Ke Name Insrucor Phsics 1210 Exam 1 Sepember 26, 2013 Please wrie direcl on he exam and aach oher shees of work if necessar. Calculaors are allowed. No noes or books ma be used. Muliple-choice problems have onl one correc answer. You ma choose o circle wo answers on a muliple-choice problem and, if one of hem is correc, receive half credi. Circle hree and if one is correc, 1/3 credi, ec. For worked problems, be complee and show all work, beginning wih diagrams and fundamenal, general equaions used. Kinemaics v avg = x 2 x 1 2 1 x 1 = x 0 + v 0 + 1 2 a 2 v 1 2 = v 0 2 + 2a(x 1 x 0 ) = Δx Δ v 1 = v 0 + a v = d r d x = 1 2 (v + v 0) + x 0 a avg = v 2 v 1 2 1 = Δv Δ a rad = v2 R = 4π 2 R T 2 a = d v d F = ma w = mg
Each of he problems 1 11 is worh 5 poins. 1. A ca sis on he floor for a few seconds, crouches, jumps up o a window ledge, sis here for a few seconds, and hen jumps back down o he floor. Which posiion-ime graph below bes describes he ca s verical posiion over ime? A. B. C. D. E. 2. To he righ is he x- graph of he moion of a paricle. Of he four poins A, B, C, and D, he veloci v x is greaes (mos posiive) a A. A B. B C. C D. D E. No enough informaion in he graph o deermine Veloci is he slope of he (angen o he) posiionime plo. I is posiive onl a A. 3. Which vecor has he larges componen in he -direcion, as defined b he axes o he righ? + +x 4. Assume ou are driving a car around a circular rack a a consan speed. Which of he following saemens is correc? A. Your speed is consan and our veloci is consan. B. Your speed changes and our veloci is consan. C. Your speed changes and our veloci changes. D. Your speed is consan and our veloci changes. We are old he speed is consan. Bu since he car is driving in a circle, he direcion of he veloci is consanl changing. Ever half-circle, he car s veloci reverses!
5. A ball moves in a circular pah wih consan speed on a horizonal, fricionless surface. I is aached b a rope o a verical pos se a he cener of he circle. If he rope breaks, A. he ball will keep moving in a circle. B. he ball will move on a curved pah, bu no a circle. C. he ball will follow a curved pah for a while, hen move in a sraigh line. D. he ball will move in a sraigh line. E. None of he above will happen. Wih he acceleraing force of ension removed, he ball will move a consan veloci. 6. A large ruck and a small compac car have a head-on collision. Which of he following saemen is correc? A. During he collision, he force exered on he ruck b he car is larger han he force exered on he car b he ruck. B. During he collision, he force exered on he ruck b he car is equal o he force exered on he car b he ruck. C. During he collision, he force exered on he ruck b he car is smaller han he force exered on he car b he ruck. D. None of he above is correc. Newon s hird law. Ineracion forces are equal in magniude, opposie in direcion. 7. A rock flies hrough he air wihou air resisance. During is rajecor, which one of he saemens below is rue? A. Is acceleraion is consan and is speed is consan. B. Is acceleraion changes and is speed is consan. C. Is acceleraion is consan and is speed changes. D. Is acceleraion changes and is speed changes. 8. A hiker finds wo abandoned mine shafs, and drops a rock from res down each. In shaf 1, he rock falls for ime T before i his he boom. In shaf 2, he rock falls for ime 2T before i his he boom. Wha can ou conclude abou he dephs of hese wo mine shafs? A. Shaf 2 is deeper han shaf 1, bu less han wice as deep: D 1 < D 2 < 2D 1. B. Shaf 2 is wice as deep as shaf 1: D 2 = 2D 1. C. Shaf 2 is more han wice as deep as shaf 1: D 2 > 2D 1. D. Shaf 2 is deeper han shaf 1, bu ha s all ou know for cerain: D 2 > D 1. In free-fall, disance fallen as a funcion of ime is Δ = v0 + g 2 /2. From res, v0 = 0, so Δ = g 2 /2. Disance fallen is proporional o 2, so doubling quadruples he fall disance.
9. The graph o he righ shows he veloci of an objec as a funcion of ime. Which of he graphs below bes shows he ne force versus ime for his objec? A. B. C. D. E. A downward acceleraion means a downward ne force. A consan veloci means zero ne force. 10. A curve on he highwa has a radius of 400 m. A consrucion crew widens he urn so ha is radius is 600 m. If vehicles ravel on he urn a he same speed as before, how does heir cenripeal acceleraion a compare o heir previous cenripeal acceleraion a 0? A. a/a 0 = 400/600. B. a/a 0 = 600/400. C. a/a 0 < 400/600. D. a/a 0 > 600/400. Cenripeal acceleraion is v 2 /r. Boh cases have he same v bu differen r : a = v 2 /(600 m) and a0 = v 2 /(400 m); hus a/a0 = 400/600. 11. Alice and Bob are boh on a merr-go-round ha is roaing uniforml. The op-down view of he merr-go-round o he righ shows he relaive posiions of Alice (A) and Bob (B). Who is acceleraing he mos? A. Alice. B. Bob. C. The are boh acceleraing a he same rae. D. Need more informaion. Alice and Bob boh reverse heir veloci in he same ime period, bu Alice s speed is greaer, so her acceleraion is correspondingl of larger magniude. B formulas, cenripeal acceleraion a = v 2 /R = 4π 2 R/T 2. Alice and Bob roae wih he same period T, bu Alice wih a larger radius R, so her acceleraion is proporionall larger.
12. (15 ps) You drive 50 miles from Laramie o Cheenne in 57 minues. The firs 22 miles of he rip, from Laramie o Buford, was on ic roads wih heav raffic. The las par of he rip, from Buford o Cheenne, was in more favorable condiions, so ou were able o ravel a 70 mi/h. Wha was our average speed raveling from Laramie o Buford? x Cheenne Δx 2 slope = 70 mi/h Buford slope = v1 =? Δx 1 Laramie Δ 1 Δ 2 Δ = Δ1 + Δ2 = 57 min (h/60 min) = 0.95 h Δx = Δx1 + Δx2 = 50 mi Δx1 = 22 mi v2 = 70 mi/h Par 2: Buford o Cheenne Δx2 = Δx Δx1 = 50 mi 22 mi = 28 mi v2 = Δx2/Δ2, so Δ2 = Δx2/v2 = (28 mi)/(70 mi/h) = 0.4 h Par 1: Laramie o Buford Δx1 = 22 mi Δ1 = Δ Δ2 = 0.95 h 0.4 h = 0.55 h v1 = Δx1/Δ1 = (22 mi)/(0.55 h) = 40 mi/h
13. (15 ps) As a pical scene in an acion movie, a car is moving owards a bridge and an acion hero jumps off he bridge and lands on he op of he car a he momen i is passing hrough. Assuming ha he car is moving a a consan speed of 45 mph (20.1 m/s), he bridge is 6.0 m above he car, and he acor is in free fall, calculae where he car should be when he acor sars his acion. You can rea boh he car and he acor as poin objecs. +x h + v D h = 6 m v = 20.1 m/s We wan o find D raveled b he car in he ime he acor drops h Fall ime = g 2 /2 = h, so 2 = 2h/g and = 2h/g Travel disance D = x= v = v 2h/g = (20.1 m/s) 2(6 m)/(9.8 m/s 2 ) = (20.1 m/s)(1.11 s) = = 22.24 m
14. (15 ps) Sepember 19 was Inernaional Talk Like a Pirae Da. To celebrae, ou and our friends sage a mock bale wih boas in he Corbe Pool, firing waer balloons a each oher. Your enem s boa is 10m awa. You aim our waer balloon launcher a a 45 angle wih respec o he horizonal, and ou overshoo he enem s boa b 4m. Assume ha he launcher is a he same level as he boas, and ha air resisance is negligible. a) Wha is he iniial speed a which he waer balloon was fired? b) Assume ha he launcher fires all balloons a he same iniial speed. A wha angle mus ou aim he launcher in order o make a direc hi? + +x v 0 α D S D = 10 m; S = 14 m v0 = v0 sin(α) vx = v0 cos(α) Disance raveled as a funcion of v0 and α = v0 g 2 /2 A landing, = 0 = v0 g 2 /2 = (v0 g/2) = 0 is launch; 0 = v0 g/2 = 2v0/g x = vx = vx 2v0/g = v0 cos(α) 2v0 sin(α)/g = (v0 2 /g) 2 sin(α) cos(α) There is a rigonomeric ideni in here: sin(2α) = 2 sin(α) cos(α), so x = v0 2 sin(2α)/g. This is he range equaion. Par a: α = 45 and x = S, find v0 S = x = v0 2 sin(2α)/g = v0 2 (1)/g = v0 2 /g v0 2 = Sg, so v0 = Sg = (14 m)(9.8 m/s 2 ) = 137.2 m/s = 11.71 m/s Par b: x = D, find α D = x = v0 2 sin(2α)/g sin(2α) = Dg/v0 2 2α = arcsin(dg/v0 2 ) or 2α = 180 arcsin(dg/v0 2 ), so Dg/v0 2 = (10 m)(9.8 m/s 2 )/(137.2 m 2 /s 2 ) = 0.7143 (uniless) Thus 2α = 45.59 or 2α = 180 45.59 = 134.41, so α = 45.59 /2 = 22.79 or α = 134.41 /2 = 67.21