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Prelim2_Spring09_Sol

Prelim2_Spring09_Sol - Physics 1116 Prelim#2 7:30 9:00 PM...

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Unformatted text preview: Physics 1116 : Prelim #2 7:30 - 9:00 PM, April 2, 2009 Name : C/ ‘9 (Tc/M7???“ 0 Don’t forget to write down your name 0 Breathe deeply and relax. Nothing can save you now. 0 Calculators are permitted, but laptops or communication devices are not. 0 There are 6 problems (+1 bonus) on the exam and 14 pages, not including the equation sheets. Please check that you have all of them. Feel free to remove the equation sheets from the back of the exam. 0 Please show your reasoning clearly on all problems. Partial credit will not be awarded if we do not understand what you are trying to do. Feel free to use the back of pages for your working (but let us know that you are doing so). 0 Don’t panic if you have difficulty with a problem, and don’t get hung up on any one particular problem. This test is designed to be sufficiently challenging that you do not have to be perfect in order to get a good score. 1- 3 30 pts 25 pts Bonus 5 pts TOTAL DO NOT GO TO THE NEXT PAGE UNTIL INSTRUCTED TO DO SO Short Questions Question 1 10 pts Batman (mass M B) hangs from a building ledge at a height 3h above the ground. A BatPlat— form consists of a massless ideal spring with spring constant k attached vertically between the ground and a platform of mass M p. The BatPlatform is initially a height h above the ground and is directly below Bruce Wayne. Batman loses his grip on the ledge and falls onto the platform, causing the spring to compress initially. The spring extends again, and when the platform once again reaches h, Batman leaps off the BatPlatform. If Batman wants to reach the ledge again, what is the minimum velocity that Batman needs to leap off the platform with? Please give this velocity relative to the platform. ' J / e7Z m #7514 “MM ene7/ ,& mmm74m/ 7m 4514 a// ‘ 54““ 4? you vfové Via/«1‘ reAt'Ki/erg jrewm/ MAUI 3474‘” ’5" C“? éq ’Vflén &‘”‘b )dA' ”15 +Wllo Minimum velocity : #— M“ 2 Question 2 : True or False (no partial credit) 2 pts each, 10 pts Each of these true / false questions is only worth 2 points. If you get stuck, you may want to move on, and come back to some of these questions later. 0 If the external forces acting on a system are conservative, the momentum of the system is conserved. o In an elastic collision, the speed of an object can change. False 0 To find the kinetic energy of an object in the center of mass frame KCM one can add the kinetic energy in the lab frame KM to %mvg~M where ’UCM is the center of mass velocity. 0 The speed of a rocket can never exceed the speed of its exhaust (exhaust speed taken relative to rocket) —a o F = A6 9 is a conservative force, where A is a constant. . Question 3 10 pts An observer on a train is moving at a constant velocity :10 forward relative to the ground. Looking out the window, she sees an accelerating car of mass m on the ground. At t = to, she sees the car moving a velocity —v0 (backwards) relative to the train. At t 2 151, she sees the car stationary relative to the train, and at £2, she sees the car moving at +210 forwards relative to the train. Given that the car burns C’ liters of fuel between to and t1, how much fuel does the car burn between t1 and t2, assuming that we can neglect air resistance and internal friction? What is the average power delivered by the engine between t1 and t2? Vela-lib; Jfo ammo“ :1 r:::I—> [:3———-> 06f Wfih-o “U053 , no W12) -.- 9110‘ i=6, 7.. 7' . . AKEéfl—tu)= ém (rub '" gimCO) = Jimflpl 4—"? C lulu; (E rpuj n. ’L a. ‘ Am; (were): Jilv'W) ~5Emtvo) = 333ml, ,_,., 36, Has GEM rpm = AK‘E “gm’lél —,__.———-— .. A15 Liters of fuel burned : 30 3mm Average Power : 2 £131: 3, ) END OF SHORT QUESTIONS 7.. Question 4 20 pts Brainiac has trapped the last Kryptonian (mass M3) in a pair of fields which have the form U1: A62” and U2: Ae‘” , such that UTOT — U1 + U2, as shown below. ( 3.) Find the position of S Jperman meg, assuming that he IS at an equilibrium point. {3119 where d—Lo—s “(A9, +Ae,x—) 4,43," -er‘j so .—\ (b) Given UToT, find the resulting force from this potential, FTO’I‘ and sketch it below. _. ch! 1:: """ =A(é:x .— 2329 F , _. ~75 2.96 FTOT 2 44(6 “'26 (c) Using his heat vision, Superman is able to displace himself slightly by an amount +Am away from meg. However, drained of his solar reserves, he is no longer able to use his heat vision, and slides back towards equilibrium Find Supermahs period of oscillation, assuming that he utilizes none of his remaining superpowers, and 13 subject only to FTOT kefilé $= A("‘°1 ear-1) ——a M's fem. answer is 4“” if: a, “'1 at? T-._..- yin: 7'43 -x 2:: ‘1» ‘1 Period: if A (6 "' ‘le, ) (cl) Now, accounting for air resistance ( fair oc ——v' ), Superman’s oscillations will be very gradually damped. We do not know the damping constant, b, but we do know the quality factor of the system, Q. Find the amount of time, T, it takes for the magnitude of Superman’s oscillations to be reduced to 1/3 of his initial displacement, Act. You may assume that the air resistance is very small, and hence the damping is very light. _:‘Q__: Xel’ X Lu fig amFli-l-ude. -—-‘, X'Ct)= X0 CL 251“- X69 .1. -1 .3115. Question 5 25 pts Iron Man is fighting Crimson Dynamo (both of mass M), and both are flying on a collision course towards one another. Iron Man has an initial velocity vector of 27 = v0[2, 0, 2] and Crimson Dynamo has an initial velocity vector of 23' = v0[2, 4, 2]. (a) Find the Avelocity of each combatant 1n the center— of—mass frame. —‘ ‘Pa 2“ o[2°9-]+E1421 au [.2 a a] CM :2. o I I ”M131- lpa’ .3 I,“ = “D:- in z wad—2 oil- [.2 2 fl) = v, [or—4,0] W)c_ = “1—7me = RAID 41]” [111]) = 'Uo[0,l,o] .1 'CM vIr/LCM I ‘1)‘[0, ~11 0] ’UCD,CMI en” 0 9‘ D] (b) If the two collide in a. completely inelastic collision, is it possible given only this information, to determine their resultant velocity vector immediately after the coliision? If yes, give the velocity; if no, explain why or why not. —9 41nd, [III Ill A ..——\ (P1311 . = Re I“ I Tl fi‘M‘ as}? Collision ,dm‘vw ’ Since. ‘fl‘e7 are, STUCK $39M (”[7 3': unkmms). (c) Now assume that the collision was completely elastic, and after the collision, Crimson Dynamo is observed to be traveling completely in the 5: — y plane (1);, = O) in the lab frame at an unknown Speed. Find the final velocity of Iron Man (in the lab frame). rflméa') Veg: 1) ”1)“ ——)'Vo[€l’a,0] = U°[c,d 21+ v,b,;1,;71 a = C..+ Q a = d"? Q 0 = 64' Q ———9 €=“‘Q. hr! VLJ‘ m [at 41c.— CJOM New ms mm; c I; e=wflr ‘ltm (944:0, if waned Wag“; l c4 .' ’Uco’: MED: 01’2“] 7131“?- “Mama? ._.. .4 .._3 ‘r-UIMI: IMI+FUCM= Wotoforaj+fuotg( 9‘,91 (d) Now assume a third possible collision, where after the collision, Crimson Dynamo is ob— served to be traveling with a final velocity of [2%, 2'00, 2%]. How much kinetic energy was lost in this collision? ,4 ___‘ / ‘Marnznium artworks“: flab-.- =E/ +191” ‘U W 'F/Vo 'Vq, hi0 VIM fl). . 9— 7. '1' 1r ‘7— .. k]; m , Ell—m'vm +—;-m’U a“.31((o+21)+ (.2 +412) ,4 14, M190 7'- ’L (‘L ? ‘1‘! l1. Kg; 3. Jimmie-[P1, + "lilting, tang (9.1‘11+911>+C3~ 4-; +32) : ll mVD AK‘E "‘ KEDE -’ Ken AKEzQ: M Question 6 25 pts A block of mass m rests on a frictionless surface. Two hoses are pointed at the block — one from the left and one from the right. Assume that both hoses spray water at a Speed 110, and both hoses spray water a constant rate ff? = c. However, the water collides elastically with the left side of the block, but inelastically with the right side of the block (after colliding, the water just drips down the right side of the block). Assume that the block is initially at rest, and the hoses turn on right at t : 0. 4-K »——9 V0 (a) Find the force FL that the water exerts on the left side of the block, as well as the force FR that the water exerts on the right side of the block at t = 0. APTOT =0 9 Milan MM +13»: v0 + A AV 93m. , 2A 1/ MA____\_/__ do 41’9”" AV m a = 'E —_""' At 'FIJ 0') ‘H- "‘ At Am 9 5m PL: QCVO before. __ Am AV Am A'PT¢T=O -) —AmU = —MA\I+ M -v, -' Arm}. r: _.{_N M Ti; Al M 1+: -c'U, —a"1’e d‘w FL I 4' QCTUO FRI —CV (b) Assume that at some moment :51, the block is traveling with a known velocity 121. Find the force F L at time 151 as well as the force FR at t1 -, _ ’ = P. ' 1’ + - ~ [email protected])‘ - '1). (Qt-M13411) 1).+AV 0 A TT9AM ° M1)! " WI+®+M(QVI “WED a 01—5 66.0% MVoWWMN~AmV°+JAmfl|+W sew = Ive-i! W :. v of An = av wow SPQA 4 afinaok L‘M—V‘l I '1} 1) —'vo—) Mao / kW“? A’m Wm'fldfid’l’oa a": “PL—LEI ' A I E I at run ¥ (”1 a 0 I bmac/ / '1. H " .. 'L 50 9%(fi°~V|)-QC (”03:10 ~— Mffl 10—? F1." «26 (Va—Wu) Same, ‘03}, ’91“ R APTW=0§ Mil-Mm = (M-i-Amfll—A’i fl}. ‘1), ’IJFH’XJ y—Amti = M+Mfllr MAW) +W 5 0 fig mfiw) = -Mm 7.. 4. $22. (”0"“) mm) : 'Vo —- CL CV0 +1]! )1 FR(tl) I “V0 (d) Find the “terminal” (1.6. equilibrium) velocity of the block as t —> 00. m4 mm “0:47 “st Eor=o 20¢ “—11: JFK! 2/5614, WV-r zo/(V +'V-.-)L M 24/ )5 " 'S f fi—ifim (U0! (1)434?) slim m% s a“ U(t~—>oo): (U0 Bflfi 11 (e) Find the rate of change of kinetic energy of the entire block / water system as a function of the velocity of the block, 1). For simplicity, do not worry about what happens to the water well after it collides with the block. 0rd Yawn-7 am‘l (R143 4 Hog, 13““ LHS (MM e/(axh‘c. COH'SIW, So no KE €14”!ij an ‘J‘Ltq‘l‘ Suite (PM ”gm PH3 —9 862.0 {3'0 (SW BlocJL dqm__g AKE . At RHJ‘ ' f1) 9. q- 1 MW J-m “Ill-"D = 3% “C; 14:32" )nt) on, .1; RH: AKrHs= LCUJV‘, )(1‘ +1) )("Uo ~19 At ave '3. v 3 .. c .00 __ AKToT = Tquo‘fv-D (fl) '1‘ a .. java Bonus Problem 5 pts Superman is able to “leap tall buildings in a single bound”. Assume that Superman’s vastly superior physiology is purely a result of the planet Krypton’s much stronger gravitational field (and not the different spectral properties of the Kryptonian sun). Provide a rough estimate of how many more times denser Krypton is than the Earth, assuming that it has a comparable radius to EZE'ERZQSMWL ‘SLWM C—w ()qu 0L I’m/1614+ lxfian M'l‘k. lru‘s [M'h‘al Welou'l7: fuzz: 231m. W' ’2 blunt (S _l1~°~ rm Elba} Superman's [288 can Merl“ _. 191E __ MS'vg—O) S _ At At 6 I S albeit $5M (La/W ( 0... (IR—— 7:10:01: (Lfc :Qal +3 a Far—“U? S'lflngf‘t‘37W3kf‘E m g ) a 7 6f 3 P: w (1)323le ah ~05,“ s. 720 ~ m/s , “was; 4/1000 N 80 1' Su mom can affly 1:: wdumrm ,"l'Lus IS «a 7 eziuaf~lw ms Wa‘ahl‘ (mag-K7lylm. a T N [0'0 £2 (2.2u300m) N /‘/000 N. 3 0-3-9: -/u// [cu/X147 M 300m «vino 414,50, If mm- a... am 3:: ~M‘ 74% mil a7! ewe. ghoég-qgifilr—e alrdfk. Nil-f For 1 point, Superman’s Kryptonian name is : Kfl/‘Efl, Sow 070 def—E1 pKrypton 3 PS. : assam' he. has 79%” Jam 7% weflu' 0—1.) ...
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