This preview shows pages 1–9. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem 1 (25 Points) Two blocks with the same mass M slide on a low—friction surface and are connected by a spring of unknown
stiffness. Initially the blocks are moving to the left, each with speed 111, and the spring has its runstretched
length Lo. A lowmass string goes through a hole in the right—hand block and winds around a lowmass
spool in the left—hand block. You slow the blocks down (and compress the spring to a length L) by pulling
to the right with a constant force F. At a later instant both blocks have a smaller (unknown) speed 122 to
the left, and at that instant your hand has moved a distance d to the right from where you started pulling.
All known distances are shown on the diagram. ‘ (a lOpts) Determine the ﬁnal speed 112 in terms of the given quantities. USe. poKni— particle. sgstlem
o AESBS= Waxt +ﬁext AK,= FAEM = ~Fw isz (vﬁ v3) = — Fw (b 15pts) Determine the spring stiffness kg in terms of the given quantities. Go {0 red sgﬁi‘em (mocks +Spr'm3) o
AE535 =~ Waxt "baat AKb‘ocks + Auswins = F. Ar ~Fw + '7':K$(LL°)z= Fd r m‘ Problem 2 (25 Points) In 1909, Ernest Rutherford explained the structure of the nucleus. He did so by ﬁring helium nuclei (alpha
particles) at gold foil and measuring the angles at which alpha particles scattered after striking the foil (i.e.
Rutherford scattering). The alpha particle and the gold nucleus interact through the electric force. Fill in
the missing code below to compute the forces acting on each particle, update each particle’s momentum
and position to simulate Rutherford’s experiment. I I i from __future__ import division ## treat integers as real numbers in division from visual import * ## Constants and data b = 1e14 ## Impact parameter mAlpha = 4*1.67e27 ## Mass of the alpha particle in kg
mGold = 197*1.67e27 ## Mass of the gold nucleus in kg
qupha = 2*1.6e19 ## Charge of the alpha particle qGold = 79*1.6e19 ## Charge of the gold nucleus oofpez * 9e9 ## Objects \ Alpha = sphere(pos=(2e—13,b,0), radius=2e~15, color=color.cyan) Gold = sphere(pos=(0,0,0), radius=8e—15, color=color.yellow) ## Initial values pAlpha = vector(1.46e19,0,0) ## Change this to the initial momentum of the alpha particle
pGold = vector(0,0,0) deltat = 1e—23 t = O ## Calculation loop while t < 2e—20: ## (a 5pts) Calculate the force on the alpha particle exerted by the gold nucleus.
r = Alpha. p05  Go\A.pos
rmag = 310.30% p 4i? or (r,x**2_+r,\j*%2+r,a*~%2)*ako.5 I Falplnq = outpei: * 1A\pha.* CLGolcl'* r /rmo.3** 3 ## (b 5pts) Calculate the force on the gold nucleus exerted by the alpha particle. Fgolcl = " Falpha ## (c 5pts) Update the momenta of both the alpha particle and the gold nucleus. PA\p\'\o. *= Fo.\p\'\o. * Adiad
at ca 1 pA\p\\q = pA\p\1a + Fa.\p\\0t*de\\a+ p Gro\d += Fgok'} * deHod‘
#090 pGoH 7 pG—OH + Fgold * data"? ## (d 5pts) Calculate the parallel and perpendicular components of the
net force on the alpha particle PJm'd' = mo.<3(PA\P\noQ
4* Needs +0 be be¥ore momen*um update. For a\p\\o. pox“den “4* Now «Her momentum updot‘ve '
chxr o.\\e\ = ( mag<pMphd3 " P_\n'\‘\'3 5" norm (pA\p\\oA / de.\‘\'ar\— 4*=C)F{ ,1:\5*110«A c)? 4*‘e; odacv¢€L1 Fpara\\e.\ = dot < Folpha , norm(PA\phd))*norm(pA\pth
F perpendkdqr = Fo.\p\'m —' Fpard\\e\ ## (e 5pts) Update the positions of both the alpha particle and the gold nucleus. Alpha. pos = A\pho.. pos + p Mth * de\\o:\‘ /m A\p\'\d. n .
#OR use ".+= no+a‘\'\on Go\d. pos =’Gold’. pos + peola *deHod' / mGoH \ 
#OR use. “+=‘ no+a+xon ## update time
t=t+deltat Problem 3 (25 Points) In outer space, far from other objects, a small block of mass m traveling at speed 121 approaches an object
initially at rest that consists of two small spheres each of mass M connected by a lowmass rod, with
centerto—center distance L. The block is observed to bounce off the upper sphere and proceed with speed
112 at an angle to the original direction. (a 10pts) Just after the collision determine the velocity (ow, vy) of the center of the rod. §gsiem " Blockp“ Barbe.“ Surrounding ' N eghgiHe. Apsgs = Fined“ $535.4: = mvz + ‘2M<\J*x\13> 3 V1: V2 <c°$(5), Sln(e), 0) > <vx.va> = ~zTA<V“—‘723 {\Jx N3) = 7237\ <v‘— \chos'(e)’vzs\n(e)> (b 15pts) but after the collision determine the angular speed 0.) of the rod. —. .—\ — N AL535 = lynchext At = O a LSSS'L = L535"? MeaSure. L m\'\'\'\ rasped' *0 CM 0'? barbe.“ q‘\. ‘\:\me. o? Co\\‘$'\on. '— . ~J—K .5 A
L $35.3. Lirans, blockﬁ’ r x PUMb‘ockﬁ. l? xﬁcM.b\ock\ = rJ mV‘ = dm‘lu
A
“shag rRSh'\'\\o.na ru\e, Hue. ereQ+ion is ‘Z (‘m‘\'o page)
A L, 535.1,: ~di, 2. ——I § —. —& ——A ._
= + =
L’ $35,; Lo block; Lad; Lfrans, Mka + L trans, rod, ; + Lroi, rodiF __L 3
= ﬂ POMyrodﬁ + r2 x PCM.b\ock.ﬁ 4’ Iraqi. 60663;;
r2 = < ‘ILz/q ‘ <52 72 d > Wad Fembmkx— = m <V2C°5<53Nz$‘m(93> .4 —\ z ’ ‘ ' ‘ A'
‘3" Pcn‘bbck‘? = LIL/q dz Sm (e)  dcosceﬂ mvz a ' A .——L ‘ ,. . I
03m; 4: = I d ‘dh'N. "JENdz s\n(93mvz+dcos(e)mvz] z Irod = M“ eff} NW???" “‘7: MC. ._\ . A ‘ We. exped and; 4:0 be. m 2 dwec‘non. Then ‘H'xe. magm'tude. is amen b3: 'Lm ‘ ‘0; = M L} LN. " m S'n(9Wz”‘ dc°sceW11 Problem 4 (25 Points) Two blocks with masses m1 and m2
(where m1 > m2) are tied to disk by
strings of negligible mass as shown in '
the ﬁgure. The disk is supported by an
axle with frictionless bearings so that
it is free to rotate clockwise or coun—
terclockwise. The disk is solid and
uniform with mass M and radius R.
Initially the disk and the two masses
are held motionless so that the points
where the string is tied to the disk is
in line with the axle and the blocks are
a distance h below (see the ﬁgure for
clariﬁcation) . (a 10pts) The disk and the blocksare released from rest. At this instant, what is the total torque exerted
on the disk with respect to a point located at the center of the disk? Give both the magnitude and the
direction of this torque. ' ‘ "End: = R (T. rm 1% v '1 ext‘ForceS 36‘
Now we need +0 ﬁnd 'er. +enslon$3
Fuck“ = ("‘18 " Tn) clown m3 Fineta: (13." “1233 'i’P Assume. m. and mi are. sma\\ enough (<< M} 50 {MA Fﬂémx ‘5 ._.\ . . 4 p ,
and FndﬂxO. Then, ‘Tlﬂm‘g and T296. N123, 7Enel: 2" RﬁCMI— m2» 2 (b 10pts) What is the angular velocity of the disk a? after after 0.1 seconds have passed?
—L ——b
.3 A
A L ro+,d'sk ~ 33 (“’H" "ID i (0453
o . A IJESKC‘O; ‘29 z R3(M.MDCO.\53’Z\: (c 5pts) After the ﬁrst 0.1 secorlds have passed will torque oh the disk change? Justify your answer,
preferably ywith a diagram. Yes, i—k mus+ change. As +he. disk ro+a\\es, ‘Hne ongk bedmean r and T. changes which Means 21"“, changes. (:ﬁ won‘+ change. Sub$+an+§q“xj). "z A For examp’Ye‘ whéh has. r‘o‘ka‘ked '90“, {he {oriueh‘trom—i is O. 0e red 9; er 4*”: ——\ .4
FM“ 0nd FMMZ canno+ be Zero; oerrwxse. noH{ms wou‘d move. S'mce '“xe mes ave. +ied +0 +he disk, H~.e +an3en‘r'\q\ acce\e.rq‘\'30n
of Po‘m‘\‘s on +‘ne r‘wn is +he same. as +he, acce\er,o.\ion 0? \—he. masses (Assuraan +he. 5+r‘m3 can‘+ sIrel'ch). Ema =Q‘=0; No+e= In \+\e \imﬂ’ M. << M ma mz<<M, we. SeE our previous resuHs. ...
View
Full
Document
This note was uploaded on 03/06/2011 for the course BIOL 1510 taught by Professor Jungh.choi during the Spring '07 term at Georgia Institute of Technology.
 Spring '07
 JUNGH.CHOI

Click to edit the document details