101f04 - Physics 171.101 Final Exam Solutions May 10, 2004...

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Unformatted text preview: Physics 171.101 Final Exam Solutions May 10, 2004 Prof. Barnett Take 7r 2 3, g = 10 m/secg, sin(3'70) = 0.6, cos(37°) : 0.8., R = 8 Joule/(mole—OK) G = 6.7 X10"11 NVmZ/kg2 1. l_\'.> Use a pen. You may use material written on four 3” X 5” index cards, . i 1 , J . .; i r1:hree masses, MI :_'1 kg, “AI? :A 2 kg, A43? 3 kg: are located at :‘ R1 = (42' + 5]") m, R2 = (51' — 2j) m, and R3 : (11+ 1]) m. _ GMM F13—*—fi—‘1Ri 7"13 Hi3 = U153 — fillZ : 25 m2 613 = RIB-If! : —0.6i — 0.85 1313 = (4.8i + 6.43) x 10-12 N B) (10 pts) What is the gravitational force, F23, of M2 on M3? 1323 = Rfi an R3, =1R'3d— is}? = 25 m2 $23 = = ~—0.8i’ + 0.65" 1323 = (12.8% — 9.63) x 10-12 N C) (10 pts) What is the acceleration, d3, of Mg? 1310),) = (17.6% — 3.23) x 10-12 N a = flow/M3 = (17.6% — 3.23) x 10‘12 5% D) (10 pts) Find a unit vector, (1, parallel to (is. a = d/ld] : 0.9842‘ — 0.1793' A basketball court has a coordinate system with the origin (0, 0, 0) on the floor at a corner, Michael Jordan runs down the court with a velocity relative to the court of VMJ 2 4i m/sec. He shoots the ball from the position R30” 2 (52 + 35' + 219) m in the coordinate system of the court at time t = O with velocity 173a” : (3)" + 10k) m/sec relative to himself. (Michael’s reference frame axes are parallel to the court’s axes. z" and j are horizontal and k is vertical.) The ball goes (swish) down through the basket located 3 meters above the court’s floor. A) (15 pts) What is the ball’s velocity when it is shot by Michael, i.e. at t = 0, as measured in the basketball court coordinate system? unrt : VMJ + VBaH = (41' + 3j+10k) m/sec. B) (15 pts) When does the ball go down through the basket? Z(t) = 2(0) + VZ(0) + §azt2 3m 2 2m +10?t— 57:2 ‘ t = 1.894 sec C) (15 pts) What is the location of the basket, égasket, in the court’s coordinate system? X(1.894) = X(O)+1.894VX = 12.576 m Y(1.894) = Y(O) + 1.89414» = 8.682 m REM, = (12.576i + 8.6823 + sic) meters 3. A cat, Moat = 5 kg, swings in a circular path on a massless rOpe tied to a pole. The angle between the pole and rope is 6 : 37'". A spring, spring constant k = 100 N/cm, with a normal, unstretched length of 20 cm is tied to the rOpe. The total length of the rope and (stretched) spring is 1 meter. A1011? 5 jprina; I ' 1 meter. A) (15 pts) Draw a picture showing all the forces acting on the cat. Using a coordinate system With k vertical and i and j horizontal, write out each force in terms “ “ K of its (i,j, k) components and 6. A 3v .2 _ r r Fgravity “ —m9k L 2: i J .a T = T(— sin(0)i + cos(0)lAc) F3 _ Mimi B) (10 pts) Find the magnitude of the tension in the rope. Tcos(l9) 2' mg —% T : mg/ cos(6) 2 62.5 N W ,,,__ C) (10 pts) Find the stretched length of the spring. IT! 2 lkrl ——> a: = T/k = 0.625 cm D) (10 pts) Given the figure showing that the cat is going “into the page find the angular momentum Fmdial : = 37.5 N I mamdial [andsz = 7-5 “1/5662 = Vz/Rmdiaz R/L : sin(9) : 0.6 —-> R = 0.6 meters V : 2.12 m/sec = RAJV = 6.364 kg mQ/sec ——> E = 6.364 is kg m2/sec 7’ (l away from you” , 4. A violin string has a fundamental frequency of 200 Hertz. Its length is 30 cm and its mass is 0.50 grams. A) (20 pts) VWhat is the string’s tension? fA = Vlahase = \/T/# u = 0.5 grams/30 cm 2 16.67 ><10‘4 leg/m f = 200 560‘1 )x = 60 cm —> Vphase = 120 m/sec T = H V2 2 24 N B) (20 pts) Another violin is mistuned by having a tension 1% higher. What beat frequency would one hear if both of these violins are played simultaneously? T = 24.24 N —> an = 200.979 sec—1 Fbem‘, : fnem _ fold : Hertz 5. A train’s boxcar, M = 3 X 104 kg, has an initial speed of 1.50 m/s. 1t rolls down a hill with height 2.00 m. At the bottom of the hill it has a speed of 0.50 m/s. Ignore the spinning motion of the Wheels. A) (15 pts) How much work does gravity do on the boxcar? {fl/gravity : Z 6 X 105 J B) (10 pts) How much work did the Normal Force do on the boxcar? W'Al'ormal : C) (10 pts) How much work did friction do on the boxcar".7 VVFriction = M-E-fmat — A’f-E'initial AJ.E.initml Z émvEmu-al ‘i— Z X 105 J A'I.E.final : %m"t)}2:mal : X 105 J Wfriction = —6.3 X 105 J D) (10 pts) What impulse did the sum of all of these forces apply to the boxcar? 1; = P final — Piniticil = m(Vfinal — Vinitial) A I : m(0.5 —1.5) 1' kg m/sec : ~3 X104ikg 771/560 6. Suppose water flows over a waterfall at 107 liters/minute. The height of the fall is 50 meters. The ambient air temperature, and initial water temperature, is 700 Fahrenheit. The specific heat of water is 4200 J/(kg - Its density is p : lg/cms. A) (15 pts) What is the temperature of the water immediately after it splashes to the bottom of the fall and stabilizes. Consider 1 kg of water. Q : mgh : 500 J : mC'AT ——> AT 2 01190 0K Ti-nmal Z 0K -—'> Tfmul Z 0K I 0F B) (15 pts) After a short while the water returns to the ambient temperature of the air. What is the rate of entropy change for the universe during this entire process? AS : Q/T : 500 J/298.15 0K = 1.6770 J/OK - kg for 1 kg of water. The rate for the water going over the fall is 107 leg/min AS/At = 1.667 X 107 J/OK min. wherit=0.250 s. _ . h .I j -_ , 7. The adjacent figure shows the waveform of a traveling transverse sinusoidal wave at two times, when t = 0 s and 0.250 seconds later. The wave is traveling toward larger values of X .' A) (15 pts) Find Y(X,t). l/ghase : 1.5 m/0.25 sec : 6 m/sec A = 2.0 m T = 0.333 sec Y(X, t) = 0.5 cos(27r(X/)\ — t/T) m = 0.5 cos(27r(X/2 — 375) m The units of K and t are meters and seconds. B) (15 pts) Draw a graph of the velocity of the particles in the wave at t = 0 as a function of X. " ' v = % = 37r sin(27r(X/2 ~ 3t) m/sec 8. Two moles of an ideal monatomic gas are used in an engine with a three process cycle. One process, A —> B, is isobaric, one‘process, B —+ C, is isochoric, one process, C —> A, is a linear decrease of pressure versus volume, as shown in the diagram. The pressure varies from 1 X 105 Pascals to 2 X 105 Pascals. The volume varies from 2 liters to 3 liters. A) (5 pts) What is the temperature, T4, at point A? PV : nRT —~> T,4 = (105) (3 x10‘3)/(2) (8) 218.75 0K B) (5 pts) What is the temperature, T3, at point B? PV : nRT ~> TB 2 (105) (2 ><10‘3)/(2) (8) 212.5 0K C) (5 pts) What are the temperature, T0, at point C? PV 2 nRT —> T4 = (2 x105) (2 ><10‘3)/(2) (8) = 25.0 0K D) (10 pts) What are the molar Specific heats of constant volume, CV, and constant pressure, Cp, for this gas? There are 3 degrees of freedom for a monatomic atom. CV=%R=12 J/mole 0K . Cp 2 CV +R 2 3R: 20 J/mole 0K E) (15 pts) How much heat ENTERS the engine during each process? Consider the A——>B process. Q : AU + W AU = nCVAT : (2)(12)(12.5 — 18.75) = —150 J W = PAV = —(105)(10-3) = w100 J QAB = —250 J Consider the B—>C_ process. Q 2 AU + W AU 2 nCVAT : (2)(12)(25.0 — 12.5) = +300 J W : PAV : 0 J Q43 2 +300 J Consider the C—>A process. Q : AU + W AU : nCVAT : (2)(12)(18.75 — 25) = —150 J W = PAV = (1.5 x105)(10_3)= +150 J QAB 2 OJ F) (15 pts) How much work is done BY the engine during each process? W43 2 —100 J WBC I O WCA Z +150 J G) (10 pts) What is the efficiency of this engine? 6 = W/QIN = 50 J/300 J 216.67% H) (10 pts) What is the maximum efficiency one could get for an engine operating between baths with temperatures of T4, TB and/or TC? _ TQQLQ __ 12.5 __ €Ca'rnot — 1 _ THOT — — —25 — liters 2 3 V ...
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This test prep was uploaded on 04/15/2008 for the course PHYSICS 171.101 taught by Professor Professorhenry during the Fall '08 term at Johns Hopkins.

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101f04 - Physics 171.101 Final Exam Solutions May 10, 2004...

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