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Constant Speed - Mechanics - Solved Past Paper, Exams of Mechanics

This is the Solved Past Paper of Mechanics which includes Sum of Impulses, Final Velocity, Quantities in Specified Units, Propeller Angular Acceleration, Total Impulse etc. Key important points are: Constant Speed, Constant Acceleration, Negative Acceleration, Entire Time Period Shown, Initial Velocity, Mid-Air Platform, Constant During Flight, Minimum Horizontal Force, Static Friction

Typology: Exams

2012/2013

Uploaded on 02/21/2013

salu-salman
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Download Constant Speed - Mechanics - Solved Past Paper and more Exams Mechanics in PDF only on Docsity! SOuction $ Physics 151 Score: 100 pts. possible May 16, 2008 PHYS 151 Final Exam (with errors corrected) Exam time limit: 120 minutes. You may use a calcalator and both sides of TWO sheets of notes, handwritten only. Closed book; no collaboration. Complete ALL questions. For each multiple choice question, choose the ONE best answer. There is no penalty for guessing. Ignore friction and air resistance in all problems, unless told otherwise. On your BUBBLE SHEET, please fill in: “Last Name, First Name” = your name “Identification Number” = your ROSTER number for PHYS 151 NO other fields (birthdate, grade, etc.) are necessary. Physical constants: 2 = 9.80 mis? G = 6.674 x 101! N-mikg? N, = 6.022 x 107 mol" key = 1.381 x 10 YK R= 8.314 Ji(mol-K) = 0.0821 L-atm/(mol-K) Useful conversions: 1 year = 3.156 x 10’s 1m? = 1000 L 1 atm = 1.013 x 10° Pa 1 cal = 4.186 J O°C = 273.15K masses radii orbital distances orbital periods Mou, = 1.99 x 10° kg Rou, = 6.96 x 10° m Mean = 5.97 x 10% kg Reon = 6.38 x 10° Tearh-Sen = 1.50 x 10" m Tray = 1.00 year Mysoon = 7.35 x 107 kg Reon = 1.74 x 106m Tath-Moon = 3-84 x 10% m Tytoon = 27.3 days (2 pts. each} Convert the following quantities into the specified units: 1. 660 kW = aw 0, A. 6.6.x 10° GW @ 66x 10 cw (+eoses Le mar \ 2 bbei0 qw B, 6.6 x 107 GW E. 0.66 GW C. 66x 10° GW 2.720 nL = em? 7 (1s 10° en? z A. 7.2% 10% em? "2, Tem 20 yt) Tote /* O.T2 em 7.2 x 107 cm® E, 720.cm* I ye — 0.72 em? 33.0km = a? on A. 3.0 x 10° mm? 3.0 x 10% mm’ (2 oud ) (tse, ww : . fi B. 3.0.x 10° mm? ©3.0x 10" mm = 3.0 B.0¥10% yam yur? C. 3.0 x 10 mm* 4. 55,000 km/s = pm/ns . 3 = A 55x10 ums _D. 5.5% 107 umnins (58 ooo Ba\f 10% Vl pe ( 10 6\. 55000 ft 000K || 1D A VT Hm = @ 55 «10° umins E. 5.5 x 10° wm/ns -6 C. 55 um/ns ¥, ib io ye LAER Each of the graphs shown here represents the VELOCITY for a different car as a function of time. (Each car is constrained to move only along the x-axis.) 5. (1 pt.) Which of the cars has/have a constant speed during the entire time period shown? aonly Da&b Bont baf rep of coustant o> herizentel GQ only. . 3 mh (store . we 6. (1 pt.) Which of the cars has/have a constant acceleration (including zero) during te or time period shown? A. aonly Da&b accel = slope of ve vs. B bony Danas 4. Constant a => graph coil Neastant ¢ lope C. fonly Coiewigee (ue) 7. (2 pts.) Which of the cars has/have negative acceleration during the entire time period shown? A. e only D. de &f = rey B. fonly E. none aceel = clope of v us. t graph Ores ie negate a > grap wth negate Slope 8. (2 pts.) Which of the Ove slowing down during the entire time period shown? ‘A. dont é only ey, Flowing down 1 gets, clesee 10 2600 closer to zero B. eonly Cdke at all trues. Visiting the circus... A “human cannonball” acrobat (m= 55 kg) is launched straight up, starting from ground level. (Ignore air resistance.) The cannonballer wants to rise to a height of exactly 15.0 m, so that he can grab onto a mid-air platform at the very peak of his arc, 9. (2 pts.) What initial velocity should the cannonballer use at launch? yf a Tv. * + 2 aA-& A. 13.3 m/s D. 19.5 mis ° B. 15.7 mis E, 21.9 mis oF peat f) = Vit 2 (4) by Ori ms > Vit 72- ‘444 22 (4% 40% )(i600) 10. (2 pts.) How much time does it take the cannonballer to ascend? ELTA ls 7s D.2.7s : . 2.08 E. 3.15 suetat C. 24s at peek: O =U, +g JE > £2 Ve. (TI = L152 axe he 11. (2 pts.) Af! of the following quantities are constant during flight (from just after launch wen just {before grabbing onto the platform) EXCEPT which one? A. weight of cannonballer # m-4, * Consteuct B. net force acting on cannonballer = 4-9 * Constance C. acceleration of cannonballer = + comsfout momentum of cannonballer = wt.47 ap decreases yr total mechanical energy of cannonballer — t 2 Zhe slimy tmgy = . K Uae 5 q y cou Sfanct A 45-gram air-hockey puck glides frictionlessly on an air-hockey table. Before being struck by a player’s paddle, the puck’s speed is 0.55 m/s to the left, As it is hit, the puck receives a total impulse of 0.15 kg-m/s to the right. {Assume that the puck moves only along the one-dimensional x-axis.) L Torpelee Let) be fo right. 2A. (2 pts.) What is the puck’s velocity (to the right) after the impact? ap “ te . r ‘ A. 1.1 mis . 2.3 mis bos wYy - Ay; B. 1.5 m/s 2.8 m/s teg he C. 1.9 mvs ONS TSE = (0.045 by) Ve (00% \-0.55% > Uy = 2,16 Be 25, (2 pis.) If the paddle and puck are in contact for 2.0 ms, what is the average force that acts on the puck? A 75N D. 7500 N og kay Orn E. 75,000 N FE, = aE 2 ts Ig = 15 N . 750N 0.0020 & =—_— 26.(1 pt.) While the paddle strikes the puck, the paddle experiences a force that is... &. weaker than the force felt by the puck , . equal in strength to the force felt by the puck 4— Newiemt 2 (aw: per of fores ave equol (woquctade) . stronger than the force felt by the puck TT and opposite (aivection) D. zero E, cannot determine from the information given An ice skater starts spinning (frictionlessly) at an angular speed of w). As she pulls her arms and legs tightly against her body, her angular speed gradually increases to 3u. 27, (2 pts.) The skater’s final period of rotation was equal to times her initial period. Ag D. 3 9 Zn T, an 1 =. oo @: E9 T A ee C1 Th 3a 3 wo =—= 28, (2 pts.) The skater’s final moment of inertia was equal to times her initial moment of inertia, AS D.3 fons of be Lee li As lar Momtntim: 1 ©; E9 “ Tew, 7 T,-e} ci >Hi wo. We =. = = = { . . . ate Bwo 3 29, (2 pts.) Which one of the following statements is TRUE about the skater? SS A, Both her angular momentum and her rotational kinetic energy increased. B. Her angular momentum increased, but her rotational kinetic energy remained constant. kK = a wr Her angular momentum remained constant, but her rotational kinetic energy increased. ret 2 . Both her angular momentum and her rotational kinetic energy remained constant. > Ke . dnuds . Ke =— longer . 30. (2 pts.) If the skater’s angular speed increased uniformly from 6.0 rad/s to 18.0 rad/s over a span of 7.0 s, what was her angular acceleration? A. 0.32 radis? 17 radis? ke, = 18.0% - 6.0% 4 B. 0.85 rad/s 2.1 radis* Fl, 7.05 s* C, 1.2 rad/s* 31. (2 pts.) During this 7.0 s of acceleration, how many revolutions did the skater make? A. 8.3 rev D. 31 rev z 2 Ope E, 38 rev oe 2 wo +2008 oot (i8.0%)"= (4008S + 2-(L1%) 06 18 rev > 40 = 84.0 cad (es = 1B4 coy. An rod Returning ta the circus... 32. (2 pts.) Suppose that 14 clowns, each averaging 75 kg, all squeeze into a circus car whose interior volume is 5500 L. a is the average density of the packed clowns? 190 kg/m? D. 610 kg/m’? m f4-(95 ka B. 310 kg/m? E. 740 kg/m? * (1545) = {4 4 C. 480 kg/m? Vv 5500 (#4) —_—** 33. (1 pt. Would a chunk of pure “packed clowns” (from the Previous, question) float in water? (Pyaer = 1 g/cm’) yes B. no Flontma, condition: coms < [Pwetee OU Gey EA yt, $0 Obs __ 34. (3 pts.) During a particular balancing act, al! eight of the Amazing Octuplets climb atop a single chair, and the chair stands on a single leg on the ground. If each of the eight acrobats has a mass of 50. kg, and the bottom of the chair leg has an area of 4.0 cm’, then what is the pressure beneath the leg of the chair? (Ignore the chair’s mass.) A, 38 atm 97 atm B. 52 atm Ora, Fol me. G-5bkq (4802) C. 77 atm P= A A (howY fey < 6 of Latu . P= 28tO & Late = 46. 7 atm. 35. (1 pt.) During this spectacle under the Big Top tent, a moderate steady breeze begins to blow outside the tent, but the aipinside the tent remains still. How would you expect the soft roof of the tent to behave? =. ___, Y70 6 the roof will bulge upward/outward se Peli? > B. there will be no change (other than a little random motion of the tent material) . . wee F C. the roof will bulge downward/inward : veo upward "lo Caleta on root 36. (2 pts.) While watching the spectacle, you drink 750 g of chilled water at 10°C. How much heat must your body give to the water to warm it up to 37°C (body temperature)? Note: Cao, = 1000 cal/(kg-K) = 4186 J/(kg:K} A. 1.2 kcal D. 13 kcal Tez ° "227% 2 BT = BIC- 10% 227°C B. 3.0 kcal ©) 2 kcal Qe meg dT 227 Ke. C. 75 keal = (0. 15044 (1000 fe (27 K) ~~ = 20,250 cal 37. (1 pt.) Next to you, your little cousin is slurping down some shavé-1ce: assume that he eats 750 g of solid ice at a temperature of -5.0°C. His body will also warm it all the way up to liquid water at 37°C. In addition to the heat you calculated in the previous question, your cousin's body will also need to add all of the following amounts of heat EXCEPT: A. heat of fusion = meat “Coranitien from solid —» Wrquid) yes. ® heat of vaporization = bei li (te utd > vapor) “no, . increasing temperature of solid ice from -5°C to 0°C -yes. D. increasing temperature of liquid water from 0°C to 10°C “yes . 38. (1 pt. As the ice melts, its entropy... 2 Q 9 increases 4 $ . 7 - As tue melts , it absorbs Q: m Le . remains unchanged oy Cheat of Fusion) wh - C. decreases ) the Te 273K (eonstenct ) 40 48 i positive => entopy meresses. wire A: - ° wire B: Two high-wire acrobats are walking along different high wires: both wire A and wire 8 have the same length L and same tension force, but wire A is made of a relatively thin, lightweight cable, while wire B is thicker and heavier. Both have fixed ends. As the acrobats shuffle their feet along the wires, the wires begin to vibrate up and down, exciting standing-wave modes along the wires. (Ignore any effect of the acrobats’ own weight or position.) 39. (2 pts.) If L= 15 m, what is the wavelength of the fundamental standing-wave mode in both wires? A. 38m 30.m yy) . \ B. 75m E. 60.m <— By imspection, Cc. 15m Ben eee” Rok 22(5m)= 30m wy, Fy (m/L)¢ “up 40. (2 pts.) Which one of the following is TRUE about the fundamentals of the two wires? Recall: v= Wire A has a higher-frequency fundamental than wire B. vv Y Wire B has a higher-frequency fundamental than wire A. Wer 60 Fe C. Both wires have the same fundamental frequency, » Wire B has larger j2 Han wire A. (Lond Fy are same for £,1f%. both. 7 Be 41. (i pt.) The standing-wave patterns that oscillate along these wires are... erie 4. longitudinal waves (8) transverse waves ~ diplacement ef wire % pecpendialor to ar of waves. The following two questions pertain only to Wite A: 42. (2 pts.) If you observe the wire’s fundamental to have a frequency of 2.0 Hz, what is the speed of the waves along the wire? x. question #37. A. 3.8 mis D. 30. m/s r £ zy ee v= (0.m)(2.0H2) B. 7.5 m/s re) 60. m/s Hat c. 15 m/s = 60." —— 43, (2 pts.) Which one of the following is a frequency of one of the wire’s overtones? A. 1.0 Hz 10. Hz Fr: of 4 B, 2.5 Hz LAS Hz requency =y- =f2.0He Oe GOH: C. 5.0 Hz Ww overtone S40 f, = ED %, f $0 Ha, 10.0 He, ... $ 44. (2 pts.) While watching a 52-kg trapeze artist high overhead, you see that she swings back and forth at the end of a long swing, completing one cycle every 5.3 s. (Ignore air resistance. Assume that she can be approximated as an ideal pendulum.) How longesfher swing’s cables? A. 5.0 D. 6.5 : L -a/L: me \ [53 8\~ B 55m Orn Tex 2 Leg z) «(as (22) C. 6.0m = OFT] an 45, (1 pt.) While in mid-swing, she catches an equal-mass team member, and the two of them form a tight ball (m = 104 kg) at the end of the swing. How does thjg affect the period of her swings? A. new period is longer period is unchanged C. new period is shorter => doe, oT Aepend om mass m of perde tam bob f
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