TheLucidSTEM · Physics olympiad feeder round mock · 80 minutes · 50 marks · Calculator allowed
Name: ______________________ Date: ____________ · MARK SCHEME
Data sheet
Data
| gravitational field strength at the Earth's surface | $g = 9.81$ N kg$^{-1}$ |
| gravitational constant | $G = 6.67\times10^{-11}$ N m$^2$ kg$^{-2}$ |
| speed of light in vacuum | $c = 3.00\times10^{8}$ m s$^{-1}$ |
| density of water | 1000 kg m$^{-3}$ |
| one day | 86 400 s |
Formulae you may use: $F = \dfrac{GMm}{r^2}$; $v_{esc} = \sqrt{2GM/r}$; $T = 2\pi\sqrt{L/g}$; $v = \sqrt{T/\mu}$ for a stretched string; $\frac1u + \frac1v = \frac1f$; $I = I_0\cos^2\theta$; $n_1\sin\theta_1 = n_2\sin\theta_2$; $E = mc^2$; $\varepsilon = BLv$; $F = BIL$.
Section A: multiple choice
Suggested time 20 minutes. One mark each, no penalty for wrong answers. Choose one option for each question.
Ball P is thrown vertically upwards at 20.0 m s$^{-1}$. Exactly 1.00 s later ball Q is thrown vertically upwards from the same point, also at 20.0 m s$^{-1}$. Air resistance is negligible. How far above the point of projection do the two balls pass each other?
A block of wood floats in water with 60% of its volume below the surface. The same block is put in oil of density 800 kg m$^{-3}$. What fraction of its volume is now below the surface? (Density of water 1000 kg m$^{-3}$.)
Planet X has the same mean density as the Earth but twice the Earth's radius. How does the escape speed from the surface of X compare with the escape speed from the Earth's surface?
A spacecraft is in a circular orbit around the Earth. It fires its engine for a very short time, pushing forwards along its direction of motion. By what percentage must its speed increase so that it just escapes from the Earth?
A simple pendulum has a period of 2.0 s on the Earth. Its length is made four times as long and it is taken to a moon where the gravitational field strength is one quarter of the Earth's. What is its new period?
A stretched string fixed at both ends has a fundamental frequency of 200 Hz. The tension is increased by 44% while the length stays the same. What is the new fundamental frequency?
A small source sends sound out equally in all directions, with no absorption or reflection. The sound level is defined as $L = 10\log_{10}(I/I_0)$ dB, where $I$ is the intensity. A listener walks from 2.0 m to 20 m from the source. By how much does the sound level fall?
A thin converging lens of focal length 20 cm forms a real image of an object that is three times as tall as the object. How far is the object from the lens? (Use $\frac1u + \frac1v = \frac1f$ with real distances positive.)
Unpolarised light of intensity $I_0$ passes through two ideal polarising filters. The transmission axis of the second is at 60° to that of the first. What intensity emerges from the second filter?
The Sun radiates energy at a rate of $3.8\times10^{26}$ W. By about how much does the Sun's mass decrease each second because of this radiation?
An air bubble of volume 2.0 cm$^3$ is released at a depth of 20 m in a lake. The water is at the same temperature at all depths, its density is 1000 kg m$^{-3}$ and the atmospheric pressure is $1.0\times10^5$ Pa. What is the volume of the bubble just before it reaches the surface?
A student of mass 60 kg runs up a flight of stairs that rises 3.0 m vertically, taking 4.0 s. Which is the best estimate of the useful power the student develops?
Section B: structured problems
Suggested time 60 minutes, about 15 minutes per problem. Marks for each part are shown; later parts can often be done even if an earlier part is not finished.
13Weighing a planet by its moon[10 marks]
Astronomers find a rocky planet with one small moon. The moon moves in a circular orbit of radius $2.00\times10^{8}$ m with a period of 4.00 days. The radius of the planet is $1.20\times10^{7}$ m. Take $G = 6.67\times10^{-11}$ N m$^2$ kg$^{-2}$ and ignore the mass of the moon compared with the planet.
(a) Show that the mass of the planet is $M = \dfrac{4\pi^2 r^3}{G T^2}$, where $r$ and $T$ are the radius and period of the moon's orbit, and calculate $M$.
Answer: $M = 3.96\times10^{25}$ kg
Gravity provides the centripetal force: $\dfrac{GMm}{r^2} = m\omega^2 r$ with $\omega = 2\pi/T$ [1].
Rearranging gives $M = 4\pi^2r^3/(GT^2)$ [1].
$T = 4.00\times86400 = 3.456\times10^5$ s, so $M = \dfrac{39.48\times(2.00\times10^8)^3}{6.67\times10^{-11}\times(3.456\times10^5)^2} = 3.96\times10^{25}$ kg [1]. Accept 3.9 to 4.0 $\times10^{25}$ kg. Leaving $T$ in days or hours loses the last mark.
(b) Calculate the mean density of the planet and the gravitational field strength at its surface.
Answer: $\rho = 5.5\times10^{3}$ kg m$^{-3}$; $g_s = 18.4$ N kg$^{-1}$
$\rho = M/(\frac43\pi R^3) = 3.96\times10^{25}/7.24\times10^{21} = 5.5\times10^3$ kg m$^{-3}$ [1] (about the Earth's density, so a rocky planet is believable).
$g_s = GM/R^2 = 6.67\times10^{-11}\times3.96\times10^{25}/(1.20\times10^7)^2 = 18.4$ N kg$^{-1}$ [1]. Allow error carried forward from (a).
(c) A probe orbits the planet just above its surface (ignore any atmosphere). Calculate its orbital speed and its orbital period.
Answer: 14.8 km s$^{-1}$; 5080 s (85 min)
$v = \sqrt{g_sR} = \sqrt{18.4\times1.20\times10^7} = 1.48\times10^4$ m s$^{-1}$ (or $\sqrt{GM/R}$) [1].
$T = 2\pi R/v = 7.54\times10^7/1.48\times10^4 = 5.08\times10^3$ s, about 85 minutes [1]. Accept 5000 to 5150 s.
(d) Imagine (it is only a thought experiment) that the planet has uniform density and that a straight, airless, frictionless tunnel is drilled through its centre. A capsule is released from rest at one end. Inside a uniform sphere, only the mass closer to the centre than the capsule pulls on it. Show that the capsule moves with simple harmonic motion, and find how long it takes to reach the far end of the tunnel.
Answer: $a = -(g_s/R)\,x$; time $= \pi\sqrt{R/g_s} = 2540$ s (42 min)
At distance $x$ from the centre the mass inside is $M x^3/R^3$, so the field is $g(x) = GMx/R^3 = g_s\,x/R$ [1].
The acceleration is $a = -(g_s/R)\,x$, towards the centre and proportional to $x$: SHM with $\omega^2 = g_s/R$ [1].
Period $2\pi\sqrt{R/g_s}$ (the same as the skimming orbit in (c)); the far end is reached after half a period, $2540$ s or about 42 minutes [1]. Accept 2500 to 2580 s. Giving the full period (85 min) as the answer loses the last mark.
14Light in a glass fibre[9 marks]
An optical fibre has a glass core of refractive index 1.48 surrounded by a cladding of refractive index 1.46. Light enters the flat end of the fibre from air (refractive index 1.00) at an angle $\theta$ to the axis, as shown, and is guided along the core by total internal reflection at the core-cladding boundary. The speed of light in vacuum is $3.00\times10^8$ m s$^{-1}$.
(a) Calculate the critical angle at the core-cladding boundary.
Answer: 80.6°
$\sin c = 1.46/1.48 = 0.9865$, so $c = 80.6$° [1]. Accept 80.5° to 80.6°.
(b) Show that the largest value of $\theta$ for which the light is still guided satisfies $\sin\theta_{max} = \sqrt{n_1^2 - n_2^2}$, where $n_1$ and $n_2$ are the core and cladding indices, and calculate $\theta_{max}$.
Answer: $\sin\theta_{max} = 0.242$, $\theta_{max} = 14.0$°
At the end face, $1.00\sin\theta = n_1\sin r$, where $r$ is the angle of the refracted ray to the axis [1].
The ray meets the side wall at angle $90^\circ - r$ to the normal, which must be at least $c$; at the limit $\sin r = \cos c = \sqrt{1 - n_2^2/n_1^2}$ [1].
So $\sin\theta_{max} = n_1\cos c = \sqrt{n_1^2 - n_2^2} = \sqrt{2.1904 - 2.1316} = 0.242$, $\theta_{max} = 14.0$° [1]. Accept 13.9° to 14.1°. Using $90^\circ - c$ directly as the angle in air (9.4°) loses the last two marks.
(c) A very short pulse of light enters a 1.0 km length of the fibre. Part of it travels straight along the axis; part zig-zags at the critical angle. Calculate the time each part takes, and the difference between them.
Answer: 4.933 μs and 5.001 μs; difference 68 ns
Speed in the core $c/n_1 = 2.027\times10^8$ m s$^{-1}$; axial time $= n_1L/c = 4.933$ μs [1].
The zig-zag ray travels a path $L/\sin c$ (for each metre along the axis it covers $1/\sin c$ metres of glass), so its time is $n_1L/(c\sin c) = n_1^2L/(n_2c) = 5.001$ μs [1].
Difference $= 6.8\times10^{-8}$ s, about 68 ns [1]. Accept 66 to 70 ns. Working each time to only two or three significant figures and getting zero or a wild difference loses the last mark.
(d) Digital data are sent as separate pulses. Estimate the largest number of pulses per second that can be sent along 1.0 km of this fibre if neighbouring pulses must not overlap at the far end. Suggest one change to the fibre that would allow a higher rate.
Answer: about $1.5\times10^7$ pulses per second; e.g. make $n_1$ and $n_2$ closer, or a much thinner core
Each pulse is spread out by about 68 ns, so pulses must be at least this far apart: rate $\approx 1/(6.8\times10^{-8}) = 1.5\times10^7$ s$^{-1}$ [1]. Accept 1 to 2 $\times10^7$.
Any one sensible change: a smaller difference between core and cladding index, a graded-index core (outer rays travel faster), or a core so thin that light travels only along the axis (single-mode fibre) [1]. A longer fibre or brighter source gets no mark.
15A rod pulled along magnetic rails[10 marks]
Two long, straight, horizontal metal rails are 0.50 m apart and are joined at one end by a 2.0 Ω resistor. A uniform vertical magnetic field of flux density 0.40 T acts over the whole region. A metal rod of mass 0.10 kg lies across the rails at right angles to them. A light string tied to the middle of the rod runs parallel to the rails, over a frictionless pulley at the edge of the table, to a hanging mass of 0.020 kg. The rod is released from rest. Ignore friction and the resistance of the rod and rails. Take $g = 9.81$ m s$^{-2}$.
(a) The rod moves at speed $v$. Explain why an emf is induced in the circuit, and show that the current is $I = \dfrac{BLv}{R}$, where $L$ is the distance between the rails.
Answer: $\varepsilon = BLv$, $I = BLv/R$
The area of the circuit (and so the flux through it) increases by $Lv$ each second as the rod moves; by Faraday's law the emf is the rate of change of flux, $\varepsilon = BLv$ (or: the rod cuts field lines) [1].
The only resistance is $R$, so $I = BLv/R$ [1].
(b) Show that the magnetic force on the rod has size $\dfrac{B^2L^2v}{R}$, and explain why it must act against the motion.
Answer: $F = BIL = B^2L^2v/R$, opposing $v$ (Lenz's law)
Force on a current in a field: $F = BIL = B^2L^2v/R$ [1].
Lenz's law (or energy conservation): the induced current's effect opposes the change causing it, so the force opposes the motion; if it helped, the rod would speed up and gain energy from nowhere [1].
(c) Calculate the terminal speed of the rod.
Answer: 9.8 m s$^{-1}$
At terminal speed the acceleration is zero, so the string tension equals the hanging weight and $mg = B^2L^2v_t/R$ [1].
$v_t = mgR/(B^2L^2) = 0.020\times9.81\times2.0/(0.40^2\times0.50^2) = 0.392/0.040 = 9.8$ m s$^{-1}$ [1]. Including the rod's mass in the driving weight loses the second mark.
(d) Show that, at the terminal speed, the rate at which the hanging mass loses gravitational potential energy equals the power dissipated in the resistor.
Answer: both 1.9 W
Rate of loss of GPE $= mgv_t = 0.020\times9.81\times9.81 = 1.92$ W [1].
$I = BLv_t/R = 0.40\times0.50\times9.81/2.0 = 0.981$ A, and $I^2R = 0.962\times2.0 = 1.92$ W, the same [1]. The kinetic energy is constant, so all the GPE becomes internal energy in the resistor.
(e) Treating the rod and hanging mass as one system of total mass $M_{tot}$, the motion obeys $\dfrac{dv}{dt} = \dfrac{v_t - v}{\tau}$. Find $\tau$, and use $v = v_t\,(1 - e^{-t/\tau})$ to find how long the rod takes to reach 90% of its terminal speed.
Answer: $\tau = 6.0$ s; about 14 s
$M_{tot}\,\dfrac{dv}{dt} = mg - \dfrac{B^2L^2v}{R} = \dfrac{B^2L^2}{R}(v_t - v)$, so $\tau = \dfrac{M_{tot}R}{B^2L^2} = \dfrac{0.12\times2.0}{0.040} = 6.0$ s [1].
$0.90 = 1 - e^{-t/\tau}$ gives $t = \tau\ln10 = 13.8$ s [1]. Accept 13.5 to 14 s. Using only the rod's mass (0.10 kg, $\tau = 5.0$ s) loses the first mark; allow the second mark carried forward.
16A leaking bottle[9 marks]
A tall plastic bottle of internal diameter 8.0 cm stands upright on a bench. A small round hole of diameter 4.0 mm is made in its side close to the bottom, and the bottle is filled with water. A student records the height $h$ of the water surface above the hole as the water runs out.
| $t$ / s | $h$ / cm |
|---|---|
| 0 | 20.0 |
| 20 | 14.2 |
| 40 | 9.3 |
| 60 | 5.5 |
| 80 | 2.7 |
| 100 | 0.9 |
Ignore viscosity and assume the water surface in the bottle falls slowly compared with the speed of the jet. You will need graph paper or squared paper for part (c).
(a) Use conservation of energy to show that the water leaves the hole at speed $v = \sqrt{2gh}$.
Answer: $v = \sqrt{2gh}$
When a small mass $\Delta m$ leaves through the hole, the net effect is that $\Delta m$ disappears from the top surface (height $h$, almost at rest) and appears in the jet at the hole: $\Delta m\,gh = \frac12\Delta m\,v^2$, so $v = \sqrt{2gh}$ [1]. Pressure is atmospheric at both places, so it does no net work.
(b) The hole lets water out as if its area were $a$ (an effective area). The bottle has cross-sectional area $A$. Show that $\sqrt{h}$ should fall linearly with time, with gradient $-\dfrac{a}{A}\sqrt{\dfrac{g}{2}}$.
Answer: $\sqrt h = \sqrt{h_0} - \frac{a}{A}\sqrt{\frac g2}\,t$
Volume flow out $= av$, so $A\dfrac{dh}{dt} = -a\sqrt{2gh}$ [1].
Since $\dfrac{d\sqrt h}{dt} = \dfrac{1}{2\sqrt h}\dfrac{dh}{dt} = -\dfrac{a}{A}\dfrac{\sqrt{2g}}{2} = -\dfrac aA\sqrt{\dfrac g2}$, a constant, $\sqrt h$ falls linearly with $t$ [1].
(c) Plot a graph of $\sqrt{h}$ (with $h$ in metres) against $t$. Use your graph to find the gradient, and hence find the effective area $a$ and the effective diameter of the hole.
Answer: gradient $-3.5\times10^{-3}$ m$^{1/2}$ s$^{-1}$; $a = 8.0\times10^{-6}$ m$^2$; diameter 3.2 mm
Values of $\sqrt{h/\mathrm{m}}$: 0.447, 0.377, 0.305, 0.235, 0.164, 0.095; points plotted and a straight line of best fit drawn [1].
Gradient $= -3.5\times10^{-3}$ m$^{1/2}$ s$^{-1}$ [1]. Accept 3.4 to 3.6 $\times10^{-3}$. (In cm$^{1/2}$ s$^{-1}$ the gradient is $-0.035$; this must be converted.)
$A = \pi(0.040)^2 = 5.03\times10^{-3}$ m$^2$, so $a = A\times3.53\times10^{-3}/\sqrt{9.81/2} = 8.0\times10^{-6}$ m$^2$, diameter $\sqrt{4a/\pi} = 3.2$ mm [1]. Accept 3.1 to 3.3 mm; allow error carried forward from the gradient.
(d) Use your graph to predict when the bottle stops leaking.
Answer: about 127 s
$\sqrt h = 0$ when $t = 0.447/3.53\times10^{-3} \approx 127$ s (or extrapolate the line to the $t$ axis) [1]. Accept 120 to 132 s. Extrapolating the curved $h$ against $t$ graph by eye gets no mark.
(e) The effective diameter from (c) is smaller than the real 4.0 mm. Find the ratio (effective area)/(real area), and suggest a physical reason why it is less than 1.
Answer: about 0.64; the jet narrows just outside the hole
Ratio $= (3.2/4.0)^2 = 0.64$ (or $8.0\times10^{-6}/1.26\times10^{-5}$) [1]. Accept 0.60 to 0.68.
Water arrives at the hole from all directions inside the bottle and cannot turn sharply at the edge, so the jet keeps narrowing for a short distance after the hole; its narrowest cross-section is smaller than the hole (some energy is also lost to viscosity at the edge) [1]. Simply saying 'experimental error' gets no mark.
TheLucidSTEM practice paper. Questions are original and written in the style of a first national round; this is not an official olympiad paper.