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.I recommend solving this by inspection.

2. Apply $Q=It$ and $F=BIl$.

3. Multiply by $4$, expand, rearrange and solve. I recommend keeping quantities on both sides positive.

4. Calculate the resistance of one of the resistors by applying $P=\frac{V^2}{R}$ then apply this again to find the power dissipated then multiply by the time.

5. Let Sunday's earnings be $X$ then use fractions to recreate the percentage changes then solve.

6. Use the graph to work out the total travel time, realising that this consists of a downward and upward leg of the path taken by the pulse.

7. Factorise and simplify. Factors can be predicted/spotted by looking at the factors you have already used, such as $(5x+3)$.

8. Subtract their percentage from $100$ and use this to work out the energy wasted per second then multiply by the time.

9. Form an equation from the area of the larger rectangle and use this to solve for $x$.

10. Eliminate half the answers with information about the current through the lamp at $6V$. Now read off the potential difference across the resistor for the second circuit and subtract from $6V$.

11. Work out $RQ$ using trigonometry then the horizontal part of $PQ$ using familiar triangles.

12. Calculate $f$ using $c=f\lambda$. From this, deduce the time period then work out what fraction of a cycle the wave has undergone by the time $7ms$ has elapsed.

13. Form an equation from the information given and make $h$ the subject.

14. Apply $\Delta Q=mc\Delta T$, simplify and solve.

15. Make $2y^3$ the subject first then $y$.

16. Use the idea that a reversal in the movement of the magnet would give an opposite direction for the induced emf. Think also about whether the magnetic field lines are more concentrated or less concentrated if you move further away from the magnet.

17. Search for triples of $1$,$2$,$3$,$4$ and $5$ which work such that when you add two of these numbers you get a result which is one of the other numbers. Work systematically, say try $1$ and $2$ and see if there is another number available which is the sum of these (in this case, $3$). Also determine how many possible choices there are of the three numbers.

18. Apply $EPE=\frac{1}{2}kx^2$ and $GPE=mgh$ and work out what multiplicative effect the changes have on the height of the slider.

19. Equate the horizontal effect of the translations to $4$ and the vertical effect to $-3$ and solve the resulting equations.

20. Apply $Pressure=\frac{Force}{Area}$ to calculate the required pressure in the cylinder (remembering to add the atmospheric pressure on to this). Now use $PV=constant$ to work out how the volume has changed.

21. Write as two separated terms with fractional powers and carefully perform the integration.

22. There are $3$ wavelengths pictured. Work out the wavelength then apply $c=f\lambda$.

23. From $\frac{1}{2}bh$ the area is $5y$. Find $y$ as a function of $x$ from the equation of the circle.

24. Apply conservation of energy or its SUVAT equation, namely $v^2=u^2+2as$ and compare the situations.

25. Work out all the linear scale factors connecting $W$, $X$, $Y$ and $Z$. Then assign a volume $V$ to $W$ and determine the other volumes in terms of $V$.

26. Work out the angle of refraction in terms of $\theta$ and apply Snell's law.

27. Cancel as many factors of $(1+\sqrt5)$ as possible then rationalise the denominator.

28. Notice that this motion would occur as a result of constant acceleration and use this information to work out which statements are true. For statement $3$, consider the change in kinetic energy from $0$ to $t$ seconds compared with the change in kinetic energy from $t$ to $2t$.

29. Apply the sum of an infinite geometric series three times and simplify.

30. Use $R=\frac{\rho L}{A}$ to work out the effect of the changes and then calculate the resistance of the parallel combination.

31. It may help to introduce a new variable here, such as $y=x^2$.

32. Reading the question carefully, calculate the useful energy obtained when the bucket is lifted by one metre and the work done required to achieve this, then divide by the total energy supplied.

33. Let the radius of the circle by $r$ and the angle subtended by the sector be $x$. Write an equation bringing in $24$ as the perimeter and use this to rewrite the area of the sector ($\frac{1}{2}r^2x$).

34. Place the pivot at an unknown distance, say $x$ from one end, conserve moments and solve for $x$.

35. Sketch the parabola and work out where its turning point is.

36. Apply Kirchhoff's laws to determine the current in the upper and lower parts of the circuit.

37. Sketch the line and the point $P$ on Cartesian axes. Include the perpendicular from $P$ to the line and use similar triangles to determine the distance.

38. Conserve momentum then work out the energy before and after the collision.

39. Define a new variable such as $y=(\sqrt2)^x$, solve the resulting quadratic in $y$ then convert to $x$.

40. Apply $Y=\frac{Fl}{Ae}$ to a small part of the cable with length $dx$, work out the extension of this small part $dx$ then sum up all these different extensions for the entire length of the cable.

Section 2

1. Try to do this in your head for practice. Think of the force $F$ as the hypotenuse of a scaled-up $3$,$4$, $5$ triangle therefore its vertical component is $48N$.

2. Think about the effect of the lift's acceleration on the acceleration experienced by the mass. Take this new $g$ to be the acceleration due to gravity which the mass is subjected to and apply Hooke's law as normal.

3. Knowledge of the transformer and its turns ratio allow us to work out the voltage of the transmission line. From this and its power we can determine the current which we can use with the resistance of the cables to calculate the rate of energy lost.

4. The phase difference given is $\frac{1}{6}$ of the cycle. The fact that the two points are separated by less than a wavelength means that the distance given must represent $\frac{1}{6}$ of a wavelength. From the other information, the time period and hence the frequency can be determined then $c=f\lambda$ can be applied.

5. Combining two springs in series halves their spring constant since you get twice the extension for the same load. Cutting a spring into four pieces multiplies the spring constant by $4$ since each piece will extend one quarter or the original for the same load. Placing two springs in parallel doubles the spring constant. These rules can be used to reach a relationship between the two loads.

6. Applying $Y=\frac{Fl}{Ae}$ we can see that the Young's modulus is the gradient of the graph times $\frac{l}{A}$. From here the dimensions of the sample can be used to proceed.

7. Avoiding a gatepost error, we can work out the speed of the motorbike to be $12ms^{-1}$. From here apply SUVAT to calculate the distance travelled by the car and equate it to $12t$.

8. $Energy=Power\times Time$ is applicable to constant Power but what mathematical operation should be applied to add up $Power\times time$ for a Power that is constantly changing? Once you are able to answer this then the way forward should be more clear.

9. Work backwards from the end of the apparatus. Apply Snell's law at the boundary between the cylinder and air to find the critical angle. This is related to the angle of refraction from the first refraction (into the cylinder) where Snell's law can be applied again.

10. Differentiate to find the velocity as a function of time. From $t=0$ to $t=1$ we can now work out the change in velocity and hence the impulse.

11. Conserve energy to find out the speed of $2m$ then conserve momentum to find the speed of the combined object, then conserve energy again to find the height that it rises to.

12. Spotting a familiar triangle, work out the length of wire $Y$. From here the lengths of each path can be found and hence the relationship between their resistances and finally the ratio of currents.

13. Consider the tension in the rope. This acts at an angle to the vertical because the centre of the rope is below the horizontal. Work out the extended length of the rope using familiar triangles and from this you can determine the angle from the vertical at which the tension acts. Double this tension and resolve vertically; this is the weight of the object. The tension is $F$ in $Y=\frac{Fl}{Ae}$ which enables you to calculate $Y$.

14. Work out which graphs are related to the others by differentiation.

15. Use the information in the question to determine $p$. In this question, $Work done=Force\times Distance$ has to be upgraded due to the variable force. The question is how? What mathematical operation do you perform if you are adding a quantity that is changing as you change the variable you are adding over?

16. Apply Kirchhoff's Laws to all three loops and solve.

17. The time change in this question is $\frac{1}{4}$ of a cycle. Therefore particles that were at a maximum/minimum are now at equilibrium and those that were at equilibrium are now at a maximum or minimum. If particle $n$ (where $n$ is an integer) has just undergone a positive displacement and particle $n+1$ is at equilibrium, then particle $n+1$ will undergo this positive displacement in the next quarter cycle since the wave is moving to the right.

18. One approach for this question is to use the potential divider to work out the potential difference across $r$ and the external resistance, $R$ and then the power dissipated in the external resistance and the whole circuit using $P=\frac{V^2}{R}$ and find the fraction as a function of $r$ and $R$. Alternatively you can consider what would happen as $R\rightarrow\infty$. In this case all the potential difference would be across the external resistor and no power would be dissipated in the internal resistor.

19. Determine the initial horizontal and vertical component of the launch velocity. At the top of the flight only the horizontal component remains which can be used in a conservation of momentum calculation to find the new (horizontal) speed of $Q$ after the split. When $Q$ falls to the ground it has the same vertical component of velocity as the initial projectile did. Now apply Pythagoras to find out the magnitude of the velocity as $Q$ returns to earth.

20. Consider a hollow shell of matter of radius $x$. The shell has surface area $4\pi x^2$. Give the shell a small thickness $dx$ and from this you can work out the volume of this thin shell and hence its mass by multiplying by the density. We now have to add up the mass of all these thin shells as $dx\rightarrow 0$ from $x=0$ to $x=R$ but the mass depends on $x$. Therefore we are required to integrate. Remember that the density is not a constant and is given in the question. The integration itself is straightforward compared to the challenge of setting it up.

Full solutions

Section 1

Part A

1.I recommend solving this by inspection.

$\frac{9xz^2}{y^5}$

2. Apply $Q=It$ and $F=BIl$.

$I=\frac{20}{90}=\frac{2}{9}$

$F=BIl=0.15\times\frac{2}{9}\times0.18=0.3\times0.02$

3. Multiply by $4$, expand, rearrange and solve. I recommend keeping quantities on both sides positive.

$15-3x-12+2x-4x<0$

$3<5x$

4. Calculate the resistance of one of the resistors by applying $P=\frac{V^2}{R}$ then apply this again to find the power dissipated then multiply by the time.

$\frac{6^2}{R}=\frac{0.15}{2}$ from $P=\frac{V^2}{R}$ applied to just one of the resistors.

Therefore $\frac{12^2}{R}=0.3W$

$E=Pt=18\times6=108J$

5. Let Sunday's earnings be $X$ then use fractions to recreate the percentage changes then solve.

$\frac{X}{2}\times\frac{6}{5}\times\frac{7}{10}=84$

6. Use the graph to work out the total travel time, realising that this consists of a downward and upward leg of the path taken by the pulse.

Total journey distance $=0.08ms\times5000ms^{-1}=0.4m$. Halve because we want the distance from the crack to the base, not the whole journey.

7. Factorise and simplify. Factors can be predicted/spotted by looking at the factors you have already used, such as $(5x+3)$.

$\frac{(5x+3)(x-4)}{(5x+3)(5x-3)}\div\frac{(x+4)(x-3)}{(x-3)(x+2)}$

$\frac{(x-4)(x+2)}{(5x-3)(x+4)}$

8. Subtract their percentage from $100$ and use this to work out the energy wasted per second then multiply by the time.

$800W$

$320W$ wasted

$320\times60\times30$

9. Form an equation from the area of the larger rectangle and use this to solve for $x$.

$(x+4)(x+1)=180$

By inspection, $x=11$.

$12:21$

10. Eliminate half the answers with information about the current through the lamp at $6V$. Now read off the potential difference across the resistor for the second circuit and subtract from $6V$.

In the parallel combination we know that the lamp has a potential difference of $6V$ across it. Reading this off the graph, the lamp must have $0.35A$ through it which eliminates half of the answers.

For the series circuit, reading off $0.18A$ gives a potential difference of less than $2V$ which pinpoints the answer.

11. Work out $RQ$ using trigonometry then the horizontal part of $PQ$ using familiar triangles.

From $RSQ$, $QR=25$.

From familiar triangles, the horizontal component of $PQ$ is $40\sqrt3$.

12. Calculate $f$ using $c=f\lambda$. From this, deduce the time period then work out what fraction of a cycle the wave has undergone by the time $7ms$ has elapsed.

$\lambda=2$ so $f=250Hz$.

Therefore the time period, $T=4ms$.

Therefore the wave has undergone $1\frac{3}{4}$ cycles and so $x$ will be at the equilibrium position.

13. Form an equation from the information given and make $h$ the subject.

$2\pi rh+2\pi r^2=4\times6\times9r^2$

$\pi h=108r-\pi r$

14. Apply $\Delta Q=mc\Delta T$, simplify and solve.

$50\times350(T-26)=100\times4200\times6$

$5(T-26)=2\times60\times6$

$T=26+144$

15. Make $2y^3$ the subject first then $y$.

$\frac{2y^3+1}{2y^3-1}=x-5$

$2y^3+1=2y^3(x-5)-(x-5)$

$2y^3(1-x+5)=-x+5-1$

$2y^3=\frac{4-x}{6-x}$

16. Use the idea that a reversal in the movement of the magnet would give an opposite direction for the induced emf. Think also about whether the magnetic field lines are more concentrated or less concentrated if you move further away from the coil.

The induced voltages should reverse over the two motions since they are in opposite directions. This eliminates $B$, $C$ and $E$. Since the magnetic field lines for the coil become more spaced out at further distances from the coil, the magnitude of the induced voltage will decrease as the magnet moves from $X$ to $Y$ which eliminates $A$ and $F$.

17. Search for triples of $1$,$2$,$3$,$4$ and $5$ which work such that when you add two of these numbers you get a result which is one of the other numbers. Work systematically, say try $1$ and $2$ and see if there is another number available which is the sum of these (in this case, $3$). Also determine how many possible choices there are of the three numbers.

$1$,$2$,$3$

$1$,$3$,$4$

$1$,$4$,$5$

$2$,$3$,$5$ are the only triples which work. The number of triples is $^5C_3=\frac{5!}{3!2!}=10$ so the probability is $\frac{4}{10}=\frac{2}{5}$.

18. Apply $EPE=\frac{1}{2}kx^2$ and $GPE=mgh$ and work out what multiplicative effect the changes have on the height of the slider.

The mass has been multiplied by $\frac{2}{3}$ so the height should be multiplied by $\frac{3}{2}$. The compression of the spring is three times what it was, therefore the energy stored has increased by a factor of $9$.

$20\times9\times\frac{3}{2}=270cm$.

19. Equate the horizontal effect of the translations to $4$ and the vertical effect to $-3$ and solve the resulting equations.

$3p+q=4$

$-4p-2q=-3$

Multiply the first equation by $2$.

$6p+2q=8$

Add the second equation and the third equation.

$2p=5$

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

$\frac{15}{2}+q=4$

$q=-\frac{7}{2}$

20. Apply $Pressure=\frac{Force}{Area}$ to calculate the required pressure in the cylinder (remembering to add the atmospheric pressure on to this). Now use $PV=constant$ to work out how the volume has changed.

$F=50g=500N$

$P=\frac{F}{A}=\frac{500}{0.02}=25000Pa$

$PV=constant$

The pressure above has to be added on to the atmospheric pressure to obtain the pressure on the cylinder (since the piston is also affected by the atmospheric pressure). Therefore we have $1.25Atm$ in the cylinder.

The pressure has been multiplied by $\frac{5}{4}$ therefore the volume has been divided by this, therefore the density has been multiplied by this.

21. Write as two separated terms with fractional powers and carefully perform the integration.

$\int^4_1 2x^{\frac{1}{2}}-3x^{-\frac{3}{2}}dx$

$[\frac{4x^{\frac{3}{2}}}{3}+6x^{-\frac{1}{2}}]_1^4$

$\frac{32}{3}+3-\frac{4}{3}-6$

$\frac{28}{3}-3$

22. There are $3$ wavelengths pictured. Work out the wavelength then apply $c=f\lambda$.

$300=f\times0.5$

23. From $\frac{1}{2}bh$ the area is $5y$. Find $y$ as a function of $x$ from the equation of the circle.

$(x-5)^2+y^2=25$

$y^2=10x-x^2$

24. Apply conservation of energy or its SUVAT equation, namely $v^2=u^2+2as$ and compare the situations.

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

$0=u^2-2gs$

$s\alpha\frac{u^2}{g}$

25. Work out all the linear scale factors connecting $W$, $X$, $Y$ and $Z$. Then assign a volume $V$ to $W$ and determine the other volumes in terms of $V$.

In the next line, the number between each of the solids is the linear scale factor between that solid and the next.

$W$, $2$, $X$, $\frac{1}{\sqrt2}$, $Y$, $^3\sqrt2$, $Z$.

If $W$ has volume $V$ then $X$, $Y$ and $Z$ have these volumes respectively:

$8V$, $\sqrt2V$, $2\sqrt2V$.

26. Work out the angle of refraction in terms of $\theta$ and apply Snell's law.

Going on a short angle chase reveals the angle of refraction to be $90-\theta$.

$1\times sin\theta=nsin(90-\theta)$

$sin\theta=ncos\theta$

27. Cancel as many factors of $(1+\sqrt5)$ as possible then rationalise the denominator.

$\frac{4}{1+\sqrt5}$

$\frac{4-4\sqrt5}{-4}$

28. Notice that this motion would occur as a result of constant acceleration and use this information to work out which statements are true. For statement $3$, consider the change in kinetic energy from $0$ to $t$ seconds compared with the change in kinetic energy from $t$ to $2t$.

Constant acceleration reveals $1$ to be true and $2$ to be false. For $3$, imagine the time taken to increase the speed from $0$ to $v$; this has increased the kinetic energy from $0$ to $\frac{1}{2}mv^2$. now if the same amount of time elapses again, the kinetic energy is now $\frac{1}{2}m(2v)^2$ which is $4$ times what it was previously, not double therefore $3$ is true.

29. Apply the sum of an infinite geometric series three times and simplify. If it helps then write out the first nine terms or so in order to see the pattern.

$\frac{\frac{1}{4}}{1-\frac{1}{2}}+\frac{\frac{1}{9}}{1-\frac{1}{3}}+\frac{\frac{1}{16}}{1-\frac{1}{4}}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}=\frac{6+2+1}{12}$

30. Use $R=\frac{\rho L}{A}$ to work out the effect of the changes and then calculate the resistance of the parallel combination.

The second wire has resistance $8R$.

$R_{Total}=\frac{R}{\frac{1}{8}+\frac{1}{1}}$

31. It may help to introduce a new variable here, such as $y=x^2$.

$y^2+4y-21<0$

$(y-3)(y+7)<0$

$-7

32. Reading the question carefully, calculate the useful energy obtained when the bucket is lifted by one metre and the work done required to achieve this, then divide by the total energy supplied.

$Mass=Volume\times Density=15kg$

Here don't add the mass of the empty bucket as this is (correctly) interpreted as wasted energy.

Over $1$ metre we have useful work $150J$.

Considering the ratio of the radii of the cylinder and the handle, the handle moves $3m$ at a cost of $750J$.

$\frac{150}{750}\times100=20$

33. Let the radius of the circle by $r$ and the angle subtended by the sector be $x$. Write an equation bringing in $24$ as the perimeter and use this to rewrite the area of the sector ($\frac{1}{2}r^2x$).

$2r+rx=24$

$\frac{1}{2}r^2x=\frac{1}{2}r^2(\frac{24}{r}-2)=r(12-r)$

Can you see by inspection that the maximum of this occurs when $r=6$?

34. Place the pivot at an unknown distance, say $x$ from one end, conserve moments and solve for $x$.

Placing the pivot a distance $x$ from the $20g$ mass we have, working in grams and ignoring $g$:

$20x=100(50-x)+80(100-x)$

$x=250-5x+400-4x$

$x=65$

35. Sketch the parabola and work out where its turning point is.

Factorising the quadratic we can see that the $x$ values of its intersection with the $x$ axis are at $-2$ and $3$. The mean of these is $\frac{1}{2}$ and the corresponding $y$ value is $-6\frac{1}{4}$. Therefore the are is $5\times6\frac{1}{4}=31\frac{1}{4}$.

36. Apply Kirchhoff's laws to determine the current in the upper and lower parts of the circuit.

In the upper loop we have $3A$ clockwise and in the lower loop we have $1A$ clockwise. Applying conservation of charge to the junction at $X$ we have $2A$ from right to left.

37. Sketch the line and the point $P$ on Cartesian axes. Include the perpendicular from $P$ to the line and use similar triangles to determine the distance.

Draw a rectangle using the points $(0,0)$, $(a,0)$, $(a,2a)$ and $(0,2a)$. Connect $(0,2a)$ and $(a,0)$ then construct a perpendicular from this line to the point $(a,2a)$. The three triangles created are similar and the scale factor from the left triangle to the right triangle is $\frac{2}{\sqrt5}$ therefore the shortest length of the right hand triangle is $\frac{2a}{\sqrt5}$.

38. Conserve momentum then work out the energy before and after the collision.

The momentum before the collision is $2mv$ to the right. Since the combined object has mass $3m$ then its velocity must be $\frac{2v}{3}$ to the right. The kinetic energy before is $9v^2$ and after is $\frac{2v^2}{3}$.

39. Define a new variable such as $y=(\sqrt2)^x$, solve the resulting quadratic in $y$ then convert to $x$.

$y^2-8y+12=0$

$(y-2)(y-6)=0$

$\sqrt2^x=2$ or $6$

$2^{\frac{x}{2}}=2$ or $6$

$x=2$ or $2log_{2}6=2+2log_{2}3$

40. Apply $Y=\frac{Fl}{Ae}$ to a small part of the cable with length $dx$, work out the extension of this small part $dx$ then sum up all these different extensions for the entire length of the cable.

Let the small part of the cable with length $dx$ be at a distance $x$ from the top. Therefore the mass of the cable below this is $\frac{40-x}{40}\times64$ and its weight is this multiplied by $g$. Applying $Y=\frac{Fl}{Ae}$ to this part with $l=dx$ we have:

$Y=\frac{(40-x)\times\frac{64g}{40}dx}{Ae}$

and so the extension of this small part $dx$ is:

$e=\frac{(40-x)\times\frac{64g}{40}dx}{AY}$

Now we are required to add up all these extensions for each part of the cable which will require an integral from the top of the cable to the other end.

$Total$ $e=\frac{16}{2\times10^{-4}Y}\int_0^{40}40-x dx$

$=\frac{8\times10^4}{Y}[40x-\frac{x^2}{2}]_0^{40}=4\times10^{-7}\times40^2(1-\frac{1}{2})$

Section 2

1. Try to do this in your head for practice. Think of the force $F$ as the hypotenuse of a scaled-up $3$,$4$, $5$ triangle therefore its vertical component is $48N$.

$Fcos\theta=36$

$Fsin\theta=48$

Therefore the downward forces on the object are $48N$ and $24N$.

2. Think about the effect of the lift's acceleration on the acceleration experienced by the mass. Take this new $g$ to be the acceleration due to gravity which the mass is subjected to and apply Hooke's law as normal.

The fact that the lift is accelerating downwards at $2ms^{-2}$ means that the acceleration due to gravity experienced in the lift is $2ms^{-2}$ less than it otherwise would be, so $8ms^{-2}$.

Applying Hooke's law,

$30\times8=kx=3000x$

$x=0.08$

3. Knowledge of the transformer and its turns ratio allow us to work out the voltage of the transmission line. From this and its power we can determine the current which we can use with the resistance of the cables to calculate the rate of energy lost.

Voltage before $=1250V$

Current before $=\frac{40000}{1250}=32A$

$P=I^2R=32^2\times5=1024\times 5$

4. The phase difference given is $\frac{1}{6}$ of the cycle. The fact that the two points are separated by less than a wavelength means that the distance given must represent $\frac{1}{6}$ of a wavelength and not a multiple of the wavelength plus $\frac{1}{6}$ of a wavelength. From the other information, the time period and hence the frequency can be determined then $c=f\lambda$ can be applied.

$n\lambda\pm\frac{\lambda}{6}=5$

$\lambda>10$ so $n=0$ and $\lambda=30$

$T=0.24$ because a particle would go through the equilibrium position twice in each cycle.

$f=\frac{1}{T}=\frac{100}{24}=\frac{25}{6}$

$c=f\lambda=\frac{25\times30}{6}$

5. Combining two springs in series halves their spring constant since you get twice the extension for the same load. Cutting a spring into four pieces multiplies the spring constant by $4$ since each piece will extend one quarter or the original for the same load. Placing two springs in parallel doubles the spring constant. These rules can be used to reach a relationship between the two loads.

If the spring constant if a single spring is $k$ then the spring constant of the first combination is $\frac{k}{2}$.

Each of the lone springs in the second combination has spring constant $4k$ so each pair in parallel has spring constant $8k$ and the whole combination has spring constant $4k$. Thus for the same extension, $F=8T$

6. Applying $Y=\frac{Fl}{Ae}$ we can see that the Young's modulus is the gradient of the graph times $\frac{l}{A}$. From here the dimensions of the sample can be used to proceed.

$Y=\frac{Fl}{Ae}=gradient\times\frac{l}{A}=\frac{gradient}{0.04}$

$Y=\frac{150\times100}{5\times10^{-6}}=30\times10^8$

7. Avoiding a gatepost error, we can work out the speed of the motorbike to be $12ms^{-1}$. From here apply SUVAT to calculate the distance travelled by the car and equate it to $12t$.

$speed=\frac{240}{20}=12ms^{-1}$

Using $s=ut+\frac{1}{2}at^2$ for the car's displacement:

$3t+3t^2=12t$

$3t^2=9t$

8. $Energy=Power\times Time$ is applicable to constant Power but what mathematical operation should be applied to add up $Power\times time$ for a Power that is constantly changing? Once you are able to answer this then the way forward should be more clear.

$E=\int_2^33t^2+4t=[t^3+2t^2]_2^3=27+18-8-8=29$

One tenth of this is wasted.

9. Work backwards from the end of the apparatus. Apply Snell's law at the boundary between the cylinder and air to find the critical angle. This is related to the angle of refraction from the first refraction (into the cylinder) where Snell's law can be applied again.

$sin\theta_c=\frac{1}{n}=\frac{\sqrt3}{2}$

$\theta_c=60^{\circ}$

The critical angle plus the angle of refraction into the cylinder equals $90^{\circ}$ because the face of the cylinder is at right angles to the curved surface.

$1\times sin\theta=nsin30=\frac{1}{\sqrt3}$

10. Differentiate to find the velocity as a function of time. From $t=0$ to $t=1$ we can now work out the change in velocity and hence the impulse.

$v=-3t^2-6t$

From $t=0$ to $t=1$, $v$ goes from $0$ to $-9$. $m\Delta v=20\times9$.

11. Conserve energy to find out the speed of $2m$ then conserve momentum to find the speed of the combined object, then conserve energy again to find the height that it rises to.

$2mgl=\frac{1}{2}mv^2$ so $v=\sqrt{4gl}$.

Applying conservation of momentum:

$2m\sqrt{4gl}=3mv_2$

$v_2=\frac{2}{3}\sqrt{4gl}$

Apply conservation of energy

$\frac{1}{2}3mv_2^2=3mgh$

$h=\frac{4l}{9}$

12. Spotting a familiar triangle, work out the length of wire $Y$. From here the lengths of each path can be found and hence the relationship between their resistances and finally the ratio of currents.

The triangle is an enlargement of a $3,4,5$ triangle with scale factor $3$, so $Y=9$.

The resistances of $X$ and $Y$ are quartered (all else being equal) due to their doubled diameter ($R=\frac{\rho L}{A}$). Otherwise their resistance is proportional to their length.

Therefore if $R_Z=R$ then $R_X=\frac{12R}{4\times15}$ and $R_Y=\frac{9R}{4\times15}$.

The upper path had resistance $\frac{7R}{20}$ and the lower path has resistance $R$.

The current in $X$ is greater since this path has a lower resistance.

13. Consider the tension in the rope. This acts at an angle to the vertical because the centre of the rope is below the horizontal. Work out the extended length of the rope using familiar triangles and from this you can determine the angle from the vertical at which the tension acts. Double this tension and resolve vertically; this is the weight of the object. The tension is $F$ in $Y=\frac{Fl}{Ae}$ which enables you to calculate $Y$.

Draw a right-angled triangle with base $0.12m$ which is half the unstretched length and height of $0.05m$ dropping down from the base which is the displacement of the mass. By the familiar $(5,12,13)$ triangle, the hypotenuse is $0.13m$ which is half of the stretched length of the rope. Therefore the total extension is $0.02m$.

Let $T$ be the tension in the string and $\theta$ be the angle between the string and the vertical.

$2Tcos\theta=mg=10$

$T=\frac{g}{2cos\theta}=\frac{g}{2\times\frac{5}{13}}=13N$

$Y=\frac{13\times0.24}{6\times10^{-8}\times0.02}$ which begins 26... and only one answer begins 26...

14. Work out which graphs are related to the others by differentiation.

Navigating via the stationary points, $\frac{dP}{dt}=R$ and $\frac{dR}{dt}=Q$. therefore $P$ is the displacement and $R$ is the velocity.

15. Use the information in the question to determine $p$. In this question, $Work done=Force\times Distance$ has to be upgraded due to the variable force. The question is how? What mathematical operation do you perform if you are adding a quantity that is changing as you change the variable you are adding over?

$2400=p\times(0.2)^2$

$p=2400\times25=60000$

$Work$ $done=60000\int_1^2x^2dx=60000[\frac{x^3}{3}]_1^2=20000(8-1)$

16. Apply Kirchhoff's Laws to all three loops and solve.

Define $I_1$ as leaving the $18V$ battery.

Define $I_2$ as leaving the $20V$ battery.

Define $I_3$ as the current left to right through the $4\Omega$ resistor.

$18=4I_3+20I_1$

$20=8I_2+4I_3$

$2=8I_2-20I_1$

$I_3=I_1+I_2$

Substituting the fourth equation into the second:

$20=12I_2+4I_1$

Multiply this equation by $2$

$40=24I_2+8I_1$

Multiply the third equation by $3$

$6=24I_2-60I_1$

Subtract these last two equations:

$34=68I_1$

$I_1=\frac{1}{2}$

17. The time change in this question is $\frac{1}{4}$ of a cycle. Therefore particles that were at a maximum/minimum are now at equilibrium and those that were at equilibrium are now at a maximum or minimum. If particle $n$ (where $n$ is an integer) has just undergone a positive displacement and particle $n+1$ is at equilibrium, then particle $n+1$ will undergo this positive displacement in the next quarter cycle since the wave is moving to the right.

$2$ is in equilibrium and $0$ and $1$ have positive displacements therefore we expect $2$ to have a positive maximum at the later time.

$6$ is in equilibrium and $4$ and $5$ have negative displacements therefore we expect $6$ to have a negative maximum at the later time.

Since only $2$ and $6$ have zero displacement at the earlier time we don't expect any other maxima/minima.

18. One approach for this question is to use the potential divider to work out the potential difference across $r$ and the external resistance, $R$ and then the power dissipated in the external resistance and the whole circuit using $P=\frac{V^2}{R}$ and find the fraction as a function of $r$ and $R$. Alternatively you can consider what would happen as $R\rightarrow\infty$. In this case all the potential difference would be across the external resistor and no power would be dissipated in the internal resistor.

$V_r=\frac{rV}{R+r}$ and $V_R=\frac{RV}{R+r}$

$Efficiency=\frac{V_R^2}{R}\div\frac{V^2}{R+r}=\frac{R^2V^2}{R(R+r)^2}\times\frac{r+R}{V^2}=\frac{R}{R+r}$

This starts at $0$ and gradually tends to $1$ as $R$ increases.

Alternatively you can deduce that if $R\rightarrow\infty$ then this resistor will have all the power dissipated in it since there is no potential difference across the internal resistor. There is only one graph that tends to $1$ as $R$ tends to $\infty$.

19. Determine the initial horizontal and vertical component of the launch velocity. At the top of the flight only the horizontal component remains which can be used in a conservation of momentum calculation to find the new (horizontal) speed of $Q$ after the split. When $Q$ falls to the ground it has the same vertical component of velocity as the initial projectile did. Now apply Pythagoras to find out the magnitude of the velocity as $Q$ returns to earth.

Using trigonometry, the horizontal component of velocity is $6$ and the vertical component is $6\sqrt3$.

At the top of the flight the speed is $6$ and so the momentum is $60$. After the collision the momentum of $P$ is $70$ and therefore the momentum of $Q$ is $-10$ and its speed is $2$ horizontally. Since its vertical speed was $6\sqrt3$ in the first instance then (since air resistance is neglected) the vertical speed on return to the ground must also be $6\sqrt3$.

By Pythagoras, the magnitude of the velocity is $\sqrt{36\times3+4}=\sqrt{112}$.

20. Consider a hollow shell of matter of radius $x$. The shell has surface area $4\pi x^2$. Give the shell a small thickness $dx$ and from this you can work out the volume of this thin shell and hence its mass by multiplying by the density. We now have to add up the mass of all these thin shells as $dx\rightarrow 0$ from $x=0$ to $x=R$ but the mass depends on $x$. Therefore we are required to integrate. Remember that the density is not a constant and is given in the question. The integration itself is straightforward compared to the challenge of setting it up.

The mass of the thin shell is $4\pi\rho x^2dx$.

$Total$ $Mass=\int_0^R4\pi\rho_0(x^2-\frac{x^3}{2R})dx$

$=4\pi\rho_0[\frac{x^3}{3}-\frac{x^4}{8R}]^R_0$

$=4\pi\rho_0R^3(\frac{1}{3}-\frac{1}{8})=\frac{5\pi\rho_0R^3}{6}$