[User]
Theoretical Question 3
Part A
                          Neutrino Mass and Neutron Decay
    A free neutron of mass mn decays at rest in the laboratory frame of reference into three
non-interacting particles: a proton, an electron, and an anti-neutrino. The rest mass of the
proton is mp, while the rest mass of the anti-neutrino mv is assumed to be nonzero and much
smaller than the rest mass of the electron me. Denote the speed of light in vacuum by c. The
measured values of mass are as follows:
         mn=939.56563 MeV/c2, mp= 938.27231 MeV/c2, me=0.5109907 MeV/c2
In the following, all energies and velocities are referred to the laboratory frame. Let E be the
total energy of the electron coming out of the decay.
(a) Find the maximum possible value Emax of E and the speed vm of the anti-neutrino when E
   = Emax. Both answers must be expressed in terms of the rest masses of the particles and the
   speed of light. Given that mv < 7.3 eV/c2, compute Emax and the ratio vm /c to 3 significant
  digits.                                                                           (4.0 points)

                                               9

Part B
                                     Light Levitation
    A transparent glass hemisphere with radius R and mass m has an index of refraction n. In
the medium outside the hemisphere, the index of refraction is equal to one. A parallel beam of
monochromatic laser light is incident uniformly and normally onto the central portion of its
                                                                  K
planar surface, as shown in Figure 3. The acceleration of gravity g is vertically downwards.
The radius δ of the circular cross-section of the laser beam is much smaller than R . Both the
glass hemisphere and the laser beam are axially symmetric with respect to the z-axis.
     The glass hemisphere does not absorb any laser light. Its surface has been coated with a
thin layer of transparent material so that reflections are negligible when light enters and leaves
the glass hemisphere. The optical path traversed by laser light passing through the
non-reflecting surface layer is also negligible.
(b) Neglecting terms of the order (δ /R)3 or higher, find the laser power P needed to balance
   the weight of the glass hemisphere.                                               (4.0 points)
Hint: cosθ ≈ 1 − θ 2 / 2 when θ is much smaller than one.

                                            z
                                                              glass hemisphere

                                      n                   R

                                 K                        laser beam
                                 g
                                                                             Figure 3
                                                2δ

                                                     10

 [Answer Sheet]                 Theoretical Question 3

  Wherever requested, give each answer as analytical expressions followed by numerical
   values and units. For example: area of a circle A = π r 2 = 1.23 m2.

                          Neutrino Mass and Neutron Decay

(a) (Give expressions in terms of rest masses of the particles and the speed of light)
   The maximum energy of the electron is (expression and value)

        Emax =

   The ratio of anti-neutrino’s speed at E = Emax to c is (expression and value)

        vm /c =

                                     Light Levitation

(b) The laser power needed to balance the weight of the glass hemisphere is

        P=

                                               11

Solution:
Solution toTheoretical Question 3
Part A
                               Neutrino Mass and Neutron Decay

                     K                   K                        K
(a) Let ( c 2 E e , cqe ) , ( c 2 E p , cq p ) , and ( c 2 E v , cqv ) be the energy-momentum 4-vectors of the
   electron, the proton, and the anti-neutrino, respectively, in the rest frame of the neutron.
                                K K K
   Notice that E e , E p , Eν , qe , q p , qν are all in units of mass. The proton and the anti-neutrino
   may be considered as forming a system of total rest mass M c , total energy c 2 Ec , and
                    G
   total momentum cqc . Thus, we have
                                                 K    K     K
                        Ec = E p + Ev ,          qc = q p + qv ,        M c2 = E c2 − qc2               (A1)
                                         K
   Note that the magnitude of the vector qc is denoted as qc. The same convention also
   applies to all other vectors.
     Since energy and momentum are conserved in the neutron decay, we have
                                   E c + E e = mn                                                       (A2)
                                        K      K
                                       qc = − qe                                                        (A3)
   When squared, the last equation leads to the following equality

                                           qc2 = qe2 = E e2 − me2                                       (A4)

   From Eq. (A4) and the third equality of Eq. (A1), we obtain

                                           E c2 − M c2 = E e2 − me2                                     (A5)

   With its second and third terms moved to the other side of the equality, Eq. (A5) may be
   divided by Eq. (A2) to give
                                             1
                                   Ec − Ee =    ( M c2 − me2 )                         (A6)
                                             mn
   As a system of coupled linear equations, Eqs. (A2) and (A6) may be solved to give
                                          1
                                   Ec =      (m 2 − me2 + M c2 )                                        (A7)
                                        2mn n
                                                 1
                                         Ee =       (m 2 + me2 − M c2 )                                 (A8)
                                                2m n n
   Using Eq. (A8), the last equality in Eq. (A4) may be rewritten as
                  1
          qe =          (mn2 + me2 − M c2 ) 2 − (2mn m e ) 2
                 2m n
                                                                                (A9)
                 1
              =         (mn + m e + M c )(mn + me − M c )(mn − me + M c )(mn − me − M c )
                2mn

        Eq. (A8) shows that a maximum of Ee corresponds to a minimum of M c2 . Now the
   rest mass M c is the total energy of the proton and anti-neutrino pair in their center of
   mass (or momentum) frame so that it achieves the minimum

                                                     27

                                             (M c )min = M = m p + mv                                    (A10)
   when the proton and the anti-neutrino are both at rest in the center of mass frame. Hence,
   from Eqs. (A8) and (A10), the maximum energy of the electron E = c2Ee is
              E max =
                       c2
                                 [                                  ]
                           m 2 + me2 − (m p + mv ) 2 ≈ 1.292569 MeV ≈ 1.29 MeV
                      2m n n
                                                                                (A11)

      When Eq. (A10) holds, the proton and the anti-neutrino move with the same velocity vm
   of the center of mass and we have
                      vm     q          qp            q           q
                         = v         =            = c          = e                   (A12)
                       c     Ev E =E    Ep            Ec E =E     E c M =m + m
                                           max           E = Emax            max             c   p   v

   where the last equality follows from Eq. (A3). By Eqs. (A7) and (A9), the last expression
   in Eq. (A12) may be used to obtain the speed of the anti-neutrino when E = Emax. Thus,
   with M = mp+mv, we have

               vm       ( mn + me + M )(mn + me − M )(mn − me + M )(mn − me − M )
                  =
                c                                   mn2 − me2 + M 2                                      (A13)
                    ≈ 0.00126538 ≈ 0.00127

------------------------------------------------------------------------------------------------------
[Alternative Solution]

      Assume that, in the rest frame of the neutron, the electron comes out with momentum
  K                                      K                                          K
cqe and energy c2Ee, the proton with cq p and c 2 E p , and the anti-neutrino with cqv and
                                      K
c 2 Ev . With the magnitude of vector qα denoted by the symbol qα, we have

                          E 2p = m 2p + q 2p ,     E v2 = mv2 + qv2 ,     E e2 = me2 + qe2                (1A)

Conservation of energy and momentum in the neutron decay leads to
                                             E p + Ev = mn − E e                                          (2A)

                                                 K     K      K
                                                 q p + qv = − qe                                          (3A)

When squared, the last two equations lead to

                                     E 2p + Ev2 + 2 E p Ev = (mn − Ee ) 2                                 (4A)

                                                   K K
                                     q 2p + qv2 + 2q p ⋅ qv = qe2 = E e2 − me2                            (5A)

Subtracting Eq. (5A) from Eq. (4A) and making use of Eq. (1A) then gives
                                                      K K
                            m 2p + mv2 + 2( E p E v − q p ⋅ qv ) = mn2 + me2 − 2mn E e                    (6A)

or, equivalently,

                                                         28

                                                                            K K
                            2mn E e = mn2 + me2 − m 2p − mv2 − 2( E p E v − q p ⋅ qv )                          (7A)
                          K       K            K K
If θ is the angle between q p and qv , we have q p ⋅ qv = q p qv cos θ ≤ q p qv so that Eq. (7A)
leads to the relation

                            2mn E e ≤ mn2 + me2 − m 2p − mv2 − 2( E p Ev − q p qv )                             (8A)

Note that the equality in Eq. (8A) holds only if θ = 0, i.e., the energy of the electron c2Ee takes
on its maximum value only when the anti-neutrino and the proton move in the same direction.
     Let the speeds of the proton and the anti-neutrino in the rest frame of the neutron be
 cβ p and cβ v , respectively. We then have q p = β p E p and qv = β v E v . As shown in Fig.
A1, we introduce the angle φ v ( 0 ≤ φ v < π / 2 ) for the antineutrino by

           qv = mv tan φ v ,           Ev = mv2 + qv2 = mv sec φ v ,             β v = qv / Ev = sin φ v        (9A)

                                                     Ev
                                                                   qv

                                                 φv                             Figure A1
                                                      mv

Similarly, for the proton, we write, with 0 ≤ φ p < π / 2 ,

          q p = m p tan φ p ,         E p = m 2p + q 2p = m p sec φ p ,         β p = q p / E p = sin φ p      (10A)
Eq. (8A) may then be expressed as
                                                                          1 − sin φ p sin φv
                     2mn E e ≤ mn2 + me2 − m 2p − mv2 − 2m p mv                                                (11A)
                                                                           cos φ p cos φv
The factor in parentheses at the end of the last equation may be expressed as

      1 − sin φ p sin φ v        1 − sin φ p sin φ v − cos φ p cos φ v          1 − cos(φ p − φ v )
                             =                                           +1 =                         +1 ≥ 1   (12A)
        cos φ p cos φ v                      cos φ p cos φ v                      cos φ p cos φ v

and clearly assumes its minimum possible value of 1 when φ p = φ v, i.e., when the
anti-neutrino and the proton move with the same velocity so that β p = β v. Thus, it follows
from Eq. (11A) that the maximum value of Ee is
                                             1
                            ( E e ) max =       ( mn2 + me2 − m 2p − mv2 − 2m p mv )
                                            2mn
                                                                                (13A)
                                        =
                                           1
                                          2mn
                                                 [
                                              mn2 + me2 − ( m p + mv ) 2    ]
and the maximum energy of the electron E = c2Ee is

                            E max = c 2 ( Ee ) max ≈ 1.292569 MeV ≈ 1.29 MeV                                   (14A)

                                                           29

    When the anti-neutrino and the proton move with the same velocity, we have, from Eqs.
(9A), (10A), (2A) ,(3A), and (1A), the result

                                 qp  q    q p + qv   qe      E e2 − me2
                   βv = β p =       = v =          =       =                                               (15A)
                                 E p E v E p + E v mn − Ee   mn − E e

Substituting the result of Eq. (13A) into the last equation, the speed vm of the anti-neutrino
when the electron attains its maximum value Emax is, with M = mp+mv, given by

               vm                    ( E e ) 2max − me2        (mn2 + me2 − M 2 ) 2 − 4mn2 me2
                  = ( β v ) max Ee =                    =
                c                    mn − ( Ee ) max             2mn2 − (mn2 + me2 − M 2 )
                        (mn + me + M )(mn + me − M )(mn − me + M )(mn − me − M )
                    =                                                                                (16A)
                                                   mn2 − me2 + M 2
                    ≈ 0.00126538 ≈ 0.00127
------------------------------------------------------------------------------------------------------
Part B
                                             Light Levitation

(b) Refer to Fig. B1. Refraction of light at the spherical surface obeys Snell’s law and leads to
                                           n sin θ i = sin θ t                               (B1)
   Neglecting terms of the order (δ /R)3or higher in sine functions, Eq. (B1) becomes
                                            nθ i ≈ θ t                                                      (B2)
   For the triangle ∆FAC in Fig. B1, we have
                                                                                z
                  β = θ t − θ i ≈ nθ i − θ i = (n − 1)θ i            (B3)
                                                                                F

       Let f 0 be the frequency of the incident light. If n p
   is the number of photons incident on the plane surface per                            θt
   unit area per unit time, then the total number of photons                                  β
   incident on the plane surface per unit time is n pπδ 2 . The
                                                                                A
   total power P of photons incident on the plane surface is                                  θi
    (n pπδ 2 )(hf 0 ) , with h being Planck’s constant. Hence,
                                         P                                                             n
                               np =                                  (B4)                 θi
                                      πδ 2 hf 0
                                                                                C
   The number of photons incident on an annular disk of
                                                                                δ
   inner radius r and outer radius r +dr on the plane surface
   per unit time is n p (2πrdr ) , where r = R tan θ i ≈ Rθ i .
                                                                                Fig. B1
   Therefore,

                                      n p (2πrdr ) ≈ n p (2πR 2 )θ i dθ i                                   (B5)

   The z-component of the momentum carried away per unit time by these photons when
                                                       30

refracted at the spherical surface is

                             hf o                      hf               β2
                 dFz = n p        ( 2πrdr ) cos β ≈ n p 0 ( 2πR 2 ) 1 −    θ i dθ i
                              c                         c               2
                                                                                (B6)
                            hf 0               ( n − 1) 2 3
                       ≈ np      ( 2πR ) θ i −
                                      2
                                                         θ i dθ i
                             c                     2

so that the z-component of the total momentum carried away per unit time is

                                            hf 0 θim       ( n − 1) 2 3
                                             c ∫0
                          Fz = 2πR 2 n p             θ i −           θ i dθ i
                                                               2
                                                                                (B7)
                                           hf 0 2        ( n − 1) 2 2
                             = πR n p
                                   2
                                                θ im 1 −           θ im
                                            c                4

where tan θ im =
                   δ
                       ≈ θ im . Therefore, by the result of Eq. (B5), we have
                   R

                        πR 2 P hf 0 δ 2     ( n − 1) 2 δ 2   P     ( n − 1) 2 δ 2
                Fz =                    1 −                =   1 −                      (B8)
                       πδ 2 hf 0 c R 2          4R 2         c         4R 2

The force of optical levitation is equal to the sum of the z-components of the forces exerted
by the incident and refracted lights on the glass hemisphere and is given by

                       P             P P    ( n − 1) 2 δ 2   ( n − 1) 2 δ 2 P
                         + ( − Fz ) = −  1−                =                            (B9)
                       c             c c        4R 2             4R 2       c

Equating this to the weight mg of the glass hemisphere, we obtain the minimum laser
power required to levitate the hemisphere as

                                                    4mgcR 2
                                             P=                                        (B10)
                                                   (n − 1) 2 δ 2

                                                   31

                                   Marking Scheme

                              Theoretical Question 3
                         Neutrino Mass and Neutron Decay
 Total      Sub
                                 Marking Scheme for Answers to the Problem
Scores     Scores
Part A       (a)    The maximum energy of the electron and the corresponding speed of the
                    anti-neutrino.
4.0 pts.    4.0     3⁄4 0.5 use energy-momentum conservation and can convert it into
                           equations.
                    3⁄4 0.5 obtain an expression for E e that allows the determination of its
                           maximum value.
                    3⁄4 (0.5+0.2) for concluding that proton and anti-neutrino must move
                           with the same velocity when E e is maximum. (0.2 for the same
                           direction)
                                                                            G G
                    3⁄4 0.6 for establishing the minimum value of ( E p Ev − q p ⋅ qv ) to be
                          m p mv or a conclusion equivalent to it.
                    3⁄4 (0.5+0.1) for expression and value of Emax.
                    3⁄4 0.5 for concluding β v = E e2 − me2 /(mn − Ee ) .
                    3⁄4 (0.5+0.1) for expression and value of vm /c.

                                    Light Levitation
Part B      (b)     Laser power needed to balance the weight of the glass hemisphere.
                    3⁄4 0.3 for law of refraction n sin θ i = sin θ t .
4.0 pts     4.0     3⁄4 0.3 for making the linear approximation nθ i ≈ θ t .
                    3⁄4 0.4 for relation between angles of deviation and incidence.
                    3⁄4 0.3 for photon energy ε = hν.
                    3⁄4 0.3 for photon momentum p = ε /c.
                    3⁄4 0.3 for momentum of incident photons per unit time = P/c.
                    3⁄4 0.6 for momentum of photons refracted per unit time as a function of
                           the angle of incidence.
                    3⁄4 0.4 for total momentum of photons refracted per unit time =
                           [1-(n-1)2δ 2/(4R2)]P/c.
                    3⁄4 0.4 for force of levitation = sum of forces exerted by incident and
                           refracted photons.
                    3⁄4 0.4 for force of levitation = (n-1)2δ 2P/(4cR 2).
                    3⁄4 0.3 for the needed laser power P = 4mgcR 2/(n-1)2δ 2.

                                            32

Part A


[User]
41st International Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010               1/3

1. Image of a charge in a metallic object
Introduction – Method of images
A point charge q is placed in the vicinity of a grounded metallic sphere of radius R [see Fig. 1(a)], and
consequently a surface charge distribution is induced on the sphere. To calculate the electric field
and potential from the distribution of the surface charge is a formidable task. However, the
calculation can be considerably simplified by using the so called method of images. In this method,
the electric field and potential produced by the charge distributed on the sphere can be represented
as an electric field and potential of a single point charge q ' placed inside the sphere (you do not have
to prove it). Note: The electric field of this image charge q ' reproduces the electric field and the
potential only outside the sphere (including its surface).

(a)                                                    (b)

Fig 1. (a) A point charge q in the vicinity of a grounded metallic sphere. (b) The electric field of the
charge induced on the sphere can be represented as electric field of an image charge q ' .

Task 1 – The image charge
The symmetry of the problem dictates that the charge q ' should be placed on the line connecting the
point charge q and the center of the sphere [see Fig. 1(b)].

      a) What is the value of the potential on the sphere? (0.3 points)
      b) Express q ' and the distance d ' of the charge q ' from the center of the sphere, in terms of
         q , d , and R . (1.9 points)
      c) Find the magnitude of force acting on charge q . Is the force repulsive? (0.5 points)

Task 2 – Shielding of an electrostatic field
Consider a point charge q placed at a distance d from the center of a grounded metallic sphere of
radius R. We are interested in how the grounded metallic sphere affects the electric field at point A
on the opposite side of the sphere (see Fig. 2). Point A is on the line connecting charge q and the
center of the sphere; its distance from the point charge q is r.

      a) Find the vector of the electric field at point A. (0.6 points)

 41st International Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010            2/3

      b) For a very large distance r  d , find the expression for the electric field by using the
         approximation (1+a)-2 ≈ 1-2a, where a  1 . (0.6 points)
      c) In which limit of d does the grounded metallic sphere screen the field of the charge q
         completely, such that the electric field at point A is exactly zero? (0.3 points)

Fig 2. The electric field at point A is partially screened by the grounded sphere.

Task 3 – Small oscillations in the electric field of the grounded metallic
sphere
A point charge q with mass m is suspended on a thread of length L which is attached to a wall, in the
vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the
wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is
attached to the wall is at a distance l from the center of the sphere. Assume that the effects of
gravity are negligible.

    a) Find the magnitude of the electric force acting on the point charge q for a given angle  and
       indicate the direction in a clear diagram (0.8 points)

    b) Determine the component of this force acting in the direction perpendicular to the thread in
       terms of l , L, R, q and  . (0.8 points)

    c) Find the frequency for small oscillations of the pendulum. (1.0 points)

Fig 3. A point charge in the vicinity of a grounded sphere oscillates as a pendulum.

Task 4 – The electrostatic energy of the system
For a distribution of electric charges it is important to know the electrostatic energy of the system. In
our problem (see Fig. 1a), there is an electrostatic interaction between the external charge q and the
induced charges on the sphere, and there is an electrostatic interaction among the induced charges

    41st International Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010                3/3

on the sphere themselves. In terms of the charge q, radius of the sphere R and the distance d
determine the following electrostatic energies:

                a) the electrostatic energy of the interaction between charge q and the induced
                    charges on the sphere; (1.0 points)
                b) the electrostatic energy of the interaction among the induced charges on the
                    sphere; (1.2 points)
                c) the total electrostatic energy of the interaction in the system. (1.0 points)
Hint: There are several ways of solving this problem:

(1) In one of them, you can use the following integral,


          xdx          1         1
 x  R  = 2 d  R .
d
      2         2 2          2        2

(2) In another one, you can use the fact that for a collection of N charges q i located at points

ri ,i = 1,, N ,           the       electrostatic   energy   is   a   sum   over   all   pairs   of   charges:
          1 N N 1 qi q j
V                          .
          2 i 1 j 1 4 0 ri  r j
                i j

Solution:
41st international Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010    1/8

Solution - Image of a charge
Solution of Task 1
Task 1a)

As the metallic sphere is grounded, its potential vanishes, V=0.

Task1b)

Let us consider an arbitrary point B on the surface of the sphere as depicted in Fig. 1.

Fig 1. The potential at point B is zero.

The distance of point B from the charge q' is

                                                                                (1)

whereas the distance of the point B from the charge q is given with the expression

                                                                                (2)

The electric potential at the point B is

                                                                                (3)

This potential must vanish,

                                                                                (4)

 i.e. its numerical value is 0 V.

 41st international Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010           2/8

Combining (1), (2) and (3) we obtain

                                                                                (5)

As the surface of the sphere must be equipotential, the condition (5) must be satisfied for every
angle α what leads to the following results

                                                                                (6)

and

                                                                                (7)

By solving of (6) and (7) we obtain the expression for the distance d' of the charge q' from the center
of the sphere

                                                                                (8)

and the size of the charge q'

                                                                                (9)

Task 1c)

Finally, the magnitude of force acting on the charge q is

        1       q 2 Rd
 F
            
      4 0 d 2  R 2 2                                                                            (10)

The force is apparently attractive.

Solution of Task 2
Task 2a)

The electric field at the point A amounts to

 41st international Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010          3/8

                                                                                (11)

Task 2b)

For very large distances r we can apply approximate formula (1+a)-2 ≈ 1-2a to the expression (11)
what leads us to

                                                                                (12)

In general a grounded metallic sphere cannot completely screen a point charge q at a distance d
(even in the sense that its electric field would decrease with distance faster than 1/r2) and the
dominant dependence of the electric field on the distance r is as in standard Coulomb law.

Task 2c)

In the limit d → R the electric field at the point A vanishes and the grounded metallic sphere screens
the point charge completely.

Solution of Task 3

Task 3a)

Let us consider a configuration as in Fig. 2.

 Fig 2. The pendulum formed by a charge near a grounded metallic sphere.
The distance of the charge q from the center of the sphere is

 41st international Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010        4/8

                                                                                (13)

The magnitude of the electric force acting on the charge q is

                                                                                (14)

From which we have

         1         q 2 R l 2  L2  2lL cos                                                     (15)
 F
               
       4 0 l 2  L2  2lL cos   R 2 2      

Task 3b)

The direction of the vector of the electric force (17) is described in Fig. 3.

 Fig 3. The direction of the force F.

The angles α and β are related as

                                                                                (16)

whereas for the angle γ the relation γ=α+β is valid. The component of the force perpendicular to the
thread is F sin γ, that is ,

           1       q 2 R l 2  L2  2lL cos 
F                                                 sin(   )
               
        4 0 l 2  L2  2lL cos   R 2 2      
where
                                                                                (17)
                            L
  arcsin(                                 sin  )
                    L2  l 2  2 Ll cos 

 41st international Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010            5/8

Task 3c)

The equation of motion of the mathematical pendulum is

 mL   F                                                                                (18)

As we are interested in small oscillations, the angle α is small, i.e. for its value in radians we have α
much smaller than 1. For a small value of argument of trigonometric functions we have approximate
relations sin x ≈ x and cos x ≈ 1-x2/2. So for small oscillations of the pendulum we have
   L /(l  L) and   l /(l  L) .

Combining these relations with (13) we obtain

     d 2   1     q 2 Rd  L                                                                       (19)
mL                         1    0
                              
     dt 2 4 0 d 2  R 2 2  d 

Where d  l  L what leads to

        q            Rd 1  L                                                                      (20)
                           1   
     d  R2
      2
                    4 0 mL  d 
                                        .
        q                 Rl    1

  (l  L) 2  R 2        4 0 mL

Solution of Task 4
First we present a solution based on the definition of the electrostatic energy of a collection of
charges.

Task 4a)

The total energy of the system can be separated into the electrostatic energy of interaction of the
external charge with the induced charges on the sphere, Eel,1, and the electrostatic energy of mutual
interaction of charges on the sphere, Eel,2, i.e.

                                                                                (21)

Let there be N charges induced on the sphere. These charges q j are located at points

r j , j  1,, N on the sphere. We use the definition of the image charge, i.e., the potential on the
surface of the sphere from the image charge is identical to the potential arising from the induced
charges:

 41st international Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010            6/8

   q'     N      qj
           ,
 r  d ' j 1 r j  r                                                                              (22)

                                                                                                    
where r is a vector on the sphere and d ' denotes the vector position of the image charge. When r
                                   
coincides with some ri , then we just have

   q'      N      qj
      .
         
 ri  d ' j 1 r j  ri                                                                            (23)
                   j i

From the requirement that the potential on the surface of the sphere vanishes we have

   q'     q
       0,
 r  d' r  d                                                                                (24)
                                                       
where d denotes the vector position of the charge q ( r is on the sphere).

For the interaction of the external charge with the induced charges on the sphere we have

               q          N
                              qi1    qq'    1    qq'          1     q2R
Eel ,1                             
           4 0 i 1 ri  d 4 0 d 'd 4 0 d  d '
                                                         
                                                            4 0 d 2  R 2                        (25)

Here the first equality is the definition of this energy as the sum of interactions of the charge q with
each of the induced charges on the surface of the sphere. The second equality follows from (21).

In fact, the interaction energy E el ,1 follows directly from the definition of an image charge.

Task 4b)

The energy of mutual interactions of induced charges on the surface of the sphere is given with

    41st international Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010     7/8

              1 1 N N qi q j
E el , 2               
              2 4 0 i 1 j 1 ri  r j
                                       j i
                       N
     1 1             q'
               qi   
     2 4 0 i 1 ri  d '
                                                                                (26)
                           N
         1 1             q
                  qi   
         2 4 0 i 1 ri  d

         1 1      qq'   1 1      qq '    1 1      q2R
                                 
         2 4 0 d 'd  2 4 0 d  d ' 2 4 0 d 2  R 2

 Here the second line is obtained using (22). From the second line we obtain the third line applying
(23), whereas from the third line we obtain the fourth using (22) again.

Task 4c)

Combining expressions (19) and (20) with the quantitative results for the image charge we finally
obtain the total energy of electrostatic interaction

                       1 1       q2R
Eel (d )  
                       2 4 0 d 2  R 2                                                        (27)

An alternative solution follows from the definition of work. By knowing the integral

                                                                                (28)
             xdx               1    1
 x  R 
d
         2         2   2
                         
                               2 d  R2
                                  2

We can obtain the total energy in the system by calculating the work needed to bring the charge q
from infinity to the distance d from the center of the sphere:

                       d                                                                       (29)
                                 
E el (d )    F ( x )dx   F ( x )dx 
                                             d
                                  2
                   1           q Rx
  ( )                                          dx 
     d                     
              4 0 x  R 2 2  2
                                              
         1 1       q2R

         2 4 0 d 2  R 2
This solves Task 4c).

The electrostatic energy between the charge q and the sphere must be equal to the energy between
the charges q and q’ according to the definition of the image charge:

 41st international Physics Olympiad, Croatia – Theoretical Competition, July 19th 2010       8/8

              1     qq'           1     q2R                                                       (30)
Eel ,1                      
            4 0 (d  d ' )    4 0 d 2  R 2

This solves Task 4a).

From this we immediately have that the electrostatic energy among the charges on the sphere is:

            1 1       q2R                                                                         (31)
Eel , 2                      .
            2 4 0 d 2  R 2

This solves Task 4b).

1. Image of a charge in a metallic object


[User] A coin of mass $m_c=\SI{10}{g}$ and density $\rho_c=\SI{8900}{kg/m^3}$ has frozen inside a piece of ice. The temperature of the ice and the coin is $\SI{0}{\degreeCelsius}$. The piece of ice without the coin weighs $m_i=\SI{130}{g}$. This piece of ice is thrown into a vessel in where there is $V_w=\SI{400}{ml}$ of water with an initial temperature $T$. How big has to be the water’s minimal initial temperature $T$ so that the piece of ice together with the coin would sink to the bottom after the thermal equilibrium is achieved? Do not account for the heat exchange with the surrounding environment. The specific heat of water is $c=\SI{4200}{J/ (kg\cdot\degreeCelsius)}$ and the ice’s enthalpy of fusion $\lambda=\SI{330}{kJ/kg}$. The ice’s density $\rho_i=\SI{900}{kg/m^3}$ and the water’s density $\rho_w=\SI{1000}{kg/m^3}$.

Solution:
Jäätükk koos mündiga hakkab uppuma siis, kui sellele mõjub raskusjõud on võrdne üleslükkejõuga. Tähistame uppumise hakkamise hetkel mündi ümber oleva jää massi $m$ ning ruumala $V$. Sellisel juhul mõjub jäätükile enne uppuma hakkamist raskusjõus $F_r=(m_m+m)g$ ning üleslükkejõud
\[ F_y=\rho_v g(V + V_m)=\rho_v g\left(\frac{m}{\rho_j} + \frac{m_m}{\rho_m}\right). \]

Kuna raskusjõud ja üleslükkejõud on uppumise hakkamise hetkel võrdsed, saame avaldada mündi ümber olnud jää massi $m$:
\[ m = \frac{\rho_j m_m(\rho_m - \rho_v)}{\rho_m(\rho_v - \rho_j)} = \SI{79,9}{g}. \]
Sulanud jää mass $m_s$ on seega
\[ m_s = m_j - m = m_j - \frac{\rho_j m_m(\rho_m - \rho_v)}{\rho_m(\rho_v - \rho_j)}. \]
Jää sulamiseks vajaminev energia $Q=\lambda m_s$ saadakse vee jahtumisel eraldunud energiast $Q=cm_v\Delta T$. Võrdsustades viimased avaldised ning avaldades $\Delta T$, saame
\[ \Delta T = \frac{\lambda m_s}{cm_v}. \]
Asendades siia sulanud jää massi $m_s$, saame temperatuuri muutuseks
\[ \Delta T \approx \SI{9,8}{\degreeCelsius}. \]
Kuna vee lõpptemperatuur pärast soojusvahetuse lakkamist on \SI{0}{\degreeCelsius}, peab vee algtemperatuur olema \SI{9,8}{\degreeCelsius}.

Münt jääs

Jäätüki sisse on jäätunud münt massiga $m_m=\SI{10}{g}$ ja tihedusega $\rho_m=\SI{8900}{kg/m^3}$. Jäätüki ja mündi temperatuur on \SI{0}{\degreeCelsius}. Jäätükk ilma mündita kaalub $m_j=\SI{130}{g}$. See jäätükk visatakse anumasse, milles on $V_v=\SI{400}{ml}$ vett algtemperatuuriga $T$. Kui suur peab olema vee minimaalne algtemperatuur $T$, et jäätükk koos mündiga vajuks pärast soojusliku tasakaalu saabumist põhja? Soojusvahetust väliskeskkonnaga mitte arvestada. Vee erisoojus $c=\SI{4200}{J/ (kg\cdot\degreeCelsius)}$ ning jää sulamissoojus $\lambda=\SI{330}{kJ/kg}$. Jää tihedus $\rho_j=\SI{900}{kg/m^3}$ ja vee tihedus $\rho_v=\SI{1000}{kg/m^3}$.

Jäätükk koos mündiga hakkab uppuma siis, kui selle keskmine tihedus on võrdne vee tihedusega. Jää sulamiseks vajaminev energia saadakse vee jahtumisel eraldunud energiast.


[User]
\begin{wrapfigure}[9]{r}{0.4\textwidth}
	\begin{center}
		\vspace{-25pt}
		\includegraphics[width=\linewidth]{2006-v2g-07-yl}
	\end{center}
\end{wrapfigure}
Joonisel toodud elektriskeemil on vooluallikas elektromotoorjõuga $\mathcal E$ ja sisetakistusega $r$, kolm takistit takistustega $R_1 = R_2 = R_3 = R$ ning kondensaatorid mahtuvustega $C_1$ ja $C_2$. Arvutage, kui suured on elektrilaengud kondensaatoritel pärast pika aja möödumist, kui:\\
\osa lüliti 1 on suletud, lüliti 2 on avatud;\\
\osa mõlemad lülitid on suletud;\\
\osa eelmisest seisust avatakse mõlemad lülitid üheaegselt.

Solution:
\osa Vooluallikas laeb mõlemad kondensaatorid elektromotoorjõuga võrdse pingeni, seega on $q_{a1} = C_1\mathcal E$ ja $q_{a2} = C_2\mathcal E$.\\
\osa Leiame, kui suure pingeni kondensaatorid laaduvad. Voolutugevus ahelas
on
\[
I = \frac{\mathcal E}{R_1+R_2+R_3+r}.
\]
Kondensaatoril $C_1$ on laeng $q_{b1} = C_1I(R_1 + R_2)$ ja kondensaatoril $C_2$ on laeng $q_{b2} = C_2IR_2$.\\
\osa Kehtib laengu jäävus, paralleelses ühenduses on pinged kondensaatoritel võrdsed. Summaarne laeng $q = q_{b1} + q_{b2}$ ning mahtuvus $C = C_1 + C_2$. Pinge kondensaatoril on $U = q/C$. Laengud kondensaatoritel on $q_{c1} = C_1U$ ja $q_{c2} = C_2U$.

Kondensaatorid

\begin{wrapfigure}[9]{r}{0.4\textwidth}
	\begin{center}
		\vspace{-25pt}
		\includegraphics[width=\linewidth]{2006-v2g-07-yl}
	\end{center}
\end{wrapfigure}
Joonisel toodud elektriskeemil on vooluallikas elektromotoorjõuga $\mathcal E$ ja sisetakistusega $r$, kolm takistit takistustega $R_1 = R_2 = R_3 = R$ ning kondensaatorid mahtuvustega $C_1$ ja $C_2$. Arvutage, kui suured on elektrilaengud kondensaatoritel pärast pika aja möödumist, kui:\\
\osa lüliti 1 on suletud, lüliti 2 on avatud;\\
\osa mõlemad lülitid on suletud;\\
\osa eelmisest seisust avatakse mõlemad lülitid üheaegselt.

\osa\osa Peale pikka aega on kondensaatoreid läbiv vool null, sest nende pinged, ja seega laengud, on stabiliseerunud. Seega võib kondensaatorid efektiivselt süsteemist välja lõigata.\\
\osa Elimineerides süsteemist vooluallika, peab kehtima laengu jäävuse seadus.


[User]
Using similar reasoning, design a machine that finds the point X that minimizes
    (AX)2 + (BX)2 + (CX)2 . What geometrical property can you conclude about this point?

       Example 9

       Show that the incenter of a triangle (i.e. the meeting point of the angle bisectors) exists.

       Solution
       Apply six forces at the vertices of a triangle as shown.

       These forces clearly balance, and also produce no net torque on the triangle. Now combine
       the forces applied at each vertex, yielding three forces that point along the angle bisectors.
       By the principles of M2, the torques of these forces can only balance if their lines of action
       meet at a point. Therefore the angle bisectors are concurrent, so the incenter exists.

       Example 10

       Let AB be a diameter of a circle, and let a mass be free to slide on the circle. The mass is
       connected to two identical straight springs of zero rest length, which are in turn connected
       to points A and B. At what points C can the mass be in static equilibrium?

       Solution

       The potential energy of the system is proportional to (AC)2 + (BC)2 . Since ABC is a right
       triangle, this is just equal to (AB)2 by the Pythagorean theorem. Since the potential energy
       doesn’t depend on where the mass is, it can be at static equilibrium at any point on the
       circle. Alternatively, you can show that the mass is in static equilibrium by force balance,
       and use the reasoning in reverse to derive the Pythagorean theorem.

Solution:
Attach a mass to springs at each of A, B, and C each with spring constant k. Then in
    equilibrium, the mass has the minimum possible potential energy, so it is at point X. Balancing
    forces gives k(X − A) + k(X − B) + k(X − C) = 0, so X is the centroid of ABC.

       Example 9

       Show that the incenter of a triangle (i.e. the meeting point of the angle bisectors) exists.

       Solution
       Apply six forces at the vertices of a triangle as shown.

       These forces clearly balance, and also produce no net torque on the triangle. Now combine
       the forces applied at each vertex, yielding three forces that point along the angle bisectors.
       By the principles of M2, the torques of these forces can only balance if their lines of action
       meet at a point. Therefore the angle bisectors are concurrent, so the incenter exists.

       Example 10

       Let AB be a diameter of a circle, and let a mass be free to slide on the circle. The mass is
       connected to two identical straight springs of zero rest length, which are in turn connected
       to points A and B. At what points C can the mass be in static equilibrium?

       Solution

       The potential energy of the system is proportional to (AC)2 + (BC)2 . Since ABC is a right
       triangle, this is just equal to (AB)2 by the Pythagorean theorem. Since the potential energy
       doesn’t depend on where the mass is, it can be at static equilibrium at any point on the
       circle. Alternatively, you can show that the mass is in static equilibrium by force balance,
       and use the reasoning in reverse to derive the Pythagorean theorem.

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User]
6. RUNNING ON ICE (4 points)

A boy is
running on a large field of ice with velocity
v=5m/s toward the north. The coefficient of
friction between his feet and the ice is μ=0.1.
Assume as a simplification that the reaction
force between the boy and the ice stays con-
stant (in reality it varies with every push,
but the assumption is justified by the fact
that the value averaged over one step stays
constant).
i) (2 points) What is the minimum time ne-
cessary for him to change his moving direc-
tion to point towards the east so that the final
speed is also v=5m/s?
ii) (2 points) What is the shape of the optimal
trajectory called?

Solution:
6. Running on ice

1) During the process, the velocity vector needs
to change its direction by 90 degrees. Let us
consider the this graphically using the vx − vy-
plane: we need to move from the point A with
coordinates (0,v) to a point B with coordinates
(v,0) while having a constant speed. Indeed,
the velocity of a point in the vx − vy-plane is
the acceleration of the body, which has here a
constant modulus μg. Obviously, the fastest
path is a straight line of length v
√
2, so that
t = v
√
2/μg ≈ 7.2s.
2) Since the direction of the acceleration is con-
stant, the trajectory is the same as for a body
in the Earth's field of gravity  a parabola.

RUNNING ON ICE


[User]
In an ethanol factory, with the help of the optical device shown in the figure, the volume percentage of ethanol is measured in a spirit and water mixture. The mixture is located in a canal of width $d=\SI{1}{cm}$. By what distance does the location of the ray on the device's photosensitive element change if the ethanol with a volume percentage 40 \% is replaced by an ethanol with a volume percentage 85 \%? You can assume that the refractive index $n\idx{mix}$ of the mixture is a linear combination of the refractive indexes of water and spirit
$$
n\idx{mix}=Xn\idx{ethanol}+(1-X)n\idx{water},
$$
where X is ethanol’s volume content ranging from 0 to 1, $n\idx{ethanol}=\num{1,3615}$ and $n\idx{water}=\num{1,3330}$.
\begin{center}
	\vspace{-0pt}
	\includegraphics[width=0.5\linewidth]{2017-v2g-04-Piiritusetehas_ing}
	\vspace{-10pt}
\end{center}

Solution:
Murdumisseaduse järgi kehtib valguse murdumisel
$$n_0\sin(\alpha_0)=n_1\sin(\alpha_1),$$
kus $n_0$ ja $n_1$ on vastavalt esimese ja teise keskkkonna murdumisnäitajad ning $\alpha_0$ ja $\alpha_1$ on vastavalt langemis- ja murdumisnurk. Seda valemit saab iteratiivselt jätkata järgmise murdumise jaoks kolmanda keskkonna piiril, kui keskkonnad on paralleelsete kihtidena:
$$n_0\sin(\alpha_0)=n_1\sin(\alpha_1)=n_2\sin(\alpha_2) = \dots$$
Selle abil näeme, et kehtib $\sin(\alpha_j)=\frac{n_0}{n_j}\sin(\alpha_0)$. See tähendab seda, et kiire murdumisnurk sõltub ainult hetke keskkonna murdumisnäitajast ja mitte sellest, kas ja kui mitu erinevat kihti varasemalt on läbitud. Seetõttu saame ignoreerida kanalit ümbritsevat klaasist või muust materjalist kihti, arvutamaks kiire nurka kanalis. Samuti, kuna ainult piirituse ja vee segu murdumisnäitaja muutub, siis ainult kanalis on kiirte liikumisnurgad erinevad. Mujal on kiire liikumine täpselt sama, ainult pärast segu läbimist on kiir nihutatud, kui segu murdumisnäitajat muuta. Segus kehtib:
$$\sin(\alpha_{\text{segu}})=\frac{n_0}{n_{\text{segu}}}\sin{\alpha_0} = \frac{1}{\sqrt{2}n_{\text{segu}}},$$
kus $\sin(\alpha_0) = 1/\sqrt{2}$ ja õhus $n_0=1$. Kiire nihe kanali sees piki kanali suunda on lihtsa geomeetria abil $$y=d\tan(\alpha_{\text{segu}}).$$
Kasutades seost $\tan(\phi) = \frac{\sin(\phi)}{\cos(\phi)} = \frac{\sin(\phi)}{\sqrt{1-\sin(\phi)^2}}$ saame
$$y = d \frac{\sin(\alpha_{\text{segu}})}{\sqrt{1-\sin(\alpha_{\text{segu}})^2}} = \frac{d}{\sqrt{2}\sqrt{n_{\text{segu}}^2-\frac12}}.$$
Nüüd leiame $y$ kahel juhul, kui $X=\num{0.40}$ siis $y=\SI{6.184}{\milli\meter}$ ja kui $X=\num{0.85}$ siis $y=\SI{6.104}{\milli\meter}$. Saame $y$ erinevuseks
$$|\Delta y| \approx \SI{80}{\micro\meter}.$$

Piiritusetehas

Piiritusetehases mõõdetakse piirituse ja vee segus oleva piirituse mahuprotsenti joonisel näidatud optilise seadme abil, kus segu paikneb kanalis laiusega $d=\SI{1}{cm}$.
Kui suure vahemaa võrra muutub seadme fototundlikul elemendil kiire asukoht, kui piiritus mahuprotsendiga \SI{40}{\percent} asendada piiritusega mahuprotsendiga \SI{85}{\percent}?
Võib eeldada, et segu murdumisnäitaja $n\idx{segu}$ on lineaarne kombinatsioon vee ja piirituse murdumisnäitajatest
$$
n\idx{segu}=Xn\idx{piiritus}+(1-X)n\idx{vesi},
$$
kus X on piirituse mahuline sisaldus vahemikus 0 -- 1 ja $n\idx{piiritus}=\num{1,3615}$ ning $n\idx{vesi}=\num{1,3330}$.

\begin{center}
	\vspace{-0pt}
	\includegraphics[width=0.5\linewidth]{2017-v2g-04-Piiritusetehas.jpg}
	\vspace{-10pt}
\end{center}

Ainus muutus kiire teekonnas toimub segu sees. Murdumisnäitaja muutuse tõttu hakkab kiir segus teise nurga all liikuma. Vastavat nurga muutust ning seejärel ka $y$-suunalist nihet on võimalik leida Snelli seaduse abil.


[User]
Juku goes flying with a spherical hot air balloon of radius $r = \SI{8.7}{\metre}$. The mass together with the passengers is $M_0= \SI{390}{\kg}$ and in addition $M_k = \SI{20}{\kg}$ of propane is brought along for fuel. How long can Juku's balloon flight last?\\
The balloon is covered with a cover that decreases thermal conductivity and thermal radiation to negligible values. In working state the air permeates through the balloon's shell with a speed $\lambda = \SI{500}{g\per\s}$. The air pressure and temperature in the flight's height is $p_0 = \SI{100}{\kilo\Pa}$ and $T_0 = \SI{10}{\degreeCelsius}$. Propane's calorific value is $k = \SI{50}{MJ/kg}$. Air's average molar mass is $\mu = \SI{29}{g\per\mole}$ and the heat capacity at constant pressure $C_p = \SI{1.0}{\frac{kJ}{K\cdot kg}}$. Universal gas constant is $R = \SI{8.3}{\frac{J}{K\cdot mol}}$.

Solution:
Ideaalse gaasi seadusest avaldub õhu tihedus sõltuvalt temperatuurist kujul $\rho = \frac{p \mu}{R T}$. Raskusjõu ning üleslükkejõu tasakaalust saame
\[
M g = V g (\rho_0 - \rho) = \frac{p \mu V g}{R} (\frac{1}{T_0} - \frac{1}{T}),
\]
kus $M = M_0 + \frac{1}{2} M_k$ on õhupalli keskmine mass lennu vältel ning $V = \frac{4}{3} \pi r^3$ on õhupalli ruumala. Kuna õhupalli pooridest imbub välja soe õhk temperatuuril $T$, kuid sisenev õhk on väliskeskonna temperatuuril $T_0$, tuleb sees olevat õhku pidevalt soojendada võimsusega
\[
P = \lambda C_p (T - T_0).
\]
Selle võimsuse saavutamiseks tuleb põletada propaani kiirusega $\frac{P}{k}$ ning kütuse lõppemiseks kuluv aeg on
\[
t = \frac{M_k k}{P} = \frac{M_k k}{\lambda C_p (T - T_0)} = \frac{M_k k (p \mu V - M R T_0)}{\lambda C_p M R T_0^2} = \SI{15}{h}.
\]

Kuumaõhupall

Juku läheb lendama kerakujulise kuumaõhupalliga, mille raadius $r = \SI{8.7}{\metre}$, mass koos reisijatega $M_0= \SI{390}{\kg}$ ning lisaks on kütusena kaasas $M_k = \SI{20}{\kg}$ propaani. Kui kaua saab kesta Juku õhupallilend?

Õhupall on kaetud kattega, mis vähendab soojusjuhtivust ning soojuskiirgust tühiste väärtusteni. Tööolukorras imbub õhk läbi õhupalli kesta kiirusega $\lambda = \SI{500}{g\per\s}$. Õhurõhk ja temperatuur lennukõrgusel on $p_0 = \SI{100}{\kilo\Pa}$ ja $T_0 = \SI{10}{\degreeCelsius}$. Propaani kütteväärtus $k = \SI{50}{MJ/kg}$. Õhu keskmine molaarmass $\mu = \SI{29}{g\per\mole}$ ning soojusmahtuvus konstantsel rõhul $C_p = \SI{1.0}{\frac{kJ}{K\cdot kg}}$. Universaalne gaasikonstant on $R = \SI{8.3}{\frac{J}{K\cdot mol}}$.

Selleks, et kuumaõhupall õhuks püsiks, peab sees olev õhk olema piisavalt madala tihedusega, et üleslükkejõud tasakaalustaks kuumaõhupalli raskusjõu. Kui õhupalli sees on õhk temperatuuril $T$, siis õhupalli pooridest imbub välja soe õhk temperatuuril $T$, samas kui õhupalli siseneb õhk temperatuuril $T_0$. Propaani põletamine peab vastava soojuskao kompenseerima.


[User]
In this problem we’ll verify some of the basic facts stated above.

      (a) Prove the statement of idea 5 for the case of elliptical orbits.

     (b) Using this result, prove the vis-viva equation

                                                  2            2 1
                                                 v = GM         −
                                                               r a
          which is often used in rocketry.

      (c) Prove Kepler’s third law.

       Remark: Scaling Symmetry

       There’s a variant of Kepler’s third law for unbound orbits. Suppose a planet is right next
       to the Sun at time t = 0, but has a large initial radial velocity, so that it has zero total
       energy. Then its distance to the Sun evolves as r(t) ∝ t2/3 , like how a ∝ T 2/3 for bound orbits.

       Both of these results come from the scaling symmetry of inverse square force laws: any
       solution to Newton’s second law remains a solution if you multiply all distances by 4 and
       all times by 8. The widest-reaching application of this idea is to the whole universe itself.
       If it contains only matter, which started at the origin at time t = 0, and it expands under
       gravity with zero total energy, then its “scale factor” evolves as a(t) ∝ t2/3 . This was a good
       description of our universe for most of its lifetime, but in the past few billion years the effects
       of dark energy took over, accelerating the expansion. We’ll revisit cosmology in X3.

Solution:
(a) Let the closest approach distance be r1 , farthest be r2 . For simplicity, let’s define
         the specific angular momentum J = L/m and specific energy ϵ = 2E/m. Then angular
         momentum conservation and energy conservation give
                                                                      2GM         2GM
                                J = r1 v1 = r2 v2 ,       ϵ = v12 −       = v22 −     .
                                                                       r1          r2

          From these facts, we see that the equation

                                                        J 2 2GM
                                                   ϵ=      −
                                                        r2    r
          is satisfied for r = r1 and r = r2 . This equation is equivalent to the quadratic ϵr2 + 2GM r −
          J 2 = 0, and applying Vieta’s formulas gives r1 + r2 = −2GM/ϵ = −GM m/E. Using
          r1 + r2 = 2a and rearranging gives the result.

     (b) This follows immediately from rearranging the statement of energy conservation,

                                                GM m  1       GM m
                                            −        = mv 2 −      .
                                                 2a   2        r

      (c) This follows from the geometrical facts stated in idea 6. First, the area swept out per unit
          time is L/2m, so (L/2m)T = πab.
          To show that
                  √ T only depends on a, we need to eliminate L and b. We  know that a = (r1 +r2 )/2
                                                                         √
          and a = b2 + d2 , where d = (r1 − r2 )/2. This implies that b = r1 r2 , and applying Vieta’s
          formulas to the quadratic in part (a) gives
                                                r
                                                   −J 2    L/m
                                           b=           =p          .
                                                    ϵ      −2E/m

          Plugging this into our initial result, we have

                                                  L         L/m
                                                    T = πa p     ,
                                                 2m         GM/a

          which implies T 2 = 4π 2 a3 /GM ∝ a3 , as desired.

       Remark: Scaling Symmetry

       There’s a variant of Kepler’s third law for unbound orbits. Suppose a planet is right next
       to the Sun at time t = 0, but has a large initial radial velocity, so that it has zero total
       energy. Then its distance to the Sun evolves as r(t) ∝ t2/3 , like how a ∝ T 2/3 for bound orbits.

       Both of these results come from the scaling symmetry of inverse square force laws: any
       solution to Newton’s second law remains a solution if you multiply all distances by 4 and
       all times by 8. The widest-reaching application of this idea is to the whole universe itself.
       If it contains only matter, which started at the origin at time t = 0, and it expands under
       gravity with zero total energy, then its “scale factor” evolves as a(t) ∝ t2/3 . This was a good
       description of our universe for most of its lifetime, but in the past few billion years the effects
       of dark energy took over, accelerating the expansion. We’ll revisit cosmology in X3.


[User]
Basic moment of inertia computations.
      (a) Compute the moment of inertia for an Lx × Ly rectangular plate about an axis passing
          perpendicular to it through the center.
     (b) Compute the moment of inertia for a uniform disc of radius R and mass M , about an axis
         perpendicular to it through its center. What about an axis lying in the disc, passing through
         its center?
        (c) Compute the moment of inertia of a uniform solid cone of mass M , with height H and a base
            of radius R, about its symmetry axis.
[2] Problem 5 (F = ma 2016 24). The moment of inertia of a uniform equilateral triangle with
    mass m and side length a about an axis through one of its sides and parallel to that side is ma2 /8.
    What is the moment of inertia of a uniform regular hexagon of mass m and side length a about an
    axis through two opposite vertices?

    3      Rotational Dynamics
    In this section we’ll consider some dynamic problems involving rotation.

         Idea 3: Angular Momentum

         For a system of particles we define the angular momentum and torque
                                     X                 X                dL
                                 L=      ri × pi , τ =    ri × Fi , τ =    .
                                                                        dt
                                      i                    i

         Using the first part of idea 1, we may write the angular momentum of a rigid body as

                                            L = I ω,   K = Iω 2

  where I is the moment of inertia about the instantaneous axis of rotation. Alternatively,
  using the second part,
                                                       1         1
                     L = ICM ω + rCM × M vCM ,      K = ICM ω 2 + M vCM

                                                       2         2
  where M is the total mass. Both forms are useful in different situations. Systems cannot
  exert torques on themselves, provided they obey the strong form of Newton’s third law: the
  force between two objects is equal and opposite, and directed along the line joining them.

  Idea 4
  The idea above refers to taking torques about a fixed point, but often it is easier to consider
  a moving point P . Let L be the angular momentum about point P in the frame of P , i.e. the
  frame whose axes don’t rotate, but whose origin follows P around. Working in this frame
  will produce fictitious forces, since P can accelerate. Such forces act at the center of mass,
  just like gravity.

  The upshot is that if P is the center of mass, then the fictitious force in the frame of P will
  produce no “fictitious torque”. So it’s safe to use τ = dL/dt about either a fixed point, or in
  the frame of the center of mass.

  Idea 5
  There is a third, more confusing way of applying τ = dL/dt that you might rarely see:
  taking torques about the instantaneous center of rotation. In general, this doesn’t work,
  because the instantaneous center of rotation can accelerate, producing an extra fictitious
  torque as mentioned above.

  However, it turns out this procedure gives the correct answer if the object is instantaneously
  at rest. That’s why taking torques about the contact point for the spool in M2 to find the
  initial angular acceleration was valid. It wouldn’t have been valid at any instant afterward,
  after the spool had picked up some velocity.

  For more discussion of this subtlety, which isn’t mentioned in any textbooks I know of, see
  the paper Moments to be cautious of .

  Example 3: KK 6.13

  A mass m is attached to a post of radius R by a string. Initially it is a distance r from the
  center of the post and is moving tangentially with speed v0 . In case (a) the string passes
  through a hole in the center of the post at the top. The string is gradually shortened by
  drawing it through the hole. In case (b) the string wraps around the outside of the post.
  Ignore gravity.

       For each case, find the final speed of the mass when it hits the post.

       Solution
      In case (a), the energy isn’t conserved, since work is done on the mass as it moves inward.
      (Physically, we can see this by noting there could be a weight slowly descending on the other
      end of the string.) However, angular momentum conservation says Rv = rv0 , so v = rv0 /R.

       If you don’t believe in angular momentum conservation yet, it’s not too hard to show this
       with F = ma as well. Let the tangential and radial speeds of the mass be vt and vr , where
       vr ≪ vt . Since vr is nonzero, there is a component of acceleration parallel to the velocity,

                                               T        v 2 vr
                                                 sin θ ≈ t
                                               m         r vt
       and this is equal to the rate of change of speed, which to first order in vr /vt is dvt /dt. Thus,

                                            dvt   vr vt    vt dr
                                                =       =−
                                            dt     r       r dt
       from which we conclude rvt is constant, as expected. (As mentioned in M2, you never need
       ideas like torque and angular momentum. Life is just harder without them.)

       In case (b), the angular momentum about the axis of the pole isn’t conserved, since the
       tension force has a lever arm about that axis. However, the mass’s energy is conserved. A
       simple physical way to see this is to note that the massless string can’t store any energy,
       and the post doesn’t do work on the string, which means the string can’t do any work on
       the mass. Thus, the final speed is just v = v0 . (Of course, if you don’t believe in energy
       conservation, you could get the same result by showing that the trajectory of the mass is
       always perpendicular to the string, though this takes more work.)

[2] Problem 6 (KK 6.9). A heavy uniform bar of mass M rests on top of two identical rollers which
    are continuously turned rapidly in opposite directions, as shown.

    The centers of the rollers are a distance 2ℓ apart. The coefficient of friction between the bar and
    the roller surfaces is µ, a constant independent of the relative speed of the two surfaces. Initially
    the bar is held at rest with its center at distance x0 from the midpoint of the rollers. At time t = 0
    it is released. Find the subsequent motion of the bar.

[2] Problem 7 (BAUPC). A mass is connected to one end of a massless string, the other end of which
    is connected to a very thin frictionless vertical pole. The string is initially wound completely around
    the pole, in a very large number of small horizontal circles, with the mass touching the pole. The
    mass is released, and the string gradually unwinds. What angle does the string make with the pole
    when it becomes completely unwound? (Though the setup is similar to that of example 3, you can’t
    ignore gravity here.)

       Example 4: MPPP 49

       A uniform cylinder of mass M and radius R is attached to two identical strings. The strings
       are wound around the cylinder as shown, and their free ends are fastened to the ceiling.

       A third cord is attached to and wound around the middle of the cylinder, and a mass M is
       attached to the other side. There is sufficient friction so that the strings do not slip. Find
       the acceleration of the mass immediately after release.

       Solution
       Let a be the downward acceleration of the center of mass of the rod, let T1 be the total
       tension in the first two strings, and let T2 be the tension in the third. The rod rolls without
       slipping about its contact axis with the first two strings, which means the downward
       acceleration of the mass is amass = 2a.

       The Newton’s second law equations are thus

                                  M a = T2 + M g − T1 ,    2M a = M g − T2

  for the rod and mass. Taking torques about the axis of the rod gives

                                      (T1 + T2 )R = M R2 α

  and using a = αR converts this to

                                        M a = 2T1 + 2T2 .

  We now have three equations in three unknowns, so we can straightforwardly solve to find
  a = (6/11)g. This implies that the acceleration of the mass is

                                          amass =       g.

  Done, right? No, this is the wrong answer! Since the acceleration is greater than free fall,
  the tension T2 must be negative. But a string can’t support a negative tension, so it instead
  goes slack. The mass thus free falls, so amass = g.

  In retrospect, we could have seen this conclusion with less work. Suppose the mass were
  not attached. Then the acceleration of the rod can be computed with the standard rolling
  without slipping formula,
                                    g sin θ    g
                               a=           =     ,      I = βM R2 .
                                    1+β       1+β

  For any (axially symmetric) mass distribution in the rod, we have 0 ≤ β ≤ 1. The acceleration
  of the part where the mass would have been attached is hence
                                                  2g
                                       amass =       ≥ g.
                                                 1+β
  This implies that any string we attach there must go slack immediately after release.

  Example 5

  If you’re riding a bike and need to stop quickly, what are the advantages and disadvantages
  of using the front brake versus the rear brake?

  Solution
  Work in the reference frame moving with the bike. In this frame, the backward friction force
  is balanced by a forward friction force on the center of mass; the combination of the two
  produces a torque that tends to lift the rear wheel off the ground. If you use the front brake,
  you can stop more quickly, because the normal force on the front tire stays higher. But if
  you brake too hard with the front brake, you could flip yourself over the handlebars. This
  can’t happen when using the rear brake alone, because the brake stops doing anything the
  moment the rear wheel lifts off the ground.

        Idea 6
        It is often useful in rotational dynamics to treat the rotational and linear motion of a rigid
        body conceptually separately.

        Example 6: F = ma 2018 B23

       Two particles with mass m1 and m2 are connected by a massless rigid rod of length L and
       placed on a horizontal frictionless table. At time t = 0, the first mass receives an impulse
       perpendicular to the rod, giving it speed v. At this moment, the second mass is at rest.
       When is the next time the second mass is at rest?

        Solution
       The motion is the superposition of two motions: uniform translation of both masses with
       speed m1 v/(m1 + m2 ) and circular motion about the common center of mass, where the two
       masses have speeds m2 v/(m1 + m2 ) and m1 v/(m1 + m2 ), respectively. This ensures that
       the second mass begins at rest and the first mass has speed v.

       The circular part of the motion determines when the second mass will be at rest again.
       The radius of the circle the second mass makes is its distance from the center of mass,
       Lm1 /(m1 + m2 ). This gives a period of

                                            2πLm1 /(m1 + m2 )   2πL
                                       t=                     =     .
                                             m1 v/(m1 + m2 )     v

[2] Problem 8 (KK 6.14). A uniform stick of mass M and length ℓ is suspended horizontally with
    end B on the edge of a table, while end A is held by hand.

    Point A is suddenly released. Right after release, find the vertical force at B, as well as the downward
    acceleration of point A. You should find a result greater than g. Explain how this can be possible,

                   01T
    given that gravity is the only downward external force in the problem.

Solution:
(a) For a uniform cylinder of length L, I = −L/2 x2 (M/L)dx = 12      M L2 . So by the

          perpendicular axis theorem, the answer is          M (L2x + L2y ) .

                           RR 2      2            2
     (b) We see that I =    0 r (M/πR )2πrdr = M R /2 . For an axis lying in the disc, the answer

          is half as much, M R2 /4 , by the perpendicular axis theorem.

       (c) First off, we know the height H doesn’t matter, because we can stretch the cone alone its
           symmetry axis without changing the answer. Letting the density be ρ, we can integrate over
           the discs making up the cone,
                                        Z          Z                Z H
                                                       1                  1
                                   I=       dI =         (dm)r2 =           (ρπr2 dh)r2
                                                       2             0    2

          where the radius of the disc at height h is, in some set of coordinates, r(h) = R(h/H). Plugging
          this in, we get                      Z H
                                                    π 4 h4            πρR4 H
                                          I=          ρR 4 dh =              .
                                                0   2      H            10

          It remains to find ρ, by noting that
                                           Z H
                                                          R2 h 2      πρR2 H
                                  Z                          Z

                            M = dm =           ρπr dh = ρπ 2   h dh =        .
                                            0             H 0           3
          Plugging in the result for ρ gives

                                                      I= M R2 .

          This makes sense, as it’s somewhat less than the moment of inertia of a uniform disc; a cone
          has comparatively more of its mass closer to the axis.
[2] Problem 5 (F = ma 2016 24). The moment of inertia of a uniform equilateral triangle with
    mass m and side length a about an axis through one of its sides and parallel to that side is ma2 /8.
    What is the moment of inertia of a uniform regular hexagon of mass m and side length a about an
    axis through two opposite vertices?
    Solution.√Four of the triangles
                               √    contribute (m/6)a2 /8, and the other two contribute (m/6)a2 /8 −
    (m/6)(a/2 3)2 + (m/6)(a/ 3)2 = 3(m/6)a2 /8 by two applications of the parallel axis theorem.
    Thus, the total is (m/6)a2 (4/8 + 3/4) = 5ma2 /24 .

    3    Rotational Dynamics
    In this section we’ll consider some dynamic problems involving rotation.

        Idea 3: Angular Momentum

        For a system of particles we define the angular momentum and torque
                                     X                     X                      dL
                                L=        ri × pi ,   τ=         ri × Fi ,   τ=      .
                                                                                dt
                                      i                      i

        Using the first part of idea 1, we may write the angular momentum of a rigid body as

                                              L = I ω,     K = Iω 2

        where I is the moment of inertia about the instantaneous axis of rotation. Alternatively,
        using the second part,
                                                                    1         1
                           L = ICM ω + rCM × M vCM ,             K = ICM ω 2 + M vCM

                                                                    2         2
        where M is the total mass. Both forms are useful in different situations. Systems cannot
        exert torques on themselves, provided they obey the strong form of Newton’s third law: the
        force between two objects is equal and opposite, and directed along the line joining them.

        Idea 4
        The idea above refers to taking torques about a fixed point, but often it is easier to consider
        a moving point P . Let L be the angular momentum about point P in the frame of P , i.e. the
        frame whose axes don’t rotate, but whose origin follows P around. Working in this frame

  will produce fictitious forces, since P can accelerate. Such forces act at the center of mass,
  just like gravity.

  The upshot is that if P is the center of mass, then the fictitious force in the frame of P will
  produce no “fictitious torque”. So it’s safe to use τ = dL/dt about either a fixed point, or in
  the frame of the center of mass.

  Idea 5
  There is a third, more confusing way of applying τ = dL/dt that you might rarely see:
  taking torques about the instantaneous center of rotation. In general, this doesn’t work,
  because the instantaneous center of rotation can accelerate, producing an extra fictitious
  torque as mentioned above.

  However, it turns out this procedure gives the correct answer if the object is instantaneously
  at rest. That’s why taking torques about the contact point for the spool in M2 to find the
  initial angular acceleration was valid. It wouldn’t have been valid at any instant afterward,
  after the spool had picked up some velocity.

  For more discussion of this subtlety, which isn’t mentioned in any textbooks I know of, see
  the paper Moments to be cautious of .

  Example 3: KK 6.13

  A mass m is attached to a post of radius R by a string. Initially it is a distance r from the
  center of the post and is moving tangentially with speed v0 . In case (a) the string passes
  through a hole in the center of the post at the top. The string is gradually shortened by
  drawing it through the hole. In case (b) the string wraps around the outside of the post.
  Ignore gravity.

  For each case, find the final speed of the mass when it hits the post.

       Solution
       In case (a), the energy isn’t conserved, since work is done on the mass as it moves inward.
       (Physically, we can see this by noting there could be a weight slowly descending on the other
       end of the string.) However, angular momentum conservation says Rv = rv0 , so v = rv0 /R.

       If you don’t believe in angular momentum conservation yet, it’s not too hard to show this
       with F = ma as well. Let the tangential and radial speeds of the mass be vt and vr , where
       vr ≪ vt . Since vr is nonzero, there is a component of acceleration parallel to the velocity,

                                                T        v 2 vr
                                                  sin θ ≈ t
                                                m         r vt
       and this is equal to the rate of change of speed, which to first order in vr /vt is dvt /dt. Thus,

                                            dvt   vr vt    vt dr
                                                =       =−
                                            dt     r       r dt
       from which we conclude rvt is constant, as expected. (As mentioned in M2, you never need
       ideas like torque and angular momentum. Life is just harder without them.)

       In case (b), the angular momentum about the axis of the pole isn’t conserved, since the
       tension force has a lever arm about that axis. However, the mass’s energy is conserved. A
       simple physical way to see this is to note that the massless string can’t store any energy,
       and the post doesn’t do work on the string, which means the string can’t do any work on
       the mass. Thus, the final speed is just v = v0 . (Of course, if you don’t believe in energy
       conservation, you could get the same result by showing that the trajectory of the mass is
       always perpendicular to the string, though this takes more work.)

[2] Problem 6 (KK 6.9). A heavy uniform bar of mass M rests on top of two identical rollers which
    are continuously turned rapidly in opposite directions, as shown.

    The centers of the rollers are a distance 2ℓ apart. The coefficient of friction between the bar and
    the roller surfaces is µ, a constant independent of the relative speed of the two surfaces. Initially
    the bar is held at rest with its center at distance x0 from the midpoint of the rollers. At time t = 0
    it is released. Find the subsequent motion of the bar.

    Solution. Let N1 be the normal force from the right roller, and N2 be the one from the left roller.
    Since there is no acceleration in the y-direction, we have that N1 + N2 = M g. Also, since the bar
    is not rotating, we have that the torque about the center is 0, so N1 (ℓ − x0 ) = N2 (ℓ + x0 ). One
    quickly sees that the solution to this system is

                                         M g(ℓ + x0 )                M g(ℓ − x0 )
                                  N1 =                ,       N2 =                .
                                             2ℓ                          2ℓ

    Now, the friction force from the right roller points to the left with magnitude N1 µ, and the one
    from the left roller points to the right with magnitude N2 µ. Therefore, the total net force on this
    system is
                                                                x0
                                             N1 µ − N2 µ = M gµ
                                                                 ℓ
                                                                             p
    to the left. This is simple harmonic motion with angular frequency ω = µg/ℓ. This neat system
    is called a “friction oscillator”, or “Timoshenko oscillator”.

[2] Problem 7 (BAUPC). A mass is connected to one end of a massless string, the other end of which
    is connected to a very thin frictionless vertical pole. The string is initially wound completely around
    the pole, in a very large number of small horizontal circles, with the mass touching the pole. The
    mass is released, and the string gradually unwinds. What angle does the string make with the pole
    when it becomes completely unwound? (Though the setup is similar to that of example 3, you can’t
    ignore gravity here.)

    Solution. Let the string have length ℓ, a final angle of θ with the pole, and final angular velocity
    ω = v/ℓ sin θ. As it unwinds, there is no source of energy loss so energy is conserved.

                                                gℓ cos θ = v 2

    The components of the force on the mass from the string is a horizontal component for a centripetal
    force, and a vertical component to balance gravity,

                                    T sin θ = mω 2 ℓ sin θ,   T cos θ = mg.

    Solving yields
                                                        v2        2
                                           tan θ =            =       .
                                                     gℓ sin θ   tan θ
                     √
    Thus, θ = arctan( 2) ≈ 54.74◦ .

       Example 4: MPPP 49

       A uniform cylinder of mass M and radius R is attached to two identical strings. The strings
       are wound around the cylinder as shown, and their free ends are fastened to the ceiling.

       A third cord is attached to and wound around the middle of the cylinder, and a mass M is
       attached to the other side. There is sufficient friction so that the strings do not slip. Find
       the acceleration of the mass immediately after release.

  Solution
  Let a be the downward acceleration of the center of mass of the rod, let T1 be the total
  tension in the first two strings, and let T2 be the tension in the third. The rod rolls without
  slipping about its contact axis with the first two strings, which means the downward
  acceleration of the mass is amass = 2a.

  The Newton’s second law equations are thus

                            M a = T2 + M g − T1 ,      2M a = M g − T2

  for the rod and mass. Taking torques about the axis of the rod gives

                                      (T1 + T2 )R = M R2 α

  and using a = αR converts this to

                                        M a = 2T1 + 2T2 .

  We now have three equations in three unknowns, so we can straightforwardly solve to find
  a = (6/11)g. This implies that the acceleration of the mass is

                                          amass =        g.

  Done, right? No, this is the wrong answer! Since the acceleration is greater than free fall,
  the tension T2 must be negative. But a string can’t support a negative tension, so it instead
  goes slack. The mass thus free falls, so amass = g.

  In retrospect, we could have seen this conclusion with less work. Suppose the mass were
  not attached. Then the acceleration of the rod can be computed with the standard rolling
  without slipping formula,
                                    g sin θ    g
                               a=           =     ,       I = βM R2 .
                                    1+β       1+β

  For any (axially symmetric) mass distribution in the rod, we have 0 ≤ β ≤ 1. The acceleration
  of the part where the mass would have been attached is hence
                                                  2g
                                       amass =       ≥ g.
                                                 1+β
  This implies that any string we attach there must go slack immediately after release.

  Example 5

  If you’re riding a bike and need to stop quickly, what are the advantages and disadvantages
  of using the front brake versus the rear brake?

      Solution
      Work in the reference frame moving with the bike. In this frame, the backward friction force
      is balanced by a forward friction force on the center of mass; the combination of the two
      produces a torque that tends to lift the rear wheel off the ground. If you use the front brake,
      you can stop more quickly, because the normal force on the front tire stays higher. But if
      you brake too hard with the front brake, you could flip yourself over the handlebars. This
      can’t happen when using the rear brake alone, because the brake stops doing anything the
      moment the rear wheel lifts off the ground.

      Idea 6
      It is often useful in rotational dynamics to treat the rotational and linear motion of a rigid
      body conceptually separately.

      Example 6: F = ma 2018 B23

      Two particles with mass m1 and m2 are connected by a massless rigid rod of length L and
      placed on a horizontal frictionless table. At time t = 0, the first mass receives an impulse
      perpendicular to the rod, giving it speed v. At this moment, the second mass is at rest.
      When is the next time the second mass is at rest?

      Solution
      The motion is the superposition of two motions: uniform translation of both masses with
      speed m1 v/(m1 + m2 ) and circular motion about the common center of mass, where the two
      masses have speeds m2 v/(m1 + m2 ) and m1 v/(m1 + m2 ), respectively. This ensures that
      the second mass begins at rest and the first mass has speed v.

      The circular part of the motion determines when the second mass will be at rest again.
      The radius of the circle the second mass makes is its distance from the center of mass,
      Lm1 /(m1 + m2 ). This gives a period of

                                           2πLm1 /(m1 + m2 )   2πL
                                      t=                     =     .
                                            m1 v/(m1 + m2 )     v

[2] Problem 8 (KK 6.14). A uniform stick of mass M and length ℓ is suspended horizontally with
    end B on the edge of a table, while end A is held by hand.

   Point A is suddenly released. Right after release, find the vertical force at B, as well as the downward
   acceleration of point A. You should find a result greater than g. Explain how this can be possible,

    given that gravity is the only downward external force in the problem.

    Solution. We take torques about B, applying idea 5. Note that τ = M gℓ/2 = Iα = 13 M ℓ2 α,
    so α = 32 gℓ . Thus, the instantaneous acceleration of the CM is αℓ/2 = 34 g down. Therefore,
    M g − F = 3M g/4, so F = M g/4 . Furthermore, the acceleration of point A is 3g/2 down. This is
    indeed greater than g.
       The reason this is possible is that the stick is a rigid object, so it supports internal shear stresses,
    which keep the whole body moving as one piece. If you consider a small piece of the rod near the
    end, gravity provides a downward acceleration g, while a downward shear stress from the rest of
    the rod provides the remaining downward acceleration g/2.


[User]
9. This entire booklet must be returned.

                            Table of Information

            Speed of light in vacuum c    =   3.00 × 108 m·s−1
                   Planck’s constant h
   Universal constant of Gravitation G

                                  C1
       Magnitude of electron charge e
                   Mass of electron me
                                          =
                                          =
                                          =
                                          =
                                                  6
                                              6.63 × 10−34 J·s
                                              6.67 × 10−11 N·m2 ·kg
                                              1.60 × 10−19 C
                                                                   −2

                                              9.11 × 10−31 kg = 0.51 MeV·c−2
         Stefan-Boltzmann constant σ      =   5.67 × 10−8 W·m−2 · K−4
                       Value of 1/4π0    =   9.00 × 109 N·m2 ·C−2
              Permeability constant µ0    =   4π × 10−7 H·m−1

                HB
             Universal Gas Constant R
                    Molar mass of air
               Rest mass of proton mp
                1 atomic mass unit 1 u
                                          =
                                          =
                                          =
                                          =
                                              8.31 J· K−1 ·mole−1
                                              29.0 kg·kmol−1
                                              1.67 × 10−27 kg
                                              1.661 × 10−27 kg

           HOMI BHABHA CENTRE FOR SCIENCE EDUCATION
                  Tata Institute of Fundamental Research
              V. N. Purav Marg, Mankhurd, Mumbai, 400 088

                         INPhO - 2016 - Final Solutions
    Please note that alternate/equivalent methods and different way of
                   expressing final solutions may exist.

Solution:
9. This entire booklet must be returned.

                            Table of Information

            Speed of light in vacuum c    =   3.00 × 108 m·s−1
                   Planck’s constant h
   Universal constant of Gravitation G

                                  C1
       Magnitude of electron charge e
                   Mass of electron me
                                          =
                                          =
                                          =
                                          =
                                                  6
                                              6.63 × 10−34 J·s
                                              6.67 × 10−11 N·m2 ·kg
                                              1.60 × 10−19 C
                                                                   −2

                                              9.11 × 10−31 kg = 0.51 MeV·c−2
         Stefan-Boltzmann constant σ      =   5.67 × 10−8 W·m−2 · K−4
                       Value of 1/4π0    =   9.00 × 109 N·m2 ·C−2
              Permeability constant µ0    =   4π × 10−7 H·m−1

                HB
             Universal Gas Constant R
                    Molar mass of air
               Rest mass of proton mp
                1 atomic mass unit 1 u
                                          =
                                          =
                                          =
                                          =
                                              8.31 J· K−1 ·mole−1
                                              29.0 kg·kmol−1
                                              1.67 × 10−27 kg
                                              1.661 × 10−27 kg

           HOMI BHABHA CENTRE FOR SCIENCE EDUCATION
                  Tata Institute of Fundamental Research
              V. N. Purav Marg, Mankhurd, Mumbai, 400 088

                         INPhO - 2016 - Final Solutions
    Please note that alternate/equivalent methods and different way of
                   expressing final solutions may exist.

INPhO 2016 Problem 9