This exam is formed of four exercises in four pages numbered from 1 to 4. The use of non-programmable calculator is recommended,

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1 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات الرسمية امتحانات الشهادة الثانوية العامة الفرع : علوم عامة مسابقة في مادة الفيزياء المدة ثالث ساعات االسم: الرقم: دورة العام 6 العادي ة االثنين 3 حزيران 6 This exam is formed of four exercises in four pages numbered from to 4. The use of non-programmable calculator is recommended, First exercise: (7.5 points) Variation of the kinetic energy of a system The aim of this exercise is to verify the theorem of kinetic energy of a system. The skier (S) of mass M = 8 kg, moves down from O to, with a constant velocity v vi, where v = 3m/s along the line of greatest slope of a track inclined by an angle α = 3 with the horizontal. The track exerts on the skier a constant force of friction f fi. The motion of the skier is represented by the motion of its center of mass on x'x where i is a unit vector along this axis (figure ). Neglect the air resistance on the skier. Take: the horizontal plane through as a gravitational potential energy reference for the system (skier, Earth). g = m/s. ) Name and represent the external forces acting on along the path O. ) a) Show that the linear momentum P of the skier is constant. b) pply Newton s second law on the skier, between the points O and, deduce the magnitude of f. 3) The skier, upon reaching, starts exerting a constant braking force f f i to stop at. The skier covers the distance during a time interval Δt = 3 s. P dp a) Determine the magnitude of f, assuming that. t b) The mechanical energy of the system (skier, Earth) decreases from to. Name the forces that are responsible of this decrease. c) Determine the distance covered by the skier during the time interval Δt. 4) a) Determine between and : i. the variation of the gravitational potential energy PEg of the system (skier, Earth) ; ii. the work done by the weight W. mg x' i O Figure b) ompare PEg and W mg. 5) ΔKE and W Fext are respectively the variation of the kinetic energy of the skier and the algebraic sum of the work done by the external forces between and. Verify, between and, the work-energy theorem: ΔKE = W Fext. = 3 x

2 Second exercise: (7.5 points) The characteristics of RL series circuit onsider: a generator delivering an alternating sinusoidal voltage : um = u = u =U cos t (u in V and t in s), where U = 5 V and ω = πf with adjustable frequency f; a coil of inductance L and of negligible resistance; a capacitor of capacitance ; a resistor of resistance R = 5 ; an oscilloscope; a milli-ammeter of negligible resistance; a switch K and connecting wires. In order to determine L and, we perform the following experiments: - First experiment We perform successively the setup of figure and of figure. For f = 5 Hz, the effective current I, indicated by the milli-ammeter, has the same value I= 5m in both setups. Take.3. ) The coil is connected across the terminals of (figure ). The circuit carries a current i of expression I = I cos( t ). (i in and t in s) a) Determine the expression of the voltage ud = ucoil in terms of L, ω, I and t. b) Deduce the value of L. ) The capacitor is connected across the terminals of (figure ). The circuit carries a current i of expression i = I cos( t ). a) Determine the expression of the voltage ud = u in terms of, ω, I and t. b) Deduce the value of. - Second experiment To verify the values obtained for L and in the first experiment, we perform the setup of the circuit shown in Figure 3. This circuit contains the generator, the coil, the capacitor, and the resistor of resistance R = 5. The oscilloscope, displays on channel (), the voltage um across the generator, and on channel (), the voltage udm across the resistor. Figure (4) shows the waveforms representing um and udm. The circuit carries a current i = I cos( t ). ) Redraw figure 3 and indicate the connections of the oscilloscope. ) pply the law of addition of voltages and give t a particular L value, show that: tan. R 3) Referring to the waveform of Figure 4 observed on the screen of the oscilloscope, determine: a) the frequency f; b) the phase difference φ between u and i. 4) The effective voltage U being kept constant and we vary f. We observe that um and udm become in phase when f takes the value f = 5 Hz. a) Name the phenomenon that takes place. b) ive the relation giving ω in terms of L and. 5) Determine L and. um udm M M M i + K Figure i Figure Figure 4 Horizontal Sensitivity :.5 ms/div + i + K K q q R Figure 3 D D L D L

3 Third exercise: (7.5 points) orpuscular aspect of light The aim of this exercise is to study the emission spectrum of the hydrogen atom and use the emitted light to produce photoelectric effect. iven: Planck s constant: h = J.s ; Speed of light in vacuum: c = 3 8 m/s ; ev =.6-9 J ; Elementary charge: e =.6-9 ; nm = -9 m.. Hydrogen atom The emission spectrum of the hydrogen atom constituted in its visible part of four radiations denoted by Hα, H, H and H of respective wavelengths, in vacuum, 656.7nm, 486.3nm, 435.5nm and 4.7nm. I. In 885, almer noticed that the wavelengths of these four radiations verify the empirical formula n where = nm where n is a non-zero positive whole number. n 4 ) The smallest value of n is 3. Justify. ) alculate the wavelength corresponding to this radiation. 3) Deduce the values of n corresponding to the wavelengths of the other three visible radiations in the emission spectrum of the hydrogen atom. II. The quantized energy levels of the hydrogen atom are given by the formula: 3.6 En (in ev) where n is a whole non-zero positive number. n Using the expression of En, determine the energy of the atom when it is: ) in the ground state. ) in each of the first five excited levels. 3) ionized state.. Photoelectric effect hydrogen lamp of power PS = W, emits uniformly radiation in all directions in a homogeneous and non- absorbing medium. This lamp Radiation illuminates a potassium cathode of a photoelectric cell of work functionw =. ev and of a surface area s = cm placed at a distance D =.5m from the lamp (figure). ) alculate the threshold wavelength of the potassium cathode. ) mong the rays of almer series, specify the radiation that can produce photoelectric emission. () () 3) Using a filter we illuminate the cell by a blue light H of wavelength = 486.3nm. The generator is adjusted so that the anode () captures all the emitted electrons by the cathode of quantum efficiency r =.875%. Figure a) Show that the received power of the radiation P of the cell is.4-5 W. b) Determine the number N of the incident photons received by the cathode in one second. c) Determine the current in the circuit. 3

4 Fourth exercise: (7.5 points) ompound Pendulum The aim of this exercise is to study the motion of a compound pendulum. onsider a compound pendulum (P) consists: of a straight and homogeneous rod (R) of length = and of mass m ; of a solid (S), taken as a particle of mass m, free to slide along the part O of the rod, O being the midpoint of the rod. We fix (S) at a point such that O = x (x > ). (P) can oscillate, in a vertical plane, around a horizontal axis () perpendicular to the rod at O ( figure ). ( P) is shifted from its equilibrium position by a small angle θm then released without initial velocity at the instant t =, the pendulum oscillate then, without friction, around its equilibrium position. t the instant t, the angular elongation of the pendulum is and its angular velocity is iven: moment of inertia of the rod about the axis of rotation (): I = m, m = 3m, d. =.5 m, g = m/s and =. For small : cos and sin ( in rd). is the center of inertia of the pendulum and the horizontal plane passing through O is taken as reference level of the gravitational potential energy. ) Show that: a) O = x 4 ; m b) The expression of the moment of inertia of the pendulum is: I = ( 4x ). ) Determine the expression the mechanical energy of the system (pendulum, Earth) in terms of, ', m,x and. 3) a) Establish the second order differential equation in which governs the oscillations of the pendulum. 4x b) Deduce that the expression of the proper period of the pendulum is: T. x 4) a) Determine the value of x for which T is minimum. b) Deduce that T(min) =.4 s. 5) Using a coupling device, the pendulum (P) plays the role of an exciter for a simple pendulum (P) of length = 65 cm. The oscillations of (P) and (P) are slightly damped. a) Knowing that the proper period of the simple pendulum, for small oscillation, is T, g alculate the value of the proper period T of (P). b) i) (P) oscillates now with its minimum period. It is noticed that (P) does not enter in amplitude resonance with (P). Justify. ii) We move (S) between O and. For a value x of x, we notice that (P) oscillates with large amplitude. alculate the value of x. (R) Figure O (S) PE= 4

5 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات مشروع معيار التصحيح امتحانات الشهادة الثانوية العامة الفرع : علوم عامة مسابقة في مادة الفيزياء المدة ثالث ساعات االسم: الرقم: دورة العام 6 العادي ة االثنين 3 حزيران 6 First exercise (7.5 points) Part of the Q The forces acting on the skier : Normal reaction N ; Weight mg ; The frictional force f Diagram..a.b 3.a nswer Mark P = MV since V = te P = te. 3/4 d P = Mg + N + f = project along x x: Mgsinα f = f = Mgsinα = 4 N. d P = Mg + f + N + f P = t MV MV MV f f= 8 N. t t P POP Fext Fext t t MV Mg sin f f f f f 8N t Project along x x Or : Project along x'x : 3.b ecause friction and braking forces / 3.c ΔM.E = W( f ) + W ( f ) M.E M.E = W( f ) + W ( f ) -/ MV Mg sinα = - f. f. ( 4 9) + (4 ) = = 45 m. 4.a.i ΔPE = PE PE Mg sin = - Mg sinα = - 8 J 3/4 4.a.ii W( Mg ) = Mgh = Mg sinα = 8 J / 4.b Δ(PE) = - W ( Mg ). /4 5 ΔM.E = ΔK.E + ΔP.E = W( f ) + W ( f ) ΔK.E = W( Mg )+ W( f ) + W( f ) since W( N ) = ΔK.E = W Fext Or : ΔM.E = ΔK.E + ΔP.E = W( f ) + W ( f ) ΔK.E. = W( f ) + W( f ) ΔP.E = W( f ) + W( f ) + W( Or W( N ) = Δ K.E. = W Fext Mg ) 3/4

6 Second exercise (7.5 points) Part of the Q..a..b..a..b nswer di ud = ul = L LI sin( t ) um ud LI sin( t ) U cos t LI cos( t ) U cos t y comparison: U LI L =.3 H = 3 mh. Or: LI sin( t ) =U cos t For t = : U LI L =.3 H = 3 mh. du i I u i sin( t ) um ud U cos t = I sin( t ) U cos t = I cos( t ) I 6 y comparison : U 3. F 3. F Or : um ud U cos t = I sin( t ) For t = : U = I sin( ). onnections of the oscilloscope 6 3. F 3. F Y i K L q Mark ¼ M R D Y. um u ud udm I U cost LI sin( t ) sin( t ) RI cos( t ) L I For t LI cos cos RI sin tan R.3.a T = 4 ms f 5 Hz. T.3.b rad a urrent resonance ¼.4.b tan L L..5 4 R L L L = = 3 mh L R =3. -6 F = 3. F

7 Third exercise (7.5 points) Part of the Q.I. nswer, and n are positive n - 4 > n > the smallest value is n = 3. Mark.I..I.3 n = nm. n 4 In these conditions: n = 4 gives = nm n = 5 gives = nm n = 6 gives = 4.7 nm.ii..ii. round state n = : E = ev. st 3.6 energy level (excited): n = : E = 3.4eV E3 = -.5 ev; E4 = -.85 ev; E5 = -.54 ev and E6 =.38 ev ¼.II.3 The atom is ionized when n E =. W = hc = hc W m = 565nm. The radiations of almer series that can produce photoelectric emission verifies the relation λ λ ; H, H and H produce this emission because λ λ.3.a P = S s 4D P 5.4 W.3.b N P P 3 s Ephoton hc 4.99 photons / s.3.b The number of effective photons = number of emitted electrons N e Ne = r N = 4.37 electrons/s q Ne e 8 I 6.99 t t 3

8 Fourth exercise (7.5 points) Part of the Q.a nswer ( m + m ) O = m OO + m OM O = 4 x Mark.b m m I(sys) = I(rod) + I(S) I = m x (4x ) 3 ME = I (m m m m )go cos (4x ) gx cos a dme m m ME = te (4x ) gxsin, 3 4gx For small angle sin (rd). 4x 3.b This differential equation has the form: 4gx 4x T 4.a dx x x dt 4x 4x, 4x 4.b T.4 s 5.a 5.b.i 5.b.ii T 4x x. dt ; T is minimum when dx = for x ; then T is minimal for 4x = x. =.6 s g The phenomenon of amplitude resonance will take place when the proper period of the exciter becomes equal (very close) of that of the resonator. s T =.4 s of (P) is smaller than T =.6 S of (P), therefore the phenomenon of resonance does not take place (P) oscillates with large amplitude, therefore it is in resonance of amplitude with (P); and then the proper period of (P) is equal to T. =.6 s. 4x (,6) x The solution of this quadratic equation gives; x = 53 cm (rejected because it is than = 5 cm) and x =.75 cm (accepted) 4

This exam is formed of three exercises in three pages. The Use of non-programmable calculators is allowed.

This exam is formed of three exercises in three pages. The Use of non-programmable calculators is allowed. 008 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات امتحانات الشهادة الثانوية العامة الفرع : علوم الحياة مسابقة في مادة الفيزياء المدة ساعتان االسم: الرقم: الدورة اإلستثنائية للعام