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Theoretical Question – T1
                                                                                Page 1 of 3

                  Mechanics of a Deformable Lattice                                (Total Marks : 20)

Here we study a deformable lattice hanging in gravity which acts as a deformable physical
pendulum. It has only one degree of freedom, i.e. only one way to deform it and the
configuration is fully described by an angle . Such structures have been studied by famous
physicist James Maxwell in 19th century, and some surprising behaviors have been
discovered recently.

As shown in the figure 1, N2 identical triangular plates (red triangle) are freely hinged by
identical rods and form an N × N lattice (N > 1). The joints at the vertices are denoted by
small circles. The sides of the equilateral triangles and the rods have the same length l. The
dashed lines in the figure represent four tubes; each tube confines N vertices (grey circles) on
the edge and the N vertices can slide in the tube, i.e. the tube is like a sliding rail.

The four tubes are connected in a diamond shape with two angles fixed at 60◦ and another
two angles at 120◦ as shown in Figure 1. Each plate has a uniform density with mass M, and
the other parts of the system are massless. The configuration of the lattice is uniquely
determined by the angle α, where 0° ≤ 𝛼 ≤ 60° (please see the examples of different angle
α in Figure 1). The system is hung vertically like a “curtain” with the top tube fixed along
the horizontal direction.

The coordinate system is shown in Figure 2. The zero level of the potential energy is
defined at y = 0. A triangular plate is denoted by a pair of indices (m,n), where m, n = 0, 1, 2,
· · ·, N − 1 representing the order in the x and y directions respectively. A(m,n), B(m,n) and
C(m,n) denote the positions of the 3 vertices of the triangle (m,n). The top-left vertex, A(0,
0) (the big black circle), is fixed.

The motion of the whole system is confined in the x-y plane. The moment of inertia of a
uniform equilateral triangular plate about its center of mass is I = Ml 2 /12. The free fall
acceleration is g. Please use Ek and Ep to denote kinetic energy and potential energy
respectively.

                                              Figure 1

                                           Theoretical Question – T1
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                                                                   l

                                                 Figure 2

Section A: When N=2 (as shown in figure 3):

                                                 Figure 3

A1   What is the potential energy Ep of the system for a general angle α when N = 2?                 2 points

A2   What is the equilibrium angle αE of the system under gravity when N = 2?                        1 point

     The system follows a simple harmonic oscillation under a small perturbation from
A3   equilibrium. Calculate the kinetic energy of this system in terms of ∆α̇ ≡d(∆α)/dt. Calculate   5 points
     the oscillation frequency fE when N = 2.

                                             Theoretical Question – T1
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Section B: For arbitrary N :
B1    What is the equilibrium angle 𝛼E′ under gravity when N is arbitrary?                             3 points

      Consider the case when 𝑁 → ∞ . Under a small perturbation of angle α, the change of
B2    potential energy of the system is ∆𝐸p ∝ 𝑁 𝛾1 , the kinetic energy of the system is 𝐸k ∝ 𝑁 𝛾2 ,   3 points
      and the oscillation frequency is 𝑓E′ ∝ 𝑁 𝛾3 . Find the values of γ1, γ2 and γ3.

Section C: A force is exerted on one of the 3N2 triangle vertices so that the system maintains at m = 60°.

         y
                                                          x

g

                                                  Figure 4

C1    Which vertex should we choose to minimize the magnitude of this force?                           1 point

      What are the direction and magnitude of this minimum force? Describe the direction in
C2                                                                                5 points
      terms of the angle θF defined in Figure 4.

Solution:
Marking Scheme – T1
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                                                                                (Full Mark = 20)
Part                                             Model Answer                                       Marks

A1     The potential energy for N = 2 is:                                                             2

                             𝐸p (𝛼) = 𝑀𝑔 ∙ 𝑦c∙m.(0,0) × 4 + 𝑀𝑔 ∙ ∆𝑦 × 2 (0.5 points)    - Eq. (1)

       where

                                                            √3𝑙       𝜋
                                        𝑦c∙m.(0,0) = − 3 sin ( 6 + 𝛼) (0.5 points)      - Eq. (2)

       is the y coordinate of center of mass of triangle (0,0), and

                                 ∆𝑦 = 𝑦A(0,1) − 𝑦A(0,0)
                                                     𝜋                𝜋
                                     = −𝑙 [sin ( 3 + 𝛼) + sin ( 3 − 𝛼)]

                                     = −√3𝑙 cos 𝛼 (0.5 points)                          - Eq. (3)

       is the translational difference of two neighbouring triangles in y-direction. Solving Eqs.
       (1), (2) and (3), we obtain
                                            2
                              𝐸p (𝛼) = − 3 𝑀𝑔𝑙(4√3 cos 𝛼 + 3 sin 𝛼) (0.5 points)        - Eq. (4)

A2                                                                                1

       At equilibrium, the potential energy reaches a minimum, which gives:

                                            𝑑𝐸p (𝛼)
                                                      |          = 0 (0.5 points)       - Eq. (5)
                                                𝑑𝛼        𝛼=𝛼E

                               √3 sin 𝛼E + 3 cos 𝛼E = 0                                 - Eq. (6)

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     or

                                                                 √3
                                         𝛼E = tan−1                      (0.5 point)                    - Eq. (7)
                                                                 4

A3   If the total energy of the oscillation has the following form                                                   5

                                                         1                   1
                     𝐸(∆𝛼, ∆𝛼̇ ) = 𝐸p + 𝐸k = 2 𝐾(∆𝛼)2 + 2 𝐼(∆𝛼̇ )2 , (0.5 points)                       - Eq. (8)

     where Ep and Ek are the potential and kinetic energies of the system respectively, then the
     motion is a simple harmonic oscillation with angular frequency 𝜔 = √𝐾/𝐼. Here  =
      − 𝛼E . Under a small perturbation, the potential energy change is:

                                      1 𝑑2 𝐸p
                          ∆𝐸p ≈               |  (∆𝛼)2
                                      2 d𝛼 2 𝛼=𝛼
                                                     E

                                     1 2
                                  = ( ) ( 𝑀𝑔𝑙) (4√3 cos 𝛼E + 3 sin 𝛼E )(∆𝛼)2
                                     2 3
                                      √57
                                  =       𝑀𝑔𝑙(∆𝛼)2 (1 point)                                           - Eq. (9)
                                       3

     The total kinetic energy of the system includes the translational kinetic energy of every
     plate and the rotational kinetic energy of every plate relative to its center of mass

                                 𝐸k = ∑ 𝐸ktrans + ∑ 𝐸krot                                              - Eq. (10)

     The rotational kinetic energy is
                                             1 𝑀𝑙2                       1
                           ∑ 𝐸krot = 4 ×             (∆𝛼̇ )2 = 𝑀𝑙 2 (∆𝛼̇ )2 (0.5 points)               - Eq. (11)
                                             2 12             6

     𝐸ktrans can be obtained by considering the motion of the center of mass of each triangle and
     setting N = 2.

                                                                                𝜋   𝑙         𝜋
                  𝑥c.m.(𝑚,𝑛) = 𝑚(2𝑙 cos 𝛼) + 𝑛(2𝑙 cos 𝛼) cos                       +   cos (𝛼 + ),
                                                                                3 √3          6

                                                             𝜋       𝑙            𝜋
                  𝑦c.m.(𝑚,𝑛) = −𝑛(2𝑙 cos 𝛼) sin 3 −                      sin (𝛼 + 6 ).   (0.5 point)
                                                                  √3

     Differentiating and substituting

                          √3                 4                   𝜋           7           𝜋     3√3
                sin 𝛼 =         , cos 𝛼 =         , sin (𝛼 + 6 ) = 2√19 , cos (𝛼 + 6 ) = 2√19,
                          √19               √19

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                                     7 3                                               3(2𝑛 − 1)
           𝑥̇ c.m.(𝑚,𝑛) = − (2𝑚 + 𝑛 + )    𝑙∆𝛼̇ ,                     𝑦̇ c.m.(𝑚,𝑛) =               𝑙∆𝛼̇ .
                                     6 √57                                               2√19

          2              2              2               (12𝑚+6𝑛+7)2 +27 2
         𝑣c.m.(𝑚,𝑛) = 𝑥̇ c.m.(𝑚,𝑛) + 𝑦̇ c.m.(𝑚,𝑛) =                    𝑙 (∆𝛼̇ )2 ,              (1 point)
                                                             228

          trans      𝑀 2              2            2            2            164 2
         𝐸c.m.,k =     [𝑣c.m.(0,0) + 𝑣c.m.(0,1) + 𝑣c.m.(1,0) + 𝑣c.m.(1,1) ]=    𝑀𝑙 (∆𝛼̇ )2 .
                     2                                                       57
                                                        347
                      𝐸ktrans = 𝐸c.m.,k
                                 trans
                                        + 𝐸krot = 114 𝑀𝑙 2 (∆𝛼̇ )2 .               (1 point)

Alternatively, another way to get 𝐸𝑘trans is based on the center of mass of the whole system:
                                      trans       rot
                          𝐸k = ∑ 𝐸c.m.,k    + ∑ 𝐸r.c.,k (0.5 points)                - Eq. (12)

where

                            trans     𝑀  2            2            2            2
                           𝐸r.c.,k = 2 [𝑣r.c.(0,0) + 𝑣r.c.(1,0) + 𝑣r.c.(0,1) + 𝑣r.c.(1,1) ]             - Eq. (13)

is the translational kinetic energy relative to the center of mass of the system and

                           trans     4𝑀2
                          𝐸c.m.,k = 2 𝑣c.m.                                                             - Eq. (14)

is the translational kinetic energy of the center of mass of the system.

The center of mass of each of the 2×2 = 4 triangles always form diamond shape with
lateral length 2l cos α. The center of mass of the whole system is at the center of the
diamond shape. Hence

                                                     𝑑(√3𝑙 cos 𝛼)
                        𝑣r.c.(0,0) = 𝑣r.c.(1,1) =                 |     ∆𝛼̇
                                                         𝑑𝛼         𝛼=𝛼        E

                                                     𝑑(𝑙 cos 𝛼)
                         𝑣r.c.(1,0) = 𝑣r.c.(0,1) =                |      ∆𝛼̇                            - Eq. (15)
                                                        𝑑𝛼        𝛼=𝛼E

Substituting Eqs. (14) and (15) into Eq. (13), we obtain

                                  trans
                                 𝐸r.c.,k = 4 sin 𝛼E2 𝑀𝑙 2 (∆𝛼)̇2                                        - Eq. (16)

     trans
For 𝐸c.m.,k ,

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                                   d𝑥         2    d𝑦        2
                     𝑣c.m. = √( d𝛼
                                 c.m.        c.m.
                                      ) + ( d𝛼    ) |                   ∆𝛼̇           - Eq. (17)
                                                                 𝛼=𝛼E

is the velocity of the center-of-mass of the four triangular plates, with

                                         1
                     𝑥c.m. = 𝑥c.m.(0,0) + (𝑥B(0,0) + 𝑥A(1,0) )
                                         2
                             √3𝑙          𝜋             3
                          = 3 cos (6 + 𝛼) + 2 𝑙 cos 𝛼                                 - Eq. (18)

                                         1
                     𝑦c.m. = 𝑦c.m.(0,0) + ∆𝑦
                                         2
                              √3𝑙             𝜋             √3
                         = − 3 sin ( 6 + 𝛼) − 2 𝑙 cos 𝛼                               - Eq. (19)

Substituting Eqs. (17), (18) and (19) and into Eq. (14), we obtain
                               2
                     trans
                    𝐸c.m.,k = (3 + 10 sin2 𝛼𝐸 ) 𝑀𝑙 2 (∆𝛼)̇2 (0.5 points)              - Eq. (20)

Combining Eqs. (12), (16) and (20), we obtain

                            𝐸k = 𝐸krot + 𝐸r.c.,k
                                          trans     trans
                                                 + 𝐸c.m.,k

                                  5
                               = ( + 14 sin2 𝛼𝐸 ) 𝑀𝑙 2 (∆𝛼)̇2
                                  6
                                    347
                               = 114 𝑀𝑙 2 (∆𝛼)̇2                 (1.5 points)        - Eq. (21)

According to Eqs. (8), (9) and (21),

                                        √57
                               1            𝑀𝑔𝑙         1
                         𝑓=         √    3
                                        347        =         √38√57 𝑔 (0.5 points)    - Eq. (22)
                               2𝜋           𝑀𝑙 2        2𝜋         347    𝑙
                                        114

[Note 1: 0.5 point should be deducted if there are numerical mistakes, but all steps are
correct.

                                         𝑔
Note 2: A rough estimate of 𝑓~√ can get 0.5 points out of 5 points.]
                                          𝑙

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B1   For arbitrary N, the total potential energy                                                           3

                                       𝐸p = ∑𝑁−1
                                             𝑚,𝑛=0 𝐸p (𝑚, 𝑛)                                - Eq. (23)

     where

                                             1
                                𝐸p (𝑚, 𝑛) = 3 𝑀𝑔[𝑦A(𝑚,𝑛) + 𝑦B(𝑚,𝑛) + 𝑦C(𝑚,𝑛) ]              - Eq. (24)

                                                                    (0.5 points for Eqs. (23) and (24))

     and

                           𝜋               𝜋
        𝑦A(𝑚,𝑛) = −𝑛𝑙 sin ( − 𝛼) − 𝑛𝑙 sin ( + 𝛼) = −√3𝑛𝑙 cos 𝛼
                           3               3

        𝑦B(𝑚,𝑛) = 𝑦A(𝑚,𝑛) − 𝑙 sin 𝛼 = −√3𝑛𝑙 cos 𝛼 − 𝑙 sin 𝛼

                                         𝜋                                   𝜋
           𝑦C(𝑚,𝑛) = 𝑦A(𝑚,𝑛) − 𝑙 sin (3 + 𝛼) = −√3𝑛𝑙 cos 𝛼 − 𝑙 sin ( 3 + 𝛼)                 - Eq. (25)

                                                        (0.5 points for all three correct coordinates)

     Thus,

                                1                                    𝜋
              𝐸p (𝑚, 𝑛) = − 3 𝑀𝑔𝑙 [3√3𝑛 cos 𝛼 + sin 𝛼 + sin (3 + 𝛼)]                        - Eq. (26)

     and
                          𝑁−1
              𝐸p = ∑             𝐸p (𝑚, 𝑛)
                          𝑚,𝑛=0

                      1                                                  𝜋
                 = − 3 𝑀𝑔𝑙 ∑𝑁−1
                            𝑚,𝑛=0 [3√3𝑛 cos 𝛼 + sin 𝛼 + sin ( 3 + 𝛼)] (0.5 points) - Eq. (27)

     Using the mathematical relations
                                                 𝑁−1         𝑁−1
                                             ∑         1=∑         1=𝑁
                                                 𝑚=0         𝑛=0

     and

                                                                    𝑁(𝑁−1)
                                             ∑𝑁−1      𝑁−1
                                              𝑚=0 𝑚 = ∑𝑛=0 𝑛 =                              - Eq. (28),
                                                                      2

     Eq. (27) becomes

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                            1          3√3(𝑁 − 1) cos 𝛼                𝜋
                      𝐸p = − 𝑁 2 𝑀𝑔𝑙 [                  + sin 𝛼 + sin ( + 𝛼)]
                            3                2                         3

                                  1                √3(3𝑁−2) cos 𝛼   3
                 or       = − 3 𝑁 2 𝑀𝑔𝑙 [                         + 2 sin 𝛼] (1 points)            - Eq. (29)
                                                         2

                       d𝐸p
     At equilibrium, d𝛼 = 0, therefore

                                                    ′
                                      3√3(𝑁−1) sin 𝛼E                           𝜋
                                  −                       + cos 𝛼E′ + cos ( 3 + 𝛼E′ ) = 0           - Eq. (30)
                                               2

                                                                   √3
                                                 𝛼E′ = tan−1 (3𝑁−2) (0.5 points)                    - Eq. (31)

     [Remark: Increasing α lowers each triangle relative to its vertex A, but globally raises the
     system, i.e. the bottom tube is raised higher. When 𝑵 → ∞ , the global displacement
     dominates, consequently 𝜶 → 𝟎.]

B2   Under a small perturbation, the potential energy change, according to Eq. (29) is                            3

                                       1 𝑑2 𝐸p
                         ∆𝐸p ≈ 2 d𝛼2 |              ′
                                                         (∆𝛼)2 ~𝑁 3 or 𝛾1 = 3 (0.5 points)         - Eq. (32)
                                                 𝛼=𝛼E

     [Remark: There are N2 triangles and the y coordinate of the total center of mass is
     proportional to N, hence 𝑬𝐩 ~𝑵𝟑 and 𝜸𝟏 = 𝟑 . Using this argument to derive the
     correct 𝜸𝟏 can also get 0.5 points.]

     The kinetic energy of a triangle includes the translational energy of its center of mass and
     the rotational energy about its center of mass. Hence the total kinetic energy of the N2
     triangles is

                                       𝐸k = ∑𝑚,𝑛 𝐸c.m.(𝑚,𝑛) + ∑𝑚,𝑛 𝐸r.c.(𝑚,𝑛)                       - Eq. (33)

     where

                                                 1 𝑀𝑙2             1
                              𝐸r.c.(𝑚,𝑛) = 2 12 (∆𝛼̇ )2 = 24 𝑀𝑙 2 (∆𝛼̇ )2 ~1                        - Eq. (34)

     and

                       𝑀 2
        𝐸c.m.(𝑚,𝑛) =    𝑣
                       2 c.m.(𝑚,𝑛)

                       𝑀(∆𝛼̇ )2       d𝑥c.m.(𝑚,𝑛) 2        d𝑦c.m.(𝑚,𝑛) 2
                  =               [(               ) +(                ) ]          (0.5 points)    - Eq. (35)
                          2               d𝛼                  d𝛼                ′
                                                                             𝛼=𝛼E

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Since

                                              √3𝑙      𝜋
             𝑥c.m.(𝑚,𝑛) = 𝑥A(𝑚,𝑛) +               cos ( + 𝛼)
                                               3       6

                                         𝑙        √3𝑙
                      = (2𝑚 + 𝑛)𝑙 cos 𝛼 + cos 𝛼 −     sin 𝛼
                                         2         6

and

                                          √3𝑙      𝜋
            𝑦c.m.(𝑚,𝑛) = 𝑦A(𝑚,𝑛) +            sin ( + 𝛼)
                                           3       6

                                                √3𝑙               𝑙
                     = √3𝑛𝑙 cos 𝛼 + 6 cos 𝛼 + 2 sin 𝛼                                          - Eq. (36)

                                                                          (0.5 points for correct x and y)

                   d𝑥c.m.(𝑚,𝑛)                     1        √3
                               = [−(2𝑚 + 𝑛) sin 𝛼 − sin 𝛼 −    cos 𝛼] 𝑙
                       d𝛼                          2         6

                   d𝑦c.m.(𝑚,𝑛)                 √3        1
                               = [−√3𝑛 sin 𝛼 −    sin 𝛼 + cos 𝛼] 𝑙
                       d𝛼                       6        2

we have

                      1
                                              (4𝑚2 + 4𝑛2 + 4𝑚𝑛 + 2𝑚 + 2𝑛) sin2 𝛼E′
                              2 (∆𝛼̇ )2
          𝐸c.m.(𝑚,𝑛) = 2 𝑀𝑙               [           2√3                           1     ]    - Eq. (37)
                                                  + 3 (𝑚 − 𝑛) sin 𝛼E′ cos 𝛼𝐸′ + 3

             1
Since 𝛼E′ ~ 𝑁 in Eq. (31), we have

                                                 1                    1
                  𝐸c.m.(𝑚,𝑛) = 𝐴 ∙ 𝑁 2 ∙ 𝑁2 + 𝐵 ∙ 𝑁 ∙ 𝑁 + 𝐶~1 (0.5 points)                     - Eq. (38)

According to Eqs. (33), (34) and (38), we have

                  𝐸k = ∑            𝐸c.m.(𝑚,𝑛) + ∑            𝐸r.c.(𝑚,𝑛) ~𝑁 × 𝑁 × 1~𝑁 2
                              𝑚,𝑛                           𝑚,𝑛

                                                or 𝛾2 = 2 (0.5 points)                         - Eq. (39)

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     [Remarks: 𝑬𝐤 ~𝑵𝟐 because there are N2 triangles, each contribute 𝑬r.c. (𝒎, 𝒏)~𝟏
     (relative-to-center-of-mass kinetic energy) and 𝑬c.m. (𝒎, 𝒏)~𝟏 (center-of-mass kinetic
     energy).]

     Note that 𝐸r.c. (𝑚, 𝑛)~1 is true for arbitrary α while 𝐸c.m. (𝑚, 𝑛)~1 is only true for the
     special case of 𝛼E′ → 0 or 𝑁 → ∞.

     Therefore

                                                   𝐸p
                                            𝑓E′ ~√𝐸 ~√𝑁
                                                    k

                                        or 𝛾3 = 0.5 (0.5 points)                       - Eq. (40)

C1   The minimum force should act on the farthest triangle (N − 1, N − 1), whose motion can be         1
     decomposed into the motion of the center of mass and the rotation around the center of
     mass: 𝑣⃗ = 𝑣⃗c.m. + 𝑣⃗rot . As shown in the figure, 𝑣⃗rot of vertex C makes the smallest angle
     relative to the direction of 𝑣⃗c.m. near 𝛼m ≡ 𝜋/3. Hence its displacement is the largest and
     its corresponding force is minimum, i.e. the minimum force should act on vertex C(N − 1,
     N − 1). (1 point)

     [Remarks: A rigorous calculation is given in Appendix 3.]

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C2   At 𝛼 = 𝛼m ≡ 𝜋/3, a small change in α will change the potential energy by:                 5

                           𝑑𝐸p
             ∆𝐸p (𝛼m ) =       |  ∆𝛼
                           𝑑𝛼 𝛼=𝛼
                                      m

                           1 2       3√3𝑁               3
                       =     𝑁 𝑀𝑔𝑙 [(     − √3) sin 𝛼m − cos 𝛼m ] ∆𝛼
                           3           2                2
                           3
                        = 4 (𝑁 − 1)𝑁 2 𝑀𝑔𝑙∆𝛼 (1 point)                           - Eq. (41)

     The displacement of C(m,n) point is

                                                           𝜋
                      ∆𝑥C(𝑚,𝑛) = − [(2𝑚 + 𝑛) sin 𝛼m − sin ( + 𝛼m )] 𝑙∆𝛼
                                                           3
                                      (2𝑚+𝑛+1)√3
                                  =                     𝑙∆𝛼 (0.5 points)
                                                2

                                                       𝜋
                       ∆𝑦C(𝑚,𝑛) = − [√3𝑛 sin 𝛼m − cos ( + 𝛼m )] 𝑙∆𝛼
                                                       3
                                      (3𝑛+1)
                                  =                 𝑙∆𝛼 (0.5 points)
                                           2

     For C(N-1,N-1), ∆𝑟 = √(∆𝑥)2 + (∆𝑦)2 = (3𝑁 − 2)(𝑙∆𝛼) . (1 point)

     Hence

                                          ∆𝐸p (𝛼m )      3(𝑁−1)𝑁 2
                               𝐹min =                 = 4(3𝑁−2) 𝑀𝑔 (1 point)     - Eq. (42)
                                           ∆𝑟max

     and

                                                          ∆𝑦C(𝑁−1,𝑁−1)
                                  𝜃𝐹min = tan−1 [                      ]+𝜋
                                                          ∆𝑥C(𝑁−1,𝑁−1)

                                                            √3         5𝜋
                                               = − tan−1 3 + 𝜋 = 6 (1 point)     - Eq. (43)

     [Remarks: This 𝜽𝑭min is not perpendicular to the C(N-1,N-1)–A(0,0) direction because
     of the constraints of the tunes, e.g. A(1,0), A(2,0), A(3,0), ⋯ , are also the holding
     points.]

                                          Marking Scheme – T1
                                                                                Page 10 of 15

Appendix 1:                                                                                N/A
(a) Calculation of the exact 𝐸p , 𝐸k and 𝑓E′ in Parts (C), (D) and € for arbitrary N
Under a small perturbation, the potential energy change is

                       1 𝑑 2 𝐸p
                 ∆𝐸p ≈          |  (∆𝛼)2
                       2 𝑑𝛼 2 𝛼=𝛼′
                                         𝐸

                        1 2     3√3𝑁 − 2√3      ′
                                                   3     ′
                                                             (∆𝛼)2
                       = 𝑁 𝑀𝑔𝑙 (           cos 𝛼E + sin 𝛼E )
                        3           2              2           2

                           √3(3𝑁−2)2 +9
                       =                      𝑁 2 𝑀𝑔𝑙(∆𝛼)2                             - Eq. (44)
                                 12

The kinetic energy of a triangle includes the translational energy of its center of mass and
the rotational energy around its center of mass. Hence the total kinetic energy of the N2
triangles is

                     𝐸k = ∑𝑚,𝑛 𝐸c.m.(𝑚,𝑛) + ∑𝑚,𝑛 𝐸r.c.(𝑚,𝑛)                            - Eq. (45)

where

                                  1 𝑀𝑙2                 1
                    𝐸r.c.(𝑚,𝑛) = 2 12 (∆𝛼̇ )2 = 24 𝑀𝑙 2 (∆𝛼̇ )2                        - Eq. (46)

and

                               𝑀 2
               𝐸c.m.(𝑚,𝑛) =     𝑣
                               2 c.m.(𝑚,𝑛)

                               𝑀(∆𝛼)̇2       d𝑥c.m.(𝑚,𝑛) 2      d𝑦c.m.(𝑚,𝑛) 2
                           =             [(            ) +(               ) ]          - Eq. (47)
                                 2              d𝛼                 d𝛼              ′
                                                                                𝛼=𝛼E

Since

                                               √3𝑙          𝜋
                 𝑥c.m.(𝑚,𝑛) = 𝑥A(𝑚,𝑛) + 3 cos ( 6 + 𝛼)

                                                 𝑙        √3𝑙
                              = (2𝑚 + 𝑛)𝑙 cos 𝛼 + cos 𝛼 −     sin 𝛼
                                                 2         6

and

                                               √3𝑙      𝜋
                 𝑦c.m.(𝑚,𝑛) = 𝑦A(𝑚,𝑛) −            sin ( + 𝛼)
                                                3       6

                                                  Marking Scheme – T1
                                                                                Page 11 of 15

                                                                √3𝑙                𝑙
                               = −√3𝑛𝑙 cos 𝛼 − 6 cos 𝛼 − 2 sin 𝛼                               - Eq. (48)

Hence,

                   d𝑥c.m.(𝑚,𝑛)                     1        √3
                               = [−(2𝑚 + 𝑛) sin 𝛼 − sin 𝛼 −    cos 𝛼] 𝑙
                       d𝛼                          2         6

                    d𝑦c.m.(𝑚,𝑛)                 √3        1
                                = [−√3𝑛 sin 𝛼 +    sin 𝛼 − cos 𝛼] 𝑙
                        d𝛼                       6        2

We have

                     1
                                             (4𝑚2 + 4𝑛2 + 4𝑚𝑛 + 2𝑚 + 2𝑛) sin2 𝛼E′
                             2 (∆𝛼̇ )2
         𝐸c.m.(𝑚,𝑛) = 2 𝑀𝑙               [                2√3                          1   ]   - Eq. (49)
                                                     + 3 (𝑚 − 𝑛) sin 𝛼E′ cos 𝛼E′ + 3

and

                  𝐸𝑘 = ∑           𝐸c.m.(𝑚,𝑛) + ∑                     𝐸r.c.(𝑚,𝑛)
                               𝑚,𝑛                              𝑚,𝑛

                        1                            5
                     = [ (11𝑁 − 1)(𝑁 − 1) sin2 𝛼E′ + ] 𝑁 2 𝑀𝑙 2 (∆𝛼̇ )2
                        6                           24
                          (11𝑁−1)(𝑁−1)                5
                     =[                          +        ] 𝑁 2 𝑀𝑙 2 (∆𝛼̇ )2                   - Eq. (50)
                          2(3𝑁−2)2 +6                24

With Eqs. (44) and (50), we have

                                        √3(3𝑁 − 2)2 + 9 2
                                 1                     𝑁 𝑀𝑔𝑙
                          𝑓E′ =    √           12
                                2𝜋 (11𝑁 − 1)(𝑁 − 1) 5
                                     [                + 24] 𝑁 2 𝑀𝑙 2
                                       2(3𝑁 − 2)2 + 6

                                     1           2√3(3𝑁−2)2 +9          𝑔
                               = 2𝜋 √ 12(11𝑁−1)(𝑁−1)                                           - Eq. (51)
                                             [                   +5] 𝑙
                                                  (3𝑁−2)2 +3

(b) Center of mass movement of the whole system
    According to Eq. (48), we have

                                     ∑𝑚,𝑛 𝑥c.m.(𝑚,𝑛)
               𝑥c.m.(sys.) (𝛼) =
                                          𝑁2

                                         Marking Scheme – T1
                                                                                Page 12 of 15

                                                       𝑙        √3𝑙
                                ∑𝑚,𝑛 [(2𝑚 + 𝑛)𝑙 cos 𝛼 + cos 𝛼 −
                                                       2         6 sin 𝛼]
                            =
                                                         𝑁2

                                 3𝑁 − 2             √3𝑙
                            =(          ) 𝑙 cos 𝛼 −     sin 𝛼
                                   2                 6

and

                                ∑𝑚,𝑛 𝑦c.m.(𝑚,𝑛)
             𝑦c.m.(𝑚,𝑛) (𝛼) =
                                     𝑁2

                                                         √3𝑙        𝑙
                                  ∑𝑚,𝑛 [√3𝑛𝑙 cos 𝛼 +
                                                          6 cos 𝛼 + 2 sin 𝛼]
                            =−
                                                         𝑁2
                                   3𝑁−2                  𝑙 sin 𝛼
                            = −(         ) √3𝑙 cos 𝛼 −                              - Eq. (52)
                                     6                        2

Eq. (52) is the trajectory of the center of mass for the whole system, which is not a straight
line.

Appendix 2: Calculation of the moment of inertia of a triangular plate                            N/A

An equilateral triangle with lateral length l can be divided into four small equilateral
triangles with lateral length l/2. For the central small triangle centered at c1, its moment of
inertia is

                                              𝑀    𝑙 2
                                     𝐼1 = 𝛽       ( )                               - Eq. (53)
                                              4    2

For the non-central small triangle centered at𝑐2 , 𝑐2′ and 𝑐2′′ ,

                                                  𝑀
                                    𝐼2 = 𝐼1 + 4 𝑑 2                                 - Eq. (54)

where 𝑑 = √3𝑙/6 is the distance between the centers of triangles 1 and 2. The second term
is from the parallel-axis theorem. The moment of inertia of the whole triangle is the sum of
the moment of inertia of the four sub-triangles:

          Marking Scheme – T1
                                              Page 13 of 15

                     𝑀   𝑙 2     𝑀
       𝛽𝑀𝑙 2 = 4 × 𝛽 4 (2) + 3 × 4 𝑑 2   - Eq.(55)

Thus
               1
           𝛽 = 12                        - Eq. (56)

                                     Marking Scheme – T1
                                                                                Page 14 of 15

Appendix 3: The minimum force corresponds to the maximum displacement of the                  N/A
exerting point of this force.

Consider the position of vertices A, B, C of a triangle (m,n) :

                         𝑥A(𝑚,𝑛) = (2𝑚 + 𝑛) cos 𝛼m 𝑙

                         𝑦A(𝑚,𝑛) = −√3𝑛 cos 𝛼m 𝑙

                         𝑥B(𝑚,𝑛) = (2𝑚 + 𝑛 + 1) cos 𝛼m 𝑙

                         𝑦B(𝑚,𝑛) = −(√3𝑛 cos 𝛼m + sin 𝛼m )𝑙

                                                           𝜋
                         𝑥C(𝑚,𝑛) = [(2𝑚 + 𝑛) cos 𝛼m + cos ( + 𝛼m )] 𝑙
                                                           3
                                                            𝜋
                         𝑦C(𝑚,𝑛) = − [√3𝑛 cos 𝛼m + sin ( 3 + 𝛼m )] 𝑙             - Eq. (57)

Taking derivatives on α on the above coordinates we get

                                                      (2𝑚 + 𝑛)√3
           ∆𝑥A(𝑚,𝑛) = −(2𝑚 + 𝑛) sin 𝛼m 𝑙∆𝛼 = −                   𝑙∆𝛼
                                                          2
                                             3𝑛
           ∆𝑦A(𝑚,𝑛) = √3𝑛 sin 𝛼m (𝑙∆𝛼) =        𝑙∆𝛼
                                              2

                                                         (2𝑚 + 𝑛 + 1)√3
           ∆𝑥B(𝑚,𝑛) = −(2𝑚 + 𝑛 + 1) sin 𝛼m 𝑙∆𝛼 = −                      𝑙∆𝛼
                                                               2
                                                           3𝑛 − 1
           ∆𝑦B(𝑚,𝑛) = −(−√3𝑛 sin 𝛼m + cos 𝛼m )𝑙∆𝛼 =               𝑙∆𝛼
                                                             2

                                             𝜋              (2𝑚 + 𝑛 + 1)√3
         ∆𝑥C(𝑚,𝑛) = [−(2𝑚 + 𝑛) sin 𝛼m − sin ( + 𝛼m )] 𝑙∆= −                𝑙∆𝛼
                                             3                    2
                                                 𝜋                (3𝑛+1)
           ∆𝑦C(𝑚,𝑛) = − [−√3𝑛 sin 𝛼m + cos ( 3 + 𝛼m )] 𝑙∆𝛼 =               𝑙∆𝛼   - Eq. (58)
                                                                    2

For ∆𝑟 = √(∆𝑥)2 + (∆𝑦)2, we have
           ∆𝑟A(𝑚,𝑛) = √3𝑚2 + 3𝑛2 + 3𝑚𝑛(𝑙∆𝛼)
           ∆𝑟B(𝑚,𝑛) = √3𝑚2 + 3𝑛2 + 3𝑚𝑛 + 3𝑚 + 1(𝑙∆𝛼)
           ∆𝑟C(𝑚,𝑛) = √3𝑚2 + 3𝑛2 + 3𝑚𝑛 + 3𝑚 + 3𝑛 + 1(𝑙∆𝛼)                        - Eq. (59)

                                      Marking Scheme – T1
                                                                                Page 15 of 15

Thus we ﬁnd

                                ∆𝑟C(𝑚,𝑛) > ∆𝑟B(𝑚,𝑛) > ∆𝑟A(𝑚,𝑛)                 - Eq. (60)

Therefore, we should choose point C of the triangle (N − 1, N − 1) to obtain

                                   ∆𝑟max = (3𝑁 − 2)𝑙∆𝛼                         - Eq. (61)

so that the force is minimal.

APhO 2016 Theory T1


[User]
6. STRıNG‐COUPLED MAssEs (5 points)

—
Aleksi Kononen.
Two small masses m are connected and
hung between walls by weightless strings
as pictured below. The acceleration due to
gravityisg. Alloscillatorymotionisassumed
to have a small amplitude and the system is
always in one of its normal modes (i.e. the
oscillations are sinusoidal).
i) (2 points) Find ω1, the angular frequency
of in‐phase oscillations (by which the oscil‐
lation phases of the both masses are always
equal), perpendicular to the plane of the fig‐
ure.
ii) (3 points) Find ω2, the angular frequency
ofanti‐phaseoscillations(bywhichtheoscil‐
lation phases of the both masses are always
opposite), in terms of ω1.

Solution:
6. STRıNG‐COUPLED MAssEs

—
Solution by Aleksi Kononen and Jaan Kalda,
grading schemes by Aleksi Kononen.
i) (2 points) For in‐phase oscillations, the
stringconnectingthemassesalwaysremains
paralleltothex‐axis. Inthatcase,themotion
is confined to the yz‐plane and is identical to
apendulumwithlengthrsinθ hungfromthe
axis of rotation:
ω1 =
√
g
rsinθ
Grading:
• Stating
∑
Fx = 0, or stating that the con‐
nectingstringremainsparalleltothex‐axis
(0.6pts)
• Realising the effective pendulum length is
rsinθ, i.e. the axis of rotation is the line
connecting the wall points (0.8pts)
• Correct expression for ω1 (0.6pts)
ii) (3 points) Now suppose one of the balls
moves along the y axis to a distance ym and
the other ball — by −y. The length of the ho‐
rizontal string remains constant, hence y2
+
(r − x)2
= r2
, where x denotes the displace‐
ment of the ball in x‐direction. For small
displacements, we can neglect the smallest
term x2
, hence x = y2
/2r. The length of the
slanted string remains also constant, hence
(rsinθ−z)2
+(rcosθ+x)2
+y2
= r2
, where
z denotestheverticaldisplacement. Neglect‐
ing here the smallest terms z2
and x2
, we ob‐
tain 2rz sinθ = 2rxcosθ+y2
= y2
(1+cosθ),
hence
z =
y2
2r
1 + cosθ
sinθ
.
Now we can write down energy conservation
law: 1
2ẏ2
+ gz = const, where dot denotes
the time derivative. Differentiating this over
time we obtain
ÿ = −
gy
r
1 + cosθ
sinθ
so that
ω2 =
√
|ÿ/y| = ω1
√
1 + cosθ.
Grading:
• Equation for a cylinder surface (0.4pts)
• Equation for a sphere surface (0.4pts)
• Both surface equations correct and co‐
ordinates consistent (0.2pts)
• Realising x2
‐ and z2
‐terms are negligible,
and not discarding y2
(0.2pts)
• Solving z in terms of y (0.5pts)
• Energy conservation between z and ẏ
(0.3pts)
• Solving relation between ÿ and y (0.6pts)
• Correct expression for ω2 (0.3pts)
• ω2 correctly in terms of ω1 (0.1pts)
Alternative solution: Let one of the
masses (mass A) be displaced by y and the
other by −y. Let T′
be the tension in the
two outermost strings (equal magnitudes by
symmetry) and T in the connecting string.
Newton’slawalongthey‐axisformassAthen
yields
−T′ y
r
− T
y
r
= mÿ,
i.e.
ÿ = −
T′
+ T
mr
y.
Without oscillations, T′
0 = mg/sinθ and
T0 = T′
0 cosθ. Let us write T′
= T′
0 + ∆T′
and T = T0 + ∆T, i.e.
T′
+ T = mg
1 + cosθ
sinθ
+ ∆T′
+ ∆T.
Let us make some estimates. Clearly, |∆T| ∼
|∆T′
|. The maximum of ∆T′
occurs when
the mass swings by its lowest point with ve‐
locity ẏ, where its trajectory has a radius of
curvature R ∼ rsinθ. Moreover, small nor‐
malmodeoscillationsimplyy = Acosω2tfor
some amplitude A ≪ rsinθ, i.e. |ẏ| ≤ Aω2.
Finally, assume ω2 = kω1 holds for some
k ∼ 1. We have
|∆T′
| ∼
mẏ2
R
∼
mk2
A2
g
r2 sin2
θ
= k2
mg
(
A
rsinθ
)2
,
i.e. quadratically smaller in (A/rsinθ) than
T′
0. Thus, T′
+ T ≈ mg1+cosθ
sinθ and
ω2 =
√
T′ + T
mr
= ω1
√
1 + cosθ.
(The result shows k ≤
√
2, i.e. the assump‐
tion holds.) Grading:
• Forceequation,eitheralongy orfullvector
form (0.4pts)
• Solving relation between ÿ and y in terms
of tension forces (0.4pts)
• Unperturbed values for T′
and T (0.4pts)
• Reasonable estimate of |∆T′
| (0.6pts)
• Showing |∆T′
| is quadratically smaller
than T′
(0.8pts)
• Correct expression for ω2 (0.3pts)
• ω2 correctly in terms of ω1 (0.1pts)
• Remark: Accept solution with assumption
∆T′
/T′
≪ 1 without estimates

STRıNG‐COUPLED MAssEs


[User]
01hIPhO 2000, problem 2. A solid question on the Lorentz force with real-world
    relevance. Requires a little relativity, namely the expressions for relativistic momentum/energy.

Solution: The official files are a mess; the solutions to this particular problem are here and here.

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


[User]
Joonisel on kujutatud pealtvaates kahte nööpi $A$ ja $B$, mis on omavahel ühendatud niidiga, mille pikkus $l=\SI{10}{cm}$. Süsteem asub pöörleval alusel. Nööpide ja aluse vaheline hōõrdetegur on $\mu$. Ühe nööbi külge on kinnitatud veel teine niit, mille abil hoitakse nööpe paigal (vt. joonist). On teada, et pöörlemiskese asub klotsist $A$ kaugusel $a=\SI{10}{cm}$ ja mōlemad niidid on horisontaalsed. Kui suur on pöörlemiskeskme ja nööbi $B$ vahekaugus ning nööbi $B$ mass $m_{B}$ ? Nööbi $A$ mass $m_{A}=\SI{1,0}{g}$ ja niidid võib lugeda kaalututeks. Nööpide joonis on ära toodud ka eraldi lehel, mida vōib kasutada lahenduse esitamiseks.
\begin{center}
	\includegraphics[width=0.8\linewidth]{2004-v3g-08-yl.pdf}
\end{center}

Solution: Klotsile $A$ mõjuva hõõdejõu kompenseerib nööri pinge, sestap on need kaks jõudu samasihilised ning klotsi $A$ juures peab pöörleva aluse kiirusvektor olema nööri sihiline. Pöörlemiskese peab lebama punktist $A$ tōmmatud lõigu $A B$ ristsirgel, pöörlemiskeskme $O$ leiame tingimusest $|\mathrm{OA}|=|\mathrm{AB}|$. Edasi on jooniselt lihtne mõõta $|OB|$ ning leida $b=a|O B| /|O A|=\SI{14}{cm}$. Klotsile $B$ mõjub hõõrdejõud, mille siht on risti sirgega $O B$. Samuti mõjub talle mõlema niidi pinged, mis on paralleelsed vastavate niitidega. Märgime punkti $C$ teisel niidil ning jagame vektori $\overrightarrow{B C}$ (mis kujutab teatud mõõtkavas selle niidi pinget) kaheks komponendiks, mis on paralleelsed kahe ülejäänud jõuga, $\overrightarrow{B C}=\overrightarrow{B D}+\overrightarrow{B E}$. Siinjuures lõik $B D$ on risti lõiguga $O B$ ja lõik $B E$ on paralleelne lõiguga $A B$. Et niidi $A B$ pinge on võrdne klotsile $A$ mõjuva hõõrdejõuga ja hõõrdejõudude suhe annab masside suhte, siis $m_{B}=m_{A}|B D| /|B E| .$ Jooniselt leiame $m_{B}=\SI{1,8}{g}$

Nööbid

Joonisel on kujutatud pealtvaates kahte nööpi $A$ ja $B$, mis on omavahel ühendatud niidiga, mille pikkus $l=\SI{10}{cm}$. Süsteem asub pöörleval alusel. Nööpide ja aluse vaheline hōõrdetegur on $\mu$. Ühe nööbi külge on kinnitatud veel teine niit, mille abil hoitakse nööpe paigal (vt. joonist). On teada, et pöörlemiskese asub klotsist $A$ kaugusel $a=\SI{10}{cm}$ ja mōlemad niidid on horisontaalsed. Kui suur on pöörlemiskeskme ja nööbi $B$ vahekaugus ning nööbi $B$ mass $m_{B}$ ? Nööbi $A$ mass $m_{A}=\SI{1,0}{g}$ ja niidid võib lugeda kaalututeks. Nööpide joonis on ära toodud ka eraldi lehel, mida vōib kasutada lahenduse esitamiseks.
\begin{center}
	\includegraphics[width=0.8\linewidth]{2004-v3g-08-yl.pdf}
\end{center}


[User] Let there be a positively charged ball of mass $m$. It is known that if the ball would move with a speed $v$ in a perpendicular magnetic field of induction $B$ then the ball’s trajectory would be a circle of radius $r$. How big should be the charge $q$ of another ball with the same mass so that the first ball would move with the same speed on the same trajectory in the electric field of the second ball? Assume that no external forces are applied to this system of two balls. All the movements are observed in a laboratory frame of reference.

Solution:
Olgu positiivse laenguga kuulikese laeng $Q$. Kuulikesele mõjuks kirjeldatud magnetväljas Lorentzi jõud suurusega $F_L=QvB$. See jõud oleks kuulikesele mõjuvaks kesktõmbejõuks $F_k=\frac{mv^2}{r}$. Saame $QvB=\frac{mv^2}{r}$. Kaks isoleeritud võrdse massiga kuulikest tiirlevad ümber ühise masskeskme, seega on nende omavaheline kaugus $d=2r$. Kuulikesi ringorbiidil hoidvaks kesktõmbejõuks on kuloniline jõud suurusega
\[
F_C=\frac{kqQ}{d^2}=\frac{kqQ}{4r^2}.
\]
Kuna teineteise ümber tiirlevad laengud peavad olema vastasmärgilised, siis $|F_C|=-F_C$. Võrdsustades kesktõmbejõu ning kulonilise jõu suuruse, saame
\[
\frac{mv^2}{r}=-\frac{kqQ}{4r^2}
\]
ehk
\[
QvB=-\frac{kqQ}{4r^2}.
\]
Lõpptulemuseks saame
\[
q=-\frac{4vBr^2}{k}.
\]

Tiirlev kuulike

Olgu meil positiivselt laetud kuulike massiga $m$. On teada, et kui kuulike
liiguks kiirusega $v$ sellega ristuvas magnetväljas induktsiooniga $B$,
siis oleks selle trajektooriks ringjoon raadiusega~$r$. Kui suure laenguga $q$ peab olema teine sama massiga kuulike, et esimene
kuulike liiguks teise kuulikese elektriväljas sama kiirusega samal trajektooril? Eeldage, et
kahe kuulikese süsteemile ei mõju väliseid jõude. Kõiki liikumisi vaadeldakse
laboratoorses taustsüsteemis.
\pagebreak

Magnetväljas ringliikumisel käitub Lorentzi jõud kesktõmbejõuna. Analoogselt on teisel juhul kesktõmbejõud kuloniline jõud kahe kuuli vahel, kusjuures peab tähelepanu pöörama sellele, et kuulid tiirlevad ümber ühise massikeskme.


[User]
Question B1
The electric potential at the center of a cube with uniform charge density ρ and side length a is

                                                       0.1894ρa2
                                                  Φ≈             .
                                                           0

You do not need to derive this.1
    For the entirety of this problem, any computed numerical constants should be written to three
significant figures.

  a. What is the electric potential at a corner of the same cube? Write your answer in terms of
     ρ, a, 0 , and any necessary numerical constants.

  b. What is the electric potential at the tip of a pyramid with a square base of side length a,
     height a/2, and uniform charge density ρ? Write your answer in terms of ρ, a, 0 , and any
     necessary numerical constants.

  c. What is the electric potential due to a square plate with side length a of uniform charge
     density σ at a height a/2 above its center? Write your answer in terms of σ, a, 0 , and any
     necessary numerical constants.

  d. Let E(z) be the electric field at a height z above the center of a square with charge density
     σ and side length a. If the electric potential at the center of the square is approximately
     0.281aσ
             , estimate E(a/2) by assuming that E(z) is linear in z for 0 < z < a/2. Write your
        0
     answer in terms of σ, a, 0 , and any necessary numerical constants.

  1
      See https://arxiv.org/pdf/chem-ph/9508002.pdf for more details if you are interested.

                       Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                 Part B                                                8

Solution:
Question B1
The electric potential at the center of a cube with uniform charge density ρ and side length a is

                                                       0.1894ρa2
                                                  Φ≈             .
                                                           0

You do not need to derive this.1
    For the entirety of this problem, any computed numerical constants should be written to three
significant figures.

  a. What is the electric potential at a corner of the same cube? Write your answer in terms of
     ρ, a, 0 , and any necessary numerical constants.

                                                       Solution

        By dimensional analysis, the answer takes the form

                                                                   Cρa2
                                                     Φc (a, ρ) ≈
                                                                    0
        for a dimensionless constant C. Note that a cube of side length a consists of 8 cubes of side
        length a/2, each with a corner at the center of the larger cube. Then

                                                0.1894ρa2    Cρ(a/2)2
                                                          =8
                                                    0          0
        so C = 0.1894/2 = 0.0947.

  b. What is the electric potential at the tip of a pyramid with a square base of side length a,
     height a/2, and uniform charge density ρ? Write your answer in terms of ρ, a, 0 , and any
     necessary numerical constants.

                                                       Solution

        A cube of side length a consists of 6 such pyramids. Then we simply compute 0.1894/6 for

                                                                0.0316 ρa2
                                                  Φp (a, ρ) ≈
                                                                    0
        .

  c. What is the electric potential due to a square plate with side length a of uniform charge
     density σ at a height a/2 above its center? Write your answer in terms of σ, a, 0 , and any
     necessary numerical constants.
  1
      See https://arxiv.org/pdf/chem-ph/9508002.pdf for more details if you are interested.

                       Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                   Part B                                                 13

                                                 Solution

     Let the potential due to such a square be Φs (a, σ). Note that adding a square plate of
     infinitesimal thickness dz and side length a to a square pyramid with base side length a and
     height a/2 yields a square pyramid with base side length a + 2dz and height a/2 + dz.
     The surface charge density of a square plate with thickness dz and volume charge density ρ
     is σ = ρ dz. Then by the principle of superposition,

                                                           0.0316ρ ((a + 2dz)2 − a2 )   0.126 aρ dz
          Φs (a, ρ dz) = Φp (a + 2dz, ρ) − Φp (a, ρ) ≈                                =             .
                                                                       0                   0
     so we have
                                                           0.126 aσ
                                             Φs (a, σ) ≈            .
                                                              0

  d. Let E(z) be the electric field at a height z above the center of a square with charge density
     σ and side length a. If the electric potential at the center of the square is approximately
     0.281aσ
             , estimate E(a/2) by assuming that E(z) is linear in z for 0 < z < a/2. Write your
        0
     answer in terms of σ, a, 0 , and any necessary numerical constants.

                                                 Solution

     The potential difference between height 0 and a/2 is
                                                             aσ   0.155 aσ
                                  ∆Φ = (0.281 − 0.126)          =          .
                                                             0      0
     On the other hand, we have
                                       Z a/2
                                                             a E(0) + E(a/2)
                                ∆Φ =           E(z) dz ≈
                                         0                   2       2

     where we approximated E(z) as linear, and E(0) = σ/20 by Gauss’s law. Solving for E(a/2),
                                                            0.119σ
                                              E(a/2) ≈
                                                              0
     where the last significant digit is not important. Incidentally, the actual value is exactly
     σ/60 , and this fact has a slick calculation-free proof.

                  Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                   Part B                                             14

USAPhO 2018 Question B1


[User]
01h
                      APhO 2014, problem 1. A challenging problem which builds intuition about

                  01h
    atmospheric physics.

Solution:
APhO 2014, problem 1. A challenging problem which builds intuition about

                  01h
    atmospheric physics.


[User]
A block of mass m is placed at rest on top of a frictionless wedge of mass M . The
    wedge rests on a frictionless horizontal table, and its sloped top makes an angle θ to the horizontal.

      (a) When the block is released, what is the horizontal acceleration of the wedge?

      (b) Assume the block starts a distance d above the table. Using results from part (a), what is the
          horizontal velocity of the block just before it reaches the floor?

        (c) Find the speed of the block after it reaches the floor by applying energy and momentum
            conservation to the entire process.

      (d) Your results for parts (b) and (c) should not match. What’s going on?

    2     Momentum
         Idea 2
         The momentum of a system is
                                                X
                                           P=        mi vi = M vCM .
                                                 i

         In particular, the total external force on the system is M aCM , and if there are no external
         forces, the center of mass moves at constant velocity.

         Example 2

         A massless rope passes over a frictionless pulley. A monkey hangs on one side, while a bunch
         of bananas with exactly the same weight hangs from the other side. When the monkey tries
         to climb up the rope, what happens?

  Solution
  Remarkably, the answer doesn’t depend on how the monkey climbs, whether slowly or
  quickly, or symmetrically or not! The total vertical force on the monkey is T − mg, so the
  acceleration of the center of mass of the monkey is T /m − g. But since the tension is uniform
  through a massless rope, the acceleration of the bananas is also T /m − g. Therefore, the
  monkey and bananas rise at the same rate, and meet each other at the pulley.

  Now here’s a question for you: compared to climbing up a rope fixed to the ceiling, climbing
  up to the pulley takes twice as much work, because the bananas are raised too. But in both
  cases, isn’t the monkey applying the same force through the same distance? Where does the
  extra work come from? (The answer involves the ideas at the end of this problem set.)

  Example 3: KK 3.14 / INPhO 2014.5

  Two men, each with mass m, stand on a railway flatcar of mass M initially at rest. They
  jump off one end of the flatcar with velocity u relative to the car. The car rolls in the opposite
  direction without friction. Find the final velocities of the flatcar if they jump off at the same
  time, and if they jump off one at a time. Generalize to the case of N ≫ 1 men, with a total
  mass of mtot .

  Solution
  In the first case, by conservation of momentum, we have

                                        M v + 2m(v − u) = 0

  where v is the final velocity of the flatcar, so
                                                   2mu
                                           v=            .
                                                  M + 2m
  In the second case, by a similar argument, we find that after the first man jumps,
                                                   mu
                                           v1 =          .
                                                  M + 2m
  Now transform to the frame moving with the flatcar. When the second man jumps, he
  imparts a further velocity v2 = mu/(M + m) to the flatcar by another similar argument. The
  final velocity of the flatcar relative to the ground is then

                                                     1         1
                             v = v1 + v2 = mu               +      .
                                                   M + 2m M + m

  It might be a bit disturbing that the final speeds and hence energies of the flatcar are
  different, even though the men are doing the same thing (i.e. expending the same amount of
  energy in their legs to jump) in both cases.

  The reason for the difference is that in the second case, the second man to jump ends up
  with less energy, since the velocity he gets from jumping is partially cancelled by the existing

       velocity v1 . So the extra energy that goes into the flatcar corresponds to less kinetic energy
       in the men after jumping, which would ultimately have ended up as heat after they slid to a
       stop. Accounting properly for the kinetic energy of everything in the system solves a lot of
       paradoxes involving energy, as we’ll see below.

       In the case of many men, by similar reasoning we have
                                                       mtot
                                              v=              u
                                                     M + mtot
       in the first case, while in the second case the answer is the sum
                                            N
                                            X     mtot u   1
                                       v=                           .
                                                   N M + (i/N )mtot
                                            i=1

       This can be converted into an integral, by letting x = i/N , in which case ∆x = 1/N and
                                              Z 1
                       X          mtot u                mtot u           M + mtot
                   v=      ∆x              ≈      dx             = log               u.
                               M + xmtot       0     M + xmtot               M
                            i

       Note that this is essentially the rocket equation, which we’ll derive in a different way in M6.

[2] Problem 5 (KK 4.11). A perfectly flexible chain of mass M and length ℓ is suspended vertically
    with its lowest end touching a scale. The chain is released and falls onto the scale. Find the reading
    on the scale when a length of chain x has fallen.

Solution:
(a) Applying Newton’s second law to the wedge gives

                                                    M a = N sin θ.

            Next, work in the noninertial frame of the wedge, where the block only moves parallel to the
            slope. Balancing the forces perpendicular to the slope gives

                                               N = mg cos θ − ma sin θ.

            It’s now straightforward to eliminate N and thereby solve for a, which gives
                                                       mg sin θ cos θ
                                                  a=                  .
                                                       M + m sin2 θ
            By the way, it’s easy to get confused on this problem if you focus too much on the block, because
            its motion is somewhat confusing; you have to decompose it into motion parallel to the slope,
            and motion of the wedge itself. Once you do that, the problem can be solved straightforwardly.
            The solution above is especially short because it never considers the acceleration of the block
            parallel to the slope, which isn’t required to get the answer.

 (b) Since the only horizontal forces in the problem are between the block and wedge, the horizontal
     acceleration of the block is
                                                      M
                                                 ab =    a.
                                                      m
     Thus, the relative acceleration of the two is
                                                             M +m
                                           arel = a + ab =        a.
                                                               m
       The block goes off the wedge
                               p once the two have moved a relative horizontal distance of d/ tan θ,
       which takes a time t = 2d/arel tan θ. At this point the block has a horizontal velocity
                                              p          M cos θ
                                vb = ab t =    2gd p                      .
                                                    (M + m sin2 θ)(M + m)

    (c) By momentum conservation, the final horizontal speeds obey

                                                  M vp = mvb

       while by energy conservation,
                                        1        1
                                          M vp2 + mvb2 = mgd.
                                        2        2
       Combining the two and solving gives
                                                   r
                                              p         M
                                        vb = 2gd            .
                                                     M +m

 (d) It turns out that both results are correct, but they’re the answers to different questions. Note
     that at the instant the block gets to the bottom of the wedge, its velocity isn’t horizontal, but
     right after it’s off the wedge, its velocity must be exactly horizontal. This requires a rather
     large vertical impulse. (For an illustration of this, see F = ma 2021, problems 1 and 2.)
       Depending on how the wedge and block are constructed, there are several possibilities. If the
       wedge abruptly ends, and the block immediately begins moving horizontally, then we have
       an inherently inelastic process. The vertical kinetic energy mvy2 /2 of the block is lost, so the
       answer to part (c) doesn’t apply, and the answer to part (b) is correct. (Another possibility,
       if the ball is very bouncy, is that the sign of its vy will flip, and it’ll bounce off the floor. In
       this case, the answer to part (b) is still correct. Energy is conserved now, but the answer to
       part (c) is still wrong because it assumes the final velocity of the block is horizontal.)
       On the other hand, if the wedge ends in a transition region, where θ smoothly goes to zero,
       then ideally energy remains conserved, and the answer of part (c) applies. In this region, a
       strong normal force reorients the velocity to be horizontal, supplying both a large horizontal
       and vertical impulse. As a result, the answer to part (b) doesn’t apply.
       This subtlety about how wedges end applies to lots of Olympiad problems. The usual assump-
       tion seems to be that option (c) holds, but in reality it depends sensitively on how the wedge
       and block are made. In fact, in practice you can lose a lot of energy even if there’s a smooth
       curve at the end, if that curve is not gradual enough.

2      Momentum

  Idea 2
  The momentum of a system is
                                           X
                                      P=         mi vi = M vCM .
                                             i

  In particular, the total external force on the system is M aCM , and if there are no external
  forces, the center of mass moves at constant velocity.

  Example 2

  A massless rope passes over a frictionless pulley. A monkey hangs on one side, while a bunch
  of bananas with exactly the same weight hangs from the other side. When the monkey tries
  to climb up the rope, what happens?

  Solution
  Remarkably, the answer doesn’t depend on how the monkey climbs, whether slowly or
  quickly, or symmetrically or not! The total vertical force on the monkey is T − mg, so the
  acceleration of the center of mass of the monkey is T /m − g. But since the tension is uniform
  through a massless rope, the acceleration of the bananas is also T /m − g. Therefore, the
  monkey and bananas rise at the same rate, and meet each other at the pulley.

  Now here’s a question for you: compared to climbing up a rope fixed to the ceiling, climbing
  up to the pulley takes twice as much work, because the bananas are raised too. But in both
  cases, isn’t the monkey applying the same force through the same distance? Where does the
  extra work come from? (The answer involves the ideas at the end of this problem set.)

  Example 3: KK 3.14 / INPhO 2014.5

  Two men, each with mass m, stand on a railway flatcar of mass M initially at rest. They
  jump off one end of the flatcar with velocity u relative to the car. The car rolls in the opposite
  direction without friction. Find the final velocities of the flatcar if they jump off at the same
  time, and if they jump off one at a time. Generalize to the case of N ≫ 1 men, with a total
  mass of mtot .

  Solution
  In the first case, by conservation of momentum, we have
                                        M v + 2m(v − u) = 0
  where v is the final velocity of the flatcar, so
                                               2mu
                                           v=         .
                                             M + 2m
  In the second case, by a similar argument, we find that after the first man jumps,
                                                mu
                                        v1 =           .
                                             M + 2m

       Now transform to the frame moving with the flatcar. When the second man jumps, he
       imparts a further velocity v2 = mu/(M + m) to the flatcar by another similar argument. The
       final velocity of the flatcar relative to the ground is then

                                                          1         1
                                  v = v1 + v2 = mu               +      .
                                                        M + 2m M + m

       It might be a bit disturbing that the final speeds and hence energies of the flatcar are
       different, even though the men are doing the same thing (i.e. expending the same amount of
       energy in their legs to jump) in both cases.

       The reason for the difference is that in the second case, the second man to jump ends up
       with less energy, since the velocity he gets from jumping is partially cancelled by the existing
       velocity v1 . So the extra energy that goes into the flatcar corresponds to less kinetic energy
       in the men after jumping, which would ultimately have ended up as heat after they slid to a
       stop. Accounting properly for the kinetic energy of everything in the system solves a lot of
       paradoxes involving energy, as we’ll see below.

       In the case of many men, by similar reasoning we have
                                                       mtot
                                              v=              u
                                                     M + mtot
       in the first case, while in the second case the answer is the sum
                                            N
                                            X     mtot u   1
                                       v=                           .
                                                   N M + (i/N )mtot
                                            i=1

       This can be converted into an integral, by letting x = i/N , in which case ∆x = 1/N and
                                              Z 1
                       X          mtot u                mtot u           M + mtot
                   v=      ∆x              ≈      dx             = log               u.
                               M + xmtot       0     M + xmtot               M
                          i

       Note that this is essentially the rocket equation, which we’ll derive in a different way in M6.

[2] Problem 5 (KK 4.11). A perfectly flexible chain of mass M and length ℓ is suspended vertically
    with its lowest end touching a scale. The chain is released and falls onto the scale. Find the reading
    on the scale when a length of chain x has fallen.

    Solution. Because the chain is flexible, each link just crumples when it hits the ground, without
    pulling the rest of the chain downward. In other words, the assumption of ideal flexibility implies
    the tension in the chain vanishes, so that the vertical part of the chain is always in free fall.
                                                                 √                              √
        Now, the lowest end of the chain is moving with velocity 2gx, so in time dt, a mass M 2gxdt/ℓ
    falls on to the scale, so the change in momentum of that piece is (2M gx/ℓ)dt. Thus, we need a force
    2M gx/ℓ to stop the links that are falling on the scale. In addition, there must be a force M gx/ℓ to
    balance the weight of the chain that’s already lying on the scale, for a total of 3M gx/ℓ .
        This is nice and elegant, but is it true? The result is actually pretty accurate, as you can see
    from experimental data here. The deviation from the expected result is because no chain is perfectly

   flexible. Since the chain has to bend at the spot it hits the scale, some tension is produced, which
   pulls down the rest of the chain slightly faster than free fall.
       This has a connection to the “inherently inelastic” processes mentioned later in the problem set.
   The fastest possible fall corresponds to the case where energy is conserved, i.e. when all the kinetic
   energy of each link hitting the ground is nearly transferred through tension to the still falling part
   of the chain. The answer we gave above corresponds to the slowest possible fall, where each link
   collides perfectly inelastically with the ground. For a flexible chain, the latter is closer to reality.


[User]
A monatomic ideal gas is adiabatically compressed to 1/8 of its original volume. For
    each of the following quantities, indicate by what factor they change.
      (a) The rms velocity vrms .

      (b) The mean free path λ.

      (c) The average time between collisions τ for each gas molecule.

      (d) The molar heat capacity Cv .

Solution:
A monatomic ideal gas is adiabatically compressed to 1/8 of its original volume. For
    each of the following quantities, indicate by what factor they change.

     (a) The rms velocity vrms .

     (b) The mean free path λ.

      (c) The average time between collisions τ for each gas molecule.

      (d) The molar heat capacity Cv .

    Answer. (a) 2 times larger, (b) 8 times smaller, (c) 16 times smaller, (d) same


[User] 01^      IZhO 2022, problem 1.3. A three-dimensional electrostatics problem.

Solution:
See the official solutions here. However, due to some algebraic errors, the final result is
    off by a factor of π, as pointed out in Stefan Ivanov’s errata. Referring to their rubric, the first 5
    formulas are right, but in going to formula 6, they drop the cos β factor inside dq. Starting from
    formula 5, the correct solution would be to write
                                         Q dq                σQ
                               FQ =         √     R cos β = √      cos2 β dβ dα
                                      4πϵ0 ( 2R)3          4 2 πϵ0
    and then perform the integral over β, yielding a factor of π/2, to get
                                                      σQ
                                                 FQ = √     dα.
                                                     8 2 ϵ0
    Setting this equal to mg dα yields an answer of
                                                     √
                                                    8 2 ϵ0 mg
                                                 Q=           .

                   01T
                                                       σ


[User]
Vaatleme kahe üherajalise tee, $A$ ja $B$, liitumist üherajaliseks teeks $C$. Tipptunni ajal on kõik kolm teed täidetud autodega; kahe naaberauto keskmise vahemaa võib lugeda kõigil kolmel teel ühesuguseks. Tee $A$ pikkus on $L_A = \SI{1}{km}$, tee $B$ pikkus $L_B = \SI{3}{km}$ ning tee $C$ pikkus $L_C = \SI{2}{km}$. Autode keskmine kiirus teel $A$ on $v_A = \SI{3}{km/h}$ ning tee $B$ läbimiseks kulub autol $t_B = \SI{36}{min}$. Kui kaua sõidab auto tee $A$ algusest tee $C$ lõpuni?
\begin{center}
	\includegraphics[width=0.9\linewidth]{2007-v2g-02-yl}
\end{center}

Solution:
Lõigul $C$ on teatud punkti ajaühikus läbivate autode arv $N_C$ võrdne lõikude $A$ ja $B$ vastavate arvude summaga: $N_C = N_A + N_B$. Olgu autode vahemaa $a$ ja vaadeldav ajavahemik $\tau$. Siis $N_i = v_i\tau /l$, ehk
\[
\frac{v_{C} \tau}{l}=\frac{v_{A} \tau}{l}+\frac{v_{B} \tau}{l} \quad \Rightarrow \quad v_{C}=v_{A}+v_{B}.
\]
Et $v_B = L_B/t_B$, siis toodud arvude põhjal leiame
\[
v_{B}=\frac{\SI{3}{km}}{\SI{36}{min}}=\SI{5}{km/h}
\]
ning seega $v_C = \SI{8}{km/h}$. Lõpetuseks, $t_A = L_A/v_A=\SI{20}{min}$ ja $t_C = L_C/v_ = \SI{15}{min}C$. Niisiis kulub autol aega $T = t_A + t_C = \SI{35}{min}$.

Ummik

Vaatleme kahe üherajalise tee, $A$ ja $B$, liitumist üherajaliseks teeks $C$. Tipptunni ajal on kõik kolm teed täidetud autodega; kahe naaberauto keskmise vahemaa võib lugeda kõigil kolmel teel ühesuguseks. Tee $A$ pikkus on $L_A = \SI{1}{km}$, tee $B$ pikkus $L_B = \SI{3}{km}$ ning tee $C$ pikkus $L_C = \SI{2}{km}$. Autode keskmine kiirus teel $A$ on $v_A = \SI{3}{km/h}$ ning tee $B$ läbimiseks kulub autol $t_B = \SI{36}{min}$. Kui kaua sõidab auto tee $A$ algusest tee $C$ lõpuni?
\begin{center}
	\includegraphics[width=0.9\linewidth]{2007-v2g-02-yl}
\end{center}

Autode hulga pidevuse tõttu on lõigul $C$ teatud punkti ajaühikus läbivate autode arv võrdne lõikude $A$ ja $B$ vastavate arvude summaga.


[User]
01^      IPhO 2014, problem 1A.

[2] Problem 14 (PPP 14). A bicycle is supported so that it can move forward or backwards but
    cannot fall sideways; its pedals are in their highest and lowest positions.

    A student crouches beside the bicycle and pulls a string attached to the lower pedal, providing a
    backward horizontal force.

     (a) Which way does the lower pedal move relative to the ground?

     (b) Which way does the bicycle move?

                   01r
    To check your answer, watch this video.

Solution:
(a) The student must do positive work on the bike to let it move, so the force and
         displacement of the point where the force is applied have to be parallel. That is, the lower
         pedal moves backwards.

     (b) Technically, it depends on the gearing of the bike, but for a typical bike (and for the one
         shown in the picture) it is backwards. To see this, let’s review how a bike works.
         A bike works by moving the pedals, which are a distance rp from the pedal axle. This rotates
         a gear of radius rg,p , which is connected by a chain to another gear of radius rg,w , which
         rotates the rear wheel of radius R. By accounting for both the gearing ratios, we have
                                         forward motion of bike               R rg,p
                                                                           =         .
                                 backward motion of pedal relative to bike   rg,w rp
         The entire point of a bike’s gearing system is to make this ratio large, so that the bike can go
         fast without your legs having to move very fast. And this is indeed true in the image, which
         has rp and rg,p comparable, and R ≫ rg,w .
         Therefore, if you move the pedal back a little, the bike goes forward a lot more, so the net
         motion of the pedal (relative to the ground) is forward. From part (a), we know the opposite

                   01r
         has to happen, so the bike as a whole goes backwards.


[User]
Question 2

The two-slit electron interference experiment was first performed by Möllenstedt et al, Merli-
Missiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference
experiment, a monochromatic electron point source emits particles at S that first passes
through an electron “biprism” before impinging on an observational plane; S1 and S2 are
virtual sources at distance d. In the diagram, the filament is pointing into the page. Note that
it is a very thin filament (not drawn to scale in the diagram).

The electron “biprism” consists of a grounded cylindrical wire mesh with a fine filament F at
the center. The distance between the source and the “biprism” is ℓ, and the distance between
the distance between the “biprism” and the screen is L.

                                                                                Page 1 of 2

(a)   (2 points) Taking the center of the circular cross section of the filament as the
      origin O, find the electric potential at any point (x,z) very near the filament in
      terms of Va, a and b where Va is the electric potential of the surface of the
      filament, a is the radius of the filament and b is the distance between the center of
      the filament and the cylindrical wire mesh. (Ignore mirror charges.)

(b)   (4 points) An incoming electron plane wave with wave vector kz is deflected by
      the “biprism” due to the x-component of the force exerted on the electron.
      Determine kx the x-component of the wave vector due to the “biprism” in terms of
      the electron charge, e, vz, Va, kz, a and b, where e and vz are the charge and the
      z-component of the velocity of the electrons (kx kz). Note that ⃗         where h
      is the Planck constant.

(c)   Before the point S, electrons are emitted from a field emission tip and accelerated
      through a potential V0. Determine the wavelength of the electron in terms of the
      (rest) mass m, charge and V0,
      (i) (2 points) assuming relativistic effects can be ignored, and
      (ii) (3 points) taking relativistic effects into consideration.

(d)   In Tonomura et al experiment,
                vz     = c/2,
                Va     = 10 V,
                V0     = 50 kV,
                a      = 0.5 Pm,
                b      = 5 mm,
                ℓ      = 25 cm,
                L      = 1.5 m,
                h      = 6.6 x 10-34 Js,
                electron charge, = 1.6 x 10-19 C,
                mass of electron, m0 = 9.1 x 10-31 kg,
                and the speed of light in vacuo, c = 3 x 108 ms-1

      (i)   (2 points) calculate the value of kx,
      (ii)  (2 points) determine the fringe separation of the interference pattern on the
            screen,
      (iii) (1 point) If the electron wave is a spherical wave instead of a plane wave, is
            the fringe spacing larger, the same or smaller than the fringe spacing
            calculated in (ii)?
      (iv) (2 points) In part (c), determine the percentage error in the wavelength of
            the electron using non-relativistic approximation.
      (v) (2 points) the distance d between the apparent double slits.

                                                                              Page 2 of 2

Solution:
Question 2

The two-slit electron interference experiment was first performed by Möllenstedt et al, Merli-
Missiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference
experiment, a monochromatic electron point source emits particles at S that first passes
through an electron “biprism” before impinging on an observational plane; S1 and S2 are
virtual sources at distance d. In the diagram, the filament is pointing into the page. Note that
it is a very thin filament (not drawn to scale in the diagram).

The electron “biprism” consists of a grounded cylindrical wire mesh with a fine filament F at
the center. The distance between the source and the “biprism” is ℓ, and the distance between
the distance between the “biprism” and the screen is L.

                                                                                Page 1 of 6

(a)   (2 points) Taking the center of the circular cross section of the filament as the
      origin O, find the electric potential at any point (x,z) very near the filament in
      terms of Va, a and b where Va is the electric potential of the surface of the
      filament, a is the radius of the filament and b is the distance between the center of
      the filament and the cylindrical wire mesh. (Ignore mirror charges.)

       Writing out |E| =                     ( )

                                                       (1 point)

       Note that

                                  ( )                  (= 0 at the mesh)

       Also at the edge of the filament, Va = (          ) , so

       Giving together
                                             (     )
                                ( )                    where       √
                                             (     )

       (1 point for final expression)

                                                                              Page 2 of 6

(b)   (4 points) An incoming electron plane wave with wave vector kz is deflected by
      the “biprism” due to the x-component of the force exerted on the electron.
      Determine kx the x-component of the wave vector due to the “biprism” in terms of
      the electron charge, e, vz, Va, kz, a and b, where e and vz are the charge and the
      z-component of the velocity of the electrons (kx       kz). Note that ⃗       where
      h is the Planck constant.

 There are several ways to work out the solution:
       A charge in an electric field will experience a force and hence a change in
       momentum. Note that potential energy of the electron (charge =        ) is
             ( ) Using impulse acting on the electron due to the electric field,
       (2 points)

                                                         (       )
                                  ∫ (        )(                      )

                                    ∫
                                         (           )

                                                         (2 points for final expression)

       The alternative solution is to write down the equations of motion for the
       electrons (2 points) and determine the deflection of the electron as it passes
       through the “biprism”:

             Since                 ,

                                                         (2 points for final expression)

                                                                                Page 3 of 6

(c)   Before the point S, the electrons are emitted from a field emission tip and
      accelerated through a potential V0. Determine the wavelength of the electron in
      terms of the (rest) mass m, charge – and V0,
      (i) (2 points) assuming relativistic effects can be ignored.

        Equating the kinetic energy to eV0 (1 point)

                                         √

                                         √
        (1 point for final expression)

      (ii)   (3 points) taking relativistic effects into consideration.

        Consider

                                     (       )   (       )

                                     (       )   (       )

                                     (                       )       (    )

                                                     (               )

                                     √               (               )

        (                                                        )

                                                                              Page 4 of 6

(d)   In Tonomura et al experiment,
                vz     = c/2,
                Va     = 10 V,
                V0     = 50 kV,
                a      = 0.5 Pm,
                b      = 5 mm,
                ℓ      = 25 cm,
                L      = 1.5 m,
                h      = 6.6 x 10-34 Js,
                electron charge,     = 1.6 x 10-19 C,
                mass of electron, m = 9.1 x 10-31 kg,
                and the speed of light in vacuo, c = 3 x 108 ms-1

      (i)    (2 points) calculate the value of kx ,

        Previous equation:

        Plugging the relevant numbers into the equation gives:
                                      (1 point for plugging the correct values)

                                      (1 point for final expression)

      (ii)   (2 points) determine the fringe separation of the interference pattern on
             the screen,

        Fringe separation is given by

        (                                             )

      (iii) (1 point) If the electron wave is a spherical wave instead of a plane wave, is
            the fringe spacing larger, the same or smaller than the fringe spacing
            calculated in (ii)?

        Larger. (1 point for the correct answer)

                                                                                Page 5 of 6

    (iv) (2 points) In part (c), determine the percentage error in the wavelength of
         the electron using non-relativistic approximation.

     Non-relativistic:

                                 √

                                 √

     Relativistic:

                                 √           (             )

     Percentage error:

                                  or 2.4 percent.

(                                                                           )

    (v)   (2 points) Calculate the distance d between the apparent double slits.

     The double slit formula is given by
                                     (      )

     where m is the order and y is the distance for maximum intensity
     from the central fringe.

     In this case, since the fringe spacing is 907Å,

     (                                       )

                                                                           Page 6 of 6

APhO 2014 Theory T2


[User]
A clay ball of mass 10 g fell vertically down to a smooth horizontal ground and broke into three fragments. The fragments scattered on the ground and stopped at points that are given in the figure (top view, the cross represents the falling point). Find the masses of the fragments. You can make additional designs on the drawing (on a separate paper) and take additional measurements. You can assume that the fragments started to slide right after the fall without bouncing upwards, the air friction is insignificant and that the coefficient of kinetic friction does not depend on the speed.
\begin{center}
\includegraphics[width=0.8\linewidth]{2012-lahg-10-killud}
\end{center}

Solution:
Kildude koguimpulss on null, seega moodustavad impulsivektorid kolmnurga, mille saame sarnasusteguri täpsusega taastada kildude liikumissuundade abil.
Sellest kolmnurgast saame küljepikkuste suhete mõõtmise abil teada algimpulsside suhted $p_1 : p_2: p_3 = \num{72} : \num{24} : \num{59}$ (need numbrid on
kolmnurga külgede pikkused millimeetrites).
Kildude libisemiskaugused suhtuvad kui algkiiruste ruudud --- mõõtmiste abil saame teada
\[
v_1^2:v_2^2:v_3^2 = \num{72}:\num{30}:\num{21}
\]
ning ruutjuure võtmise järel ka
kiiruste suhted
\[
v_1 : v_2: v_3 = \num{8,5}:\num{5,5}:\num{4,6}.
\]
Jagades impulsside suhted kiiruste suhtega, saame teada masside suhted
\[
m_1:m_2:m_3=\num{8,5}:\num{4,4}:\num{12,8}
\]
ning teades kogumassi ka massid eraldi võetuna,
\[
m_1=\SI {10}g\cdot \frac{\num{8,5}}{\num{8,5}+\num{4,4}+\num{12,8}} = \SI{3,3}g\text{, }m_2=\SI {10}g\cdot \frac{\num{4,4}}{\num{25,7}}=\SI{1,7}g\text{, }
m_3=\SI {10}g\cdot \frac{\num{12,8}}{\num{25,7}}=\SI{5,0}g.
\]
\hspace*{0\columnwidth}\includegraphics[width=\columnwidth]{2012-lahg-10-killud_lah}

Killud

Savikuulike massiga \SI{10}{g} kukkus vertikaalselt alla siledale horisontaalsele
põrandale ja läks kolmeks killuks.
Killud lendasid laiali ja peatusid punktides, mis on näidatud juuresoleval
joonisel
(ülaltvaade, ristiga on märgitud kukkumiskoht). Määrake kildude massid.
Joonisel (lisalehel)
võib teha lisakonstruktsioone ja mõõtmisi.
Võite lugeda, et killud hakkasid kohe pärast kukkumist üles põrkumata libisema,
õhuhõõre on tühine ja liugehõõrdetegur ei sõltu kiirusest.
\begin{center}
\includegraphics[width=0.8\linewidth]{2012-lahg-10-killud}
\end{center}

Kildude koguimpulss on null, seega moodustavad impulsivektorid kolmnurga, mille sarnasustegurid on võimalik kildude liikumissuundadest taastada. Sarnaselt saab toimida kildude kineetilise energia ja libisemiskaugustega.


[User]
1. A glass rod of refractive index 1.50 of rectangular cross section {d × l} is bent into a
   “U” shape (see Fig. (A). The cross sectional view of this rod is shown in Fig. (B).

                                                               p1

                                                                             R+d
                                                       p2
                                                                       R

                 D        C         H         G
                                E         l
             A       B              d F

                                                            r2 r1
                              (A)
                                                       6               (B)

   Bent portion of the rod is semi-circular with inner and outer radii R and R + d re-

                                        C1
   spectively. Parallel monochromatic beam of light is incident normally on face ABCD.
   (a) Consider two monochromatic rays r1 and r2 in Fig. (B). State whether the                  [1]
        following statements are True or False.
                                        Statement                                     True/False
    If r1 is total internally reflected from the semi circular section at the

             HB
    point p1 then r2 will necessarily be total internally reflected at the point p2 . True
    If r2 is total internally reflected from the semi circular section at the
    point p2 then r1 will necessarily be total internally reflected at the point p1 . False

   (b) Consider the ray r1 whose point of incidence is very close to the edge BC. Assume
       it undergoes total internal reflection at p1 . In cross sectional view below, draw
       the trajectory of this reflected ray beyond the next glass-air boundary that it
                                                                                [1]

       encounters.
                                       p1
                                           ii
                                                               i
                                                                   i
                                                   O

                                              r1

                      INPhO - 2016 - Final Solutions
    (c) Obtain the minimum value of the ratio R/d for which any light ray entering              [2]
        the glass normally through the face ABCD undergoes at least one total internal
        reflection.

   (d) A glass rod with the above computed minimum ratio of R/d, is fully immersed              [2]
       in water of refractive index 1.33. What fraction of light flux entering the glass

                                                        6
       through the plane surface ABCD undergoes at least one total internal reflection?

                                       C1
                  HB

Solution:
1. A glass rod of refractive index 1.50 of rectangular cross section {d × l} is bent into a
   “U” shape (see Fig. (A). The cross sectional view of this rod is shown in Fig. (B).

                                                               p1

                                                                             R+d
                                                       p2
                                                                       R

                 D        C         H         G
                                E         l
             A       B              d F

                                                            r2 r1
                              (A)
                                                       6               (B)

   Bent portion of the rod is semi-circular with inner and outer radii R and R + d re-

                                        C1
   spectively. Parallel monochromatic beam of light is incident normally on face ABCD.
   (a) Consider two monochromatic rays r1 and r2 in Fig. (B). State whether the                  [1]
        following statements are True or False.
                                        Statement                                     True/False
    If r1 is total internally reflected from the semi circular section at the

             HB
    point p1 then r2 will necessarily be total internally reflected at the point p2 . True
    If r2 is total internally reflected from the semi circular section at the
    point p2 then r1 will necessarily be total internally reflected at the point p1 . False

   (b) Consider the ray r1 whose point of incidence is very close to the edge BC. Assume
       it undergoes total internal reflection at p1 . In cross sectional view below, draw
       the trajectory of this reflected ray beyond the next glass-air boundary that it
                                                                                [1]

       encounters.
                                       p1
                                           ii
                                                               i
                                                                   i
                                                   O

                                              r1

                      INPhO - 2016 - Final Solutions
    (c) Obtain the minimum value of the ratio R/d for which any light ray entering              [2]
        the glass normally through the face ABCD undergoes at least one total internal
        reflection.

          Solution: For total internal reflection

                                              sin i ≥ sin θc

          where i is the incidence angle of the ray on bent portion of the rod (see figure
          in part (b)) and θc is the critical angle for glass-air boundary. For a light ray
          close to edge BC
                                                                R       1
                                                                    ≥
                                                               R+d    nglass
                       For refractive index nglass = 1.5          R ≥ 2d

          Minimum value of R/d = 2

   (d) A glass rod with the above computed minimum ratio of R/d, is fully immersed              [2]
       in water of refractive index 1.33. What fraction of light flux entering the glass

                                                           6
       through the plane surface ABCD undergoes at least one total internal reflection?

          Solution:
                                       C1           O

                  HB                    r3
                                             x

          Let the intensity of beam be I0 . Flux entering through glass slab will be dlI0 .
          Assume that any light ray up to distance x from the edge BC undergoes at
          least one total internal reflection. Then the flux going through at least one
          total internal reflection will be (d − x)lI0 . Also

                                             R+x   nwater
                                                 =
                                             R+d   nglass

          For R/d = 2 and nwater = 1.33, x = 2d/3.
          Fraction of light = 0.33 or 0.34.

INPhO 2016 Problem 1


[User]
Tasaparalleelne plaat koosneb kahest klaaskiilust väikse nurgaga $\varphi \ll 1$ (vt joonist). Kiilude murdumisnäitajad on $n_1$ ja $n_2$ ($n_2 > n_1$). Plaadile risti tema pinnaga langeb paralleelne valgusvihk. Plaadi taga asub koondav lääts fookuskaugusega $f$. Läätse fokaaltasandis asub ekraan. Joonistage kiirte käik süsteemis. Kui palju nihkub valguslaik ekraanil, kui me eemaldame plaadi?

\emph{Vihje:} väikeste nurkade puhul kehtib ligikaudne võrdus $\tan \varphi \approx \sin \varphi \approx \varphi $.

\begin{center}
	\includegraphics[width=0.6\linewidth]{2006-lahg-06-yl}
\end{center}

Solution:
On lihtne näha, et plaadist väljub endiselt paralleelne valgusvihk. Küll aga on selle levimise suund muutunud. Plaati sisenedes murdumist ei toimu, sest kiired liiguvad risti pinnaga. Kaldpinnale langevad kiired langemisnurga $\varphi$ all, murdumisnurga $\gamma$ saame murdumisseadusest:
\[
n_1 \sin \varphi = n_2 \sin \gamma.
\]
Meil on lubatud kasutada väikeste nurkade lähendust $\sin \alpha \approx \alpha$, mistõttu $\varphi n_1 = \gamma n_2$. Murdumise tõttu muutus kiirte levimise suund nurga $\varphi -\gamma$ võrra. Ühtlasi on lihtne näha, et see on ka langemisnurgaks plaadi välistasandile, sest esialgu liikus kiir risti plaadiga. Arvestades, et õhu murdumisnäitaja on 1, saame leida kiire murdumisnurga $\delta$ plaadist väljumisel:
\[
n_{2}(\varphi-\gamma)=\delta, \quad \delta=n_{2}\left(\varphi-\frac{\varphi n_{1}}{n_{2}}\right)=\varphi\left(n_{2}-n_{1}\right).
\]
Valguslaigu nihet on nüüd lihtne leida. Vaatleme läätse optilist keskpunkti
läbivat kiirt. Et see kiir läätses ei murdu, lõikab see fokaaltasandit teljest kaugusel
\[
d = \delta f = \varphi f (n_2 - n_1).
\]

Kiilud

Tasaparalleelne plaat koosneb kahest klaaskiilust väikse nurgaga $\varphi \ll 1$ (vt joonist). Kiilude murdumisnäitajad on $n_1$ ja $n_2$ ($n_2 > n_1$). Plaadile risti tema pinnaga langeb paralleelne valgusvihk. Plaadi taga asub koondav lääts fookuskaugusega $f$. Läätse fokaaltasandis asub ekraan. Joonistage kiirte käik süsteemis. Kui palju nihkub valguslaik ekraanil, kui me eemaldame plaadi?

\emph{Vihje:} väikeste nurkade puhul kehtib ligikaudne võrdus $\tan \varphi \approx \sin \varphi \approx \varphi $.

\begin{center}
	\includegraphics[width=0.6\linewidth]{2006-lahg-06-yl}
\end{center}

On lihtne näha, et plaadist väljub endiselt paralleelne valgusvihk. Küll aga on selle levimise suund muutunud. Valguslaigu nihke leidmiseks on mugav vaadelda kiirt, mis läbib läätse optilist keskpunkti.