# Oxford Physics Interviews – Model Answers

Here we have a list of suggested solutions to the example interview questions given in the post Oxford Physics Interviews – How to Prepare. These are by no means the only way of tackling the questions given; physics interview questions tend to be fairly open-ended in order to encourage discussion. Remember not to look at these solutions until you’ve had a think about the questions for yourself.

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1. What experiment could you do to show that the Sun is not just a giant ball of molten gold?

A good place to start is: how do we know what stars are made of? Most of our information comes from spectroscopy.

If we take a sample of starlight, and pass it through a dispersive optical element like a diffraction grating or a prism, the light will be revealed to comprise a nearly continuous spectrum of light of different frequencies. I say nearly continuous, because there will be narrows gaps distributed throughout the spectrum.

This is because a star, as a so-called blackbody, emits a continuous range of frequencies; this distribution has a single maximum at a frequency determined by the surface temperature of the star. But not all of this emitted light makes it to us, the observer, because a few select frequencies are absorbed by elements in the Sun’s atmosphere. Some photons will have exactly the right frequency to cause an internal energy transition in the atom it interacts with. These photons are absorbed,  and so will appear as gaps in the starlight’s spectrum. Where these gaps occur depends on the elements present in the star’s atmosphere.

If, for example, you were to shine blackbody radiation on a sample of gaseous hydrogen, you would find the radiation emerging from the other side contains families of gaps, all of which have been extensively studied and have their own names. If you were to examine the spectrum from most any star you could name, you will find exactly the same gaps – the implication is that stars (or at the very least their atmospheres) contain a substantial amount of hydrogen. Each element has its own characteristic set of absorption lines, somewhat like a barcode.

Then in principle you could boil a sample of gold, shine blackbody radiation through it and observe the absorption spectrum. You will find that gaps present in gold’s absorption spectrum are only very weakly visible in the Sun’s absorption spectrum. It is not unthinkable that the Sun contains gold, since the Earth does, and the Sun and the Earth are ultimately derived from the same cloud of dust. But the absorption lines from other elements such as hydrogen, oxygen, iron and other metals will be far more prominent, suggesting the Sun is not primarily made of gold.

The diagram above shows the main features of the absorption spectrum of the Sun’s atmosphere (top) and a detailed absorption spectrum of gold (bottom). As you can see, there are a few coincidental absorption lines, the most striking of which are the two that occur in the orange-red part of the spectrum. In fact, the red line in the Sun’s absorption spectrum comes from the Lyman-$\alpha$ transition in hydrogen while the orange line actually comes from diatomic oxygen!

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2. Suggest how artificial gravity might be created on a space station. Estimate the parameters needed to achieve 9.8 N/kg inside the ship.

The point of artificial gravity is to create an environment in which you tend to accelerate towards a particular surface inside the space station. With this inevitably comes a perceived ‘weight’ when you come into contact with this surface. The use of artificial gravity has the benefits of i) making the craft easier to navigate and ii) reducing the effects of muscle and bone deterioration associated with a microgravity environment.

One way of generating artificial gravity is to build a surface whose normal is parallel to the direction in which the space station is accelerating. Imagine you’re initially floating inside the space station. When it starts to accelerate, one surface will appear to rush up to meet you. To you, it looks as if you’re falling towards this surface.

When you come into contact with the surface, the surface has to exert a force on you in order to keep your velocity equal to that of the space station (since you can’t fall through the floor!). From this force you will experience the sensation of weight. If you could keep the space station accelerating at a constant 9.8 ms-2, your weight in the station would be equal to your weight on Earth.

The problem with this design is that it’s quite difficult to keep a spacecraft accelerating parallel to its velocity at 9.8 ms-2 for a prolonged period of time, using modern technology. This issue can be circumvented if instead the acceleration you experience is due to circular motion, rather than linear motion.

Suppose the space station was cylindrical, and was rotating about its axis. If  you were to stand on the (inside!) curved surface of the station, you would be forced to undergo circular motion, such that you were subject to a constant centripetal acceleration. From the corresponding centripetal force you will experience the sensation of weight.

The advantage of this design is that once the spacecraft is set rotating, it takes a very long time for its rotation to decay, since there are only minuscule frictional and drag forces slowing its motion down in space.

To get some idea of the parameters required, we start off with the expression for the centripetal acceleration $a$ experienced by a point mass moving in a circle of radius $r$ at tangential speed $v$:

$\displaystyle a=\frac{v^2}{r}$

Taking $a=$ 10 ms-2  for near-Earth weight and $r\approx$ 100 m, then

$v^2\approx 1\,000\,\text{m}^2\text{s}^{-2}$

$v\approx 30\,\text{ms}^{-1}$

Artificial gravity via centripetal acceleration comes with some odd side-effects. For example, if you were to walk around the circumference of the space station in the same direction as it spins, the centripetal force needed to keep you moving in a circle will increase, since your tangential speed $v$ has increased, so you suddenly feel heavier! This effect is part of the Coriolis force.

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3. Estimate the mass of rubber worn off car tyres in the UK each year

This is an example of a Fermi problem, one whose solution relies on making a number of sensible assumptions and approximations. Here’s how you might come up with an estimate.

How much rubber needs to be worn off a tyre before it needs to be replaced? For the tread to wear away appreciably, let’s say the radius of the tyre needs to decrease by a conservative 1 cm. If the radius of the tyre, with width $w$, changes from $r_1$ to $r_2$ in this time, then

$\Delta V=\pi w(r_1^2-r_2^2)$

$\Delta V=\pi w(r_1-r_2)(r_1+r_2)$

$\Delta V\approx\pi w(\Delta r)(2\bar{r})$

$\Delta V\approx 2\pi w\bar{r}\Delta r$

where $\bar{r}$ is the mean radius of the tyre. Assuming the radius of the tyre is on average 25 cm and its width is 20 cm, then

$\Delta V\approx 2\pi\cdot 0.25\cdot 0.2\cdot 0.01$

$\Delta V\approx 3\times 10^{-3} \text{m}^3$

Assuming there are about 10 million cars in use in the UK, every time the tyres of all the cars in the UK are replaced a total volume of

$V=4\times 10^7\times \Delta V$

of rubber has been deposited on the roads of the UK. We multiply this volume by the density of rubber to get the total mass deposited. All that remains is to ask: how frequently do the tyres of a car need to be replaced? Assuming this happens about once every 5 years, the total mass deposited per year is

$\displaystyle m=\frac{\rho V}{5}$

$\displaystyle m=\frac{10^3\times(4\times 10^7\times 3\times 10^{-3})}{5}$

$m\approx 24\times 10^6\text{kg}$

Really, the significant figures aren’t important here; to quote anything more specific than just a power of 10 would be false accuracy. The emphasis in this kind of question is not so much on the numbers you use or the number you end up at, but the way you get to your answer.

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4. Sketch a speed-time graph for a projectile launched from ground level. Sketch a velocity-time graph for the vertical, then horizontal components of its velocity. Explain the shape of your original speed-time graph.

Without getting bogged down in the maths, let’s make some sketches. We’ll assume the normal ideal conditions, namely that the gravitational field is uniform throughout the projectile’s motion and it experiences no air resistance. We’ll also assume that the projectile’s initial vertical component of velocity is positive, so that it initially moves upwards.

How will the speed vary with time? At launch, the projectile will have some non-zero speed. During its ascent, its vertical component of velocity is decreasing, while its horizontal component is (by assumption) constant. This means the overall speed is decreasing. When the projectile reaches its maximum height, its vertical component of velocity is zero, meaning the projectile’s speed has reached a (non-zero) minimum. As soon as the projectile starts falling, its speed starts increasing again. From our assumption that there are no drag forces acting on the projectile, its lands on the ground at the speed at which it left the ground.

Drawing in these key points and interpolating, we end up with a sketch something like the diagram on the right. It’s worth mentioning that the curve is not a parabola. For the more mathematically inclined, we can derive the exact shape of the speed-time curve using expressions for the components of the projectile’s velocity.

Assuming the projectile’s initial speed is $v_0$ and the angle from which it is launched relative to the horizontal is $\theta$, its components of velocity are

$v_x=v_0\cos\theta$

$v_y=v_0\sin\theta-gt$,

relying on the assumptions made above. Using the definition of speed, we can say that

$v=\sqrt{v_x^2+v_y^2}$

$v^2=v_0^2\cos^2\theta+(v_0\sin\theta-gt)^2$

$(v)^2-(gt-v_0\sin\theta)^2=(v_0\cos\theta)^2$

You may recognise this equation, of the form

$\displaystyle \left(\frac{y-y_0}{a}\right)+\left(\frac{x-x_0}{b}\right)^2=c^2$,

as representing a hyperbola. Of course, knowing this doesn’t necessarily make sketching the speed-time curve any easier. It is much simpler to rely on your physical intuition to get the answer.

The velocity-time graphs are a little easier. By assumption, the horizontal component of velocity does not vary with time, since the particle does not suffer air resistance. The vertical component meanwhile starts with some non-zero positive value and decreases linearly with time, according to the equation

$v_y=v_0\sin\theta-gt$

Since the particle is launched from the same height as that at which it lands, its final vertical component of velocity will be minus its initial component of velocity. Using these ideas, we can draw the sketches like those on the right.

How can we get the speed-time graph from the velocity-time graphs? A neat way of thinking about this is using vectors. The velocity of the projectile is, in vector terms, the sum of its horizontal and vertical velocities, while its speed is simply the magnitude of its total velocity. The animation on the right sums this up quite nicely. Keep an eye on the length of the resultant vector $v$ as it varies with time.

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5. Sketch and label a force diagram showing the Earth-Moon system. Why does the Moon not fall towards the Earth?

Here’s the force diagram:

The most important features are

• the fact that the forces on each body are antiparallel
• the fact that the magnitude of the forces are equal

This is what is generally meant by forces being ‘equal and opposite’, or Newton’s third law.

More interesting is the question: why does the Moon not fall towards the Earth? Imagine how the position of the Moon changes during some small interval of time. The Moon’s tangential component of orbital velocity causes it to move away from the Earth; in that same interval, the gravitational interaction of the Earth and the Moon causes the Moon to move slightly towards Earth’s centre. For an idealised circular orbit, these two effects cancel one another out perfectly. In short, the Moon’s orbital velocity prevents it falling into the Earth.

For a more mathematical treatment, we sometimes use the idea of an effective potential. Approximating the Earth as being stationary, the total energy of the Earth-Moon system is

$\displaystyle E=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\varphi}^2-\frac{Gm_1m_2}{r}$

where $r$ is the distance of the Moon from the Earth, and $\varphi$ is an angle which describes how far ‘around’ the Moon is in its orbit. What we have done is split the kinetic energy into two parts: the first part comes from the rate at which the Moon is approaching or regressing from the Earth, and the second part comes from the Moon’s kinetic energy it has by virtue of hurtling around the Earth. The third term is the gravitational potential energy of the system.

Because the gravitational force is a central force, the angular momentum of the Moon $L$ relative to the Earth is conserved, meaning we can actually get rid of the $\dot{\varphi}$ dependence:

$L=m_2r^2\dot{\varphi}$

$\displaystyle E=\frac{1}{2}m\dot{r}^2+\frac{L^2}{2m_2r^2}-\frac{Gm_1m_2}{r}$

The second and third terms are combined into a single term called the effective potential:

$\displaystyle V_{eff}(r)=\frac{L^2}{2m_2r^2}-\frac{Gm_1m_2}{r}$

Here’s the important bit: the potential energy becomes unboundedly large and positive as $r\to 0$. Why is this? Although the gravitational term goes to negative infinity as $r\to 0$, the angular momentum term goes to positive infinity as $r\to 0$, since r^-2 increases faster than r^-1 does.

This means the Moon’s non-zero angular momentum / orbital velocity creates a kind of barrier which prevents it from getting near the Earth.

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6. Estimate the square root of 13 000

The quickest way to get to an answer is through trial and error.

First things first, all those extra zeroes do not make the problem any more difficult:

$13\,000=130\times 10^2$

So

$\sqrt{13\,000}=10\sqrt{130}$

Suddenly the problem looks a lot less intimidating, since you’re only being asked to guess the square root of 130. I’m sure you know that

$11^2=121$

$12^2=144$

Hence

$11<\sqrt{130}<12$

$\sqrt{13\,000}\approx 10\cdot 11.5=115$

It turns out that

$115^2=13\,225$

which is pretty close to the answer you want!

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7. Sketch the graph of exp x, sin x and finally exp sin x

You’ll be expected to know what ex and sin x look like; the point here is to use what you know to piece together the shape of esin x.

The first thing we can say is that it will be periodic, just as sin x is. It will also have the same period as sin x, since esin x = esin (x+2$\pi$).

A good idea is to find the values between which esin x oscillates. Here are a few example values:

$e^{\sin (0)}=e^0=1$

$e^{\sin\left(\frac{\pi}{2}\right)}=e^1=e$

$e^{\sin(\pi)}=e^0=1$

$e^{\sin\left(\frac{3\pi}{2}\right)}=e^{-1}=\frac{1}{e}$

$e^{\sin(2\pi)}=e^0=1$

So esin x oscillates between $e$ and $\frac{1}{e}$ with period $2\pi$. Based on this information, here’s a sketch:

The exact curvature of the graph between these points is going to be tricky to work out, but you can guess it isn’t going to be symmetrical. A good way of adding some detail to your plot is to notice that the distance from e-1 to 1 is smaller than the distance from 1 to e:

$e^{-1}\approx 0.3$

$e\approx 2.7$

So the peaks are higher than the troughs are deep, if you see what I mean.

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For anyone with interviews soon, the best of luck to you! Any questions on this topic are welcomed.