{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 101f04 - Physics 171.101 Final Exam Solutions Prof Barnett...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}