Exam Sv Physics 2009

‫امتحانات الشهادة الثانوية العامة‬ ‫ علوم الحياة‬: ‫الفرع‬

‫وزارة التربية والتعليم العالي‬ ‫المديرية العامة للتربية‬ ‫دائرة المتحانات‬

‫مسابقة في مادة الفيزياء‬ ‫المدة ساعتان‬

:‫السم‬ :‫الرقم‬

This exam is formed of three exercises in three pages. The Use of non-programmable Calculators is recommended. Answer the three following questions: First question (7 points) Motion of a solid under the action of a spring r

The free extremity of a spring of horizontal axis (O, i ), of negligible mass and of stiffness k = 15 N/m, is attached to a solid (S) of mass m. (S) is able to move on a horizontal table while the other extremity is fixed. Let v and x be the algebraic value of the velocity and the abscissa respectively, with respect to the axis (O, i ), of the centre of inertia G of (S) at the instant t. The forces of friction are negligible and the horizontal plane containing G is taken as level of reference of gravitational potential energy. A- Theoretical Study (S) is moved a distance x0 along the axis (O, i ) in the positive direction and then released from rest at the instant t0 = 0. 1. Give the expression of the mechanical energy of the system (S, spring, Earth) in terms of k, m, x and v. 2. Derive, from this expression, the differential equation of motion of (S). 3. a. Verify that the time equation of motion of (S) is of the form: = xm cos(

2π t + ϕ ). Deduce T0

the expressions of the constants T0, xm and ϕ . b. Write the time equation of the speed. Deduce the relation between x0 and the maximum speed Vm. B- Experimental study We record the variations of the abscissa x of G as a function of time (figure 1) and those of the speed v (figure 2). v (cm/s)

x (cm)

30

3

23.6

2

20

1

10

0

t (s) 0.2

0

0.4 0.6 0.8 1.0 1.2

t (s) 0.2

0.4 0.6 0.8 1.0 1.2

-10

-1

-17

-2

-20

-3

-30 Fig. 1

Fig. 2

1. The graphs of figures 1 and 2 show the initial conditions of motion of (S)? Justify. 2. Determine, using these graphs, the values of T0, x0 and Vm. 1

3. Verify the relation between x0 and Vm. 4. Determine the mass m of (S).

t(s) v(m/s) KE (J) x(m) EPE (J) ME (J)

0

0.2 - 0.236 6.77×10-3

0.3 -0.17

0.8

C- Energetic study 1. KE, EPE and ME are respectively 0.030 -0.021 the kinetic energy of (S) and the -3 6.77×10 elastic potential energy and the mechanical energy of the system ((S), spring). Copy and complete the table below. 2. Are the results compatible with the law of conservation of mechanical energy? Why? Second question (7 points) Measurement of the velocity of a plane This exercise shows evidence of the role of the phenomenon of electromagnetic induction in the measurement of the velocity measurement of a plane. A- Velocity of a conductor moving in a magnetic field Consider a metallic homogeneous rod MN of length ℓ, which can slide, on two metallic, horizontalr and parallel rails AA' and EE', with a constant velocity v . During this sliding motion, the rod remains perpendicular to the rails and its center of gravity G moves on axis Ox. At the instant t0 = 0, G is at O, origin of the abscissa. At the instant t, the abscissa of G is x = OG and v =

dx is the dt

algebraic measurement of its velocity. The device, formed of the rod and the rails, is placed in a uniform magnetic r field B and perpendicular to the horizontal plane of the rails. (Figure 1). 1) Determine, at the instant t, in terms of B,  and x, the expression of the magnetic flux which crosses the surface AMNE respecting the arbitrary positive direction indicated on figure 1. 2) Explain the existence of an induced emf e which is obtained between the extremities M and N. 3) Determine the expression of the induced emf e in terms of B,  and v. 4) The rod is not crossed by any electric current. Why? 5) Deduce the polarity of the points M and N of the rod. 6) Determine, in terms of B,  and v, the value of the electric voltage which appears across the extremities of the rod. B Velocity measurement of a plane A plane flies horizontally along a straight line, at the constant speed v , in the uniform terrestrial magnetic field B . The vector B , in the area of flight, has a horizontal component of magnitude Bh = 2,3 × 10 −5 T and a vertical component of magnitude Bv = 4 × 10 −5 T. The wings of the plane can be considered as with a rectilinear M horizontal conductor of length MN = 30 m, and perpendicular to the velocity vector v (Figure 2). 1. Determine, at the instant t, in terms of Bv,  and v, the expression of the emf e which appears at the extremities M and N of the wings. 2. Determine v, if the voltage across the extremities M and N has as a value 0.36 V. N Third question (6 points) Fig.2 2

Sodium vapor lamp Sodium vapor lamps are used to light roads. These lamps contain sodium vapor under very low pressure. This vapor is excited by a beam of electrons which crosses the tube containing the vapor. The electrons yield energy to the sodium atoms. At the time of the return to the ground state, energy is restored in the form of luminous radiations. Given: h = 6.62 × l0 -34 J.s; c = 3 × 108 ms -1; e = 1.60 × 10-19 C. 1. What do the quantities h, c and e represent? 2. The analysis of the emission spectrum of a sodium vapor lamp shows the presence of lines of well defined wavelengths λ . The figure below represents some lines of the emission spectrum of a sodium vapor lamp. 330.3 nm

568.8 nm 615.4 nm Doublet 589.0 and 589.6 nm

819.5 nm

1138.2 nm

a. The yellow lines doublet of wavelengths 589.0 nm and 589.6 nm is much more intense than the others. i) To which domain, visible, infra-red or ultraviolet, does each of the other lines of the spectrum belong? ii) Why do we speak about a yellow light emitted by the lamps with sodium vapor? b Is the visible light emitted by the sodium lamp Energy (eV) monochromatic or polychromatic? Justify the answer. E7 -0.85 0 3. The adjacent figure represents the simplified diagram of the E6 -1.01 energy levels of the sodium atom. -1.38 E5 E4 a. Specify from the adjacent energy diagram, an indicator that -1.51 E -1.93 justifies the discontinuity of the emission spectrum of a 3 E2 -3.03 sodium vapor lamp. b. Consider the yellow line of the sodium doublet of E1 -5.14 wavelength λ = 589.0 nm. i) Show that the energy ∆E which corresponds to the emission of this line is 2.11 eV. ii) Specify, with justification, the electronic transition corresponding to the emission of this line. Justify the answer. iii) Make a diagram showing the previous transition and that which corresponds to the emission of the radiation of wavelength 589.6 nm. 4. The sodium atom, taken in its ground state, is hit respectively by electrons of kinetic energies 1.01 eV, 3.03 eV and 6.14 eV. a) Which value(s) of the kinetic energy of these electrons can excite the sodium atom? Justify the answer. b) What becomes the kinetic energy of each electron after its collision with the sodium atom?

3

First question (7 points) Part of the Q A.1 A.2

A.3.a

Answer

Mark

Expression of the mechanical energy: ME = ½ mv2 + ½ kx2 There is conservation of the mechanical energy: ME = constant, then dE m =0 dt k  + ½ k2x x = 0  mx(  x + x ) = 0 ; as x ≠ 0 ∀ t, ½ m 2 VV m k x = 0 then: x + m 2π 2π 2π dx = x = x = xm cos( t + ϕ) ; xm sin( t + ϕ) ; T0 T0 T0 dt

0.25 0.75

1.75

2

2π  2π  d2x  = x cos( t + ϕ) By replacing in the differential = x m   T0 dt 2  T0  equation and by simplifying by x  0 , we obtain: 2

2

A.3.b B.1 B.2 B.3 B.4

 2   2  k x    x  0 , while comparing :   = m  T0   T0  m  T0 = 2 k 2π For t0 = 0, x = xm cos(ϕ) = x0 > 0 et v = x sin(ϕ) = 0 T0 m ⇒ ϕ = 0 or π. xm > 0 ⇒ cos ϕ =1 thus xm = x0 and ϕ = 0 2π 2π 2π dx x0 = x = .v= x0 sin( t) ; ⇒ Vm = T0 T0 T0 dt Yes, because for t0 = 0, x is maximum (x0) and v = 0 T0 = 0.8 s, x0 = 3 cm and Vm = 23.6 cm/s or 0.236 m/s 2π 2π Vm = x = ×0.03 = 0.2356 m/s = 0.236 m/s T0 0 0.8 2

2

 T   0.8  m T0 = 2π ⇒m=  0  k=   × 15 = 0.243 kg. k  2π   2π 

4

1.00 0.50 0.75 0.50 0.50

C.1

0.75

t(s) 0 0.2 0.3 0.8 v(m/s) 0 - 0.236 -0.17 0

KE (J) 0 6.77×10-3 3.51×10-3 0

x(m) 0.030 0 -0.021 0.030

EPE (J) 6.77×10-3 0 3.31×10-3 6.77×10-3

ME (J) 6.77×10-3 6.77×10-3 6.82×10-3 6.77×10-3

C.2

Yes, because it is found that the value of the mechanical energy remains almost the same.

0.25

Second question (7 points) Part of the Q

Answer

Mark

A.1

ϕ = BScos α = - BS = -Bℓx.

0.75

A.2

The magnetic flux varies, an induced emf e appears across the extremities N and M of the rod dϕ dx = B = Bv. e=dt dt

0.50

A.3

5

0.75

A.4

The circuit is open ⇒ i = 0.

0.50

A.5

uNM = e – ri (r = 0 and i =0) ⇒ uNM = e > 0, The point N is positive and the point M is negative. │uNM │= e = Bv.

1.00

In this case: ϕ = ϕ V + ϕ h = BvScos00 + BhScos900= Bv S dϕ dx = −B v  = −B v v. e=dt dt 0.36  300 m / s . │uNM│ = Bv v ⇒ v = 4 105  30

1.50

A.6 B.1

B.2

1.00

1.00

Third question (6 points) Part of the Q 1

Answer

Mark

h: Planck's constant; c: speed of light, e: elementary charge

0.50

330.3 nm ultraviolet domain; 568.8 nm, 589 nm and 615.4 nm visible domain; 819.5 nm and 1138.2 nm infra-red domain. Because this light is much more intense than the others

0.50

2.b

It is polychromatic because it is made of several visible radiations

0.50

3.a

The discontinuity of the emission spectrum is justified by the discontinuous levels of the energy of the sodium atom. hc 6.62 1034  3  108 E = = = 3.3710-19 J or  589 3.37 1019 E = = 2.11 eV. 1.6  1019 This transition will take place from the excited level E2 to the ground level E1; because E2 – E1 = -3.03 - (-5.14) = 2.11 eV. ' It is necessary that the level E 2 is close to E2 and E2 below because E(589.0) < E(589.6) E '2

0.50

2.a.i 2.a.ii

3.b.i

3.b.ii 3.b.iii

0.25

0.75

0.50 0.50

E1 4.a

- For 1.01 eV: the atom is not excited, because 1.01 -5.14 = -4.13 brings the sodium atom to an excited state which does not exist between the ground state and the first excited state. - For 3.03 eV: the atom is excited because 3.03 – 5.14 = - 2.11 eV, the atom passes to the level lower than -2.11 eV which is E2 by absorbing from the electron the sufficient quantity of energy to excite it which is 2.11 eV. - For 6.14 eV: the atom ionizes because 6.14 -5.14 = +1.00 which is a level higher than the state 0 eV. Thus the atom ionizes by absorbing from the electron the sufficient quantity of energy to ionize it.

1.25

4.b

- For 1.01 eV: KE remains 1.01 eV. - For 3.03 eV: the kinetic energy of the electron will be 3.03 – 2.11 = 0.92 eV. - For 6,14 eV: the kinetic energy of the electron will be 6.14 -5.14 = +1.00 eV

0.75

6