{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Exam 2 - Problem 1(25 Points Below is an incomplete VPython...

Info icon This preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 1 (25 Points) Below is an incomplete VPython script that models the interaction between two charged particles (one negatively and the other positively charged). The only interaction between the two particles is the electric force. Follow the directions below to insert code to: (a) find the force, (b) update the momentum and (c) update the position of the negatively charged particle. from __future__ import division from visual import * ## Create objects plusparticle = sphere(pos=vector(3,4,0), radius=0.5, color=color.red) minusparticle = sphere(pos=vector(-8,-4,0), radius=0.35, color=color.blue) ## Velocity of objects vplusparticle = vector(0,0,0) ## Velocity of plus charge vminusparticle = vector(0.4,0,0)## Velocity of minus charge ## Constants and other values oofpez = 9e9 ## Electric constant mplusparticle = 5e-3 ## Mass of plus charge qplusparticle = 1e~6 ## Charge of plus charge mminusparticle = 3.5e-3 ## Mass of minus charge qminusparticle = -1e-6 ## Charge of minus charge pplusparticle = mplusparticle*vplusparticle ## Momentum of plus charge pminusparticle = mminusparticle*vminusparticle ## Momentum of minus charge deltat = 1e—4 ## Time step t = 0 while t<20: ## (a 12pts) Add statements to calculate the force on the negative particle r 1 m‘muspmtic‘iepos - p\u5par+'\cle. pos rmofi m stiff (r.x**2. t ng‘mzw‘ rsfiflh #29; magbfi that = r / rmafi #fl normbr) F net = oow‘pel 9“ qypbspmtsc‘xef“ (iMUXQSPQPiEC-‘lg 9k final: /er3* at: Z ## (b 5pts) Add statements to update the momentum of the negative particle pm‘mus particle, = pm'mus‘neriida t Fmeta‘i deified” #95, pm‘musperticie +2“ Feet ’4‘ defied” ## (c 8pts) Add statements to update the position of the negative particle m'muspotrtxcie‘pos *1 m‘anSPQrficie.pos + pmimsparfigielkdaer‘ET/mm‘muspariiclfi 1% fl “Y‘U5Pwiide-e05 Jr: Pm\m5§mrt\cla* daiicfi /mm'muspar~ilcle_ 1: = t+deltat Problem 2 (25 Points) You are a visitor to the planet Mars and decide to repeat an experiment you did on Earth. Back on earth you know that when you kick the ball so that its initial velocity is 45 degrees from the ground, its maximum horizontal displacement is 37 meters. (a 15pts) Ignoring air resistance, what must be the initial speed of the ball back on Earth? . \ M“ Moe/M Hammered» D\S"JYO~V\QE, firm =~ \lx Am (3mm (“X—.2 O» WM“WW.MWM~MW-WMJ V‘W‘i’fi : M5 /A+High+ ‘3 “ark Q“ ground and ends on ground =5 Ara: O. \J‘N‘M‘l = O f— Ji(\‘Jiu5+\i-F.\Q “J‘W‘Cfi /Enei: firm : consiufi‘i’ V' *3 'V4Ig AP“: = Fmicij Atrigahi (Memen‘ium Pr‘mg‘igfla) m Aw‘j 2 ’ m a A): r\“g\n.‘% (M0653 QfififlaX-E) ) <8 = (“ls-,1) 1:"- : (\j‘l\\.8' a 145° means s‘mce, 1105(45‘”): em (45"): ("S/Z Comb'me d\\ Jrloree. circled QCBUQ¥©U$Z H '5‘7m /, ZVi/ES if: 2.\I;:2g Then \ix fl' iswch/g ‘3 l3.L‘l(om/5 Dee, either “viii: \1i7&*\lfi‘5 or \JX=HQ; (103045") “to and (b 5pts) N ow on Mars, you kick the soccer ball with the same initial velocity (speed and angle/ direction) it had on Earth. You observe that it travels 97.7 In in the horizontal direction before it touches the ground. This close to the surface of Mars, the force of gravity is approximately constant. Calculate the acceleration of the soccer ball on Mars (1. e. Mars’ gravitational constant gmars). Ignore air resistance. Lees osa the. regain” groom and: 3:} since. am {swig diger’efiae is aagmfifé ., A W, 2. Moot, Harri. Eisiamna W 'ZNx ll‘amms The“ QMars =1 2V:- /ol'l.7m and use. \iX i:er (c 5pts) If the radius of Mars is known to be 3.396 X 106 m use your result from part b to determine the mass of Mars. If you were unable to numerically calculate a number to part b, use the variable gmars in your answer. “ _— “‘ ~ Mm r ii Fine}; Farm! N W t: m ngs Mars % ... G M \"z %Ma¢s ' _T“fl‘_5 “3-) MMMS-z: gmqr's Mam rMNS 6’ ll Problem 3 (25 Points) One mole of titanium has a mass of 48 grams and a density of 4.54 grams per cubic centimeter. You have a long thin bar of titanium, 2.9 m long, with a square cross section, 0.08 cm on a side. You hang the rod vertically and attach a 41 kg mass to the bottom. You then observe that the bar becomes 1.52 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in titanium. (a 5pts) What is the spring stiffness ks of the entire bar, considered as a single macroscopic (large scale) spring? “amnesia =3» w “amt/m o M “A __. “c A“: I: O =% Ffiei w Q 2% - Vspr'mg 3 FEer Fjprlnfl 1 “mg ks * mS/M = flggwfgimls‘l = 2.60% M04 N/mi (b 5pts) What is the center—to—center (interatomic) distance between titanium atoms? Densi'i‘j = P z: Mafia =7» Was-Mafia “C immimfi Maw-ma“ \Iswma Waiter 1 Mum 43 -3 = 2.60 9H0 cm a = W ; < Ha almi. Ya newton/max: 45% film? = 2.CQOX\(3\Om i (c 5pts) What is the stiffness [63¢ of a single interatomic “spring”? k ' .4 sti’eBonds: ks“ 2 k3 Ld 5“ sending (‘59) A d2. a; (Z-MSMOH “Iszfl m}(2.eox:6‘0 m) N (Q-QB X 10‘?“ my?" 95 PM Y em deemiam and k5‘12Ydr-(tmsmd‘ifl (mom =-— 31.1% “An (d 10pts) The 41 kg mass is removed from the first rod and attached to a second bar of titanium. This bar is 1.2 m long and has a square cross section 0.04 cm on a side. If you hang the second rod vertically (with the 41 kg mass attached to the bottom) how much does this new rod stretch? Shae. W‘s mm "‘t‘v‘com‘mm) k5; is sh“ 4m same, K5 2 k; g x archakng 2 km A :(31-‘N “’"0 (O-Qootimf‘ #Eonds LC} (X.Zm3(1.<ooxsémm ks £1.59“? xaoH N/m 1 u ESpring“ 1: S \SE a?” ‘8‘ z m a z; <41i<33(9-8m/§‘\) m k5 i-59‘7xioqN/m Problem 4 (25 Points) A helicopter flies to the right (in the +x direction) at a constant speed of 20 m/s, parallel to the surface of the ocean. A 1900 kg package of supplies is suspended from the helicopter by a metal cable. The package also travels in a straight line at a constant 20 m/s to the right, as shown in the figure. (a 6pts) Draw a force diagram showing all of the forces acting on the package. Label each force with the name of the object exerting the force. You should scale the length of your force arrows so that the relative lengths are consistent. :fll our i 11m l \x l~~~-l 4.5 m x L Fade“. Fon'l’n (b 10pts) Compute the magnitude of the force exerted on the package by the cable. an: M3 —.= ‘Q‘ :5 Ffigtm Q A\\ iorgefi WUS‘l "z’mum *0 E‘irm' 3” Cll VELQJO‘QH . LL a, ngowahusé “l: mwim... -_—.. 0.9m -«l122+q.52 Or use e=are£an(H.S/Hl= 22.256 go‘NE’. $1M“ Fcabta: FQQlD\ 9. =1 (\‘iookqhici r8 “1&3 __.‘ cosflzziS") = 0.92% 0326 (c 9pts) After some time the package breaks away from the cable and soon is traveling with a velocity of ifPackage = (5, —15,0) m/s. Suddenly the package explodes into two pieces. One of the pieces has a mass of 771] = 1800 kg and a velocity 171 = (10, 2, 0) m/s. What is the velocity of the second piece immediately after the explosion? Ssgs’rem : A“ Masees gurroending = Ear—H1 Fhefl’, e‘fi‘rar 00A, 1: FEQA" firh APsgs 2' fixfilzxkema‘ [Eli 9mm we‘re, 6X53 \m’i abmfi ‘mmadlfifirelvj (NR-err Jrhe exgfloalon take. {0: l5me all: expigglmfl and A4: =1 O. A? 533 ""1 Q dorm?) axplmslzgm swag :2 Mmfix K71 : (Moog) <5,*!S,O>m/g ml fific‘sasa mm)“ + mzfizmc :2 (\800k3)<\o‘zlo>mjs +(‘qook3»agmgkg\)vm WW . __i 1 mg; Th6.“ \IzfiF MOO <5fi’l510>“i893<\t0.210>3k3‘m/ kg 5 =l<—85)-321)o>m/5 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern