[User]
5. Two friends, Amina (A) and Beena (B), are sitting at diametrically opposite points of a merry-
   go-round (taken as a circular disk in the horizontal plane) of radius R that is rotating at constant
   angular speed ω in the anticlockwise direction, when viewed from the top (see ﬁgure below).
                                                                                D
  When Amina is at the position A (as shown in the ﬁgure),                                   C
                                                                                ω
  she throws a ball with velocity ⃗u (relative to the merry-
  go-round) in such a manner that Beena catches it when
  she reaches the position C (∠BAC = α). Here ⃗u makes                                 90◦
                                                                             α
  an angle θ with respect to the horizontal, and ϕ is the           A                            B
                                                                            R      O
  angle made by the horizontal projection of ⃗u with respect
  to the line AB. Neglect air resistance, friction, and the
  eﬀect of throwing or catching the ball on the speed of
  the merry-go-round.
                                                                                Top view
   (a) [6 marks] Determine u, θ and ϕ, in terms of R, ω, α, and other relevant quantities.
   (b) [3 marks] If Amina throws the ball with ϕ = 60°, and appropriate values of θ and u such
       that Beena can catch it, what is the magnitude of the displacement, s, of the ball when it is
       caught by Beena? For this part only, take R = 1.5 m, and it is enough to state your answer
       within a range of 0.5 m.
   (c) [0.5 marks] Determine the speed of throwing uD if Beena catches the ball at the point D
       (∠BOD = 90◦ ), instead of C.
   (d) [3 marks] What should be the angular speed ωm of the merry-go-round for which the speed
       of throwing uD will be minimum for Beena to catch the ball at the position D? What is this
       minimum speed of throwing um ?
   (e) [2.5 marks] Consider the case when Amina throws the ball when she is at A, and catches it
       herself when she reaches the point B (Beena
                                              √    is not involved in this case). Take the angular
       speed of the merry-go-round to be ω = g/R. Find u, θ and ϕ in this case.
                        **** END OF THE QUESTION PAPER ****

Page 4   Code 65           Questions    Last six digits of Roll No.:

     Space for rough work — will NOT be submitted for evaluation

Solution:
5. Two friends, Amina (A) and Beena (B), are sitting at diametrically opposite points of a merry-
   go-round (taken as a circular disk in the horizontal plane) of radius R that is rotating at constant
   angular speed ω in the anticlockwise direction, when viewed from the top (see ﬁgure below).
                                                                                D
  When Amina is at the position A (as shown in the ﬁgure),                                       C
                                                                                ω
  she throws a ball with velocity ⃗u (relative to the merry-
  go-round) in such a manner that Beena catches it when
  she reaches the position C (∠BAC = α). Here ⃗u makes                                     90◦
                                                                                α
  an angle θ with respect to the horizontal, and ϕ is the               A                            B
                                                                                R      O
  angle made by the horizontal projection of ⃗u with respect
  to the line AB. Neglect air resistance, friction, and the
  eﬀect of throwing or catching the ball on the speed of
  the merry-go-round.
                                                                                Top view
   (a) [6 marks] Determine u, θ and ϕ, in terms of R, ω, α, and other relevant quantities.

         Solution:
         Point of throwing: A; Point of catching: C
         Position of C at instant of projection: B

         We take the point A as the origin and the x-axis along the diameter AB. The y-axis is in
         the horizontal plane, perpendicular to AB. The z-axis is taken along vertical direction.
                   y
                                          C                             z
                                                 ω
                       u                                                        u
                                   R
                        ϕ α          β                                A    θ           y
                  A                                   x
                           R      O     P        B
                                                                        ϕ
                                                                          x

         Given, ω = angular speed of rotation; ⃗u = velocity of throwing

Page 7     Code 65                       Questions

         θ = Projection angle with respect to horizontal
         ϕ = Projection angle with respect to diameter AB (x-axis)
         α = ∠BAC       =⇒ β = ∠BOC = 2α
                                                            Rβ     Rβ   2α
         Time of ﬂight = time taken for B to reach C = T =      =     =
                                                             vs    Rω   ω
         Equations of motion along three directions:
                  x : ux · T = AP
                                            2α
                      =⇒ (u cos θ cos ϕ) ·     = R + R cos β = R(1 + cos 2α) = 2R cos2 α
                                            ω
                                           Rω
                      =⇒ u cos θ cos ϕ =       cos2 α                                           (5.1)
                                            α
                  y : (uy − Rω) · T = CP
                                                  2α
                      =⇒ (u cos θ sin ϕ − Rω) ·       = R sin β = 2R sin α cos α
                                                   ω
                                           Rω                      Rω
                      =⇒ u cos θ sin ϕ =       sin α cos α + Rω =      [sin α cos α + α]        (5.2)
                                            α                       α
                                1
                  z : uz · T − gT 2 = 0
                                2
                                     gT     gα
                      =⇒ u sin θ =        =                                                     (5.3)
                                      2      ω
         Dividing eq. (5.2) by eq. (5.1),
                                             sin α cos α + α
                                    tan ϕ =                  = tan α + α sec2 α
                                                  cos2 α
                               =⇒ ϕ = tan−1 (tan α + α sec2 α)                                  (5.4)
         Squaring eqs. (5.1), (5.2), (5.3) and adding,

                                      Rω 2  4                                              gα 2
           u2 cos2 θ + u2 sin2 θ =            cos α + sin2 α cos2 α + α2 + 2α sin α cos α +
                                       α                                                      ω
                                         2
                                      Rω  2                                   gα 2
                        =⇒ u2 =               cos α + 2α sin α cos α + α2 +                     (5.5)
                                       α                                        ω
                                  "                                                  #1/2
                                     gα 2  Rω 2
                      =⇒ u =                +          cos2 α + 2α sin α cos α + α2             (5.6)
                                      ω          α

         From (5.3) and (5.6),
                              "                                                #−1/2 
                           gα     gα 2  Rω 2
               θ = sin−1               +         cos2 α + 2α sin α cos α + α2                 (5.7)
                           ω       ω        α

   (b) [3 marks] If Amina throws the ball with ϕ = 60°, and appropriate values of θ and u such
       that Beena can catch it, what is the magnitude of the displacement, s, of the ball when it is
       caught by Beena? For this part only, take R = 1.5 m, and it is enough to state your answer
       within a range of 0.5 m.

         Solution:
         The displacement of the ball is the length of AC = s = 2R cos α.
         Thus we need to determine α when ϕ = 60°. Equation (5.4) can be used for this. Note
         that values of θ and u are not needed.
         Putting ϕ = 60° in equation (5.4), we have
                                                                      √
                                  f (α) = tan α + α sec2 α = tan 60° = 3
         This equation cannot be solved analytically. We use trial values of α to ﬁnd the solution

Page 8     Code 65                            Questions              Last six digits of Roll No.:

         by interpolation.

                                             1   π   2 2            √
                                  f (π/6) = √ +      √     = 1.275 < 3
                                              3 6      3
                                               π √ 2            √
                                  f (π/4) = 1 + ( 2) = 2.571 > 3
                                               4
         Thus
                                         π        π
                                            <α<
                                         6
                                         √        4
                                           3            1
                                      =⇒     > cos α > √
                                          2              2
                                            √
                                              3                  1
                                      =⇒ 2R     > 2R cos α > 2R √
                                             2                    2
                                         √           √
                                      =⇒ 3R > s > 2R
         Putting R = 1.5 m,
                                              2.1 metre < s < 2.6 metre
         Any answer that encloses the actual value of 2.4 m and has a range ≤ 0.5 m is acceptable.

   (c) [0.5 marks] Determine the speed of throwing uD if Beena catches the ball at the point D
       (∠BOD = 90◦ ), instead of C.

         Solution:
                                                       π
         This is a special case of the above, where α = . Using the above results,
                                                       4
                                      "                               #
                                         gπ 2  4Rω 2  1 π π 2  1/2
                                uD =           +            + +                                     (5.8)
                                          4ω        π      2 4    16

   (d) [3 marks] What should be the angular speed ωm of the merry-go-round for which the speed
       of throwing uD will be minimum for Beena to catch the ball at the position D? What is this
       minimum speed of throwing um ?

         Solution:
         This can be determined by ﬁnding the minimum of uD , or equivalently, u2D . From (5.8),
                                        gπ 2  4Rω 2  1 π π 2
                                  2
                                 uD =         +            + +
                                         4ω         π     2 4     16

                         d(u2D )               (gπ)2 32R2 ωm 1 π π 2
                      ∴             = 0 =⇒ −      3
                                                     +           + +         =0
                          dω ωm                 8ωm     π2      2 4     16
                                                              g2π4
                                                  4
                                              =⇒ ωm =
                                                           2
                                                              1 π π2
                                                      256R      + +
                                                              2 4      16
                                                                     −1/4 r
                                                       π 1 π π2               g
                                              =⇒ ωm =        + +
                                                       4 2 4       16         R
         Also,

                              d2 (u2D )   3(gπ)2 32R2 1 π π 2
                                        =       +       + +      > 0.
                               dω 2 ωm       4
                                           8ωm    π2   2 4  16
         implying u2D is minimum at ω = ωm .
                                                        1/2                       1/2
                                              1 π π2                       1 π π2
                             ∴ u2m = gR        + +              + gR        + +
                                              2 4  16                      2 4  16

Page 9     Code 65                      Questions

                                                          1/4
                                                 1 π π2           p
                                  =⇒ um =         + +              2gR
                                                 2 4  16

         Alternative solution without calculus
         Observe that
                                                    λ
                                             u2D = 2 + µω 2
                                                    ω
                     gπ 2           2
                                      4R       1 π π2
         where λ =          > 0, µ =             + +         > 0.
                       4               π       2 4      16
         We can write                        √           !2
                                                λ     √         p
                                     u2D =        − ω µ + 2 λµ.
                                               ω
         The ﬁrst term can be made zero by the choice of
                                                      1/4
                                                       λ
                                            ω = ωm =
                                                       µ
                                                       √
         leading to the minimum value of u2D as u2m = 2 λµ. Upon substituting the values of λ
         and µ the desired expressions are obtained.

   (e) [2.5 marks] Consider the case when Amina throws the ball when she is at A, and catches it
       herself when she reaches the point B (Beena
                                              p    is not involved in this case). Take the angular
       speed of the merry-go-round to be ω = g/R. Find u, θ and ϕ in this case.

         Solution:
         This case is NOT a special case of the above.
                   τ    π
         Now T = = .
                   2 pω
         Further, ω = g/R.
         The equations of motion are:
                                                   π                     2Rω   2p
            x : ux · T = AB =⇒ (u cos θ cos ϕ) · = 2R =⇒ u cos θ cos ϕ =     =   gR
                                                   ω                      π    π
                                                                                (5.9)
                                                            p
            y : (uy − Rω) · T = 0 =⇒ u cos θ sin ϕ = Rω = gR                      (5.10)
                        1                        gT    πg   π p
            z : uz T − gT 2 = 0 =⇒ u sin θ =         =    =    gR                 (5.11)
                        2                         2    2ω   2
         Dividing eq. (5.10) by eq. (5.9),
                                           π                π
                                  tan ϕ =    =⇒ ϕ = tan−1 = 57.52◦
                                           2                2
         Squaring eqs. (5.10) and (5.9), and adding,

                                    2    2          4
                                   u cos θ = gR 2 + 1
                                                    π
                                               2         1/2
                                         p     π      4              p
                               =⇒ u = gR           + 2 +1      = 1.97 gR                  (5.12)
                                                4     π
         Using eqs. (5.11) and (5.12),
                                           −1/2
                               π π2    4
                       sin θ =       + 2 +1       =⇒ θ = sin−1 (0.80) = 52.96◦
                               2 4     π

                     **** END OF THE QUESTION PAPER ****

Page 10   Code 65          Questions    Last six digits of Roll No.:

     Space for rough work — will NOT be submitted for evaluation

INPhO 2021 Problem 5


[User]
2. An insulating uniformly charged cylindrical shell                                                  [12]
   of radius a lies with its axis along the z axis. The
   shell’s moment of inertia per unit length about the
   z axis and the surface charge density are I and σ

                                                     0
   respectively. The cylinder is placed in an external
   uniform magnetic field Bex ẑ, and is initially at rest.
   Starting at t = 0 the external magnetic field is

                                 C2
   slowly reduced to zero. What is the final angular
   velocity ω of the cylinder?

    ω=

         HB
   Detailed answers can be found on page numbers:

Solution:
2. The velocity of P is perpendicular to the string right after the impact (refer to the figure).
     We have two unknowns - ω and u.

INPhO 2020 Page 14         Questions & Summary Answers                                 Last four digits of Roll No.:

                                                                           y

                                                                                x

                                                                               ω
                                   θ             P

                                                                CM
                                             u

     We first note that the momentum of CM is preserved perpendicular to the string since the
     string applies force only along its length.

                                             ( Pi ) ⊥ = ( Pf ) ⊥
                                            mv sin θ = 2m(vf )⊥

     where (vf )⊥ is the velocity of the CM perpendicular to the string. CM is located at a distance
      4 from P . So, (vf )⊥ = 4 sin θ − u
     3L                         3Lω

                                       =⇒
                                             3Lω
                                              4
                                                         0
                                                 sin θ − u =
                                                             vo sin θ
                                                                2
                                                                                (5.1)

                               C2
     Applying conservation of angular momentum about P ,

                                    Li = Lf

                                mvo L = 2m
                                               3Lω
                                                     4
                                                         − u sin θ
                                                                      3L
                                                                       4
                                                                            + Icm ω                          (5.2)

         HB
     Solving equations 5.1 and 5.2, we get

                                                     ω=
                                                       vo
                                                        L
                                                     u= √
                                                         vo
                                                       3 2
                                                                                (5.3)

                                                                                (5.4)

                                                                           y

                                                                                x

                                                                               ω
                                   θ        T P

                                                                CM
                                             u

     Designate the tension in the string immediately after the impact as T . This gives a clockwise
     angular acceleration of the rod equal to

                                             3L T cos θ   18T cos θ
                                       α=               =                                                    (5.5)
                                              4 Icm         5mL

INPhO 2020 Page 15         Questions & Summary Answers

                                          3Lα
                                           4
                T
               2m

                                         θ
                                                                          ω 2 3L
                                                                               4
                                             P

     We now calculate the acceleration of P on the rod in the direction of the string, (aP )|| (in
                                             2
     the ground frame) and set it equal to uD .
     To this end, note that the acceleration of the CM in the direction of the string is equal to
     2m . Also, point P is instantaneously rotating about the CM with angular velocity ω in a
      T

     circle of radius 3L
                       4 - this gives a centripetal acceleration of ω × 4 directed towards the CM.
                                                                     2  3L

     Furthermore, the tangential acceleration of P along its trajectory is given by 3Lα
                                                                                4 . Accounting

                                                                 0
     for all these contributions and taking appropriate components, we can write the acceleration
     of point P in the direction of the string as
                                                  27T          3ω 2 L

     Now we know
                                (aP )|| =

                             (aP )|| =
                                         C2
                                          T
                                            +
                                              T
                                                +
                                             2m 10m

                                              27T
                                         2m 10m
                                                      cos2 θ −

                                                  cos2 θ −
                                                           3ω 2 L
                                                             4
                                                                 4
                                                                      sin θ

                                                                  sin θ =
                                                                          u2
                                                                          D
                                                                                (5.6)

                HB
     Hence we get

     Comment [not for grading purposes]:
                                         T =
                                             5mvo2 1
                                              4
                                                       "
                                                        1
                                                      + √
                                                   18D L 2
                                                                      #

     We assumed above that the string is taut. This is consistent with the solution above, as we
     found a non-zero answer for the tension. However, if you do not picture that the geometry of
     the problem requires tension in the string, you can convince yourselves that this is the correct
     assumption, by assuming that the tension is zero and arriving at a contradiction.
     Assume that the string becomes slack. Let the angular velocity of the rod be ω counter-
     clockwise, as seen from top, immediately after the impact. Then, noting that velocity of CM
     remains preserved (because the string is assumed to go slack during the impact) and applying
     conservation of angular momentum about the point of impact, one gets
                                             Li = Lf
                                                       vo L
                                                 0 = 2m     − Icm ω
                                                       2 4
                                                    mvo L
                                                 ω=
                                                    4Icm
     where Icm is the moment of inertia of the rod and the bullet system about its center of mass.
                     2  mL2   mL2     5mL2
     Now, Icm = mL 12 + 8 + 8 = 24 , thereby yielding
                                                           6vo
                                                     ω=
                                                           5L

INPhO 2020 Page 16        Questions & Summary Answers                   Last four digits of Roll No.:

                                                            y

                                                                    x
                                                       vo
                                                       2        ω
                                 θ
                                                      CM

     The velocity of the end of thestring tied to the rod (P ) immediately after the impact,
     therefore, would be given by − 3L4 × 5L − 2 ĵ = − 5 ĵ. But this would mean the string
                                          6vo   vo        2vo

     would get elongated. Contradiction!

                         Detailed answers can be found on page numbers:

                                              0
                     **** END OF THE QUESTION PAPER ****

                             C2
        HB

INPhO 2020 Problem 2


[User]
1. EsCAPE (8 points)

— Päivo Simson.
During a nuclear weapon test, a bomb is
dropped from an airplane at an altitude of
H = 9km and is set to detonate at h = 500m
above the ground. The air drag acting on the
bomb during the fall is negligible. Imme‐
diately after releasing the bomb, the plane
starts to escape the explosion. The crew is
protected from radiation burst by a protect‐
ive screen, but the plane is vulnerable to the
shockwave and needs to be as far as possible
from the detonation point.
i) (1 point) The maximum speed of the air‐
plane for a straight level flight (constant alti‐
tude) is v0. What is the maximum diving
angle so that the speed does not exceed the
speed of sound c? The mass of the airplane
is m, the air drag force is Fd = kv2
, and the
gravitational acceleration is g.
For simplicity, assume from now on that
the airplane stays at a constant altitude, flies
at a constant speed v = 190m/s, and all air
maneuvers are limited by the maximum al‐
lowed lift‐to‐weight ratio n = 2.5. The gravit‐
ational acceleration g = 9.81m/s2
.
ii) (1 point) After releasing the bomb, how
much time does the airplane have before the
radiation burst from the bomb hits it?
iii) (1 point) What is the smallest possible
curvature radius R of the plane’s trajectory
and the corresponding bank angle α (cf fig‐
ure) of the plane?
iv) (3 points)Suggest a trajectory for the fast‐
est escape. Calculate all the parameters that
define the shape of the trajectory and its po‐
sition relative to the detonation point.
v) (2 points) Based on the suggested traject‐
ory, how far is the airplane from the deton‐
ation point when the shockwave hits it? It is
estimated that the safe distance is 25 km. Is
the plane able to escape the explosion? As‐
sume that the average traveling speed of the
shock wave is u = 350m/s. For this part,
if needed, you can use justified approxima‐
tions to simplify the algebra.

Solution:
1. EsCAPE

— Solution by Päivo Sim
son.
i) Let’s denote the diving angle by φ, the
magnitude of the trust force by T and the
magnitude of the lift force by L. As we can
see below, the later is not needed in the solu‐
tion. The Forces acting on the ariplane dur‐
ing a dive are shown in the figure below.
For the level flight φ = 0, and we have, by
balancing the horizontal forces
Fd = kv2
0 = T.
For the dive we have, by balancing the forces
acting parallel to the trajectory,
Fd = kv2
= T + mg sinφ.
By eliminating T from the above equations
we get
v2
= v2
0
(
1 +
mg
kv2
0
sinφ
)
.
Themaximumangleisobtainedfromthelast
equation by equating v = c and solving for φ:
φmax = arcsin
(
k ·
(
c2
− v2
0
)
mg
)
.
Grading:
• Correct force equations (0.5pts)
• Correct expression for v or v2
(0.3pts)
• Expressing the final answer (0.2pts)
ii)Thepropagationspeedoftheradiation
can be considered infinite compared to the
speed of the airplane. Therefore, the radi‐
ation burst hits the plane at the same instant
when the bomb detonates. The time it takes
for the bomb to drop from altitude H to h,
without air resistance, is obtained from
H − h = g
t2
2
.
By solving for t we have
t =
√
2(H − h)
g
≈ 41.6s.
Grading:
• Understanding that the propagation speed
of the radiation can be considered infin‐
ite compared to the speed of the plane
(0.3pts)
• Correct free fall equation (0.2pts)
• Correct formula for t (0.3pts)
• Correct numerical answer (0.2pts)
iii)Theforcesactingperpendiculartothe
plane’s trajectory during a turn are shown in
the figure below.
Here
Fc = m
v2
R
is the centripetal force, and R is the
curvature radius of the trajectory. From the
figure we get
Lcosα = mg; Lsinα = m
v2
R
.
Sincethemaximumallowedlift‐to‐weightra‐
tio is n = 2.5, we have for this case
L =
mg
cosα
= nmg.
By solving the above equations with respect
toRandα,wegetthefollowingvaluesforthe
minimal curvature radius and for the corres‐
ponding bank angle of the airplane:
R =
v2
g tanα
=
v2
g
√
n2 − 1
=
= 1606.0m ≈ 1.6km;
α = arccos
(
1
n
)
≈ 66.4◦
.
Grading:
• Correct force equations (0.3pts)
• Correct understanding and use of n
(0.1pts)
• Correct formula for R (0.2pts)
• Correct formula for α (0.2pts)
• Correct numerical value for R (0.1pts)
• Correct numerical value for α (0.1pts)
iv) As the problem text states, the air res‐
istanceactingonthebombcanbeneglected.
This means that the bomb’s horizontal speed
is always the same as the speed of the plane,
v = 190m/s. If the airplane keeps travel‐
ling straight after releasing the bomb, it will
end up directly above the detonation point
by the time the bomb explodes, leaving only
the distance H − h = 8.5 km between the
bomb and the airplane. Hypothetically, the
best way to get as much distance between
the bomb and the airplane would be if the
plane turned around instantly and kept fly‐
ing straight after that. This is clearly not pos‐
sible, but it gives us the clue that the plane
should start turning immediately after re‐
leasing the bomb, and the turn should be as
sharp as possible, leaving us with the previ‐
ously found curvature radius R = 1.6 km.
Aftertheturn,itshouldflystraightsothatthe
detonation point is directly behind it. The
geometric construction of this trajectory is
shown in the figure below.
In the figure, d is the horizontal distance
the bomb travels before detonation,
d = vt = v
√
2(H − h)
g
= 7909.4m ≈ 7.9km,
and β is the total turning angle. From the
right triangle shown in the figure, we get
β = 2arctan
(
d
R
)
≈ 157◦
.
Grading:
• Goodandwellexplainedoverallanalysisof
the problem (even if it contains some mis‐
taken assumptions and the suggested tra‐
jectory ends up being incorrect) (0.6pts)
• Realising that the plane has to start turn‐
ing immediately with the previously found
turning radius R (0.2pts)
• Correct trajectory (0.6pts)
• Good justification for the correct trajectory
(0.5pts)
• Correct formula for d (0.2pts)
• Correct numerical value for d (0.2pts)
• Correct formula for β (0.5pts)
• Correct numerical value for β (0.2pts)
v) Solution 1. Let D be the distance from
the detonation point to the airplane at the
moment when the shockwave hits it, and let
z = H − h = 8500 m. We can divide the ho‐
rizontal distance from the detonation point
to the plane into three parts as shown in the
figure below.
In the figure, s is the distance the plane
files straight after the turn and before the
bomb detonates,
s = v · (t − tturn) = d − Rβ = 3507.3m,
and x is the distance it flies straightwhilethe
shockwavetravels. Ifthetravellingtimeist0,
then x = vt0 and D = ut0. Eliminating t0
from these equations gives
x =
v
u
D.
From the Pythagorean theorem we have
D2
= (d + s + x)2
+ z2
,
or
D2
=
(
a +
v
u
D
)2
+ z2
,
wherewehaveeliminatedxanddenoteda =
d + s = 2d − Rβ for simplicity. The dis‐
tance D is obtained by solving this quadratic
equation. The solution corresponding to our
problem is
D =
au
v
·
1 +
√
1 +
(u2
v2 − 1
)(z2
a2 + 1
)
u2
v2 − 1
≈
≈ 27.9km.
Sincethesafedistancewas25km,wecansay
that the airplane can escape the explosion.
Additional numerical values to help with the
grading. Lenght of the turn: Rβ = 4.4 km;
the time required to make the turn: tturn =
Rβ/v = 23.2 s; the time to travel the distance
s: t − tturn = 41.6 − 23.2 = 18.4 s; distance
x: x = vt0 = 15.1 km; total time from re‐
leasing the bomb to getting hit by the shock‐
wave: t + t0 = 121.3 s; Total horizontal dis‐
tance: d + s + x = 26.5 km; the parameter a:
a = d + s = 2d − Rβ = 11.4 km.
Grading:
• Correct understanding of the problem and
correctly breaking it into smaller parts
(even if the detailed calculations are incor‐
rect) (0.2pts)
• Correctexpressionand/orvalueforthedis‐
tance s (0.4pts)
• Correct expression relating x and D
(0.4pts)
• Correct Pythagorean equation or analog‐
ous equation for the suggested trajectory
(even if the trajectory found before was in‐
correct) (0.4pts)
• Correct solution of the Pythagorean or cor‐
responding equation (0.4pts)
• Correct value D = 27.9km (0.2pts)
• Verticaldistancez isnottakenintoaccount
when calculating D (‐0.3pts)
Solution 2. The beginning is the same as
in Solution 1. Instead of directly calculating
D, we can first calculate the travelling time
t0 of the shockwave. We have
x = vt0 = 190t0; D = ut0 = 350t0;
d = 7909.4m; s = 3507.3m.
Inserting these into the Pythagorean the‐
orem and simplifying, we get a quadratic
equation for t0:
t2
0 − 50.2t0 − 2344.8 = 0.
The positive solution is t0 = 79.65 s and
D = ut0 ≈ 27.9km.
Grading:
• Correct understanding of the problem and
correctly breaking it into smaller parts
(even if the detailed calculations are incor‐
rect) (0.2pts)
• Correctexpressionand/orvalueforthedis‐
tance s (0.4pts)
• Correct expressions for x and D (0.4pts)
• Correct equation for shockwave travelling
time t0 for the suggested trajectory (even if
the trajectory found before was incorrect)
(0.4pts)
• Correct expression and/or value for t0 for
the suggested trajectory (0.4pts)
• Correct numerical value D = 27.9km or
close to it if approximations were used
(0.2pts)
• Verticaldistancez isnottakenintoaccount
when calculating D (‐0.3pts)
Solution 3. Same as Solution 1. but
solving the Pythagorean equation approxim‐
ately. Forthefirstapproximationwecantake
z = 0. This gives us a simple linear equation
for D1:
D1 = a +
v
u
D1
with a = d+s = 2d−Rβ = 11.4 km, and the
solution
D1 =
a
1 − v
u
= 25.0km.
The second approximation is
D2 =
√(
a +
v
u
D1
)2
+ z2 = 26.4km.
This is enough to see that the plane is able to
escape the explosion. The third approxima‐
tion would be
D3 =
√(
a +
v
u
D2
)2
+ z2 = 27.1km.
Grading:
• Correct understanding of the problem and
correctly breaking it into smaller parts
(even if the detailed calculations are incor‐
rect) (0.2pts)
• Correctexpressionand/orvalueforthedis‐
tance s (0.4pts)
• Correct expression relating x and D
(0.4pts)
• Correct Pythagorean equation or analog‐
ous equation for the suggested trajectory
(even if the trajectory found before was in‐
correct) (0.4pts)
• Correct approximate solution method of
the Pythagorean or corresponding equa‐
tion (0.4pts)
• Correct value D = 26...28km (0.2pts)
• Verticaldistancez isnottakenintoaccount
when calculating D (‐0.3pts)

EsCAPE


[User]
Q T-1
                                                                                Page 1 of 2

                                                Particles from the Sun1                                   (Total Marks: 10)
       Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also
       confirm that the Sun shines because of nuclear reactions.
       Throughout this problem, take the mass of the Sun to be                               , its radius,
             , its luminosity (radiation energy emitted per unit time),                             , and the Earth-Sun
       distance,                      .
       Note:
          ∫             (         )

           ∫                (              )

           ∫                (                       )

    A Radiation from the sun :
       Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature,         , of the
A1                                                                                0.3
       solar surface.
       The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the
       solar energy incident on any surface on the Earth per unit time per unit frequency interval, , is given by

       where    is the frequency and     is the area of the surface normal to the direction of the incident radiation.
       Now, consider a solar cell which consists of a thin disc of semiconducting material of area,              , placed
       perpendicular to the direction of the Sun’ rays.
       Using the Wien approximation, express the total radiated solar power, , incident on the surface of the
A2                                                                                0.3
       solar cell, in terms of , ,  , and the fundamental constants , , .
       Express the number of photons,            , per unit time per unit frequency interval incident on the surface of
A3                                                                                0.2
       the solar cell in terms of , ,       ,   , and the fundamental constants , , .
       Th         du            r l f h     l r ll h       “b d       ” f     r y, . We assume the following
       model. Every photon of energy         excites an electron across the band gap. This electron contributes an
       energy, , as the useful output energy, and any extra energy is dissipated as heat (not converted to useful
       energy).
       Define                 where          . Express the useful output power of the cell,        u , in terms of    , ,
A4                                                                                1.0
         , ,       and the fundamental constants , , .
A5     Express the efficiency, , of this solar cell in terms of     .                                                        0.2
       Make a qualitative sketch of    versus     . The values at         and           should be clearly shown. What
A6                                                                                1.0
       is the slope of     at          and            ?
       Let    be the value of  for which         is maximum. Obtain the cubic equation that gives         . Estimate the
A7                                                                                1.0
       value of within an accuracy of              . Hence calculate   .
       The band gap of pure silicon is                  . Calculate the efficiency,   , of a silicon solar cell using this
A8                                                                                0.2
       value.

1
 Amol Dighe (TIFR), Anwesh Mazumdar (HBCSE-TIFR) and Vijay A. Singh (ex-National Coordinator, Science Olympiads)
were the principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and the
International Board are gratefully acknowledged.

                                                    Q T-1
                                                                                Page 2 of 2

      In the late nineteenth century, Kelvin and Helmholtz (KH) proposed a hypothesis to explain how the Sun
      shines. They postulated that starting as a very large cloud of matter of mass, , and negligible density, the
      Sun has been shrinking continuously. The shining of the Sun would then be due to the release of
      gravitational potential energy through this slow contraction.
       Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential
A9                                                                                0.3
       energy, , of the Sun at present, in terms of G,   and    .
       Estimate the maximum possible time,      (in years), for which the Sun could have been shining, according
A10                                                                                0.5
       to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period.
       The     calculated above does not match the age of the solar system estimated from studies of meteorites.
       This shows that the energy source of the Sun cannot be purely gravitational.

  B Neutrinos from the Sun :
       In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source
       of its energy. The net nuclear reaction is:

       Th “ l r          ur     ”, , produced in this reaction may be taken to be massless. They escape the Sun
       and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried
       away by the neutrinos can be neglected in this problem.
       Calculate the flux density,    , of the number of neutrinos arriving at the Earth, in units of         The
B1     energy released in the above reaction is                    . Assume that the energy radiated by the Sun is        0.6
       entirely due to this reaction.
       Travelling from the core of the Sun to the Earth, some of the electron neutrinos, , are converted to other
       types of neutrinos, . The efficiency of the detector for detecting is 1/6 of its efficiency for detecting .
       If there is no neutrino conversion, we expect to detect an average of  neutrinos in a year. However, due to
       the conversion, an average of     neutrinos ( and combined) are actually detected per year.
B2     In terms of    and     , calculate what fraction,   , of   is converted to   .                                     0.4
       In order to detect neutrinos, large detectors filled with water are constructed. Although the interactions of
       neutrinos with matter are very rare, occasionally they knock out electrons from water molecules in the
       detector. These energetic electrons move through water at high speeds, emitting electromagnetic radiation
       in the process. As long as the speed of such an electron is greater than the speed of light in water (refractive
       index, ), this radiation, called Cherenkov radiation, is emitted in the shape of a cone.
       Assume that an electron knocked out by a neutrino loses energy at a constant rate of per unit time, while it
       travels through water. If this electron emits Cherenkov radiation for a time, , determine the energy
B3                                                                                2.0
       imparted to this electron (       r d by the neutrino, in terms of ,    , n,   and . (Assume the electron to
       be at rest before its interaction with the neutrino.)
       The fusion of H into He inside the Sun takes place in several steps. Nucleus of      (rest mass, ) is
       produced in one of these intermediate steps. Subsequently, it can absorb an electron, producing
       a    nucleus (rest mass,    <     ) and emitting a . The corresponding nuclear reaction is:

       When a Be nucleus                                is at rest and absorbs an electron also at rest, the emitted
       neutrino has energy                      . However, the      nuclei are in random thermal motion due to the
       temperature at the core of the Sun, and act as moving neutrino sources. As a result, the energy of emitted
       neutrinos fluctuates with a root mean square (rms) value        .
       If      =               , calculate the rms speed of the Be nuclei,   , and hence estimate       . (Hint:
B4                                                                                2.0
       depends on the rms value of the component of velocity along the line of sight).

Solution:
Theoretical Task 1 (T-1) : Solutions                                          1 of 7

                                        Particles from the Sun1

 Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures
 and also confirm that the Sun shines because of nuclear reactions.

 Throughout this problem, take the mass of the Sun to be M = 2.00 × 1030 kg, its radius, R =
 7.00 × 108 m, its luminosity (radiation energy emitted per unit time), L = 3.85 × 1026 W, and the
 Earth-Sun distance, d = 1.50 × 1011 m.

 Note:
        Z
                   x     1
   (i)       ax
         xe dx =      −       eax + constant
                   a a2
       Z           2
          2 ax      x      2x     2
  (ii)   x e dx =       − 2 + 3 eax + constant
                     a     a     a
                           3x2 6x
       Z           3
          3 ax      x                   6
 (iii)   x e dx =       − 2 + 3 − 4 eax + constant
                     a      a      a    a

 A. Radiation from the Sun :

(A1) Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature,
     Ts , of the solar surface.                                                                                [0.3]

         Solution:
         Stefan’s law: L = (4πR2 )(σTs4 )
                                                          1/4
                                                    L
                                       Ts =                       = 5.76 × 103 K
                                                  4πR2 σ

 The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly,
 the solar energy incident on any surface on the Earth per unit time per unit frequency interval, u(ν),
 is given by
                                          R2 2πh 3
                                u(ν) = A 2         ν exp(−hν/kB Ts ),
                                          d c2
 where A is the area of the surface normal to the direction of the incident radiation.
 Now, consider a solar cell which consists of a thin disc of semiconducting material of area, A, placed
 perpendicular to the direction of the Sun’s rays.

(A2) Using the Wien approximation, express the total power, Pin , incident on the surface of the solar
     cell, in terms of A, R , d , Ts and the fundamental constants c, h, kB .                                       [0.3]
    1
     Amol Dighe (TIFR), Anwesh Mazumdar (HBCSE-TIFR) and Vijay A. Singh (ex-National Coordinator, Science
 Olympiads) were the principal authors of this problem. The contributions of the Academic Committee, Academic De-
 velopment Group and the International Board are gratefully acknowledged.

                           Theoretical Task 1 (T-1) : Solutions                                      2 of 7

        Solution:                Z ∞                 Z ∞
                                                               R2 2πh 3
                         Pin =         u(ν)dν =            A          ν exp(−hν/kB Ts )dν
                                 0                   0         d 2 c2

                   hν               kB Ts            kB Ts
        Let x =         . Then, ν =       x    dν =        dx.
                  kB Ts              h                h

                     2πhAR2 (kB Ts )4 ∞ 3 −x            2πkB4 4 R2           12πkB4 4 R2
                                        Z
              Pin =                         x e   dx =         Ts A    · 6 =        Ts A 2
                        c2 d 2    h4      0              c2 h3      d2        c2 h3     d

(A3) Express the number of photons, nγ (ν), per unit time per unit frequency interval incident on the
     surface of the solar cell in terms of A, R , d , Ts ν and the fundamental constants c, h, kB .           [0.2]

        Solution:
                                              u(ν)
                                     nγ (ν) =
                                               hν
                                                R2 2π
                                            = A 2 2 ν 2 exp(−hν/kB Ts )
                                                d c

 The semiconducting material of the solar cell has a “band gap” of energy, Eg . We assume the follow-
 ing model. Every photon of energy E ≥ Eg excites an electron across the band gap. This electron
 contributes an energy, Eg , as the useful output energy, and any extra energy is dissipated as heat (not
 converted to useful energy).

(A4) Define xg = hνg /kB Ts where Eg = hνg . Express the useful output power of the cell, Pout , in
     terms of xg , A, R , d , Ts and the fundamental constants c, h, kB .                                     [1.0]

        Solution:
        The useful power output is the useful energy quantum per photon, Eg ≡ hνg , multiplied by
        the number of photons with energy, E ≥ Eg .
                                         Z ∞
                           Pout = hνg         nγ (ν)dν
                                                νg
                                                             Z ∞
                                            R2 2π
                                     = hνg A 2 2                    ν 2 exp(−hν/kB Ts )dν
                                            d c                νg
                                                         2
                                                                               3 Z ∞
                                                 R 2π                   kB Ts
                                     = kB Ts xg A 2 2                                    x2 e−x dx
                                                 d c                     h         xg

                                         2πkB4 4 R2
                                     =         Ts A 2           xg (x2g + 2xg + 2)e−xg
                                         c2 h3     d

(A5) Express the efficiency, η, of this solar cell in terms of xg .                                           [0.2]

                          Theoretical Task 1 (T-1) : Solutions                                   3 of 7

       Solution:
                        Pout   xg 2
       Efficiency η =        =   (x + 2xg + 2)e−xg
                        Pin    6 g

(A6) Make a qualitative sketch of η versus xg . The values at xg = 0 and xg → ∞ should be clearly
     shown. What is the slope of η(xg ) at xg = 0 and xg → ∞?                                             [1.0]

       Solution:
                                            1 3
                                       η=     (x + 2x2g + 2xg )e−xg
                                            6 g
       Put limiting values, η(0) = 0      η(∞) = 0.
       Since the polynomial has all positive coefficients, it increases monotonically; the exponential
       function decreases monotonically. Therefore, η has only one maximum.
                                    dη   1
                                        = (−x3g + x2g + 2xg + 2)e−xg
                                    dxg  6

                                    dη         1         dη
                                             =                     =0
                                    dxg xg =0 3          dxg xg →∞
                      η

                                                                                xg

(A7) Let x0 be the value of xg for which η is maximum. Obtain the cubic equation that gives x0 .
     Estimate the value of x0 within an accuracy of ±0.25. Hence calculate η(x0 ).                        [1.0]

       Solution:
                                  dη   1
       The maximum will be for        = (−x3g + x2g + 2xg + 2)e−xg = 0
                                  dxg  6
                                    ⇒ p(xg ) ≡ x3g − x2g − 2xg − 2 = 0

       A Numerical Solution by the Bisection Method:
       Now,
                                 p(0) = −2
                                 p(1) = −4
                                 p(2) = −2
                                 p(3) = 10         ⇒      2 < x0 < 3
                              p(2.5) = 2.375       ⇒ 2 < x0 < 2.5
                             p(2.25) = −0.171 ⇒ 2.25 < x0 < 2.5
       The approximate value of xg where η is maximum is x0 = 2.27.

                           Theoretical Task 1 (T-1) : Solutions                                     4 of 7

         Alternative methods leading to the same result are acceptable.

         η(2.27) = 0.457

 (A8) The band gap of pure silicon is Eg = 1.11 eV. Calculate the efficiency, ηSi , of a silicon solar cell
      using this value.                                                                                [0.2]

         Solution:
                                            1.11 × 1.60 × 10−19
                                          xg =                  = 2.23
                                            1.38 × 10−23 × 5763
                                           xg
                                      ηSi = (x2g + 2xg + 2)e−xg = 0.457
                                           6

  In the late nineteenth century, Kelvin and Helmholtz (KH) proposed a hypothesis to explain how the
  Sun shines. They postulated that starting as a very large cloud of matter of mass, M , and negligible
  density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the
  release of gravitational energy through this slow contraction.

 (A9) Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational
      potential energy, Ω, of the Sun at present, in terms of G, M and R .                                    [0.3]

         Solution:
                                                                   Z M
                                                                        Gm dm
         The total gravitational potential energy of the Sun: Ω = −
                                                                    0       r
                                     3M               4
         For constant density, ρ =       3
                                                m = πr3 ρ       dm = 4πr2 ρdr
                                    4πR               3
                            Z R
                                                                        16π 2 Gρ2 R5    3 GM 2

                                          4 3                    dr
                     Ω=−          G         πr ρ       4πr2 ρ        =−              =−
                             0            3                       r         3      5    5 R

(A10) Estimate the maximum possible time τKH (in years), for which the Sun could have been shin-
      ing, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant
      throughout this period.                                                                                [0.5]

         Solution:
                                                                  −Ω
                                                        τKH =
                                                                  L
                                                3GM 2
                                          τKH =       = 1.88 × 107 years
                                                5R L

  The τKH calculated above does not match the age of the solar system estimated from studies of mete-
  orites. This shows that the energy source of the Sun cannot be purely gravitational.

                          Theoretical Task 1 (T-1) : Solutions                                     5 of 7

 B. Neutrinos from the Sun:
 In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is
 the source of its energy. The net nuclear reaction is:

                                        41 H −→ 4 He + 2e+ + 2νe

 The “electron neutrinos”, νe , produced in this reaction may be taken to be massless. They escape the
 Sun and their detection on Earth confirms the occurrence of nuclear reactions inside the Sun. Energy
 carried away by the neutrinos can be neglected in this problem.

(B1) Calculate the flux density, Φν , of the number of neutrinos arriving at the Earth, in units of
     m−2 s−1 . The energy released in the above reaction is ∆E = 4.0 × 10−12 J. Assume that the
     energy radiated by the Sun is almost entirely due to this reaction.                                     [0.6]

        Solution:
                                              4.0 × 10−12 J ↔ 2ν

                    L                     3.85 × 1026
        ⇒ Φν =       2
                         × 2 =                 11 2        −12
                                                               × 2 = 6.8 × 1014 m−2 s−1 .
                  4πd δE       4π × (1.50 × 10 ) × 4.0 × 10

 Travelling from the core of the Sun to the Earth, some of the electron neutrinos, νe , are converted to
 other types of neutrinos, νx . The efficiency of the detector for detecting νx is 1/6th of its efficiency
 for detecting νe . If there is no neutrino conversion, we expect to detect an average of N1 neutrinos in
 a year. However, due to the conversion, an average of N2 neutrinos (νe and νx combined) are actually
 detected per year.

(B2) In terms of N1 and N2 , calculate what fraction, f , of νe is converted to νx .                         [0.4]

        Solution:

                                             N1    =   N0
                                             Ne    =   N0 (1 − f )
                                             Nx    =   N0 f /6
                                             N2    =   Ne + Nx

                                                       OR

                                                        f
                                           (1 − f )N1 + N1 = N2
                                                        6

                                                    6     N2
                                             ⇒f =      1−
                                                    5     N1

                          Theoretical Task 1 (T-1) : Solutions                                     6 of 7

In order to detect neutrinos, large detectors filled with water are constructed. Although the interactions
of neutrinos with matter are very rare, occasionally they knock out electrons from water molecules in
the detector. These energetic electrons move through water at high speeds, emitting electromagnetic
radiation in the process. As long as the speed of such an electron is greater than the speed of light
in water (refractive index, n), this radiation, called Cherenkov radiation, is emitted in the shape of a
cone.

(B3) Assume that an electron knocked out by a neutrino loses energy at a constant rate of α per
     unit time, while it travels through water. If this electron emits Cherenkov radiation for a time
     ∆t, determine the energy imparted to this electron (Eimparted ) by the neutrino, in terms of
     α, ∆t, n, me , c. (Assume the electron to be at rest before its interaction with the neutrino.)         [2.0]

       Solution:
       When the electron stops emitting Cherenkov radiation, its speed has reduced to vstop = c/n.
       Its total energy at this time is

                                                m e c2        nme c2
                                     Estop = q              =√
                                                    2
                                              1 − vstop /c2    n2 − 1

       The energy of the electron when it was knocked out is

                                                              nme c2
                                              Estart = α∆t + √
                                                               n2 − 1

       Before interacting, the energy of the electron was equal to me c2 .
       Thus, the energy imparted by the neutrino is

                                                                       n
                                                      2
                         Eimparted = Estart − me c = α∆t +           √     − 1 me c2
                                                                       2
                                                                      n −1

The fusion of H into He inside the Sun takes place in several steps. Nucleus of 7 Be (rest mass, mBe )
is produced in one of these intermediate steps. Subsequently, it can absorb an electron, producing a
7
  Li nucleus (rest mass mLi < mBe ) and emitting a νe . The corresponding nuclear reaction is:
                                          7
                                              Be + e− −→ 7 Li + νe .

When a Be nucleus (mBe = 11.65×10−27 kg) is at rest and absorbs an electron also at rest, the emitted
neutrino has energy Eν = 1.44 × 10−13 J. However, the Be nuclei are in random thermal motion due
to the temperature Tc at the core of the Sun, and act as moving neutrino sources. As a result, the
energy of emitted neutrinos fluctuates with a root mean square value ∆Erms .

(B4) If ∆Erms = 5.54 × 10−17 J, calculate the rms speed of the Be nuclei, VBe and hence estimate Tc .
     (Hint: ∆Erms depends on the rms value of the component of velocity along the line of sight.)
                                                                                [2.0]

                     Theoretical Task 1 (T-1) : Solutions                                  7 of 7

Solution:
Moving 7 Be nuclei give rise to Doppler effect for neutrinos. Since the fractional change
in energy (∆Erms /Eν ∼ 10−4 ) is small, the Doppler shift may be considered in the non-
relativistic limit (a relativistic treatment gives almost same answer). Taking the line of sight
along the z-direction,

                                    ∆Erms   vz,rms
                                          =
                                     Eν        c
                                          = 3.85 × 10−4
                                            1 VBe
                                           =√
                                             3 c

            √
⇒ VBe =         3 × 3.85 × 10−4 × 3.00 × 108 m s−1 = 2.01 × 105 m s−1 .
The average temperature is obtained by equating the average kinetic energy to the thermal
energy.
                                   1      2    3
                                     mBe VBe = kB Tc
                                   2           2
                               ⇒       Tc = 1.13 × 107 K

Particles from the Sun1                                   (Total Marks: 10)