Solutions t o the problems of the 5-th 5-th Internatio nal Physics Olympiad, 1971 1971,, Sofi Sofi a, Bul garia The The prob roblem lems and the solut lution ions are adapted by Victor Ivanov Sofi Sofia a Stat State e Univ University, Facu F acullty of Phy Physics, 5 J ames Bour Bourcie cier Blv Blvd., 1164 1164 Sofi Sofia, a, Bulga Bulgarria Referenc rence e: O. F. Ka K abardin, V. V . A. A . Orlov, Orl ov, in in “Int “Inte ernation tiona al Physi Physics Olympiad Olympiads s for Hi High School School Stude Students”, nts”, eds. ds. V. V . G. G. Ra Razumovski, ovski, Mosco M oscow, w, Na Nauka, 1985. 1985. (In (I n Russian). Russian).
Theoretical Theoretical problems Question 1. The The bloc locks slid slide e relat lative ive to th the prism ism with ith acceler leration ions a1 and a2, which are parall rallel to its i ts side sides s and and have the same magnitude nitude a (see Fig. 1.1). The blocks move rela relative tive to the earth with wi th acce accellerations: rations: (1.1) w1 =a1 +a0; (1.2) w2 =a2 +a0. Now we project w1 and w2 along the x- and y-axes: y (1.3) w1x = acosα1 − a0 ; w1
(1.4)
w1y = asinα inα 1 ;
(1.5)
w2x = acosα 2 − a0 ;
(1.6)
w2y = − asinα 2 .
a1 a0
α1
w2
a2
α2
x
Fig. 1.1 The The equation ions of motion ion for for the bloc locks and for for the prism ism have the follo follow wing ing vector forms (see Fig. 1.2): (1.7) m1w1 = m1g + R 1 + T1; (1.8) m2w2 = m2g + R 2 + T2 ; (1.9) Ma0 = Mg − R 1 − R 2 + R − T1 − T2 . y R1
T 2
T 1
R2 R
m2g
m1g Mg
x
Fig. 1.2 The The for forces of te tension ion T 1 and T 2 at the ends ends of the thread are are of the same magnitude T tude T since since the masse asses s of the thread thread and that of the pull pulley are negli gligibl gible. e. Note Note that in in equa equation tion (1.9) we account for the net force –( T 1 + T 2), which which the bended thread thread exerts exerts on the
prism through the pulley. The equations of motion result in a system of six scalar equations when projected along xand y: (1.10) m1a cosα1 − m1a0 = T cosα1 − R1 sinα 1; (1.11) (1.12)
m1asinα1 = T sinα1 + R1 cosα1 − m1g ; m2a cosα 2 − m2a0 = − T cosα 2 + R2 sinα 2 ;
(1.13) (1.14)
m2asinα 2 = T sinα 2 + R2 sinα 2 − m2 g ; − Ma0 = R1 sinα 1 − R2 sinα 2 − T cosα 1 + T cosα 2 ;
(1.15) 0 = R − R1 cosα 1 − R2 cosα 2 − Mg . By adding up equations (1.10), (1.12), and (1.14) all forces internal to the system cancel each other. In this way we obtain the required relation between accelerations a and a0: M + m1 + m2 (1.16) a = a0 . m1 cosα1 + m2 cosα 2 The straightforward elimination of the unknown forces gives the final answer for a0: (m1 sin α1 − m2 sinα 2 )(m1 cosα1 + m2 cosα 2 ) (1.17) a0 = . (m1 + m2 + M )(m1 + m2 ) − (m1 cosα1 + m2 cosα 2 )2 It follows from equation (1.17) that the prism will be in equilibrium ( a0 =0) if: m1 sinα 2 (1.18) . = m2 sinα1
Question 2. We will denote by H (H =const) the height of the tube above the mercury level in the pan, and the height of the mercury column in the tube by hi. Under conditions of mechanical equilibrium the hydrogen pressure in thetube is: (2.1) PH = Pair − ρghi , 2
where ρ is the density of mercury at temperature ti: (2.2) ρ = ρ 0 (1− βt) The index i enumerates different stages undergone by the system, ρ0 is the density of mercury at t0 =0 °C, or T0 =273 K, and β its coefficient of expansion. The volume of the hydrogen is given by: (2.3) Vi =S(H – hi). Now we can write down the equations of state for hydrogen at points 0, 1, 2, and 3 of the PV diagram (see Fig. 2): m (2.4) RT0 ; (P0 − ρ 0 gh0 )S(H − h0 ) = M m (2.5) RT0 ; (P1 − ρ 0 gh1)S(H − h1) = M m (2.6) RT2 , (P2 − ρ1gh2 )S(H − h2 ) = M P T ρ0 where P2 = 1 2 , ρ1 = ( 2 − T0 )] since the process 1–3 is ≈ ρ 0 [1− β T T0 1+ β T ( 2 − T0 ) isochoric, and:
m RT3 M V H − h3 where ρ 2 ≈ ρ 0[1− β T for the isobaric process 2–3. ( 3 − T0 )] , T3 = T2 3 = T2 V2 H − h2 (2.7)
(P2 − ρ 2 gh3 )S(H − h3 ) =
P P0
0 3
2
P2
P1
1 V0
V1=V2
V3
V
Fig. 2 After a good deal of algebra the above system of equations can be solved for the unknown quantities, an exercise, which is left to the reader. The numerical answers, however, will be given for reference: H ≈ 1.3 m; m≈ 2.11×10–6 kg; T2 ≈ 364 K; P2 ≈ 1.067×105 Pa; T3 ≈ 546 K; P2 ≈ 4.8×104 Pa.
Question 3. A circuit equivalent to the given one is shown in Fig. 3. In a steady state (the capacitors are completely charged already) the same current I flows through all the resistors in the closed circuit ABFGHDA. From the Kirchhoff’s second rule we obtain: E − E1 (3.1) . I = 4 4R Next we apply this rule for the circuit ABCDA: (3.2) V1 + IR = E2 − E1 , where V1 is the potential difference across the capacitor C1. By using the expression (3.1) for I, and theequation (3.2) we obtain: E − E1 (3.3) V1 = E2 − E1 − 4 = 1V. 4
Similarly, we obtain the potential differences V2 and V4 across the capacitors C2 and C4 by considering circuits BFGCB and FGHEF: E − E1 (3.4) V2 = E4 − E2 − 4 = 5 V, 4
(3.5)
V4 = E4 − E3 −
E4 − E1 4
= 1V.
Finally, the voltage V3 across C3 is found by applying the Kirchhoff’s rule for the outermost circuit EHDAH: E − E1 (3.6) V3 = E3 − E1 − 4 = 5V. 4
The total energy of the capacitors is expressed by the formula: C (3.7) W = (V12 + V22 + V32 + V42 ) = 26 µ J . 2
C3 A E1 D
R
B
R
E2 C1
C4
F E4
C
C2
G
E E3
R
H
R
Fig. 3 When points B and H are short connected the same electric current I’ flows through the resistors in the BFGH circuit. It can be calculated, again by means of the Kirchhoff’s rule, that: E (3.8) I′ = 4 . 2R The new steady-state voltage on C2 is found by considering the BFGCB circuit: (3.9) V2′ + I ′R = E4 − E2 or finally: E (3.10) V2′ = 4 − E2 = 0V. 2
Therefore the charge q′2 on C2 in the new steady state is zero.
Question 4. In a small time interval ∆t the fish moves upward, from point A to point B, at a small distance d = v∆t. Since the glass wall is very thin we can assume that the rays leaving the aquarium refract as if there was water – air interface. The divergent rays undergoing one single refraction, as show in Fig. 4.1, form the first, virtual, image of the fish. The corresponding vertical displacement A1B1 of that image is equal to the distance d1 between the optical axis a and the ray b1, which leaves the aquarium parallel to a. Since distances d and d1 are small compared to R we can use the small-angle approximation: sinα ≈ tanα ≈ α (rad). Thus we obtain: (4.1) d1 ≈ R α; (4.2) d ≈ R γ; (4.3) α +γ =2β; (4.4) α ≈ nβ.
From equations (4.1) - (4.4) we find the vertical displacement of the first image in terms of d: n (4.5) d1 = d, 2− n and respectively its velocity v1 in terms of v: n (4.6) v1 = = 2v . 2− n
α
B1
b d
β
β
α
B γ
a
d A1
A
Fig. 4.1 The rays, which are first reflected by the mirror, and then are refracted twice at the walls of the aquarium form the second, real image (see Fig. 4.2). It can be considered as originating from the mirror image of the fish, which move along the line A’B’ at exactly the same distance d as the fish do. 4R
B a
d2 b2
d
A2 α
β
A γ d' β
B’ d
δ α
A’
B2
Fig. 4.2 The vertical displacement A2B2 of the second image is equal to the distance d2 between the optical axis a and the ray b2, which is parallel to a. Again, using the small-angle approximation we have: (4.7) d’ ≈ 4Rδ - d, (4.8) d2 ≈ Rα Following the derivation of equation (4.5) we obtain: n (4.9) d2 = d′ . 2− n Now using the exact geometric relations:
(4.10) δ =2α – 2β and the Snell’s law (4.4) in a small-angle limit, we finally express d2 in terms of d: n (4.11) d2 = d, 9n − 10 and the velocity v2 of the second image in terms of v: n 2 (4.12) v2 = v = v. 9n − 10 3 The relative velocity of the two images is: (4.13) vrel =v1 – v2 in a vector form. Since vectors v1 and v2 are oppositely directed (one of the images moves upward, the other, downward) the magnitude of the relative velocity is: 8 (4.14) vrel = v1 + v2 = v . 3
Experimental prob l em The circuit is given in the figure below:
R
A
V
E Sliding the contact along the rheostat sets the current I supplied by the source. For each value of I the voltage U across the source terminals is recorded by the voltmeter. The power dissipated in the rheostat is: P =UI provided that the heat losses in the internal resistance of the ammeter are negligible. 1. A typical P–I curve is shown below: P Pmax
I0
I
If the current varies in a sufficiently large interval a maximum power Pmax can be detected at a certain value, I 0, of I. Theoretically, the P(I) dependence is given by: (5.1) P = EI − I 2r , whereE and r are the EMF and the internal resistance of the dc source respectively. The maxim value of P therefore is: E2 (5.2) , Pmax = 4r and corresponds to a current: E (5.3) . I0 = 2r 2. The internal resistance is determined trough (5.2) and (5.3) by recording Pmax and I 0 from the experimental plot: P r = m2ax . I0 3. Similarly, EMF is calculated as: 2P E = max . I0 4. The current depends on the resistance of the rheostat as: E I = . R+ r Therefore a value of R can be calculated for each value of I: E (5.4) R= −r. I The power dissipated in the rheostat is given in terms of R respectively by: E 2R (5.5) P= . (R + r )2 TheP–R plot is given below: P E2/(4r)
R =r
R
Its maximum is obtained at R =r. 5. The total power supplied by the dc source is: E2 (5.6) Ptot = . Ptot R+ r E /r
R
6. The efficiency respectively is: P R (5.7) . η= = Ptot R + r
η
1
R
