# Exam 4 - Problem 1 (25 Points) Two blocks with the same...

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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 low-mass string goes through a hole in the right—hand block and winds around a low-mass 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= Wax-t +ﬁex-t AK,= F-AEM = ~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 =~ Wax-t "baa-t AKb‘ocks + Auswins = F. Ar ~Fw + '7':K\$(L-L°)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 = 1e-14 ## Impact parameter mAlpha = 4*1.67e-27 ## Mass of the alpha particle in kg mGold = 197*1.67e-27 ## Mass of the gold nucleus in kg qupha = 2*1.6e-19 ## Charge of the alpha particle qGold = 79*1.6e-19 ## 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(1.46e-19,0,0) ## 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. * Adi-ad at ca 1 pA\p\-\q = pA\p\1a + Fa.\p\\0t*de\-\-a+ p Gro\d += Fgok'} * deH-od‘ #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 G-o\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 low-mass rod, with center-to—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? x-ﬁcM.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, rodi-F __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. "JEN-dz 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 Fine-ta: (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, 7Ene-l: 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 bed-mean 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; o-errwxse. no-H-{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‘+ s-I-re-l'ch). Ema =Q‘=0; No+e= In -\+\e \imﬂ’ M. << M ma mz<<M, we. Se-E our previous resuHs. ...
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## 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.

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Exam 4 - Problem 1 (25 Points) Two blocks with the same...

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