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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 = 5e3 ## Mass of plus charge qplusparticle = 1e~6 ## Charge of plus charge mminusparticle = 3.5e3 ## Mass of minus charge qminusparticle = 1e6 ## 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‘U5Pwiidee05 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~MWWMJ 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+\iF.\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 QﬁfiﬂaXE) ) <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) MMMSz: 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» WasMaﬁa “C immimﬁ Mawma“
\Iswma Waiter 1 Mum 43 3
= 2.60 9H0 cm a = W ; < Ha almi. Ya newton/max: 45% ﬁlm? = 2.CQOX\(3\Om i (c 5pts) What is the stiffness [63¢ of a single interatomic “spring”? k ' .4 sti’eBonds: ks“ 2 k3 Ld
5“ sending (‘59) A
d2. a; (ZMSMOH “Iszﬂ m}(2.eox:6‘0 m) N (QQB X 10‘?“ my?" 95 PM Y em deemiam and k5‘12Ydr(tmsmd‘iﬂ (mom =— 31.1% “An (d 10pts) The 41 kg mass is removed from the ﬁrst rod and attached to a second bar of titanium. This
bar is 1.2 m long and has a square cross section 0.04 cm on a side. If you hang the second rod vertically
(with the 41 kg mass attached to the bottom) how much does this new rod stretch? Shae. W‘s mm "‘t‘v‘com‘mm) k5; is sh“ 4m same, K5 2 k; g x archakng 2 km A :(31‘N “’"0 (OQootimf‘
#Eonds LC} (X.Zm3(1.<ooxsémm ks £1.59“? xaoH N/m 1 u ESpring“ 1: S \SE a?” ‘8‘ z m a z; <41i<33(98m/§‘\) m k5 i59‘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 (NRerr
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 ...
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 Spring '07
 JUNGH.CHOI

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