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. Make a two-way table.

2. Know and apply how alpha and beta decays affect the number of protons.

3. Write an equation equating the volume and twice the surface area.

4. Apply $V=IR$ to work out the voltage of the cell and once more to determine the current.

5. Work out the gradient in terms of $p$ and equate it to the gradient of the line.

6. Apply $c=f\lambda$ whilst taking care with which side of the visible spectrum UV is on.

7. Define an unknown for the height of the rectangle.

8. Use percentage multipliers.

9. Recall that the volume scale factor is the linear scale factor cubed.

10. Use percentage multipliers and $P=IV$.

11. Sketch the lines and find their intercepts and $x$-co-ordinate of their intersection.

12. Form two equations and solve for $v$.

13. Work out the mass of the full-sized pillar, converting to $cm$. Divide this by the cube of the scale factor since mass is proportional to volume.

14. Observe the background count then take care to choose the correctly halved count rate.

15. Mark angles and equal sides on the diagram until you reach $STU$.

16. Draw a free body diagram and apply Newton's Second Law.

17. Use percentage multipliers and simplify efficiently.

18. Work out how many oscillations take place in 20 seconds and how many amplitudes worth of motion occur in each oscillation.

19. Draw a rough diagram taking care to read the question carefully.

20. Work out the difference in mass in terms of the volume and equate to 20g.

21. Mark angles on the diagram using circle theorems and triangle rules until you reach QSO

22. Locate the relevant part of the graph, use a large triangle to work out the gradient and use this in the equation.

23. Use the information given to solve for the base of the triangle then use Pythagoras' Theorem.

24. Make a reaction equation and equate number of neutrons. Then recall what happens during beta decay.

25. Work out the second differences then the nth term of the given sequence.

26. Work out $\frac{V^2}{R}$ for the whole circuit and equate to $18W$. Then the power dissipated in $X$ can be found in terms of $\frac{V^2}{R}$ which can then be evaluated.

27. Work out the probability that we have no green or one green then work from there.

28. Draw a rough diagram and be confident in you conclusions about each statement.

Part B

29. Draw a rough sketch and integrate whilst taking care with the limits.

30. Apply SUVAT and cancel the result.

31. Take care not to miss any solutions upon division by $sin\theta$ or when taking a square root.

32. Read the options rapidly and once you are confident about one of them there is no need to re-consider the others.

33. Search for a zero discriminant after equating the line and the quadratic.

34. Calculate the acceleration of the whole then assign tensions to the couplings whilst working from the left.

35. The radius of the circle and angle of $S$ are unknown - make two equations and solve for them, then use them to find the arc length of $T$.

36. Equate clockwise and anti-clockwise moments about Y.

37. Form two equations from the bullet points and solve them efficiently.

38. Use Work Done = Force $\times$ Distance and subtract this from the initial energy then divide by the initial energy.

39. Form an equation in $p$ efficiently then solve it to yield $r$.

40. Work out the velocity of $X$ before the collision then apply conservation of momentum.

41. Combine as a single logarithm, simplify and evaluate.

42. Write initial energy - final energy to work out the energy lost.

43. Recall that tangents meet radii at right angles and use Pythagoras' Theorem.

44. Apply conservation of momentum whilst being careful with signs.

45. Turn the given equation into one involving the co-ordinates and $u$ and solve.

46. Resolve and use horizontal equilibrium.

47. Simplify and predict the answer. There is no need to expand.

48. Spot a familiar triangle and equate lost Gravitational Potential Energy to a gain in Kinetic Energy.

49. Integrate and find an efficient way to eliminate $2^m$.

50. Apply conservation of momentum.

51. Integrate the first equation then apply the next two to the result.

52. Equate energy gained (taking into account that some is wasted) and equate to Kinetic Energy.

53. Work out the surface area in terms of $x$ and differentiate.

54. Apply the correct SUVAT equation to both trajectories taking account of the time difference.

Section 2

1. Work out the distances travelled by both the man and the boy as a function of time and equate.

2. Work out the mass and the volume of the mixture and divide.

3. Consider the effect this change has on the voltage across $X$ and $Y$.

4. Consider change of momentum from the very start to the very end without considering any momentum changes in the interim.

5. Think about what extra weight the pan is supporting in both scenarios.

6. Apply $c=f\lambda$ whilst taking care to determine the correct wavelength from diagram 1.

7. Use Impulse $=mv-mu=F\times t$ while realising that for a variable force we have a sum of the product of each individual force and time hence the area under the curve.

8. Make a moments calculation taking all objects into account including the plank.

9. Work out the current in each branch and hence the voltage dropped across each resistor.

10. Work out the extension of each spring then apply $EPE=\frac{1}{2}kx^2$.

11. We require the friction force on the block for which we need the acceleration which is achieved with SUVAT.

12. Work out the resistance of the combination then write the current leaving the cell in terms of $r$. Then apply Ohm's law to the internal resistance.

13. Work out how far the wave travels in one second then apply $v=f\lambda$.

14. Use the vertical equilibrium of $R$ to find the tension in the string.

15. Begin by equating the gravitational force on the cube to $kx$ for the spring.

16. Apply $R=\frac{\rho L}{A}$ taking care with the introduction of the mass of the wires and the circuit which represents the configuration of wires.

17. Work out how much velocity the fluid gains and over what time period this change takes place.

18. Work out how much energy is lost by the ball as as a result of its loss of height. A certain fraction of this must be lost by the time it reaches $P$.

Full solutions

Section 1

Part A

1. Make a two-way table.

The answer is $110$

2. Know and apply how alpha and beta decays affect the number of protons.

The alpha decays reduce the number of protons by $10$. Under beta decay a neutron turns into a proton so we have 2 more protons so the answer is $8$.

3. Write an equation equating the volume and twice the surface area.

$2\sqrt{2}x^3=4(2+2\sqrt2+\sqrt2)x^2$

$x=6+2\sqrt2$

4. Apply $V=IR$ to work out the voltage of the cell and once more to determine the current.

$V=\frac{12+15+3}{5}=6V$

$I=\frac{6}{12+15}=\frac{2}{9}=0.22A(2d.p.)$

5. Work out the gradient in terms of $p$ and equate it to the gradient of the line.

$\frac{2p-(p-1)}{1-2p}=-\frac{2}{3}$

$3p+3=-2+4p$

$p=5$

6. Apply $c=f\lambda$ whilst taking care with which side of the visible spectrum UV is on.

We are interested in the low wavelength (high energy) end of the visible spectrum for UV. It is worth thinking about whether we require a maximum or a minimum frequency at this stage - it will be a minimum because increasing the frequency will take the wave further into UV.

$f=\frac{c}{\lambda}=\frac{3\times 10^8}{400 \times 10^{-9}}=7.5\times 10^{14}Hz$

7. Define an unknown for the height of the rectangle.

Let the height of the rectangle be $x$.

$2x^2=96$

$x=\sqrt{48}$

From Pythagoras' Theorem applied with $OQ$ as the hypotenuse, $r=\frac{\sqrt5}{2}x=\frac{4\sqrt3\sqrt5}{2}=2\sqrt{15}$

8. Use percentage multipliers.

Wasted energy is the percentage waste multiplied by the power and the time $=\frac{95}{100}\times 100 \times 60 \times 10$. At this point I looked at the answers and saw that the answer must be a little below $60000$ and marked the answer $E$.

9. Recall that the volume scale factor is the linear scale factor cubed.

Large Volume $=\frac{320\times5^3}{4^3}=5\times 5^3=625cm^3$

10. Use percentage multipliers and $P=IV$.

$P=IV=400\times1250$

$E=Pt=4\times\frac{45}{100}\times400\times1250=4\times45\times5000$

At this point I realised the answer would begin with a $9$ and marked $C$.

11. Sketch the lines and find their intercepts and $x$-co-ordinate of their intersection.

Area$=\frac{1}{2}\times bh=\frac{1}{2}\times 33\times 6=99$

12. Form two equations and solve for $v$.

$mv=24$

$\frac{1}{2}mv^2=96$

$mv^2=2\times 96$

Dividing gives $v=8$ so the answer is $A$.

13. Work out the mass of the full-sized pillar, converting to $cm$. Divide this by the cube of the scale factor since mass is proportional to volume.

Mass = Density $\times$ Volume $=\frac{12\pi}{40^3}\times1000000\times\frac{4}{3}$

where the factor of $1000000$ is to convert the Volume to $cm^3$.

Mass = $\frac{1000\pi}{4}=250\pi g$

14. Observe the background count then take care to choose the correctly halved count rate.

The background count is $20$. For the half-life I took the points $(0,120)$ and $(40,70)$ to give $40$ minutes.

15. Mark angles and equal sides on the diagram until you reach $STU$.

$RSU=60^{\circ}$ because the triangle is equilateral.

$RST=108^{\circ}$ because it is the interior angle of a regular pentagon so $TSU=48^{\circ}$.

$STU$ is isosceles because $ST=SU$ and so $STU=\frac{180-48}{2}=66^{\circ}$

16. Draw a free body diagram and apply Newton's Second Law.

Your diagram should have an arrow upwards of $12N$ because this is the resistive force and $mg$ downwards.

$mg-12=2m$

$8m=12$

$m=1.5kg$

17. Use percentage multipliers and simplify efficiently.

$2.25p\times 0.6=q$

$m=\frac{2.25\times3}{5}=\frac{3\times9}{20}=\frac{27}{20}$

18. Work out how many oscillations take place in 20 seconds and how many amplitudes worth of motion occur in each oscillation.

$200$ oscillations occur in $20s$. A complete oscillation consists of $4$ amplitudes so the total distance covered is $200\times 4\times 4=32m$

19. Draw a rough diagram taking care to read the question carefully.

The diagram is fairly easy to complete, noting that triangle $PQR$ is isosceles so $QRP=\frac{180-90}{2}=45^{\circ}$. From this the required bearing is $360-25-45=290^{\circ}$.

20. Work out the difference in mass in terms of the volume and equate to 20g.

Mass $=$ Density $\times$ Volume so $1\times V-0.9\times V=290-270=20g$

$0.1V=20g$ so $V=200cm^{3}$ therefore the mass of the water is $200g$ and the mass of the cylinder is $90g$.

21. Mark angles on the diagram using circle theorems and triangle rules until you reach QSO

$SQP=75^{\circ}$ by the alternate segment theorem.

Triangle $SQP$ is isosceles so $PSQ=\frac{180-75}{2}=52.5^{\circ}$

$OST$ is a right-angle so $OSP=15^{\circ}$ and $OSQ=37.5$.

22. Locate the relevant part of the graph, use a large triangle to work out the gradient and use this in the equation.

The portion of the graph with constant gradient represents where the skydiver has terminal velocity. This is the best part of the graph to focus on because we can equate the force to $1000N$.

For the gradient I chose $(5,130)$ and $(10,380)$ to give a gradient of $\frac{380-130}{5}=50$.

$1000=k\times 50^2$ so $k=0.4 Nm^{-2}s^2$.

23. Use the information given to solve for the base of the triangle then use Pythagoras' Theorem.

$\frac{1}{2}bh=14$ so $bh=28$ and from the question, $h=b+3$ so $b(b+3)=28$.

Solve this by inspection to give $b=4$ and therefore $h=7$.

Using Pythagoras' Theorem, $s=\sqrt{53}$ which yields the answer $C$.

24. Make a reaction equation and equate number of neutrons. Then recall what happens during beta decay.

$^1_0 n + ^{235}_{92}U \rightarrow ^{88}_{35}Br + ^{145}_{57}La + ^k_0 n$

From this equation we can see that $k=3$ and that after the beta decay there will be $58$ neutrons in the Lanthanum nucleus.

25. Work out the second differences then the nth term of the given sequence.

The second differences are $10$ so the $nth$ term for this sequence begins $5n^2$.

$5n^2=5,20,45,80,125...$ so by inspection the $nth$ term is $5n^2-3$.

Therefore $p=5$ and $q=-3$ and the answer is $\frac{5+3}{5-3}=4$.

26. Work out $\frac{V^2}{R}$ for the whole circuit and equate to $18W$. Then the power dissipated in $X$ can be found in terms of $\frac{V^2}{R}$ which can then be evaluated.

Defining the emf of the battery to be $V$ and the resistance of each resistor to be $R$ we have $\frac{V^2}{\frac{3R}{2}}=18$ so $\frac{V^2}{R}=27$. By inspection (via a potential divider) the voltage across $X$ is $\frac{V}{3}$ and the power dissipated is $\frac{(\frac{V}{3})^2}{R}=\frac{V^2}{9R}=\frac{27}{9}=3W$.

27. Work out the probability that we have no green or one green then work from there.

P(More green than red)=1-P(Equal or fewer green than red)$=1-P(R)-P(GR)=1-\frac{1}{2}=\frac{1}{2}\times\frac{6}{11}=1-\frac{1}{2}-\frac{3}{11}=\frac{22-11-6}{22}=\frac{5}{22}$.

28. Draw a rough diagram and be confident in you conclusions about each statement.

The diagram shows a possible scenario (E is the epicentre). X and Y merely have to be the same distance from E - they don't have to lie on a straight line containing E. Z could be anywhere on the circle which falsifies the second statement. $240km$ is the minimum distance from X or Y to the circle - imagine if Z were at 6 o'clock then this would be much further than $240km$ so none of the statements are true.

Part B

29. Draw a rough sketch and integrate whilst taking care with the limits.

Let $9-x^2=5$ to find out where the intersections are ($x=\pm 2$).

Area$=2(\int_0^2 9-x^2dx-10)=2(18-[\frac{x^3}{3}]^2_0-10)=2(8-\frac{8}{3})=\frac{32}{3}$

30. Apply SUVAT and cancel the result.

$v^2=u^2+2as$

$u=0$ so $80^2=2\times1600\times a$

Cancelling zeros leads to $a=\frac{64}{32}=2ms^{-2}$.

31. Take care not to miss any solutions upon division by $sin\theta$ or when taking a square root.

Rearranging and factorising we have $sin\theta(2sin^2\theta-1)=0$

$sin\theta(\sqrt2 sin\theta+1)(\sqrt 2sin\theta-1)=0$

$sin\theta=0$ or $\pm \frac{1}{\sqrt2}$

After a sketch of $sin\theta$ in the given interval we can see that there will be $5$ solutions.

32. Read the options rapidly and once you are confident about one of them there is no need to re-consider the others.

Taking this approach it is clear that $F$ is the correct answer - there is no need to consider the others at this stage.

33. Search for a zero discriminant after equating the line and the quadratic.

$x+k=3x^2-2x+1$

$3x^2-3x+1-k=0$

Using $b^2-4ac=0$ we have $9=12(1-k)$ or $k=\frac{1}{4}$.

34. Calculate the acceleration of the whole then assign tensions to the couplings whilst working from the left.

Using $F=ma$ we have $a=\frac{30}{3+4+6+2}=2ms^{-2}$

For object W, we define the tension in the coupling pulling it as $T_1$.

$T_1=2\times3=6N$

If we label the tension in the next coupling as $T_2$ we have the following for object X:

$T_2-6=4\times 2=8$ so $T_2=14N$.

35. The radius of the circle and angle of $S$ are unknown - make two equations and solve for them, then use them to find the arc length of $T$.

Let $S$ have radius $r$ and angle $\theta$.

$\frac{1}{2} \theta r^2 =10\pi$

$\frac{1}{2}(\theta + \frac{\pi}{20})r^2=\frac{25\pi}{2}$

Dividing we have $\frac{\theta+\frac{\pi}{20}}{\theta}=\frac{5}{4}$

This gives $5\theta=4\theta + \frac{\pi}{5}$ so $\theta=\frac{\pi}{5}$.

This means that $r^2=100$ and $r=10$. Finally the arc length is give by radius $\times$ angle which for sector T is $10\times(\frac{\pi}{5}+\frac{\pi}{20})=\frac{5\pi}{2}$.

36. Equate clockwise and anti-clockwise moments about Y.

Denoting the reaction force at X as $F$ we have $2g\times10+3g\times40=4F$ giving $F=\frac{1400}{4}=350N$

37. Form two equations from the bullet points and solve them efficiently.

The equations are:

$a+12d=6a$ which rearranges to give $12d=5a$

$a+10d=2(a+4d)-1$ which rearranges to give $2d+1=a$

Substituting in we have $12d=10d+5$ so $d=\frac{5}{2}$ and $a=6$, giving the third term as $6+2\times\frac{5}{2}=11$.

38. Use Work Done = Force $\times$ Distance and subtract this from the initial energy then divide by the initial energy.

The initial potential energy is $mgh=mgl sin\theta$ or using $W=mg$ we have $Wlsin\theta$. The friction force is already given so to find the work done against it we multiply by $l$ and the required fraction is:

$\frac{Wlsin\theta-Wklsin\theta}{Wlsin\theta}=1-k$.

39. Form an equation in $p$ efficiently then solve it to yield $r$.

Since this is a geometric progression we can immediately write

$\frac{2p+2}{p-2}=\frac{5p+14}{2p+2}$

Multiplying by the denominators gives $4p^2+8p+4=5p^2+4p-28$

$p^2-4p-32=0$

$(p-8)(p+4)=0$

All terms are greater than $0$ so $p=8$ and hence $a=6$ and $r=3$. The fifth term is $6\times3^4=486$.

40. Work out the velocity of $X$ before the collision then apply conservation of momentum.

Using $F=ma$ we have that the acceleration of X is $\frac{5}{2}$.

Applying $v=u+at$ we have the final velocity to be $\frac{9}{2}+3\times \frac{5}{2}=12 ms^{-1}$.

Now using conservation of momentum we obtain $2\times 12=(2+3)v$ because the objects stick together. This yields $v=4.8 ms^{-1}$.

41. Combine as a single logarithm, simplify and evaluate.

When all the fractions are multiplied together we are left with $\frac{64}{4}$ so the expression is $log_2 16=4$.

42. Write initial energy - final energy to work out the energy lost.

$\frac{1}{2}mu^2+mgh-\frac{1}{2}mv^2$

$\frac{1}{2}\times0.2\times16+0.2\times0.45\times10-\frac{1}{2}\times0.2\times4=\frac{8}{5}+\frac{9}{10}-\frac{2}{5}=\frac{6}{5}+\frac{9}{10}=2.1J$

43. Recall that tangents meet radii at right angles and use Pythagoras' Theorem.

The distance required is $XP$ which is $OP$ minus the radius.

$OP^2=(5\sqrt3)^2+(\sqrt5)^2$ so $OP=\sqrt{80}=4\sqrt5$.

$XP=4\sqrt5-\sqrt5=3\sqrt5$

44. Apply conservation of momentum whilst being careful with signs.

From the diagram we can apply conservation of momentum, taking right to be positive:

$mv-4mv=-mv+2mu$

$-2mv=2mu$ so $u=-v$ and the answer is $F$.

45. Turn the given equation into one involving the co-ordinates and $u$ and solve.

$AC=2AB$

$\sqrt{2^2+(2-u)^2}=2\sqrt{4^2+1^2}$

Squaring both sides we have $4+4-4u+u^2=68$

$u^2-4u-60=0$

$(u-10)(u+6)=0$ so the answer is $D$.

46. Resolve and use horizontal equilibrium.

$Tsin30=P$ so $P=\frac{T}{2}$

47. Simplify and predict the answer. There is no need to expand.

$(1-2x)(1+2x)=(1-4x^2)$ so the expansion consists only of even powers so the answer is $D$. This is an example where the examiners can penalise candidates by wasting their time, rather than docking marks.

48. Spot a familiar triangle and equate lost Gravitational Potential Energy to a gain in Kinetic Energy.

From the $3$,$4$,$5$ triangle, the vertical distance of X below the pivot is $40cm$.

Equating lost GPE to KE we have $mgh=\frac{1}{2}mv^2$ so $2=v^2$ and the answer is $A$.

49. Integrate and find an efficient way to eliminate $2^m$.

From the two integrals we have:

$\frac{2^{m+1}}{m+1}=\frac{16\sqrt2}{7}$

$\frac{2^{m+2}}{m+2}=\frac{32\sqrt2}{9}$

Dividing we have $\frac{2(m+1)}{m+2}=2\times\frac{7}{9}$ giving $7m+14=9m+9$ so $m=\frac{5}{2}$.

50. Apply conservation of momentum

Letting the new velocity of the greater mass be v_M, $mu=-mv+Mv_M$ so $v_M=\frac{m(v+u)}{M}$

51. Integrate the first equation then apply the next two to the result.

After integrating the first equation from $2$ to $4$, apply the fundamental theorem of calculus:

$\int^4_2f'(x)dx=f(4)-f(2)=[\frac{ax^2}{2}]^4_2+\int^4_2g(x)dx$

Applying the other two equations given in the question we have:

$18=\frac{a}{2}(16-4)+12$ giving $6=6a$ so $a=1$.

52. Equate energy gained (taking into account that some is wasted) and equate to Kinetic Energy.

Only half of the GPE is available for transfer to KE so we can write:

$\frac{mgh}{2}=\frac{1}{2}\times m \times 10^2$ giving $gh=100$ so $h=10m$.

53. Work out the surface area in terms of $x$ and differentiate.

From the question, $2x^2y=576$. The surface area is $2(2x^2+3xy)$.

Using the volume equation to write this in terms of $x$ only we have:

$\frac{A}{4}=x^2+\frac{3\times144}{x}$

Differentiating this and equating the result to $0$ gives:

$2x-\frac{3\times 144}{x^2}=0$

$x^3=3\times72=8\times27$ so $x=6$, $y=8$ and the answer is $96$ because $2x$ and $y$ are the longest sides.

54. Apply the correct SUVAT equation to both trajectories taking account of the time difference.

For the first object, $s=ut+\frac{1}{2}at^2$ gives

$0=40t-5t^2$ which can be solved by inspection to give $t=8$.

This give the time of flight for the second object to be $6s$.

Applying $s=ut+\frac{1}{2}t^2$ again we have $h=\frac{1}{2}\times10\times36=180m$.

Section 2

1. Work out the distances travelled by both the man and the boy as a function of time and equate.

Apply $s=ut+\frac{1}{2}at^2$ to the man (LHS) and the boy (RHS):

$9t=5t+\frac{1}{2}\times \frac{4}{5}t^2$

$10t=t^2$ so the answer is $B$.

2. Work out the mass and the volume of the mixture and divide.

The density is the total mass divided by the total volume.

$D=\frac{\rho_PV_P+ \rho_QV_Q}{V_P+V_Q}$ so the answer is $B$.

3. Consider the effect this change has on the voltage across $X$ and $Y$.

Think of a potential divider. When the resistance of W increases, this increases the voltage across the combination of W and Y at the expense of the voltage across X. X and Y keep the same resistance throughout, so applying $P=\frac{V^2}{R}$ we see that the power dissipated in Y increases and that in X decreases.

4. Consider change of momentum from the very start to the very end without considering any momentum changes in the interim.

We have zero momentum at the start so should expect zero momentum at the end no matter how complicated the route to the end is therefore the answer is $A$.

5. Think about what extra weight the pan is supporting in both scenarios.

In the second diagram the pan supports an extra $100g$ because the scale has dropped by $1N$. In the third diagram the entire $3N$ is supported by the pan so this is an extra $300g$. Therefore the readings are $500g$ then $700g$.

6. Apply $c=f\lambda$ whilst taking care to determine the correct wavelength from diagram 1.

The diagram shows half a wavelength so the wavelength is $1m$. Now we need $f$ which is given by $c=f\lambda$ so $f=1000Hz$. The time until X in the diagram is $1.5$ cycles so this takes $1.5\times 10^{-3}s$.

7. Use Impulse $=mv-mu=F\times t$ while realising that for a variable force we have a sum of the product of each individual force and time hence the area under the curve.

$m\Delta v=Ft=\frac{1}{2}bh=\frac{750\times4}{1000}$

Using the KE to find out the mass of the ball we have $\frac{1}{2}mv^2=27$ so $m=\frac{27\times 2}{30^2}$.

Rearranging for $\Delta v$ we have:

$\Delta v=\frac{750\times4\times30^2}{27\times2\times1000}$ which cancels to give $\Delta v=50$.

8. Make a moments calculation taking all objects into account including the plank.

Take $F$ to be the force on the plank due to the support at Q.

Taking moments about P we have $(0.2\times5 + 1\times 74 + 1\times 24)g=1.5\times F$

$F=\frac{2}{3}(1+24+74)\times 10=660N$

9. Work out the current in each branch and hence the voltage dropped across each resistor.

The current in each branch is $\frac{12}{6}=2A$. Therefore the voltage of the upper terminal of the voltmeter compared to the negative terminal of the battery is $10V$ and the voltage of the lower terminal of the voltmeter compared to the negative terminal of the battery is $2V$. Therefore the potential difference is $8V$.

10. Work out the extension of each spring then apply $EPE=\frac{1}{2}kx^2$.

The extension of the lower spring is $\frac{mg}{k}$ as normal and for each upper spring it is $\frac{mg}{3k}$ because the load is shared between the three springs. Therefore the energy stored is:

$\frac{1}{2}k((\frac{mg}{k})^2+3(\frac{mg}{3k})^2)=\frac{(mg)^2}{2k}(1+\frac{1}{3})=\frac{2(mg)^2}{3k}$

11. We require the friction force on the block for which we need the acceleration which is achieved with SUVAT.

$v^2=u^2+2as$

$4=3a$ so $a=\frac{4}{3}$.

The force down the slope due to the weight of the block is $3.6gsin30=\frac{3.6g}{2}$. Applying Newton's Second Law to the block we have:

$\frac{3.6g}{2}-F=3.6\times\frac{4}{3}$ where $F$ is the friction force.

$F=18-4.8=13.2N$

The average rate of work done is the average power which is the Force $\times$ the average velocity ($1ms^{-1}$) so the average rate of work done is $13.2N$.

12. Work out the resistance of the combination then write the current leaving the cell in terms of $r$. Then apply Ohm's law to the internal resistance.

The resistance of the combination is $\frac{1}{\frac{1}{20}+\frac{1}{30}}=\frac{60}{3+2}=12\Omega$.

The current leaving the cell is $I=\frac{6}{12+r}$ but we also have that $Ir=1.2V$ because $4.8V$ is dropped across the remainder of the circuit so $6-4.8=1.2V$ must be dropped across the internal resistance. So we have:

$1.2=\frac{6r}{12+r}$

$14.4+1.2r=6r$

$r=\frac{14.4}{4.8}=3\Omega$

13. Work out how far the wave travels in one second then apply $v=f\lambda$.

Inbetween the diagrams, the wave has travelled a wavelength plus $1.5cm$ which has taken one second. Therefore this distance is numerically equal to the speed of the wave.

$v=\lambda+1.5=f\lambda=\frac{5\lambda}{4}$ so $\lambda=6cm$.

14. Use the vertical equilibrium of $R$ to find the tension in the string.

For mass $R$, we have $T=m_Rg$

Therefore the force on $Q$ to the right is also $m_Rg$.

This means that its acceleration would be $\frac{m_Rg}{m_Q}$ so this must be the acceleration of $P$ to prevent $Q$ slipping off $P$.

15. Begin by equating the gravitational force on the cube to $kx$ for the spring.

$mg=3.2N$

$\rho Vg=3.2$

$\rho\times(0.04)^3\times 10=3.2$

$\rho=\frac{32\times100^3}{100\times64}=5\times10^3 kgm^{-3}$

$Pressure=\frac{Force}{Area}=\frac{3.2}{(0.04)^2}=2\times10^3 Nm^{-2}$

16. Apply $R=\frac{\rho L}{A}$ taking care with the introduction of the mass of the wires and the circuit which represents the configuration of wires.

The wires have different resistivities, densities and masses, so working generally we have:

$R_{wire}=\frac{\rho_{wire}L}{A}$ but the volume of the wire is given by $AL$ but also $\frac{M_{wire}}{d_{wire}}$.

Rearranging this we have $\frac{A}{L}=\frac{M_{wire}}{d_{wire}L^2}$ or after taking a reciprocal, $\frac{L}{A}=\frac{d_{wire}L^2}{M_{wire}}$.

Substituting in we have $R_{wire}=\frac{\rho_{wire}d_{wire}L^2}{M_{wire}}$.

To determine $M_{wire}$ for each of the Copper and Aluminium wires, note that the Copper wire is $3$ times as dense as the Aluminium. There are $6$ Aluminium wires and $1$ Copper wire which is $3$ times as heavy so the total mass of the wire is $9$ times that of the Aluminium wire. Therefore the mass of each Aluminium wire is $\frac{M}{9}$ and the Copper wire is $\frac{M}{3}$.

Substituting this and the resistivities and densities in from the table gives:

$R_{Copper}=\frac{2\rho \times 3dL^2}{\frac{M}{3}}=\frac{18\rho dL^2}{M}$ and

$R_{Aluminium}=\frac{3\rho \times dL^2}{\frac{M}{9}}=\frac{27\rho dL^2}{M}$.

The wires make a parallel combination of $6$ Aluminium wires and $1$ Copper wire, giving a resistance of:

$R_{Total}=\frac{1}{\frac{6}{R_{Aluminium}}+\frac{1}{R_{Copper}}}=\frac{1}{\frac{6}{27}+\frac{1}{18}}\times \frac{\rho dL^2}{M}=\frac{54}{15}\times\frac{\rho dL^2}{M}=\frac{18\rho dL^2}{5M}$

17. Work out how much velocity the fluid gains and over what time period this change takes place.

The fluid is incompressible which means that if a certain volume of it is pushed through a pipe then this volume remains constant - it is neither squashed nor expanded. Imagine a cylinder of fluid approaching the narrowing of the pipe. Its volume is $A_1\times l_1$ where $A_1$ is the cross-sectional area of the pipe on the left and $l_1$ is the length of the cylinder of fluid from the left. The volume on the right is $A_2\times l_2$ where the symbols have the same meaning but for the fluid on the right and we have $A_1l_1=A_2l_2$. The lengths of the cylinders are proportional to the speed that the fluid travels at so we have $\frac{v_1}{v_2}=\frac{A_2}{A_1}$ and the velocity after the narrowing is $v_2=\frac{5\times 0.6}{0.25}=12ms^{-1}$. The change in velocity therefore is $7ms^{-1}$. We seek to use Force as the rate of change of Momentum so we now need a mass of fluid and a time over which the narrowing takes place. It doesn't matter how much fluid we use because we are interested in the rate of change, not the overall amount of fluid passing. If we work with $1m^3$ of fluid, this has length $\frac{5}{3}m$ as it approaches the narrowing. At $5ms^{-1}$ this takes $\frac{1}{3}s$. To check, the same $1m^3$ of fluid has a length of $4m$ after the narrowing and at a speed of $12ms^{-1}$ the fluid takes $\frac{1}{3}ms^{-1}$ to leave the narrowing as expected.

$F=\frac{m\Delta v}{\Delta t}=\frac{800\times7}{\frac{1}{3}}=7\times2400=16800N$

18. Work out how much energy is lost by the ball as as a result of its loss of height. A certain fraction of this must be lost by the time it reaches $P$.

Let O be the centre of the sphere. The vertical height lost is $OQcos45=\frac{r}{\sqrt2}$ and so the energy lost to friction up to Q is $\frac{mgr}{\sqrt2}$.

Since the friction force is constant, $\frac{2}{3}$ of this must be lost by the point P. The ball starts with potential energy $mgr$ and the KE is given by this minus the loss due to friction.

KE$=mgr-\frac{2mgr}{3\sqrt2}=mgr(1-\frac{\sqrt2}{3})$.

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. Make a two-way table.

2. Know and apply how alpha and beta decays affect the number of protons.

3. Write an equation equating the volume and twice the surface area.

4. Apply $V=IR$ to work out the voltage of the cell and once more to determine the current.

5. Work out the gradient in terms of $p$ and equate it to the gradient of the line.

6. Apply $c=f\lambda$ whilst taking care with which side of the visible spectrum UV is on.

7. Define an unknown for the height of the rectangle.

8. Use percentage multipliers.

9. Recall that the volume scale factor is the linear scale factor cubed.

10. Use percentage multipliers and $P=IV$.

11. Sketch the lines and find their intercepts and $x$-co-ordinate of their intersection.

12. Form two equations and solve for $v$.

13. Work out the mass of the full-sized pillar, converting to $cm$. Divide this by the cube of the scale factor since mass is proportional to volume.

14. Observe the background count then take care to choose the correctly halved count rate.

15. Mark angles and equal sides on the diagram until you reach $STU$.

16. Draw a free body diagram and apply Newton's Second Law.

17. Use percentage multipliers and simplify efficiently.

18. Work out how many oscillations take place in 20 seconds and how many amplitudes worth of motion occur in each oscillation.

19. Draw a rough diagram taking care to read the question carefully.

20. Work out the difference in mass in terms of the volume and equate to 20g.

21. Mark angles on the diagram using circle theorems and triangle rules until you reach QSO

22. Locate the relevant part of the graph, use a large triangle to work out the gradient and use this in the equation.

23. Use the information given to solve for the base of the triangle then use Pythagoras' Theorem.

24. Make a reaction equation and equate number of neutrons. Then recall what happens during beta decay.

25. Work out the second differences then the nth term of the given sequence.

26. Work out $\frac{V^2}{R}$ for the whole circuit and equate to $18W$. Then the power dissipated in $X$ can be found in terms of $\frac{V^2}{R}$ which can then be evaluated.

27. Work out the probability that we have no green or one green then work from there.

28. Draw a rough diagram and be confident in you conclusions about each statement.

Part B

29. Draw a rough sketch and integrate whilst taking care with the limits.

30. Apply SUVAT and cancel the result.

31. Take care not to miss any solutions upon division by $sin\theta$ or when taking a square root.

32. Read the options rapidly and once you are confident about one of them there is no need to re-consider the others.

33. Search for a zero discriminant after equating the line and the quadratic.

34. Calculate the acceleration of the whole then assign tensions to the couplings whilst working from the left.

35. The radius of the circle and angle of $S$ are unknown - make two equations and solve for them, then use them to find the arc length of $T$.

36. Equate clockwise and anti-clockwise moments about Y.

37. Form two equations from the bullet points and solve them efficiently.

38. Use Work Done = Force $\times$ Distance and subtract this from the initial energy then divide by the initial energy.

39. Form an equation in $p$ efficiently then solve it to yield $r$.

40. Work out the velocity of $X$ before the collision then apply conservation of momentum.

41. Combine as a single logarithm, simplify and evaluate.

42. Write initial energy - final energy to work out the energy lost.

43. Recall that tangents meet radii at right angles and use Pythagoras' Theorem.

44. Apply conservation of momentum whilst being careful with signs.

45. Turn the given equation into one involving the co-ordinates and $u$ and solve.

46. Resolve and use horizontal equilibrium.

47. Simplify and predict the answer. There is no need to expand.

48. Spot a familiar triangle and equate lost Gravitational Potential Energy to a gain in Kinetic Energy.

49. Integrate and find an efficient way to eliminate $2^m$.

50. Apply conservation of momentum.

51. Integrate the first equation then apply the next two to the result.

52. Equate energy gained (taking into account that some is wasted) and equate to Kinetic Energy.

53. Work out the surface area in terms of $x$ and differentiate.

54. Apply the correct SUVAT equation to both trajectories taking account of the time difference.

Section 2

1. Work out the distances travelled by both the man and the boy as a function of time and equate.

2. Work out the mass and the volume of the mixture and divide.

3. Consider the effect this change has on the voltage across $X$ and $Y$.

4. Consider change of momentum from the very start to the very end without considering any momentum changes in the interim.

5. Think about what extra weight the pan is supporting in both scenarios.

6. Apply $c=f\lambda$ whilst taking care to determine the correct wavelength from diagram 1.

7. Use Impulse $=mv-mu=F\times t$ while realising that for a variable force we have a sum of the product of each individual force and time hence the area under the curve.

8. Make a moments calculation taking all objects into account including the plank.

9. Work out the current in each branch and hence the voltage dropped across each resistor.

10. Work out the extension of each spring then apply $EPE=\frac{1}{2}kx^2$.

11. We require the friction force on the block for which we need the acceleration which is achieved with SUVAT.

12. Work out the resistance of the combination then write the current leaving the cell in terms of $r$. Then apply Ohm's law to the internal resistance.

13. Work out how far the wave travels in one second then apply $v=f\lambda$.

14. Use the vertical equilibrium of $R$ to find the tension in the string.

15. Begin by equating the gravitational force on the cube to $kx$ for the spring.

16. Apply $R=\frac{\rho L}{A}$ taking care with the introduction of the mass of the wires and the circuit which represents the configuration of wires.

17. Work out how much velocity the fluid gains and over what time period this change takes place.

18. Work out how much energy is lost by the ball as as a result of its loss of height. A certain fraction of this must be lost by the time it reaches $P$.

Full solutions

Section 1

Part A

1. Make a two-way table.

The answer is $110$

2. Know and apply how alpha and beta decays affect the number of protons.

The alpha decays reduce the number of protons by $10$. Under beta decay a neutron turns into a proton so we have 2 more protons so the answer is $8$.

3. Write an equation equating the volume and twice the surface area.

$2\sqrt{2}x^3=4(2+2\sqrt2+\sqrt2)x^2$

$x=6+2\sqrt2$

4. Apply $V=IR$ to work out the voltage of the cell and once more to determine the current.

$V=\frac{12+15+3}{5}=6V$

$I=\frac{6}{12+15}=\frac{2}{9}=0.22A(2d.p.)$

5. Work out the gradient in terms of $p$ and equate it to the gradient of the line.

$\frac{2p-(p-1)}{1-2p}=-\frac{2}{3}$

$3p+3=-2+4p$

$p=5$

6. Apply $c=f\lambda$ whilst taking care with which side of the visible spectrum UV is on.

We are interested in the low wavelength (high energy) end of the visible spectrum for UV. It is worth thinking about whether we require a maximum or a minimum frequency at this stage - it will be a minimum because increasing the frequency will take the wave further into UV.

$f=\frac{c}{\lambda}=\frac{3\times 10^8}{400 \times 10^{-9}}=7.5\times 10^{14}Hz$

7. Define an unknown for the height of the rectangle.

Let the height of the rectangle be $x$.

$2x^2=96$

$x=\sqrt{48}$

From Pythagoras' Theorem applied with $OQ$ as the hypotenuse, $r=\frac{\sqrt5}{2}x=\frac{4\sqrt3\sqrt5}{2}=2\sqrt{15}$

8. Use percentage multipliers.

Wasted energy is the percentage waste multiplied by the power and the time $=\frac{95}{100}\times 100 \times 60 \times 10$. At this point I looked at the answers and saw that the answer must be a little below $60000$ and marked the answer $E$.

9. Recall that the volume scale factor is the linear scale factor cubed.

Large Volume $=\frac{320\times5^3}{4^3}=5\times 5^3=625cm^3$

10. Use percentage multipliers and $P=IV$.

$P=IV=400\times1250$

$E=Pt=4\times\frac{45}{100}\times400\times1250=4\times45\times5000$

At this point I realised the answer would begin with a $9$ and marked $C$.

11. Sketch the lines and find their intercepts and $x$-co-ordinate of their intersection.

Area$=\frac{1}{2}\times bh=\frac{1}{2}\times 33\times 6=99$

12. Form two equations and solve for $v$.

$mv=24$

$\frac{1}{2}mv^2=96$

$mv^2=2\times 96$

Dividing gives $v=8$ so the answer is $A$.

13. Work out the mass of the full-sized pillar, converting to $cm$. Divide this by the cube of the scale factor since mass is proportional to volume.

Mass = Density $\times$ Volume $=\frac{12\pi}{40^3}\times1000000\times\frac{4}{3}$

where the factor of $1000000$ is to convert the Volume to $cm^3$.

Mass = $\frac{1000\pi}{4}=250\pi g$

14. Observe the background count then take care to choose the correctly halved count rate.

The background count is $20$. For the half-life I took the points $(0,120)$ and $(40,70)$ to give $40$ minutes.

15. Mark angles and equal sides on the diagram until you reach $STU$.

$RSU=60^{\circ}$ because the triangle is equilateral.

$RST=108^{\circ}$ because it is the interior angle of a regular pentagon so $TSU=48^{\circ}$.

$STU$ is isosceles because $ST=SU$ and so $STU=\frac{180-48}{2}=66^{\circ}$

16. Draw a free body diagram and apply Newton's Second Law.

Your diagram should have an arrow upwards of $12N$ because this is the resistive force and $mg$ downwards.

$mg-12=2m$

$8m=12$

$m=1.5kg$

17. Use percentage multipliers and simplify efficiently.

$2.25p\times 0.6=q$

$m=\frac{2.25\times3}{5}=\frac{3\times9}{20}=\frac{27}{20}$

18. Work out how many oscillations take place in 20 seconds and how many amplitudes worth of motion occur in each oscillation.

$200$ oscillations occur in $20s$. A complete oscillation consists of $4$ amplitudes so the total distance covered is $200\times 4\times 4=32m$

19. Draw a rough diagram taking care to read the question carefully.

The diagram is fairly easy to complete, noting that triangle $PQR$ is isosceles so $QRP=\frac{180-90}{2}=45^{\circ}$. From this the required bearing is $360-25-45=290^{\circ}$.

20. Work out the difference in mass in terms of the volume and equate to 20g.

Mass $=$ Density $\times$ Volume so $1\times V-0.9\times V=290-270=20g$

$0.1V=20g$ so $V=200cm^{3}$ therefore the mass of the water is $200g$ and the mass of the cylinder is $90g$.

21. Mark angles on the diagram using circle theorems and triangle rules until you reach QSO

$SQP=75^{\circ}$ by the alternate segment theorem.

Triangle $SQP$ is isosceles so $PSQ=\frac{180-75}{2}=52.5^{\circ}$

$OST$ is a right-angle so $OSP=15^{\circ}$ and $OSQ=37.5$.

22. Locate the relevant part of the graph, use a large triangle to work out the gradient and use this in the equation.

The portion of the graph with constant gradient represents where the skydiver has terminal velocity. This is the best part of the graph to focus on because we can equate the force to $1000N$.

For the gradient I chose $(5,130)$ and $(10,380)$ to give a gradient of $\frac{380-130}{5}=50$.

$1000=k\times 50^2$ so $k=0.4 Nm^{-2}s^2$.

23. Use the information given to solve for the base of the triangle then use Pythagoras' Theorem.

$\frac{1}{2}bh=14$ so $bh=28$ and from the question, $h=b+3$ so $b(b+3)=28$.

Solve this by inspection to give $b=4$ and therefore $h=7$.

Using Pythagoras' Theorem, $s=\sqrt{53}$ which yields the answer $C$.

24. Make a reaction equation and equate number of neutrons. Then recall what happens during beta decay.

$^1_0 n + ^{235}_{92}U \rightarrow ^{88}_{35}Br + ^{145}_{57}La + ^k_0 n$

From this equation we can see that $k=3$ and that after the beta decay there will be $58$ neutrons in the Lanthanum nucleus.

25. Work out the second differences then the nth term of the given sequence.

The second differences are $10$ so the $nth$ term for this sequence begins $5n^2$.

$5n^2=5,20,45,80,125...$ so by inspection the $nth$ term is $5n^2-3$.

Therefore $p=5$ and $q=-3$ and the answer is $\frac{5+3}{5-3}=4$.

26. Work out $\frac{V^2}{R}$ for the whole circuit and equate to $18W$. Then the power dissipated in $X$ can be found in terms of $\frac{V^2}{R}$ which can then be evaluated.

Defining the emf of the battery to be $V$ and the resistance of each resistor to be $R$ we have $\frac{V^2}{\frac{3R}{2}}=18$ so $\frac{V^2}{R}=27$. By inspection (via a potential divider) the voltage across $X$ is $\frac{V}{3}$ and the power dissipated is $\frac{(\frac{V}{3})^2}{R}=\frac{V^2}{9R}=\frac{27}{9}=3W$.

27. Work out the probability that we have no green or one green then work from there.

P(More green than red)=1-P(Equal or fewer green than red)$=1-P(R)-P(GR)=1-\frac{1}{2}=\frac{1}{2}\times\frac{6}{11}=1-\frac{1}{2}-\frac{3}{11}=\frac{22-11-6}{22}=\frac{5}{22}$.

28. Draw a rough diagram and be confident in you conclusions about each statement.

The diagram shows a possible scenario (E is the epicentre). X and Y merely have to be the same distance from E - they don't have to lie on a straight line containing E. Z could be anywhere on the circle which falsifies the second statement. $240km$ is the minimum distance from X or Y to the circle - imagine if Z were at 6 o'clock then this would be much further than $240km$ so none of the statements are true.

Part B

29. Draw a rough sketch and integrate whilst taking care with the limits.

Let $9-x^2=5$ to find out where the intersections are ($x=\pm 2$).

Area$=2(\int_0^2 9-x^2dx-10)=2(18-[\frac{x^3}{3}]^2_0-10)=2(8-\frac{8}{3})=\frac{32}{3}$

30. Apply SUVAT and cancel the result.

$v^2=u^2+2as$

$u=0$ so $80^2=2\times1600\times a$

Cancelling zeros leads to $a=\frac{64}{32}=2ms^{-2}$.

31. Take care not to miss any solutions upon division by $sin\theta$ or when taking a square root.

Rearranging and factorising we have $sin\theta(2sin^2\theta-1)=0$

$sin\theta(\sqrt2 sin\theta+1)(\sqrt 2sin\theta-1)=0$

$sin\theta=0$ or $\pm \frac{1}{\sqrt2}$

After a sketch of $sin\theta$ in the given interval we can see that there will be $5$ solutions.

32. Read the options rapidly and once you are confident about one of them there is no need to re-consider the others.

Taking this approach it is clear that $F$ is the correct answer - there is no need to consider the others at this stage.

33. Search for a zero discriminant after equating the line and the quadratic.

$x+k=3x^2-2x+1$

$3x^2-3x+1-k=0$

Using $b^2-4ac=0$ we have $9=12(1-k)$ or $k=\frac{1}{4}$.

34. Calculate the acceleration of the whole then assign tensions to the couplings whilst working from the left.

Using $F=ma$ we have $a=\frac{30}{3+4+6+2}=2ms^{-2}$

For object W, we define the tension in the coupling pulling it as $T_1$.

$T_1=2\times3=6N$

If we label the tension in the next coupling as $T_2$ we have the following for object X:

$T_2-6=4\times 2=8$ so $T_2=14N$.

35. The radius of the circle and angle of $S$ are unknown - make two equations and solve for them, then use them to find the arc length of $T$.

Let $S$ have radius $r$ and angle $\theta$.

$\frac{1}{2} \theta r^2 =10\pi$

$\frac{1}{2}(\theta + \frac{\pi}{20})r^2=\frac{25\pi}{2}$

Dividing we have $\frac{\theta+\frac{\pi}{20}}{\theta}=\frac{5}{4}$

This gives $5\theta=4\theta + \frac{\pi}{5}$ so $\theta=\frac{\pi}{5}$.

This means that $r^2=100$ and $r=10$. Finally the arc length is give by radius $\times$ angle which for sector T is $10\times(\frac{\pi}{5}+\frac{\pi}{20})=\frac{5\pi}{2}$.

36. Equate clockwise and anti-clockwise moments about Y.

Denoting the reaction force at X as $F$ we have $2g\times10+3g\times40=4F$ giving $F=\frac{1400}{4}=350N$

37. Form two equations from the bullet points and solve them efficiently.

The equations are:

$a+12d=6a$ which rearranges to give $12d=5a$

$a+10d=2(a+4d)-1$ which rearranges to give $2d+1=a$

Substituting in we have $12d=10d+5$ so $d=\frac{5}{2}$ and $a=6$, giving the third term as $6+2\times\frac{5}{2}=11$.

38. Use Work Done = Force $\times$ Distance and subtract this from the initial energy then divide by the initial energy.

The initial potential energy is $mgh=mgl sin\theta$ or using $W=mg$ we have $Wlsin\theta$. The friction force is already given so to find the work done against it we multiply by $l$ and the required fraction is:

$\frac{Wlsin\theta-Wklsin\theta}{Wlsin\theta}=1-k$.

39. Form an equation in $p$ efficiently then solve it to yield $r$.

Since this is a geometric progression we can immediately write

$\frac{2p+2}{p-2}=\frac{5p+14}{2p+2}$

Multiplying by the denominators gives $4p^2+8p+4=5p^2+4p-28$

$p^2-4p-32=0$

$(p-8)(p+4)=0$

All terms are greater than $0$ so $p=8$ and hence $a=6$ and $r=3$. The fifth term is $6\times3^4=486$.

40. Work out the velocity of $X$ before the collision then apply conservation of momentum.

Using $F=ma$ we have that the acceleration of X is $\frac{5}{2}$.

Applying $v=u+at$ we have the final velocity to be $\frac{9}{2}+3\times \frac{5}{2}=12 ms^{-1}$.

Now using conservation of momentum we obtain $2\times 12=(2+3)v$ because the objects stick together. This yields $v=4.8 ms^{-1}$.

41. Combine as a single logarithm, simplify and evaluate.

When all the fractions are multiplied together we are left with $\frac{64}{4}$ so the expression is $log_2 16=4$.

42. Write initial energy - final energy to work out the energy lost.

$\frac{1}{2}mu^2+mgh-\frac{1}{2}mv^2$

$\frac{1}{2}\times0.2\times16+0.2\times0.45\times10-\frac{1}{2}\times0.2\times4=\frac{8}{5}+\frac{9}{10}-\frac{2}{5}=\frac{6}{5}+\frac{9}{10}=2.1J$

43. Recall that tangents meet radii at right angles and use Pythagoras' Theorem.

The distance required is $XP$ which is $OP$ minus the radius.

$OP^2=(5\sqrt3)^2+(\sqrt5)^2$ so $OP=\sqrt{80}=4\sqrt5$.

$XP=4\sqrt5-\sqrt5=3\sqrt5$

44. Apply conservation of momentum whilst being careful with signs.

From the diagram we can apply conservation of momentum, taking right to be positive:

$mv-4mv=-mv+2mu$

$-2mv=2mu$ so $u=-v$ and the answer is $F$.

45. Turn the given equation into one involving the co-ordinates and $u$ and solve.

$AC=2AB$

$\sqrt{2^2+(2-u)^2}=2\sqrt{4^2+1^2}$

Squaring both sides we have $4+4-4u+u^2=68$

$u^2-4u-60=0$

$(u-10)(u+6)=0$ so the answer is $D$.

46. Resolve and use horizontal equilibrium.

$Tsin30=P$ so $P=\frac{T}{2}$

47. Simplify and predict the answer. There is no need to expand.

$(1-2x)(1+2x)=(1-4x^2)$ so the expansion consists only of even powers so the answer is $D$. This is an example where the examiners can penalise candidates by wasting their time, rather than docking marks.

48. Spot a familiar triangle and equate lost Gravitational Potential Energy to a gain in Kinetic Energy.

From the $3$,$4$,$5$ triangle, the vertical distance of X below the pivot is $40cm$.

Equating lost GPE to KE we have $mgh=\frac{1}{2}mv^2$ so $2=v^2$ and the answer is $A$.

49. Integrate and find an efficient way to eliminate $2^m$.

From the two integrals we have:

$\frac{2^{m+1}}{m+1}=\frac{16\sqrt2}{7}$

$\frac{2^{m+2}}{m+2}=\frac{32\sqrt2}{9}$

Dividing we have $\frac{2(m+1)}{m+2}=2\times\frac{7}{9}$ giving $7m+14=9m+9$ so $m=\frac{5}{2}$.

50. Apply conservation of momentum

Letting the new velocity of the greater mass be v_M, $mu=-mv+Mv_M$ so $v_M=\frac{m(v+u)}{M}$

51. Integrate the first equation then apply the next two to the result.

After integrating the first equation from $2$ to $4$, apply the fundamental theorem of calculus:

$\int^4_2f'(x)dx=f(4)-f(2)=[\frac{ax^2}{2}]^4_2+\int^4_2g(x)dx$

Applying the other two equations given in the question we have:

$18=\frac{a}{2}(16-4)+12$ giving $6=6a$ so $a=1$.

52. Equate energy gained (taking into account that some is wasted) and equate to Kinetic Energy.

Only half of the GPE is available for transfer to KE so we can write:

$\frac{mgh}{2}=\frac{1}{2}\times m \times 10^2$ giving $gh=100$ so $h=10m$.

53. Work out the surface area in terms of $x$ and differentiate.

From the question, $2x^2y=576$. The surface area is $2(2x^2+3xy)$.

Using the volume equation to write this in terms of $x$ only we have:

$\frac{A}{4}=x^2+\frac{3\times144}{x}$

Differentiating this and equating the result to $0$ gives:

$2x-\frac{3\times 144}{x^2}=0$

$x^3=3\times72=8\times27$ so $x=6$, $y=8$ and the answer is $96$ because $2x$ and $y$ are the longest sides.

54. Apply the correct SUVAT equation to both trajectories taking account of the time difference.

For the first object, $s=ut+\frac{1}{2}at^2$ gives

$0=40t-5t^2$ which can be solved by inspection to give $t=8$.

This give the time of flight for the second object to be $6s$.

Applying $s=ut+\frac{1}{2}t^2$ again we have $h=\frac{1}{2}\times10\times36=180m$.

Section 2

1. Work out the distances travelled by both the man and the boy as a function of time and equate.

Apply $s=ut+\frac{1}{2}at^2$ to the man (LHS) and the boy (RHS):

$9t=5t+\frac{1}{2}\times \frac{4}{5}t^2$

$10t=t^2$ so the answer is $B$.

2. Work out the mass and the volume of the mixture and divide.

The density is the total mass divided by the total volume.

$D=\frac{\rho_PV_P+ \rho_QV_Q}{V_P+V_Q}$ so the answer is $B$.

3. Consider the effect this change has on the voltage across $X$ and $Y$.

Think of a potential divider. When the resistance of W increases, this increases the voltage across the combination of W and Y at the expense of the voltage across X. X and Y keep the same resistance throughout, so applying $P=\frac{V^2}{R}$ we see that the power dissipated in Y increases and that in X decreases.

4. Consider change of momentum from the very start to the very end without considering any momentum changes in the interim.

We have zero momentum at the start so should expect zero momentum at the end no matter how complicated the route to the end is therefore the answer is $A$.

5. Think about what extra weight the pan is supporting in both scenarios.

In the second diagram the pan supports an extra $100g$ because the scale has dropped by $1N$. In the third diagram the entire $3N$ is supported by the pan so this is an extra $300g$. Therefore the readings are $500g$ then $700g$.

6. Apply $c=f\lambda$ whilst taking care to determine the correct wavelength from diagram 1.

The diagram shows half a wavelength so the wavelength is $1m$. Now we need $f$ which is given by $c=f\lambda$ so $f=1000Hz$. The time until X in the diagram is $1.5$ cycles so this takes $1.5\times 10^{-3}s$.

7. Use Impulse $=mv-mu=F\times t$ while realising that for a variable force we have a sum of the product of each individual force and time hence the area under the curve.

$m\Delta v=Ft=\frac{1}{2}bh=\frac{750\times4}{1000}$

Using the KE to find out the mass of the ball we have $\frac{1}{2}mv^2=27$ so $m=\frac{27\times 2}{30^2}$.

Rearranging for $\Delta v$ we have:

$\Delta v=\frac{750\times4\times30^2}{27\times2\times1000}$ which cancels to give $\Delta v=50$.

8. Make a moments calculation taking all objects into account including the plank.

Take $F$ to be the force on the plank due to the support at Q.

Taking moments about P we have $(0.2\times5 + 1\times 74 + 1\times 24)g=1.5\times F$

$F=\frac{2}{3}(1+24+74)\times 10=660N$

9. Work out the current in each branch and hence the voltage dropped across each resistor.

The current in each branch is $\frac{12}{6}=2A$. Therefore the voltage of the upper terminal of the voltmeter compared to the negative terminal of the battery is $10V$ and the voltage of the lower terminal of the voltmeter compared to the negative terminal of the battery is $2V$. Therefore the potential difference is $8V$.

10. Work out the extension of each spring then apply $EPE=\frac{1}{2}kx^2$.

The extension of the lower spring is $\frac{mg}{k}$ as normal and for each upper spring it is $\frac{mg}{3k}$ because the load is shared between the three springs. Therefore the energy stored is:

$\frac{1}{2}k((\frac{mg}{k})^2+3(\frac{mg}{3k})^2)=\frac{(mg)^2}{2k}(1+\frac{1}{3})=\frac{2(mg)^2}{3k}$

11. We require the friction force on the block for which we need the acceleration which is achieved with SUVAT.

$v^2=u^2+2as$

$4=3a$ so $a=\frac{4}{3}$.

The force down the slope due to the weight of the block is $3.6gsin30=\frac{3.6g}{2}$. Applying Newton's Second Law to the block we have:

$\frac{3.6g}{2}-F=3.6\times\frac{4}{3}$ where $F$ is the friction force.

$F=18-4.8=13.2N$

The average rate of work done is the average power which is the Force $\times$ the average velocity ($1ms^{-1}$) so the average rate of work done is $13.2N$.

12. Work out the resistance of the combination then write the current leaving the cell in terms of $r$. Then apply Ohm's law to the internal resistance.

The resistance of the combination is $\frac{1}{\frac{1}{20}+\frac{1}{30}}=\frac{60}{3+2}=12\Omega$.

The current leaving the cell is $I=\frac{6}{12+r}$ but we also have that $Ir=1.2V$ because $4.8V$ is dropped across the remainder of the circuit so $6-4.8=1.2V$ must be dropped across the internal resistance. So we have:

$1.2=\frac{6r}{12+r}$

$14.4+1.2r=6r$

$r=\frac{14.4}{4.8}=3\Omega$

13. Work out how far the wave travels in one second then apply $v=f\lambda$.

Inbetween the diagrams, the wave has travelled a wavelength plus $1.5cm$ which has taken one second. Therefore this distance is numerically equal to the speed of the wave.

$v=\lambda+1.5=f\lambda=\frac{5\lambda}{4}$ so $\lambda=6cm$.

14. Use the vertical equilibrium of $R$ to find the tension in the string.

For mass $R$, we have $T=m_Rg$

Therefore the force on $Q$ to the right is also $m_Rg$.

This means that its acceleration would be $\frac{m_Rg}{m_Q}$ so this must be the acceleration of $P$ to prevent $Q$ slipping off $P$.

15. Begin by equating the gravitational force on the cube to $kx$ for the spring.

$mg=3.2N$

$\rho Vg=3.2$

$\rho\times(0.04)^3\times 10=3.2$

$\rho=\frac{32\times100^3}{100\times64}=5\times10^3 kgm^{-3}$

$Pressure=\frac{Force}{Area}=\frac{3.2}{(0.04)^2}=2\times10^3 Nm^{-2}$

16. Apply $R=\frac{\rho L}{A}$ taking care with the introduction of the mass of the wires and the circuit which represents the configuration of wires.

The wires have different resistivities, densities and masses, so working generally we have:

$R_{wire}=\frac{\rho_{wire}L}{A}$ but the volume of the wire is given by $AL$ but also $\frac{M_{wire}}{d_{wire}}$.

Rearranging this we have $\frac{A}{L}=\frac{M_{wire}}{d_{wire}L^2}$ or after taking a reciprocal, $\frac{L}{A}=\frac{d_{wire}L^2}{M_{wire}}$.

Substituting in we have $R_{wire}=\frac{\rho_{wire}d_{wire}L^2}{M_{wire}}$.

To determine $M_{wire}$ for each of the Copper and Aluminium wires, note that the Copper wire is $3$ times as dense as the Aluminium. There are $6$ Aluminium wires and $1$ Copper wire which is $3$ times as heavy so the total mass of the wire is $9$ times that of the Aluminium wire. Therefore the mass of each Aluminium wire is $\frac{M}{9}$ and the Copper wire is $\frac{M}{3}$.

Substituting this and the resistivities and densities in from the table gives:

$R_{Copper}=\frac{2\rho \times 3dL^2}{\frac{M}{3}}=\frac{18\rho dL^2}{M}$ and

$R_{Aluminium}=\frac{3\rho \times dL^2}{\frac{M}{9}}=\frac{27\rho dL^2}{M}$.

The wires make a parallel combination of $6$ Aluminium wires and $1$ Copper wire, giving a resistance of:

$R_{Total}=\frac{1}{\frac{6}{R_{Aluminium}}+\frac{1}{R_{Copper}}}=\frac{1}{\frac{6}{27}+\frac{1}{18}}\times \frac{\rho dL^2}{M}=\frac{54}{15}\times\frac{\rho dL^2}{M}=\frac{18\rho dL^2}{5M}$

17. Work out how much velocity the fluid gains and over what time period this change takes place.

The fluid is incompressible which means that if a certain volume of it is pushed through a pipe then this volume remains constant - it is neither squashed nor expanded. Imagine a cylinder of fluid approaching the narrowing of the pipe. Its volume is $A_1\times l_1$ where $A_1$ is the cross-sectional area of the pipe on the left and $l_1$ is the length of the cylinder of fluid from the left. The volume on the right is $A_2\times l_2$ where the symbols have the same meaning but for the fluid on the right and we have $A_1l_1=A_2l_2$. The lengths of the cylinders are proportional to the speed that the fluid travels at so we have $\frac{v_1}{v_2}=\frac{A_2}{A_1}$ and the velocity after the narrowing is $v_2=\frac{5\times 0.6}{0.25}=12ms^{-1}$. The change in velocity therefore is $7ms^{-1}$. We seek to use Force as the rate of change of Momentum so we now need a mass of fluid and a time over which the narrowing takes place. It doesn't matter how much fluid we use because we are interested in the rate of change, not the overall amount of fluid passing. If we work with $1m^3$ of fluid, this has length $\frac{5}{3}m$ as it approaches the narrowing. At $5ms^{-1}$ this takes $\frac{1}{3}s$. To check, the same $1m^3$ of fluid has a length of $4m$ after the narrowing and at a speed of $12ms^{-1}$ the fluid takes $\frac{1}{3}ms^{-1}$ to leave the narrowing as expected.

$F=\frac{m\Delta v}{\Delta t}=\frac{800\times7}{\frac{1}{3}}=7\times2400=16800N$

18. Work out how much energy is lost by the ball as as a result of its loss of height. A certain fraction of this must be lost by the time it reaches $P$.

Let O be the centre of the sphere. The vertical height lost is $OQcos45=\frac{r}{\sqrt2}$ and so the energy lost to friction up to Q is $\frac{mgr}{\sqrt2}$.

Since the friction force is constant, $\frac{2}{3}$ of this must be lost by the point P. The ball starts with potential energy $mgr$ and the KE is given by this minus the loss due to friction.

KE$=mgr-\frac{2mgr}{3\sqrt2}=mgr(1-\frac{\sqrt2}{3})$.