Physics Fall 2008 Common Hour Exam 1 Solutions

Physics Fall 2008 Common Hour Exam 1 Solutions - 9. 10....

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Unformatted text preview: 9. 10. Physics 123 — Analytical Physics FIRST COMMON HOUR. EXAM Monday, October 6, 2008 Professor RA. Bartynski Mm :> <2 Your name sticker with exam code. SIGN HERE: The exam will last from 9:40 — 11:00 pm. Use a #2 pencil to make entries on the answer sheet. Enter the following ID information now, before the exam starts. In the section labeled NAME (Last, First, MI.) enter your last name, then fill in the empty circle for a blank, then enter your first name, another blank, and finally your middle initial. Under STUDENT # enter your 9—digit RUID Number. Enter 123 under COURSE, and your section number (see label above) under SEC. Under CODE enter the exam code given above. During the exam, you may use pencils. a calculator, and one 8.5 x 11 inch sheet (both sides) with formulas and notes. There are 15 multiple—choice questions on the exam. For each question, mark only one answer on the answer sheet. There is no deduction of points for an incorrect answer. so even if you cannot work out the answer to a question, you should make an educated guess. At the end of the exam, hand in the answer sheet and the cover page. Retain this question paper for future referente and study. When you are asked to open the exam, make sure that your copy contains all 15 questions. Raise your hand if this is not the case, and a proctor will help you. Also raise your hand during the exam if you have a question. Please SIGN the cover sheet under your name sticker and have your student ID ready to show to the proctor during the exam. If needed, the acceleration due to gravity on earth may be take as g : 9.81 m/s2. A student drops a pebble from the top of a water well, and the ”splash” is heard 2 s later. What is the approximate depth of the well? The speed of sound can be ignored. , __ + l/ *1 40111 3— 9/1, + 3t L66 b 2111 0 0 20111 2 2) CD 10111 L3 : Jig-t1“: _.J—l¢‘l\8lx 2 m:-— Cm e) 5m 4 A bicycle travels 308.5 In in a straight line in 40.3 s. The correct way to state the magnitude of its average velocity is: 7.7Iri/s 303, Q" m _4 b) 7.6111/5 ’0’: C) 7.655111/5 40:5 A“ @ 7.66 rn/s (mg 3 e) 7.6550868 In/s A toy rocket, launched from rest on the ground7 rises vertically with an acceleration which is constant at 20 ni/s2 and which lasts for 6.0 s until its motor stops, Disregarding any airiresistance, what maximum height above M the ground will the rocket achieve? BQer-Q ‘TKL {W W) 6L = L0 /$L \ 7. _ \ 2. _ _ ~ t @1.ll<111 SSZK‘L — E>‘LO><(_6,0) _ 340nm ‘ b) 0.73krn W W W410 4/- Th4 c) 1.9km “gate 41‘ a v: a 1) 0.36krn ‘ Edi—VFWAHMMM.W§§L6A~ e) 1.5km nfdtro “31‘ 210—; T=4L/3=\2.2A_7_ 3" file‘mfi" Y = «MT 4587 = 120x121 ~ Consider the displacement vectors A = 3.00i + 3.003, B = 1.00i — 4.0037 and C = -2.00i + 5.003. Find the magnitude and direction (the angle counterclockwise from the r—axis) of the vector E : dA — 2.00B + C. l 8 4 7 A A ' : ‘00” M a . 9‘ 135° " - ~ - ’ : *EJJU L ——- 3, 00 I "" b) 3.00; 0.000 E L ,r J )0 \ol ‘QW‘ @ 12.2;125° +(—Z.00L + 8.05 d ) 17.5;1140 ,3 A > e) 6.70;2430 + ("2.00 x + 51003 A particle starts from the origin at t 2 0 with a velocity of (16i — 123) 111 /s and moves in the x — y plane with a constant acceleration of 51’ : (3.0i — 6.03) 111/s2. What is the speed of the particle at t = 2.0 s? A A I a) 52111/s a}? : CUM + aft) “" (0)35 + a3t 3») b) 39 111/s " M T m T T (M c 46111/s l6? 3'03—2 ~lZ.D£:’ -—é.0-—{§ 33111/5 it: 2.04 e) 43111/s ’4’ A P 1m - '4 NW qr: (221 a NJ 1; )l’U‘it33T At time t = 0 an object has a velocity of +10 In/s. It has a constant acceleration given by —2.0 n1/s2. The total path length traveled by the object between t = 0andt38s is NO+Q TM :9} I}; 1 :2 y? m W a) 25111 04“ \OJLCW a; 17., b) 5.0 111 M: Ma +Qt| 1' O "=7 ‘5' :7 EOI‘L . [Luz til/WEE? I \ i c) 9.0111 “9‘91 _2'01§—1 7h: Untrk ‘12“ (Ce/‘2. :ZSM Z @34m aamtw, Tm 1L; MW' far does it move horizontally and vertically? Horizontally Vertically a) 86.8 ft; 103 ft @ 10311.; 86.8 11. - 3 .357. c) 135 ft; 135 ft W ( d) 95.50.; 95.5 ft will—Audit? e) 103 ft; 53.7 ft A particle with V0 2 5.00i 111/s and 5 = 3.003 111/s2 is at the origin at t=0 s. Find the position and speed of the particle at t:2.00 s A A a) (5.00, 3.00); 5.83 m/s 2"} : (0/071 + can?) 1 +0153 + 4346 @ (10.0, 0.00); 7.81 In/s u H u I/ I, c) (—500.600); 4.00111/s 5.005 '0 0 9 Don d) (10.0 1.50)v 6.32 m/s at t _ ' f" 7 7 __ 2‘00 e) (7.50, 4.50); 4.37111/s _' 5 ) > /\ ’i m a; r 5303: +6.00) A 2T3) I‘m/é %: (UM—t +Jiax~tl 210.0 'm ‘3: (Uoj'i: +JiaJtL =é‘>\00’m 9. The position of a model train locomotive was observed as a function of time and the results are shown in the table. Find the average velocity of the car for the first second7 the last three seconds, and the entire period of observation. A/x X(1n) 0 2.3 12.3 20.7 15.2 6.3 MM = 73;; 0 t(s) 0 1.0 2.0 3.0 4.0 5.0 ._ 213’ _ 51'; a) 0 43 m/ 2 0 m/ 13 n/ g'wyt Atjl 5/ Arm 7- 2. A r . s; . 's; --1 S - _é.3—sz$ m b 2.3 m/s; —2.0 m/s; 6.3 rn/s M At“ 3 0') A/Z‘W" : ‘2'0 A 2.3 m/s; —2.U iii/s; 1.3111/s _ 5 g 3'0 d) 1.0 m/s; —2.0 m/s; —1.3m/s AQQ At fi ’0’) ’V— I ’3 #0: L312.- e) 2.3 m/s; ~48 m/s; 1.3 m/s 5 0 10. A Ferris wheel has a radius of 15 In and rotates counterclockwise around its axis so that the cars on its circumference have a speed of 1.5 m/s. What is the instantaneous velocity vector (v1.7 vy) of a car the instant its position vector is (0, —15 111)? [Note: The origin of the coordinate system coincides with the axis of the wheel and the X—axis is horizontal] 4‘ ‘1' a) (—1.5111/s, O) 1 ‘F' {V )) (0.1 0 m/s) 0; M @ (1.5 in/s, U) .a m d) (1.0 m/s, 1.0 m/s) ,U :(q; q)“ }:<L5) 0)? e) (0. —1.5m/s) 7‘ V i 3 L79 .3):C O) 4an) . V(m/s) 11. The velocity of a particle as a function of time is shown in the figure. Determine the instanta- neous acceleration of the particle at t = 4 s. +0375 m/s2 —0.375 m/s2 0 0 in/s2 a) b) c) d +1.5 m/s2 '2 <5 -—1.5In/s2 4 The. position of a particle moving along the X axis is given by = (21 + 22t — 6.0152) m, where t is time in What is the average velocity during the time interval t;)1l0:6l3:11753'05? [02” : 9E6 fl ; l'OL) tL=3,0/9~ 16) :3: m .: 2‘ + 7.th ~ 6.0155: 33 a d) —8.0nl/S Kl 5 Z. + 2H2 Fé‘o’blz :37 m e) 8.0111/s [Um : ’Xz-9Yr _: d2 o 31 152-13“! ' 3 A mason plans to make a building that is 15 in wide, 10 m deep and 7 m tall having brick walls. A typical brick is 0.33 meter wide and 0.1 m thick. About how many bricks should the mason buy for the building? a) 100000 For fowl, QSJ—jmgl-e/ 7‘9 CW allm rflkees 9F b) iii "Ea/pr" ff {/5401 /Om X 7m) Witt (0:35mXOJm) . 7 br‘ flu; as PM 3,000 ’Ck) m 0" V’ 5 “39"”: a 2,0» 10* brie/25 e) 350 Marx/0H0” +I5x73/(0t33xod) :— 12“ he bwifls only Tie Ms, 6am 4M2><on7 NED/bfixal) A stone is thrown from the top of a building with an initial velocity of 20 g I OX (07‘ m/s downward. The top of the building is 60 In above the ground. How ~ brick 8 much time elapSes between the instant of the release and the instant of ' r 't '"tl tl " l? 1111p‘tC WI 1 1e groum MI, “’07. + 1% A“ .. z 1"; @ 2.0g u vr/ 7 V:3‘t’l S r. M b; s ‘5' GD m c . s l .‘1 d>1.6s fir-“WM? 2) t=3——:2.oa e) 1.0 s n ” N 9.8\ 30m 3» 2 0 ’g For an object moving in a circular path at constant speed, which of the My following is true? a \ ’0" CL, a) velocity constant7 acceleration constant b) velocity constant, acceleration changing C) velocity changing, acceleration constant 0/ 5) velocity changing7 acceleration changing ) e The average velocity has magnitude 1 / 2 of its maximum. OI ...
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Physics Fall 2008 Common Hour Exam 1 Solutions - 9. 10....

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