Translational Motion - Physics - Solved Past Exam, Exams of Physics

Download Translational Motion - Physics - Solved Past Exam and more Exams Physics in PDF only on Docsity! Vector mechanics PHY1021: Jan 2011 exam- Hints and tips Lecturer: Dr. Stavroula Foteinopoulou 1 (i) W < 0 if force causes a !KE < 0 contribution. Friction is not possible to do negative work in translational motion. W = !KE W = ! B A Fxdx = ! B A maxdx = ! B A mdvx dt dx = ! B A vxdvx = KB ! KA 1 (ii) minimum velocity at the bottom of the circle vA is: 0 ! 1 2 mv2A = mg(2L) " vA = 5.6 ms!1 Force at the top of the circle is: FRod = mg = 4.9 N First find velocity at this point vB from work energy theorem, when mass has completed a quarted of a circle. 1 2 mv2B ! 12mv 2 A = !mgL " vB = # 2gL Force of rod, when the mass has compelted a quarter of a circle is: FRod = mv2 B L " FRod = 9.8 N 2(i) 1 (a)v = (4i + 4j) ms!1 (b)a = dv dt = (2i + 4j) ms!2 (c) â = a|a| = 1" 20 (2i + 4j) (d) F = ma = (2i + 4j) N (e) v1 = |v(2s)| = 4 ! 2 ms!1 v2 = |v(4s)| = ! 320 ms!1 From work-energy theorem: W = 1 2 m(v22 " v21) = 144 J (f) v1 = v(2s) = (4i + 4j) ms!1 v2 = v(4s) = (8i + 16j) ms!1 Magnitude of impulse is: |J| = |mv2 " mv1| = 12.65 Kgms!1 Direction- find angle ! between J and x-axis. cos(!) = J·i|J| # ! = 71.6 0 2 (ii) The location of the support should be the center of gravity, which is the same as the center of mass (constant g), and has coordinate xCM : xCM = 1 M1+M2+M3 (M1 L 4 + mL 2 + M2 3L 4 ) # xCM = 1.714 m 2 Ki ! Kf = 12mv 2 ! 21 2 mv2f = 1 2 mv2 ! m( v! 2 )2 = 0 Which means initial and final kinetic energy are equal i.e. collision is ellastic. 4(ii) (a) ! = RF = 1N · m (b) W = !!" = 4# J (c) Work energy theorem for rotational motion: 1 2 I$2 ! 0 = W which gives $ = 7.1rads"1 (d) $ = $0 + %t with $0 = 0, and % = ! I which gives t = 1s 5 (i) (a) The work of a force must be indepedent of the path for the force to be conservative. (b) K + U = constant K: Kinetic energy U: Potential energy (c) No, only when only conservative forces are present. In the case that non-conservative forces are present mechanical energy is still conserved if the latter contribute zero work (e.g. friction in rolling without slipping motion, normal force) (d) 5 Work energy theorem: WA!B = KB ! KA Conservative force: WA!B = UA ! UB which indicates that work depends only on start and end point but not on the particular path taken. So: KB ! KA = UA ! UB which gives: UA + KA = UB + KB which expresses conservation of mechanical energy 5 (ii) Fr = !dUdr = = 12U0 R0 [(R0 r )13 ! (R0 r )7] Fr = 0 which occurs ar: r = R0 6 6(i) proper time: Time interval between two events measured in a reference frame where the events took place at the same location. proper length: The length of an object in a reference frame where its end points are not moving, or the distance between two points in a reference frame where they are stationary. 7 PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code : PHY1022 Name of module introductions to Astrof=y}sics Date of examination January 2010 1,@ B= nf = © p=m,y= =0.37eV “2m, 2m,A? E= ate, +6 (h) BY = Af” aa +6 Lily he| —-— |= BL -E, dz 1) Hk, so E, =2.15eV 20 © OF a3x10" Hy AtAE > hi 2+. MAf 21/2m-. At =107s 3. @ BB, =if == ae | and A= 650m 0! 49 5.2.08 AU, 15x 10’ m, 9.6.x 10° kg, 70000km, 696 km m°, gas giant H/He. 6.m—-M =Slogd—5.. M =-125 1 1) 46 AU =-GM a a17 x 107 = 01M * (aan isi) e P=54s Pulsar. PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code PHY1025 Name of module Mathematics Skills Date of examination 2011 Ys (i rs us + (7) Gi) 4/3 Gi) y=landy=e (iv) | p= 4, (Hint: use long division and then find a value for p which would make the reminder zero.) 6sin@x) | 9sin? (3x) 1 x ) - > = i+ cos?(3x} cos" (3x) xe ak? Vek? Gi) Using both methods, ® =o ax os ou 1 ou 1 &. > “(0,-1)== a dx y~-2z ax ( ) 2 ou x ou l Be -; “to0-1)=-- dy {y= 22) ay | ) 4 oH 2k Hagel az z 2 @ [aver ddr = = (x +4)" @xr—-8)+C Gi) xtetdy=e'(x? -2x+2}+€ b 2 3 Gi) Use @= f 1+(2) dx =(Vi-+sinh* xdx= 8.84 a ag 1 2 3 x Xx (a) in(it x)= anaes (b) toe ee? T+x 2 (c) fait gyata sake 9 14 (a) A+4B= 24 29 © warele Tle lls aS 3) worl Me as JS 3 () At partsalo vi _l 3 pet f7 3)? pa | 2x7-3x6(-6 2) 4l-6 2) | 3. 2 ii) 2. 3. 4 @ jw Gi) 5. (i) | Gi) (iii) PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code PHY2204 Name of module Statistical Physics Date of examination Jan 2014 aszeversible -- Q, Melting ice: AS = mb, /T = 1220 JK". Heating water: AS = ky ln(f,/T,) = 131UK'. 6Q=0> rar +3av + Pav > GS —RX = Infrer(y — B)*] = const => (RT) VV — BY = const a(P+ 4)" (VB) = const = (p+) (V — BY’ = const, where y = 4, See course notes. Isotropy implies p(—u,)= p(,) See course notes, Note, that the number of states with the speed from uto w+ du is 2au du. e=mu?/2= udu= sas => plu)du = 22 exp(- Qaelm)de = Bexp fejde. m m Boltzmann factor: exp(- 8 €). (e) . . high t ot ensity of states in 2D does not depend on energy. = = =(e-t)e. Density of states in 2D does not depend wy. Pe <kgT) =| ple)de fetdt=(e-IVe oO See course notes. pmol - 2 | / Sew(- 8 £4). Z= el is) ensures 2p =0. 'B’ s_ Lt _& _1dZ__1din(Z) T= Yew, -LEeem( fs a} Ek Sewl-A,) pH a7 SH). Z= aen[-2}ra00(-Zi:]+a° on(-f)- 14+3e745e7? ~ 2.78. 8: ‘Be: G=— LS ee, 00{- n= kgf (0+ Lx3e! +2x5e? YZ = 088K 7. ky? H=U+PV .dH=TAS+VAP = dH =TdS =8Q (heat flow) for P = const. See course notes: F =U —TS; extensive; dF =—PdV —-SdT > dF =—SW for T = const. AS, total 4 B c D & Tr 7 .T 0 fF, y0C[nesingeam geome emZt)- Cuff? (z,7,7-7,T,)|=0 oT) [(CLoTeTy Ts) => T, = CTsTeT ote)”. $= k, InQ; see course notes. Og4y =24'Q, PS yg hy NQ, +h MQ, =S,+Sz 2) Q,=18=0; Q,=N=10>S5=h, MN =k, ln10~3.18x10 IK; _ . NI 10! ~ 23 yet 0) Shy hn0, = ig of aA = k, 5H 3)- k, In(210) « 7.3810 JK, 4) Most probable: Njeaas = Miaes = 5 => S = ky INQ, = ky n( 2 i) keg in(252) = 7.63 x 107 JK". See course notes. E, £, Classical timit: oof ; — | Therefore, {(E, Jweon{— Ze} Boltzmann's factor. Be: B UNIVERSITY OF EXETER - SCHOOL OF PHYSICS SOLUTION TO EXAMINATION QUESTION: Name of setter CP Saws Paper No | Question No Name of module Puy 3 I30 32 p/. [a 7 ~ Year of examination Qo tI an “x p Ss initials of checker ) \ \ O (3) Conse worst. Pmswer dw bank bat ‘ Ex 4k . 2m i) Cringe Wit Homewrel (deur gibt in Lary) 2 (1) Course wort, We Ward out pe Stuns Hh » 6 Aw @), Ob hy (4); ti» so BY ah BG env nre_ a Les oh 4 4h [> 7 6&>. (¢) ai-= 4 pez 82 aon Le Le ( +le ) CM enviliws . £ at a ‘ a} 75 Wes ra Mei) + bw ~ Ue Le. Ole Come non. u .~2wi |ep * 4, A CO) SR = xy) \e2 L, bw =22 tu Cit) Come wort PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code PHY3102/PHY3305 Name of module Solid State Physics 1 / Solid State Physics 1 (1S) Date of examination January 2011 dv v , or 5 ma —eE —m,— whereris relaxation time, ¢ is electron charge r Steady state solution: y =———£ =v, m, 2 janev, _ ne Et op a T t=mine’p £=6.3x10"y at room temperature, and ¢=1.0x10""s at T=0 l=v,T 7 = 69 nm at room temperature, and /= 114 ym at T= 0 These mean free paths are much longer than the interatomic spacings (0.Inm), and we can conclude that the electrons are not scattered by the atoms themselves. Gi) (a) m7 =5.6K10 "ke . # afd (b) e, =-107"(10°)’ =10°°J (0.62 ev) (c) Pe = “AK =-1.1x 107 kgm si k, (@ Vn = dalidk = -2 x 10°7Wh =-1.9 x 10° ms", For free electrons (a) = e4; p(a) = ¥(a)P*(a) = eb tenthe = 1 For Bloch electrons ¥,(a) = e!4u,(a); p(a) = ef Ou, (ale "uj (a) = ug which has the periodicity of the laitice, since 4, (a) is lattice periodic.  The common base circuit has low input and high output impedances and is commonly used for impedance matching. Base transport factor, emitter efficiency – see lecture notes. Ic/Ib = 33. 3.(i) Hard disk drive operation – see lecture notes. Relaxation time obeys and Arrhenius relation ! " = "0 exp µB kBT( ) where the magnetic moment µ is the product of the magnetization and the particle volume, and !0 = 1/fFMR. Setting ! = 10 years, critical diameter = 9.6 nm. Reduced grain sizes require larger anisotropy for thermal stability, which then requires larger write fields. 3(ii) Operation of MESFET – see lecture notes. ID = 0 when Ld = 0.15 µm, so Nd = 2.8 " 10 22 m#3 Begin from ! Q = AeNdLd and differentiate. 4(i) Derivation of guiding equation – see lecture notes. ($1)min = 80.8°, 12 allowed modes (counting both possible polarization states) Intermodal dispersion (see lecture notes) yields a pulse duration of 675 fs. Shorter pulse duration may be achieved by reducing the core thickness so that the waveguide supports only a single mode. 4(ii) MOSFET and CMOS – see lecture notes In 3 years the rate of heat dissipation will increase by a factor of 4 under constant voltage scaling. Hints and Tips for examination of PHY3142 ” STARS FROM BIRTH TO DEATH” January 2011 (i) Energy liberated = 2 mp - Macuterium - Mz = 0.420 Mev Total energy = Energy liberated + 2m, = 1.441 Mev Gi) 4He + “He > 8Be + 4He >?C, Fraction of rest mass energy X= 6.5 10-4. Total energy liberated: 0.25 6.5 10-4 3.2 10° 3 108)? = 8.8 10% J. t= 8.8 10%/10% = 8.8 10" s ~ 2.8108 yr, (iii) See lecture VI @® At ~ 1.8M,. CNO transform essentially into “N. See lecture V (iii} see lecture II1.5.b @ R=11610'm B= 3M/(4aR?). p ='1.53 10" kg m> and p, = 9.18104 kg m3. P = Kp?" 59 P= 2.8 10% Pa. Te = 2.2x 108K 1 PHY3142 Gi) f= 5.4x 10°. T=1.1x 10K atr = 0.98R T=1.3 x 10° K atr = 0.8R Variation of T is very rapid see lecture VIII dp = ~S8dT + a8dP Adiabatic gradient implies that dq=0 => Vea = az Perfect gas, region stabilised because Cy < 0 Degenerate gas, region destabilised because Cy > 0 PHY3142 UNIVERSITY OF EXETER - SCHOOL OF PHYSICS SGLUTION TO EXAMINATION QUESTION: Name of setter CP Syvaetena PaperNo — { Question No Name of module Puy Sly | Year of examination 2ote—l : ~ “~ > Ss initials of checker - \ \ - 7RINS < L= 2 wy> = -y + Myx Sued =A ager , X= 29 ME S see. x = B17 mM os . —_ . W= ce 22; & = k GR = 2M DM 7 k +2*) oy 2) Come wnt _ X> = OR J Cae x, éF = os a Se @) Course unt las t part 6 Stunw G) Camee wnt Qabraqg , = Sata, (R-a4) u . xo \ = omar a ot “ey get cot OF) 4 B > XS OX car, (£2) thet EXSbI=4, Cy, A}=4, xy hl=s Vy, Ped = 4R AGG mde 2. Gi) PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code PHYM423 Name of module Classical and Quantum Fluids Date of examination January 2011 Horizontal pipe > gravitational force = 0, steady and uniform —+ inertial force = 0, p depends only on x, whilst v depends only on r > — 2 + ola =0 a r or\ or 2 2 (rm) 2% = rh (pra |B aa ar\ or} wax or Ox\" Bb ax\2u 2 Whenr=0,A=0 > po BE ar ox Qu anf p2 | WPT ye Bf hay + y- 2 yal ar dx Qu ax” 2u ax\4u Hl 2 Whenr=R, per > yah (pp?) Au ox 4u Upthrust = Sak p98 Dias force =C,,Al Lal w= 2 (nu Poe 1 aR? rag force = Cp Ay | = 7 | 7 OAR Poe mass * acceleration = upthrust — weight — drag after some integration and using initial conditions: w= 2 {i-e*") where p_o= ele and c= 380 Cc Py 8Rp, ig = = 8kleo~PLle Cc 3apo a C 3apo 9.8ms™ x 0.075 Ms sop-lyp ao sara tt -28ms X00Bm oa at p s 85 Jkg K™ V2 3 ii) For dilution refrigerator assume ideal heat exchangers (i.e. 7\,, = 7.) and uo heat leak Q=N; (aut. - af) ~ Nq(aq ~ a¢)Tine = 72,84 J mot) K™ x 50 pmol s! = 4.272, mW K™ O13 nW =4.272,.mWK™ => Those =56mK Oop =4.2(0.1K)” mW K-? -13 pW =29 pW For 3He refrigerator assume ideal vapour at pressure: P=Py exp(-5 K/T) Pp = exp(+5 K/3.2K) x10 Nm? =4.8x10° Nm? 48 x10° Nm™ x25 x 10 m? s“exp(-5 K/T} Q = 5K Q=5K 300K = 200exp(-5 K/T) W Oreu = 200exp(-5 K/350 mK) W =125 pW Spherical cavity, radius 7, hence Apyht~ h/2 =» Ap*~3(h/2r) => Eglr)~ Ap?/am, ~{30°n?/8m,r?) Total energy of ion is then Ein {t) = (3074 /8m,r?) +4m0r? + P4sr3/3 = =0 => (-6n°A?/8m,r°) +8zor + PAnr? =0 i = =0 = (-6n°9?/8m,y9)+8nar+ Pam =O 2. P= (at? /32m,r°) -20/r P bar) 25] \ 204 \ LS pe 107 T ot r trace 2 Xt.28687 ¥:34.875 P0285 nm)=15.0bar and P(.784 nm) = 0.00 bar 212