Exam1solutions

Exam1solutions - Problem 1 The graph below represents the...

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Unformatted text preview: Problem 1 The graph below represents the position of an object as a function of time: For each labeled point, indicate what the graph tells you about the velocity and acceleration at that time by circling the appropriate choices: Velocity Acceleration fi Point A negative zero negative zero E PointB negative zero zero positive (k PointC zero positive negative positive 8‘ PointD negative positive negative zero LN ..__,~; whad- If' downy/art é ~ «pa/an} downs/ml gm; 1, M wan} “fans; slope ) “or’z slope cuvvfime 9 coffin/«(we The graph on the left below show velocity as a function of time, i.e. v(t). Make a sketch on the right showing a possible graph of position as a function of time which is consistent with that v(t). x X @ ® .___ 0 pie f v: :5: g: in <?: a \E Q) 6'tawrs 140V} Zaniar/ , '6 C [WI/es prm/o‘ CL ® M 1%.; Wm: games 545/ ,» 81¢ 55,44 keg/)5 Same upward slope. 59 @ fit-mfgw with (gmstarw' “PM” 5/9/06 ' ' 5+0V’f5' CMFW‘VI downWarJ, @ 44’ mgfioi‘figiafly 5+2” has pa (Hi/e slape @ Ends up MW: fl 17690‘“ V8 5/0199 Problem 2 (answer this and the following problems on the blank paper provided) After the first big snowfall of the season, you take your sled out to a nearby sledding hill, which has a slope of 28°. You set up your sled 10 m up from the base of the hill (measured diagonally), as shown here: The sled and you together have a mass of 75 kg. Once you are seated in the sled at rest, you pick up your feet and the sled starts accelerating down the hill. As it goes, there is a constant friction force of 65 N between the snow and the sled. 1,, mg (a) Draw a free—body diagram for the sled. I, Pqfl, (b) What is the magnitude of the downhill acceleration? (c) How much time does it take the sled to reach the bottom of the hill? (d) What is the sled’s speed when it reaches the bottom of the hill? At the bottom of the hill, the sled transitions smoothly to a horizontal field, initially keeping the same speed. There is now a constant friction force of 75 N as it glides across this field. (9) Measuring from the base of the hill, how far does the sled travel horizontally before coming to a stop? Paw/em Z - bf pail/(#5 per park (C!) Free~booLy OLIOTQMI/M ’FQV Me s/ed: F Nah: OVI/y arch/W fat/C65 Séow/o’f ” _ E? {7Q show/I 014 a diagram like 7%!5". W3 Look a+ ‘Forc’é campowenfs gal/«Mel 1’? Me é/W: fl E: .' F I a “.230 9514(5) N6+1£0V€€ down We fix/fl 64+: [:9 5’”(Zga> "" EC 2 (75 k3)(9.81 m/sésmfigfi —— (55 N) 3: (’5’1‘5N)*(65N) : 280 N 50 0r =5€+/m = (280 N)/{75k3) :1 B7 “7/529 (C > Mafia/4 wflé (0/454’M4L acce/em’h‘O/g SWIM/g From M57“: Ax = La?” 2. (mm>ciz{3.7w/59 2:7“ =9 Cc") M0790” W5” COMSWML acre/eating: V65) 3 V0 “Fat ’P zero 60 0% (70Mom 0? WW, V ZONE = (3~7W/52)(Z.3S) -= (62) 6/60} MOW firm/d5 horizon had/y IW/Wa/ Speec} = 8.5 m/s from 5W” 0’ Ne’r «Col/Ce : 7149f Me ‘Ffl'chFM F0 V66, +0me leg- 5: E F; = —75 N F9 . I: + (’75 N) t? {70 .‘ a '3: ne 5 :: v" 7‘ ACC km 14 m (75/9) [,0 m/5 Time H’ ’fakes +0 Come 1‘?) a Sivp: V =-" V0 Tai $ 0 z [8.5 m/5>+(-/.0m/5§'é $ 73 I 8.5 6 DIS/karma), 3+ ’h/aVe/S M Mm‘ 731443: AX : v0 t figaiz : (85 m/s)(85 5) +é(-/«0 "r/s’XM 531 ll (—‘\ \x N é + ’7 \N 0\ § \/ 8pm 6 pow; 6mm Problem 3 Standing on top of a wall, you toss a tennis ball to your dog who is waiting on the ground. The vertical distance from your hand to your dog’s mouth is 3.4 m, and the horizontal distance is 5.5 m. (a) Assuming that you toss the ball with an initial downward angle of 15" relative to the horizontal, HS and you want your dog to catch the ball in its mouth without having to move, what speed should you give the ball when you toss it? (b) What angle does the ball’s velocity vector have (relative to the horizontal) at the moment your dog catches it? (c) Suppose that instead of the speed you found in part (a), you throw the ball with a speed of 13 11115, but with the same initial downward angle of 15°. Your dog jumps straight up to catch the ball in its mouth. How high did she have to jump? (That is, when she catches the ball in her mouth, how high is her mouth above its initial position before the jump.) pa-_. (This Is like p m blew: 3 "86 33”” gram We hOMQM/Ofk) *7 L,“ fem Wop/4 7%6 7 M00 8 an x, Coordr'waré sys (00 gr} at" you); hafio‘. [Other Choices are posst'é/e. Catlrthe bat/1's Miflal speeo‘ 1/0. The” i/x =3 VOCOSKIE") and remains (Maw/HE HOW 10149 PS the WW M Hal/1+7 We know fogfyhf : 5.5 M 5.5m P 5.51% , 57"" 50 : \/K It %(¢5([5°) .— V0 [at/er) IMhL/‘m/ VeMim/ Velocf+y Fs Megwh't/é’ P1447495 (oodimrke syshm: vyo :- —Vo Sit/’05") : (-026) v0 Vet/Hm! egMorch 01C wok/ow I V 5 yo ”“ Vyafi did)”; WWW (fog car’ft/Ies éa/l flH’ é: é¢/,:9;,+: (“3% m) z O + (wig) Va éfifigw "L (-026) v0 I 2— 47.8] Vx : Vowsm") I 6’8 W/s Vy = \40 +ar£+7ighf :— osMOso) + (—9.8/ m/59(5-l:om : (~2H m/s)+(—é.l “4/5) 3 *85 147/5 ‘hme :— J—IW : 85"4/5 :— 097 [yd 8.8141/5 #W (C) W Int‘h'af V6 1052+}; : 13 m/s 2 Vx = V5039 :- (Ism/s3rosfl6°)= I16 “4/; Vyo=-Va 5M6 = "(ram/skiwafi"): *BH W/s HOW {0’49 (foes 50‘” ’f‘mke “99+ o‘iredé/ ov’er dog? Ax: vx’t 5-5 WI = (12.6 Viv/s) 7542/9“ :’> tr/gh/r = 0W s VQVHCaf posffion 6h" ’dldfl' 'Hmé : v= yo r vyat + éqyfiz 2 = O + (“39 W/5)(0.W s) + 3' ("7.8! m/s”) (0.41% s) : (v1.5 m) + (-057 m) :- “Zfl m 60 dog has 7%; JIM/1p Verfi'm/O/ (“2H "7) “(J-‘1’”) -= IO M Problem 4 The Earth follows a slightly elliptical orbit around the Sun, However, for this problem, assume it follows a uniform circular orbit with a radius equal to the average Earth~Sun distance. The period of the orbit is, of course, equal to one year. 6 POMIB (a) In what direction is the Earth accelerating? 7 Pow; (b) What is the magnitude of the Earth’s acceleration? Write your answer with units of m/sz. 7 (6) Where is the center of mass of the Sun-Earth system? CG) RWOH’J 7%6 50M. (6) Cefll'w‘pefaI/ acre/erah‘on: :L GIG R Wheat/'5 v.7 We know V*(Iy€ar):=2vR v=(2rr3(/.50x10”m), Iyéar K Ida Mm— X—._______ l year 365,245.91}! Z‘f haw 36005 1 2.6!? X IO” "7/5 50 ac: :)6.0x/0"3m/s1 1.50 ><I0"m [or 0.005 lav/6‘) Sui/I (c) Q Ear +x For font/ememe, choose a Coordr‘HmLQ Sysfem Carmel/ac} «M Me SW4. [0mg choices we possré/ej _ m,X,+W2Xz : (1.9?z/030kgl/0)+(5fl7x/o’15) max/0”». Xm, — w “L- . 0.7?x1030+5.77«102q)k9 : m from fie (Egret/(HQ W SWfl Ll paints f gr? ((1) System Problem 5 Two blocks are held on opposite sides of a triangular wedge, connected with a string that runs over a pulley at the top of the wedge, as shown below. Block A is also connected to a spring that attaches it to the base of the wedge. Block A has a mass of 5 kg, block B has a mass of 4 kg, and the Spring has a spring constant of 3.5 N/cm. Assume that there is no friction between the blocks and the wedge, and that the string and pulley are massless and frictionless. The system is initially at rest. (a) What is the tension in the string? (b) Is the spring stretched or compressed? By how much? Suddenly, the spring comes apart from block A, freeing it to move. (c) What is the acceleration of block A? In which direction? (d) What is the tension in the string now? (3) What is the acceleration vector for the center of mass of the block—string—block system? {6 (1+ rest so 146% ’EM’CQ on Hock B tab/54- 59 Zero. ‘r/ L“ Horizontal Forces: 0 : F” Sin/70°) - Tcos (70") 2; F” :- ‘;;;g:) T: {0.35) T veV'h'mf Forces -' 0 : TSMUO") + 5, 605/70 ’3 #1449 = (mm + (0.12” —- [‘fk View/5‘) 33? T: (3‘31 m/51)/(r.og) : 37 N T. 5, m8 ((7) Look (2% 1CorC€S 0/4 é/odc r4 -' F” T Spw‘ng 'Force E WSW E: e p05;+2ve or neanrn/e “‘3 Look 0H“ force CompofléM’s perm/M “(0 We inc/me: F; = 3 N /T 7 mgsmflo") z (5k3)(‘7.8/ m/s”)(0.£‘f) : 31.5 N For Hock r4 7% be ad, mag #42 5;; ring {are mam!— bé ofr‘red'eo‘ down “like 2146/2146 and provide 37N—315N:5I5,V Nofe aboW’ We Vemmmng Eon/*5 097W Broé/em 4% We emd— Mshamf fiécfl'fi’ve SPW'Mg o‘emc'ées 79/0/47 block A #29 black musk Feel #12 Mr fine: #19 Same as {ye—90;?) (,e. or net «force 01¢ 37Nv3/.5/\/ =55 N HP We (“m/me and or rem/M9 “(CE/QMHMOF Etziw : 1.! "1/51, However, a 144004er faker, fire fermion M 7%? S—m'ng goes dowry am) {704+} blocks accelercma 0% Me same raw. 7145 flange 26 a f€6MH‘ 0-? We 5479149 VE/axfng -—*a rear! shah/:9 wow/d have beg/1 5+ye fiched Slighfly M We M21951“! a+—resf’ S‘I—a‘f-e) arc-117W3 like or 6pr£n3 WW, a large k. Am ideaf Mass/€55 firing relaxes Mac/Wkly rapid/y. IP 24 arm/arm’es Mp 7+5 Side 04; My Wecfge, B 14M$+ arc‘ce /e rflE MW! Me Same m Mfflm’g Glow/M #5 wk 01‘: e Wedge Force COMpOMefif’S on /4 MP Me weolge: F" 7' § 7. i; £4009 ._ 9 By, .— War t; T -— (5 kg) 9 5;:1[£/o”):(5k3) a T —— (3/.5 N) -_— {5 kg)a Force compomem's on 8 down Me wedge.- 1’ T 5, Beat 3’ WOT % (4k9)95r‘r4{7c°) ~T :(‘fk3JGr F5, (35-7 N) “ T 2 (W3) or NOW neeaHU Soft/e «Cor or WM T. AOL} Me eqflmln‘or/S: T» (3!.5N) 21519)“ + {am/v) ~T=~Mk93a (359 «31.5) N : (7 kg) or @ q -: 060 “1/57” (4)9 516 inc/Me (<0me MOI/e) T*{3l.5N)={5kg)(0.50 141/52,) g a?) a :050 m/sz @5 k3 Mid/£3 ot20.60 147/5?“ Acce/emh‘om Vecfw For Me ce/zfizr’oFerrss IS jus+ +he Wer‘ghv‘ed Sam 0? Me accre/QVMHOH V€C+0rs 01C 4%@ +Wo War/<3 __& —.._'h 6“ ,_ onrA +1449 0:3 cm _ 1414+ m8 CaICH/mLE X and y compo/4914*: Sepan‘E/y: (5 k (0.50m/51)cos@09 + (9 kg) (0.60 m/gz) 695(700) 0(6me : : “4/57. 5 k3+ Ll k3 : “0.0‘7’ M/sz 60 I 06m 3 {0.35141/51) 4' (“0.0V ...
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This note was uploaded on 11/11/2008 for the course PHYS PHYS171 taught by Professor Shawhan during the Fall '07 term at Maryland.

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Exam1solutions - Problem 1 The graph below represents the...

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