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1. Fundamental and derivate quantities and units  Fundamental Physical quantities are those that can be directly defined, and for which the units are chosen arbitrarily, independent of others physical quantities.

No. Fundamental Quantity Unit Symbol Dimension 1 Length Meter m L 2 Time Second s T 3 Mass Kilogram kg M 4 Current intensity Ampere A I 5 Light intensity Candela Cd J 6 Temperature Kelvin K  7 Solid angle Ste-radian Sr 8 Plane angle Radian rad  The derivative physical quantities are those that are defined indirectly. They have the measurement units’ functions of fundamental units. If we have:  U = U(L, T, M, I, J, )  [U] = LTMIJ 2 The homogeneity of physical equations:  The physical formulas are invariant to the measurement units’ transformation. 3 Examples:

3.1 Velocity:

x dx x m 1 1 v   v   m s  L T (1) lim t dt t0 t s 3.2 Acceleration:

v dv v m 2 2 a   a   m s  L T (2) lim t dt 2 t0 t s 3.3 Impulse: m p  m  v p mv kg  kg  m s1  M  L T1 (3) s

1 3.4 Force: p dp dmv dv F     m  m a lim t dt dt dt t0

p kg  m 2 2 kg  m F   kg  m s  M  L T F1 1N (4) t s2 s2 3.5 Pressure:

dF F kg 1 2 1 2 N P  P   kg  m s  M  L T P1 1Pa (5) dS S s2  m m2 3.6 Mechanical work: m2 dL  Fds L Fs kg  kg  m2 s2  M  L2 T2 L1N  m 1J (6) s2 3.7 Kinetic energy: m  v2 E  ; c 2 2 2 kg  m 2 2 2 2 2 2 E  mv   kg  m s  M  L T ; Ec 1kg  m s 1J (7) c s2 3.8 Gravitational potential Energy:

Ep  m g  h 2 m 2 2 2 2 2 2 E  mgh kg  kg  m s  M  L T ; Ep 1kg  m s 1J (8) p s2 3.9 Elastically potential Energy: k  x 2 E  p,e 2

2 2 kg  m 2 2 2 2 2 2 E  kx   kg  m s  M  L T ; Ep,e 1kg  m s 1J (9) p,e s2 3.10 Power: dE P  dt

2 E kg  m 2 3 2 3 J P   kg  m s  M  L T P1 1W (10) t s3 s 3.11 Density:

dm m kg 3 3      kg  m  M  L (11) dV V m3 3.12 Angular velocity:

2 d  rad 1 1      rad s  T (12) dt t s 3.13 Solid angle:

2 dA A m 0 d     L 1 1L0 1steradian (13) R 2 R2 m2

4 Establishes into a constant approximation the formula of a gravitational (mathematical) pendulum, T using the homogeneity of the dimensional equations. R: From direct observation the physical quantities that can occur are: the length (l), mass of the pendulum (m), gravitational acceleration (g). T = k l m g

 T1  k 0 L M LT 2   T1  L M T 2 , then by identifying the upper coefficients:  1       0  2   1/ 2 0 1/ 2 l   0    0 T  kl m g  T  k g  2  1  1      2 5 The mass of a parachute with jumper (parachutist) is m = 100 kg, and is launched from a tower completely open with no initial velocity. Find the velocity expression, v(t) and the velocity limit if we know that the resistance force is proportional with the velocity, R = kv, where k = 500 Ns/m.   R: If the force is defined as: F  ma dv Along vertical direction can be rewritten as: F  m dt The resultant (net) force is F  G  R , then:

vt  vt  dv dv t m  k  v00  k  k m  mg  kv   dt'   lng  v  t  ln1 vt   t dt  k  k m  mg  m v0 g  v 0   v0   m

k mg   t   m   vt  1 e  . k  

3 If we will replace the physical quantities with values we get:

500 10010   t  m  100  5t vt  1 e   vt  21 e    500    s  For the time goes to infinity one obtains:

k mg   t  mg 10010 v  v t  1 e m   v    v  2 m  7.2 km lim lim   lim   lim lim s h t t  k   k 500 Finally the velocity expression can be written as:

k   t  v t  v 1 e m    lim     6. A skier with the weight G descends a hill that make an angle  with the horizontal plane. The motion equations are x = Agt2 along the hill land y = 0, where g is the gravitational acceleration and A a constant coefficient. How much is the friction force between skier and snow on the hill. (Particular case: the skier mass m = 70 kg, angle  = 300; A = 0.1). R: The second principle of dynamics can be written as:     G  R  Ff  ma The projections along x and y axes are:

Gsin Ff  max

R  Gcos  ma y The component of accelerations can be calculated with the following equations: d2x d2 Agt2 a      2Ag x dt 2 dt 2 d2y d2 0 a    0 y dt 2 dt 2 With those:

Ff  Gsin 2Amg  Gsin 2A

Ff  mgsin 2A  70 9.81sin30 2  0.1  686.7  0.3

Ff  206.01 N 7. A material point with mass m is moving along a trajectory given by the Cartesian components: xt  A coskt and yt  Bcoskt. Characterize the force F that produces this type of motion if we know that the force depends only by the material point position. Give some examples.

4 R: The force components along Ox and Oy axes are: d d F  m  a  m  ˙x˙  F  m  x˙  mk  Asinkt  mk2Acoskt  mk2x x x x dt dt d d F  m  a  m  ˙y˙  F  m  y˙  mk  Bsinkt  mk2Bcoskt  mk2y y y y dt dt The force modulus is:

2 2 2 2 2 2 F  Fx  Fy  mk x  y  mk r  where r is the material point position vector. The ratio:

F  mk2x  x    x    cos x,F F mk2r r   are the direction cosine of the force F. 2  Fy  mk y  y      cos y,F F mk2r r   In the same time are the angles between the position vector and OX and OY axes, respectively. From here we can observe that the force F is anti-parallel with the position vector. Then the force F is pointing to the centre of the reference frame. Such a force is called attraction central force.

8. On a body with a mass m = 2 kg are acting two forces, F1 = 3 N and F2 = 4 N, which are

0 0 characterized by the angles 1 = 60 , and 2 = 120 respectively, with the direction of velocity  v0 . Find the body acceleration, a, velocity, v and the distance covered into a time t = 10 s

starting from the beginning of motion (Particular case v0 = 20 m/s). R: If we are considering a reference frame related to earth then the acceleration components along OX and OY axes are: F cos  F cos 3cos60 4cos120 3 0.5  4  0.5 1 a  1 1 2 2      0.25 m /s2 x m 2 2 4 3 3 3  4 F sin   F sin  7 3 a  1 1 2 2  2 2   3.03 m /s2 y m 2 4 and the acceleration modulus:

2 2 2 2 2 a  a x  a y  0.25  3.03  0.0625  9.1809  9.2434  3.04 m/s

5 We observe that the acceleration is constant in time (a = const), the velocity equation can be calculated from:

vt  dv t a   dv  a  dt  dv  a  dt'  vt v  at  vt  v  at dt   0 0 v0 0 or by components:

vx t  v0x  a x t cu v0x  v0

vy t  v0y  a yt cu v0  0 and from here:

vx t  v0  a x t  20  0.2510  17.5 m /s

vy t  a yt  3.0310  30.3 m /s and the velocity modulus:

2 2 2 2 v  vx  vy  17.5  30.3  306.25  918.09  1224.34  34.99 m /s   The velocity, v at time t makes an angle  cu v0 given by the equation:

vy  vy   30.3 o tg     arctg     arctg   arctg1.731  59 59'28". vx  vx  17.5  In order to calculate the distance we have: ds st  t t v   ds  v dt  dv  v dt'  st s  v  atdt' dt   0  0 s0 0 0 2 2 at s0 0 at 3.04100  st  s  v t   st  v t   2010   200 152  352 m 0 0 2 0 2 2 or by components: a t 2 0.25100 xt  v t  x  20 10   200 12.5  187.5 m 0 2 2 a t2 3.03100 yt  y  151.5 m 2 2

9. A tractor is traveling with a velocity of v0 = 36 km/h. If the radius of the wheel is R = 0.5 m find out: a) The parametric motion equations of a point from the external wheel circumference. b) The tangential velocity components and the value of velocity. c) The path distance by a point between two contacts with the road.

6 v rad R: a) v  R    0  20 R s

x  v0t  x W x w  xCW  x R xCW  0 x R  R sint     y  0 t  yw yw  yCW  yR yCW  R  yR yR  R cost

x R  v0t  R sint 10t  0.5sin20t  yR  R  R cost  0.5  0.5cos20t this is the equations of a cycloid. dx b) v   v  R cost  v 1 cost101 cos20t x dt 0 0 dy v   R sint  v sint 10sin20t y dt 0 t v  v2  v2  v2 1 cost2  v2 sin 2 t  v 21 cost  2v sin x y 0 0 0 0 2 v  20sin10t ds c) v  dt t t ddt  ds  vdt  ds  2R sin dt  ds  2R sin d 2 2

2 s 2      ds'  2R sin d  s  4R cos   8R  4 m 0 0 2  2  0 10. A body with a mass m = 5 kg can slide with friction on a horizontal surface if is pushed by a spiral spring, with an elastic constant k = 200 N/m, and which was compressed at half of his length l = 20 cm. Find out: a) the mechanical work of friction force during expansion. b) how much must is the friction coefficient if in the final position the spring in not stressed.

R: a) Fe  kx

0 W 0 0 x 2 l dW  Fdx   dW'   Fdx  W    kx  dx  k  k 0 l 2 l 2 2 l 2 8

202 104 W  200 1J 8

7 l2 kl b) E  L E  E  L  0  k  Nl     Nl t f f i f 2 2mg 2000.1    0.204 259.81 11. A concrete cube with the side of a = 0.8 m and a density of 2500kg/m3 must be tipped around an edge. Calculate: a) Point of application direction and value of the minimal force needed to turn turn-over; b) The expression of a horizontal force that can turn-over the concrete block if this is applied on superior edge as function of rotation angle. c) The mechanical work spent for turning over the concrete block.       R: a) M  r  F  r  F  cosrF  F  

    cosrF 1     r  a 2     MFmax  r  Fmax cosrF  a 2Fmax   

a a Vg ga3 M  M  a 2F  G  F  mg   Fmax G max 2 max 2a 2 2 2 2 2

25009.810.83 F   4439.5 N max 2 2 b)

MF  aF  a  F cos 2    2    MF  MG  aFcos  Ga cos   MG  G  a cos   2  4  2  4 

3 3 2         a g F  a g cos cos sin sin  F  1 tg 2cos   4   4   2 0.83  25009.81 F  1 tg 6278.41 tg N 2

c) W  MF 

W  4  4 a3g dW  M  d  dW '  M  d  W  1 tg   d F   F     0 0 0 2

8  4  4 2  2  2  2  4   4   4   W   a g cos   d  a g sin    a g 1  0 2  4  2  4  0 2  2  a 4g 0.84  25009.81 W   2 1 W    2 1 2080.48 J 2 2 12. On an inclined plane with inclination  = arcsin(3/5) and length l = 2.1 m is rolling with no sliding friction an homogeneous sphere of mass m = 40 g and radius R = 10 mm. How much is the transversal velocity, angular velocity and frequency at the base of the plane.

R: Et,A  Et,B

Et,A  Ep,A  mgh  mglsin mv2 I2 E t,B  Ekin,tr,B  Ekin,rot,B E  E  kin,tr,B 2 kin,rot,B 2 2 v I  mR 2 v  R    5 R mv2 I2 mv2 2 v2 mv2 2 mv2 7 mv2 E     mR 2    t,B 2 2 2 5 2R 2 2 5 2 5 2 7 mv2 10 10 3 m mglsin   v  glsin  v  9.81 2.1  4.20 5 2 7 7 5 s v 4.2 rad     420.2 R 0.01 s  420.2     66.88 Hz 2 2 13. Demonstrate that after a perfect elastic collision of two hokey pucks of the same mass initial one being at rest the angle between the directions of pucks is 900. How much are the velocities

0 and scatter angle 2 if the initial velocity of projectile puck is v = 40 km/h and 1 = 60 ? R:

2 2 2 2 2 2 2 2 2 Ec,i  Ec,f mv  mv1  mv 2 m v  m v1  m v 2           pi  pf p  p1  p2 p  p1  p2

2 2 2 p  p1  p2   2 2 2   2 2     p1  p2   p1  p2  2p1p2  p1  p2  2p1p2 cos1  2   0 p  p1  p2

0   cos1  2   0  1  2  90  p1  p2

0 0 0 0 2  90  1  90  60  30

9 mv v 1 m cos   1  1  v  vcos   40  20 1 mv v 1 1 2 s

mv v 3 m cos   2  2  v  vcos   40  34.64 2 mv v 2 2 2 s 14. On the deflection plates of a cathodic oscilloscope with sensibility of 2 cm/V on both axes Ox and Oy it is applied simultaneously the electric voltage

2 U x  2cos100t and U y  4sin 50t. Calculate: i. The spot equations of motion on Ox and Oy axes. ii. Equation of the trajectory. x cm R: a) Sx   x  Sx Ux  2 2Vcos100t  x  4cos100t cm Ux V

y cm 2 2 Sy   y  SyU y  2 4Vsin 50t  y  8sin 50t cm Uy V

x  cos100t x x  4cos100t 4   cos100t b)  2    4 x  8sin 50t 1 cos100t y  8 y  41 cos100t  2  x  y  4 1  4  x  x  y  4  4    15. A material point execute a motion described by the equation: xt  2sin 2 3t   . Show that  2  this is a harmonic oscillatory motion. Find the amplitude and period of motion, the velocity and acceleration of the material point.

 2    xt  2sin 3t     2  1 cos6t      R:   xt  2  1 cos6t  1 sin6t   1 cos2 2  2  sin 2    2

xt  x'Asint  0  x' 1 A  1   2 1        6     6  T   0.33 s xt 1 sin6t     T 3  2   0   2

dx d    v   x'Asint  0  Acost  0   6cos6t   dt dt  2 

10 dv    a   362 sin6t   dt  2  16. A material point with the mass m = 10 g is oscillating under the law:

    xt 10cos2  t    5 . Calculate: 12 4 

a. The time t1 necessary to reach the maximum velocity, and the time t2 necessary to reach the maximum acceleration; b. The maximum value of the elastic force that act on the material point; c. Expression of kinetic, potential and total energy. R:

 2         xt 10cos  t    5 1 cos t    12 2   6 2     a)   xt 10  5  5sin t 1 cos2 2  6  cos2    2   vmax  5 dx      6  v   5 cos t     t1  n  t1  6n s dt 6  6      6 cos t1   1   6 

 2 2 a max  5 dv      36   a   5 sin t    t 2  2n 1  t 2  32n 1 s dt 36  6      6 2 sin t 2   1   6 

b) Fe  kx  kAsint  0   Fmax sint  0 

2 F  kA  m2A  10103  5102  1.37104 N max 36

2 2 2 2 4 2 mv mA  cos t 102510  36 2    6 2    c) Ec    cos  t   3.4310 cos  t  J 2 2 2  6   6 

2 2 2 2 2 4 kx mA  sin t 10 36 2510 2    6 2    E p    sin  t   3.4310 sin  t J 2 2 2  6   6  mv2 kx 2 E    3.43106 J t 2 2

11 17. A harmonic oscillator which oscillates with the amplitude, A of 8 mm is after   0.01 s from the beginning of motion (from equilibrium position) at the distance 4 mm measured from equilibrium position. Calculate: a. oscillatory pulsation; b. oscillatory period; c. oscillatory frequency; d. velocity of oscillation in the given position; e. acceleration of oscillation in the given position.

R: a) yt  Asint  0 

y0  0  Asin0  0   sin0   0  0  0 A 1    50 y  4   Asin  sin             2 2 6 6 60.01 3 2 2 2    T    T  0.12 s b) 50 T   3 1 1 c)       8.33 Hz T 0.12 dy 50  50     v   Acost  v  8103 cos 0.01  0.419cos   d) dt 3  3   6  v  0.36 m s dv 2500  50  m e) a   2Asint  a  2 8103 sin 0.01   a 10.97 dt 9  3  s2 18. The sonic boom created by an airplane it is heard in the moment in which the direction observer–airplane makes an angle  = 600 with the vertical. How much

is the airplane speed? (vs = 340 m/s) R:   90  

A0 v  t v v 340 m sin    s  v  s  s   2340  680  2 Mach a 1 AA' va  t sin  cos s 2 19.How much is the fundamental frequency of vibration of a copper wire ( = 8.9 kg/dm3) with section of 2 mm2 and the length of 1 m when is stress with a force of 17.8 daN.

12 v v R:   vT        

 T v    T   v  ;   2l m V Sl S      S  l l l

1 T 1 17.810 100      50 Hz 2l S 21 8900 2106 2

20.Calculate the wavelength value of a sound of 440 Hz in air (va = 340 m/s) and into a tram line

3 (vt = 5000 m/s). Calculate the steel Young modulus if the density is (t = 8000 kg/m ). v 340 R:   a   0.773 m  77.3 cm a  440 v 5000   t  11.36 m t  440

E 2 N v   E  v2  8103 5103   21011 t  t m2

21.In the liquid Helium (below 4.2 K) the sound velocity is 220 m/s. If we know the He density  = 0.15 g/cm3 find the compressibility modulus. Compare this result with the water compressibility modulus if we know that the sound velocity is 1460 m/s.

 N R: v  He    v2 150 2202  7.26106 He  He He m2

N   v2 100014602  2.1316109 W W m2

9 W 2.131610  6  293.6 He 7.2610 22.By a copper wire with the length l = 160 cm and diameter  = 1 mm is suspended a body with mass 10 kg. If the copper wire elongation is 2.56 mm find out: F l a. The stress,   , strain,   and elastic constant, k; S l b. The Young modulus, E;

o c. How much is the inter-atomic distance if in the unstressed material is ; R 0  2.56 A

13 F mg 4mg 4109.81 N R: a)       1.25108 S r 2 d 2 106 m2 l 2.56103    1.6103  0.16% l 1.6 mg 109.81 N F  F  kl  mg  k    3.83104 e l 2.56103 m F l  1.25108 N b)  E    E  E    7.81010 S l  1.6103 m2

R l l 3 10 13 c)   R  R 0  R 0  1.610 2.5610  4.09610 m . R 0 l l 23. In front of a microphone a sound has the pressure of 0.441 Pa (N/m2). How much we have to

2 2 amplify the sound in order to reach the pain threshold, Ipain = 10 W/m ? We know the air

2 acoustic impedance Za = c = 441 kg/m s. How much is the pressure and the sound level?

R: Nm  NA  NL  NA  NL  Nm

 I   I  N 10lg m  N 10lg L  m   L  I   I0   0 

 I   I   I   L   m   L  NA  NL  Nm  NA 10lg  10lg  10lg   I0   I0   Im 

IL  Ipain  2 2 2  p p 0.441 4 W Im     4.4110 2  c Za 441 m

 I   102   1  N 10lg L  10lg   60lg A    4     Im   4.4110   4.41

NA  53.5 dB 24. Find the relation between the intensity of wave into an absorbed media as a function of depth of penetration and absorption coefficient. R:

dI Ix  dI x  Ix dI  Idx   dx    dx  ln   x  Ix  I ex I  I   I  0 I0 0  0  25. Hydrogen atom is constituted from a proton and an electron with equal electrical charges but opposite sign, and absolute value e = 1.610-19 C. In the fundamental state, the radius of electron

14 -10 first orbit around proton is r0 = 0.5310 m. Calculate the attraction force between proton and electron. R: 2 1 e2 1.61019 1.62 F    9109    9 1093822  0.82107 C 2 11 2 2 40 r0 5.310  5.3 8 FC  8.210 N 26. What is the value of equal electrical charges, whit which two identical balls, having the same mass 1 kg, must be charged, if these are situated in air at 1 mm distance, and the Coulombian force that acts on each ball is equal in value with the ball weight into a place in which the gravitational acceleration is g = 9.8 m/s2. R:

27. Two small identical spheres placed in void, having the same mass m = 0.1 kg are suspended from a single point with the help of two isolated wires, not extensible, of neglected mass and with the same length l = 20 cm. What is the value of the equal charges with which the two spheres must be charged is such way that the angle between the fires should be 2 = 900? The gravitational accelerations is considered g = 9.8 m/s2. (sin 45 = 0.707; tg (45) = 1). R:

28. Calculate the dependence of electrical field intensity and electrical potential on the center of a sphere having the electrical charge uniform distributed in the entire volume. The total charge is denoted by Q and the sphere radius by R. R:

15 29. Calculate the electric field intensity at a distance a from a rod, uniformly charged with a total charge Q. R:

30. Let’s consider a coaxial cable constituted from two conductors having the radius a and b, with a < b. Use the Gauss theorem to determine the dependence of electrical field intensity and electric potential from the coaxial cable axes. Calculate the coaxial cable capacity. The cable length is labeled by l is charged with an electrical charge q.

R:

31. Use the Gauss theorem to calculate the expression of a plane capacitor capacity. R:

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