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 merely set out a series of hints below - try and do the questions with the help of the hints alone and then click on "Full solution" if you are still having any difficulty.

Section 1

Part A

1. Write an equation equating the area of the metal (100) plus the area of the semicircle to the area of the square noticing that the side of the square is twice the radius. Don't define the side of the square to be $a$ or $x$, just call it $2r$.

Full solution

2. Ignore all superfluous information and concentrate on the forces acting then apply Newton's second law to the resultant.

Full solution

3. Find the area of the triangle in two different ways.

Full solution

4. To work out the speed of a wave, use $c=f\lambda$. Now you need to focus on the frequency and the wavelength. The wavelength is given in the question and the frequency is $\frac{1}{period}$ which is read from the graph.

Full solution

5. Use Pythagoras' Theorem to find the diagonal of the the base of the cube and divide by two, This is the base of the second triangle that you have to apply Pythagoras' Theorem to in order to find the length of the dashed line.

Full solution

6. Keeping the mass as $m$ and the velocity as $v$, form two equations in $m$ and $v$ from momentum = $mv$ and kinetic energy = $\frac{1}{2}mv^{2}$. Solve for $m$.

Full solution

7. The minute hand points to the 9 and the hour hand is three quarters of the way between the 9 and the 10. The fraction of the circle covered by this angle is $\frac{3}{4}$ of $\frac{1}{12}$.

Full solution

8. For this style of question, recall an equation which is relevant to each answer. For instance, answer A brings $V=IR$ to mind but this doesn't work as this would yield amp ohm, not amp per ohm. $P=IV$ is the relevant equation for answer E.

Full solution

9. Apply $\frac{1}{2}\times$base$\times$height and expand.

Full solution

10. In the time period given, source X undergoes five half-lives and source Y undergoes three. The activity is proportional to the number of nuclei present so the activity of source X goes down by a factor of 32 and source Y by a factor of 8.

Full solution

11. Write the volume of each shape in terms of $r$ and divide the volume of the sphere by the cylinder.

Full solution

12. The loss in gravitational potential energy should not present a problem. For the resistive force, use work done = force $\times$ distance moved in the direction of the force realising that the frictional force acts up the slope, so the relevant distance is that which the cyclist moves down the slope.

Full solution

13. It helps to think of the shape of graphs such as $y=x$, $y=x^2$ and $y=\sqrt{x}$ for $0\leq x \leq 1$. The higher the power, the smaller the end result because multiplying numbers between 0 and 1 together makes the product smaller than the original numbers. Alternatively, substitute $x=\frac{1}{2}$ into each expression.

Full solution

14. Equate kinetic energy to gravitational potential energy.

Full solution

15. There are two triangles in the diagram which are similar to each other. Divide DE by BC, equate to AD divided by AB and solve for $x$.

Full solution

16. Equate kinetic energy to work done = force $\times$ distance.

Full solution

17. Using percentage multipliers, multiply $Q$ by $1.4$. This has the effect of dividing $P$ by $1.4^{2}=1.96$. This gives a result just above half of the original $P$.

Full solution

18. The $R-2$ is the hallmark of an alpha decay and an alpha particle has four nucleons. In the second decay, the number of nucleons remains the same so this is a beta decay.

Full solution

19. Write the two sentences as equations and rearrange to give two equations in $z$. Equate and raise to the power of 6.

Full solution

20. Interestingly, some information has been given here which is not required. Ignore the frequency and use $Time=\frac{Distance}{Speed}$, remembering the return leg of the journey.

Full solution

21. Let QX be 1, work out PX and XR then use PR to find PM and hence MX.

Full solution

22. Firstly, spot that graphs R and S are distance-time graphs exhibiting zero acceleration. Make a quick calculation of the gradients of P and Q.

Full solution

23. Rearrange and factorise then sketch and interpret the resulting quadratic.

Full solution

24. It is my opinion that there is a typographical mistake here. In the original version of this question, there were six answers. One has been removed for the purposes of creating the ENGAA specimen, changing the identity of the correct answer which was then not updated in the solution. The solution should read D not E. These mistakes are possible since the questions are written in code and I know from experience that errors are difficult to eradicate. Should this happen in your exam then keep calm, correct the mistake and move on.

Full solution

25. Turn the first sentence into an equation and rearrange for $h$.

Full solution

26. This is a potential divider. Work out the current using $V_{total}=I_{total}\times R_{total}$ and then apply Ohm's law once more to the resistor in question to find the voltage across this resistor. Finally apply $P=\frac{V^{2}}{R}$ or $P=I^{2}R$.

Full solution

27. Draw a set of axes and label the quadrants P Q R and S. Work out what happens to each of these points under the transformations given and draw a new set of axes with a new P Q R and S. It is clear from here which transformation is required to restore the square.

Full solution

28. Work out the quantities mentioned in P, R and S. Note that the speed of sound in air is roughly $340ms^{-1}$. If you think of the amplitude of a longitudinal wave as the maximum displacement of a molecule in a medium from its mean position, 5mm is too large for this as this would be a peak to peak disturbance.

Full solution

Part B

29. Factorise the numerator and denominator and cancel.

Full solution

30. Use Constant acceleration = $\frac{Change\, in\, velocity}{Time}$ then Resultant force = Mass $\times$ Acceleration.

Full solution

31. Take logarithms to base 10 of both sides and use the laws of logarithms to bring $x$ down from the superscript. Take $x$ out as a factor then combine the three logarithms into one. Finally make $x$ the subject.

Full solution

32. Momentum is conserved. Draw a diagram of the situation before and after taking care of the signs of the velocities in your equations.

Full solution

33. Evaluate these efficiently. It helps to spot that a number between 0 and 1 which is raised to the power 10 is going to be small. For answer D, consider that this must be between 3 and 4 because $2^{3}=8$ and $2^4=16$.

Full solution

34. Air resistance cannot be negative - it only hinders motion. This eliminates B and D. When the parachutist is travelling at a terminal velocity, the air resistance equals his weight both before the parachute is opened and after, so the horizontal portions of the graph should be at the same height.

Full solution

35. Be on the lookout for disguised quadratic equations. Let $2^{x}=y$ and solve the resulting quadratic.

Full solution

36. Use graph Z to eliminate three of the options then graphs Y and X to eliminate another one each. Briefly check that the remaining option fits the graphs.

Full solution

37. Multiplying an inequality by a negative number reverses its sense. Consider $(a+b)^{2}$; this must be greater than or equal to zero so rearranging this gives statement 2. For statement 3, we have no assurance that $c$ is positive, so this is not always true.

Full solution

38. Frictional forces balance the applied force up to a maximum point, so the block must start out stationary. Once it is moving the force applied is still increasing and hence so is the acceleration.

Full solution

39. Work out some values of $a_{n}$ to spot that this sequence is very repetitive. It repeats every two terms, so add these two terms up and multiply by 50.

Full solution

40. Momentum is conserved. Draw a diagram of the situation before and after taking care of the signs of the velocities in your equations.

Full solution

41. The factors of this equation are not whole numbers. It is crucial to spot this quickly and move onto a different method than factorisation. Try differentiating to discover that the turning points are $(0,-10)$, $(1,-9)$ and $(2,-10)$. Alternatively consider adding 10 to the function, sketching it and noting that upon subtraction of 10 all the turning points are well below the $x$-axis.

Full solution

42. Equate kinetic energy to potential energy and rearrange for $g_{p}\times h$. Note that only one of the answers gives the correct product.

Full solution

43. Use Pythagoras' Theorem to find PR, halve this and construct a triangle consisting of T, P and a vertical line downwards from T to apply trigonometry to.

Full solution

44. A $50\%$ loss of kinetic energy equates to a $50\%$ loss of gravitational potential energy and hence height. Therefore the height after each bounce is half what it was before that bounce.

Full solution

45. Take logarithms (the base doesn't matter) of each equation, preferably in your head. Any with $x$ in the power won't work because after taking logarithms the $x$ won't be inside a logarithm which is what we require. The right hand side of equation C cannot be simplified after taking a logarithm.

Full solution

46. The elevator must be accelerating downward regardless of the direction of its velocity. It would also be true to say that the elevator is moving downwards with increasing speed, but this isn't one of the options.

Full solution

47. The discriminant must be greater than 0. This yields $(a-2)^{2}>-8a$. Expand this, take everything to the left hand side and factorise. The resulting inequality is satisfied for all values of $a$ apart from -2.

Full solution

48. This is a fairly standard question involving the equations of motion.

Full solution

49. Don't be afraid to use trial and improvement. Start with a 4, 4, 4 equilateral triangle and make perturbations from there. Discard any triangles which don't close and you will reach an answer relatively quickly. Note the large difference between three triangles and the next answer up (ten triangles). This should help you settle on three quicker than if the next answer up were say four because of the large gap between three and ten.

Full solution

50. Resolve the diagonal force horizontally and vertically. Equate the vertical component to the weight of the particle and the horizontal component to the tension in the horizontal rope.

Full solution

51. A minimum for QR requires a minimum for PQ (3.5) and a maximum for PR (2.5) since $PR^{2}$ is subtracted from $PQ^{2}$. Use Pythagoras' Theorem with these values to find QR. Use $\sqrt{(3.5^{2}-2.5^{2})}=\sqrt{(3.5+2.5)(3.5-2.5)}$ to save time.

Full solution

52. Since the velocity of the mass is constant at this time, all of the power applied is going into increasing the potential energy of the mass. Therefore $Power=\frac{Energy}{Time}=\frac{mgh}{t}=mgv$.

Full solution

53. Solve each equation to give $0\leq x \leq \frac{\pi}{4}$, $\frac{3\pi}{4} \leq x \leq \pi$ and $\frac{\pi}{12} \leq x \leq \frac{5\pi}{12}$. Both are satisfied for $\frac{\pi}{12} \leq x \leq \frac{\pi}{4}$.

Full solution

54. Work out the acceleration of the entire ensemble. For carriage 2 this must be achieved as a result of the force in the coupling, T which is the mass of the carriage multiplied by its acceleration.

Full solution

Section 2

1a. Reaction forces are perpendicular to the surface and frictional forces act to oppose motion.

1b. The block is not accelerating off or into the slope, but it is accelerating down the slope and has components of this acceleration which are vertical and horizontal.

1c. Find the resultant force down the slope after resolving, equate to Mass $\times$ Acceleration to give $mgsin\alpha - \mu mgcos\alpha = ma$ then use $v=at$ for constant acceleration.

1d. The presence of this drag force means that the above equation is modified such that the resultant force is reduced by $kv$ to give $mgsin\alpha - \mu mgcos\alpha -kv = ma$. When the velocity is terminal, $a=0$. Substitute this in and rearrange for $v$ to give the terminal velocity.

Full solution

2a. Apply $R=\frac{\rho L}{A}$.

2b. To do this question, it helps to have an idea of what is actually happening to the current. It leaves the power station along one of the cables, travels through some kind of network which we don't need to know about then returns along the other cable. Working out this current is crucial to this question and can be found using $P=IV$. Once you have this current, use $P=I^{2}R$ to find the power dissipated in each cable and multiply by 2 because there are two of them. Alternatively, you can use $V=IR$ to work out the voltage dropped across each cable and then use $P=\frac{V^{2}}{R}$. Don't be tempted to use the $V$ from the question in $P=\frac{V^{2}}{R}$ because 400kV is the voltage dropped across the cables $and$ the city, not just the cables.

2c. A kilowatt hour is a unit of energy equal to the energy output of 1kW of power over a time of one hour. This is easily scaled up to work out a GWh in Joules.

2d. Answers A,B,D and E are simple to eliminate as follows:

A: The efficiency of the power station is of no consequence after you quote its power output which would have already had the efficiency taken into account if necessary.

B: Similarly, the appliances consume a certain power (as quoted in the question) and still use this power whether they are efficient or not.

D. This assumes that the energy can't be lost in wasted forms.

E. This also assumes that the energy can't be lost in wasted forms.

Full solution

3a. The function given is positive over the given interval, has a single root at the origin and a repeated root at $x=T$. This narrows the choices down to C and D. Note that $v=0$ for all $t>T$ and $t<0$.

3b. Differentiating an expression for velocity yields the acceleration of the particle. Factorise the result.

3c. The acceleration is a quadratic and hence is symmetrical and has a minimum value when $t$ is the mean of its roots. Therefore set $t=\frac{2T}{3}$.

3d. Look at the graph - this must be for $t=0$. Substitute into the equation for $a$.

3e. The object doesn't move after $t=T$. Integrate the velocity between $t=0$ and $t=T$ to find the area under the curve and hence the distance travelled.

Full solution

4a. Equate gravitational potential energy to kinetic energy.

4b. Do the same as for 4a except that the change in height is now $H-2R$.

4c. Circular motion requires a force towards the centre of the circle - think of planetary motion - there is no way the force could be in any other direction.

4d. If the car is to just make it round the loop. not only does it need to have enough gravitational energy to do so, it also needs some velocity to stay on the track. In this case, the reaction force at the top would be zero and the only force acting on the car would be the gravitational force. Equating this to the $\frac{mv^{2}}{R}$ in the question gives $v=\sqrt{gR}$ but from earlier in the question we have that $v=\sqrt{2g(H-2R)}$. Equating these yields $H=2.5R$ and the height needs to be greater than this because a reaction force of exactly zero means that the car has left the track.

Full solution

I will merely set out a series of hints below - try and do the questions with the help of the hints alone and then click on "Full solution" if you are still having any difficulty.

Section 1

Part A

1. Write an equation equating the area of the metal (100) plus the area of the semicircle to the area of the square noticing that the side of the square is twice the radius. Don't define the side of the square to be $a$ or $x$, just call it $2r$.

Full solution

2. Ignore all superfluous information and concentrate on the forces acting then apply Newton's second law to the resultant.

Full solution

3. Find the area of the triangle in two different ways.

Full solution

4. To work out the speed of a wave, use $c=f\lambda$. Now you need to focus on the frequency and the wavelength. The wavelength is given in the question and the frequency is $\frac{1}{period}$ which is read from the graph.

Full solution

5. Use Pythagoras' Theorem to find the diagonal of the the base of the cube and divide by two, This is the base of the second triangle that you have to apply Pythagoras' Theorem to in order to find the length of the dashed line.

Full solution

6. Keeping the mass as $m$ and the velocity as $v$, form two equations in $m$ and $v$ from momentum = $mv$ and kinetic energy = $\frac{1}{2}mv^{2}$. Solve for $m$.

Full solution

7. The minute hand points to the 9 and the hour hand is three quarters of the way between the 9 and the 10. The fraction of the circle covered by this angle is $\frac{3}{4}$ of $\frac{1}{12}$.

Full solution

8. For this style of question, recall an equation which is relevant to each answer. For instance, answer A brings $V=IR$ to mind but this doesn't work as this would yield amp ohm, not amp per ohm. $P=IV$ is the relevant equation for answer E.

Full solution

9. Apply $\frac{1}{2}\times$base$\times$height and expand.

Full solution

10. In the time period given, source X undergoes five half-lives and source Y undergoes three. The activity is proportional to the number of nuclei present so the activity of source X goes down by a factor of 32 and source Y by a factor of 8.

Full solution

11. Write the volume of each shape in terms of $r$ and divide the volume of the sphere by the cylinder.

Full solution

12. The loss in gravitational potential energy should not present a problem. For the resistive force, use work done = force $\times$ distance moved in the direction of the force realising that the frictional force acts up the slope, so the relevant distance is that which the cyclist moves down the slope.

Full solution

13. It helps to think of the shape of graphs such as $y=x$, $y=x^2$ and $y=\sqrt{x}$ for $0\leq x \leq 1$. The higher the power, the smaller the end result because multiplying numbers between 0 and 1 together makes the product smaller than the original numbers. Alternatively, substitute $x=\frac{1}{2}$ into each expression.

Full solution

14. Equate kinetic energy to gravitational potential energy.

Full solution

15. There are two triangles in the diagram which are similar to each other. Divide DE by BC, equate to AD divided by AB and solve for $x$.

Full solution

16. Equate kinetic energy to work done = force $\times$ distance.

Full solution

17. Using percentage multipliers, multiply $Q$ by $1.4$. This has the effect of dividing $P$ by $1.4^{2}=1.96$. This gives a result just above half of the original $P$.

Full solution

18. The $R-2$ is the hallmark of an alpha decay and an alpha particle has four nucleons. In the second decay, the number of nucleons remains the same so this is a beta decay.

Full solution

19. Write the two sentences as equations and rearrange to give two equations in $z$. Equate and raise to the power of 6.

Full solution

20. Interestingly, some information has been given here which is not required. Ignore the frequency and use $Time=\frac{Distance}{Speed}$, remembering the return leg of the journey.

Full solution

21. Let QX be 1, work out PX and XR then use PR to find PM and hence MX.

Full solution

22. Firstly, spot that graphs R and S are distance-time graphs exhibiting zero acceleration. Make a quick calculation of the gradients of P and Q.

Full solution

23. Rearrange and factorise then sketch and interpret the resulting quadratic.

Full solution

24. It is my opinion that there is a typographical mistake here. In the original version of this question, there were six answers. One has been removed for the purposes of creating the ENGAA specimen, changing the identity of the correct answer which was then not updated in the solution. The solution should read D not E. These mistakes are possible since the questions are written in code and I know from experience that errors are difficult to eradicate. Should this happen in your exam then keep calm, correct the mistake and move on.

Full solution

25. Turn the first sentence into an equation and rearrange for $h$.

Full solution

26. This is a potential divider. Work out the current using $V_{total}=I_{total}\times R_{total}$ and then apply Ohm's law once more to the resistor in question to find the voltage across this resistor. Finally apply $P=\frac{V^{2}}{R}$ or $P=I^{2}R$.

Full solution

27. Draw a set of axes and label the quadrants P Q R and S. Work out what happens to each of these points under the transformations given and draw a new set of axes with a new P Q R and S. It is clear from here which transformation is required to restore the square.

Full solution

28. Work out the quantities mentioned in P, R and S. Note that the speed of sound in air is roughly $340ms^{-1}$. If you think of the amplitude of a longitudinal wave as the maximum displacement of a molecule in a medium from its mean position, 5mm is too large for this as this would be a peak to peak disturbance.

Full solution

Part B

29. Factorise the numerator and denominator and cancel.

Full solution

30. Use Constant acceleration = $\frac{Change\, in\, velocity}{Time}$ then Resultant force = Mass $\times$ Acceleration.

Full solution

31. Take logarithms to base 10 of both sides and use the laws of logarithms to bring $x$ down from the superscript. Take $x$ out as a factor then combine the three logarithms into one. Finally make $x$ the subject.

Full solution

32. Momentum is conserved. Draw a diagram of the situation before and after taking care of the signs of the velocities in your equations.

Full solution

33. Evaluate these efficiently. It helps to spot that a number between 0 and 1 which is raised to the power 10 is going to be small. For answer D, consider that this must be between 3 and 4 because $2^{3}=8$ and $2^4=16$.

Full solution

34. Air resistance cannot be negative - it only hinders motion. This eliminates B and D. When the parachutist is travelling at a terminal velocity, the air resistance equals his weight both before the parachute is opened and after, so the horizontal portions of the graph should be at the same height.

Full solution

35. Be on the lookout for disguised quadratic equations. Let $2^{x}=y$ and solve the resulting quadratic.

Full solution

36. Use graph Z to eliminate three of the options then graphs Y and X to eliminate another one each. Briefly check that the remaining option fits the graphs.

Full solution

37. Multiplying an inequality by a negative number reverses its sense. Consider $(a+b)^{2}$; this must be greater than or equal to zero so rearranging this gives statement 2. For statement 3, we have no assurance that $c$ is positive, so this is not always true.

Full solution

38. Frictional forces balance the applied force up to a maximum point, so the block must start out stationary. Once it is moving the force applied is still increasing and hence so is the acceleration.

Full solution

39. Work out some values of $a_{n}$ to spot that this sequence is very repetitive. It repeats every two terms, so add these two terms up and multiply by 50.

Full solution

40. Momentum is conserved. Draw a diagram of the situation before and after taking care of the signs of the velocities in your equations.

Full solution

41. The factors of this equation are not whole numbers. It is crucial to spot this quickly and move onto a different method than factorisation. Try differentiating to discover that the turning points are $(0,-10)$, $(1,-9)$ and $(2,-10)$. Alternatively consider adding 10 to the function, sketching it and noting that upon subtraction of 10 all the turning points are well below the $x$-axis.

Full solution

42. Equate kinetic energy to potential energy and rearrange for $g_{p}\times h$. Note that only one of the answers gives the correct product.

Full solution

43. Use Pythagoras' Theorem to find PR, halve this and construct a triangle consisting of T, P and a vertical line downwards from T to apply trigonometry to.

Full solution

44. A $50\%$ loss of kinetic energy equates to a $50\%$ loss of gravitational potential energy and hence height. Therefore the height after each bounce is half what it was before that bounce.

Full solution

45. Take logarithms (the base doesn't matter) of each equation, preferably in your head. Any with $x$ in the power won't work because after taking logarithms the $x$ won't be inside a logarithm which is what we require. The right hand side of equation C cannot be simplified after taking a logarithm.

Full solution

46. The elevator must be accelerating downward regardless of the direction of its velocity. It would also be true to say that the elevator is moving downwards with increasing speed, but this isn't one of the options.

Full solution

47. The discriminant must be greater than 0. This yields $(a-2)^{2}>-8a$. Expand this, take everything to the left hand side and factorise. The resulting inequality is satisfied for all values of $a$ apart from -2.

Full solution

48. This is a fairly standard question involving the equations of motion.

Full solution

49. Don't be afraid to use trial and improvement. Start with a 4, 4, 4 equilateral triangle and make perturbations from there. Discard any triangles which don't close and you will reach an answer relatively quickly. Note the large difference between three triangles and the next answer up (ten triangles). This should help you settle on three quicker than if the next answer up were say four because of the large gap between three and ten.

Full solution

50. Resolve the diagonal force horizontally and vertically. Equate the vertical component to the weight of the particle and the horizontal component to the tension in the horizontal rope.

Full solution

51. A minimum for QR requires a minimum for PQ (3.5) and a maximum for PR (2.5) since $PR^{2}$ is subtracted from $PQ^{2}$. Use Pythagoras' Theorem with these values to find QR. Use $\sqrt{(3.5^{2}-2.5^{2})}=\sqrt{(3.5+2.5)(3.5-2.5)}$ to save time.

Full solution

52. Since the velocity of the mass is constant at this time, all of the power applied is going into increasing the potential energy of the mass. Therefore $Power=\frac{Energy}{Time}=\frac{mgh}{t}=mgv$.

Full solution

53. Solve each equation to give $0\leq x \leq \frac{\pi}{4}$, $\frac{3\pi}{4} \leq x \leq \pi$ and $\frac{\pi}{12} \leq x \leq \frac{5\pi}{12}$. Both are satisfied for $\frac{\pi}{12} \leq x \leq \frac{\pi}{4}$.

Full solution

54. Work out the acceleration of the entire ensemble. For carriage 2 this must be achieved as a result of the force in the coupling, T which is the mass of the carriage multiplied by its acceleration.

Full solution

Section 2

1a. Reaction forces are perpendicular to the surface and frictional forces act to oppose motion.

1b. The block is not accelerating off or into the slope, but it is accelerating down the slope and has components of this acceleration which are vertical and horizontal.

1c. Find the resultant force down the slope after resolving, equate to Mass $\times$ Acceleration to give $mgsin\alpha - \mu mgcos\alpha = ma$ then use $v=at$ for constant acceleration.

1d. The presence of this drag force means that the above equation is modified such that the resultant force is reduced by $kv$ to give $mgsin\alpha - \mu mgcos\alpha -kv = ma$. When the velocity is terminal, $a=0$. Substitute this in and rearrange for $v$ to give the terminal velocity.

Full solution

2a. Apply $R=\frac{\rho L}{A}$.

2b. To do this question, it helps to have an idea of what is actually happening to the current. It leaves the power station along one of the cables, travels through some kind of network which we don't need to know about then returns along the other cable. Working out this current is crucial to this question and can be found using $P=IV$. Once you have this current, use $P=I^{2}R$ to find the power dissipated in each cable and multiply by 2 because there are two of them. Alternatively, you can use $V=IR$ to work out the voltage dropped across each cable and then use $P=\frac{V^{2}}{R}$. Don't be tempted to use the $V$ from the question in $P=\frac{V^{2}}{R}$ because 400kV is the voltage dropped across the cables $and$ the city, not just the cables.

2c. A kilowatt hour is a unit of energy equal to the energy output of 1kW of power over a time of one hour. This is easily scaled up to work out a GWh in Joules.

2d. Answers A,B,D and E are simple to eliminate as follows:

A: The efficiency of the power station is of no consequence after you quote its power output which would have already had the efficiency taken into account if necessary.

B: Similarly, the appliances consume a certain power (as quoted in the question) and still use this power whether they are efficient or not.

D. This assumes that the energy can't be lost in wasted forms.

E. This also assumes that the energy can't be lost in wasted forms.

Full solution

3a. The function given is positive over the given interval, has a single root at the origin and a repeated root at $x=T$. This narrows the choices down to C and D. Note that $v=0$ for all $t>T$ and $t<0$.

3b. Differentiating an expression for velocity yields the acceleration of the particle. Factorise the result.

3c. The acceleration is a quadratic and hence is symmetrical and has a minimum value when $t$ is the mean of its roots. Therefore set $t=\frac{2T}{3}$.

3d. Look at the graph - this must be for $t=0$. Substitute into the equation for $a$.

3e. The object doesn't move after $t=T$. Integrate the velocity between $t=0$ and $t=T$ to find the area under the curve and hence the distance travelled.

Full solution

4a. Equate gravitational potential energy to kinetic energy.

4b. Do the same as for 4a except that the change in height is now $H-2R$.

4c. Circular motion requires a force towards the centre of the circle - think of planetary motion - there is no way the force could be in any other direction.

4d. If the car is to just make it round the loop. not only does it need to have enough gravitational energy to do so, it also needs some velocity to stay on the track. In this case, the reaction force at the top would be zero and the only force acting on the car would be the gravitational force. Equating this to the $\frac{mv^{2}}{R}$ in the question gives $v=\sqrt{gR}$ but from earlier in the question we have that $v=\sqrt{2g(H-2R)}$. Equating these yields $H=2.5R$ and the height needs to be greater than this because a reaction force of exactly zero means that the car has left the track.

Full solution