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IPhO 2000, Теория — Задача 1. Theoretical Problem 1

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Theoretical Problem 1

Условие

A

A bungee jumper is attached to one end of a long elastic rope. The other end of the elastic rope is fixed to a high bridge. The jumper steps off the bridge and falls, from rest, towards the river below. He does not hit the water. The mass of the jumper is $m$ , the unstretched length of the rope is $L$ , the rope has a force constant (force to produce 1 m extension) of $k$ and the gravitational field strength is $g$ .

You may assume that

the jumper can be regarded as a point mass $m$ attached to the end of the rope,
the mass of the rope is negligible compared to $m$ ,
the rope obeys Hooke’s law,
air resistance can be ignored throughout the fall of the jumper.

Obtain expressions for the following and insert on the answer sheet:

(a) the distance $y$ dropped by the jumper before coming instantaneously to rest for the first time,

(b) the maximum speed $v$ attained by the jumper during this drop,

(c) the time $t$ taken during the drop before coming to rest for the first time.

B

A heat engine operates between two identical bodies at different temperatures $T_A$ and $T_B$ ( $T_A > T_B$ ), with each body having mass $m$ and constant specific heat capacity $s$ . The bodies remain at constant pressure and undergo no change of phase.

(a) Showing full working, obtain an expression for the final temperature $T_0$ attained by the two bodies A and B, if the heat engine extracts from the system the maximum amount of mechanical work that is theoretically possible.

Write your expression for the final temperature $T_0$ on the answer sheet.

(b) Hence, obtain and write on the answer sheet an expression for this maximum amount of work available.

The heat engine operates between two tanks of water each of volume 2.50 m $^3$ . One tank is at 350 K and the other is at 300 K.

(c) Calculate the maximum amount of mechanical energy obtainable. Insert the value on the answer sheet.

Specific heat capacity of water = $4.19 \times 10^{3}\ \mathrm{J\ kg^{-1}\ K^{-1}}$
Density of water = $1.00 \times 10^{3}\ \mathrm{kg\ m^{-3}}$

C

It is assumed that when the earth was formed the isotopes $^{238}\mathrm{U}$ and $^{235}\mathrm{U}$ were present but not their decay products. The decays of $^{238}\mathrm{U}$ and $^{235}\mathrm{U}$ are used to establish the age of the earth, $T$ .

(a) The isotope $^{238}\mathrm{U}$ decays with a half-life of $4.50 \times 10^{9}$ years. The decay products in the resulting radioactive series have half-lives short compared to this; to a first approximation their existence can be ignored. The decay series terminates in the stable lead isotope $^{206}\mathrm{Pb}$ .

Obtain and insert on the answer sheet an expression for the number of $^{206}\mathrm{Pb}$ atoms, denoted $^{206}n$ , produced by radioactive decay with time $t$ , in terms of the present number of $^{238}\mathrm{U}$ atoms, denoted $^{238}N$ , and the half-life time of $^{238}\mathrm{U}$ . (You may find it helpful to work in units of $10^{9}$ years.)

(b) Similarly, $^{235}\mathrm{U}$ decays with a half-life of $0.710 \times 10^{9}$ years through a series of shorter half-life products to give the stable isotope $^{207}\mathrm{Pb}$ .

Write down on the answer sheet an equation relating $^{207}n$ to $^{235}N$ and the half-life of $^{235}\mathrm{U}$ .

(c) A uranium ore, mixed with a lead ore, is analysed with a mass spectrometer. The relative concentrations of the three lead isotopes $^{204}\mathrm{Pb}$ , $^{206}\mathrm{Pb}$ and $^{207}\mathrm{Pb}$ are measured and the number of atoms are found to be in the ratios 1.00 : 29.6 : 22.6 respectively. The isotope $^{204}\mathrm{Pb}$ is used for reference as it is not of radioactive origin. Analysing a pure lead ore gives ratios of 1.00 : 17.9 : 15.5.

Given that the ratio $^{238}N : {}^{235}N$ is 137 : 1, derive and insert on the answer sheet an equation involving $T$ .

(d) Assume that $T$ is much greater than the half lives of both uranium isotopes and hence obtain an approximate value for $T$ .

(e) This approximate value is clearly not significantly greater than the longer half life, but can be used to obtain a much more accurate value for $T$ . Hence, or otherwise, estimate a value for the age of the earth correct to within 2%.

D

Charge $Q$ is uniformly distributed in vacuo throughout a spherical volume of radius $R$ .

(a) Derive expressions for the electric field strength at distance $r$ from the centre of the sphere for $r \le R$ and $r > R$ .

(b) Obtain an expression for the total electric energy associated with this distribution of charge.

Insert your answers to (a) and (b) on the answer sheet.

E

A circular ring of thin copper wire is set rotating about a vertical diameter at a point within the Earth's magnetic field. The magnetic flux density of the Earth's magnetic field at this point is 44.5 μT directed at an angle of 64° below the horizontal. Given that the density of copper is $8.90 \times 10^{3}\ \mathrm{kg\ m^{-3}}$ and its resistivity is $1.70 \times 10^{-8}\ \Omega\,\mathrm{m}$ , calculate how long it will take for the angular velocity of the ring to halve. Show the steps of your working and insert the value of the time on the answer sheet. This time is much longer than the time for one revolution.

You may assume that the frictional effects of the supports and air are negligible, and for the purposes of this question you should ignore self-inductance effects, although these would not be negligible.

Решение

A Bungee Jumper

(a) Comes to rest when loss of gravitational potential energy = strain energy stored in rope.

mgy=\tfrac12k(y-L)^2.

Rearranging:

ky^2-2y(kL+mg)+kL^2=0.

Quadratic solution:

y=\frac{2(kL+mg)\pm\sqrt{4(kL+mg)^2-4k^2L^2}}{2k} =\frac{kL+mg\pm\sqrt{2mgkL+m^2g^2}}{k}.

Need positive root. This gives the lower rest position (other root after initial rise).

(b) Maximum speed when acceleration is zero, i.e. $mg=kx$ .

Kinetic energy = loss of potential energy − strain energy:

\tfrac12mv^2=mg(L+x)-\tfrac12kx^2.

Putting $x=mg/k$ ,

v^2=2g\left(L+\frac{mg}{k}\right)-\frac{mg^2}{k},\qquad v=\sqrt{2gL+\frac{mg^2}{k}}.

(c) Time to come to rest = time of free fall + time for SHM to stop stretching.

Free fall: $L=\tfrac12gt_f^2$ , so $t_f=\sqrt{2L/g}$ .

The jumper enters the SHM with free fall velocity $gt_f=\sqrt{2gL}=v_\tau$ .

Period of SHM $=2\pi\sqrt{m/k}=T$ .

We represent a full SHM cycle by

Full SHM velocity cycle, down/up directions, v, v_tau, tau, T/2 and T.

The jumper enters the SHM at time $\tau$ given by

\tau=\frac1\omega\sin^{-1}\frac{v_\tau}{v}=\frac1\omega\sin^{-1}\frac{\sqrt{2gL}}v.

Jumper comes to rest at one half cycle of the SHM at total time given by

t=t_f+(T/2-\tau) =\sqrt{\frac{2L}{g}}+\pi\sqrt{\frac mk}-\frac1\omega\sin^{-1}\frac{\sqrt{2gL}}v.

t=\sqrt{\frac{2L}{g}}+\pi\sqrt{\frac mk}-\frac1\omega\sin^{-1}\frac{\sqrt{2gL}}{\sqrt{2gL+mg^2/k}}.

t=\sqrt{\frac{2L}{g}}+\sqrt{\frac mk}\left\{\pi-\sin^{-1}\frac{\sqrt{2gL}}{\sqrt{2gL+mg^2/k}}\right\}.

Right triangle with angle phi and sides square root of 2gL and square root of mg squared over k.

This is the same as

t=\sqrt{\frac{2L}{g}}+\sqrt{\frac mk}\left\{\frac\pi2+\cos^{-1}\frac{\sqrt{2gL}}{\sqrt{2gL+mg^2/k}}\right\}.

t=\sqrt{\frac{2L}{g}}+\sqrt{\frac mk}\tan^{-1}\left\{-\sqrt{\frac{2kL}{mg}}\right\}.

B Heat Engine Question

Heat-engine flow diagram between T1 initially TA and T2 initially TB, heat flows Delta Q1 and Delta Q2, work Delta W.

In calculating work obtainable, we assume no loss (friction etc.) in engine working.

$\Delta Q_1=$ energy from body A $=-ms\Delta T_1$ ( $\Delta T_1$ −ve).

$\Delta Q_2=ms\Delta T_2$ ( $\Delta T_2$ +ve).

(a) For maximum amount of mechanical energy assume Carnot engine.

\frac{\Delta Q_1}{T_1}=\frac{\Delta Q_2}{T_2}

throughout operation (second law).

But $\Delta Q_1=-ms\Delta T_1$ and $\Delta Q_2=ms\Delta T_2$ .

-ms\int_{T_A}^{T_0}\frac{dT_1}{T_1}=ms\int_{T_B}^{T_0}\frac{dT_2}{T_2}.

\ln\frac{T_A}{T_0}=\ln\frac{T_0}{T_B},\qquad T_0^2=T_AT_B,\qquad T_0=\sqrt{T_AT_B}.

Q_1=-ms\int_{T_A}^{T_0}dT_1=ms(T_A-T_0).

Q_2=ms\int_{T_B}^{T_0}dT_2=ms(T_0-T_B).

W=Q_1-Q_2.

W=ms(T_A-T_0-T_0+T_B)=ms(T_A+T_B-2T_0)=ms(T_A+T_B-2\sqrt{T_AT_B})

or $ms(\sqrt{T_A}-\sqrt{T_B})^2$ .

(d) Numerical example:

Mass = volume × density.

W=2.50\times1.00\times10^3\times4.19\times10^3(350+300-2\sqrt{350\times300})\ \mathrm J

=20\times10^6\ \mathrm J=20\ \mathrm{MJ}.

C Radioactivity and age of the Earth

(a) $N=N_0e^{-\lambda t}$ , $N_0=$ original number.

n=N_0(1-e^{-\lambda t}).

Therefore

n=Ne^{\lambda t}(1-e^{-\lambda t})=N(e^{\lambda t}-1).

So $n=N(2^{t/T}-1)$ where $T$ is half-life,

or as $\lambda=\ln2/T=0.6931/T$ ,

n=N(e^{0.6931t/T}-1).

{}^{206}n={}^{238}N(2^{t/4.50}-1)\quad\text{or}\quad{}^{206}n={}^{238}N(e^{0.1540t}-1),

where time $t$ is in $10^9$ years.

(b)

{}^{207}n={}^{235}N(2^{t/0.710}-1)\quad\text{or}\quad{}^{207}n={}^{235}N(e^{0.9762t}-1).

(c) In mixed uranium (i.e. containing Pb of both natural and radioactive origin)

	204 : 206 : 207
Proportions	1.00 : 29.6 : 22.6
In pure lead (no radioactivity)	1.00 : 17.9 : 15.5

Therefore for radioactively produced lead by subtraction

	204 : 206 : 207
Proportions	1.00 : 29.6 : 22.6
In pure lead (no radioactivity)	1.00 : 17.9 : 15.5

Therefore for radioactively produced lead by subtraction

206:207=11.7:7.1.

Dividing equations from (a) and (b) gives

\frac{{}^{206}n}{{}^{207}n}=\frac{{}^{238}N}{{}^{235}N}\left\{\frac{2^{t/4.50}-1}{2^{t/0.710}-1}\right\}\quad\text{or}\quad\frac{{}^{206}n}{{}^{207}n}=\frac{{}^{238}N}{{}^{235}N}\left\{\frac{e^{0.1540t}-1}{e^{0.9762t}-1}\right\}.

\frac{11.7}{7.1}=137\left\{\frac{2^{T/4.50}-1}{2^{T/0.710}-1}\right\}\quad\text{or}\quad\frac{11.7}{7.1}=137\left\{\frac{e^{0.1540T}-1}{e^{0.9762T}-1}\right\}.

0.0120(2^{T/0.710}-1)=(2^{T/4.50}-1),

0.0120(e^{0.9762T}-1)=(e^{0.1540T}-1).

(d) Assume $T\gg4.50\times10^9$ and ignore 1 in both brackets:

0.0120\{2^{T/0.710}\}=\{2^{T/4.50}\}\quad\text{or}\quad0.0120\{e^{0.9762T}\}=\{e^{0.1540T}\}.

0.0120=\{2^{T/4.50-T/0.710}\}=2^{T(0.222-1.4084)}=2^{-1.1862T}.

T=-\frac{\log0.0120}{\log2\times1.1862}=5.38,\qquad T=5.38\times10^9\ \text{years}.

Or $0.0120=e^{-0.8222T}$ ,

T=\frac{\ln0.0120}{-0.8222}=\frac{-4.4228}{-0.8222}=5.38,\qquad T=5.38\times10^9\ \text{years}.

(e) $T$ is not $\gg4.50\times10^9$ years but is $>0.71\times10^9$ years.

We can insert the approximate value for $T$ (call it $T^*=5.38\times10^9$ years) in the $2^{T/4.50}$ term and obtain a better value by iteration (in the rapidly changing $2^{T/0.710}$ term). We now leave in the −1’s, although the −1 on the right-hand side has little effect and may be omitted.

Either

0.0120(2^{T/0.710}-1)=2^{T^*/4.50}-1.

2^{T/0.710}-1=\frac{2^{1.1956}-1}{0.0120}=\frac{2.2904-1}{0.0120}=107.5.

T=0.710\frac{\log108.5}{\log2}=4.80(0).

Put $T^*=4.80(0)\times10^9$ years

2^{T/0.710}=\frac{2^{1.0668}-1}{0.0120}=\frac{2.0948-1}{0.0120}=91.2.

T=0.710\frac{\log91.2}{\log2}=4.62(3).

Further iteration gives 4.52.

So more accurate answer for $T$ to be in range $4.6\times10^9$ years to $4.5\times10^9$ years (either acceptable).

D Spherical charge

(a) Charge density

\rho=\frac{Q}{\frac43\pi R^3}

within sphere.

$x\le R$ : Field at distance $x$ :

E=\frac{\frac43\pi x^3\rho}{4\pi\varepsilon_0x^2}=\frac{Qx}{4\pi\varepsilon_0R^3}.

$x>R$ : Field at distance $x$ from the centre:

E=\frac{Q}{4\pi\varepsilon_0x^2}.

(b) Method 1

Energy density is $\tfrac12\varepsilon_0E^2$ .

$x\le R$ : Energy in a thin shell of thickness $\delta x$ at radius $x$ is given by

\tfrac12\varepsilon_0E^2\,4\pi x^2\delta x=\tfrac124\pi\varepsilon_0\frac{Q^2x^2}{(4\pi\varepsilon_0)^2R^6}x^2\delta x.

Energy within the spherical volume

=\frac12\frac{Q^2}{(4\pi\varepsilon_0)R^6}\int_0^R x^4dx=\frac1{40}\frac{Q^2}{\pi\varepsilon_0}\frac1R.

$x>R$ : Energy within spherical shell

=\tfrac12\varepsilon_0E^2\,4\pi x^2\delta x=\tfrac124\pi\varepsilon_0\frac{Q^2}{(4\pi\varepsilon_0)^2x^4}x^2\delta x.

Energy within the spherical volume for $x>R$

=\frac12\frac{Q^2}{4\pi\varepsilon_0}\int_R^\infty\frac1{x^2}dx=\frac18\frac{Q^2}{\pi\varepsilon_0}\frac1R.

Total energy associated with the charge distribution

=\frac1{40}\frac{Q^2}{\pi\varepsilon_0}\frac1R+\frac18\frac{Q^2}{\pi\varepsilon_0}\frac1R=\frac3{20}\frac{Q^2}{\pi\varepsilon_0}\frac1R.

Method 2

A shell with charge $4\pi x^2\delta x\rho$ moves from $\infty$ to the surface of a sphere radius $x$ where the electric potential is

\frac{\frac43\pi x^3\rho}{4\pi\varepsilon_0x}=\frac{x^2\rho}{3\varepsilon_0}

and will therefore gain electrical potential energy

\left(\frac{x^2\rho}{3\varepsilon_0}\right)(4\pi x^2\rho)\delta x.

Total energy of complete sphere

=\int_0^R\frac{4\pi\rho^2x^4}{3\varepsilon_0}dx=\frac4{15}\frac{\pi\rho^2R^5}{\varepsilon_0}.

Putting $Q=$ charge on sphere $=\tfrac43\pi R^3\rho$ , $\rho=3Q/(4\pi R^3)$ ,

\text{total energy}=\frac4{15}\pi\left(\frac{9Q^2}{16\pi^2R^6}\right)\frac{R^5}{\varepsilon}=\frac3{20}\frac{Q^2}{\pi\varepsilon_0R}.

(c) Binding energy $E_{\mathrm{binding}}=E_{\mathrm{electric}}-E_{\mathrm{nuclear}}$ .

Binding energy is a negative energy.

Therefore $-8.768=E_{\mathrm{electric}}-10.980$ MeV per nucleon.

$E_{\mathrm{electric}}=2.212$ MeV per nucleon.

Radius of cobalt nucleus is given by

R=\frac3{20}\frac{Q^2}{\pi\varepsilon_0E_{\mathrm{electric}}^{\mathrm{total}}}

=\frac{3\times27^2\times(1.60\times10^{-19})^2}{20\pi\times8.85\times10^{-12}\times2.212\times10^6\times57\times1.60\times10^{-19}}\ \mathrm m

=5.0\times10^{-15}\ \mathrm m.

T=\frac{4\rho d\ln2}{B^2}=\frac{4\times1.70\times10^{-8}\times8.90\times10^3\times0.6931}{(44.5\times10^{-6}\times0.4384)^2}\ \mathrm s

=1.10(2)\times10^6\ \mathrm s\quad(=306\ \mathrm{hr}=12\ \mathrm{days}\ 18\ \mathrm{hr}).

E E.M. Induction

Method 1 Equating energy

Horizontal component of magnetic field $B$ inducing emf in ring:

B=44.5\times10^{-6}\cos64^\circ.

Magnetic flux through ring at angle $\theta=B\pi a^2\sin\theta$ , where $a=$ radius of ring.

Instantaneous emf

=\frac{d\phi}{dt}=B\pi a^2\frac{d\sin\omega t}{dt}=B\pi a^2\omega\cos\omega t=B\pi a^2\omega\cos\theta,

where $\omega=$ angular velocity.

R.m.s. emf over 1 revolution $=B\pi a^2\omega/\sqrt2$ .

Average resistive heating of ring $=B^2\pi^2a^4\omega^2/(2R)$ .

Moment of inertia $=\tfrac12ma^2$ .

Rotational energy $=\tfrac14ma^2\omega^2$ , where $m=$ mass of ring.

Power producing change in $\omega$

=\frac{d}{dt}\left\{\tfrac14ma^2\omega^2\right\}=\tfrac14ma^2\,2\omega\frac{d\omega}{dt}.

Equating:

\tfrac12ma^2\omega\frac{d\omega}{dt}=-\frac{B^2\pi^2a^4\omega^2}{2R}.

\frac{d\omega}{\omega}=-\frac{B^2\pi^2a^2}{mR}dt.

If $T$ is time for angular velocity to halve,

\int_\omega^{\omega/2}\frac{d\omega}{\omega}=-\int_0^T\frac{B^2\pi^2a^2}{mR}dt,

\ln2=\frac{B^2\pi^2a^2}{mR}T.

But $R=2\pi a\rho/A$ , where $A$ is cross-sectional area of copper ring;

$m=2\pi adA$ ( $d=$ density).

\ln2=\frac{B^2\pi^2a^2T}{\frac{2\pi a\rho}{A}\,2\pi adA}=\frac{B^2T}{4\rho d}.

T=\frac{4\rho d\ln2}{B^2}=\frac{4\times1.70\times10^{-8}\times8.90\times10^3\times0.6931}{(44.5\times10^{-6}\times0.4384)^2}\ \mathrm s

=1.10(2)\times10^6\ \mathrm s\quad(=306\ \mathrm{hr}=12\ \mathrm{days}\ 18\ \mathrm{hr}).

Method 2 Back Torque

Horizontal component of magnetic field $=B=44.5\times10^{-6}\cos64^\circ$ .

Cross-section of area of ring is $A$ . Radius of ring $=a$ . Density of ring $=d$ . Resistivity $=\rho$ . $\omega=$ angular velocity ( $\omega$ positive when clockwise).

Resistance $R=\rho\,2\pi a/A$ .

Mass of ring $m=2\pi aAd$ .

Moment of inertia $=M=\tfrac12ma^2$ .

Magnetic flux through ring at angle $\theta=B\pi a^2\sin\theta$ .

Instantaneous emf

=\frac{d\phi}{dt}=B\pi a^2\frac{d\sin\omega t}{dt}=B\pi a^2\omega\cos\omega t=B\pi a^2\omega\cos\theta.

Induced current $I=B\pi a^2\cos\theta/R$ .

Torque opposing motion

=(B\pi a^2\cos\theta)I=\frac1R(B\pi a^2)^2\omega\cos^2\theta.

Work done in small $\delta\theta$ :

\frac1R(B\pi a^2)^2\omega\,\tfrac12(\cos2\theta+1)\delta\theta.

Average torque = (work done in $2\pi$ revolution)/ $2\pi$

=\frac1{2\pi R}(B\pi a^2)^2\omega\,\tfrac12\,2\pi=\frac1{2R}(B\pi a^2)^2\omega.

This equals $M\,d\omega/dt$ so that

M\frac{d\omega}{dt}=-\frac{B(\pi a^2)B(\pi a^2)\frac12}{(\rho/A)(2\pi a)}\omega.

\tfrac12(2\pi aAd)a^2\frac{d\omega}{dt}=-\frac{B^2(\pi a^2)^2A}{4\rho\pi a}\omega.

\frac{d\omega}{dt}=-\frac{B^2}{4\rho d}\omega.

\int_\omega^{\omega/2}\frac{d\omega}{\omega}=\int_0^T\frac{B^2}{4\rho d}dt.

\ln2=\frac{B^2T}{4\rho d}.

T=\frac{4\rho d\ln2}{B^2}=\frac{4\times1.70\times10^{-8}\times8.90\times10^3\times0.6931}{(44.5\times10^{-6}\times0.4384)^2}\ \mathrm s

=1.10(2)\times10^6\ \mathrm s=306\ \mathrm{hr}=12\ \mathrm{days}\ 18\ \mathrm{hr}.

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