BPhO · Round 2 · Long answer

Mechanics

17 questions — reveal the final answers. Full worked solutions: the book.

2003 Q42003Mechanics · Model aeroplane lift; water turbine
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Figure 4.1 shows a diagram of a model aeroplane. The wing may be regarded as a thin rigid sheet; lift is obtained by the air hitting the angled wing. When the aeroplane is flying horizontally the wing is angled at $10^{\circ}$ to the horizontal. The mass of the aeroplane is $m$ and the density of air is $\rho_{\text {air }}$.
(i) Find an expression for $v$ the speed of the aircraft, relative to the air, if it is to maintain horizontal flight.
(ii) Find the minimum power needed to maintain this speed if $m=1 \mathrm{~kg}, \rho_{\text {air }}=1.2 \mathrm{kg} \mathrm{m}^{-2}, d=2 \mathrm{~m}$ and $w=0.2 \mathrm{~m}$.
(iii) Solar panels produce about $60 \mathrm{~W} \mathrm{~m}^{-2}$. Comment on the wing area if the plane is to be driven by solar power alone.

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Figure 4.2 shows two views of a water turbine. The blades move at a mean tangential velocity $v_{b}$. The water from the fixed nozzle, velocity $v_{\text {in }}$ makes an angle of $\alpha$ with the direction of the blades and the angle at which the water leaves the blades relative to the blades is $\beta$, Figure 4.3.

Figure 4.3

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Copy the two vector diagrams in Figure 4.3 and complete the triangle of vectors. Explain the added vectors and their effects.
(c)

Assume that the magnitude of the velocity of the water in the passage between the blades does not change. Write down an expression for the change in momentum of the water in the direction X in terms of $v_{\text {out }}, v_{\text {in }}, \alpha, \beta$ and the mass flow per second $\mu$. Show that this leads to the expression:

\begin{equation*} F v_{b}=\mu\left(\frac{v_{\text {in }}^{2}-v_{\text {out }}^{2}}{2}\right) \end{equation*}

where $F$ is the force on a blade. Explain the significance of this equation.

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Final answers · 最终答案
  • (i)$v = \frac{1}{\sin\theta}\sqrt{\frac{mg}{\rho_{\mathrm{air}} \, d \, w}}$
  • (ii)$P \approx 45\;\mathrm{W}$
  • (iii)The wing area $A = d \times w = 0.4\;\mathrm{m^2}$ generates about 24 W from solar panels.
  • (c)$F v_b = \mu\left(\frac{v_{\text{in}}^2 - v_{\text{out}}^2}{2}\right)$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2004 Q42004Mechanics · Rolling and sliding polygons on slopes; bouncing spheres
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(a) Figure 4.1 shows the cross-section of a slope on which a pencil rests. The pencil is hexagonal in cross-section. Find the angle, $\theta$, of the slope, at which the pencil will tend to roll down the slope.
(b) The pencil could slip. Find the value of $\theta$ at which the pencil will slide if the coefficient of friction between the surfaces is $\mu$.
(c) Consider a cylindrical object that has a cross-section in the form of a polygon with $N$ equal sides. Find the algebraic condition such that if $\mu$ is greater than a certain value the object will roll rather than slip.

(d)
Figure 4.2

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Two layers of identical spheres are arranged so that each sphere forming the upper layer sits on top of four other identical touching spheres arranged with their centres at the corners of a square, Figure 4.2. Calculate the separation of the planes through the centres of the spheres, $h$, in terms of the diameter of the spheres, $d$.
(e)

In practice it is found that piles of rough objects are conical and the angle that the face of the cone makes with the horizontal is about $33^{\circ}$.

Consider a simple two layer model as shown in (d). A slope, (angle $\theta$ with the horizontal), is constructed of two layers of spherical particles, Figure 4.3. A spherical particle is displaced and lands on the slope such that it drops symmetrically between four other particles, as in Figure 4.3, at a speed $v$. On impact it loses $50 \%$ of its KE. Calculate the smallest value of $v$ so that it continues to bounce down the slope. What criticism can be made of this model and its ability to predict the behaviour of non-attracting particles?
(f) Clay slopes are never stable. Suggest a reason for this.

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Final answers · 最终答案
  • (a)$\theta = 30°$
  • (b)$\theta_{\text{slide}} = \arctan(\mu)$
  • (c)$\mu > \tan\!\left(\frac{\pi}{N}\right)$
  • (d)$h \approx 0.707\,d$
  • (e)$v_{\min} = \sqrt{2\sqrt{2}\,gd\cos\theta}$
  • (f)See the book for the worked solution.

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2005 Q22005Mechanics · Ping-pong ball accelerated into vacuum: momentum, gas pressure, piston
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In the 2004 International Physics Olympiad in Korea a demonstration showed how a ping pong ball (table tennis ball) could make a hole in three ceramic plates, Figures 2.1 and 2.2. The ceramic plates were placed between two glass tubes. The tubes were then evacuated so that there was a vacuum on each side of the plates. Suddenly the tube was opened on the side of the ball adjacent to the end of the tube, Figure 2.1. The atmospheric pressure on that side shot the ball into the plates making a hole in the plates. The ball made a virtually air tight fit with the glass tubes with little friction.
a) (i) Write down the maximum speed with which the ball could hit the plates.
(ii) The ball had a mass of 0.010 kg. Calculate its maximum KE.
b) Consider a cubic box, side $l$, volume $V$, containing $n$ molecules of mass $m$ all moving in random directions at constant speed $v$. All collisions are to be assumed to be perfectly elastic and the total volume of the molecules is very much less than $V$.
(i) What is the change in momentum when a molecule, moving in an arbitrary direction, hits a face of the box with speed $u$ normal to the face of the box?
(ii) Calculate the average time it takes this molecule to traverse the box.
(iii) Hence, or otherwise, show that the pressure in the box, $p$, is related to $v$, $m$, $n$ and $V$ by $$p=\frac{1}{3V}nmv^{2},$$ when the molecules are moving in random directions with speed $v$.
c)

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Figure 2.3 shows a simplified diagram of the experiment where the ping pong ball, diameter $d$, has been replaced by a gas tight piston of mass $M$ diameter $d$. The distance from the end of the tube, length $l$, is $x$ and its speed is $u_{x}$ at time $t$.
(i) Assuming that $v$, the speed of the molecules, remains the same, derive a formula that relates the pressure on the piston to $n$, $m$, $v$, $l$ and $d$ when the piston is moving at speed $u_{x}$.
(ii) Find an expression for the acceleration of the piston when it is at a distance $x$ from the end of the tube. Assume that $u_{x}\ll c_{s}$, where $c_{s}$ is the speed of sound of air at the local temperature.

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Final answers · 最终答案
  • (a)(i)$v_{\max} = 3.30\times10^2\ \mathrm{m\,s^{-1}}$
  • (a)(ii)$K_{\max} = 5.45\times10^2\ \mathrm{J}$
  • (b)(i)$2mu$
  • (b)(ii)$t_{\mathrm{cross}}=\frac{l}{u}$
  • (b)(iii)$p=\frac{1}{3V}nmv^2$
  • (c)(i)$p(u_x)=\frac{4nmv^2}{3\pi d^2l} \left(1-\frac{u_x}{v}\right)^3$
  • (c)(ii)$a_x=\frac{a_0}{(1+q)^3}$
  • (c)(ii)$a_0=\frac{nmv^2}{3Ml}$
  • (c)(ii)$q=\sqrt{\frac{2a_0x}{v^2}}$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2006 Q32006Mechanics · Rotational collision of ball bearings
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Figure 3.1 shows an experiment involving a rotational collision between two steel ball bearings, A and D. They rest in a very shallow spherical dish on a horizontal table. A has diameter of 5 mm and is permanently magnetised, but not strongly. D has a diameter of 2 mm and is not permanently magnetised. The probe C can be either a Hall effect probe with output p.d. $V_{\text {out }}$ proportional to the magnetic field $\boldsymbol{B}$, or a small search coil. $V_{\text {out }}$ has the form shown in Figure 3.2. A line joining the North and South poles is parallel to the plane on which the dish is resting.

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a) Use the data displayed in Figure 3.2 to calculate the angular velocity of the ball A .
b) Reproduce Figure 3.2 for a single period beginning at Time $=0$ and superpose a sketch of the output expected from a search coil
c) Initially the ball bearing A is given an angular velocity about a vertical axis when D is not present. After a short period of time ball D is placed in the dish and allowed to roll slowly towards A . Ignore the rotation of D about a horizontal axis. A collision takes place and D , which is not permanently magnetised, bounces away. Using the hint and the Table below, or otherwise, sketch, with numerical values, the trace observed on the computer, using the Hall probe, before and after the collision. Assume that no slippage occurs between the contact surfaces when the balls collide.

Hint: You may assume that the laws for angular motion are similar to those for linear motion. This results in similar equations of motion.

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Final answers · 最终答案
  • (a)$\omega = 3.93\times10^2\ \mathrm{rad\,s^{-1}}$
  • (b)$V_{\text{coil}} = -K\frac{dV_{\text{Hall}}}{dt}$
  • (c)$\omega_A' = 3.85\times10^2\ \mathrm{rad\,s^{-1}}$
  • (c)$T' = 16.3\ \mathrm{ms}$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2007 Q32007Mechanics · Rope friction around a bollard (capstan equation)

Hint: In this question you may find it a help to consider a small length of rope length $\delta l$ and the forces on this element of rope.

a)
Figure 3.1

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Figure 3.1 shows a bollard B with a rope wrapped round it. $\quad T_{1}$ is the tension caused by the rope being attached to a boat. $T_{2}$ is the tension due this rope being held by a sailor on the quayside. The angle between the ropes is $\theta$. and the rope is just slipping pulling the sailor along.
Find an expression relating $\theta, T_{2}, T_{1}$ and the coefficient of friction, $\mu$, between the rope and the bollard.
b)

Figure 3.2

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Figure 3.2 shows a long, heavy, flexible rope with mass per unit length $\rho$ that is kept tight by a force $F$. A sudden movement causes a circular loop radius $R$ to form. This loop moves along the rope in a horizontal plane at a transverse velocity of $v$.
(i) Calculate the speed $v$ of the loop in terms of the given variables.
(ii) Determine the energy, momentum and angular momentum of the loop.

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Final answers · 最终答案
  • (a)$T_1=T_2e^{\mu(\pi+\theta)}$
  • (b)(i)$v=\left(\frac{F}{\rho}\right)^{1/2}$
  • (b)(ii)$E=2\pi RF$
  • (b)(ii)$p = 2\pi R\rho v$
  • (b)(ii)$L = 2\pi\rho R^2v$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2011 Q12011Mechanics · Fluid flow in pipes, hydraulic jump, energy changes
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(a) Figure 1.a shows a schematic drawing of part of a water system. The cross sectional area of the in pipe is $a_{\mathrm{i}}$ and the cross sectional area of the out pipe is $a_{\mathrm{o}}$. What characteristics distinguish this system from similar pipes that contain a gas rather than water? Bernoulli's equation gives:

$$p+\frac{1}{2} \rho v^{2}+\rho g h=\text { constant } \quad p, \text { pressure } \rho, \text { density }$$

Write down an equation that relates $p_{\text {in }}, v_{\text {in }}, h_{\text {in }}$ to $p_{\text {out }}, v_{\text {out }}, h_{\text {out }}$, What does Bernoulli's equation assume?
(b)

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Figure 1.b. 2

Figure 1.b. 1 and Figure 1.b. 2 show the action of water flowing under an obstacle (dam).

Figure 1.b. 1 shows a photograph of a hydraulic jump with a turbulent region where the water rises and Figure 1.b. 2 is a diagram that shows the principal features of a hydraulic jump. Assuming that the width of the channel is constant, write down equations that predict the flow speed immediately before (region A) and after the turbulent region (region B).
(c) Describe in detail the energy changes that occur to the water in the channel.

Part (d) follows on page 3
(d) Read the following passage and answer the questions at the end.

D-D fuel cycle
N.B. D is the same as ${ }^{2} \mathrm{H}$

Though more difficult to facilitate than the deuterium-tritium reaction, fusion can also be achieved through the reaction of deuterium with itself. This reaction has two branches that occur with nearly equal probability:

$$\begin{array}{ll} { }^{2} \mathrm{H}+{ }^{2} \mathrm{H} \rightarrow{ }^{3} \mathrm{H}+\mathrm{p}+4.0 \mathrm{MeV} & (\mathrm{p} \text { is a proton }) \\ { }^{2} \mathrm{H}+{ }^{2} \mathrm{H} \rightarrow{ }^{3} \mathrm{He}+\mathrm{n}+3.3 \mathrm{MeV} & (\mathrm{n} \text { is a neutron }) \end{array}$$

The optimum temperature for this reaction is 15 keV , only slightly higher than the optimum for the D-T reaction. The first branch does not produce neutrons, but it does produce tritium ( ${ }^{3} \mathrm{H}$ ), so that a D-D reactor will not be completely tritium free, even though it does not require an input of tritium ( ${ }^{3} \mathrm{H}$ ) or lithium.

$${ }^{3} \mathrm{H}+{ }^{2} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+\mathrm{n}+18.3 \mathrm{MeV}$$

Most of the tritium produced will be burned before leaving the reactor, which reduces the tritium handling required, but also means that more neutrons are produced and that some of these are very energetic. The neutron from the second branch has an energy of only $2.45 \mathrm{MeV}(0.393 \mathrm{pJ})$, whereas the neutron from the D-T reaction has an energy of 14.1 MeV ( 2.26 pJ ), resulting in a wider range of isotope production and material damage.

The primary advantage of the D-D fuel cycle is that tritium breeding is not required. Other advantages are independence from limitations of lithium resources and a somewhat softer neutron spectrum. The price to pay compared to D-T is that the energy confinement (at a given pressure) must be 30 times better and the power produced (at a given pressure and volume) is 68 times less.

Questions
(i) Explain what is meant by "The optimum temperature for this reaction is 15 keV ".
(ii) What is this temperature in Kelvin?
(iii) Paraffin wax is good at stopping neutrons. Why?
(iv) Explain why the neutron from the D-T reaction has an energy of 14.1 MeV .
(v) Why is tritium difficult to handle?
(vi) What does "a somewhat softer neutron spectra" mean?

One way of bringing the deuterium to the point of criticality is to compress a small spherical capsule of it using multiple laser beams. Derive an expression for the momentum of light quanta of frequency $f$. Find the pressure produced on a 1 mm diameter sphere using a laser of power 1 MW with multiple beams that are spread around the sphere giving a uniform symmetrical pressure, see Figure 1.d.a.

Figure1.d.a

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Final answers · 最终答案
  • (a)$p_{\mathrm{in}}+\frac{1}{2}\rho v_{\mathrm{in}}^2 +\rho g h_{\mathrm{in}} =p_{\mathrm{out}}+\frac{1}{2}\rho v_{\mathrm{out}}^2 +\rho g h_{\mathrm{out}}$
  • (b)$v_B=\left(\frac{g h_A(h_A+h_B)}{2h_B}\right)^{1/2}$
  • (c)$\left(h_A+\frac{v_A^2}{2g}\right) -\left(h_B+\frac{v_B^2}{2g}\right) =\frac{(h_B-h_A)^3}{4h_Ah_B}>0$
  • (d)(i)$k_BT=15\ \mathrm{keV}$
  • (d)(ii)$T = 1.74\times10^8\ \mathrm{K}$
  • (d)(iii)See the book for the worked solution.
  • (d)(iv)$K_{\mathrm{n}} = 14.1\ \mathrm{MeV}$
  • (d)(v)See the book for the worked solution.
  • (d)(vi)$p_{\mathrm{light}} = 1.1\times10^3\ \mathrm{Pa}$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2013 Q22013Mechanics · Wave splash forces on a bridge
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The photo, Figure 2.1, was taken at Broadhaven, Pembrokeshire. The road that is damaged crossed a small stream flowing into the sea. The tidal range is about 8 m . The previous day at another location a wave hitting a harbour wall was observed to be producing a splash that rose 6 m high.
(i) Calculate the speed at which the water was moving vertically upwards when it was about 3.5 m above the normal surface of the stream - that is about the level of the under surface of the bridge. Assume that an unrestricted splash would have been 6 m high.
(ii) Find the force on a square stone in the under surface of the bridge that is $0.5 \mathrm{~m} \times 0.5 \mathrm{~m}$ and with a thickness $\approx 100 \mathrm{~mm}$ as illustrated in Figure 2.2.

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(iii) The density of the stone is $4000 \mathrm{~kg} \mathrm{~m}^{-3}$. What is the lowest speed of the water splash at which you might expect the bridge to fail?
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Final answers · 最终答案
  • (i)$v_{3.5} \cong 7.0\ \mathrm{m\,s}^{-1}$
  • (ii)$F_{ws} = 12,250\ \mathrm{N}$
  • (iii)$v \cong 2.0\ \mathrm{m\,s}^{-1}$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2013 Q42013Mechanics · Skier turning on a slope: friction and circular motion
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a) Figure 4.1 shows a skier, mass 70 kg , going down a snow slope of $5^{\circ}$ that is covered in firm hard snow, temperature $-4^{\circ} \mathrm{C}$. His skis are initially parallel, close together and going down the line of steepest slope. He intends to go down the slope in a series of sweeping curves. His first turn is $30^{\circ}$ to the steepest slope and his next turn is through $60^{\circ}$. You are asked to answer the following questions as quantitatively as possible using estimated numerical values where appropriate. Ski jargon is not appropriate!
(i) How can he reduce the normal reaction $R$ ? His mass is 70 kg . Give a numerical estimate of the maximum percentage of $R$ it would be reasonable to achieve.
(ii) How can he rotate his skis in a clockwise direction? How can he stop the rotation?
b) Another way of turning (carving) depends on the skis bending to form a curve which enables the skier to follow a curved path. He decides to test the stiffness of the skis Figure 4.3. He uses a weight of 700 N .

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(i) The skis are tested using the setup illustrated in Figure 4.3. A force of 700 N deflects the centre of the ski 50 mm . What speed should the skier be going to achieve a turn with a radius of curvature of 25 m ? Comment.
(ii) Illustrate your solution with a force diagram. Is this the only consideration that concerns the skier? Ignore the presence of other skiers!

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Final answers · 最终答案
  • (a)(i)$\frac{R_{0}-R}{R_{0}}\times100\% =100\%$
  • (a)(ii)This allows the skis to be set first at $30^{\circ}$ and subsequently through the required $60^{\circ}$.
  • (b)(i)$v \cong 27\ \mathrm{km\,h}^{-1}$
  • (b)(ii)$mg\sin5^{\circ}\sin30^{\circ} \cong 30\ \mathrm{N}$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2014 Q42014Mechanics · Wind loading and toppling of a concrete wall

It has been reported that the wind speed reached $200 \mathrm{mph}\left(90 \mathrm{~ms}^{-1}\right)$ in the Philippines. This is capable of toppling walls.

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Force from the ground

Figure 4.1

A concrete wall is made from two sections $A$ and $B$. Section $A$ is 2 m high and section $B$ is 1.3 m long. The thickness of the concrete is 0.7 m . The cross section is made of concrete. The orange lines within the concrete show the position of reinforcing steel bars with a twisted cross section of rather rusty steel. Why are these steel bars embedded in the wall? It may be assumed that there is a linear increase in wind speed with height.
i) How does the pressure of the wind vary with the height of the wall? Hence, or otherwise, find how the pressure on the ground varies with position. Hence estimate what wind speed is likely to topple the wall.
ii) How else could you help make the wall more stable?

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Final answers · 最终答案
  • (i)The ideal rigid-wall estimate is therefore about $2\times10^{2}\text{ m s}^{-1}$, with loss of full ground contact beginning near $1.5\times10^{2}\text{ m s}^{-1}$.
  • (ii)See the book for the worked solution.

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2015 Q22015Mechanics · Skier turning in a gully: banked circular motion

Figure 2a shows the cross section of a gully that a cross country skier is skiing into.
Figure 2b shows an aerial view. She wishes to turn left at the gully bottom and continue down the mountain. There is very little wind and air resistance is negligible. However, if she is going at a speed greater than $5 \mathrm{~ms}^{-1}$ she will not make the turn and will hit the rubbly snow. (Cross country skis are only attached to the boot at the toe and it is very difficult to execute a small radius turn at speed). If she continues to go up the other side of the gully, shown in Figure 1a, she can then turn very slowly (or stop) in order to turn through $180^{\circ}$ and go back. By repeating this procedure up and down the slopes on each side, she will be skiing backwards and forwards until she loses enough speed to turn safely left (or right) down the centre of the gully, as shown in Figure 1b.

The slopes shown are not to scale. Take the maximum slope as $5^{\circ}$. You may assume that the mass of the skier plus equipment is 65 kg . The effective coefficient of friction of the snow is 0.08 .

What is the maximum height, $h$, (Figure 1a) that the she can start from in her final descent so that she is able to make the turn without going up the slope on the other side?

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Final answers · 最终答案
  • Final resultThus the maximum starting height is about $15\text{ m}$ above the gully centre.

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2018 Q32018Mechanics · Phase-space trajectories of particles and pendulums

In cases involving multiparticle systems, it can be an advantage to consider an abstract space called phase space. In a particle accelerator such as the LHC at CERN, managing to produce a finer and denser beam of particles at a discrete energy, reducing the phase space of the beam, has been one of the keys to its success. The approach known as stochastic cooling won Simon van der Meer a Nobel Prize in Physics.

In 1-D, for a moving particle, the phase space trajectory is a line on a graph. The line shows the momentum $p_{x}$, versus position, $x$, on a $p_{x}$ versus $x$ graph. For a free particle of mass $m$, moving at constant speed $v_{x}$ in the positive $x$-direction, the trajectory in phase space as it passes through the origin is shown in the $p_{x}-x$ graph of Fig. 6. Included is an arrow on the trajectory to show its direction in time along the trajectory.

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(a). By calculation, or otherwise, determine the behaviour of a particle with the trajectory in phase space given by the graph in Fig. 7.

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(b). A simple pendulum consists of a light, rigid rod of length $\ell$, with mass $m$ attached at the end, as shown in Fig. 8. It swings freely under gravity.
(i) Write down an expression for the total energy $E$ of the pendulum in terms of $p$, the magnitude of the momentum and $\theta$, the angle of displacement of the rod from the vertical.
(ii) For small amplitude oscillations, it executes simple harmonic motion (SHM). Using a small angle approximation for $\theta$, and expressing the total energy in terms of angular variables $p_{\theta}$ and $\theta$, where $p_{\theta}=\ell p$, sketch the phase space trajectory on a $p_{\theta}-\theta$ graph. Include arrows on the trajectory. Show the limits on your axes, in terms of $m, E, \ell, g$.
(iii) On your graph, sketch a trajectory corresponding to a pendulum which has twice the initial amplitude of swing (but still SHM). How will this change the area of phase space enclosed by the trajectory?
(iv) If the pendulum loses $0.002 \%$ of its energy each swing, what will be the percentage area of phase space remaining enclosed by the trajectory after 400 swings?
(v) For the phase space trajectory of a simple harmonic oscillator, give a physical interpretation of the angles at which the trajectory crosses the horizontal and vertical axes.
(vi) If the pendulum has a small amount of viscous damping due to air resistance, what will be the trajectory in phase space as the pendulum slows down to very small amplitudes? Can it be said that the pendulum will ever stop swinging?
(vii) If the pendulum is damped by contact friction at the support, sketch the phase space trajectory as the pendulum reduces to very small amplitude swings.
(viii) If the pendulum has a large total energy, its motion will not be oscillatory but continuous rotation, with the kinetic energy term greater than the the potential energy term. Sketch a new $p_{\theta}-\theta$ graph which has trajectories for the pendulum (small KE term) and the circulating rod (large KE term).

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(c). A linear, elastic collision takes place between particles of mass $m_{1}$ and $m_{2}$, with initial velocities $u_{1}$ and $u_{2}$, resulting in final velocities of $v_{1}$ and $v_{2}$ respectively.
(i) Show that the velocities after collision can be expressed in terms of those before, in the form $v_{1}=a u_{1}+b u_{2}$ and $v_{2}=c u_{1}+d u_{2}$
(ii) Determine the coefficients $a, b, c$ and $d$ in terms of the masses $m_{1}$ and $m_{2}$.
(iii) If many repeated collisions are made using the same masses, but with various initial speeds in the ranges $u_{1}+\Delta u_{1}$ and $u_{2}+\Delta u_{2}$, then final speeds will lie in the ranges $v_{1}+\Delta v_{1}$ and $v_{2}+\Delta v_{2}$. On graph axes of $u_{2}$ against $u_{1}$, the velocity domain of the collisions are represented by the scatter of points within the rectangle, shown in Fig. 9a. If a graph of $v_{2}$ against $v_{1}$ is plotted, as in Fig. 9b, how does the velocity domain (the rectangular area of Fig. 9a) transform onto the new axes?

Hint: for two vectors $f=f_{1} \mathrm{X}+f_{2} \mathrm{Y}$ and $g=g_{1} \mathrm{X}+g_{2} \mathrm{Y}$ on $\mathrm{X}-\mathrm{Y}$ axes, which form the two sides of a triangle, the area of the triangle is given by the modulus (ignore any sign) of $\frac{1}{2}\left(f_{1} g_{2}-f_{2} g_{1}\right)$.

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(d). The time development of particles in phase space can be seen if a set of particles, each of mass $m$, have a range of momenta, and are distributed in the shaded, square region of phase space shown in the graph of Fig. 10. After a time $t$ later, what will be the possible distribution of the particles in phase space if they do not interact? Sketch your result on a copy of the graph of $p$ vs. $x$.

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Final answers · 最终答案
  • (a)$p_x=-kx_0e^{-kt/m}$
  • (b)(i)$E=\frac{p^2}{2m}+mg\ell(1-\cos\theta)$
  • (b)(ii)At the upper intercept $p_{\theta}>0$, $\theta$ increases, so the arrows run clockwise around the ellipse.
  • (b)(iii)See the book for the worked solution.
  • (b)(iv)Thus about $99\%$ of the original area remains.
  • (b)(v)At $p_{\theta}=0$ it is at a turning point with maximum potential energy and zero speed.
  • (b)(vi)See the book for the worked solution.
  • (b)(vii)See the book for the worked solution.
  • (b)(viii)$p_{\theta}=\mathord{\pm}\ell \sqrt{2m\left[E-mg\ell(1-\cos\theta)\right]}$
  • (c)(i)$d=\frac{m_2-m_1}{m_1+m_2}$
  • (c)(iii)$|ad-bc| = 1$
  • (d)It is therefore a right-sheared parallelogram: the $p=0$ edge stays fixed and the $p=P$ edge moves right by $Pt/m$.

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2018 Q52018Mechanics · Zero-momentum frame and collisions

The Zero Momentum Frame (ZMF) is a frame of reference in which the total momentum of a collection of particles is zero.
(a). Consider two particles of masses $m_{1}$ and $m_{2}$ with velocities $u_{1}$ and $u_{2}$ along a line.
(i) Show that the addition of a velocity

$$u^{\prime}=-\frac{m_{1} u_{1}+m_{2} u_{2}}{m_{1}+m_{2}}$$

to both particles transforms the system to a new frame of reference in which the total momentum is zero.
(ii) A particle of mass $m$, moving with velocity $u$ in a straight line, collides elastically head on with an identical stationary particle. Using the ZMF, or otherwise, find the velocities of the two particles after the collision.
(iii) In a game of pool, the white cue ball collides elastically with a red ball. At the instant of collision, the cue ball is moving with speed $u$ at an angle $\theta$ to the line connecting the centres of the two balls as shown in Figure 11. By transforming to a ZMF parallel to the line of centres, or otherwise, find the angle between their directions of motion after the collision. You may assume that the two pool balls are perfectly smooth.

figure

(b). The energy and momentum of a particle travelling at speeds approaching the speed of light, $c$, may be written as

$$E=\gamma m c^{2} \quad \& \quad p=\gamma m v$$

where $m$ is the mass of the particle, $v$ is its velocity and $\gamma$ is the relativistic factor

$$\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

(i) Show that

$$E^{2}=p^{2} c^{2}+m^{2} c^{4}$$

and check that this gives the usual formulas both for massive particles at rest, and for massless particles.
(ii) When an electron annihilates with a positron (the antiparticle of an electron, having all the same properties as an electron but opposite charge), photons are produced. Explain why it is not possible for this process to produce a single photon only, and if exactly two photons are produced, how will they be emitted in the ZMF?
(iii) A positron moving with velocity $v$ in the positive $x$-direction annihilates with a stationary electron, producing two identical photons. By using the ZMF, or otherwise, find the angle between the emitted photons in terms of $\gamma_{v}=\left(1-v^{2} / c^{2}\right)^{-1 / 2}$. What is the angle in the limit $v \rightarrow 0$ ?

You may find the following information helpful: When transforming to a new frame of reference with velocity $v^{\prime}$ relative to the original frame, the momentum of a particle in the new frame, $p^{\prime}$, is related to the momentum $p$ and energy $E$ in the original frame by

$$p^{\prime}=\gamma_{v^{\prime}}\left(p+\frac{v^{\prime}}{c^{2}} E\right)$$

where $\gamma_{v^{\prime}}=\left(1-v^{\prime 2} / c^{2}\right)^{-1 / 2}$.
Hint: In the ZMF consider the photons to be emitted parallel to the $y$-direction.
(iv) Fluorine-18 undergoes beta-plus decay and is commonly used in positron emission tomography. Positrons are emitted with a range of kinetic energies up to a maximum of 635 keV . If a positron were to immediately annihilate with an electron, producing two photons, what would be the minimum angle between them?
(v) Pair production is the process by which a photon may interact with a nucleus to produce an electron positron pair. Show that this process will not occur in free space, i.e. that the process depicted in Figure 12 is forbidden.

Hint: Consider the relationship between Energy and Momentum for the photon.

figure
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Final answers · 最终答案
  • (a)(i)$u'=-\frac{m_1u_1+m_2u_2}{m_1+m_2}$
  • (a)(ii)$v_1 = u$
  • (a)(iii)$90^{\circ}$
  • (b)(i)$E^2=p^2c^2+m^2c^4$
  • (b)(ii)In the electron--positron ZMF the initial total momentum is zero, whereas any single photon of non-zero energy has momentum $p=E/c\neq0$.
  • (b)(iii)$\theta=2\cos^{-1}\sqrt{\frac{\gamma_v-1}{\gamma_v+1}}$
  • (b)(iv)$\theta_{\rm min} \approx 104^{\circ}$
  • (b)(v)$E_{+}+E_{-}>c\left(|p_{+}|+|p_{-}|\right) \geq c|p_{+}+p_{-}| = E_{\gamma}$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2019 Q22019Mechanics · Balance toy: potential energy and stability

A simple balance toy is made as shown in Fig. 3. Two masses, $m$, are mounted at the end of symmetrical arms of length $\ell$ each at an angle $\theta$ to the body of the toy at a height $h$ above the pivot point. Initially we shall take the body of the toy to be of zero mass.

figure

(a). Taking the zero of potential energy as the point where the arms connect to the body of the toy, find an expression for the potential energy, $U$, of the system if the toy is tilted clockwise through an angle $\phi$. A diagram will be helpful.
(b). For the system to be stable, we expect $\frac{\mathrm{d} U}{\mathrm{~d} \phi}$ to be zero. Find the angle for which this is true and explain why you reject other solutions.
(c). Further, we require $\frac{\mathrm{d}^{2} U}{\mathrm{~d} \phi^{2}}$ to be positive.
(i) Explain why the second derivative must be positive.
(ii) Using this condition, explain where the masses must be relative to the toy for stability.
(d). For a small sideways displacement of the point of contact, $\phi \approx x / h$. Use this and the small angle approximation to find the potential and kinetic energy of the toy in terms of $x$ and $\dot{x}$.
(e). Compare your expressions to the equivalent terms for the potential and kinetic energies of a mass on a spring undergoing SHM. Explain why you would expect the toy to undergo SHM for small displacements. Using the expression for the angular frequency of a mass on a spring, $\omega=\sqrt{\frac{k}{m}}$, find an equivalent expression for the toy, by analogy.
(e). Repeat the analysis but allow the body of the toy itself to have a mass $M$ and a centre-of-mass level with the point where the arms contact it. How will the new oscillation frequency compare with the previous one, assuming the system is still stable?

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Final answers · 最终答案
  • (a)$U = 2mg(h-\ell\cos\theta)\cos\phi$
  • (b)$\frac{\mathrm dU}{\mathrm d\phi} =-2mg(h-\ell\cos\theta)\sin\phi$
  • (c)(i)A small displacement must raise the potential energy, so the curvature $\mathrm d^2U/\mathrm d\phi^2$ must be positive.
  • (c)(ii)The position $\phi=\pi$ is then a maximum of $U$ and is unstable.
  • (d)$K = \frac{m}{h^2}\!\left[\ell^2\sin^2\theta+(h-\ell\cos\theta)^2\right]\dot{x}^{\,2}$
  • (e)$\omega=\left[\frac{g(\ell\cos\theta-h)} {\ell^2\sin^2\theta+(h-\ell\cos\theta)^2}\right]^{1/2}$
  • (e)$0<\omega_M^2<\omega^2$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2022 Q32022Mechanics · Runner's pace function on slopes: optimal routes

The pace function of a runner specifies how much time she takes to run a certain horizontal distance, according to the slope of the ground on which she travels.

A particular runner has the following pace function:

\begin{equation*} p(m)=3.500+0.210 m+0.035 m^{2} \tag{3} \end{equation*}

where $p$ is measured in $\min \mathrm{km}^{-1}$, and $m$ is the gradient of the slope expressed as a percentage. For example: if $m=+1.5$, the ground rises 1.5 m vertically for every 100 m of horizontal distance travelled; if $m=-4$, the ground falls 4 m vertically for every 100 m of horizontal distance travelled.

figure

(a) Suggest why $p$ increases with increasing $m$ for all $m>0$.
(b) Determine the speed of the runner on flat ground in $\mathrm{ms}^{-1}$.
(c) Determine the gradient on which she runs the fastest, and the corresponding pace in $\operatorname{min~}^{-1}$.
(d) Sketch the pace function for $-15 \leq m \leq 15$, including numerical detail. Would you expect this pace function to be valid for all $m$ ?
(e) Figure 4 shows the elevation profile of a particular route, which indicates the height above sea level of any particular point along the route. Determine the time the runner takes to run the route and her average pace in $\min \mathrm{km}^{-1}$.
(f) Determine the gradient that allows the runner to gain vertical height most rapidly. Mark this critical gradient and the corresponding pace on your sketch in part (d).
(g) Another runner has a pace function which is linear, rather than quadratic, in $m$. For routes that start and finish at the same place, explain the impact that the route elevation profile has in determining their total run time. Does a critical gradient exist in this case?

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Final answers · 最终答案
  • (a)$\frac{dp}{dm}=0.210+0.070m>0$
  • (b)$v \simeq 4.76\ \mathrm{m\,s^{-1}}$
  • (c)Thus the runner is fastest on a $3.0\%$ downhill gradient, at a pace of approximately $3.19\ \mathrm{min\,km^{-1}}$.
  • (d)$p(15)=14.525$
  • (e)$p_{\mathrm{av}} \simeq 3.70\ \mathrm{min\,km^{-1}}$
  • (f)The corresponding maximum rate of height gain is $100/9.10\simeq11.0\ \mathrm{m\,min^{-1}}$.
  • (g)$\frac{d}{dm}\left(\frac{m}{a+bm}\right) =\frac{a}{(a+bm)^2}>0$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2024 Q22024Mechanics · Aircraft lift and drag from Bernoulli's principle

Bernoulli's principle for incompressible and inviscid (zero viscosity) fluids states that

\begin{equation*} \frac{1}{2} \rho v^{2}+p=\mathrm{constant} \tag{1} \end{equation*}

throughout the flow, where $\rho$ is the fluid density, $v$ the fluid speed and $p$ the fluid pressure.
(a) In normal operational flight, air flows over the upper surface of an aircraft wing faster than the bottom surface, as shown in Fig. 2. Use Bernoulli's principle to explain how this generates lift and drag forces on the wing.

figure

The lift force $L$ and drag force $D$ on an aircraft can be well modelled by

\begin{align*} L & =\frac{1}{2} C_{L} \rho A v^{2} \tag{2}\\ D & =\frac{1}{2} C_{D} \rho A v^{2} \tag{3} \end{align*}

where $A$ is the wing area, $\rho$ is the air density and $v$ is the speed of the aircraft relative to still air. The lift coefficient $C_{L}$ and drag coefficient $C_{D}$ both depend upon the wing shape and the angle of attack $\alpha$, which is the angle the wing makes to the undisturbed air-flow direction. This is illustrated in Fig. 3. Typical data for the lift and drag coefficients are shown in Fig. 4.

figure

Figure 3: The angle of attack $\alpha$ of a wing is defined as the acute angle subtended by the undisturbed flow direction and the chord of the wing.

figure

Figure 4: Upper panel: lift coefficient $C_L$ versus angle of attack $\alpha$ for a typical wing shape. Lower panel: drag coefficient $C_D$ versus angle of attack $\alpha$ for the same wing shape.

The Airbus A380 is the world's largest passenger aircraft. It has a wing area $A=845 \mathrm{~m}^{2}$ and take-off mass $m=5.60 \times 10^{5} \mathrm{~kg}$. In reality the aircraft mass decreases over time as fuel is burnt but, for simplicity, you should assume this to remain constant.
(b) Shortly after take-off the A380 is moving at $90 \mathrm{~m} \mathrm{~s}^{-1}$ with the wings at an angle of attack of $15^{\circ}$. Determine the vertical acceleration of the A380. Take the air density at sea-level to be $1.3 \mathrm{~kg} \mathrm{~m}^{-3}$ and use data from Fig. 4. Suggest how this vertical acceleration is experienced by a passenger.
(c) At a higher altitude the air density has dropped to $0.80 \mathrm{~kg} \mathrm{~m}^{-3}$. With an angle of attack of $8^{\circ}$, the A380 flies at constant speed and constant altitude (a condition known to aviators as steady and level flight). Determine the engine thrust and speed of the aircraft, and hence the useful power developed by the A380 jet engines.
(d) At its cruising altitude of $11,000 \mathrm{~m}$ the A 380 flies at a steady speed of $250 \mathrm{~m} \mathrm{~s}^{-1}$. The air density at this height is $0.36 \mathrm{~kg} \mathrm{~m}^{-3}$. Estimate the necessary angle of attack and the useful power developed by the engines.
(e) An A380 is in steady and level flight. Explain, without calculation, why the drag force can be expected to be large for both very small and very large angles of attack.
(f) Use data from Fig. 4 to estimate the angle of attack that minimises the drag for an A380 in steady and level flight. Suggest the importance to airline companies in knowing this value.

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Final answers · 最终答案
  • (a)$p+\frac{1}{2}\rho v^2=\text{constant}$
  • (b)$a_y \approx 1.55\;\text{m s}^{-2}\ \text{upward}$
  • (c)$P \approx 9.8\times10^6\;\text{W}$
  • (c)$v\approx134\;\text{m s}^{-1}$
  • (c)$F_T\approx7.3\times10^4\;\text{N}$
  • (d)$\alpha\approx5.4^\circ$
  • (d)$P\approx2.20\times10^7\,\mathrm{W}$ Corrected — the printed 2.0×10⁷ W uses the wrong drag coefficient from Fig. 4; see the interactive walkthrough.
  • (e)To retain $L=mg$, the aircraft must fly very quickly, so the $v^2$ factor makes the drag large even though $C_D$ itself is small.
  • (f)$\alpha\approx8^\circ$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2025 Q32025Mechanics · Trinity test: dimensional analysis of a blast wave

This question is about the world's first atomic bomb test, which took place in New Mexico, USA, in 1945. The operation - part of the Manhattan Project - was code-named Trinity and the bomb itself was nicknamed The Gadget.

The radius $R$ of the shockwave from an atomic bomb depends upon the density of the atmosphere $\rho$, the time after the explosion $t$ and the energy $E$ released in the explosion.
(a) Using arguments of dimensional homogeneity (i.e., consistency of units), determine the values of the dimensionless unknowns $\alpha, \beta$ and $\gamma$ if

\begin{equation*} R \propto \rho^{\alpha} t^{\beta} E^{\gamma} . \tag{5} \end{equation*}

It is reasonable to assume the dimensionless constant of proportionality $k$ implied in Eq. 5 is of order unity. Its precise value depends upon the circumstances of the explosion. Assume for the remainder of the question, and for simplicity, that $k=1$.
(b) Prove that the atmospheric density $\rho$ is given by

\begin{equation*} \rho=\frac{M_{R} p}{R T}, \tag{6} \end{equation*}

where $M_{R}$ is the (representative) molar mass of an air molecule. Show that, under typical sea-level conditions, $\rho \approx 1.3 \mathrm{~kg} \mathrm{~m}^{-3}$.
(c) The speed of sound $c_{s}$ is related to the ambient pressure $p$ and density $\rho$ by a simple formula:

\begin{equation*} c_{s}=\sqrt{\frac{\gamma p}{\rho}}, \tag{7} \end{equation*}

where $\gamma=1.40$ is the adiabatic index for air. Calculate the speed of sound under typical sea-level conditions.

figure

(d) Use information in Fig. 4 to show the the energy released by The Gadget was approximately 100 TJ .
(e) Calculate the radius of the shockwave generated by The Gadget after (i) 1 ms , (ii) 10 ms , and (iii) 0.1 s .
(f) Calculate the speed of the shockwave after (i) 1 ms , (ii) 10 ms , and (iii) 0.1 s .
(g) Eventually, the shockwave stalled due to dissipation of the shockwave energy to the surroundings. Briefly explain why energy dissipation increases dramatically when the shockwave speed drops below the speed of sound in the external atmosphere.
(h) Estimate the radius at which The Gadget's shockwave stalled and the corresponding time, after detonation, that this occurred.
(i) Following the test, scientists recorded structural damage to buildings at a distance of 3 km from the detonation site. Broken glass damage was observed as far as 30 km away. Reconcile these figures with your answer to part (h).

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Final answers · 最终答案
  • (a)$\alpha=-\frac15$
  • (a)$\beta=\frac25$
  • (a)$\gamma=\frac15$
  • (a)$R=\left(\frac{Et^2}{\rho}\right)^{1/5}$
  • (b)$\rho=\frac{M_Rp}{RT}$
  • (b)$\rho \approx 1.3\ \mathrm{kg\,m^{-3}}$
  • (c)$c_s=3.30\times10^2\ \mathrm{m\,s^{-1}}$
  • (d)$E = 100\ \mathrm{TJ}$
  • (e)$R(1\ \mathrm{ms})=37.8\ \mathrm{m}$
  • (e)$R(10\ \mathrm{ms})=94.9\ \mathrm{m}$
  • (e)$R(0.1\ \mathrm{s})=238\ \mathrm{m}$
  • (f)$v_s(1\ \mathrm{ms})=1.51\times10^4\ \mathrm{m\,s^{-1}}$
  • (f)$v_s(10\ \mathrm{ms})=3.80\times10^3\ \mathrm{m\,s^{-1}}$
  • (f)$v_s(0.1\ \mathrm{s})=9.53\times10^2\ \mathrm{m\,s^{-1}}$
  • (g)See the book for the worked solution.
  • (h)$t_{\mathrm{stall}}=0.59\ \mathrm{s}$
  • (h)$R_{\mathrm{stall}} \approx 0.48\ \mathrm{km}$
  • (i)It can therefore still damage structures at $3\ \mathrm{km}$ and, because glass is especially vulnerable to overpressure, break windows as far away as $30\ \mathrm{km}$ even though the original strong shock has ceased.

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)

2025 Q42025Mechanics · Large-amplitude pendulum period

A mass $m$ is suspended from a frictionless pivot $P$ by a light string of length $l$. It is released from rest at an angle $\theta_{0}$ from the vertical.
(a) Prove, given certain assumptions that you should clearly state, that the time period $\tau$ of the subsequent oscillations is

\begin{equation*} \tau=2 \pi \sqrt{\frac{l}{g}} \tag{8} \end{equation*}

(b) By considering the principle of conservation of energy, prove that the angular velocity $\frac{d \theta}{d t}$ at angular displacement $\theta<\theta_{0}$ satisfies

\begin{equation*} \left(\frac{d \theta}{d t}\right)^{2}=\frac{2 g}{l}\left(\cos \theta-\cos \theta_{0}\right) \tag{9} \end{equation*}

(c) Hence argue that the time period of the subsequent oscillations can be written in integral form as

\begin{equation*} T=\sqrt{\frac{8 l}{g}} \int_{0}^{\theta_{0}} \frac{d \theta}{\sqrt{\cos \theta-\cos \theta_{0}}} \tag{10} \end{equation*}

(d) By introducing an auxillary angle $\phi$ such that

\begin{equation*} \sin \phi=\frac{\sin (\theta / 2)}{\sin \left(\theta_{0} / 2\right)} \tag{11} \end{equation*}

prove that

\begin{equation*} T=\sqrt{\frac{16 l}{g}} \int_{0}^{\pi / 2} \frac{d \phi}{\sqrt{1-k^{2} \sin ^{2} \phi}} \tag{12} \end{equation*}

and express the constant $k$ in terms of $\theta_{0}$. Hint: $\cos \theta=1-2 \sin ^{2}(\theta / 2)$.
(e) By inserting a relevant value for $k$ into the integral expression above, show clearly how the result from part (a) can be recovered.
(f) For a general value of $\theta_{0}$ (and therefore $k$ ) there is no closed-form solution for the integral above. However, the result can be expressed in terms of the arithmetic-geometric mean function, agm:

\begin{equation*} T=\frac{\tau}{\operatorname{agm}\left[\cos \left(\theta_{0} / 2\right), 1\right]} \tag{13} \end{equation*}

where $\tau$ is the result derived in part (a). The arithmetic-geometric mean function agm is recursively defined as the large $n$ limit of two simple sequences $a_{n}, b_{n}$ : to find agm $[a, b]$, set $a_{0}=a, b_{0}=b$ and then:

\begin{align*} a_{n+1} & =\frac{1}{2}\left(a_{n}+b_{n}\right) \tag{14}\\ b_{n+1} & =\sqrt{a_{n} b_{n}} \end{align*}

for $n=0,1,2, \ldots$. If $\ell=1.00 \mathrm{~m}$ and the mass is released from the same level as the pivot, with the string taut, determine the oscillatory time period to 3 significant figures.
(g) A more complex pendulum is composed of two masses, each $m / 2$, respectively at distances $\ell(1-\alpha)$ and $\ell(1+\alpha)$ from the pivot, where $0<\alpha<1$. The masses are connected with the pivot and to one another by co-linear, light rods. Draw the arrangement and explain, for a given value of $\theta_{0}$, whether the oscillatory time period is shorter than, equal to, or longer than the simple pendulum of mass $m$ and length $\ell$.

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Final answers · 最终答案
  • (a)$\tau = 2\pi\sqrt{\frac{\ell}{g}}$
  • (b)$\left(\frac{d\theta}{dt}\right)^2 =\frac{2g}{\ell}\left(\cos\theta-\cos\theta_0\right)$
  • (c)$T=\sqrt{\frac{8\ell}{g}} \int_0^{\theta_0}\frac{d\theta} {\sqrt{\cos\theta-\cos\theta_0}}$
  • (d)$T=\sqrt{\frac{16\ell}{g}} \int_0^{\pi/2}\frac{d\phi} {\sqrt{1-k^2\sin^2\phi}}$
  • (d)$k=\sin\left(\frac{\theta_0}{2}\right)$
  • (e)$2\pi\sqrt{\frac{\ell}{g}}=\tau$
  • (f)$T = 2.37\ \mathrm{s}$
  • (g)$T_{\mathrm{two\ masses}} =\sqrt{1+\alpha^2}\,T_{\mathrm{simple}}>T_{\mathrm{simple}}$

Full worked solutions: BPhO Round 2: Past Questions and Solutions (book)