Sum of Impulses - Mechanics - Solved Past Paper - Mechanics - Solved Past Paper, Exams of Mechanics

Download Sum of Impulses - Mechanics - Solved Past Paper - Mechanics - Solved Past Paper and more Exams Mechanics in PDF only on Docsity! Physics 151 Roster vo; SOLUTIONS Due: Friday, April 24, 2009 - Score: Part A 95 pts. (+4 ply. Bonus) Take-Home Midterm Exam #3, Part A Total: 1S pts. NO exam time Simit. Calculator required. All books and notes are allowed, and you may obtain help from others. Complete all of Part A AND Part B. For multiple-choice questions, circle the letter of the one best answer (unless more than one answer is asked for), For fill-in-the-blank and multiple-choice questions, you do NOT need to show your work Show your work on all free-rcsponse questions. Be sure to use proper units and significant figures in your final answers. Ignore friction and air resistance in all problems, unless told otherwise. Physical constants: Ws an open-book test, so you can look them up in your textbook! Usefirl conversions: It’s an open-book test, so you can look them up in your textbook! L. A three-part rocket begins intact as a single object in distant outer space, traveling to the right at 4.0 km/s. The first “stage” (#2, = 950,000 kg) explodes away from the rear of the rocket, with an unknown final velocity (¥,)). Later, the second stage (my = 550,000 kg) explodes away from the rear, with a final velocity to the right at 6.0 km/s. ‘The third stage (mn, ,000 kg) ends up with a final velocity to the right of 13.0 km/s. (All three masses move only along the x-axis. Ignore gravity throughout this problem. Assume that all parts of the racket have constant masses.) For ALL answers to this problem, use positive values for “to the right,” and negative values for “to the left” : v,=4+4.0 km/s L vis? Vor = +6,0 kin/s Va = 413.0 kin/s km fem, - — “ff Fee a. (3 pts.) What is the final velocity v,, of the first stage? 0.5 Bee Conservation of Mementunt Ep cir = Ep tral Cita tg) a7 = mi Vig HHL Vig +My Vee ( 150,000 ka + 50,000 4 260 cc0l4Y4.0 *2) = (9Se,ceoUy) vp + (sr9p0ols Yt.ohs ) + fsx ceoty) (.0'g) o . TA x10 ges > Gstpe0 4) vig 4 330" bybee + 56 vero gk F tag. bok Poeniveley, lone ob one HO XI HE Rsneco's) Vee Sig ity. here due to Sebtrectt vig 7 O47 SE ; -4.3 410 © bg-bw b. (2 pts.) Kind the impulse received by the first rocket SHEETS ES losion): * s Supulse: 4p, (Pe- Pas . = 450,000 "ke —3 gop eno'g-e =-4 250,000 tyke a u " = MU MV = (450,00014\(~0.47 4) — (450.000 40%) b “ bes c.(2 pts) Find the total oe received by the second men {after both explosions): Le X10 te Ap (pe ~ Pe M2 V2¢ - MV (Peee44 Yeo te’) - (550,000 )(4.0%8) = 3,300,000 ke -2,200,060 tpn = 1,100,000 legen [eet a Ww 6 bye d. (2 pis.) Find the total impulse reccived by the third rocket stage (after both explosions): 3.2 % iQ & ARs (Pe ~ Pde = Mg Vag - Ms V5; = (350,000 aX 9.088) - (350,0c0lg)(4.0 s 2 4,550,000 'B ~ 1,400,000'%FE" = 3,150,000 ¢. (1 pt.) Find the sum of these three impulses: oO Hint: According to conservation of momentum, what should the sum of all impulses be equal to? Coutersaton ZprEpe > (49) OD. Tas agrees wits toe au a te 20 of momentum above values: BP. + AP2 +6Ps : f.(1 pt.) During the second explosigg, the force acting on m, was the force acting on m;. A. less than oO the same as C. greater than Auton’ Law: he impulbive force of my en uy 1s equal aud oppotite to the Goun [taneous Tax peelive. ferre of WMzoK m,. 2. (1 pl.) After both explosions, the total kinetic energy of the system... increased B. decreased C. remained unchanged ak=, -2k: : zat 2,1 zl L > [bmn + oe + oy? 5 Cn ty tg) V; =| 5leseamolg)(-aTy + Slice cog Veeco’ £ goo byNinco0 ¢)°] - zlysse cory | 4oc03) U1. 07 mo" T + 44x07 T+ 246K0" T]- LHSKOPT aks 249 x10"? J + Aku positive, oo Ky, increases . leg “hd pl) After both explosions, what is the final velocity of the center-of-mass of the three parts? 4.0 % Belere ony explosions, dhe rocker’ View = 140. / Smce romentim 1 Conserved, EP suit * Zz Perel =” (M Ven); > (NVia 2\ Vi) (Ven): soe : 2 MMe t MU, + WGy, 450,200 v5 (+ 0.474 &) + (580,060) t s oe: (Vez 1 L 3 Use =f vs f) +( Me trom) Wart My 48t,000 kg + SG50,000 las Bez, " ? oo, > (Vou) TH00,000 44:8, ore coaeg = 40 # % t 2. A simple Jever gives the user a méhanicafadvantage: pushing down with a small force F,, on the long end of the lever creates a large upward force &,,, on the short end. uppose thar you want lo support mass M al rest by using a lever of lolal length L whose fulcrum is located at a distance d from the short end of the lever. (Assume that the mass of the lever itself is negligibly small.) a, (2 pis.) What is the mechanical advantage (7, + /,) of this lever? Express your answer ONLY in terms of L,d,and 6 fever and M are. at vest ; ok TSE, fleen: tnathematical constants: L-d Ereo. —-d—_—— Tu - Tp ce (cow postive, CW negative) ee ee Fu Oe oct zO fed - Fi. {L-4) =O > fu, bd. 3900 5 or 3.40210" 4 £. (3 pts.) Calculate Mars’s density, including MKS units: Volume of 47,4007 2 Ha y= AUK (Onn? ephect VY anRk,, 3" 3. 4O KO mn L GHEXIO_m U2 deusity pur Me 2 BHP be = 3.900 «0° He Vv. {64x10 w3 _— am g. (1 pt) A typical density for rock is ~2500 kg/m’, while iron (the most common melal in the solar system) has a density of ~9000 kg/m’, We can conclude that Mars’s interior mass consists, very roughly, of a mixture of... A. 99% rock and 1% metal 2 P= 2600 4Au? 80% rock and 20% metal > ’ 2 3300 ". 50% rock and 50% metal pp x F8OO D. 20% rock and 80% metal > p * 7T70D E, 1% rock and 99% metal = PX FVOO beg /,3 4. A water polo ball (which looks fike a waterproof yellow volleyball) is a hollow sphere of radius 11.1. cm and mass 450. grams. Assume that the density of water is 1.000 g/cm’. -3 3 a. (2 pts.) Find the volume of the ball: 5.73 ¥10_w oc 5.73 L 4% cf 3 Sphere + Vetnr = Ea(o.utm) 25,129 x10 b.(1 pt) While floating, what mass of water does the ball displace? = a _O4SO kes c-(2 pts.) While floating, what volume of water does the ball displace? O.4SO Ly oc ue cm? {b Arciwedes principle: A Hleadove object displaces a wasy of Fluid or HSOX10 “4 ya equal tothe obieds mass = HSV. Cc) Le 450, 4 _ 3 NU Ww 1B00 A/a 50. em oe: Elon te => Fe wb AND? M,, wv Va wé © A Non =m nag Vas * JL , & a (2) seag m me Vac Le je00 het © 150, eu? mC d. (3 pts.) Suppose that a polo player starts ering downward on the floating floating ball, gradually increasing his force. What is the magnitude of the athlete’s downward force when the ball has cxactly half of its volume submerged below the surface? 23.1 N = o seek og . Fp moyen) ZFy = mG «equilibrium shod ic Feo Fy FF, = © ~ > R= Fe, 2 = pe Vang Meg, (attclete) = (1000 eC 440 “el (e340 2.3)] ~ fo.us0 ly )(280%h) Z48.0TN ~ Hai N = 23.66N 3 it = &  e. (3 pls.) Later, the polo player completely submerges the ball, pushing ita short depth below the surface of the water. Then he releases the ball from rest, underwater, /mmediately aflerward, what is the ball’s upward 1s “% acceleration? (Neglect any dragforee.) _.€@ © ary = MOy Fa- Fa = mM ay Vaso = Was fo Nag a a = Way [:1o004..)(4.20%)(s.15 910 oS] - [oso bg) asonle) = (oso) “ay SEIN N= HHI N= (ogo ky) -ay 51.73 N = (0.4509) ay » ay = HG.0 ee Suppose (he water polo ball has an internal pressure (absolute pressure, not “gauge pressure”) of 185 kPa. 2 f.(2 pls.) Find the surface area of the ball: 0. (S> m aphere: Ae dn” = 4 (om) = OATHS mm g. (2 pis.) While the ball sils at rest in 1.00-atm air, what is the net outward force acting on the inside of the ball? 130*10'N Fes = Feat - Foyt = Ba A - Bora’ A (his is almost 3000 Lost) Fn 4.00 atm ro z (1.260105 Pe )(0.154¢ 2) — (1.013 «10 Pa) (0.154) = a = 4 = 2.964 x04 N - f.s6exi0'N = 1,246 ¥10'N h. (3 pts.) AU what depth in a swimming pool would the absolute pressure on the outside of the ball Pent 185 kPa. causing the balll to start to collapse? 3.5m (this is approx, 28 Feet deep.) (Assume that the pool is focated near sea level, so thal there is 1.00 aim of atmospheric pressure at the water’s surface. Your answer will be deeper than most swimming pools... but not by much!) vr Punter + Tae tor Pree 1.8510" Fa W iy ‘ate (100, '§,)(490%) h t+ Lome B B.32110'p. = (1060. "%,)(4.G0%)-h peak + Bow (Puc = 100 ofm = Lorsnc*t. ) nw /& he 3.54 Note: (0% dh sig fi. due fo subtraction. Physics 151 Roster No DOLAATIONS Spring 2009 Score:_Yart B: 20 pi: pesseble Take-Home Midterm Exam #3, Part B L. A bowling ball (solid, uniform-density sphere of mass M and radius R) rolls without slipping toward a hill of maximum height /7 and varying slope. a. (5 pts.) If the howling bal! starts with lincar specd v, at the bottom of the hill, what is its linear speed as it rounds the crest of the hill? Show your work completely. Express your final answer algebraically ONLY in terms of “known” variables M, R, H, Vo, , and any necessary numerical constants (but NOT @ or any other variables!), Simplify your final answer as much as possible! cy Coservation of Energy t 7 Vy E; = E, (Kees t Kay + Uy.) = ( Kenous + eet Use). “7 H t 1 oO oMy,*+! ri ela, 2 L 3 te ‘Myf shy +50 e + Mg LMyi 2 tb 2 Nap NL Rolla withost -x | 2 gM =o Mul + 5 Tle uxf) - MgH- "8 Say: c ee 2 L » Sokd Spl Te% > POE ae ABP” phere: Me 2 gubstitude to eltumote TL and ao. b. (1 pt.) Suppose someone makes a “hollow” bowling ball whosc center is mostly empty, but whose outer edge is loaded with dense metal. Overall, it has the same mass M and outer radius R& as a standard bowling ball. If you roll the Aaliow bowling ball with the same initial lincar speed vy toward the same hill H, the hollow ball "Or the top of the hill with linear speed than the standard bowling baif. eee Hotlow sphere: 2 ZMp™ (iustead of soled ophere: a) B. less C. the same Same method as above gives: 6 v, = [vir- Egham hollow 2 eS <[eobe ee > [ud S Vicitow