Phys 0174 - Exam _2 Master

Phys 0174 - Exam _2 Master - MASTE R PHYSICS 0174 Fall 2008...

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Unformatted text preview: MASTE R PHYSICS 0174 Fall 2008 P. Koehler Second Hour Examination November 10, 2008 (2200-2250pm) This sheet of equations and a calculator are the only aids you may use during the examination. 0 cos 9 = b/c a sin 9 = a/c 90° 9 tan 9 = a/b b a2 + b2 = c2 Vector algebra: IfZ=Aj+Aj+Aj mm E=Bj+Bj+BJ max 21:3:(4 in)f+(Ay iBy)}+(Az iB:)/€ Em? AXBX + AyBy + A23: = (El-lélcosa HxB=(AyBZ—Asz)f+(Asz—AxB:)j+(AxBy—AyBx)l€ and |Zx1§l=lfillélsm¢ Eguations of Motion with constant acceleration Linear motion: Rotational motion: Equ. (1): vx = VOX + at w = we + at Equ. (2): x - x0 = vext + 1Azaxt2 9 - 90 = wot + 1/20lt2 Equ. (3): vx2 = vexz + 2ax(x — x0) w2 = woz + 2a(9 — Go) Equ. (4): x - x0 = 1/2(vox + vx)t 6 - 60 = 1/2(ooo + w)t Relationships between linear and angular kinematic variables: 3 = R*G vtan = R*w atan = R*0. CONTINUED ON THE OTHER SIDE OF THIS SHEET Newton’s 2"d Law for rotational motion: 2?“, = I «)7 Torque: E=FxF |fl=|F|~|F|-sin¢ Moment of inertia: general definition: I = Emir,2 Kinetic energy: translational K =§mv2 rotational K =%Ia)2 Work: W = F or? = |F1~t31~c056 Work — Kinetic Energy Theorem: Wm, = AK Gravitational Potential Energy: Ug = mgh Elastic Potential Energy: Ue. = 1/2 kx2 Conservation of Mechanical Energy: E = Ef if only conservative forces act on an object If a force other than a conservative force does work on an object, then E1 +Wmher = E, Frictional force magnitude: static friction: f3 3 ,us -N kinetic friction: fk = ,uk ~N Linear momentum: f) = rm"; Linear momentum is conserved if 217;" = 0 Impulse: :7 = A}? = E ~13, = [Fm 61’ = 13W 'N Center of Mass position: xCM = Em’x’ and yCM = Zm’ ’ / 2m: Zmi Equilibrium Conditions: Z(Fen)x = O 2(Fw)y = 0 Z? = 0 . . I ml ' m7 _ _11 N ' m2 GraVItational Force. FG =G 2 - where G—6.67x10 2 r12 kg Magnitude of gravitational acceleration near earth's surface: g = G% = 9.80m/s2 E TEAR OFF THIS SHEET AND USE FOR REFERENCE DURING TEST Physics 0174 (Koehler) Second Hour Examination Fall 2008 November 10, 2008 Print Your Name: I l E Page 1 of 7 ***|MPORTANT NOTICE*** Before you start the test, print your name on each page. Use the blank space provided after each question to work out the problem. If you need more space, use the back of the sheet. WRITE NEATLY! No partial credit will be given if your work is illegible. SHOW YOUR WORK! No credit will be given for a numerical answer that is correct but not accompanied by sufficient written work that shows how you arrived at your result. Be sure to enter your final answer into the marked box whenever such a box is provided. Express numerical answers with the appropriate number of significant figures and give units when appropriate. DO NOT WRITE IN THIS BOX! Problem (points) #3 (25) 7. ? Total Test(100) Problem #1 (25 points): (a) Which one of the following statements correctly describes the mechanical energy of any system? (Circle your choice.) (A) The work done on the system by the gravitational force (B) The difference between the system’s kinetic and potential energy at any point. @ The sum of the system’s kinetic and potential energy at any point. (D) The sum of the system’s translational and rotational kinetic energies at any point. (E) The potential energy of a spring at any displacement. (Problem #1 continues on the next page) Physics 0174 (Koehler) Second Hour Examination Page 2 of7 Fall 2008 November 10, 2008 Print Your Name: M g (b) Which of the following statements is true if a body moves in such a way that its linear momentum is constant? (Circle your choice.) (A) The body’s kinetic energy is zero. The sum of all external forces acting on the body must be zero. (C) The body’s acceleration is greater than zero and is constant. (D) The body's center of mass remains at rest. (E) The sum of all external forces acting on the body is constant and not zero. (c) A disk is free to rotate about a fixed axis. A force F applied at a distance d from the axis causes the disk to undergo an angular acceleration of magnitude at. What would be the magnitude of the disk's angular acceleration produced if the same force F were applied at a distance 2d from the axis of rotation? (Circle your choice. ) (A) (1/4 (B) (1/2 (C) (1 2a (E) 4a (d) The moment of inertia of a solid disk of mass M and radius R is l=1/2MR2. Consider a solid disk that is rolling without slipping along a level table top. Which of the following statements regarding the relative magnitudes of the disk’s translational kinetic energy and rotational kinetic energy is true? (Circle your choice.) lts translational kinetic energy IS greater than its rotational kinetic energy. % (M V .= K E t B lts rotational kinetic energy is greater than its translational kinetic energy. ( ) 7. L - .. (C) lts translational kinetic energy is equal to its rotational kinetic energy. 3' E LO ‘ KC f (D) The answer depends on the value of the radius R. = i HR: ,' w = «£— l t l (E) The answer depends on the value of the mass M. g Kfif : z 2 N a Nat: 2. (e) The diagram below shows five satellites in circular orbits around the earth. They differ in 7;“ N V mass (the larger dot indicates twice the mass of the smaller dot) and orbital radius. Which _ i {re satellite has the greatest orbital speed? (Circle your choice.) ’ 3 1-” 2.. (A) Satellite 1. Hg 9* y_ (B) Satellite 2 G 1-)“ ‘ )7 (C) Satellite 3 x E D 3 ll' 4 — 6 Ma 3 Mat ( ) ate ite \/ - j: er/ean on @ Satellite 5 WI \/ CS greatest (F «r is Swalltst Physics 0174 (Koehler) Second Hour Examination Page 3 of7 ‘ Fall 2008 November 10, 2008 Print Your Name: A STE, & ___l Problem #2 (25 points): (a) The adjacent diagram shows a small block of mass m=0.480kg on a frictionless ramp which is inclined at an angle 6=35.0° above the horizontal. lf the block is released from rest, how long will it take to slide a distance d=5.00m down the ramp? (5 points) UZ=O My answer is: _ ol ~ oi _ M- (b) Starting from rest as before, how long would the block take to slide an equal distance d down this ramp if the coefficient of sliding friction were uk=0.300? (5 pornts) E; 'l- Wokker = /‘ wo+k‘r=/aK.Ffi/.0l.(m((go‘7‘=wK-Wg.d.&ve Wgexg ‘flKW3 dme : £541??- ?)(12 2 25d(5m 9 WK (me) My answer is: I); : t/ZSDQ(§CMe‘/uu (we): 5:6? (“/5 tL.‘ 716$ 20( I t = w : I. ?6 s 11‘; Q (c) Now take a solid sphere of equal mass m=0.480kg and radius r=0.0750m. Starting from rest, how long will the sphere take to roll without slipping a distance d=5.00m down this ramp? [A solid sphere of mass m and radius r has l=(2/5)mr2.] (5 points) / -——‘— oz ‘1: 2- ~(fi>mvl-w~i‘i E;: :> M§Q‘~ZW‘1~Z wF/ I- r] / 4rd,, 2. 2. L L -l z q; ‘)(__t£c-i J— =l’ “Mt-sway; H” «r ~2W£c+emic WW;- U-g: W: Iéggd‘sm‘e7, % Myansweris: t: 1,23,: L53; t: “is ( roblem #2 continues on the next page) Physics 0174 (Koehler) Second Hour Examination Page 3 of 7 Fall 2008 November 10, 2008 Print Your Name: Q Problem #2 (25 points): m (a) The adjacent diagram shows a small block of " ‘ mass m=0.480kg on a frictionless ramp which is inclined at an angle 6=35.0° above the horizontal. If the block is released from rest, how long will it take to slide a distance d=5.00m down the ramp? (5 points) Afterna-Hue, foiutfiout (Age Ncwéau's 2H aw ZFX = mgiwe=max => ax: 3-w9= $.62 iii/5a Zr, Mag ' C096 +N= O = 0/. X‘Xoad/ My answer is: - 0=V -t+"i‘qxtL-‘> 4': ZLQ’Vt _ X )( X0 z or t: E: L335 t—/.33_S 1. (b) Starting from rest as before, how long would the block take to slide an equal distance d down this ramp if the coefficient of sliding friction were uk=0.300? (5 points) Her Vra-i-tue. So/u {Tom in: Macaw/“Ah max => ax= g-swe‘flafi'we =3.le44/5Z 2?, = N-M3-cm6=0 9 N=Wl3fw€ 2 2. X'Xo = 0i: Vox +31;th = £51K t Myansweris: twig: = (.96; #1955 (c) Now take a solid sphere of equal mass m=0.480kg and radius r=0.0750m. Starting from rest, how long will the sphere take to roll without slipping a distance d=5.00m down this ramp? [A solid sphere of mass m and radius r has |=(2/5)mr2.] (5 points) flierua‘i L‘UQ SOlu‘f'l'vkfl 27:“ : mg's‘he ‘fls'fifl‘ max :>MQ’<= mgm‘efi Iq-a’zxzmjfi'ué’éwax ‘Z’r‘y = N‘wgcosé‘zo =>N=MJCUD§ ,/’7 4X: égsmatqeozays; Myansweris: = N. T =I<><~= I 55 a M w (Problem #2 continues on the next page) __ t- I = I. :8 5 «. i Physics 0174 (Koehler) Second Hour Examination Page4 of? I 1 Fall 2008 November 10, 2008 i i Print Your Name: MAST E Q (d) What is the magnitude of the angular acceleration of the sphere? (5 points) town a); = Hi: 2 OH: aitcmart-Veiyz (x z 9% f z 4.02Vsz_53‘éfld/ 04 2 2 Sgél'ra4/5z “wing/“answer is: 57' g; (e) What must be the minimum value of the coefficient of static friction between the sphere and the ramp in order for the sphere to roll without slipping? (5 points) ~ =¥vf= "m ‘E..J:ok ; #5 L Id émT-d Zd’aL = ——-—--——————- I #:0524170 W5 WJCWE‘T' wgcmfi'fi’ 5.09.“)9 My answer is: Problem #3 25 oints : g.- 0.200 The diagram shows puck A of mass mA=O.125kg i” that is sliding on a frictionless, level surface with 5 (‘va speed v0A=5l50mls in the positive x-direction. At mA the origin of the x-y coordinate system it collides ' with puck B of mass mB=O.250kg that is initially at ' rest. Because the collision is slightly off center, mA VOA 9“ puck A is deflected by an angle 6A=65.0° above the """""""" "* positive x-axis and puck B recoils at an angle mef x 65:37.O° below the positive x-axis. i m, VfB (a) Determine the magnitude of the final velocity m of puck A. (5 points) fan: PM +9193 9 MAVM: ""4 V€A“59A*mfivf"8'@91 ' O 3 ‘MAV‘A Sm 9p, ~‘M3V;5 442493 =7 M3V43: W‘Avfiitéme‘q) Sow/m = W! [Case + [Stuart Myansweris SW33 ‘ r q (a A (St'ugfl)m65 V s V‘GA =( :1 Esme )"mfl‘ : v'gfl S (Problem #3 continues on the next page) Physics 0174 (Koehler) Second Hour Examination Page 4 of7 Fall 2008 November 10, 2008 Print Your Name: A S TEK (d) What is the magnitude of the angular acceleration of the sphere? (5 points) My answer is: (e) What must be the minimum value of the coefficient of static friction between the sphere and the ramp in order for the sphere to roll without slipping? (5 points) My answer is: Problem #3 (25 points): The diagram shows puck A of mass mA=0.125kg that is sliding on a frictionless, level surface with speed v0A=5.50m/s in the positive x-direction. At the origin of the x-y coordinate system it collides : with puck B of mass mB=O.250kg that is initially at 3 rest. Because the collision is slightly off center, mA VOA 9A puck A is deflected by an angle 6A=65.0° above the """""""" "* positive x-axis and puck B recoils at an angle mBi X 63:37.O° below the positive x—axis. 1 m3 VfB (a) Determine the magnitude of the final velocity m of puck A. (5 points) Altcrmq-HVQ Sam-Hoot: App/y Me. /a,Lu affirms A '4 A 304C; cans. of) (5:4ch Mama-fa»: mare; {4442‘ @334! +555 , yanswerls: We 3 Mamet: +KM vectors 014qu {brwl 6L WWW/file: SC“ 04 5C“ 65 SUM 993 SM} V '4' 3 33 ___.._ = : —> z t B M ‘ PfA 5\'u 935 (Problem #3 continues on the next page) : 0.723 now/s : 30-94—95: is {lance m = 1%: 3.33 “Y; Physics 0174 (Koehler) Second Hour Examination Page 5 of? i Fall 2008 November 10, 2008 i Print Your Name: TE (b) Determine the magnitude of the final velocity we of puck B. (5 points) 56M 9 {M = W V { 4 B V915 A FA 5(‘k9 a.) \/(5 = VéA 5““ 6‘4): 2,5;“4/5 My answer is: SCu 9} (453455” s (c) is the collision between these two pucks elastic or inelastic? You must provide a numerical justification for your answer. (5 points) :9 K6: = KEFZ K J- m v L 89 E; ‘ 2 l9 0: - .3— .l l 2 ~ ~— KE; : 2 W‘DV43 4 Z WAU‘FA ‘ ' Myansweris: fit‘bcuz. KE; 75 K5; / Wreck L'S €K€(¢${CIL— (d) What is the magnitude of the impulse that puck A exerted on puck B during the collision? (5 points) .A 3‘: P3§'?Bi= “43%.3 "O 6 0 i?i= Webb wave - .63? tar-"Vs My answer is: Ewes? k5‘W/5 (e) What were the magnitude and direction (relative to the positive x-axis) of the velocity of the center of mass of the two pucks before they collided? (5 points) WW.W0(L: ch: VOA = 'MAl-Wlfi dict/eel (0 K : (05‘ X - A‘L'Vech'ak My answer is: A V¢M=(.83 (5W 6'0 WA) 05. X ! Physics 0174 (Koehler) f Fall 2008 l l Second Hour Examination November 10, 2008 Print Your Name: (I! A 5 T E K Page 6 of 7 Problem #4125 points): The adjacent diagram shows a uniform beam of mass m=35.0kg and length L which is held at rest in the following way: its upper end is attached to a vertical wall by means of a hinge H, and a point C that is (2/3)*L down the beam from H is attached to the same wall by means of a light cable which makes an angle 0L=35.0° with the wall and an angle 6:250" with the beam. A crate of mass M=125kg is suspended at rest by a cable attached to the far lower end of the beam. (a) Draw a “stick diagram" of the beam showing all of the forces that act on it at the locations where they act; clearly label all of the forces. (5 points) \\ es l80—(l5’0-d~/3> : 4%: ea“ T (b) Calculate the magnitude of the tension T in the cable between the beam and the wall. (10 points) ZMfl T1 sx‘vxol Y+lY—\M5—Hg/ 7;:T'C‘79“ T-(%>'5l'hfl,-H5'L‘5Ck8"m<7[$)%9 .. g-mefl” %>’3 = my My answer is: (Problem #4 continues on the next page) Physics 0174 (Koehler) Second Hour Examination Page 7 of 7 Fall 2008 November 10, 2008 Print Your Name: iTE: é (c) Calculate the horizontal force Hx that the hinge H exerts on the beam. A positive answer will be interpreted to mean that Hx points to the right. (5 points) ll TSL'MOC=K%6Z N ..—.__. HX: TX My answer is: (d) Calculate the vertical force Hy that the hinge H exerts on the beam. A positive answer will be interpreted to mean that Hy points up. (5 points) HY = mg + HJ—Ty =(wr+M)5 -T-cmo<: — (QC/(WV #7 is WgQ/J’L.U*L/ 5:9 it F,.—-—-———'_‘-> My answer is: N END OF TEST ...
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Phys 0174 - Exam _2 Master - MASTE R PHYSICS 0174 Fall 2008...

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