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.

Hints

Section 1

Part A

1. Solve the inequality as normal remembering to reverse the sense of the inequality if you multiply by a negative number.

2. Recall how alpha and beta decays affect the composition of a nucleus. Crucially, beta decay does not change the number of nucleons.

3. Expand using ordinary methods.

4. Write equations for each scenario. This will eliminate one of the options. Substituting numbers will confirm which options agree with the graph.

5. Re-write both ratios with $R$ the same size. This makes it easy to compare $Q$ and $S$.

6. The number of protons and neutrons must be conserved.

7. Form an equation with the sum of the ages of the two new members as an unknown and solve.

8. Let the resistance of the resistor be $R$ and find an expression for $I$. Use $P=I^2R$, deduce what value of $R$ will maximise this and substitute it in to find $P$.

9. Use percentage multipliers to find out the value after two years and subtract from the initial value.

10. Ignore $k$. The changes detailed in the question have the effect of multiplying the power by a number and dividing it by another.

11. Draw a clear diagram and spot a familiar triangle.

12. In a transverse wave, particles don't move in the direction of propagation of the wave, they only move at right angles to it. Ignore the wavelength and work out how many amplitudes the particle must travel through.

13. Write the proportionality as an equation with a constant of proportionality $k$. Use the boundary conditions to evaluate $k$ then substitute for $x$ in order to find $y$.

14. The tension in the rope is enough to support the weight and furthermore give it an acceleration. Write an equation for $T$ and remember to double this because the question is asking for the tension in the coupling, not the rope.

15. Express the area of the trapezium in terms of $x$ and equate to 120. The resulting equation can be solved by inspection.

16. Use $V=IR$ to work out the current in the circuit then $Q=It$ for the charge delivered. Equate the energy dissipated in the heater to $IVt$ and use the current to work out $V$ for the heater.

17. Eliminate fractions, place all terms involving $b$ on one side, factorise then isolate $b$.

18. Use $Density=\frac{Mass}{Volume}$ whilst being careful with units.

19. Work out the radius of the cylinder from the circumference and then use $Volume=\pi r^2 h$

20. Convection can only occur if there is a temperature difference and the rate of energy loss by thermal emission of a black body depends only on the surface area of it and its absolute temperature $T$. Black bodies are the best emitters/absorbers of radiation.

21. Factorise first then cancel.

22. The area under a velocity-time graph is the displacement. Be careful to add distances and subtract displacements where appropriate. The average speed is the total distance travelled divided by the time taken.

23. Make a two-way table - for example place the activities as columns and the genders as rows and fill in how many of each gender chose each activity.

24. Write an expression for the total mass. Divide the mass of tin by this and simplify.

25. Note all the powers of 3. Re-write everything as a power of 3 and apply the laws of indices.

26. Use $Kinetic Energy=\frac{1}{2}mv^2$ and re-write the velocity of the thorium nucleus using conservation of momentum. Rearrange for the square of the velocity of the alpha particle and use $\frac{1}{2}mv^2$.

27. There are $n$ equal exterior angles and they sum to 360. Use this to find one angle of the isosceles triangle.

28. From the distances deduce that one path is $\frac{5}{3}$ times as long as the other. Therefore the short path takes the same time to traverse as three clicks. Use this to work out the time taken for one click and hence the frequency.

Part B

29. Substitute the two values of $x$ which will give simultaneous equations in $p$ and $q$. Eliminate $q$.

30. Newton's Third Law can only be applied between two bodies where the forces are equal and opposite. This eliminates all but one equation.

31. Find the centre and radius of the circle and expand the resulting equation.

32. Define a tension $T$ in the rope and apply Newton's second law to the crate in isolation.

33. Find the common ratio of the series (preferably by inspection) and apply the sum to infinity formula.

34. Resolve the pushing force horizontally and vertically using the approximations given then invoke vertical equilibrium and $Work=Force\times Distance$.

35. Use Pythagoras or familiar triangles to work out the length of the tangents. Once you have this it is straightforward to calculate the area of the two triangles and subtract the area of the sector from it.

36. Define the centre of mass of the plank to be an unknown distance from its centre, say $x$ to the right. Use this and moments to form an equation involving $x$ and solve it. The force on the pivot is the combined weight of the people and the plank.

37. Rewrite the equation in terms of $cos\theta$ using trigonometric identities. Solve the resulting quadratic equation.

38. Use initial energy + kinetic energy gained due to fall - energy lost due to friction = final kinetic energy. Solve for the energy lost due to friction.

39. Rearrange and form an inequality for the discriminant then solve.

40. The only forces apart from friction acting on the block are at right angles or parallel to each other. Investigating whether the perpendicular distance from the pivot is $d$ will help to ascertain the moment due to $P$.

41. From the equations of the lines, the distances $LO$ and $OM$ can be found in terms of $m$ and $p$. Equate their sum to 5 and use the condition for two gradients to be perpendicular to eliminate $m$ or $p$.

42. Use $GPE=mgh$ being careful to locate the centres of mass of the blocks correctly.

43. Differentiate and set the gradient to be greater than zero.

44. Use change in KE = change in GPE to find $h$. Knowing that solving for $t$ in the second part will lead to two solutions should point you to the correct equation of constant acceleration to use.

45. Find the co-ordinates of the vertex of the new parabola and write its equation in completed square form.

46. Use conservation of momentum to find out the velocity of $P$ after the collision and work out the kinetic energy of all the balls before and after the collision.

47. Spot that this is a disguised quadratic. Let $y=x^2$ and solve for $y$ then $x$.

48. Work out the magnitude of the resultant force (either by Pythagoras or (better) familiar triangles) and subtract the maximum friction force available from it. Divide by the mass of the object.

49. Differentiate the expression and find the height of the turning point which has a negative $x$ value.

50. Define a tension, $T$ in the string. Form two equations using Newton's second law for each mass. Eliminate $T$ and solve for the frictional force.

51. Note that the answer to this question is identical under reflection, interchange $x$ and $y$ then integrate as normal.

52. Form a before and after conservation of momentum equation. This is best done from the perspective of the spacecraft.

53. This is an occasion where it is best to use stationary points to ascertain the behaviour of a graph. Find the turning points and rough shape. This should give you an idea of which horizontal lines, when drawn through the curve, intersect it in four places.

54. Make a triangle of forces with the weight, tension and external force. Use familiar triangles to work out the distance between the object and the ceiling and then the change in height and hence GPE.

Section 2

1a. The stiffest sample will deform least for the same force applied.

1b. Hooke's law is a straight line relationship between force and extension. Although it doesn't look like it, this is not the case for one of the samples over the range indicated.

1c. Use Young's modulus $=\frac{Fl}{Ae}$.

1d. The work done on the sample is the area under the graph.

2a. Series combinations deliver the same current to all components. Parallel combinations deliver the same voltage to all components.

2b. We need the current for the whole circuit which is $\frac{V}{R_{Total}}$. $R_{Total}$ can be found using the rules for adding resistances in series/parallel.

2c. Use $P=\frac{V^2}{R}$ where this time, $V$ is the voltage across $R_3$. Let $R_1=R_2$.

2d. Differentiate your expression with respect to $R_3$ and set equal to zero or follow the hint and take the reciprocal first.

3a. Use $Distance=Speed\times Time$.

3b. This can be thought of simply by realising that the refractive index means that the pulse will take longer to reach the other end of the fibre.

3c. Similarly to 3b., the greater the refractive index, the longer the journey will take.

4a. Work out the speed of the cyclist at B then resolve this down the slope.

4b. Work out the horizontal and vertical distances travelled in terms of t. Use trigonometry to link them.

4c. Resolve initial velocity and acceleration parallel to the slope then apply the equations of constant acceleration.

4d. This time work perpendicular to the slope and minimise the height attained

Full solutions

Section 1

Part A

1. Solve the inequality as normal remembering to reverse the sense of the inequality if you multiply by a negative number.

Subtract 6 from both sides to give

$-14<-\frac{x}{2}$

$\frac{x}{2}<14$

$x<28$

2. Recall how alpha and beta decays affect the composition of a nucleus. Crucially, beta decay does not change the number of nucleons.

Looking at the mass numbers, the reduction by 4 implies that there must be only one alpha decay, so the answer must be D

3. Expand using ordinary methods.

$(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})$

$3-2\sqrt{2}\sqrt{3}+2=5-2\sqrt{2}\sqrt{3}$

4. Write equations for each scenario. This will eliminate one of the options. Substituting numbers will confirm which options agree with the graph.

$KE=\frac{1}{2}mv^2$ which would be a parabola, not a straight line.

$GPE=mgh$ but if you lift a mass of 20kg by a height of 2m then it doesn't have a GPE of 10J.

$v=at$ and $a=\frac{F}{m}=5$ which agrees with the graph.

$Work = Force \times Distance$ and 10J of work would be done over 2m which agrees with the graph. The 12kg is superfluous information.

5. Re-write both ratios with $R$ the same size. This makes it easy to compare $Q$ and $S$.

$Q:R=5:2=15:6$

$S:R=10:3=20:6$

$Q:S=15:20=3:4$

6. The number of protons and neutrons must be conserved.

1. $1+235\neq144+1+1+89$ and so is not possible.

2. $1+235=137+1+1+1+96$ and $92=54+38$ so this is possible.

3. $1+235\neq87+1+1+1+145$ so this is not possible.

7. Form an equation with the sum of the ages of the two new members as an unknown and solve.

Let $x$ be the ages of the current members and $y$ be the ages of the new members.

$\frac{\Sigma x}{20}=28$

$\frac{\Sigma x +\Sigma y}{22}=30$

$\Sigma y = 30\times22-20\times28$

Mean age of new members$=\frac{\Sigma y}{2}=15\times22-10\times28=330-280=50$

8. Let the resistance of the resistor be $R$ and find an expression for $I$. Use $P=I^2R$, deduce what value of $R$ will maximise this and substitute it in to find $P$.

$I=\frac{V}{5+R}. P=\frac{24^2\times5}{(5+3)^2}=3^2\times5=45W$. Time saving tip: Don't evaluate $24^2$! Look for cancellations where possible.

9. Use percentage multipliers to find out the value after two years and subtract from the initial value.

$15000\times(0.8)^2=\frac{4^2}{5^2}\times15000=4^2\times600=9600$

$15000-9600=5400$

10. Ignore $k$. The changes detailed in the question have the effect of multiplying the power by a number and dividing it by another.

The changes are a multiplication by $100^2=10000$ and a division by $2^4=16$. Note that $4\div16=\frac{1}{4}$. The result is $0.25\times10^{30}$.

11. Draw a clear diagram and spot a familiar triangle.

$ACB$ is a right-angle. It is best to spot that we have half an equilateral triangle with side $4$. This means that the height is $2\sqrt3$. Alternatively, $cos30=\frac{h}{4}$ therefore $h=4\times\frac{\sqrt{3}}{2}=2\sqrt{3}$

12. In a transverse wave, particles don't move in the direction of propagation of the wave, they only move at right angles to it. Ignore the wavelength and work out how many amplitudes the particle must travel through.

There are 5 oscillations per second which means that a point on the wave travels 20 amplitudes in one second. $20\times60\times3=3600$.

13. Write the proportionality as an equation with a constant of proportionality $k$. Use the boundary conditions to evaluate $k$ then substitute for $x$ in order to find $y$.

$x=\frac{k}{\sqrt{y}}$

$8=\frac{k}{3}$, therefore $k=24$

Substitute $x=6$

$6=\frac{24}{\sqrt{y}}$

$y=16$

14. The tension in the rope is enough to support the weight and furthermore give it an acceleration. Write an equation for $T$ and remember to double this because the question is asking for the tension in the coupling, not the rope.

$T-mg=ma$

$T=5(10+0.8)$

$2T=10\times10.8=108$

15. Express the area of the trapezium in terms of $x$ and equate to 120. The resulting equation can be solved by inspection.

$\frac{x}{2}(x+5+x-1)=120$

$x(x+2)=120$

$x=10$

16. Use $V=IR$ to work out the current in the circuit then $Q=It$ for the charge delivered. Equate the energy dissipated in the heater to $IVt$ and use the current to work out $V$ for the heater.

$I=\frac{V}{R}=\frac{6}{15}=0.4A$

$Q=It=0.4\times3\times60=72C$

$180=IVt$ and $t=180$ so $IV=1$ and $V=\frac{1}{0.4}=2.5$

17. Eliminate fractions, place all terms involving $b$ on one side, factorise then isolate $b$.

$3b^2a-a=b^2+2$

$b^2(3a-1)=a+2$

$b=\pm\sqrt\frac{a+2}{3a-1}$

18. Use $Density=\frac{Mass}{Volume}$ whilst being careful with units.

$Mass=3kg$

$Density=\frac{3}{1000-10\times25}=\frac{3}{750}=\frac{1}{250}=4\times10^{-3}kgm^{-3}$

19. Work out the radius of the cylinder from the circumference and then use $Volume=\pi r^2 h$

$r=\frac{5}{2\pi}$

$V=\pi r^2\times10=\pi\times\frac{25}{4\pi^2}\times10=\frac{125}{2\pi}$

20. Convection can only occur if there is a temperature difference and the rate of energy loss by thermal emission of a black body depends only on the surface area of it and its absolute temperature $T$. Black bodies are the best emitters/absorbers of radiation.

21. Factorise first, then cancel.

$4+\frac{(2+x)(2-x)}{x(x-2)}=4-\frac{2+x}{x}=3-\frac{2}{x}$

22. The area under a velocity-time graph is the displacement. Be careful to add distances and subtract displacements where appropriate. The average speed is the total distance travelled divided by the time taken.

$Large Area = 80$

$Small Area = 10$

$Distance = 80+10=90$

$Displacement = 80-10=70$

$Average Speed = \frac{90}{30}=3$

23. Make a two-way table - for example place the activities as columns and the genders as rows and fill in how many of each gender chose each activity.

The probability is $\frac{32}{74}=\frac{16}{37}$

24. Write an expression for the total mass. Divide the mass of tin by this and simplify.

Total mass=Mass of tin + mass of copper

$M_{Tot}=\frac{V}{10}\times Y + \frac{9V}{10}\times X$

$Percentage Tin=\frac{\frac{VY}{10}}{\frac{VY}{10}+\frac{9VX}{10}}\times 100=\frac{100VY}{VY+9VX}=\frac{100Y}{Y+9X}$

25. Note all the powers of 3. Re-write everything as a power of 3 and apply the laws of indices.

$\frac{3^{4n+2}\times 3^{4-3n}}{3^{6-3n}}=\frac{3^{n+6}}{3^{6-3n}}=3^{4n}$

26. Use $Kinetic Energy=\frac{1}{2}mv^2$ and re-write the velocity of the thorium nucleus using conservation of momentum. Rearrange for the square of the velocity of the alpha particle and use $\frac{1}{2}mv^2$.

Conservation of momentum

$234V_T=4V_{\alpha}$

Conservation of energy $E=2V_{\alpha}^2+117V_{T}^2=2V_{\alpha}^2+117\times(\frac{2}{117})^2V_{\alpha}^2=2V_{\alpha}^2+\frac{4}{117}V_{\alpha}^2=\frac{238V_{\alpha}^2}{117}$

$KE_{alpha}=\frac{1}{2}mV_{\alpha}^2=2V_{\alpha}^2=\frac{117E}{119}$

27. There are $n$ equal exterior angles and they sum to 360. Use this to find one angle of the isosceles triangle.

$RQT=\frac{360}{n}$

$x=180-2\times RQT=180-\frac{720}{n}$

$n=\frac{720}{180-x}$

28. From the distances deduce that one path is $\frac{5}{3}$ times as long as the other. Therefore the short path takes the same time to traverse as three clicks. Use this to work out the time taken for one click and hence the frequency.

$96\div3=32m$

Time taken for 1 click = $\frac{32}{320}=\frac{1}{10}$ therefore the frequency is $10Hz$.

Part B

29. Substitute the two values of $x$ which will give simultaneous equations in $p$ and $q$. Eliminate $q$.

$8+4p+2q+p^2=0$

$1+p+q+p^2=-3.5$

Multiplying the second equation by 2 gives

$2+2p+2q+2p^2=-7$

Subtracting this from the first equation gives

$6+2p-p^2=7$

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

$p=1$

30. Newton's Third Law can only be applied between two bodies where the forces are equal and opposite. This eliminates all but one equation.

Newton's Third Law states that "If body A exerts a force on body B then body B exerts an equal and opposite force on body A". This can't therefore be true for any equation with more than two unknowns. The air resistance on the parachute and the parachutist won't be equal which leaves us with only one equation which follows from Newton's Third Law.

31. Find the centre and radius of the circle and expand the resulting equation.

The $x$ co-ordinate of the centre is the midpoint of PQ (-2). The square has side 6 so the radius of the circle is 3 and the $y$ co-ordinate of its centre is 3.

$(x+2)^2+(y-3)^2=9$

$x^2+y^2+4x-6y+4+9=9$

$x^2+y^2+4x-6y+4=0$

32. Define a tension $T$ in the rope and apply Newton's second law to the crate in isolation.

$T-mg=ma$

$T=800(10+2)=9600N$

33. Find the common ratio of the series (preferably by inspection) and apply the sum to infinity formula.

$r=\frac{1}{\sqrt{2}}$

$S_{\infty}=\frac{a}{1-r}=\frac{8}{1-\frac{1}{\sqrt{2}}}=\frac{8(1+\frac{1}{\sqrt{2}})}{(1-\frac{1}{\sqrt{2}})(1+\frac{1}{\sqrt{2}})}=\frac{8(1+\frac{1}{\sqrt{2}})}{\frac{1}{2}}=8(2+\sqrt{2})$

34. Resolve the pushing force horizontally and vertically using the approximations given then invoke vertical equilibrium and $Work=Force\times Distance$.

$Vertical component = 50sin37 = 30N$

This combines with the weight of the trolley to increase the normal reaction to $350+30=380$

Horizontally, work done = force $\times$ distance $= 50cos37\times15 = 40\times 15 = 600J$

35. Use Pythagoras or familiar triangles to work out the length of the tangents. Once you have this it is straightforward to calculate the area of the two triangles and subtract the area of the sector from it.

Let the length of each tangent to the circle be $h$

$h=\sqrt{20^2-10^2}=10\sqrt{3}$

Area of two triangles $=2\times\frac{1}{2}\times 10h=100\sqrt{3}$

The triangles are each half an equilateral triangle, so the angle subtended by the sector is $120^{\circ}$, therefore the sector is one third of the circle.

Shaded Area $=100\sqrt{3}-\frac{\pi 10^2}{3}$

36. Define the centre of mass of the plank to be an unknown distance from its centre, say $x$ to the right. Use this and moments to form an equation involving $x$ and solve it. The force on the pivot is the combined weight of the people and the plank.

$60\times80=35\times120+15\times x$

$x=\frac{60\times80-35\times120}{15}=320-280=40cm$

Combined mass of people plus plank $=35+60+15=110kg$ therefore the force is $1100N$

Note - don't perform any unnecessary multiplications.

37. Rewrite the equation in terms of $cos\theta$ using trigonometric identities. Solve the resulting quadratic equation.

$7cos\theta-\frac{3sin^2\theta}{cos\theta}=1$

$7cos^2\theta-3+3cos^2\theta=cos\theta$

$10cos^2\theta-cos\theta-3=0$

$(5cos\theta-3)(2cos\theta+1)=0$

38. Use initial energy + kinetic energy gained due to fall - energy lost due to friction = final kinetic energy. Solve for the energy lost due to friction.

$\frac{1}{2}mv^2+mgh-Fr=\frac{1}{2}m\times 9^2$

$Fr=\frac{1}{2}m\times 5^2 + 6mg - \frac{1}{2}m \times 9^2=200(\frac{25}{2}-\frac{81}{2}+60)=200\times32=6400$

39. Rearrange and form an inequality for the discriminant then solve.

$b^2-4ac>0$

$(a+2)^2-9\times4>0$

$(a+2)^2>36$

$-6>a+2, a+2>6$

$-8>a, a>4$

40. The only forces apart from friction acting on the block are at right angles or parallel to each other. Investigating whether the perpendicular distance from the pivot is $d$ will help to ascertain the moment due to $P$.

P is the only horizontal force acting on the block apart from friction, so there has to be a friction force to the left to counter P if the block is to remain in equilibrium. This helps us decide between 1,2 and 3. The top of the cube is a distance $d(cos30+sin30)$ from the table which is not the same as $d$ so 4 is not the case.

41. From the equations of the lines, the distances $LO$ and $OM$ can be found in terms of $m$ and $p$. Equate their sum to 5 and use the condition for two gradients to be perpendicular to eliminate $m$ or $p$.

$LO=\frac{3}{m}$ and $OM=\frac{-2}{p}$

$LO+OM=\frac{3}{m}-\frac{2}{p}=5$

$\frac{3p-2m}{mp}=5$

Let $p=\frac{-1}{m}$

$\frac{\frac{-3}{m}-2m}{-1}=5$

$2m^2-5m+3=0$

$(2m-3)(m-1)=0$ therefore $m=1$ or $\frac{3}{2}$.

From the question, $m\neq 1$ therefore $m=\frac{3}{2}$ and $n=\frac{-2}{3}$ so $m+n=\frac{5}{6}$.

42. Use $GPE=mgh$ being careful to locate the centres of mass of the blocks correctly.

1. $mg(\frac{b}{2} + b+\frac{b}{2})=2mgb$

2. $mg(\frac{a}{2} + a+\frac{a}{2})=2mga$

$2. - 1. = 2mg(a-b)$

43. Differentiate and set the gradient to be greater than zero.

$3x^2-a^2>0$

$x>\frac{a}{\sqrt{3}}$ or $x<\frac{-a}{\sqrt{3}}$

44. Use change in KE = change in GPE to find $h$. Knowing that solving for $t$ in the second part will lead to two solutions should point you to the correct equation of constant acceleration to use.

$\frac{1}{2}m\times 8^2 - \frac{1}{2}m\times 2^2 = mgh$

$h=\frac{32-2}{10}=3$

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

$3=8t-5t^2$

$5t^2-8t+3=0$

$(5t-3)(t-1)=0$

Since the object is on the way down, select the larger value, so $t=1$.

45. Find the co-ordinates of the vertex of the new parabola and write its equation in completed square form.

The $x$ co-ordinate is 4 and the $y$ co-ordinate is 3 before the reflection. After the reflection, the vertex has moved down by 8 units since it was originally 4 units above the line $y=-1$.

$y=-5-(x-4)^2$

46. Use conservation of momentum to find out the velocity of $P$ after the collision and work out the kinetic energy of all the balls before and after the collision.

$4\times10=4v+2\times10$ therefore $v=5ms^{-1}$

$KE_{Initial}=\frac{1}{2}\times 4 \times 10^2 = 200J$

$KE_{Final}=\frac{1}{2}\times 4 \times 5^2 + \frac{1}{2}\times 2 \times 10^2= 150J$

47. Spot that this is a disguised quadratic. Let $y=x^2$ and solve for $y$ then $x$.

$2y^2-9y+4>0$

$(2y-1)(y-4)>0$

Either $2y-1>0$ and $y-4>0$ or $2y-1<0$ and $y-4<0$

$y>4$ gives $x<-2$, $x>2$

$y<\frac{1}{2}$ gives $\frac{-1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

48. Work out the magnitude of the resultant force (either by Pythagoras or (better) familiar triangles) and subtract the maximum friction force available from it. Divide by the mass of the object.

Using knowledge of 3,4,5 triangles, deduce that the resultant of these forces is 15N

$ma=F-\mu R=15-0.25\times2\times10=10$ therefore $a=\frac{10}{2}=5$

49. Differentiate the expression and find the height of the turning point which has a negative $x$ value.

$12x^2-24x-36=0$

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

$(x+1)(x-3)=0$

We are interested in the negative $x$ value. Let $x$=-1.

$y=-4-12+36-15=5$ so this point is 5 away from the path of the cursor

50. Define a tension, $T$ in the string. Form two equations using Newton's second law for each mass. Eliminate $T$ and solve for the frictional force.

$30g-T=30a$

$T-20gsin30-Fr=20a$

Add the two equations.

$30g-10g-Fr=50a$

$Fr=20g-50a=75N$

51. Note that the answer to this question is identical under reflection, interchange $x$ and $y$ then integrate as normal.

By inspection, the intersections of the line and curve occur for $x=0$ and $2$.

$\int_{0}^{2}7-3x^2+6x-7dx=[-x^3+3x^2]_{0}^{2}=-8+12=4$

52. Form a before and after conservation of momentum equation. This is best done from the perspective of the spacecraft before the ejection.

In the frame of the spacecraft before the ejection the momentum is $0$. After the ejection of fuel, we have an unknown mass travelling to the left at $1425ms^{-1}$ and $4000$ minus this mass travelling to the right at $75ms^{-1}$. The reason the spacecraft's velocity is now $75ms^{-1}$ is because the spacecraft is now travelling $75ms^{-1}$ faster than it previously was.

$1425m=75\times4000-75m$

$1500m=75\times4000$

$m=200$

53. This is an occasion where it is best to use stationary points to ascertain the behaviour of a graph. Find the turning points and rough shape. This should give you an idea of which horizontal lines, when drawn through the curve, intersect it in four places.

$12x^3-12x^2-24x=0$

$x(x^2-x-2)=0$

$x(x+1)(x-2)=0$ therefore $x=-1, 0$ or $2$.

When $x=-1$, $y=15$

When $x=0$, $y=20$

When $x=2$, $y=-12$

From the sketch, notice that a horizontal line drawn through the curve would only intersect in 4 different places if the height of that line were between (but not including) 15 and 20.

54. Make a triangle of forces with the weight, tension and external force. Use familiar triangles to work out the distance between the object and the ceiling and then the change in height and hence GPE.

From the diagram, deduce that we have a 3,4,5 triangle, so the vertical line in the triangle on the right must have length $0.28$. The change in height is $0.35-0.28=0.07$. Multiply by $mg$ to give $2.8J$.

Section 2

1a. The stiffest sample will deform least for the same force applied.

Stiffness is a measure of how little the sample deforms after a given tension is applied. This is true for the sample with the largest gradient. Answer=A

1b. Hooke's law is a straight line relationship between force and extension. Although it doesn't look like it, this is not the case for one of the samples over the range indicated.

$Strain=\frac{Extension}{Natural Length}$, so the extension in question is 2mm. Hold your pencil up to each of the lines and you will see that S2 is not linear in this interval. Answer=B

1c. Use Young's modulus $=\frac{Fl}{Ae}$.

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

Reading the values from the graph:

$E=\frac{250\times0.1}{(5\times10^{-3})^2\times0.005}=\frac{25\times10^9}{125}=\frac{10^9}{5}=2\times10^8$. Answer = A

1d. The work done on the sample is the area under the graph.

$\int_0^{0.01}ax-bx^2dx=[\frac{ax^2}{2}-\frac{bx^3}{3}]_0^{0.01}=\frac{a}{2}\times10^{-4}-\frac{b}{3}\times10^{-6}$. Answer=A

2a. Series combinations deliver the same current to all components. Parallel combinations deliver the same voltage to all components.

$R_2$ and $R_3$ form a parallel combination. Answer=F

2b. We need the current for the whole circuit which is $\frac{V}{R_{Total}}$. $R_{Total}$ can be found using the rules for adding resistances in series/parallel.

$I=\frac{V}{R_{Total}}$

$R_{Total}=R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}}=R_1+\frac{R_2R_3}{R_2+R_3}=\frac{R_1R_2+R_1R_3+R_2R_3}{R_2+R_3}$

$I=\frac{V(R_2+R_3)}{R_1R_2+R_1R_3+R_2R_3}$. Answer=A

2c. Use $P=\frac{V^2}{R}$ where this time, $V$ is the voltage across $R_3$. Let $R_1=R_2$.

$P=\frac{V_3^2}{R_3}$

$V_3=V-IR_1=V-\frac{VR_1(R_2+R_3)}{R_1R_2+R_1R_3+R_2R_3}=\frac{VR_2R_3}{R_1R_2+R_1R_3+R_2R_3}$

Use $R_1=R_2$

$V_3=\frac{VR_1R_3}{R_1^2+2R_1R_3}$ therefore $P=\frac{V^2R_1^2R_3}{(R_1^2+2R_1R_3)^2}=\frac{V^2R_3}{(R_1+2R_3)^2}$. Answer=D

2d. Differentiate your expression with respect to $R_3$ and set equal to zero or follow the hint and take the reciprocal first.

$P\alpha\frac{R_3}{(R_1+2R_3)^2}$

$\frac{1}{P}\alpha\frac{R_1^2+4R_1R_3+4R_3^2}{R_3}$

$\frac{d\frac{1}{P}}{dR_3}\alpha\frac{-R_1^2}{R_3^2}+4=0$

$R_3=\frac{R_1}{2}$. Answer=B (negative resistances don't make sense).

3a. Use $Distance=Speed\times Time$.

$3\times10^8\times1\times10^{-9}=0.3m$. Answer=C

3b. This can be thought of simply by realising that the refractive index means that the pulse will take longer to reach the other end of the fibre.

$Time=\frac{Distance}{Speed}=1.5\times\frac{9000}{3\times10^8}=\frac{3\times 3000\times10^{-8}}{2}=\frac{9}{2}\times10^{-5}$. Answer=F

3c. Similarly to 3b., the greater the refractive index, the longer the journey will take.

Nominal is the slowest because the speed is divided by the greatest $n$. The next slowest is blue, followed by red. Answer=A

4a. Work out the speed of the cyclist at B then resolve this down the slope.

$E+mgh=\frac{1}{2}mV^2$

$V=\sqrt{(\frac{2E}{m}+2gh)}$

$V_a=\sqrt{(\frac{2E}{m}+2gh)}cos\theta$. Answer=E

4b. Work out the horizontal and vertical distances travelled in terms of t. Use trigonometry to link them.

$x=Vt$, $y=\frac{1}{2}gt^2$

$tan\theta=\frac{y}{x}=\frac{\frac{1}{2}gt^2}{Vt}$

$t=\frac{2Vtan\theta}{g}$. Answer=B

4c. Resolve initial velocity and acceleration parallel to the slope then apply the equations of constant acceleration.

$t=\frac{2Vtan\theta}{g}$, $a=gsin\theta$, $u=Vcos\theta$

$s=ut+\frac{1}{2}at^2=Vcos\theta\times\frac{2Vtan\theta}{g}+\frac{1}{2}\times gsin\theta\times \frac{4V^2tan^2\theta}{g^2}=\frac{2V^2sin\theta}{g}+\frac{2V^2sin\theta tan^2\theta}{g}=\frac{2V^2sin\theta}{g}(1+tan^2\theta)=\frac{2V^2sin\theta}{gcos^2\theta}$. Answer=C

4d. This time work perpendicular to the slope and minimise the height attained.

$a=gcos\theta$, $u=Vsin\theta$

$s=ut+\frac{1}{2}at^2=Vsin\theta t - \frac{1}{2}gcos\theta t^2$

$\frac{ds}{dt}=Vsin\theta -gcos\theta t=0$

$t=\frac{Vtan\theta}{g}$

Let $t=\frac{Vtan\theta}{g}$ in the equation for $s$ down the plane used in part c).

$s=Vcos\theta\times\frac{Vtan\theta}{g}+\frac{1}{2}gsin\theta\times\frac{V^2tan^2\theta}{g^2}=\frac{V^2}{g}sin\theta(1+\frac{tan^2\theta}{2})$. Answer=D

I will firstly set out a series of hints. Try to do the questions from these alone and if you are still having difficulty then scroll down to the full solutions.

Hints

Section 1

Part A

1. Solve the inequality as normal remembering to reverse the sense of the inequality if you multiply by a negative number.

2. Recall how alpha and beta decays affect the composition of a nucleus. Crucially, beta decay does not change the number of nucleons.

3. Expand using ordinary methods.

4. Write equations for each scenario. This will eliminate one of the options. Substituting numbers will confirm which options agree with the graph.

5. Re-write both ratios with $R$ the same size. This makes it easy to compare $Q$ and $S$.

6. The number of protons and neutrons must be conserved.

7. Form an equation with the sum of the ages of the two new members as an unknown and solve.

8. Let the resistance of the resistor be $R$ and find an expression for $I$. Use $P=I^2R$, deduce what value of $R$ will maximise this and substitute it in to find $P$.

9. Use percentage multipliers to find out the value after two years and subtract from the initial value.

10. Ignore $k$. The changes detailed in the question have the effect of multiplying the power by a number and dividing it by another.

11. Draw a clear diagram and spot a familiar triangle.

12. In a transverse wave, particles don't move in the direction of propagation of the wave, they only move at right angles to it. Ignore the wavelength and work out how many amplitudes the particle must travel through.

13. Write the proportionality as an equation with a constant of proportionality $k$. Use the boundary conditions to evaluate $k$ then substitute for $x$ in order to find $y$.

14. The tension in the rope is enough to support the weight and furthermore give it an acceleration. Write an equation for $T$ and remember to double this because the question is asking for the tension in the coupling, not the rope.

15. Express the area of the trapezium in terms of $x$ and equate to 120. The resulting equation can be solved by inspection.

16. Use $V=IR$ to work out the current in the circuit then $Q=It$ for the charge delivered. Equate the energy dissipated in the heater to $IVt$ and use the current to work out $V$ for the heater.

17. Eliminate fractions, place all terms involving $b$ on one side, factorise then isolate $b$.

18. Use $Density=\frac{Mass}{Volume}$ whilst being careful with units.

19. Work out the radius of the cylinder from the circumference and then use $Volume=\pi r^2 h$

20. Convection can only occur if there is a temperature difference and the rate of energy loss by thermal emission of a black body depends only on the surface area of it and its absolute temperature $T$. Black bodies are the best emitters/absorbers of radiation.

21. Factorise first then cancel.

22. The area under a velocity-time graph is the displacement. Be careful to add distances and subtract displacements where appropriate. The average speed is the total distance travelled divided by the time taken.

23. Make a two-way table - for example place the activities as columns and the genders as rows and fill in how many of each gender chose each activity.

24. Write an expression for the total mass. Divide the mass of tin by this and simplify.

25. Note all the powers of 3. Re-write everything as a power of 3 and apply the laws of indices.

26. Use $Kinetic Energy=\frac{1}{2}mv^2$ and re-write the velocity of the thorium nucleus using conservation of momentum. Rearrange for the square of the velocity of the alpha particle and use $\frac{1}{2}mv^2$.

27. There are $n$ equal exterior angles and they sum to 360. Use this to find one angle of the isosceles triangle.

28. From the distances deduce that one path is $\frac{5}{3}$ times as long as the other. Therefore the short path takes the same time to traverse as three clicks. Use this to work out the time taken for one click and hence the frequency.

Part B

29. Substitute the two values of $x$ which will give simultaneous equations in $p$ and $q$. Eliminate $q$.

30. Newton's Third Law can only be applied between two bodies where the forces are equal and opposite. This eliminates all but one equation.

31. Find the centre and radius of the circle and expand the resulting equation.

32. Define a tension $T$ in the rope and apply Newton's second law to the crate in isolation.

33. Find the common ratio of the series (preferably by inspection) and apply the sum to infinity formula.

34. Resolve the pushing force horizontally and vertically using the approximations given then invoke vertical equilibrium and $Work=Force\times Distance$.

35. Use Pythagoras or familiar triangles to work out the length of the tangents. Once you have this it is straightforward to calculate the area of the two triangles and subtract the area of the sector from it.

36. Define the centre of mass of the plank to be an unknown distance from its centre, say $x$ to the right. Use this and moments to form an equation involving $x$ and solve it. The force on the pivot is the combined weight of the people and the plank.

37. Rewrite the equation in terms of $cos\theta$ using trigonometric identities. Solve the resulting quadratic equation.

38. Use initial energy + kinetic energy gained due to fall - energy lost due to friction = final kinetic energy. Solve for the energy lost due to friction.

39. Rearrange and form an inequality for the discriminant then solve.

40. The only forces apart from friction acting on the block are at right angles or parallel to each other. Investigating whether the perpendicular distance from the pivot is $d$ will help to ascertain the moment due to $P$.

41. From the equations of the lines, the distances $LO$ and $OM$ can be found in terms of $m$ and $p$. Equate their sum to 5 and use the condition for two gradients to be perpendicular to eliminate $m$ or $p$.

42. Use $GPE=mgh$ being careful to locate the centres of mass of the blocks correctly.

43. Differentiate and set the gradient to be greater than zero.

44. Use change in KE = change in GPE to find $h$. Knowing that solving for $t$ in the second part will lead to two solutions should point you to the correct equation of constant acceleration to use.

45. Find the co-ordinates of the vertex of the new parabola and write its equation in completed square form.

46. Use conservation of momentum to find out the velocity of $P$ after the collision and work out the kinetic energy of all the balls before and after the collision.

47. Spot that this is a disguised quadratic. Let $y=x^2$ and solve for $y$ then $x$.

48. Work out the magnitude of the resultant force (either by Pythagoras or (better) familiar triangles) and subtract the maximum friction force available from it. Divide by the mass of the object.

49. Differentiate the expression and find the height of the turning point which has a negative $x$ value.

50. Define a tension, $T$ in the string. Form two equations using Newton's second law for each mass. Eliminate $T$ and solve for the frictional force.

51. Note that the answer to this question is identical under reflection, interchange $x$ and $y$ then integrate as normal.

52. Form a before and after conservation of momentum equation. This is best done from the perspective of the spacecraft.

53. This is an occasion where it is best to use stationary points to ascertain the behaviour of a graph. Find the turning points and rough shape. This should give you an idea of which horizontal lines, when drawn through the curve, intersect it in four places.

54. Make a triangle of forces with the weight, tension and external force. Use familiar triangles to work out the distance between the object and the ceiling and then the change in height and hence GPE.

Section 2

1a. The stiffest sample will deform least for the same force applied.

1b. Hooke's law is a straight line relationship between force and extension. Although it doesn't look like it, this is not the case for one of the samples over the range indicated.

1c. Use Young's modulus $=\frac{Fl}{Ae}$.

1d. The work done on the sample is the area under the graph.

2a. Series combinations deliver the same current to all components. Parallel combinations deliver the same voltage to all components.

2b. We need the current for the whole circuit which is $\frac{V}{R_{Total}}$. $R_{Total}$ can be found using the rules for adding resistances in series/parallel.

2c. Use $P=\frac{V^2}{R}$ where this time, $V$ is the voltage across $R_3$. Let $R_1=R_2$.

2d. Differentiate your expression with respect to $R_3$ and set equal to zero or follow the hint and take the reciprocal first.

3a. Use $Distance=Speed\times Time$.

3b. This can be thought of simply by realising that the refractive index means that the pulse will take longer to reach the other end of the fibre.

3c. Similarly to 3b., the greater the refractive index, the longer the journey will take.

4a. Work out the speed of the cyclist at B then resolve this down the slope.

4b. Work out the horizontal and vertical distances travelled in terms of t. Use trigonometry to link them.

4c. Resolve initial velocity and acceleration parallel to the slope then apply the equations of constant acceleration.

4d. This time work perpendicular to the slope and minimise the height attained

Full solutions

Section 1

Part A

1. Solve the inequality as normal remembering to reverse the sense of the inequality if you multiply by a negative number.

Subtract 6 from both sides to give

$-14<-\frac{x}{2}$

$\frac{x}{2}<14$

$x<28$

2. Recall how alpha and beta decays affect the composition of a nucleus. Crucially, beta decay does not change the number of nucleons.

Looking at the mass numbers, the reduction by 4 implies that there must be only one alpha decay, so the answer must be D

3. Expand using ordinary methods.

$(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})$

$3-2\sqrt{2}\sqrt{3}+2=5-2\sqrt{2}\sqrt{3}$

4. Write equations for each scenario. This will eliminate one of the options. Substituting numbers will confirm which options agree with the graph.

$KE=\frac{1}{2}mv^2$ which would be a parabola, not a straight line.

$GPE=mgh$ but if you lift a mass of 20kg by a height of 2m then it doesn't have a GPE of 10J.

$v=at$ and $a=\frac{F}{m}=5$ which agrees with the graph.

$Work = Force \times Distance$ and 10J of work would be done over 2m which agrees with the graph. The 12kg is superfluous information.

5. Re-write both ratios with $R$ the same size. This makes it easy to compare $Q$ and $S$.

$Q:R=5:2=15:6$

$S:R=10:3=20:6$

$Q:S=15:20=3:4$

6. The number of protons and neutrons must be conserved.

1. $1+235\neq144+1+1+89$ and so is not possible.

2. $1+235=137+1+1+1+96$ and $92=54+38$ so this is possible.

3. $1+235\neq87+1+1+1+145$ so this is not possible.

7. Form an equation with the sum of the ages of the two new members as an unknown and solve.

Let $x$ be the ages of the current members and $y$ be the ages of the new members.

$\frac{\Sigma x}{20}=28$

$\frac{\Sigma x +\Sigma y}{22}=30$

$\Sigma y = 30\times22-20\times28$

Mean age of new members$=\frac{\Sigma y}{2}=15\times22-10\times28=330-280=50$

8. Let the resistance of the resistor be $R$ and find an expression for $I$. Use $P=I^2R$, deduce what value of $R$ will maximise this and substitute it in to find $P$.

$I=\frac{V}{5+R}. P=\frac{24^2\times5}{(5+3)^2}=3^2\times5=45W$. Time saving tip: Don't evaluate $24^2$! Look for cancellations where possible.

9. Use percentage multipliers to find out the value after two years and subtract from the initial value.

$15000\times(0.8)^2=\frac{4^2}{5^2}\times15000=4^2\times600=9600$

$15000-9600=5400$

10. Ignore $k$. The changes detailed in the question have the effect of multiplying the power by a number and dividing it by another.

The changes are a multiplication by $100^2=10000$ and a division by $2^4=16$. Note that $4\div16=\frac{1}{4}$. The result is $0.25\times10^{30}$.

11. Draw a clear diagram and spot a familiar triangle.

$ACB$ is a right-angle. It is best to spot that we have half an equilateral triangle with side $4$. This means that the height is $2\sqrt3$. Alternatively, $cos30=\frac{h}{4}$ therefore $h=4\times\frac{\sqrt{3}}{2}=2\sqrt{3}$

12. In a transverse wave, particles don't move in the direction of propagation of the wave, they only move at right angles to it. Ignore the wavelength and work out how many amplitudes the particle must travel through.

There are 5 oscillations per second which means that a point on the wave travels 20 amplitudes in one second. $20\times60\times3=3600$.

13. Write the proportionality as an equation with a constant of proportionality $k$. Use the boundary conditions to evaluate $k$ then substitute for $x$ in order to find $y$.

$x=\frac{k}{\sqrt{y}}$

$8=\frac{k}{3}$, therefore $k=24$

Substitute $x=6$

$6=\frac{24}{\sqrt{y}}$

$y=16$

14. The tension in the rope is enough to support the weight and furthermore give it an acceleration. Write an equation for $T$ and remember to double this because the question is asking for the tension in the coupling, not the rope.

$T-mg=ma$

$T=5(10+0.8)$

$2T=10\times10.8=108$

15. Express the area of the trapezium in terms of $x$ and equate to 120. The resulting equation can be solved by inspection.

$\frac{x}{2}(x+5+x-1)=120$

$x(x+2)=120$

$x=10$

16. Use $V=IR$ to work out the current in the circuit then $Q=It$ for the charge delivered. Equate the energy dissipated in the heater to $IVt$ and use the current to work out $V$ for the heater.

$I=\frac{V}{R}=\frac{6}{15}=0.4A$

$Q=It=0.4\times3\times60=72C$

$180=IVt$ and $t=180$ so $IV=1$ and $V=\frac{1}{0.4}=2.5$

17. Eliminate fractions, place all terms involving $b$ on one side, factorise then isolate $b$.

$3b^2a-a=b^2+2$

$b^2(3a-1)=a+2$

$b=\pm\sqrt\frac{a+2}{3a-1}$

18. Use $Density=\frac{Mass}{Volume}$ whilst being careful with units.

$Mass=3kg$

$Density=\frac{3}{1000-10\times25}=\frac{3}{750}=\frac{1}{250}=4\times10^{-3}kgm^{-3}$

19. Work out the radius of the cylinder from the circumference and then use $Volume=\pi r^2 h$

$r=\frac{5}{2\pi}$

$V=\pi r^2\times10=\pi\times\frac{25}{4\pi^2}\times10=\frac{125}{2\pi}$

20. Convection can only occur if there is a temperature difference and the rate of energy loss by thermal emission of a black body depends only on the surface area of it and its absolute temperature $T$. Black bodies are the best emitters/absorbers of radiation.

21. Factorise first, then cancel.

$4+\frac{(2+x)(2-x)}{x(x-2)}=4-\frac{2+x}{x}=3-\frac{2}{x}$

22. The area under a velocity-time graph is the displacement. Be careful to add distances and subtract displacements where appropriate. The average speed is the total distance travelled divided by the time taken.

$Large Area = 80$

$Small Area = 10$

$Distance = 80+10=90$

$Displacement = 80-10=70$

$Average Speed = \frac{90}{30}=3$

23. Make a two-way table - for example place the activities as columns and the genders as rows and fill in how many of each gender chose each activity.

The probability is $\frac{32}{74}=\frac{16}{37}$

24. Write an expression for the total mass. Divide the mass of tin by this and simplify.

Total mass=Mass of tin + mass of copper

$M_{Tot}=\frac{V}{10}\times Y + \frac{9V}{10}\times X$

$Percentage Tin=\frac{\frac{VY}{10}}{\frac{VY}{10}+\frac{9VX}{10}}\times 100=\frac{100VY}{VY+9VX}=\frac{100Y}{Y+9X}$

25. Note all the powers of 3. Re-write everything as a power of 3 and apply the laws of indices.

$\frac{3^{4n+2}\times 3^{4-3n}}{3^{6-3n}}=\frac{3^{n+6}}{3^{6-3n}}=3^{4n}$

26. Use $Kinetic Energy=\frac{1}{2}mv^2$ and re-write the velocity of the thorium nucleus using conservation of momentum. Rearrange for the square of the velocity of the alpha particle and use $\frac{1}{2}mv^2$.

Conservation of momentum

$234V_T=4V_{\alpha}$

Conservation of energy $E=2V_{\alpha}^2+117V_{T}^2=2V_{\alpha}^2+117\times(\frac{2}{117})^2V_{\alpha}^2=2V_{\alpha}^2+\frac{4}{117}V_{\alpha}^2=\frac{238V_{\alpha}^2}{117}$

$KE_{alpha}=\frac{1}{2}mV_{\alpha}^2=2V_{\alpha}^2=\frac{117E}{119}$

27. There are $n$ equal exterior angles and they sum to 360. Use this to find one angle of the isosceles triangle.

$RQT=\frac{360}{n}$

$x=180-2\times RQT=180-\frac{720}{n}$

$n=\frac{720}{180-x}$

28. From the distances deduce that one path is $\frac{5}{3}$ times as long as the other. Therefore the short path takes the same time to traverse as three clicks. Use this to work out the time taken for one click and hence the frequency.

$96\div3=32m$

Time taken for 1 click = $\frac{32}{320}=\frac{1}{10}$ therefore the frequency is $10Hz$.

Part B

29. Substitute the two values of $x$ which will give simultaneous equations in $p$ and $q$. Eliminate $q$.

$8+4p+2q+p^2=0$

$1+p+q+p^2=-3.5$

Multiplying the second equation by 2 gives

$2+2p+2q+2p^2=-7$

Subtracting this from the first equation gives

$6+2p-p^2=7$

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

$p=1$

30. Newton's Third Law can only be applied between two bodies where the forces are equal and opposite. This eliminates all but one equation.

Newton's Third Law states that "If body A exerts a force on body B then body B exerts an equal and opposite force on body A". This can't therefore be true for any equation with more than two unknowns. The air resistance on the parachute and the parachutist won't be equal which leaves us with only one equation which follows from Newton's Third Law.

31. Find the centre and radius of the circle and expand the resulting equation.

The $x$ co-ordinate of the centre is the midpoint of PQ (-2). The square has side 6 so the radius of the circle is 3 and the $y$ co-ordinate of its centre is 3.

$(x+2)^2+(y-3)^2=9$

$x^2+y^2+4x-6y+4+9=9$

$x^2+y^2+4x-6y+4=0$

32. Define a tension $T$ in the rope and apply Newton's second law to the crate in isolation.

$T-mg=ma$

$T=800(10+2)=9600N$

33. Find the common ratio of the series (preferably by inspection) and apply the sum to infinity formula.

$r=\frac{1}{\sqrt{2}}$

$S_{\infty}=\frac{a}{1-r}=\frac{8}{1-\frac{1}{\sqrt{2}}}=\frac{8(1+\frac{1}{\sqrt{2}})}{(1-\frac{1}{\sqrt{2}})(1+\frac{1}{\sqrt{2}})}=\frac{8(1+\frac{1}{\sqrt{2}})}{\frac{1}{2}}=8(2+\sqrt{2})$

34. Resolve the pushing force horizontally and vertically using the approximations given then invoke vertical equilibrium and $Work=Force\times Distance$.

$Vertical component = 50sin37 = 30N$

This combines with the weight of the trolley to increase the normal reaction to $350+30=380$

Horizontally, work done = force $\times$ distance $= 50cos37\times15 = 40\times 15 = 600J$

35. Use Pythagoras or familiar triangles to work out the length of the tangents. Once you have this it is straightforward to calculate the area of the two triangles and subtract the area of the sector from it.

Let the length of each tangent to the circle be $h$

$h=\sqrt{20^2-10^2}=10\sqrt{3}$

Area of two triangles $=2\times\frac{1}{2}\times 10h=100\sqrt{3}$

The triangles are each half an equilateral triangle, so the angle subtended by the sector is $120^{\circ}$, therefore the sector is one third of the circle.

Shaded Area $=100\sqrt{3}-\frac{\pi 10^2}{3}$

36. Define the centre of mass of the plank to be an unknown distance from its centre, say $x$ to the right. Use this and moments to form an equation involving $x$ and solve it. The force on the pivot is the combined weight of the people and the plank.

$60\times80=35\times120+15\times x$

$x=\frac{60\times80-35\times120}{15}=320-280=40cm$

Combined mass of people plus plank $=35+60+15=110kg$ therefore the force is $1100N$

Note - don't perform any unnecessary multiplications.

37. Rewrite the equation in terms of $cos\theta$ using trigonometric identities. Solve the resulting quadratic equation.

$7cos\theta-\frac{3sin^2\theta}{cos\theta}=1$

$7cos^2\theta-3+3cos^2\theta=cos\theta$

$10cos^2\theta-cos\theta-3=0$

$(5cos\theta-3)(2cos\theta+1)=0$

38. Use initial energy + kinetic energy gained due to fall - energy lost due to friction = final kinetic energy. Solve for the energy lost due to friction.

$\frac{1}{2}mv^2+mgh-Fr=\frac{1}{2}m\times 9^2$

$Fr=\frac{1}{2}m\times 5^2 + 6mg - \frac{1}{2}m \times 9^2=200(\frac{25}{2}-\frac{81}{2}+60)=200\times32=6400$

39. Rearrange and form an inequality for the discriminant then solve.

$b^2-4ac>0$

$(a+2)^2-9\times4>0$

$(a+2)^2>36$

$-6>a+2, a+2>6$

$-8>a, a>4$

40. The only forces apart from friction acting on the block are at right angles or parallel to each other. Investigating whether the perpendicular distance from the pivot is $d$ will help to ascertain the moment due to $P$.

P is the only horizontal force acting on the block apart from friction, so there has to be a friction force to the left to counter P if the block is to remain in equilibrium. This helps us decide between 1,2 and 3. The top of the cube is a distance $d(cos30+sin30)$ from the table which is not the same as $d$ so 4 is not the case.

41. From the equations of the lines, the distances $LO$ and $OM$ can be found in terms of $m$ and $p$. Equate their sum to 5 and use the condition for two gradients to be perpendicular to eliminate $m$ or $p$.

$LO=\frac{3}{m}$ and $OM=\frac{-2}{p}$

$LO+OM=\frac{3}{m}-\frac{2}{p}=5$

$\frac{3p-2m}{mp}=5$

Let $p=\frac{-1}{m}$

$\frac{\frac{-3}{m}-2m}{-1}=5$

$2m^2-5m+3=0$

$(2m-3)(m-1)=0$ therefore $m=1$ or $\frac{3}{2}$.

From the question, $m\neq 1$ therefore $m=\frac{3}{2}$ and $n=\frac{-2}{3}$ so $m+n=\frac{5}{6}$.

42. Use $GPE=mgh$ being careful to locate the centres of mass of the blocks correctly.

1. $mg(\frac{b}{2} + b+\frac{b}{2})=2mgb$

2. $mg(\frac{a}{2} + a+\frac{a}{2})=2mga$

$2. - 1. = 2mg(a-b)$

43. Differentiate and set the gradient to be greater than zero.

$3x^2-a^2>0$

$x>\frac{a}{\sqrt{3}}$ or $x<\frac{-a}{\sqrt{3}}$

44. Use change in KE = change in GPE to find $h$. Knowing that solving for $t$ in the second part will lead to two solutions should point you to the correct equation of constant acceleration to use.

$\frac{1}{2}m\times 8^2 - \frac{1}{2}m\times 2^2 = mgh$

$h=\frac{32-2}{10}=3$

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

$3=8t-5t^2$

$5t^2-8t+3=0$

$(5t-3)(t-1)=0$

Since the object is on the way down, select the larger value, so $t=1$.

45. Find the co-ordinates of the vertex of the new parabola and write its equation in completed square form.

The $x$ co-ordinate is 4 and the $y$ co-ordinate is 3 before the reflection. After the reflection, the vertex has moved down by 8 units since it was originally 4 units above the line $y=-1$.

$y=-5-(x-4)^2$

46. Use conservation of momentum to find out the velocity of $P$ after the collision and work out the kinetic energy of all the balls before and after the collision.

$4\times10=4v+2\times10$ therefore $v=5ms^{-1}$

$KE_{Initial}=\frac{1}{2}\times 4 \times 10^2 = 200J$

$KE_{Final}=\frac{1}{2}\times 4 \times 5^2 + \frac{1}{2}\times 2 \times 10^2= 150J$

47. Spot that this is a disguised quadratic. Let $y=x^2$ and solve for $y$ then $x$.

$2y^2-9y+4>0$

$(2y-1)(y-4)>0$

Either $2y-1>0$ and $y-4>0$ or $2y-1<0$ and $y-4<0$

$y>4$ gives $x<-2$, $x>2$

$y<\frac{1}{2}$ gives $\frac{-1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

48. Work out the magnitude of the resultant force (either by Pythagoras or (better) familiar triangles) and subtract the maximum friction force available from it. Divide by the mass of the object.

Using knowledge of 3,4,5 triangles, deduce that the resultant of these forces is 15N

$ma=F-\mu R=15-0.25\times2\times10=10$ therefore $a=\frac{10}{2}=5$

49. Differentiate the expression and find the height of the turning point which has a negative $x$ value.

$12x^2-24x-36=0$

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

$(x+1)(x-3)=0$

We are interested in the negative $x$ value. Let $x$=-1.

$y=-4-12+36-15=5$ so this point is 5 away from the path of the cursor

50. Define a tension, $T$ in the string. Form two equations using Newton's second law for each mass. Eliminate $T$ and solve for the frictional force.

$30g-T=30a$

$T-20gsin30-Fr=20a$

Add the two equations.

$30g-10g-Fr=50a$

$Fr=20g-50a=75N$

51. Note that the answer to this question is identical under reflection, interchange $x$ and $y$ then integrate as normal.

By inspection, the intersections of the line and curve occur for $x=0$ and $2$.

$\int_{0}^{2}7-3x^2+6x-7dx=[-x^3+3x^2]_{0}^{2}=-8+12=4$

52. Form a before and after conservation of momentum equation. This is best done from the perspective of the spacecraft before the ejection.

In the frame of the spacecraft before the ejection the momentum is $0$. After the ejection of fuel, we have an unknown mass travelling to the left at $1425ms^{-1}$ and $4000$ minus this mass travelling to the right at $75ms^{-1}$. The reason the spacecraft's velocity is now $75ms^{-1}$ is because the spacecraft is now travelling $75ms^{-1}$ faster than it previously was.

$1425m=75\times4000-75m$

$1500m=75\times4000$

$m=200$

53. This is an occasion where it is best to use stationary points to ascertain the behaviour of a graph. Find the turning points and rough shape. This should give you an idea of which horizontal lines, when drawn through the curve, intersect it in four places.

$12x^3-12x^2-24x=0$

$x(x^2-x-2)=0$

$x(x+1)(x-2)=0$ therefore $x=-1, 0$ or $2$.

When $x=-1$, $y=15$

When $x=0$, $y=20$

When $x=2$, $y=-12$

From the sketch, notice that a horizontal line drawn through the curve would only intersect in 4 different places if the height of that line were between (but not including) 15 and 20.

54. Make a triangle of forces with the weight, tension and external force. Use familiar triangles to work out the distance between the object and the ceiling and then the change in height and hence GPE.

From the diagram, deduce that we have a 3,4,5 triangle, so the vertical line in the triangle on the right must have length $0.28$. The change in height is $0.35-0.28=0.07$. Multiply by $mg$ to give $2.8J$.

Section 2

1a. The stiffest sample will deform least for the same force applied.

Stiffness is a measure of how little the sample deforms after a given tension is applied. This is true for the sample with the largest gradient. Answer=A

1b. Hooke's law is a straight line relationship between force and extension. Although it doesn't look like it, this is not the case for one of the samples over the range indicated.

$Strain=\frac{Extension}{Natural Length}$, so the extension in question is 2mm. Hold your pencil up to each of the lines and you will see that S2 is not linear in this interval. Answer=B

1c. Use Young's modulus $=\frac{Fl}{Ae}$.

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

Reading the values from the graph:

$E=\frac{250\times0.1}{(5\times10^{-3})^2\times0.005}=\frac{25\times10^9}{125}=\frac{10^9}{5}=2\times10^8$. Answer = A

1d. The work done on the sample is the area under the graph.

$\int_0^{0.01}ax-bx^2dx=[\frac{ax^2}{2}-\frac{bx^3}{3}]_0^{0.01}=\frac{a}{2}\times10^{-4}-\frac{b}{3}\times10^{-6}$. Answer=A

2a. Series combinations deliver the same current to all components. Parallel combinations deliver the same voltage to all components.

$R_2$ and $R_3$ form a parallel combination. Answer=F

2b. We need the current for the whole circuit which is $\frac{V}{R_{Total}}$. $R_{Total}$ can be found using the rules for adding resistances in series/parallel.

$I=\frac{V}{R_{Total}}$

$R_{Total}=R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}}=R_1+\frac{R_2R_3}{R_2+R_3}=\frac{R_1R_2+R_1R_3+R_2R_3}{R_2+R_3}$

$I=\frac{V(R_2+R_3)}{R_1R_2+R_1R_3+R_2R_3}$. Answer=A

2c. Use $P=\frac{V^2}{R}$ where this time, $V$ is the voltage across $R_3$. Let $R_1=R_2$.

$P=\frac{V_3^2}{R_3}$

$V_3=V-IR_1=V-\frac{VR_1(R_2+R_3)}{R_1R_2+R_1R_3+R_2R_3}=\frac{VR_2R_3}{R_1R_2+R_1R_3+R_2R_3}$

Use $R_1=R_2$

$V_3=\frac{VR_1R_3}{R_1^2+2R_1R_3}$ therefore $P=\frac{V^2R_1^2R_3}{(R_1^2+2R_1R_3)^2}=\frac{V^2R_3}{(R_1+2R_3)^2}$. Answer=D

2d. Differentiate your expression with respect to $R_3$ and set equal to zero or follow the hint and take the reciprocal first.

$P\alpha\frac{R_3}{(R_1+2R_3)^2}$

$\frac{1}{P}\alpha\frac{R_1^2+4R_1R_3+4R_3^2}{R_3}$

$\frac{d\frac{1}{P}}{dR_3}\alpha\frac{-R_1^2}{R_3^2}+4=0$

$R_3=\frac{R_1}{2}$. Answer=B (negative resistances don't make sense).

3a. Use $Distance=Speed\times Time$.

$3\times10^8\times1\times10^{-9}=0.3m$. Answer=C

3b. This can be thought of simply by realising that the refractive index means that the pulse will take longer to reach the other end of the fibre.

$Time=\frac{Distance}{Speed}=1.5\times\frac{9000}{3\times10^8}=\frac{3\times 3000\times10^{-8}}{2}=\frac{9}{2}\times10^{-5}$. Answer=F

3c. Similarly to 3b., the greater the refractive index, the longer the journey will take.

Nominal is the slowest because the speed is divided by the greatest $n$. The next slowest is blue, followed by red. Answer=A

4a. Work out the speed of the cyclist at B then resolve this down the slope.

$E+mgh=\frac{1}{2}mV^2$

$V=\sqrt{(\frac{2E}{m}+2gh)}$

$V_a=\sqrt{(\frac{2E}{m}+2gh)}cos\theta$. Answer=E

4b. Work out the horizontal and vertical distances travelled in terms of t. Use trigonometry to link them.

$x=Vt$, $y=\frac{1}{2}gt^2$

$tan\theta=\frac{y}{x}=\frac{\frac{1}{2}gt^2}{Vt}$

$t=\frac{2Vtan\theta}{g}$. Answer=B

4c. Resolve initial velocity and acceleration parallel to the slope then apply the equations of constant acceleration.

$t=\frac{2Vtan\theta}{g}$, $a=gsin\theta$, $u=Vcos\theta$

$s=ut+\frac{1}{2}at^2=Vcos\theta\times\frac{2Vtan\theta}{g}+\frac{1}{2}\times gsin\theta\times \frac{4V^2tan^2\theta}{g^2}=\frac{2V^2sin\theta}{g}+\frac{2V^2sin\theta tan^2\theta}{g}=\frac{2V^2sin\theta}{g}(1+tan^2\theta)=\frac{2V^2sin\theta}{gcos^2\theta}$. Answer=C

4d. This time work perpendicular to the slope and minimise the height attained.

$a=gcos\theta$, $u=Vsin\theta$

$s=ut+\frac{1}{2}at^2=Vsin\theta t - \frac{1}{2}gcos\theta t^2$

$\frac{ds}{dt}=Vsin\theta -gcos\theta t=0$

$t=\frac{Vtan\theta}{g}$

Let $t=\frac{Vtan\theta}{g}$ in the equation for $s$ down the plane used in part c).

$s=Vcos\theta\times\frac{Vtan\theta}{g}+\frac{1}{2}gsin\theta\times\frac{V^2tan^2\theta}{g^2}=\frac{V^2}{g}sin\theta(1+\frac{tan^2\theta}{2})$. Answer=D