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.

If you find this page useful please consider recommending my tuition to anyone who is due to sit the ENGAA, NSAA, STEP or Oxford PAT/MAT or would like to improve their performances in Maths/Physics Challenges or Olympiads.

Hints

Section 2 - Physics

1 a) It is likely that you are familiar with Hooke's law and elastic potential energy.

1 b) Use Hooke's law to find the spring constant and then use this to find the elastic potential energy.

1 c) Note that the force due to her weight is always present. The force due to the rope is absent until it becomes taut and then gradually increases as it extends.

1 d) Consider the GPE lost and the EPE stored. The GPE goes into KE and EPE.

1 e) Make the height fallen an unknown quantity and equate GPE lost to EPE stored.

1 f) The maximum speed will be when her weight balances the force due to the rope; after this the rope starts to arrest her motion and her speed will decrease.

1 g) The maximum acceleration will either be during the free fall or at the lowest point of her fall where the rope is pulling up the strongest. Use Hooke's law to work out the force due to the rope and offset this against her weight.

1 h) The graph starts at $g$ and remains there until the rope starts to have an effect. The force due to the rope increases linearly with distance.

2 a) Give a full explanation of the existence of the interference pattern and note that this can only occur if the light is thought of as a wave.

2 b) The sketch should consist of dark and light fringes, possibly with a slight decrease in amplitude towards the edges.

2 c) Mark a point $X$ on the line $P2$ so that $PX=P1$.

2 d) Define an angle $\theta$ between the rays of light and the horizontal dotted line. Write this angle in two different ways using trigonometry then apply the small angle approximation $sin\theta \approx tan\theta \approx \theta$.

2 e) The instructions given are full and clear - do your best to follow them and don't be afraid to use a few steps.

2 f) The contents of the bracket containing $\Delta L$ have to be such that their cosine is equal to $0$.

2 g) Substitute the values into your answer for part d)

Full solutions Section 2 - Physics

1 a) It is likely that you are familiar with Hooke's law and elastic potential energy.

The extension of the rope is proportional to the force applied to it. Elastic potential energy is the energy stored in the rope when it is stretched beyond its natural length.

1 b) Use Hooke's law to find the spring constant and then use this to find the elastic potential energy.

$F=kx$

$500=16k$

$k=31.25Nm^{-1}$

$E=\frac{1}{2}kx^2=\frac{500}{32}\times16^2=4000J$

1 c) Note that the force due to her weight is always present. The force due to the rope is absent until it becomes taut and then gradually increases as it extends.

Taking the upward direction to be positive, initially the acceleration is $-g$. When the rope becomes taut, as Alice moves further down her acceleration increases. When her acceleration is $0$ she still has a considerable negative velocity. Therefore her acceleration continues to increase until it reaches a maximum at the lowest point of the fall.

1 d) Consider the GPE lost and the EPE stored. The GPE goes into KE and EPE.

GPE lost=$15mg$

EPE stored=$\frac{1}{2}\times31.25\times5^2=390.625$

$\frac{1}{2}mv^2=7500-390.625$

$v=16.86ms^{-1} (2d.p.)$

1 e) Make the height fallen an unknown quantity and equate GPE lost to EPE stored.

Let the height fallen be $h$.

GPE lost=$mgh$

EPE stored=$\frac{1}{2}k(h-10)^2$ because the rope has natural length $10$ therefore the extension is $h-10$.

$mgh=\frac{1}{2}k(h-10)^2$

$500h=\frac{500}{32}(h-10)^2$

$32h=h^2-20h+100$

$h^2-52h+100=0$

$(h-2)(h-50)=0$

For $h=2$ the rope is not even stretched so we discard this solution and the distance below the bridge is $50m$.

1 f) The maximum speed will be when her weight balances the force due to the rope; after this the rope starts to arrest her motion and her speed will decrease.

From earlier in the question, this occurs when the extension is $16m$ or when Alice is $26m$ below the bridge.

GPE lost =$26mg$

EPE stored =$\frac{1}{2}k\times16^2$

$\frac{1}{2}mv^2=26mg-\frac{1}{2}\times31.25\times16^2$

$v=18.97 ms^{-1} (2d.p.)$

1 g) The maximum acceleration will either be during the free fall or at the lowest point of her fall where the rope is pulling up the strongest. Use Hooke's law to work out the force due to the rope and offset this against her weight.

At the lowest point, $F=kx=40k=1250N$ upwards. The resultant force is therefore $750N$ upwards. This is larger than the $500N$ due to gravity with no rope. The maximum force is $750N$ so the maximum acceleration is $15ms^{-1}$ upwards.

1 h) The graph starts at $g$ and remains there until the rope starts to have an effect. The force due to the rope increases linearly with distance.

2 a) Give a full explanation of the existence of the interference pattern and note that this can only occur if the light is thought of as a wave.

Light arriving at $P$ has travelled different distances depending on whether it went through slit 1 or 2. This path difference gives rise to a phase difference. If the two waves arrive in phase (that's to say the path difference is an integer number of wavelengths) then they interfere constructively giving a bright fringe. If the two waves arrive out of phase (that's to say their path difference is an integer number of wavelengths plus half a wavelength) then the waves destructively interfere and give a dark fringe. The distance of $P$ from $x=0$ determines the path difference and hence the phase difference and brightness of the light at that point.

2 b) The sketch should consist of dark and light fringes, possibly with a slight decrease in amplitude towards the edges.

2 c) Mark a point $X$ on the line $P2$ so that $PX=P1$.

2 d) Define an angle $\theta$ between the rays of light and the horizontal. Write this angle in two different ways using trigonometry then apply the small angle approximation $sin\theta \approx tan\theta \approx \theta$.

Path difference=$dsin\theta$

Also, $tan\theta=\frac{x}{L}$

$\theta\approx tan\theta\approx sin\theta$

$\frac{dx}{L}=\frac{\lambda}{2}$

$x=\frac{\lambda L}{2d}$

2 e) The instructions given are full and clear - do your best to follow them and don't be afraid to use a few steps.

$A=2A_0cos(\frac{2\pi\Delta L}{\lambda})cos(\omega t-\frac{2\pi L}{\lambda})$

2 f) The contents of the bracket containing $\Delta L$ have to be such that their cosine is equal to $0$.

Now, $A=2A_0cos(\frac{2\pi\Delta L}{\lambda})cos(-\frac{2\pi L}{\lambda})$

For the smallest values of $\Delta L$ which reduce this to zero we require

$\frac{2\pi\Delta L}{\lambda}=\frac{\pi}{2}$ or $\frac{3\pi}{2}$ which gives

$\Delta L=\frac{\lambda}{4}$ or $\frac{3\lambda}{4}$

2 g) Substitute the values into your answer for part d)

$0.015=\frac{\lambda L}{2d}$

$\Delta L=\frac{d\times 0.015}{2L}=1.5\times10^{-7}m$

$\lambda=600nm$

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.

If you find this page useful please consider recommending my tuition to anyone who is due to sit the ENGAA, NSAA, STEP or Oxford PAT/MAT or would like to improve their performances in Maths/Physics Challenges or Olympiads.

Hints

Section 2 - Physics

1 a) It is likely that you are familiar with Hooke's law and elastic potential energy.

1 b) Use Hooke's law to find the spring constant and then use this to find the elastic potential energy.

1 c) Note that the force due to her weight is always present. The force due to the rope is absent until it becomes taut and then gradually increases as it extends.

1 d) Consider the GPE lost and the EPE stored. The GPE goes into KE and EPE.

1 e) Make the height fallen an unknown quantity and equate GPE lost to EPE stored.

1 f) The maximum speed will be when her weight balances the force due to the rope; after this the rope starts to arrest her motion and her speed will decrease.

1 g) The maximum acceleration will either be during the free fall or at the lowest point of her fall where the rope is pulling up the strongest. Use Hooke's law to work out the force due to the rope and offset this against her weight.

1 h) The graph starts at $g$ and remains there until the rope starts to have an effect. The force due to the rope increases linearly with distance.

2 a) Give a full explanation of the existence of the interference pattern and note that this can only occur if the light is thought of as a wave.

2 b) The sketch should consist of dark and light fringes, possibly with a slight decrease in amplitude towards the edges.

2 c) Mark a point $X$ on the line $P2$ so that $PX=P1$.

2 d) Define an angle $\theta$ between the rays of light and the horizontal dotted line. Write this angle in two different ways using trigonometry then apply the small angle approximation $sin\theta \approx tan\theta \approx \theta$.

2 e) The instructions given are full and clear - do your best to follow them and don't be afraid to use a few steps.

2 f) The contents of the bracket containing $\Delta L$ have to be such that their cosine is equal to $0$.

2 g) Substitute the values into your answer for part d)

Full solutions Section 2 - Physics

1 a) It is likely that you are familiar with Hooke's law and elastic potential energy.

The extension of the rope is proportional to the force applied to it. Elastic potential energy is the energy stored in the rope when it is stretched beyond its natural length.

1 b) Use Hooke's law to find the spring constant and then use this to find the elastic potential energy.

$F=kx$

$500=16k$

$k=31.25Nm^{-1}$

$E=\frac{1}{2}kx^2=\frac{500}{32}\times16^2=4000J$

1 c) Note that the force due to her weight is always present. The force due to the rope is absent until it becomes taut and then gradually increases as it extends.

Taking the upward direction to be positive, initially the acceleration is $-g$. When the rope becomes taut, as Alice moves further down her acceleration increases. When her acceleration is $0$ she still has a considerable negative velocity. Therefore her acceleration continues to increase until it reaches a maximum at the lowest point of the fall.

1 d) Consider the GPE lost and the EPE stored. The GPE goes into KE and EPE.

GPE lost=$15mg$

EPE stored=$\frac{1}{2}\times31.25\times5^2=390.625$

$\frac{1}{2}mv^2=7500-390.625$

$v=16.86ms^{-1} (2d.p.)$

1 e) Make the height fallen an unknown quantity and equate GPE lost to EPE stored.

Let the height fallen be $h$.

GPE lost=$mgh$

EPE stored=$\frac{1}{2}k(h-10)^2$ because the rope has natural length $10$ therefore the extension is $h-10$.

$mgh=\frac{1}{2}k(h-10)^2$

$500h=\frac{500}{32}(h-10)^2$

$32h=h^2-20h+100$

$h^2-52h+100=0$

$(h-2)(h-50)=0$

For $h=2$ the rope is not even stretched so we discard this solution and the distance below the bridge is $50m$.

1 f) The maximum speed will be when her weight balances the force due to the rope; after this the rope starts to arrest her motion and her speed will decrease.

From earlier in the question, this occurs when the extension is $16m$ or when Alice is $26m$ below the bridge.

GPE lost =$26mg$

EPE stored =$\frac{1}{2}k\times16^2$

$\frac{1}{2}mv^2=26mg-\frac{1}{2}\times31.25\times16^2$

$v=18.97 ms^{-1} (2d.p.)$

1 g) The maximum acceleration will either be during the free fall or at the lowest point of her fall where the rope is pulling up the strongest. Use Hooke's law to work out the force due to the rope and offset this against her weight.

At the lowest point, $F=kx=40k=1250N$ upwards. The resultant force is therefore $750N$ upwards. This is larger than the $500N$ due to gravity with no rope. The maximum force is $750N$ so the maximum acceleration is $15ms^{-1}$ upwards.

1 h) The graph starts at $g$ and remains there until the rope starts to have an effect. The force due to the rope increases linearly with distance.

2 a) Give a full explanation of the existence of the interference pattern and note that this can only occur if the light is thought of as a wave.

Light arriving at $P$ has travelled different distances depending on whether it went through slit 1 or 2. This path difference gives rise to a phase difference. If the two waves arrive in phase (that's to say the path difference is an integer number of wavelengths) then they interfere constructively giving a bright fringe. If the two waves arrive out of phase (that's to say their path difference is an integer number of wavelengths plus half a wavelength) then the waves destructively interfere and give a dark fringe. The distance of $P$ from $x=0$ determines the path difference and hence the phase difference and brightness of the light at that point.

2 b) The sketch should consist of dark and light fringes, possibly with a slight decrease in amplitude towards the edges.

2 c) Mark a point $X$ on the line $P2$ so that $PX=P1$.

2 d) Define an angle $\theta$ between the rays of light and the horizontal. Write this angle in two different ways using trigonometry then apply the small angle approximation $sin\theta \approx tan\theta \approx \theta$.

Path difference=$dsin\theta$

Also, $tan\theta=\frac{x}{L}$

$\theta\approx tan\theta\approx sin\theta$

$\frac{dx}{L}=\frac{\lambda}{2}$

$x=\frac{\lambda L}{2d}$

2 e) The instructions given are full and clear - do your best to follow them and don't be afraid to use a few steps.

$A=2A_0cos(\frac{2\pi\Delta L}{\lambda})cos(\omega t-\frac{2\pi L}{\lambda})$

2 f) The contents of the bracket containing $\Delta L$ have to be such that their cosine is equal to $0$.

Now, $A=2A_0cos(\frac{2\pi\Delta L}{\lambda})cos(-\frac{2\pi L}{\lambda})$

For the smallest values of $\Delta L$ which reduce this to zero we require

$\frac{2\pi\Delta L}{\lambda}=\frac{\pi}{2}$ or $\frac{3\pi}{2}$ which gives

$\Delta L=\frac{\lambda}{4}$ or $\frac{3\lambda}{4}$

2 g) Substitute the values into your answer for part d)

$0.015=\frac{\lambda L}{2d}$

$\Delta L=\frac{d\times 0.015}{2L}=1.5\times10^{-7}m$

$\lambda=600nm$