Average Velocity - Mechanics - Solved Past Paper, Exams of Mechanics

Download Average Velocity - Mechanics - Solved Past Paper and more Exams Mechanics in PDF only on Docsity! SOLUTIONS Score: Port A 20 pts possi ble. Entre Craw: SO pis. total (+24. Exam time limit: 50 minutes. You may use a calculator and both sides of ONE sheet of notes, handwritten Bonus) only, Closed book; no collaboration. Ignore friction and air resistance in all problems, unless told otherwise. Physics 151 Roster No. February 16, 2007 Midterm Exam #1, Part A Part A: For each question, fill in the letter of the one best answer on your bubble answer sheet. Physical constants: g=9.80 m/s? (2 pts. each) Convert the following quantities into the given units: 1. 550nm = m om) al A. 55x10 m D. 55x 10m 550 awe Tp J * FF RIO 5.5 x 107 m E. 5.5x10'm C. 5.5.x 10° m 2.13% 10% kg = ng -t 1080 I A. 13% 10% ng D. 13x10 ug beet J i ae = 13 B. 0.013 ug z KS . =F 3x 10" pe 13 pg 3. 4.4m = cm? 0 2 ry 2 A. 4x 10 en? © 44x lo%em? 44 | 100 ew = 44 KIO tm B. 0.044 cm? E. 4.4 x 10° cm? Lye C. 440 cm* You know that your friend’s car starts at UH Manoa at 1:30 p.m. and finally arrives at Aloha Tower at 2:15 p.m., but you don’t know the route that your friend took. (It is not necessarily the shortest or most efficient route, and il may have included stops, detours, or side trips.) The straight-line distance (“as the bird flies”) between UH Manoa and Aloha Tower is 4.8 km, but the car’s final odometer reading shows thal the car traveled a total distance of 8.5 km from start to finish. 4, (2 pis.) What is the magnitude of the car’s average velocity over the entire ee eat fom orig A. 5.5 km/h D. 9.1 km/h a bie @ 64m E. tl km/h ae DRO 48 km 64 es ou = = eS sl64 >. 7.7 kinfh at ors hk h 5. (2 pts.) What was the car’s average speed over the entire trip? A. 5.5 kavh D. 9.1 km/h B. 64 km/h Oiikmn wh: total dslance traveled — 8.5 Ion 1B es C. 7.7 kev BEERS, = at OIE a 6, (1 pt.) A sealar is a physical quantity which... A. never has a negative value B. never has a zero value “ . a . ; C. has only one digit Vectors, have. bet, wage tude, (vole e, iachuding unity) . has no units AW Atrectren: 6 has no direction ‘uP ) scalars have oly a value, but Mo directional sense. 2 Ps 2 rem tat, reker tt Lat® y*sv,7+ 2a0n 7d "6: 3 main equations of kinematics CANNOT be used for which one of the following cases? An object whose acceleration varies over time + tad sek over B. An object whose velocity varies over time These ous COMA be uw C. An object whose acceleration is zero at all times time tatervels dunng whith accelerate is D, An object with a force acting on it “At E. Aw object in freefall (near the surface of the Earth) C@USF Bee unchanging), $0 "A" is Not vain. Valid cases, Bt Coustant non-tere accel, “P chenging velocity! $0, bv /ot is ke Ci tmstant accelerntio lacludes a7 O (0 constant value of zero). Di touglant accel. » FF = couslant, so there can be one or more forces acting mm. Es freefall Ay = 3, & constont. 8.(1 pt.) Which one of the following is equal to a newton? [force] = [nave |: [acceleration | A. keg/s* D. kg-m?/s” kgs E. kgm? > {wh = [ig J. | 6 kg-m/s” F. scrumptious dried fruit with a delightful cookie coating 9. (1 pt.) Qn Earth, which one of the following statements about weight is TRUE? An object’s weight always has a magnitude of m-g — ey! Weight always points perpendicular to the surface of contact, - Ne, ways penets toward center of Earth. C. Valid units for weight include pounds (British) and kilograms (MKS system), = Alo, pounds and ngurtons. D. An object’s weight depends on the velocity of the object. — No, tad. ependitact of F, A 2350-kg motorboat is at rest. Standing on a nearby dock, its owner pulls with a constant horizontal force on a rope tied to the boat. As a result, the boat accclerates horizontally at a slow but constant 0.025 mis’. A 19. (2 pts.) What is the magnitude of the tension force in the rope? & > A. 41N D. 89N SE, = may fay E @ son E. 110N vx beaks po r Cc 71N Fyre mean —— a 2 (0.025"%,») 2 [F, = 63.6 0] ‘ F, 11. (2 pts.) Starting from rest, how much time will it take for the boat to travel 8.0 m? A. 198 D. 28s 42 Xe azte Lat® B. 22s E. 3Ls 2 2 Oo 38 $.0mn-O+ O + L(o02s%) + 2 |b ©2536 12. (2 pis.) Instead of “normal” force, the supportive force of the water acting upward on the boat is called the “buoyancy” force. If the boat ngver has any vertical motion, what is the magnitude of the buoyancy force? A, 240N 23,000 N Fy = A = As B. 9800 N F 180.00N 2 PY TM Ay C. 12,000 N F,- Fe * m{o) a Fy = ma = Garo by ) (4.802) 2? [F, = 23,030 N + RS ww a a 2, A small spacecraft is constrained to move only along the x-axis. It faces in the +x direction. The spacecraft has two rocket thrusters that can be used to accelerate it along the x-axis with either constant positive or constant negative acceleration. (Ignore friction/air resistance. Assume that all numerical values have af least 2 significant figures.) The spacecraft’s VELOCITY as a function of time is shown below: 1s time [s] velocity [m/s] 2 “5 -15 a7 Parts (0),(b), (2): -L pt. for each error, up to -Z pte. max. a. (2 pts.) During which time interval(s) is the spacecraft at rest? Circle ONE or MORE: A. 0107.58 D. 20s to 258 B. 7.50 15s E, 255 to 35s Niwtervals when we. C. 15810 20s ©3550405 b. (2 pts.) During which interval(s) is the spacecraft moving backwards (i.c., “in reverse”)? Circle ONE or MORE: Oo75s D. 20sto25s & , 7.5 sto 15s B. 25s to 35s matervele when VSO (vis negative) C. 15s to20s F. 35s to40s ¢. (2 pts.) During which interval(s) is the spacecraft slowing down (i.e., speed decreasing)? Circle ONE or MORE: (Note: This one can be tricky!) 00755 D. 20st025s N intervals when is getting closer te Zero, 75510158 25510355 15s to 20s . 35s tod0s d. (4 pts.) Carefully and accurately, graph the spacecraft’s ACCELERATION vs, time on the axes provided. below. (You do NOT need to show your work.) At alt mes, accelerotion 16 tHe slope of v(t) graph (show) For exe, seyment (whenever slope changes), re caletlate: Aw . ay = at’ For Skample : time[s} 4= © ts -t, : lon - (~ tors) (So -Os a=1O3%. Continued next page... acceleration [m/s*] ° 2 2. continued: ‘The spacecraft’s VELOCITY as a function of time is repeated here for your reference: 15 time |s] yelocity {m/s} BONUS (2 pis.): Calculate the spacecraft’ s{net displacement from its starting position)at t= 40 s. Show your work completely and/or explain your reasoning: bX + cumulative area of v(t) graphs (down above). Dwrdmg graph mato segues AF: Ze OKT Slbese-heinht), => (t54)(-104) «375m. OF gop = OKA t ort OX bys 4 (rase height), zs (ts) (10 F) = 375m. * 2-37.54 37. 5m BK. = Case); (merge bengal), = (55/1 (10% + 53) > BT.Sem $37.5m + 25m bx, 2 (base. hetg it), = (5s(5%) = 25m. +25m + O phe © § Coase. height), = $loss ss 25m. DX por = STS om big 7 Om. (ord Over intervals when ateeleration 16 Cousteuh, we tan apply! X= Xo +v5t +tot® = s 2 ao Qfet-O- 1582 Kye XorVelot) + Salet)’ = x, +(0§N 153)+3 (139s) © xX, —Is0% + [50m t lg 208+ 5 L re 7 = Xo. (Otte Me Mist Malet) eS lat) = (40) +(108) (5) +1 (1B) 58)" 2 Ky 4 50m las > = kot 315m, (D) t+ 20-4254 Kay * Brot Ty (ot) 4 5A (ob) =[K, 1375) + 6S Ysi)o1 (05) y Ke + BTS t25im 2 = X,.4+62.5m. (E) b= 25-2354: Kyo = Nay + Hp lOh) +i a,(64) = [Xgt 625m) +(5)lios) ot (-05 B Yio.) = Kot 625m + 50m -25a = kot BTLTm, (F) EF 35 7HOs? Kuo Kae + 1G, E) + aglot) = (x,t 815) + (5) + $(0)(S8) "= X 481 Sm. 2 Key § Xuo7 Xo = (Kot BL Sm)- XQ = (475m 0 Oe