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$^{-1}$ |
| speed of light in vacuum | $c = 3.00\times10^{8}$ m s$^{-1}$ |
| Planck constant | $h = 6.63\times10^{-34}$ J s |
| elementary charge | $e = 1.60\times10^{-19}$ C |
| electron mass | $m_e = 9.11\times10^{-31}$ kg |
| specific heat capacity of water | 4200 J kg$^{-1}$ K$^{-1}$ |
Formulae you may use: $v^2 = u^2 + 2as$; $T^2\propto r^3$ for circular orbits; $E=\frac12CV^2$; $\sin c = n_2/n_1$; $\lambda = h/p$; $\Delta Q = mc\,\Delta T$.
Section A: multiple choice
Suggested time 20 minutes. One mark each, no penalty for wrong answers. Choose one option for each question.
A car travels the first half of a journey's distance at 30 m s$^{-1}$ and the second half of the distance at 20 m s$^{-1}$. What is its average speed for the whole journey?
A stone is thrown horizontally at 8.0 m s$^{-1}$ from the top of a cliff 20 m high. Air resistance is negligible. With what speed does it hit the ground below?
A trolley of mass 0.20 kg moving at 3.0 m s$^{-1}$ hits a stationary trolley of mass 0.10 kg and they stick together. What fraction of the original kinetic energy is transferred to other forms in the collision?
A satellite is moved from one circular orbit to another whose radius is four times as large. By what factor does its orbital period change?
0.50 kg of water at 80 °C is mixed with 0.30 kg of water at 20 °C in an insulated flask of negligible heat capacity. What is the final temperature?
A battery has emf 12 V and internal resistance 2.0 Ω. A variable resistor is connected across it. What is the largest power that can be delivered to the variable resistor?
A 100 μF capacitor charged to 10 V is disconnected from its supply and then connected across an identical uncharged 100 μF capacitor. What is the total energy stored in the two capacitors once the charge has settled?
Two small loudspeakers driven in phase by the same signal are 5.00 m and 5.85 m from a listener. The speed of sound is 340 m s$^{-1}$. What is the lowest frequency at which the listener hears a minimum of loudness?
Light travels in glass of refractive index 1.50 towards a flat boundary with water of refractive index 1.33. What is the critical angle in the glass at this boundary?
The activity of a radioactive sample falls from 800 Bq to 50 Bq in 24 hours. What is the half-life?
An electron, initially at rest, is accelerated through a potential difference of 150 V. What is its de Broglie wavelength, to one significant figure?
A small liquid drop can wobble. Its wobble frequency $f$ depends only on its radius $r$, its density $\rho$ and the surface tension $\gamma$ of the liquid (unit N m$^{-1}$). If a drop of the same liquid with twice the radius is used, $f$ is multiplied by about
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.
13A parcel on a rough slope[10 marks]
A parcel of mass 2.0 kg is released from rest at the top of a straight slope 4.0 m long, inclined at 30° to the horizontal. The coefficient of kinetic friction between the parcel and the slope is 0.20. At the bottom it moves smoothly onto a horizontal floor with the same coefficient of friction and slides until it stops. Treat the parcel as a particle and ignore any effect of the corner. Take $g = 9.81$ m s$^{-2}$.
(a) Show that the acceleration of the parcel down the slope is about 3.2 m s$^{-2}$.
Answer: $a = g(\sin 30^\circ - \mu\cos 30^\circ) = 3.21$ m s$^{-2}$
Normal force $N = mg\cos 30^\circ$ (17.0 N) and friction $\mu N$ (3.40 N) up the slope [1].
$ma = mg\sin 30^\circ - \mu mg\cos 30^\circ$, so $a = 9.81(0.500 - 0.20\times0.866) = 3.21$ m s$^{-2}$ [1].
Common error: using $\mu mg$ for the friction (gives 2.94 m s$^{-2}$), no mark for the answer.
(b) Calculate the speed of the parcel at the bottom of the slope.
Answer: 5.1 m s$^{-1}$
$v = \sqrt{2as} = \sqrt{2(3.21)(4.0)} = 5.06$ m s$^{-1}$ [1]. Accept 5.0 to 5.1 m s$^{-1}$; allow error carried forward from (a).
(c) Calculate how far the parcel slides along the floor before it stops.
Answer: 6.5 m
Deceleration on the floor $\mu g = 1.96$ m s$^{-2}$ [1].
$d = v^2/(2\mu g) = 25.6/3.92 = 6.5$ m [1]. Accept 6.4 to 6.6 m; allow error carried forward from (b).
(d) Show that the total work done against friction, on the slope and the floor together, equals the loss of gravitational potential energy.
Answer: both are 39 J
Loss of GPE $= mgh = 2.0\times9.81\times(4.0\sin30^\circ) = 39.2$ J [1].
Friction work: slope $0.20\times17.0\times4.0 = 13.6$ J, floor $0.20\times19.6\times6.5 = 25.6$ J, total 39.2 J [1]. Accept a statement that the parcel starts and ends at rest, so all the GPE goes to friction, only if both friction terms are also calculated.
(e) The slope is replaced by a steeper or a shallower straight slope that starts at the same height $h$ and has the same coefficient of friction (the parcel still slides). Show that the horizontal distance from the starting point to where the parcel stops does not depend on the angle of the slope, and find its value.
Answer: $x = h/\mu = 10.0$ m
Friction on a slope of length $L$ at angle $\theta$: work $= \mu mg\cos\theta\,L = \mu mg\,x_1$, where $x_1 = L\cos\theta$ is the horizontal length of the slope [1].
Energy: $mgh = \mu mg\,x_1 + \mu mg\,d$, so $x_1 + d = h/\mu$, which contains no $\theta$ [1].
$x = 2.0/0.20 = 10.0$ m (check: $3.46 + 6.54$) [1].
A numerical check at a second angle alone earns at most 1 mark.
14A ladder that never ends[9 marks]
A very long ladder network is made from identical resistors, each of resistance $R = 10$ Ω. Each section has one resistor $R$ in series along the top rail, followed by one shunt resistor $R$ joined from the top rail to the bottom rail. The sections repeat without end. A 12 V supply of negligible internal resistance is connected between the start of the top rail and the bottom rail, as shown.
(a) The resistance of the whole ladder is $R_{eq}$. Removing the first section leaves a ladder that is still infinitely long. Use this to show that $R_{eq}^2 - R\,R_{eq} - R^2 = 0$, and hence find $R_{eq}$.
Answer: $R_{eq} = \frac{1+\sqrt5}{2}R = 16.2$ Ω
The rest of the ladder (resistance $R_{eq}$) is in parallel with the first shunt, and this pair is in series with the first series resistor: $R_{eq} = R + \dfrac{R R_{eq}}{R + R_{eq}}$ [1].
Multiplying out: $R_{eq}^2 - R R_{eq} - R^2 = 0$ [1].
Positive root $R_{eq} = \frac{1+\sqrt5}{2}R = 16.2$ Ω [1]. The negative root must be rejected; an answer of $-6.2$ Ω loses the last mark.
(b) Calculate the current drawn from the supply.
Answer: 0.74 A
$I = 12/16.18 = 0.742$ A [1]. Allow error carried forward from (a).
(c) Show that the potential difference across each shunt resistor is the same fraction $k$ of the potential difference across the one before it (for the first shunt, of the supply), find $k$, and hence find the potential difference across shunt 3.
Answer: $k = \frac{3-\sqrt5}{2} = 0.382$; $V_3 = 0.67$ V
The network to the right of the first series resistor is the first shunt in parallel with an identical infinite ladder; seen from any node, the rest of the circuit looks the same, so each stage divides the voltage in the same ratio [1].
$k = \dfrac{R_{eq}-R}{R_{eq}} = \dfrac{6.18}{16.18} = 0.382$ (or from (b): $V_1 = 12 - 0.742\times10 = 4.58$ V, $4.58/12 = 0.382$) [1].
$V_3 = 12k^3 = 12\times0.0557 = 0.67$ V [1]. Accept 0.66 to 0.68 V.
(d) Calculate the total power dissipated in the ladder.
Answer: 8.9 W
$P = V^2/R_{eq} = 144/16.18 = 8.9$ W (or $VI = 12\times0.742$) [1].
(e) A real ladder has a finite number of sections. Which is the first shunt resistor whose potential difference is less than 1% of the supply? Explain briefly why a ladder of about ten sections behaves almost exactly like the infinite one.
Answer: shunt 5
$12k^4 = 0.26$ V but $12k^5 = 0.098$ V, less than 0.12 V, so shunt 5, and the far end of a ten-section ladder carries almost no current, so cutting it off hardly changes anything [1]. Both the number and the reason are needed for the mark.
15How fast does coffee cool?[10 marks]
A student pours hot coffee into a mug in a room at a steady 20.0 °C and records its temperature $T$ at times $t$.
| $t$ / min | $T$ / °C |
|---|---|
| 0 | 80.0 |
| 5 | 69.1 |
| 10 | 60.2 |
| 15 | 52.9 |
| 20 | 47.0 |
| 30 | 38.1 |
One model says that the rate of loss of heat is proportional to the temperature excess $\theta = T - 20.0$ °C, so that $\dfrac{d\theta}{dt} = -k\theta$ with $k$ constant. You will need graph paper or a sketch with a ruler for part (b).
(a) Show that, if the model is correct, a graph of $\ln\theta$ against $t$ is a straight line. State what its gradient tells you.
Answer: $\ln\theta = \ln\theta_0 - kt$; gradient $= -k$
Solving (or separating variables) gives $\theta = \theta_0 e^{-kt}$ [1].
So $\ln\theta = \ln\theta_0 - kt$: straight line, gradient $-k$, intercept $\ln\theta_0$ [1].
(b) Calculate $\ln\theta$ for each reading, plot the graph and use it to find $k$ in min$^{-1}$.
Answer: $k = 0.040$ min$^{-1}$
Values of $\ln(\theta/^\circ\mathrm{C})$: 4.09, 3.89, 3.69, 3.49, 3.30, 2.90 [1].
Points plotted to within half a small square and a straight line of best fit drawn (or, without graph paper, gradient from two widely separated points) [1].
Gradient $=(2.90-4.09)/30 = -0.040$ min$^{-1}$, so $k = 0.040$ min$^{-1}$ [1]. Accept 0.038 to 0.042 min$^{-1}$. Using $\ln T$ instead of $\ln\theta$ loses the first and last marks.
(c) Use the model to predict the time at which the coffee reaches 30.0 °C.
Answer: about 45 min
$\theta = 10.0$ °C, so $10.0 = 60.0e^{-kt}$ [1].
$t = \ln 6/k = 1.79/0.040 = 45$ min [1]. Accept 43 to 47 min; allow error carried forward from $k$. Reading off a straight line on a $T$ against $t$ graph gets no marks.
(d) The mug holds 0.30 kg of coffee of specific heat capacity 4200 J kg$^{-1}$ K$^{-1}$. Ignoring the mug, estimate the rate of heat loss from the coffee at $t = 0$.
Answer: about 50 W
Rate of cooling at $t=0$: $k\theta_0 = 0.040\times60.0 = 2.4$ K min$^{-1} = 0.040$ K s$^{-1}$ [1].
$P = mc\,\dfrac{dT}{dt} = 0.30\times4200\times0.040 = 50$ W [1]. Accept 48 to 53 W. Forgetting to change minutes to seconds (3000 W) loses the second mark.
(e) The mug itself has a heat capacity of 250 J K$^{-1}$ and stays at the same temperature as the coffee. Does your answer to (d) overestimate or underestimate the true rate of heat loss? Give a corrected value.
Answer: underestimate; about 60 W
Underestimate: the mug also cools at the same rate, so $P = (1260 + 250)\times0.040 = 60$ W [1]. Both the direction and the value are needed.
16Pushing with light[9 marks]
Light carries momentum: a photon of wavelength $\lambda$ has momentum $p = h/\lambda$ and energy $E = hc/\lambda$. A green laser pointer emits a beam of power 5.0 mW at a wavelength of 532 nm.
(a) Calculate the energy of one photon in the beam.
Answer: $3.7\times10^{-19}$ J (2.3 eV)
$E = hc/\lambda = (6.63\times10^{-34})(3.00\times10^{8})/(532\times10^{-9}) = 3.74\times10^{-19}$ J [1].
(b) Calculate the number of photons emitted per second.
Answer: $1.3\times10^{16}$ s$^{-1}$
$N = P/E = 5.0\times10^{-3}/3.74\times10^{-19} = 1.3\times10^{16}$ s$^{-1}$ [1].
(c) The beam hits a perfectly reflecting mirror at normal incidence. Show that the force on the mirror is $F = 2P/c$, where $P$ is the beam power, and calculate it.
Answer: $3.3\times10^{-11}$ N
Each photon's momentum changes from $+h/\lambda$ to $-h/\lambda$, a change of $2h/\lambda$; force = rate of change of momentum $= N\cdot 2h/\lambda = 2(NE)/c = 2P/c$ [1].
$F = 2(5.0\times10^{-3})/(3.00\times10^8) = 3.3\times10^{-11}$ N [1]. Using $P/c$ (absorbing surface) loses the second mark.
(d) A solar sail spacecraft of total mass 100 kg has a flat, perfectly reflecting sail facing the Sun. Near the Earth the intensity of sunlight is 1.36 kW m$^{-2}$. What sail area gives the craft an acceleration of 1.0 mm s$^{-2}$ from radiation pressure alone?
Answer: $1.1\times10^4$ m$^2$ (a square about 105 m across)
Force on the sail $F = 2IA/c$ [1].
$A = mac/(2I) = (100)(1.0\times10^{-3})(3.00\times10^8)/(2\times1360) = 1.1\times10^4$ m$^2$ [1]. Using mm s$^{-2}$ as m s$^{-2}$ ($1.1\times10^7$ m$^2$) loses the second mark.
(e) The Sun's gravitational field strength at the Earth's distance is $5.93\times10^{-3}$ N kg$^{-1}$. Explain why the ratio (force from sunlight) / (gravitational pull of the Sun) on a given sail is the same at every distance from the Sun. Hence find the mass per unit area of a sail (with its payload) for which the two forces balance.
Answer: ratio fixed because both forces fall as $1/r^2$; about $1.5\times10^{-3}$ kg m$^{-2}$ (1.5 g m$^{-2}$)
Intensity spreads over a sphere, $I\propto1/r^2$, and gravity $\propto 1/r^2$, so their ratio does not depend on $r$ [1].
Balance: $2IA/c = M g_S$, so $M/A = 2I/(c\,g_S)$ [1].
$M/A = 2(1360)/(3.00\times10^8\times5.93\times10^{-3}) = 1.5\times10^{-3}$ kg m$^{-2}$ [1]. Accept 1.5 g m$^{-2}$.
TheLucidSTEM practice paper. Questions are original and written in the style of a first national round; this is not an official olympiad paper.