Centripetal Force - Physics - Solved Past Exam, Exams of Physics

Download Centripetal Force - Physics - Solved Past Exam and more Exams Physics in PDF only on Docsity! Vector mechanics PHY1021: Jan 2012 exam- Hints and tips Lecturer: Dr. Stavroula Foteinopoulou 1 (i) C = A ! B = ! ! ! ! ! ! i j k 3 "4 2 "3 2 3 ! ! ! ! ! ! = "16i " 15j" 6k (1) A · C = 0 (A # C) A · B = "9 " 8 + 6 = "11 |A| = $ 9 + 16 + 4 = $ 29 |B| = $ 9 + 4 + 9 = $ 22 cos! = A · B |A||B| = "0.43 (2) Thus: ! = 115.80 (3) ii) The centripetal force is not a new force. It equals to the net force in the radial direction pointing inwards, keeping the particle in circular motion. !Frad = " mv2 R r̂ (4) !Frad: Net force along the radial direction r̂: radial unit vector R: Radius of circle m: mass v:linear velocity A is not physically possible (net force in the radial directions points out- ward) B is the physical choice. Angular velcoity vector is along the "z direction 1 2. (i) µs = F mg = 0.153 (5) µk = 0.80µs = 0.122 (6) a = F " µkmg m = 0.3ms!2 (7) (ii) !F = dp dt (8) J = " !Fdt = "p (9) (iii) 1 2 Mv2 = Mgh (10) Thus v = 9.9ms!1 pi = mvj pf = 0 So "p = "544.5kgms!1j "p = " 4s 0 F(t)dt = " 2s 0 (""tj)dt + " 4s 2s ["(t " 4s)j]dt = ""4s2j (11) So: " = 136.1 Ns!1 So Fmax =-272.2 N j 3 (i) ax = d2x dt2 = 6#t Fx = max = 6m# 2 3 x 1 3 2 (ii) O represents the event (at x = 0, t = 0). The black lines represent x = ct and x = "ct "s2 = c2"t2 " "x2 (iii) E = pc E2 = p2c2 + m20c 4 (22) So m0 = 0 iv) m = (m0 & ( = 1.5 ( = 1q 1! v 2 c2 So v = 0.745c So p = mu =0.571 MeV/c2 So: P = (E/c,p) = (E/c, px, 0, 0) = (0.766MeV/c, 0.571MeV/c, 0, 0) 5 6. (i) W in a closed loop is zero (indepedent of path) W = " (2) (1) Fxdx = U(1) " U(2) = " " (2) (1) dU (23) Thus by comparison: Fx = " dU dx (24) (ii) x= 5cm Stable because d 2U dx2 > 0 0 ' x ' 10 cm U + K = E so: K =8.4 J So: v = $ 2K m = 12.96ms!1 (25) U increases with x at x=7 cm. So Fx < 0, i.e. force points in the negative x direction. 6 PHYSICS HINTS AND TIPS Module Code PHY1022 Name of module Introduction to Astrophysics Date of examination January 2012 1. The first parts of this question (i) are bookwork. For the final part: E = hc ! = 6.62 "10#34 " 3"108 6.3"10#7 = 6.62 " 3 6.3 "10#19 J = 1.97eV For part (ii) Use !p!x ! " and take 2r ! !x (but would accept r ! !x also) so E ! !p 2 2me ! "2 8mer 2 = 2.4 "10 #20 J = 0.15eV . 2. Part (i) bookwork. For the last question of part (ii) p 2m = !2k2 2m = eV so k = 1 ! (2meV )1/2 = 5.1!1010m"1 3. (i) The first part is bookwork. One wave is forward going, one wave is backward going. ! =! 0 cos(kx "#t) +! 0 cos(kx +#t) ! =! 0[coskx cos#t + sin kx sin#t + coskx cos#t " sin kx sin#t] ! = 2! 0 coskwcos#t. This is a standing wave. (ii) The first part is bookwork. The total mass available for nuclear burning is 0.01! 0.1! 2 !1030 = 2 !1027 kg. The energy is given by E = mc2 = 1.8 !1044 J and the timescale is 1.8 !1044 4 !1026 = 4.5 !1017 s = 1.5 !1010 years. This is comparable to the age of the Universe (1/H). 4. (i) This is bookwork. Part (ii) for the supernova the distance modulus is m ! M = +42.8 and then use m ! M = 5 logd ! 5 to give d = 3631Mpc . 5. The first part on Kepler’s laws is bookwork. For the exoplanet question we have P2 = a3 so a = 0.049AU The star’s orbital radius is 100 ! 4 ! 24 ! 3600 / 2" = 5.5 !106m The planet’s mass is 1.5 !1027 kg The planet’s radius is 7 !107m The planet’s density is 1044kgm-3 These are approximately the parameters of a hot Jupiter. 6. The first parts of this question are bookwork. The galaxy mass is 7 !1010 solar masses, the luminous mass is much smaller and the discrepancy is due to dark matter, which contributes to the gravitational mass estimate but not the luminosity-based estimate. PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY _Module Code PHY2021 | Name of module Electromagnetism! _ January 2012 1, The definitions and derivations are bookwork. Use cylindrical form of Laplace’s equation and boundary conditions ® =0 at r=a and © = ®, at r=b to derive ® = B, In(r/a)/In(b/a) -®, fF In(b/a) r The potential is 2.925 V and the electric field is 2404 V/m directed radially inward. Use electric field is negative the gradient of the potential to obtain E= 2. The definitions and sketchs are bookwork. The derivations are bookwork. The potential is 1.286x10* V. The radial component of the electric field is E, = 1.248 x 10° V/m and the theta component is E, =2.49x10° Vim. 3, (a) E =0 z7_ 2 > (0) E= 4me gr? r (©) E=0 £oX 4nea? Inner surface: o, =~ £ Outer surface: 0, = qe ITE! Capacitance increased by factor ¢/e, PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY i PRAY2021 Electromagnetism | _Module Code _Name of module January 2012 T | ; ic Date ofexamination | 4, Magnetic flux: WY, = fs dS yp, = Hotline a) 2n Vv -t 1 =—2exp| —E ait e= HaVol (2 )exo at 2aR?C \a)” (RC emf = 0.89 V current = 0.178 A 5. Bookwork. Magnetic scalar potential only applies in the absence of currents, but most interesting cases involve currents. Can be used for permanent magnets. Laplace’s equation, from divB=0 and using B as a gradient of the scalar potential. 6. Bookwork, Compare equation for B to equation for Vx A, By symmetry, argue which partial derivatives J OA, . + + are zero. This leaves 4% = er which canbe integrated to give the required equation. ar vr z component of magnetic vector potential is 1.386 x10 Tm (Tesla-metres). PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code PHY2022 Name of module Quantum Mechanics 1 Date of examination January 2012 1.! ! ! 2 " probability density for a position measurement ! ! To normalize let !norm = C! , then C 2 = 1 ! 2 dV all space " . This makes ! 2 equal to, not just proportional to, the probability density for a position measurement. ! ! C = a (a)! x"<"0; Probability = 0.5 (by symmetry) (b)! ! 1 2a < x < 1 2a ; Probability = 0.632 (c)! x > 1 2a . Probability = 0.184 2.! (i)! General solution is u = Acoskx + Bsin kx . Boundary condition that u is continuous forces A"="0 and k = n! L . En = n 2! 2!2 2mL2( ) ! ! u1 has higher probability near centre because it has an antinode there, whereas u2 has a node. ! (ii)! 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(P= AR) 4 AE After soheti hy of ate @ Compe wrod. Last part ; £4, Pot l4R) - PE WR DALIT een in SL Gwe BE ae o£ [REP ere] tae ’ 1 OE 2 RPS «wa pep (SE 2B 1 PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code PHYM401 Name of module Solid State II Date of examination January 2012 1. (i) BCC lattice ! " 2 /a3 atoms per unit volume hence: ! n = 2 4.25 "10#10 m( )3 = 2.6 "1028 m3 Drude conductivity is: ! " = ne2# me = 2.6 $1028 m%3 $ 1.60 $10%19 C( )2 $ 3.2 $10%14 s 9.11$10%31 kg = 2.3 $107S m%1 Mean free path is: ! " = vF# =10 6 m s$1 % 3.2 %10$14 s = 3.2 %10$8 m (ii) Conductance vs width sketch: 2 PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code PHYM401 Name of module Solid State II Date of examination January 2012 2. Sketches of the energy dispersion for different configurations of confining potential and external magnetic field for the transverse modes: Equate Lorentz force to centripetal force to find cyclotron frequency, ! "C = eB m* . To observe QHE, we require the mean free path ! vF" to equal or exceed the cyclotron orbit, say ! 2"RC = 2"vF #C . Hence ! " = 2#m* eB and the cyclotron radius must be smaller than the sample width. Expected quantisation step is ! h e2 = 25812.806" . 1 PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code PHYM423 Name of module Classical and Quantum Fluids Date of examination January 2012 1. In practice, the speed in the vena contracta is reduced slightly by surface tension effects, etc., so we can write , and the best speed to quote is that at the orifice giving or , where Cd = CcCv (a) When l is small, A2 << 2gh and A << and (b) When l is large, A2 >> , using the Taylor expansion of 2. (i) N/A (ii) Molar entropy of liquid: ! S T( ) = dQ " T 0 T# = C " T ( )d " T " T 0 T# = m*$ 2RT 2mTF Molar entropy of solid: ! S = kB ln "( ) = kB ln 2NA( ) = R ln 2( ) At minimum, ! R ln 2( ) = 2.8" 2RT 2TF # Tmin ~ 2TF ln 2( ) 2.8" 2 = 250 mK 3. (i) Assume that ideal ( ) and steady ( ) Euler's equation (ii) ! CV "T 3 and hence S"T3 c2 = const # c2 2 = const = $sS 2T $nCV # $s T 3( ) 2 T $nT 3 = $sT 4 $n = const $s %1 so $n ~ T 4 4. ! Eatom = ER+ " E0 = 6.69 #10 "23 J $ patom = 9.47 #10 "25 N s & vatom =141m s "1 ! p|| = p0 + 2µr E r " #( )( ) $ sin 15°( ) = 6.03 $10"25 N s Hence the atom is eject at an angle ! " = arcsin 6.03 #10$25 N s 9.47 #10$25 N s( ) = 40° SOLUTION TO DEGREE EXAMINATION QUESTION for Coordinator      Name of setter David Sing Paper Question Name of module Relativity and Cosmology Y e a r o f examination 2012 PHYM432 1 Initials of checker 1. An inertial frame of reference is a frame of reference in which Newtons first law of motion holds true. X ￿µ = ΛµνX ν [2] use Λµν = ∂X ￿µ ∂Xν Λµν =   γ −γv/c 0 0 −γv/c γ 0 0 0 0 1 0 0 0 0 1   [4] UαUα = c2 PαPα = m2c2 p0 = mc(1 + 12v 2/c2 + ...) p^0 is the Relativistic Kinetic Energy [3,3,3] There are 2^3=8 combinations. Non-diagonal components of metric are zero, as well as derivatives with respect to phi, giving three non-zero combinations, Γ011 = − 12a2 ￿ ∂a2sin2θ ∂θ ￿ = −sinθcosθ Γ101 = Γ110 = − 12a2sin2θ ￿ ∂2a2sin2θ ∂θ ￿ = cotθ R0101 = sin 2 θ [10,5] SOLUTION TO DEGREE EXAMINATION QUESTION for Coordinator      Name of setter David Sing Paper Question Name of module Relativity and Cosmology Y e a r o f examination 2012 PHYM432 2 Initials of checker 2. (i) Describe what is meant by the Cosmological Principle. Discussion with keywords homogeneous and isotropic. [4] (ii) Write down and explain the purpose of the Friedmann cosmological equations. The Friedmann equations describe the evolution of the cosmic scale factor with time. 1 R d2R dt2 = − 4πG 3 ￿ ρm,0 ￿ R0 R(t) ￿3 + 2ρr,0 ￿ R0 R(t) ￿4 + 2ρΛ ￿ ￿ 1 R dR dt ￿2 = 8πG3 ￿ ρm,0 ￿ R0 R(t) ￿3 + ρr,0 ￿ R0 R(t) ￿4 + ρΛ ￿ − kc 2 R2 [6] Disussion including key words open, flat or closed for k. k < 0 and space is open, k = 0 and space is flat, k > 0 and space is closed. [3] Flat space so k=0 Friedman equation gives R = R0 ￿ 3 2 ￿ 8 3πGρm t ￿2/3 [7] Closed space so k=+1 Friedman equation gives R = c H cosh(Ht) This model does not have a big bang singularity, as R(t) never vanishes over the whole range of t. [7,3]