the angle between the hands which is $100^{\circ}$.

22. Use conservation of momentum with the number of passenger carriages as an unknown and solve.

$2(390+210)=5(140+10X)$

$2\times600=5(140+10X)$

$240=140+10X$

$X=10$

23. Form an equation in $x$ and solve.

$\frac{x}{4+x}\times\frac{x-1}{3+x}=\frac{1}{3}$

$12+7x+x^2=3x^2-3x$

$2x^2-10x-12=0$

$(x-6)(x+1)=0$

$x=6$

24. Work out the quantities in statements 1 and 2. Your recollection of Newton's third law should reveal the accuracy of statement 3.

$KE=900J$

Rate of loss of GPE$=\frac{mgh}{t}=mgv=72\times5\times10=3600$

Newton's third law relates to two bodies and their equal and opposite forces. This is not the case for statement 3 because air resistance is between the parachutist and the air molecules whereas the gravity is between the parachutist and the earth. Therefore the only correct statement is 2.

25. Write the radius of the circle in terms of $x$ using Pythagoras' theorem and subtract the area of the square from the area of the semi-circle.

$r^2=x^2+(\frac{x}{2})^2=\frac{5x^2}{4}$

Shaded area=$\frac{\pi r^2}{2}-x^2=\frac{5\pi x^2}{8}-x^2=\frac{5\pi x^2-8x^2}{8}$

26. Let there be $N$ nuclei each of $X$ and $Y$. Work out how many of each have decayed, add the results and divide by the initial number.

After six hours, there are $\frac{N}{4}$ nuclei of $X$ and $\frac{N}{8}$ nuclei of $Y$. This means there are $\frac{3N}{4}+\frac{7N}{8}$ nuclei of $Z$ out of the original $2N$. Therefore the fraction is $\frac{\frac{6N}{8}+\frac{7N}{8}}{2N}=\frac{13}{16}$.

27. Working efficiently, find the area of the front and multiply by the length and the density.

There are two time-saving tips here. Firstly, evaluate $5^2-4^2$ as $3^2=9$ using Pythagorean triples. Secondly, the answers are fairly widely spread so you don't need to work out the final multiple of $\pi$ exactly.

$M=D\times V=8\times\pi(5^2-4^2)\times16=8\pi\times9\times16=9\pi\times128\approx1000\pi$

28. Work out the factor by which the kinetic energy has changed from the information given.

The KE has been multiplied by $\frac{4}{5}$ and $1.5^2$ which is a factor of $\frac{4}{5}\times(\frac{3}{2})^2=\frac{9}{5}=1.8$.

Part B

29. Divide the numerator and denominator by $\sqrt3$ to save time then expand and simplify without writing too many steps.

$1-(\frac{\sqrt3+1}{2\sqrt3-2})^2=1-(\frac{1}{2}(\frac{\sqrt3+1}{\sqrt3-1}))^2=1-(\frac{1}{2}(\frac{4+2\sqrt3}{2}))^2=1-(1+\frac{\sqrt3}{2})^2=-\frac{3}{4}-\sqrt3$

30. Work out the moment as a result of the horizontal section of the crane first by inspection then use this in an equation balancing moments for the second scenario.

The moment of the horizontal section is $1600\times10$.

$1600\times10+15\times400=2000(10+x)$ where $x$ is the distance the counterweight moves. Cancel two zeros and expand.

$160+60=200+20x$ so $x=1$. It should be clear that the counterweight moves to the right.

31. Find and list some solutions for each equation and read off how many are in the desired range.

$cos2x=\frac{1}{2}$ gives $x=30^{\circ}, 150^{\circ}, 210^{\circ}$ etc.

$sinx=-\frac{1}{2}$ gives $x=210^{\circ}$ etc.

These agree for the first time for $x=210^{\circ}$. There are two values of $x$ below this in the range which give solutions, so three in total.

32. $g$ is a constant so eliminate any curved graphs. The direction of the initial velocity and the fact that this direction must change reveal the only possible graph.

The velocity starts positive and ends negative, changing only via a straight line, so the answer is $A$.

33. Factorise fully and solve for the contents of the bracket being zero.

$3\times(3^x)^2-6\times(3^x)=3\times3^x(3^x-2)$ so $x=log_{3}2$.

34. Determine what is useful information in the question. There is no need to read all of the answers.

The aircraft is travelling at a constant velocity so there is no resultant force. Save time by marking $C$ and moving on.

35. Work out the radius of the circle from the length of the arc. Use this to find the area of the sector and the triangle.

$22\pi=2\pi r\times\frac{11}{12}$

$r=12mm$

$Area=\frac{11}{12}\pi\times12^2+\frac{1}{2}\times 30^2\times sin30=132\pi+\frac{30^2}{2^2}=132\pi+225$

36. Briefly note the format of the answers then equate clockwise and anti-clockwise moments for the bar.

$60g\times2=Fsin60\times4$. Keep $sin60$ intact.

$30g=Fsin60$

$F=\frac{300}{sin60}$

37. Write as a single logarithm using the laws of indices.

$log_2(\frac{(2^{3p})^3}{(\frac{1}{2})^{4q}})=log_2(2^{9p+4q})$

38. Identify the correct suvat equation quickly and try to avoid any unnecessary calculations.

$v^2=u^2+2as=40^2-40\times14.4=40^2-4\times144=40^2-24^2$

Save time at this stage by noting that this is a multiple of a Pythagorean triple $(3,4,5\times 8 $gives$ 24,32,40$ so the answer is $32$.

39. Differentiate once, set equal to zero and substitute values of $x$ or differentiate twice and use a process of elimination.

Method 1 - differentiating once

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

$12+4p+q=0$

$48+8p+q=0$

$4p=-36$ so $p=-9$ and substituting in gives $q=24$

Method 2 - differentiating twice

$\frac{d^2y}{dx^2}=6x+2p$ which is negative when $x=2$ only for $p=-9$ which gives the answer $D$.

40. Focus on what is important - the required tension in the string must be equal to the weight of the supported object regardless of what holds the other end of the rope.

Ignore the left hand mass and the rough surface. The required tension is $1g$ or $10N$.

41. Use $\frac{1}{2}absinC$ and maximise the following expression efficiently.

$\frac{1}{2}absinC=\frac{(8-3x)4x\sqrt{3}}{4}$

You can proceed by differentiating but it is quicker to note that this is a negative quadratic therefore the maximum will be for $x$ is the mean of the roots. Therefore $x=\frac{4}{3}$ and $Area=\frac{(8-4)\times16\sqrt{3}}{3\times4}=\frac{16\sqrt3}{3}$.

42. Work out the initial GPE and the final KE and evaluate the discrepancy

$GPE=mgh=4$

$KE=\frac{1}{2}mv^2=\frac{0.1\times8^2}{2}=3.2$ so the difference is $0.8J$

43. Only work out the required coefficient and note the large differences between the answers.

$\frac{dy}{dx}=18(2+3x)^5$. The coefficient is $18\times^5C_3\times27\times4=270\times72$. Note that the answers are widely spread so you don't need to evaluate this exactly to get $E$.

44. Read the question carefully, try to work out the initial downward velocity efficiently and apply suvat.

Since there is no air resistance, the initial downward velocity can be taken to be $13ms^{-1}$ because no energy is lost so the stone must fall with the same speed it started with.

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

$6=13t+5t^2$

$0=5t^2+13t-6=(5t-2)(t+3)$ therefore $t=\frac{2}{5}$. Remember that you may be able to speed up your quadratic factorisation by looking at the answers.

45. Apply the sum to infinity of a geometric series formula, rearrange and solve the resulting equation.

$\frac{1}{1-\frac{1}{2}sin2x}=\frac{4}{3}$

$3=4-2sin2x$

$sin2x=\frac{1}{2}$

$x=\frac{13}{12}\pi$ or $\frac{17}{12}\pi$

46. Work out the area efficiently and equate this to $mgh$ while keeping an eye on the answers to help avoid any unnecessary calculations.

$Area=\frac{192\times\frac{2}{5}}{2}=\frac{192}{5}$

You can save time here by not evaluating this exactly and just choosing 38.4 from the answers instead.

$38.4=mgh$

$h=\frac{3.84}{0.024}=\frac{3840}{24}=160$, once again by looking at the answers.

47. Work out the first four terms of the sequence, spot a solution (the sign of which is hinted at in the question) to the cubic and perform the addition without substituting into the cubic.

The first four terms are $2, 2p+3, 2p^2+3p+3, 2p^3+3p^2+3p+3$

From the question, $2p^3+3p^2+3p+3=-7$

This can only be negative if $p$ is negative and furthermore, $p$ is an integer. This is a time-pressured exam, so $p$ is unlikely to be a large negative integer. I started (and ended) my search with $p=-2$. The sum of the first four terms is $2-1+8-6+3-7=-1$. Remember that you have already evaluated the cubic as -7, so don't substitute $p=-2$ into it for a second time.

48. Work out the maximum frictional force from the first scenario then apply it to find the resultant.

The maximum frictional force is $\mu mgcos20$. This is equal to $mgsin20$ because the book slides down at a constant speed. Therefore $\mu=tan20$. The resultant force is $mgsin25-\mu mgcos25=mg(sin25-cos25tan20)$ giving the answer $H$.

49. Factorise (preferably mostly by inspection) and interpret the inequality taking care to notice the repeated root.

$\frac{(x-1)(x^2-5x+4)}{x}>0$

$\frac{(x-1)^2(x-4)}{x}>0$

This is the case for $x$ is large and positive and for $x$ large and negative. The curve cuts the $x$-axis at $x=0$, touches at $x=-1$ and cuts at $x=4$ and so is negative or $0$ for $0\leq x \leq 4$ and positive otherwise yielding the answer $A$.

50. Recall that frictional forces increase to match any opposing forces up to a maximum (which has not been achieved in this scenario).

The suitcase is not in limiting equilibrium so the frictional force up the plane must equal the force due to gravity resolved down the plane so the force is $mgsin\theta$.

51. Change $x$ accordingly, preferably by inspection.

The first transformation means that $x$ should be multiplied by $2$. For the translation, $x$ should be replaced by $x+\frac{\pi}{4}$ so $sinx$ has become $sin(2x+\frac{\pi}{2})$.

52. Use Force=Rate of change of momentum taking care with the signs of the velocities.

The momentum change is $mv+mu$ because the velocity changes direction. This takes a time $t_2-t_1$ so the force is $\frac{mv+mu}{t_2-t_1}$ which rearranges to give the answer $B$.

53. Set the product of the gradients to $-1$ and either solve for $p$ or substitute likely candidates for $p$ from the answers.

$(2p^2-p)(p-2)=-1$

$p(2p-1)(p-2)=-1$

Approach 1 - substitute in likely values of $p$

When $p=2$, we have $0$. For $p=1.75$ we have $-1.75\times2.5\times0.25=-\frac{35}{32}$. For $p=1.5$ we have $-1.5\times2\times0.5=-1.5$ so the answer must be $p\approx 1.75$.

Approach 2 - expand and solve the resulting cubic

$2p^3-5p^2+2p+1=0$

By inspection, $p=1$ is a solution. It is a good idea to be able to factorise by inspection:

$(p-1)(2p^2-3p-1)=0$ and solving the quadratic gives $p=\frac{3\pm\sqrt{17}}{2}\approx\frac{3\pm4}{2}$ which gives $p\approx1.75$.

54. Use a suvat equation to relate the speed and height and use a process of elimination.

The flight takes $0.8$ seconds so the upward part takes $0.4$ seconds and starts with velocity $v$.

$s=ut+\frac{1}{2}at^2=0.4v-5\times(0.4)^2=0.4(v-5\times0.4)=0.4\times(v-2)$

The only answer which this works for is $A$.

Section 2

1a. $g$ is the acceleration and hence the gradient of the graph. It is changing which eliminates all but three of the possibilities. From here, substituting values for $t$ reveals the correct answer.

$a=-0.4t$

By integration, $v=-0.2t^2+c$ but $c=0$ because the ball starts at rest.

$v=-0.2t^2$

Substituting in $t=0.6$ and $t=1$ gives $0.072$ and $0.2$ respectively which gives the answer $B$.

1b. Integrate the expression for velocity to get an expression for distance.

$s=-\frac{0.2t^3}{3}+d$ where $d$ is the constant of integration/the height at $t=0$. Let $s=0$

$15d=t^3$

1c. Use an energy approach to ascertain how the peaks should change over time.

The GPE at the top of each bounce is $mgh$ but $g$ is increasing with time. Since energy is conserved and no other forces act on the ball, there cannot be any increase in GPE. Since $g$ is increasing with time this must mean that $h$ decreases over time, so the only possible graph is $P$. As an exercise you might like to try to evaluate the time interval between the first bounce and the next maximum as $(\sqrt2-1)\times^3\sqrt{15d}$ or the height of this maximum as $d(4\sqrt2-5)$, both of which are in agreement with graph $P$ which I believe may be drawn to scale which is a nice touch.

1d. Substitute in the units for mass and acceleration.

This simply reveals the answer as $C$.

1e. Rearrange for $X$ and substitute units in.

The units of $x$ are $\frac{kgms^{-2}}{kgm^{-3}m^2s^{-2}m^2}$. The numerator and denominator cancel entirely so $X$ has no units.

2a. Only one of the devices obeys Ohm's law.

$V=IR$ predicts a straight line for all currents, so device $Y$ is the resistor.

2b. As filament lamps draw more current they heat up which gradually increases their resistance.

This means that we would expect the line to curve over gradually since we gain less and less current for the same increase in voltage. Therefore the filament lamp is device $X$.

2c. Draw a circuit diagram and find the values of each of the currents through the two components.

The circuit should be a cell connected to a parallel combination of a lamp and a resistor. Label the current in the lamp as $I_l$ and in the resistor as $I_r$.

$I_r=\frac{V}{R}=\frac{6}{8}=0.75$

$I_l=1.5$ from the graph knowing that it has a voltage of $6V$ across it.

$I_{total}=I_l+I_r=2.25A$

2d. Use early knowledge of series circuits.

Components in series receive the same current (there is nowhere else for it to go but through each component).

2e. For this fairly high voltage, it should be possible to deduce the voltage across W then the voltage across the resistor followed by the current for the circuit. You then have all the information you need to use $P=IV$ for W.

From the graph, the only current for which the voltages across W and Y sum to $6V$ is $0.5A$ and in this case the voltage across W is $2V$. Using $P=IV=0.5\times2$ we have that the power dissipated is roughly $1W$.

3a. Hooke's law is obeyed for straight lines on this graph. Strain is extension divided by original length.

This means that Hooke's law is obeyed up to P. The fracture is at Q where the extension is $0.05m$. Strain=$\frac{extension}{original length}=\frac{0.05}{0.5}=0.1$ giving the answer $D$.

3b. Work out the area under the graph (I did this by counting large boxes and multiplying by the area of one large box).

There are roughly $6$ large boxes under the graph. Each large box represents $10\times0.01=0.1J$ so the total energy stored is $0.6J$.

3c. Halving the length of the rope halves its spring constant and its extension. Alternatively, you now have half the rope which is being stretched to the same limit as the whole rope.

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

The spring constant has halved but so has the extension which results in $U$ halving. Therefore $\frac{U}{2}=\frac{1}{2}mv_{max}^2$ giving the answer $B$. Alternatively, you can think of half the rope being able to store half the energy and work from there.

3d. Two ropes means you have twice the energy stored.

$U$ has now doubled which means that $v_{max}$ has increased by a factor of $\sqrt2$.

4a. Coherence is a necessary condition for interference from a double slit. The path difference determines what type of interference. An integer number of wavelengths will give rise to constructive interference.

If the path difference is an integer number of wavelengths, the waves will arrive in phase and constructively interfere. If the path difference is an integer number of wavelengths plus half a wavelength then the waves will arrive out of phase and destructively interfere. The maxima near the centre of the pattern will be brightest because a smaller proportion of the light is diffracted as the angle it is diffracted through increases.

4b. The introduction of the material introduces an extra path difference for one slit only. In order for e.g. a given maximum to recover from this change the light must now be travelling at an angle to what it was previously.

The time taken to traverse the material is $\frac{2\times300\times10^{-9}}{c}=\frac{2\times3\times10^{-7}}{3\times10^{8}}$. Without the material the time would have been $\frac{3\times10^{-7}}{3\times10^{8}}$. The difference between these is $\frac{3\times10^{-7}}{3\times10^{8}}$ so the extra path difference introduced is $300nm$. This means that the points which were maxima are now minima and vice versa so the pattern shifts.

4c. Work out the wavelength and deduce whether interference will occur and what distance there is between maxima at the receiver.

$c=f\lambda$

$3\times10^8=600\times10^6\times\lambda$ so $\lambda=\frac{1}{2}m$

Since the antennas are only $0.5m$ apart, diffraction effects will occur and could easily lead to a destructive interference at the receiver, so the answer is $E$.