[User]
Q1-1
Theory

                                                                                English (Official)

Water-Hammer Effect
Introduction

This problem studies variations of fluid pressure caused by pressure waves in a flow pipe. Proposed tasks
mainly deal with the water-hammer effect arising from both fast and slow closings of a flow-control valve
in the pipe.
We consider only nonviscous liquids and liquid flows which are essentially one-dimensional. All pipes
including their valves are assumed to be rigid, but liquids are not always considered to be incompressible.
If a liquid element of volume 𝑉0 at equilibrium under pressure 𝑃0 is subjected to a change of pressure
Δ𝑃 , the change of its volume Δ𝑉 is assumed to be proportional to Δ𝑃 so that

                                                           Δ𝑉
                                                Δ𝑃 = −𝐵                                                      (1)
                                                           𝑉0

The constant of proportionality𝐵 represents the bulk modulus of the liquid. For water, take 𝜌0 = 1.0 ×
103 kg/m3 as its equilibrium density and 𝐵 = 2.2 GPa.

Part A. Excess Pressure and Propagation of Pressure Wave (2.2 points)
In a uniform cylindrical pipe of length 𝐿, water is flowing steadily along the +𝑥 direction with horizontal
velocity 𝑣0 , density 𝜌0 , and pressure 𝑃0 . As shown in Fig. 1, the pipe is connected to a reservoir at a depth
ℎ and opens into the atmosphere at pressure 𝑃a .
Suppose the flow-control valve T at the end of the pipe is then shut instantly so that the oncoming liquid
element next to the valve suffers both a pressure change Δ𝑃s ≡ 𝑃1 −𝑃0 and a velocity change Δ𝑣 = 𝑣1 −𝑣0
with 𝑣1 ≤ 0. This causes a longitudinal wave of excess pressure Δ𝑃s to travel upstream in the −𝑥 direction
with a speed of propagation 𝑐.

                                    Fig. 1: Steady flow in a uniform pipe.

     A.1     The excess pressure Δ𝑃s is related to the velocity change Δ𝑣 by Δ𝑃s = 𝛼𝜌0 𝑐Δ𝑣.          1.6pt
             The speed of propagation 𝑐 is given by 𝑐 = 𝛽 + √𝛾𝐵/𝜌0 . Find 𝛼, 𝛽, and 𝛾.

                                                                                Q1-2
Theory

                                                                                English (Official)

     A.2       Calculate values of 𝑐 and Δ𝑃s for the case of water flow with 𝑣0 = 4.0 m/s and       0.6pt
               𝑣1 = 0.

Part B. A Model for the Flow-control Valve (1.0 points)
Fig. 2 shows a model for control valve T and the liquid flow through it. The valve is taken to be a short
section of length Δ𝐿 and inner radius 𝑅 near the end A of the pipe. Its cone-shaped outlet has an orifice
of radius 𝑟 and opens into the atmosphere at pressure 𝑃a . Effects of gravity on the efflux are to be
neglected.
The liquid is to be regarded as incompressible and the flow as steady with liquid element at the valve
inlet having velocity 𝑣in , pressure 𝑃in , and density 𝜌0 . In Fig. 2, stream lines and normal lines are drawn
only as an aid for visualizing the flow pattern.

                              Fig. 2: Valve dimensions and contraction of jet.

It is known that, after leaving the valve into the atmosphere, the cross section of the flow will contract
until it reaches a minimum where the stream lines are again parallel.
                                                                    √ At this point of minimum, the flow
velocity is 𝑣c and the cross section of the flow has a radius 𝑟c = 𝑟 𝐶 c . Here 𝐶c , called the contraction
coefficient, depends on the ratio 𝑟/𝑅 and the cone angle 𝛽 as shown in Table 1.

        𝑟/𝑅                      𝐶c (𝛽 = 45∘ )                      𝐶c (𝛽 = 90∘ )
        0.00                     0.746                              0.611
        0.20                     0.747                              0.616
        0.30                     0.748                              0.622
        0.40                     0.749                              0.631
        1.00                     1.000                              1.000
                                Table 1. Contraction Coefficients for Orifices

                                                                                Q1-3
Theory

                                                                                English (Official)

     B.1     Find the excess pressure Δ𝑃in = 𝑃in − 𝑃a at the valve inlet where the stream           1.0pt
             lines are parallel. Give your answer in terms of 𝜌0 , 𝑣in , 𝑟, 𝑅, and 𝐶c .

For all tasks in Part C and Part D, we consider the reservoir-pipe system in Fig. 1 and make the following
assumptions:
   • Speed of propagation 𝑐 and density 𝜌0 of liquid are given constants independent of flow velocity.
     The ambient atmospheric pressure 𝑃a and the acceleration of gravity 𝑔 are constant.
   • Initially, the valve is fully open and the flow in the pipe is steady with fluid pressure 𝑃0 and velocity
     𝑣0 .
   • As in Fig. 1 and Fig. 2, the pipe has length 𝐿 and radius 𝑅. The valve T is a circular opening of variable
     radius 𝑟 with angle 𝛽 = 90∘ and its length Δ𝐿 is negligible so that the valve inlet is effectively at the
     end A of the pipe. Effects of gravity on the efflux are negligible.
   • Liquid in the reservoir is quasi-static so that its pressure 𝑃ℎ near the pipe entrance B remains con-
     stant and we assume that the variation of fluid pressure across the pipe is negligible so that the
     flow is one-dimensional throughout the pipe.
   • The model outlined in Part B may be used to determine the excess pressure Δ𝑃in = 𝑃in − 𝑃a at the
     valve inlet.

Part C. Water-Hammer Effect due to Fast Closure of Flow Control Valve (1.8 points)
Refer to the reservoir-pipe system in Fig. 1. When liquid flow in the pipe is obstructed by complete or
partial closure of the valve, a pressure wave starts traveling upstream. It gets reflected at the reservoir
end of the pipe and travels back to the valve and gets reflected there. Then another pressure wave is
generated and the process just described is repeated. This causes a sequence of sudden pressure surges
and dips for liquid element next to the valve and is referred to as water-hammering.

     C.1     Refer to Fig. 1 and Fig. 2. Find the pressure 𝑃0 and velocity 𝑣0 of the steady flow    0.6pt
             in the pipe when valve T is fully open (𝑟 = 𝑅). Give answers in terms of 𝜌0 , 𝑔, ℎ,
             and 𝑃a .

     C.2     Consider the same steady flow as in task C.1 with pressure 𝑃0 and flow velocity        1.2pt
             𝑣0 . Now, at 𝑡 = 0, the valve is closed (𝑟 = 0) instantly. A pressure wave heads
             toward the reservoir with speed of propagation 𝑐. Take note 𝑃ℎ = 𝑃0 + 𝜌0 𝑔ℎ.
             Let 𝜏 = 2𝐿/𝑐. What are the fluid pressure 𝑃 (𝑡) and flow velocity 𝑣(𝑡) in the pipe
             when 𝑡 is getting very close to each of the instants 𝜏 /2 and 𝜏 ?

Part D. Water-Hammer Effect due to Slow Closure of Flow Control Valve (5.0 points)
Consider again the same steady flow as in task C.1 with pressure 𝑃0 and flow velocity 𝑣0 . Now we close
the valve slowly and adopt a finite-step approach to simulate the closing process.
Starting at time 𝑡 = 0, the instant reduction of the radius 𝑟 of the valve (see Fig. 2) is carried out sequen-
tially at a time interval 𝜏 = 2𝐿/𝑐. Immediately after each instant reduction of radius, the flow in the valve

                                                                               Q1-4
Theory

                                                                                English (Official)

region is approximated to be steady as in Part B. The pressure and velocity at the valve are then different
from those of the rest of the flow in the pipe.
For each closing step 𝑛, its duration and the radius 𝑟𝑛 of the valve opening are specified in Table 2 along
with the symbols used to represent the corresponding fluid pressure 𝑃𝑛 and flow velocity 𝑣𝑛 at the valve.

 closing step 𝑛     time interval of   ratio 𝑟𝑛 /𝑅        pressure at valve when     flow velocity at valve
                    step 𝑛                                𝑡 = (𝑛 − 1)𝜏               when 𝑡 = (𝑛 − 1)𝜏
 𝑛=0                𝑡<0                1.00               𝑃0                         𝑣0
 𝑛=1                0≤𝑡<𝜏              0.40               𝑃1                         𝑣1
 𝑛=2                𝜏 ≤ 𝑡 < 2𝜏         0.30               𝑃2                         𝑣2
 𝑛=3                2𝜏 ≤ 𝑡 < 3𝜏        0.20               𝑃3                         𝑣3
 𝑛=4                3𝜏 ≤ 𝑡 < 4𝜏        0.00               𝑃4                         𝑣4 = 0
                                       Table 2. Valve closing steps

Take fluid density 𝜌0 and speed of propagation 𝑐 as constants. Let 𝑛 = 0, 1, 2, 3, 4. Define Δ𝑃𝑛 = 𝑃𝑛 − 𝑃0
and Δ𝑣𝑛 = 𝑣𝑛 − 𝑣0 . Make sure to enforce the approximation 𝑃ℎ = 𝑃0 .

    D.1     Obtain an equation which expresses Δ𝑃𝑛 /(𝜌0 𝑐) in terms of Δ𝑃𝑛−1 /(𝜌0 𝑐), 𝑣𝑛−1 ,       3.0pt
            and 𝑣𝑛 . It must be valid for all steps 𝑛 > 0 specified in Table 2. For 𝑛 = 1, 2, 3,
            obtain also an equation which allows 𝑣𝑛 to be computed if both 𝑣𝑛−1 and
            Δ𝑃𝑛−1 /(𝜌0 𝑐) are known.

    D.2     Apply the result of task D.1 to water flow with 𝑣0 = 4.0 m/s. Use the graph paper      2.0pt
            provided in the Answer Sheet to make all plots of Δ𝑃 versus 𝜌0 𝑐𝑣. Be sure to
            draw lines and curves intersecting at points having coordinates which give the
            values of 𝜌0 𝑐𝑣𝑛 and Δ𝑃𝑛 for steps 𝑛 = 1, 2, 3, 4. On the plot, label each point of
            intersection (𝜌0 𝑐𝑣𝑛 , Δ𝑃𝑛 ) with the value of 𝑛 to which it corresponds. From the
            graph, estimate values of 𝜌0 𝑐𝑣𝑛 and Δ𝑃𝑛 (both in units of MPa) for 𝑛 = 1, 2, 3, 4.

Solution:
Solution
                                   Water Hammer
Part A. Excess Pressure and Propagation of Pressure wave

                 𝑃a
                                                                 𝑥
                                                                          T

                                              𝑣0         𝑣1
                                    𝜌0   𝑃0                     𝜌1   𝑃1

                                              𝑐                           A
                               B               wave
                      Fig. S1. Pressure wave (shaded) with speed 𝑐

A.1 (1.6 pt) Excess pressure and speed of propagation of the pressure wave
     When the valve opening is suddenly blocked, fluid pressure at the valve jumps
from 𝑃0 to 𝑃1 = 𝑃0 + ∆𝑃s , thus sending a pressure wave traveling upstream (to the
left) with speed 𝑐 and amplitude ∆𝑃s . Taking positive 𝑥 direction as pointing to
the right, the velocity of fluid particles next to the valve changes from 𝑣0 to 𝑣1
(𝑣1 ≤ 0). Thus the velocity change is ∆𝑣 = 𝑣1 − 𝑣0 .
     In a frame moving to left (along – 𝑥 direction) with speed 𝑐, i.e., riding on the
wave (see Fig. S1), velocity of fluid in the pressure wave is 𝑐 + 𝑣1 , while that of the
incoming fluid in the steady flow ahead of the wave is 𝑐 + 𝑣0 . Let 𝜌1 be the density
of fluid in the pressure wave. From conservation of mass, i.e., equation of continuity,
we have
                   𝜌0 (𝑐 + 𝑣0 ) = 𝜌1 (𝑐 + 𝑣1 )                                        (a1)
or, by letting ∆𝜌 ≡ 𝜌1 − 𝜌0 ,
                  ∆𝜌          𝜌0 𝑣0 − 𝑣1    −∆𝑣
                     =1− =               =                                           (a2)
                  𝜌1          𝜌1 𝑐 + 𝑣0    𝑐 + 𝑣0
Moreover, impulse imparted to the fluid must equal its momentum change. Thus, in
a short time interval 𝜏 after the valve is closed, we must have
                  𝜌0 (𝑐 + 𝑣0 )𝜏[(𝑐 + 𝑣1 ) − (𝑐 + 𝑣0 )] = −𝜏∆𝑃 = (𝑃0 − 𝑃1 )𝜏 (a3)
or
                        𝑣0                          𝑣0                𝑣0
     ∆𝑃s = −𝜌0 𝑐 (1 +      ) (𝑣1 − 𝑣0 ) = −𝜌0 𝑐 (1 + ) ∆𝑣 ⇒ 𝛼 = − (1 + ) (a4)
                        𝑐                           𝑐                 𝑐
If 𝑣0 /𝑐 ≪ 1, we have
                  ∆𝑃s = −𝜌0 𝑐∆𝑣                                                    (a5)
    Note that the negative sign in Eqs. (a4) and (a5) follows from the fact that the
direction of propagation is opposite to the positive direction for 𝑥 axis (and velocity).
Otherwise the sign should be positive. Note also that for a compressional wave

(∆𝑃s > 0), the velocity imparted to the fluid particle is in the direction of propagation,
while for an extensional wave (∆𝑃s < 0), the velocity imparted is in the opposite
direction of propagation.
    Eqs. (a2) and (a4) can be combined to give
                                     𝑣0 2 ∆𝜌
                 ∆𝑃s = 𝜌0 𝑐 2 (1 +     )                                            (a6)
                                     𝑐 𝜌1
From the definition of the bulk modulus 𝐵, which is assumed to be constant, it
follows
                           𝑉0 − 𝑉1    1/𝜌0 − 1/𝜌1    ∆𝜌
                 ∆𝑃s = 𝐵           =𝐵             =𝐵                                (a7)
                              𝑉0         1/𝜌0        𝜌1
From Eqs. (a6) and (a7), we obtain
                               𝑣0 2
                 𝜌0 𝑐 2 (1 +     ) =𝐵                                               (a8)
                               𝑐
Thus

                      𝐵
                 𝑐 = √ − 𝑣0                    ⇒ 𝛾=1           𝛽 = −𝑣0              (a9)
                      𝜌0

However, if in the definition of bulk modulus one uses the fractional change of
density ∆𝜌/𝜌0 instead of −∆𝑉/𝑉0 , the result is then 𝛾 = 1 + ∆𝑃s /𝐵.* Either result
is considered valid.
If 𝑣0 /𝑐 ≪ 1, we have

                        𝐵
                 𝑐=√                                                               (a10)
                        𝜌0

*The result (a7) is pointed out by Dr. Jaan Kalda.

A.2 (0.6 pt) Values of 𝑐 and ∆𝑃s for water flow
Ans:
     From Eqs. (a5) and (a10), we have
                         𝑐 = √𝐵/𝜌0
                        Δ𝑃s = 𝜌0 𝑐𝑣0 = 𝑣0 √𝜌0 𝐵
Putting in the given values 𝑣0 = 4.0 m/s, 𝑣1 = 0, 𝜌0 = 1.0 × 103 kg/m3 ,
and 𝐵 = 2.2 × 109 Pa, we have
                        𝑐 = √𝐵/𝜌0 = 1.5 × 103 m/s                                   (b1)
                       Δ𝑃s = 𝑣0 √𝜌0 𝐵 = 5.9 MPa                                     (b2)
so that Δ𝑃s is nearly 59 times the standard pressure.
     Note that 𝑣0 /𝑐~10−3 so that the use of approximate formulas (a5) and (a10) is
justified when solving tasks in this problem.

Part B. A Model for the Flow-Control Valve
(B.1) (1.0 pt) Excess pressure at valve inlet

                    𝜌0                              𝑃a

                    𝑣in
                                                                   𝑣c
                             2𝑅     2𝑟
                    𝑃in
                                                         2𝑟c = 2𝑟√𝐶c
                                             𝛽
                            Δ𝐿           A
            Fig. 2. Valve dimensions and contraction of jet.
Ans:
  The model assumes the fluid to be incompressible. Neglecting effects of gravity,
Bernoulli’s principle gives us
                  1              1
                   𝜌 𝑣 2 + 𝑃in = 2𝜌0 𝑣c2 + 𝑃a
                  2 0 in
                                                                               (c1)

Equation of continuity and definition of contraction coefficient imply that
                  𝜋𝑅2 𝑣in = 𝜋𝑟c2 𝑣c = 𝜋𝑟 2 𝐶c 𝑣c
Therefore
                       1 𝑅 2
                 𝑣c = ( ) 𝑣in                                                  (c2)
                       𝐶c 𝑟
    From Eqs. (c1) and (c2), we obtain
                                    1    2
                                              1 𝑅 4       𝑘    2
                   ∆𝑃in = 𝑃in − 𝑃a = 𝜌0 𝑣in [ 2 ( ) − 1] = 𝜌0 𝑣in              (c3)
                                    2        𝐶c 𝑟         2
This may be cast into a form involving only dimensionless variables:
                   ∆𝑃in 1 𝑣in 2 1 𝑅 4         𝑘 𝑣in 2
                         = (  ) [ 2( ) − 1] =  ( )                             (c4)
                   𝜌0 𝑐 2 2 𝑐    𝐶c 𝑟         2 𝑐
where
                          1 𝑅 4
                    𝑘 = [ 2 ( ) − 1]                                           (c5)
                         𝐶c 𝑟
Thus we see from eq. (c4) that ∆𝑃in is a quadratic function of 𝑣in .

Part C. Water-Hammer Effect due to Fast Closure of Flow-Control Valve
(C.1) (0.6 pt) Pressure 𝑃0 and velocity 𝑣0 when the valve is fully open
Ans:
     According to Bernoulli’s theorem and the definition of 𝑃ℎ , we have

                   1             1
                    𝜌 𝑣 2 + 𝑃0 = 2𝜌0 𝑣c2 + 𝑃a = 0 + 𝑃a + 𝜌0 𝑔ℎ = 𝑃ℎ
                   2 0 0
                                                                                (d1)

From the second equality in the preceding equation, it follows
                 𝑣c = √2𝑔ℎ
Furthermore, from continuity equation and 𝐶c (𝑟 = 𝑅) = 1.0, we have
                  𝜋𝑅2 𝑣0 = 𝜋(𝐶c 𝑅)2 𝑣c = 𝜋𝑅2 𝑣c ⇒ 𝑣0 = 𝑣c = √2𝑔ℎ                    (d2)
Therefore
                    𝑃0 = 𝑃a = 𝑃ℎ − 𝜌0 𝑔ℎ                                    (d3)
                                                                   𝜏 𝐿
(C.2) (1.2 pt) Pressure 𝑃(𝑡) and flow velocity 𝑣(𝑡) just before 𝑡 = = and 𝑡 = 𝜏
                                                                      2    𝑐
Ans:
     When the valve is open, the flow in the pipe is steady with velocity 𝑣0 and
pressure 𝑃0 . The sudden closure of the valve causes an excess pressure Δ𝑃𝑠 on the
fluid element next to the valve, causing it to stop with velocity 𝑣1 = 0. The velocity
change is thus ∆𝑣 = 𝑣1 − 𝑣0 = −𝑣0 . Thus, according to Eq. (a5), the excess pressure
on the fluid is given by
                  𝛥𝑃s = −𝜌0 𝑐∆𝑣 = 𝜌0 𝑐𝑣0                                           (e1)
     At time 𝑡 = 𝜏/2 = 𝐿/𝑐, the pressure wave reaches the reservoir. The velocity of
fluid in the length of the pipe has all changed to 𝑣(𝜏/2) = 𝑣1 = 𝑣0 + ∆𝑣 = 0 and
the fluid pressure is 𝑃(𝜏/2) = 𝑃1 = 𝑃0 + Δ𝑃s = 𝑃0 + 𝜌0 𝑐𝑣0 .
     At the reservoir end of the pipe, fluid pressure reduces to the constant
hydrostatic pressure 𝑃ℎ = 𝑃0 + 𝜌0 𝑔ℎ. Equivalently, we may say that the reservoir
acts as a free end for the pressure wave and, in reducing its excess pressure to 𝑃ℎ ,
causes a compression wave to be reflected as an expansion wave. Relative to the
hydrostatic pressure 𝑃ℎ , the amplitude of the incoming pressure wave is ∆𝑃1r =
𝑃1 − 𝑃ℎ , hence the reflected expansion wave will have an amplitude ∆𝑃1′ = −∆𝑃1r
and we have
 ∆𝑃1′ = −∆𝑃1r = 𝑃ℎ − 𝑃1 = (𝑃0 + 𝜌0 𝑔ℎ) − (𝑃0 + 𝜌0 𝑐𝑣0 ) = −𝜌0 𝑐(𝑣0 − 𝑔ℎ/𝑐) (e2)
(Here we allow the pressure amplitude to have both signs with negative amplitude
signifying an expansion wave.) This will cause the fluid at the reservoir end of the
pipe to suffer a velocity change (keeping in mind that the direction of propagation is
now the same as the +𝑥 axis)
                    ∆𝑣1r = +∆𝑃1′ /(𝜌0 𝑐) = −(𝑣0 − 𝑔ℎ/𝑐)
Consequently, its velocity changes to
                                                   𝑔ℎ
                    𝑣1r = 𝑣1 + ∆𝑣1r = 0 − (𝑣0 − )                                     (e3)
                                                    𝑐
Ahead of the front of the reflected wave, conditions are unchanged and the particle
velocity is still 𝑣1 = 0 and the fluid pressure is still 𝑃1 = 𝑃0 + Δ𝑃s , but behind the
wave front the particle velocity now becomes 𝑣1r = −(𝑣0 − 𝑔ℎ/𝑐) and the
pressure becomes

                                                        𝑔ℎ
               𝑃1 + ∆𝑃1′ = (𝑃0 + 𝜌0 𝑐𝑣0 ) − 𝜌0 𝑐 (𝑣0 −     ) = 𝑃0 + 𝜌0 𝑔ℎ      (e4)
                                                         𝑐
Therefore, just moment before 𝑡 = 𝜏 = 2𝐿/𝑐 when the front of the reflected wave
reaches the valve, the fluid in the whole length of the pipe will be under the
pressure 𝑃(𝜏) = 𝑃0 + 𝜌0 𝑔ℎ = 𝑃ℎ as given in Eq. (e4) , and all fluid particles in the
pipe will move, as given in Eq. (e3), with velocity 𝑣(𝜏) = 𝑣1r = −𝑣0 + 𝑔ℎ/𝑐, i.e., the
fluid in the pipe is expanding and flowing toward the reservoir.

Part D. Water-Hammer Effect due to Slow Closure of Flow-Control Valve
(D.1) (3.0 pt) Recursion relations for Δ𝑃𝑛 and 𝑣𝑛
Ans:
     Enforcing the approximation 𝑃ℎ = 𝑃0 + 𝜌0 𝑔ℎ ≈ 𝑃0 is equivalent to putting
ℎ = 0 in all of the results obtained in task (e).
(1) Partial closing 𝑛 = 1
     At the valve, immediately after partial closing 𝑛 = 1, fluid pressure jumps
from 𝑃0 to 𝑃1 , causing flow velocity to change from 𝑣0 to 𝑣1 . The pressure and
velocity changes are related by Eq. (a5):
                    1
                       (𝑃 − 𝑃0 ) = −(𝑣1 − 𝑣0 )                                        (f1)
                   𝜌0 𝑐 1
Just before reflection by the reservoir, the fluid in the entire pipe has pressure 𝑃1
and velocity 𝑣1 . After reflection by the reservoir, i.e., a free end, and before the start
of valve closure 𝑛 = 2, the fluid in the entire pipe has pressure (Eq. (e4) with ℎ = 0)
                    𝑃1 − (𝑃1 − 𝑃0 ) = 𝑃0
and velocity
                               −(𝑃1 − 𝑃0 )
                   𝑣1′ = 𝑣1 +               = 𝑣1 + (𝑣1 − 𝑣0 )
                                   𝜌0 𝑐
(2) Partial closing 𝑛 = 2
     Immediately after partial closing 𝑛 = 2, valve pressure changes from 𝑃0 to 𝑃2 ,
causing flow velocity to change from 𝑣1′ to 𝑣2 . The pressure and velocity changes are
given by Eq. (a5):
                     1
                        (𝑃2 − 𝑃0 ) = −(𝑣2 − 𝑣1′ ) = −𝑣2 + 𝑣1 + (𝑣1 − 𝑣0 )              (f2)
                   𝜌0 𝑐
Using Eq. (f1), we may rewrite the preceding equation as
                     1                                 1
                        (𝑃2 − 𝑃0 ) = −(𝑣2 − 𝑣1 ) −         (𝑃 − 𝑃0 )                   (f3)
                   𝜌0 𝑐                              𝜌0 𝑐 1
Just before reflection by the reservoir, the fluid in the entire pipe has pressure 𝑃2
and velocity 𝑣2 . After reflection by the reservoir and before valve closure 𝑛 = 3, the
fluid in the entire pipe has pressure
                   𝑃2 − (𝑃2 − 𝑃0 ) = 𝑃0
and velocity
                   𝑣2′ = 𝑣2 + (𝑣2 − 𝑣1′ )

(3) Partial closing 𝑛 = 3
     Immediately after partial closing 𝑛 = 3, valve pressure changes from 𝑃0 to 𝑃3 ,
causing flow velocity to change from 𝑣2′ to 𝑣3 . The pressure and velocity changes are
given by Eq. (a5):
                      1
                          (𝑃 − 𝑃0 ) = −(𝑣3 − 𝑣2′ ) = −𝑣3 + 𝑣2 + (𝑣2 − 𝑣1′ )          (f4)
                    𝜌0 𝑐 3
Using Eq. (f2), we may rewrite the preceding equation as
                      1                                 1
                          (𝑃3 − 𝑃0 ) = −(𝑣3 − 𝑣2 ) −       (𝑃 − 𝑃0 )                 (f5)
                     𝜌0 𝑐                             𝜌0 𝑐 2
Just before reflection by the reservoir, the fluid in the entire pipe has pressure 𝑃3
and velocity 𝑣3 . After reflection by the reservoir and before valve closure 𝑛 = 4, the
fluid in the entire pipe has pressure
                    𝑃3 − (𝑃3 − 𝑃0 ) = 𝑃0
and velocity
                    𝑣3′ = 𝑣3 + (𝑣3 − 𝑣2′ )
(4) Partial closing 𝑛 = 4
     When the valve is fully shut at valve closing 𝑛 = 4, the valve becomes a fixed
end, so the fluid velocity at the valve changes from 𝑣3′ to 𝑣4 = 0. The pressure
𝑃4 at the valve is then given by Eq. (a5):
                      1                                             1
                          (𝑃4 − 𝑃0 ) = −(𝑣4 − 𝑣3′ ) = −𝑣4 + 𝑣3 −       (𝑃 − 𝑃0 )     (f6)
                    𝜌0 𝑐                                          𝜌0 𝑐 3
     Finally, if we take note of the fact that ∆𝑃0 = 0 and 𝑣4 = 0, then all equations
obtained above relating excess pressures and velocity changes after valve closings all
have the same form:
                    ∆𝑃𝑛                       ∆𝑃𝑛−1
                          = −(𝑣𝑛 − 𝑣𝑛−1 ) −                     (𝑛 = 1,2,3,4)        (f7)
                    𝜌0 𝑐                        𝜌0 𝑐
     To solve for Δ𝑃𝑛 = 𝑃𝑛 − 𝑃0 , we note that, from Eqs. (c3) and (c5), we have
another relation between Δ𝑃𝑛 and 𝑣𝑛 :
                             1
                    ∆𝑃𝑛 = 𝑘𝑛 𝜌0 𝑣𝑛 2                            (𝑛 = 1,2,3)          (f8)
                             2
where 𝐶𝑛 represents 𝐶c for 𝑟 = 𝑟𝑛 and
                              1 𝑅 4
                    𝑘𝑛 = [ 2 ( ) − 1]                           (𝑛 = 1,2,3)          (f9)
                             𝐶𝑛 𝑟𝑛
Combining Eqs. (f7) and (f8), we have a quadratic equation for 𝑣𝑛 :
                     1      𝑣𝑛 2 𝑣𝑛       ∆𝑃𝑛−1 𝑣𝑛−1
                        𝑘𝑛 ( ) + + (              −       )=0         (𝑛 = 1,2,3) (f10)
                     2       𝑐       𝑐     𝜌0 𝑐 2     𝑐
which can be solved readily using the formula
                                              𝑣𝑛−1 ∆𝑃𝑛−1
                     𝑣𝑛 −1 + √1 + 2𝑘𝑛 ( 𝑐 − 𝜌𝑐 2 )
                         =                                           (𝑛 = 1,2,3)    (f11)
                      𝑐                    𝑘𝑛
     If both ∆𝑃𝑛−1 /(𝜌𝑐 2 ) and (𝑣𝑛−1 /𝑐) are known, Eq. (f11) may be used to
compute 𝑣𝑛 /𝑐 and then find ∆𝑃𝑛 /(𝜌𝑐 2 ) by using Eq. (f8). Therefore, Eq. (f7) may

be solved iteratively starting with 𝑛 = 1 until 𝑛 = 3. For 𝑛 = 4, we know 𝑣𝑛 = 0, so
Eq. (f7) may be used directly to find ∆𝑃𝑛 .
     Note that, from Eq. (f8), ∆𝑃𝑛−1 is a quadratic function of 𝑣𝑛−1 , so that if 𝑣𝑛−1
is known, then 𝑣𝑛 may be computed using Eq. (f11) and then ∆𝑃𝑛 may again be
computed using Eq. (f8).
(D.2) (2.0 pt) Estimating Δ𝑃𝑛 and 𝜌0 𝑐𝑣𝑛 by graphical method
Ans:
     To solve Eqs. (f7) and (f8) using graphical method, we rewrite them as follows:
                   ∆𝑃𝑛 = −(𝜌0 𝑐𝑣𝑛 − 𝜌0 𝑐𝑣𝑛−1 ) − ∆𝑃𝑛−1       (𝑛 = 1,2,3,4)        (g1)
                                         𝑘𝑗
                               ∆𝑃𝑛 =           (𝜌 𝑐𝑣 )2                                      (𝑛 = 1,2,3,4)                 (g2)
                                        2𝜌0 𝑐 2 0 𝑛
In a plot of ∆𝑃 vs. 𝜌0 𝑐𝑣, Eq. (g1) and Eq. (g2) correspond to a line passing through
the point (𝜌0 𝑐𝑣𝑛−1 , −∆𝑃𝑛−1 ) with slope −1 and a parabola passing through the
origin, respectively. Thus one may readily obtain the solutions for each step of valve
closing by locating their points of intersection, starting with 𝑛 = 1. The result is
shown in the following graph.

                                                       P-cv at valve
           2

P/MPa
         1.5
                                      𝑛=3                                 𝑛=2
           1
                                                                                𝑛=1
          𝑛=4
         0.5         0, ∆𝑃4

           0
                 0      0.5     1      1.5         2   2.5         3     3.5    4      4.5    5      5.5    6      6.5         7
         -0.5
                                                                                𝜌0 𝑐𝑣1 ,−∆𝑃1
          -1
                                    𝜌0 𝑐𝑣3 , −∆𝑃3                      𝜌0 𝑐𝑣2 , −∆𝑃2
         -1.5                                 n=3, r/R=0.2                n=2, r/R=0.3            n=1, r/R=0.4

          -2
                                                                    cv/MPa

                Excess Pressures and particle velocities at the valve for slow closing
 𝑛              𝑟𝑛 /𝑅          𝐶𝑛            𝑘𝑛        𝑣𝑛 /(m/s)           𝜌0 𝑐𝑣𝑛 /MPa        ∆𝑃𝑛 /(MPa)         ∆𝑃𝑛 /(𝜌0 c𝑣0 )
 0              1.00          1.00           0.0             4.0               6.0                 0.0                   0.0
 1              0.40          0.631       97.1               3.6               5.8                0.62               10 %
 2              0.30          0.622       318.               2.5               3.8                 1.0               17 %
 3              0.20          0.616      1646.               1.1               1.7                 1.1               18 %

4   0.00                     0.0           0.0              0.64   11 %
           𝜌0 𝑐 = 1.50 × 106 kg m−2 s −1         𝑣0 = 4.0 m/s

-----------------------------------------------------------------------------------------------------------

Appendix
(The following table and graph are for reference only, not part of the task.)
     For 𝑣0 = 4.0 m/s, 𝑐 = 1.5 × 103 m/s, and 𝜌 = 1.0 × 103 kg/m3 , the results
for 𝑣𝑛 and Δ𝑃𝑛 are shown in the following table and graph. They are computed
according to equations given in task (f). Note that for a sudden full closure of the
valve, we have Δ𝑃sudden = 𝜌c𝑣0 = 6.0 MPa.

          Excess Pressures and particle velocities at the valve for slow closing
 𝑛       𝑟𝑛 /𝑅         𝐶𝑛         𝑘𝑛       𝑣𝑛 /(m/s)        𝜌𝑐𝑣𝑛 /MPa         ∆𝑃𝑛 /(MPa)          ∆𝑃𝑛 /(𝜌c𝑣0 )
 0       1.00         1.00        0.0           4.0              6.0                0.0                0.0
 1       0.40         0.631      97.1          3.58             5.37              0.624               10 %
 2       0.30         0.622      318.          2.50             3.75              0.997               17 %
 3       0.20         0.616     1646.          1.13            1.695               1.06               18 %
 4       0.00                                   0.0              0.0              0.643               11 %

            Pn /MPa                    Excess pressures at valve
            1.5

              1

            0.5

              0
                  0               1                2                3                4                5

                                              valve closing step n

---------------------------------------------------------------------------------------------------

APhO 2021 Theory T1


[User]
\begin{wrapfigure}[10]{r}{0.4\textwidth}
\vspace{-5pt}
\begin{tikzpicture}[thick,scale=0.6, every node/.style={scale=0.7}]

    % Def. koordinaadid
    \coordinate (O) at (0,0) ;
    \coordinate (1) at (4,0) ;
    \coordinate (A) at (0,-2) ;
    \coordinate (2) at (0,-4);
    \coordinate (3) at (-1.41,-0.69);

    % Jooned, t2pid
    \draw[dash pattern=on5pt off3pt] (O) -- (1);
    \draw[dash pattern=on5pt off3pt] (A) -- (2);
    \draw[thick,dashed] (4.25,0) circle (0.25cm);
    \draw[thick,dashed] (0,-4.25) circle (0.25cm);
    \draw[thick] (O) -- (A);
    \draw[thick] (A) -- (3);
    \filldraw [black] (A) circle (1.5pt);
    \filldraw [black] (O) circle (1.5pt);
    \filldraw [black] (-1.57,-0.53) circle (6pt);

    % Nurgad ja punktid
    \draw (0,-1) arc (90:136:1);
    \draw[thick, dashed] (0.25,-4.25) arc (270:360:4);
    \draw[thick, dashed] (-0.25,-4.25) arc (270:135:2.1);
    \node[] at (-0.25,-1.4)  {$\alpha$};
    \node[] at (4.7,-0.2)  {$1$};
    \node[] at (0.3,-3.8)  {$2$};
    \node[] at (-2.0,-0.4)  {$3$};
    \node[] at (-0.2,0.2)  {O};
    \node[] at (0.3,-2)  {A};
    \node[] at (0.2,-1)  {$\frac{l}{2}$};
    \node[] at (-1,-1.6)  {$\frac{l}{2}$};
\end{tikzpicture}
\end{wrapfigure}
A thread of length $l$ is fixed to a point O and a small ball is hanging at the end of the thread. The ball is brought to the side and is released at the position 1 without additional force. When the ball reaches the position 2 the thread meets a rod at the point A, the rod is perpendicular to the drawing’s plane. In the same vertical the point A is at a length $l/2$ from the point O. Find for what value of $\alpha$ is the thread’s tension $T=0$ (position 3). Do not account for air resistance or friction.

Solution:
Niidis kaob tõmbejõud, kui raskusjõu niidisuunaline komponent saab võrdseks tsentrifugaaljõuga. Märgistame nurga niidi ning horisontaaljoone vahel $\theta$-ga. Tsentrifugaaljõud avaldub kui $F_t=mv^2/r=2mv^2/l$ ning raskusjõu niidisuunaline komponent $F_rn=mg\sin\theta$. Need peavad võrduma, seega:
$$\frac{2mv^2}{l}=mg\sin\theta \Rightarrow v^2=\frac{gl\sin\theta}{2}.$$
Võttes potentsiaalse energia nivoo nullpunktiks asendi 2, saame et kuuli energia asendis 1 on $E_1=mgl$. Asendis 3 on aga kuulikese energia $E_3=mv^2/2+mg(1+\sin\theta)l/2$. Kuna kehtib energia jäävuse seadus, peab kehtima $E_1=E_3$:
$$mgl=\frac{mv^2}{2}+mg(1+\sin\theta)\frac{l}{2} \Rightarrow v^2=gl(1-\sin\theta).$$
Pannes kaks avaldist $v^2$ jaoks omavahel võrduma, saame:
$$\frac{gl\sin\theta}{2}=gl(1-\sin\theta)\Rightarrow \frac{3}{2}\sin\theta=\frac{3}{2}\cos\alpha=1 \Rightarrow \alpha=\arccos\frac{2}{3}.$$

Kaheosaline pendel

\begin{wrapfigure}[10]{r}{0.4\textwidth}
\vspace{-5pt}
\begin{tikzpicture}[thick,scale=0.6, every node/.style={scale=0.7}]

 % Def. koordinaadid
 \coordinate (O) at (0,0) ;
 \coordinate (1) at (4,0) ;
 \coordinate (A) at (0,-2) ;
 \coordinate (2) at (0,-4);
 \coordinate (3) at (-1.41,-0.69);

 % Jooned, t2pid
 \draw[dash pattern=on5pt off3pt] (O) -- (1);
 \draw[dash pattern=on5pt off3pt] (A) -- (2);
 \draw[thick,dashed] (4.25,0) circle (0.25cm);
 \draw[thick,dashed] (0,-4.25) circle (0.25cm);
 \draw[thick] (O) -- (A);
 \draw[thick] (A) -- (3);
 \filldraw [black] (A) circle (1.5pt);
 \filldraw [black] (O) circle (1.5pt);
 \filldraw [black] (-1.57,-0.53) circle (6pt);

 % Nurgad ja punktid
 \draw (0,-1) arc (90:136:1);
 \draw[thick, dashed] (0.25,-4.25) arc (270:360:4);
 \draw[thick, dashed] (-0.25,-4.25) arc (270:135:2.1);
 \node[] at (-0.25,-1.4) {$\alpha$};
 \node[] at (4.7,-0.2) {$1$};
 \node[] at (0.3,-3.8) {$2$};
 \node[] at (-2.0,-0.4) {$3$};
 \node[] at (-0.2,0.2) {O};
 \node[] at (0.3,-2) {A};
 \node[] at (0.2,-1) {$\frac{l}{2}$};
 \node[] at (-1,-1.6) {$\frac{l}{2}$};
\end{tikzpicture}
\end{wrapfigure}

Punktis O kinnitatud niidi pikkusega $l$ otsas ripub väike kuulike. Kuulike viiakse kõrvale ja vabastatakse tõuketa asendist 1. Kuuli jõudes asendisse 2, kohtab niit joonise tasandiga risti olevat varrast punktis A, mis asub punktist O kaugusel $l/2$ sellega samal vertikaalil. Leida, millise nurga $\alpha$ väärtuse korral niidi pinge $T=0$ (asend 3). Õhutakistust ja hõõrdumist vardal arvestama ei pea.

Niidis kaob tõmbejõud, kui raskusjõu niidisuunaline komponent saab võrdseks tsentrifugaaljõuga. Kuuli kiirus asendis 3 on leitav energia jäävuse seadusest.


[User]
Question B2
An electric dipole consists of two charges of equal magnitude q and opposite sign, held rigidly apart
by a distance d. The dipole moment is defined by p = qd.
   Now consider two identical, oppositely oriented electric dipoles, separated by a distance r, as
shown in the diagram.

                        A                                                 B

                   d                                                            d

                                                  r

  a. It is convenient when considering the interaction between the dipoles to choose the zero of
     potential energy such that the potential energy is zero when the dipoles are very far apart
     from each other. Using this convention, write an exact expression for the potential energy of
     this arrangement in terms of q, d, r, and fundamental constants.

  b. Assume that d  r. Give an approximation of your expression for the potential energy
     to lowest order in d. Rewrite this approximation in terms of only p, r, and fundamental
     constants.

   c. What is the force (magnitude and direction) exerted on one dipole by the other? Continue
      to make the assumption that d  r, and again express your result in terms of only p, r, and
      fundamental constants.

  d. What is the electric field near dipole B produced by dipole A? Continue to make the assump-
     tion that d  r and express your result in terms of only p, r, and fundamental constants.

                   Copyright c 2009 American Association of Physics Teachers

Solution:
Question B2
An electric dipole consists of two charges of equal magnitude q and opposite sign, held rigidly apart
by a distance d. The dipole moment is defined by p = qd.
   Now consider two identical, oppositely oriented electric dipoles, separated by a distance r, as
shown in the diagram.

                        A                                                 B

                   d                                                              d

                                                  r

  a. It is convenient when considering the interaction between the dipoles to choose the zero of
     potential energy such that the potential energy is zero when the dipoles are very far apart
     from each other. Using this convention, write an exact expression for the potential energy of
     this arrangement in terms of q, d, r, and fundamental constants.

  b. Assume that d  r. Give an approximation of your expression for the potential energy
     to lowest order in d. Rewrite this approximation in terms of only p, r, and fundamental
     constants.

   c. What is the force (magnitude and direction) exerted on one dipole by the other? Continue
      to make the assumption that d  r, and again express your result in terms of only p, r, and
      fundamental constants.

  d. What is the electric field near dipole B produced by dipole A? Continue to make the assump-
     tion that d  r and express your result in terms of only p, r, and fundamental constants.

                                            Solution
  a. There are two +q/−q pairs separated by a distance d, each having potential energy

                                                       q2
                                                 −
                                                      4π0 d
     There are two +q/−q pairs separated by a distance r, each having potential energy

                                                       q2
                                                 −
                                                      4π0 r
                                                                              √
     There are a +q/+q pair and a −q/−q pair separated by a distance              r2 + d2 , each having
     potential energy
                                             q2
                                            √
                                       4π0 r2 + d2

                   Copyright ©2009 American Association of Physics Teachers

2009 Semifinal Exam                              Part B                                         13

     Note that the latter two terms go to zero as r becomes large, whereas the first term is not
     dependent on r. Thus the given zero convention will include only the latter two terms,

                                         q2

                                                 2       2
                                   U=          − +√              .
                                        4π0     r     r2 + d2

  b. Rearranging the result above, we have
                                                                      
                                          2q 2            1
                                    U=            q                − 1 .
                                         4π0 r               d 2

                                                       1+     r

     Using the binomial approximation (1 + x)n ≈ 1 + nx,
                                                  !
                            2q 2       1 d 2              q 2 d2       p2

                      U≈           1−          −1 =−             = −         .
                           4π0 r      2 r               4π0 r3     4π0 r3

  c. We can infer by symmetry that the force must point along the line separating the dipoles.
     Since the potential energy decreases with decreasing distance, the force is attractive. Its
     magnitude can be determined by taking the derivative of the potential energy:

                                                  dU        p2
                                        F =−         = −3
                                                  dr      4π0 r4
     with the negative sign confirming that the force is attractive.
     One can also use an approach analogous to the previous one, i.e. write down the force exactly
     and use a binomial approximation as above. One must take care to account for the fact that
     the force between like-signed charges is not exactly in the same direction as that between
     opposite-signed charges.

  d. The electric field near dipole B produced by dipole A is directed approximately downward.
     Its magnitude is exactly
                                          q       2           d
                                    E=         2       2
                                                         sin      .
                                         4π0 r + (d/2)       2r
     Using d  r and the small angle approximation sin θ ≈ θ, we have
                                                        p
                                                 E≈
                                                      4π0 r3
     where we neglected all higher-order terms in d/r.
     Note that at this level of accuracy, the force on dipole B would be exactly zero. In order to
     extend this calculation to get the answer to part c, we would have to account for the fact
     that the electric fields at the two charges in dipole B do not point exactly downward. This is
     a smaller effect than what we’ve found in this part, which is why the answer to part (c) falls
     off as 1/r4 while the answer here falls off as 1/r3 .

                  Copyright ©2009 American Association of Physics Teachers

USAPhO 2009 Question B2


[User]
A U-tube has vertical arms of radii r and 2r, connected by a horizontal tube of length l whose
radius increases linearly from r to 2r. The U-tube contains liquid up to height h in each arm. The
liquid is set oscillating, and at a given instant the liquid in the narrower arm is at a distance y
above the equilibrium level.

  1. Show that, to second order in y, the change in potential energy of the liquid is

                                             U = gρπr2 y 2 .

  2. Show that, to second order in y, the kinetic energy of the liquid is
                                                    2
                                        1   2   5     dy
                                     K = ρπr ℓ + h        .
                                        4       2     dt

     You may find it useful to integrate over slices dx, as shown in the figure. Ignore any nastiness
     at the corners, and assume ℓ ≫ r.

  3. Assuming ℓ = 5h/2, compute the period of oscillations.

  4. Explain why the assumption ℓ ≫ r is necessary to get an accurate result.

Solution:
A U-tube has vertical arms of radii r and 2r, connected by a horizontal tube of length l whose
radius increases linearly from r to 2r. The U-tube contains liquid up to height h in each arm. The
liquid is set oscillating, and at a given instant the liquid in the narrower arm is at a distance y
above the equilibrium level.

  1. Show that, to second order in y, the change in potential energy of the liquid is

                                               U = gρπr2 y 2 .

  2. Show that, to second order in y, the kinetic energy of the liquid is
                                                         2
                                         1    2       5      dy
                                   K = ρπr ℓ + h                   .
                                         4            2      dt
     You may find it useful to integrate over slices dx, as shown in the figure. Ignore any nastiness
     at the corners, and assume ℓ ≫ r.

  3. Assuming ℓ = 5h/2, compute the period of oscillations.

  4. Explain why the assumption ℓ ≫ r is necessary to get an accurate result.

Solution. This is problem 3.17 of Vibrations and Waves by French.

  1. (5) We can think of moving a chunk of liquid from one side to the other. If a volume V = πr2 y
     is added to the narrow arm, then a volume V will be removed from the other, so the water
     level will sink a distance of y2 = V /(4πr2 ) = y/4. The center of mass of the chunk of water
     will go from a height of −y2 /2 to y/2, going a distance of ∆y = y/2 + y/8 = 5y/8. Then the
     potential energy change is

                                       U = ∆mg∆y = ρπr2 gy 2 .

  2. (10) Consider points along the vertical arms first. The left arm’s fluid will have kinetic energy
     KL = 12 ρπr2 hẏ 2 and the right KR = 12 ρπ(2r)2 h(ẏ/4)2 , so the kinetic energy of the arms is
     KA = 12 ρπr2 ẏ 2 (h + h/4) = 58 ρπr2 hẏ 2 . In the horizontal part, a slice dx with radius r + rx/ℓ
     will have area A and kinetic energy dK = 12 ρAdxv 2 , and v can be found with the fact that
     the fluid is incompressible: Av = πr2 ẏ = A0 ẏ.
                                     1 A2 ẏ 2       1            dx
                              dK = ρA 0 2 dx = A20 ẏ 2
                                     2    A          2       π(r + rx/ℓ)2
                             1 2 2 ℓ
                                      Z
                                               dx          1     4 2ℓ   1   1
                         Kh = ρA0 ẏ                  2
                                                        = ρπr ẏ          −
                             2         0 π(r + rx/ℓ)       2        r r 2r

                                              = ρπr2 ℓ

                                           1             5
                                       K = ρπr ℓ + h ẏ 2 .

                                           4             2

  3. (5) The energy equation is
                               5            5           1           1
                            E = ρπr2 hẏ 2 + gρπr2 y 2 = meff ẏ 2 + keff y 2
                               4            8           2           2
     so using the standard method in M4, we have
                                                   r               s
                                                       meff            2h
                                          T = 2π            = 2π          .
                                                       keff             g

  4. (5) This ensures that our estimate of the kinetic energy above is good. There are additional
     contributions to the kinetic energy due to the complicated turnaround at the corners, and
     also the radial velocity in the horizontal part, but these are small by assumption.

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


[User]
3. Bars and rod (5 points)

Two cylindrical horizontal bars are fixed one above the
other; the distance between the axes of the bars is 4d, where d
is the diameter of the rod. Between the bars, a cylindrical rod
of diameter d is placed as shown in Figure (this is a vertical
cross-section of the system). The coefficient of friction between
the rod and the bars is μ = 1
2. If the rod is long enough, it will
remain in equilibrium in such a position. What is the minimal
length L of the rod required for such an equilibrium?
Problem

Solution:
3. Bars and rod

With μ = 1
2, the result-
ant force of the normal force
and friction force forms an angle
arctan 1
2 with the surface nor-
mal (assuming that the rod is as
short as possible and hence, at
the threshold of slipping). There
are three forces applied to the
rod — the gravity force mg ap-
plied to the centre of mass C, and
the two forces due to the bars.
At equilibrium, the three lines
s1, s2, and s3, defined by these
three forces need to intersect at
a single point Q (otherwise, with
respect to the intersection point of two lines, the torque of the
third force would cause a rotation of the rod). This configura-
tion is depicted in Figure.
Since the friction force forms angle arctan 1
2 with the sur-
face normal, hence ∠DPQ = ∠DRQ = arctan2, hence AP =
RS = 1
4d (see Figure). From the geometry of the blue tri-
angle, AS = 2
√
3d; due to AP = RS, PR = AS = 2
√
3d and
PD =
√
3d. Now, let us recall that tan∠DPQ = 2, hence
DQ = PR = 2
√
3d. From the geometry of the blue triangle,
∠DCQ = 30◦
, so that DC = DQ/tan30◦
= 6d. Now we can
finally express
AC = CD + DP − PA = 5
3
4
d +
√
3d ⇒
L = 2AC = (11,5 + 2
√
3)d ≈ 14.96d.
Problem

Bars and rod


[User]
Some questions about light energy.

     (a) Estimate the number of photons emitted per second by a standard light bulb. (The energy of
         a photon is E = hf , and the frequency of a photon is related to the wavelength by c = f λ.)

     (b) The Sun supplies power of intensity 1400 W/m2 to the Earth. The nearest star is about 4
         light years away. Assuming this star is similar to the Sun, about how many of its photons
         enter your eye per second?

Solution:
Before we continue, it’s important to note that for estimation questions, one should only
    expect an answer to within an order of magnitude. Some teachers tweak their example calculations
    until they give almost exactly the right answer. This makes them look brilliant, but it’s deceptive,
    because then when the student tries to do the same, their results will be much further off. So to
    combat this, in all solutions here, we’ve just presented our very first, simplest guesses. They can be
    up to an order of magnitude off from the real numbers, so if your numbers are within two orders of
    magnitude of ours, you’re fine!
     (a) We can estimate a standard light bulb to have around 50 W of power. The power P = N E
         where N is the number of photons emitted per second, and the wavelength of visible light is
         from 400 − 700 nm (we can use 500). Then
                                                  Pλ
                                            N=       ≈ 1020 photons/s
                                                  hc
     (b) 1 AU is about 1.5 × 1011 m. (If you forget, you can use something like GMS /r2 = (2π/T )2 r,
         where T is one year and MS ≈ 2 × 1030 kg). 1 light year is c × 1 year ≈ 9.5 × 1015 m. Then
         the intensity from the star is reduced by a factor of (1 AU/4 ly)2 due to the inverse square
         law, so I ≈ 3.5 × 10−7 W/m2 . The area of a human pupil depends on the light conditions, but
         is roughly πr2 = π(5 mm)2 . Then the number of photons that enter it per second is P λ/hc,
         which gives N ≈ 107 photons/s. That’s plenty, so it’s very easy to see such a star at night,
         while it might be difficult during the day because of the background light from the Sun.


[User]
01^USAPhO 2022, problem A3. A great problem on the rainbow caustic. For some
    brilliant rainbow graphics, see here and here.

    4     Modern

Solution:
USAPhO 2022, problem A3. A great problem on the rainbow caustic. For some
    brilliant rainbow graphics, see here and here.

    4    Modern


[User]
\begin{wrapfigure}[7]{r}{0.15\textwidth}
  \vspace{-35pt}
  \begin{center}
    \includegraphics[width=0.18\textwidth]{2015-v2g-05-toru}
  \end{center}
  \vspace*{-10pt}
\end{wrapfigure}
A water pipe with a length $L$ is fixed to a wall so that it can freely rotate on the vertical plane. The water pipe’s mass together with the water it is filled with is $M$. The cross-sectional area of the pipe’s end is $S$. The end is turned 90 degrees with respect to the rest of the pipe (see figure). Water flows out of the end of the pipe with a speed $v$ and density $\rho$. What is the angle between the vertical direction and the pipe’s axis? Gravitational acceleration is $g$.

Solution:
Kui toru otsast voolab välja vesi kiirusega $v$, siis ajaühikus torust väljunud vee hulk on $\frac{\Delta m}{\Delta t}=\rho S v$. Paneme kirja Newtoni II seaduse torust väljunud veehulga jaoks:
\[ F=\frac{\Delta p}{\Delta t}=\frac{\Delta(mv)}{\Delta t}=v \frac{\Delta m}{\Delta t}. \]
Kuna vesi saab ajaühikus sellise impulsi, peab järelikult Newtoni III seaduse tõttu mõjuma torule sama suur ja vastassuunaline jõud. Paneme kirja jõumomentide tasakaalu võrrandi toru kinnituskoha suhtes:
\[ \frac{MgL\sin \alpha}{2}=FL. \]
Asendades sisse $\frac{\Delta m}{\Delta t}$ ning avaldades $\alpha$, saame
\[ \alpha=\arcsin \left(\frac{2\rho v^2 S}{Mg}\right). \]

Veetoru

\begin{wrapfigure}[5]{r}{0.18\textwidth}
 \vspace{-30pt}
 \begin{center}
 \includegraphics[width=0.18\textwidth]{2015-v2g-05-toru}
 \end{center}
\end{wrapfigure}
Veetoru pikkusega $L$ on kinnitatud seina külge nii, et see saab vertikaaltasandis vabalt pöörelda. Veetoru mass koos seda täitva veega on $M$. Toru ots ristlõikepindalaga $S$ on ülejäänud toruga võrreldes \num{90} kraadi pööratud (vt joonist) ning sellest voolab välja vesi kiirusega $v$ ja tihedusega $\rho$. Kui suure nurga all vertikaali suhtes paikneb toru telg? Raskuskiirenduse väärtus on $g$.

Toru lõpus järsult pöörav vesi surub toru teatud jõuga külgsuunas. Antud jõud on leitav, kui vaadelda ajaühikus väljuva veehulga impulsimuutu siseneva veega võrreldes.


[User] If pressing down on a violin string with a length $L$ at a distance $\frac{3}{7}L$ from one end and dragging along the shorter part with a violin bow, a sound with fundamental frequency will be heard. When just touching the string at the same distance $\frac{3}{7}L$ (without pressing down) the sound heard is different. What is the ratio of these two fundamental frequencies?

Solution:
Tekkival seisulainel peavad olema sõlmed mõlemas keele võnkuva osa otspunktis, seega võngub alla vajutades osa pikkusega $\frac{3}{7}L$, millele vastab lainepikkus $\lambda_0=\frac{6}{7}L$. Puudutades võngub kogu keel ning on kolm tingimust: sõlmpunktid on mõlemas otsas ning lisaks puudutatavas punktis. Seega peab sellest punktist mõlemale poole mahtuma täisarv poollainepikkusi. Võnkuvate osade suhe on $\frac{3/7}{1-3/7}$ ehk $\frac{3}{4}$. $3$ ja $4$ on ühistegurita. Seega peab jääma võnkuvatele pooltele vastavalt $3$ ja $4$ poollainepikkust. Vaadeldes pikkusega $\frac{3}{7}L$ keele poolt, taipame et $\lambda=\frac{2L}{7}$ ning
\[
\frac{\nu}{\nu_0}=\frac{\lambda_0}{\lambda}=3.
\]

Viiul

Viiulikeelt pikkusega $L$ kaugusel $\frac{3}{7}L$ ühest otsast alla vajutades ning lühemal osal poognaga tõmmates kõlab mingi põhisagedusega heli. Samal kaugusel $\frac{3}{7}L$ keelt ainult puudutades (alla vajutamata), on kõlav heli erinev. Milline on nende kahe põhisageduse suhe?

Tekkival seisulainel peavad olema sõlmed mõlemas keele võnkuva osa otspunktis. Viiuli keelt puudutades peab ka keele puutepunkt sõlmpunkt olema.


[User] IPhO 1995, problem 1. A straightforward problem on gravitational redshift.

Solution:
See the official solutions, but note that there’s a numeric mistake in the very last subpart.
    They should have written m0 c2 = 3.752 × 109 eV and ∆f /f0 = 5.44 × 10−9 .


[User]
Theory

                                                                                Q2-1       English (Official)

Where is the neutrino? (10 points)
When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles
may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their
respective anti-particles. Most of these particles can be detected by the particle detector surrounding
the collision point. For example, quarks undergo a process called hadronisation in which they become a
shower of subatomic particles, called "jet''. In addition, the high magnetic field present in the detectors
allows even very energetic charged particles to curve enough for their momentum to be determined.
The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00
Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles
with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field
and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes
through the detector without interacting.
Data: Electron rest mass, m = 9.11 × 10−31 kg; Elementary charge, e = 1.60 × 10−19 C;
Speed of light, c = 3.00 × 108 m s−1 ; Vacuum permittivity, ε0 = 8.85 × 10−12 F m−1

Part A. ATLAS Detector physics (4.0 points)

     A.1     Derive an expression for the cyclotron radius, r, of the circular trajectory of an    0.5pt
             electron acted upon by a magnetic force perpendicular to its velocity, and ex-
             press that radius as a function of its kinetic energy, K; charge modulus, e; mass,
             m; and magnetic field, B. Assume that the electron is a non-relativistic classical
             particle.

Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for
the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered.

     A.2     Calculate the minimum value of the momentum of an electron that allows it to          0.5pt
             escape the inner part of the detector in the radial direction. The inner part of
             the detector has a cylindrical shape with a radius of 1.1 meters, and the electron
             is produced in the collision point exactly in the center of the cylinder. Express
             your answer in MeV/c.

When accelerated perpendicularly to the velocity, relativistic particles of charge e and rest mass m emitt
electromagnetic radiation, called synchrotron radiation. The emitted power is given by

                                                          e2 a2 γ 4
                                                  P =
                                                          6πε0 c3

where a is the acceleration and γ = [1 − (v/c)2 ]−1/2 .

     A.3     A particle is called ultrarelativistic when its speed is very close to the speed of   1.0pt
             light. For an ultrarelativistic particle the emitted power can be expressed as:

                                                       e4
                                             P =ξ            E 2 B2 ,
                                                    ε0 mk cn
             where ξ is a real number, n, k are integers, E is the energy of the charged par-
             ticle and B is the magnetic field. Find ξ, n and k.

Theory

                                                                                Q2-2        English (Official)

    A.4     In the ultrarelativistic limit, the energy of the electron as a function of time is:   1.0pt
                                                         E0
                                             E(t) =             ,
                                                      1 + αE0 t
            where E0 is the initial energy of the electron. Find α as a function of e, c, B, ε0
            and m.

    A.5     Consider an electron produced at the collision point along the radial direction        0.5pt
            with an energy of 100 GeV. Estimate the amount of energy that is lost due to
            synchrotron radiation until the electron escapes the inner part of the detector?
            Express your answer in MeV.

    A.6     Find an expression for the cyclotron frequency of the electron as a function of        0.5pt
            time in the ultrarelativistic limit.

Part B. Finding the neutrino (6.0 points)
The collision between the two protons shown in Figure 1 leads to the production of a top quark (t) and
an anti-top quark (t),̄ the heaviest elementary particles ever detected. The top quark decays into a W +
boson and a bottom quark (b), while the anti-top quark decays into a W − boson and an anti-bottom quark
(b).
  ̄ In the case depicted in Figure 1, the W + boson decays into a anti-muon (μ+ ) and a neutrino (ν), and
the W − boson decays into a quark and an anti-quark. The task of this problem is to reconstruct the full
momentum of the neutrino using the momenta of some detected particles. For simplicity, all particles
and jets in this problem will be considered massless, except for the top quark and W± bosons.
The momenta of the top quark decay products can be determined from the experiment (see Table),
except for the neutrino momentum component along the (z-axis). The total linear momentum of the
final state particles caught by the detector is only zero on the transverse plane (xy plane), and not along
the collision line (z-axis). As such, one can find the transverse momentum of the neutrino from the
missing momentum in the transverse plane.
On June 4, 2015, the ATLAS experiment at the LHC recorded a proton-proton collision at 00:21:24 GMT+1
like the one represented in Figure 1.

Theory

                                                                               Q2-3English (Official)

Figure 1. Schematic representation of the ATLAS detector coordinate system (left) and proton-proton
collision (right).
The linear momenta of the three final-state particles coming from the top quark decay, including the
neutrino, is presented below for each component.

                        Particle           px (GeV/c)   py (GeV/c)   pz (GeV/c)
                        anti-muon (μ ) +
                                             −24.7        −24.9        −12.4
                        jet 1 (j1 )          −14.2        +50.1        +94.1
                        neutrino (ν)        −104.1        +5.3          ---

    B.1    Find an equation which relates the square of the W + boson mass, m2W , with    1.5pt
           the neutrino and anti-muon momentum components presented in the table
           above. Express your answer in terms of the neutrino and anti-muon transverse
           momentum,
                               (ν)  (ν)      (ν)         (μ)  (μ)      (μ)
                            p⃗T = px ı ̂ + py ȷ ̂ and p⃗T = px ı ̂ + py ȷ,̂
                                                         (μ)      (ν)
           and their z-axis momentum components, pz and pz .

    B.2    Assuming a W + boson mass of mW = 80.4 GeV/c2 calculate the two possible so-   1.5pt
                                                                 (ν)
           lutions for the neutrino momentum along the z-axis, pz . Express your answer
           in GeV/c.

    B.3    Calculate the top quark mass for each one of the two previous solutions. Ex-   1.0pt
           press your answer in GeV/c2 .
           [If you did not obtain the two solutions in B.2, use
                                (ν)                     (ν)
                              pz = 70 GeV/c and pz = −180 GeV/c.]

The normalised number of collision events for the measurement of the top quark mass (as determined

Theory

                                                                          Q2-4      English (Official)

from the experiment), has two components: the so-called "signal" (corresponding to the decay of
top quarks) and "background" (corresponding to events from other processes that do not include top
quarks). Experimental data include both processes, see Fig. 2.

    Figure 2. Top quark mass distribution as determined from the experiment, i.e. the normalised
    number of events plotted against the top quark mass. The dots correspond to the data. The
    dashed line corresponds to the "signal" and the shade to the "background".

    B.4    According to the top quark mass distribution, which one of the two previous      1.0pt
           solutions is more likely to be the right one? Estimate the probability for the
           most likely solution.

    B.5    Calculate the distance traveled by the top quark before decaying, using the      1.0pt
           most likely solution. Assume the top quark has a mean lifetime at rest of
           5 × 10−25 s.

Solution:
Confidential
          Secretariado IPhO 2018
          Sociedade Portuguesa de Física
          Avenida da República No45 3oEsq
          1050-187 Lisboa, Portugal Secretariado         Sociedade Portuguesa de F
                                             IPhO 2018   Avenida da República No45
                                                         1050-187 Lisboa, Portugal
          +351 21 799 36 65
          info@ipho2018.pt

Solutions to Theory Problem 2

    Where is the neutrino?

  (Miguel C N Fiolhais and António Onofre)

               July 24, 2018

                   v1.2

                              Secretariado
                              IPhO 2018
                                                      Sociedade Portuguesa de Física
                                                      Avenida da República No45 3oEsq
                                                           Theory
                                                      1050-187 Lisboa, Portugal
                                                          English (UK)                     ST2-1
                                                                                +351 21 799 36 65
                                                                                info@ipho2018.pt

Where is the neutrino? (10 points)
Part A. ATLAS Detector physics (4.0 points)

A.1

The magnetic force is the centripetal force:

                                                      v2             mv
                                                  m      = evB ⇒ r =    .
                                                       r             eB
First express the velocity in terms of the kinetic energy,

                                                         1 2          2K

                                                                                Confidential
                                              K=           mv ⇒ v = √    ,
                                                         2            m
and then insert it in the expression above for the radius to get

     A.1                                                        √                                           0.5pt
                                                                    2Km
                                                          r=            .
                                                                    eB

A.2

The radius of the circular motion of a charged particle in the presence of a uniform magnetic field is
given by,
                                                         mv
                                                     r=     .
                                                         eB
This formula is valid in the relativistic scenario if the mass correction, m → γm is included:
                                                      γmv    p
                                              r=          =    ⇒ p = reB .
                                                       eB   eB
Note that the radius of the circular motion is half the radius of the inner part of the detector. One obtains
[1 MeV/c = 5.34 × 10−22 m kg s−1 ]

     A.2                                                                                0.5pt
                                                        p = 330 MeV/c .

A.3

The acceleration for the particle is a = evB  ecB
                                         γm ∼ γm , in the ultrarelativistic limit. Then,

                                                    e4 c 2 γ 4 B 2   e4 γ 2 c 4 B 2
                                             P =                   =                .
                                                   6πε0 c3 γ 2 m2    6πε0 c5 m2
                                              2
                                      E
Since E = γmc2 we can obtain γ 2 c4 = m 2 and, finally,

                              Secretariado
                              IPhO 2018
                                                   Sociedade Portuguesa de Física
                                                   Avenida da República No45 3oEsq
                                                          Theory
                                                   1050-187 Lisboa, Portugal
                                                       English (UK)                        ST2-2
                                                                                +351 21 799 36 65
                                                                                info@ipho2018.pt

                                                             e4
                                                  P =                E 2 B2 .
                                                          6πε0 m4 c5
Therefore,

       A.3                                                                                1.0pt
                                                    1
                                             ξ=       ,    n = 5 and k = 4.
                                                   6π

A.4

The power emitted by the particle is given by,

                                                                                Confidential
                                                     dE      e4
                                             P =−       =            E 2 B2 .
                                                     dt   6πε0 m4 c5

The energy of the particle as a function of time can be calculated from
                                          E(t)               t
                                                 1                 e4
                                      ∫             dE = − ∫               B2 dt ,
                                       E0
                                                 E2        0
                                                               6πε 0 m4 c5

where E(0) = E0 . This leads to,

                              1    1     e4 B 2                                         E0
                                 −   =            t                ⇒      E(t) =               ,
                             E(t) E0   6πε0 m4 c5                                    1 + αE0 t

with

       A.4                                                                                1.0pt
                                                          e4 B 2
                                                     α=            .
                                                        6πε0 m4 c5

A.5
                                                                                E
If the initial energy of the electron is 100 GeV, the radius of curvature is extremely large (r = eBc ≈ 167 m).
Therefore, in approximation, one can consider the electron is moving in the inner part of the ATLAS
detector along a straight line. The time of flight of the electron is t = R/c, where R = 1.1 m is the radius
of the inner part of the detector. The total energy lost due to synchrotron radiation is,

                                                                 E0                   R
                              ΔE = E(R/c) − E0 =                         − E0 ≈ −αE02
                                                              1 + αE0 Rc              c

and

       A.5                                                                                0.5pt
                                                     ΔE = −56 MeV .

                              Secretariado
                              IPhO 2018
                                                Sociedade Portuguesa de Física
                                                Avenida da República No45 3oEsq
                                                       Theory
                                                1050-187 Lisboa, Portugal
                                                      English (UK)                   ST2-3
                                                                                +351 21 799 36 65
                                                                                info@ipho2018.pt

A.6

In the ultrarelativistic limit, v ≈ c and E ≈ pc. The cyclotron frequency is,

                                                       c     ecB    ec2 B
                                             ω(t) =        =      =
                                                      r(t)   p(t)   E(t)

     A.6                                                                                0.5pt
                                              ec2 B        e4 B 2
                                       ω(t) =       (1 +           E t) .
                                               E0        6πε0 m4 c5 0

                                                                                Confidential

                             Secretariado
                             IPhO 2018
                                                      Sociedade Portuguesa de Física
                                                      Avenida da República No45 3oEsq
                                                              Theory
                                                      1050-187 Lisboa, Portugal
                                                             English (UK)                         ST2-4
                                                                                +351 21 799 36 65
                                                                                info@ipho2018.pt

Part B. Finding the neutrino (6.0 points)

B.1

Since the W + boson decays into an anti-muon and a neutrino, one can use principles of conservation
                                                              (ν)
of energy and linear momentum to calculate the unknown pz of the neutrino. Moreover, the anti-
muon and the neutrino can be considered massless, which implies that the magnitude of their momenta
(times c) and their energies are the same. Therefore, the conservation of linear momentum can be ex-
pressed as

                                                   p⃗ (W )     = p⃗ (μ) + p⃗ (ν) ,

and the conservation of energy as,

                                                                                Confidential
                                                   E (W )     =    cp(μ) + cp(ν) .

In addition, one can also relate the energy and the momentum of the W + boson through its mass,

                                               m2W = (E (W ) )2 /c4 − (p(W ) )2 /c2
                                                    (ν)
which leads to a quadratic equation on pz ,

                                  m2W          =   [(p(μ) + p(ν) )2 − (p⃗ (μ) + p⃗ (ν) )2 ] /c2
                                               =   (2p(μ) p(ν) − 2p⃗ (μ) ⋅ p⃗ (ν) ) /c2

      B.1                                                                                1.5pt
                                 1            (ν)      (ν)        (μ)      (ν)  (μ) (ν)
                       m2W =        (2p(μ) √(pT )2 + (pz )2 − 2pT⃗    ⋅ pT⃗ − 2pz pz ) .
                                 c2

B.2

The numerical substitution directly in the answer of B.1, using
                                                                                (ν) 2
                 p(μ) = 37.2 GeV/c           m2W c2 = 6464.2 (GeV/c)2                pT       = 10 864.9 (GeV/c)2
                               (μ)       (ν)                                   (μ)
                           pT⃗       ⋅ pT⃗     = 2439.3(GeV/c)2               pz = −12.4 GeV/c ,
leads to
                                                                      (ν) 2
                            6464.2 = 74.4√10 864.9 + pz                       − 4878.6 + 24.8p(ν) .

This is a quadratic equation, equivalent to
                                                     (ν) 2               (ν)
                                        0.88889 pz           + 101.64 pz − 12378 = 0

whose solutions are:

                                Secretariado
                                IPhO 2018
                                                           Sociedade Portuguesa de Física
                                                           Avenida da República No45 3oEsq
                                                                  Theory
                                                           1050-187 Lisboa, Portugal
                                                                 English (UK)                            ST2-5
                                                                                +351 21 799 36 65
                                                                                info@ipho2018.pt

      B.2                                                                                1.5pt
                              (ν)                                                  (ν)
                             pz = 74.0 GeV/c                      or           pz = −188.3 GeV/c.

The general solution of the equation above in B.1 leads to
                              (μ)     (ν) (μ)                     (μ)
               (ν)       2pT⃗       ⋅ pT⃗ pz + m2W c2 pz
              pz     =                      (μ)
                                        2(pT )2
                                            (μ)            (ν)               (μ)         (ν)      (μ)      (ν)
                             p(μ) √−4(pT )2 (pT )2 + 4(p⃗T                         ⋅ p⃗T )2 + 4p⃗T      ⋅ p⃗T m2W c2 + m4W c4
                         ±                                                             (μ)
                                                                              2(pT )2

                                                                                (ν)
Numerical substitution leads to the above mentioned values for pz .

                                                                                Confidential
B.3

The final state particles of the top quark decay are the anti-muon, the neutrino and jet 1. Since the
neutrino is now fully reconstructed the energy and linear momentum of the top quark can be calculated
as,

                                                E (t)      =      cp(μ) + cp(ν) + cp(j1 )
                                                p⃗   (t)
                                                           = p⃗ (μ) + p⃗ (ν) + p⃗ (j1 ) .

The top quark mass is,

                         mt     = √(E (t) )2 /c4 − (p⃗ (t) )2 /c2
                                                                                2                             2
                                =      c−1 √(p(μ) + p(ν) + p(j1 ) ) − (p⃗ (μ) + p⃗ (ν) + p⃗ (j1 ) ) .

The substitution of values leads to two possible masses:

      B.3                                                                                1.0pt
                                                             2                                               2
                                mt = 169.3 GeV/c                        or               mt = 311.2 GeV/c

B.4

According to the frequency distribution for signal (dashed line), the probability of the mt = 169.3 GeV/c2
solution is roughly 0.1 while the probability of the mt = 311.2 GeV/c2 solution is below 0.01. Therefore,

      B.4   The most likely candidate is the mt = 169.3 GeV/c2 solution.                                                        1.0pt

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B.5

The top quark energy for the most likely candidate is E (t) = cp(μ) + cp(ν) + cp(j1 ) = 272.6 GeV .

                                                                               2
                                                     p(t)           E (t)
                                   d = vt = vγt0 =        t0 = ct0 √ 2 4 − 1 .
                                                     mt             mt c

      B.5                                                                                1.0pt
                                              d = 2 × 10−16 m .

                                                                                Confidential

Q2-1       English (Official)


[User]
Consider a particle of mass m in a two-dimensional box of width and length L, with
    hard walls. This is the two-dimensional analogue of problem 3, and again α1 = α2 = 0.
      (a) Write down the energy of the state with quantum numbers n1 and n2 .
      (b) What is the lowest energy level with a degeneracy of greater than 2?

Solution:
(a) There are two quantization conditions, for the two dimensions of the box,
                                     I                    I
                                       p1 dx1 = n1 h,        p2 dx2 = n2 h

          from which we conclude that
                                                   πn1 ℏ            πn2 ℏ
                                            p1 =         ,   p2 =         .
                                                    L                L
          This yields an energy of
                                                  π 2 ℏ2 2
                                               E=       (n + n22 ).
                                                  2mL2 1
          To have a sensible standing wave, we need both n1 and n2 to be positive.
      (b) The first number that can be written as the sum of two squares in more than two ways is
                                         50 = 72 + 12 = 52 + 52 = 12 + 72
          corresponding to an energy E = 25π 2 ℏ2 /mL2 .


[User]
Question A3
A beam of muons is maintained in a circular orbit by a uniform magnetic field. Neglect energy loss
due to electromagnetic radiation.
    The mass of the muon is 1.88 × 10−28 kg, its charge is −1.602 × 10−19 C, and its half-life is
1.523 µs.

  a. The speed of the muons is much less than the speed of light. It is found that half of the
     muons decay during each full orbit. What is the magnitude of the magnetic field?

  b. The experiment is repeated with the same magnetic field, but the speed of the muons is
     increased; it is no longer much less than the speed of light. Does the fraction of muons which
     decay during each full orbit increase, decrease, or stay the same?
     The following facts about special relativity may be useful:

        • The Lorentz factor for a particle moving at speed v is
                                                      1
                                                γ=p
                                                   1 − v 2 /c2

        • The Lorentz factor gives the magnitude of time dilation; that is, a clock moving at speed
          v in a given reference frame runs slow by a factor γ in that frame.
        • The momentum of a particle is given by

                                                    p~ = γm~v

          where m does not depend on v.
        • The Lorentz force law in the form
                                               d~
                                                p     ~ + ~v × B)
                                                               ~
                                                  = q(E
                                               dt
          continues to hold.

                   Copyright c 2013 American Association of Physics Teachers

2013 Semifinal Exam                                   Part A                                         5

Solution:
Question A3
A beam of muons is maintained in a circular orbit by a uniform magnetic field. Neglect energy loss
due to electromagnetic radiation.
    The mass of the muon is 1.88 × 10−28 kg, its charge is −1.602 × 10−19 C, and its half-life is
1.523 µs.
  a. The speed of the muons is much less than the speed of light. It is found that half of the
     muons decay during each full orbit. What is the magnitude of the magnetic field?
  b. The experiment is repeated with the same magnetic field, but the speed of the muons is
     increased; it is no longer much less than the speed of light. Does the fraction of muons which
     decay during each full orbit increase, decrease, or stay the same?
     The following facts about special relativity may be useful:
        • The Lorentz factor for a particle moving at speed v is
                                                       1
                                             γ=p
                                                    1 − v 2 /c2
        • The Lorentz factor gives the magnitude of time dilation; that is, a clock moving at speed
          v in a given reference frame runs slow by a factor γ in that frame.
        • The momentum of a particle is given by
                                                        p~ = γm~v
          where m does not depend on v.
        • The Lorentz force law in the form
                                              d~
                                               p     ~ + ~v × ~
                                                 = q(E        B)
                                              dt
          continues to hold.

                                            Solution
  a. The muons perform uniform circular motion, so we have
                                                 v2
                                            a=      ,     2πr = vT
                                                 r
     where T is the period of motion. Solving for T gives
                                                          2πv
                                                  T =         .
                                                           a
     Newton’s second law is
                                       dp
                                           = qv × B = ma
                                        dt
     and taking magnitudes gives a = qvB/m. Then
                                      2πm                    2πm
                                T =           ⇒         B=         = 4.85 mT
                                       qB                    qT1/2
     where we used T = T1/2 , the half-life of the muon.

                  Copyright c 2013 American Association of Physics Teachers

2013 Semifinal Exam                                     Part A                                       7

  b. The first two lines above still hold, since they follow from ordinary geometry. Since the speed
     is constant, γ is constant, and Newton’s second law is now
                                                     dp
                                                        = qv × B = γma
                                                     dt
      so taking magnitudes gives a = qvB/γm. Then the period is a factor of γ larger,
                                                            2πm
                                                     T =γ       = γT1/2 .
                                                             qB
      However, the muons experience time dilation, so in the lab frame, half of them decay in time
      γT1/2 . Then the same fraction of muons decays per orbit, i.e. one half.

USAPhO 2013 Question A3


[User]
3. This problem is designed to illustrate the advantage that can be obtained by the use
   of multiple-staged instead of single-staged rockets as launching vehicles. Suppose that
   the payload (e.g., a space capsule) has mass m and is mounted on a two-stage rocket
   (see figure). The total mass (both rockets fully fuelled, plus the payload) is N m.

              HB                                  The mass of the second-stage rocket plus the
                                                  payload, after first-stage burnout and sepa-
                                                  ration, is nm. In each stage the ratio of con-
                                                  tainer mass to initial mass (container plus
                                                  fuel) is r, and the exhaust speed is V , con-
                                                  stant relative to the engine. Note that at the
                                                  end of each state when the fuel is completely
                                          m
                                  nm              exhausted, the container drops off immedi-
                        Nm
                                                  ately without affecting the velocity of rocket.
                                                  Ignore gravity.
    (a) Obtain the velocity v of the rocket gained from the first-stage burn, starting from       [21/2]
        rest in terms of {V, N, n, r}.

                                         Page 3

                  INPhO - 2016 - Final Solutions

(b) Obtain a corresponding expression for the additional velocity u gained from the       [1]
    second stage burn.

                                                  6
(c) Adding v and u, you have the payload velocity w in terms of N , n, and r. Taking
    N and r as constants, find the value of n for which w is a maximum. For this
    maximum condition obtain u/v.

                                 C1
                                                                                [21/2]

             HB
(d) Find an expression for the payload velocity ws of a single-stage rocket with the      [2]
    same values of N , r, and V .

(e) Suppose that it is desired to obtain a payload velocity of 10 km/s, using rockets     [2]
    for which V = 2.5 km/s and r = 0.1. Using the maximum condition of part (c)
    obtain the value of N if the job is to be done with a two-stage rocket.

                                    Page 4

                      INPhO - 2016 - Final Solutions

Solution:
3. This problem is designed to illustrate the advantage that can be obtained by the use
   of multiple-staged instead of single-staged rockets as launching vehicles. Suppose that
   the payload (e.g., a space capsule) has mass m and is mounted on a two-stage rocket
   (see figure). The total mass (both rockets fully fuelled, plus the payload) is N m.

              HB
       1st Stage           2nd Stage    Pay
                                        load
                                                  The mass of the second-stage rocket plus the
                                                  payload, after first-stage burnout and sepa-
                                                  ration, is nm. In each stage the ratio of con-
                                                  tainer mass to initial mass (container plus
                                                  fuel) is r, and the exhaust speed is V , con-
                                                  stant relative to the engine. Note that at the
                                                  end of each state when the fuel is completely
                                          m
                                  nm              exhausted, the container drops off immedi-
                        Nm
                                                  ately without affecting the velocity of rocket.
                                                  Ignore gravity.
    (a) Obtain the velocity v of the rocket gained from the first-stage burn, starting from       [21/2]
        rest in terms of {V, N, n, r}.

          Solution: Variable mass equation gives
                                        d~v                           dm
                                    m       = F~external + ~vrelative
                                        dt                            dt
          No gravity hence F~external = 0, |~vrelative | = v. Solving rocket equation
                                                          mi
                                             v = V ln                                   (1)
                                                          mf

                                             Page 3

                  INPhO - 2016 - Final Solutions

      Here

                                   initial mass mi = N m                        (2)
                        final mass mf = [N r + n(1 − r)]m                       (3)

      Answers are written in terms of speed. Answer written in terms
      of velocity with proper sign (-ve) are also acceptable.

(b) Obtain a corresponding expression for the additional velocity u gained from the       [1]
    second stage burn.

      Solution: Now mi = nm, mf = m(nr + 1 − r). Equation (1) yields
                                                   n
                                  u = V ln                                      (4)
                                               nr + 1 − r
      or
                                                   n
                                  ~u = V~ ln                                    (5)
                                               nr + 1 − r

                                                     6
(c) Adding v and u, you have the payload velocity w in terms of N , n, and r. Taking
    N and r as constants, find the value of n for which w is a maximum. For this
    maximum condition obtain u/v.

                                 C1
      Solution: From Eqs. (3 and 5)
                                                                                [21/2]

                                               Nn
                        w = V ln
                                   [N r + n(1 − r)][nr + 1 − r]

             HB             = V ln f (n)

      Maximizing w is equivalent to maximizing f (n). Differentiating and setting
      equal to zero, we obtain
                         √
                     n= N ⇒ =
                               u
                               v
                                       √     √
                                    ln[ N /{r N + (1 − r)}]
                                                √
                                    ln[N/{N r + N (1 − r)}]
                                                               =1             (6)

      where we have used Eqs. (1 and 5).

(d) Find an expression for the payload velocity ws of a single-stage rocket with the      [2]
    same values of N , r, and V .

      Solution: Here mi = N m and mf = m + r(N m − m). Using Eq. (1)

                                                  N
                                 ws = V ln
                                               Nr + 1 − r

(e) Suppose that it is desired to obtain a payload velocity of 10 km/s, using rockets     [2]
    for which V = 2.5 km/s and r = 0.1. Using the maximum condition of part (c)
    obtain the value of N if the job is to be done with a two-stage rocket.

                                    Page 4

                      INPhO - 2016 - Final Solutions

         Solution: Payload velocity
                                                          √
                                                     N
                                w = u + v = 2V ln √
                                                 r N +1−r
         For the desired value of w, N = 649.4 (645 < N < 655 accepted.)

INPhO 2016 Problem 3