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Exam 2A Solutions

# Exam 2A Solutions - Physics 53 Putt 07 Grant 2 A Name...

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Unformatted text preview: Physics 53 Putt 07 Grant 2 A Name: Section: I certify that l have abided by the rules and [he spiril Off/18 Duke Community Standard PLEASE READ CAREFULLY BEFORE YOU START Do not forget to write your name and section number above. You must do all 5 multigle choice Questions and two (2) of the three long er groblems. If you attempt to do more than two of the longer problems, indicate below which two problems you want to be graded. Remember that credit will only be given for solutions that show your work. Do not just write down your result, but show and explain how you got it. No work shown = no credit. if numbers are required, use g = .10 m/sz. Record your MC answers (by letter A — E) here. No credit will be given, ifyou do not record your answer here! MCI u Mcz E MC3 é MC4 p MC5§ Check here, which two of the long problems should be graded: P1 P2 P3 {To be ﬁlled in by grader:] Grading scores: MC P1 P2 P3 Total: Multiple Choice Problems: 4 Points Each 1. Masses of 1.00 kg and 4.00 kg are hung from opposite ends of a 2.00 m long rod, which is pivoted at a point 0.50 m away from the heavy mass. If the system is in equilibrium, what is the mass of the rod? A) The rod is massless. gig, ‘ “‘1'” ‘ test B) 0500 kg. g t “a; ) 0.750 kg. % gig“; an KM 1.00 kg. 2? l W} .3 ﬁzﬁgj ,3 E) 1.50 kg. gg’ g; ' 2. A ball moving with a speed of 3.00 m/s hits a second ball, which is initially at rest. After the collision, the second ball moves in the direction of approach ofthe first ball with a speed of 2.00 m/s, while the ﬁrst ball is reflected by 180° and moves with a speed of 1.00 m/s back to where it came from. 11" the mass of the ﬁrst ball is 3.00 kg, what is the mass ofthe second ball? [Assume that all motion is frictionless.) P 0% it 4% A)1.50kg. - a: "" teei it 13) 2.00 kg. ‘ at j 3 \$ é w 4.50kg. MW? I W .21 r” t a 6.00kg. ’ ‘”" “W ’ ‘2’ 1 Impossible to tell from the data given. 3. Consider a turntable of mass 8.00 kg and radius 2.00 m, which is uniformly rotating at an angular velocity 0). lfa 16.0 kg mass is suddenly dropped vertically down onto the turntable at a distance of 1.00 m away from the center, and it starts rotating with the same speed as the turntable, what will be the new angular velocity of the turntable? A) (0/4' : w, 2- %} *1 gees B) (9/3. (0/2. we.“ ta... D) 203/3. E) 303/4. M y a as.“ g 5: egg“ e‘fgg M iéwﬁe} «r aang 2 as Several objects of different shape roll (without slipping) down an incline, starting at rest from the same height. Which of the objects takes the least time to roll down the same length ofthe incline (as defined by the distance traveled by its center of mass)? A) A thin hoop. {5a,an 5%; B) A solid disk. . «6% la It” M C) A hollow sphere. W“? KW” is; y W 6% MK 5% (1,, may 5‘ A solid sphere. E) They all arrive at the same time. E, «,r : maﬁa amateaé %%§ ” \ ' age/4‘ ﬁ 'ﬁmt An airplane oftotal mass 25,000 kg descends from an altitude of l0,000 m to an ‘ altitude of 8,000 in without changing its speed. What is the total work done on the airplane? A)-«50MJ. g. we; 45%in get. B) n 50 Ml. gilt“? i C) -— 500 MJ. ) .. 500 Ml. 0 Mi, The fotlowing problem is worth 2+4+4+6+4 = 20 points: Pl . Consider a solid sphere with mass 1.50 kg and radius 0.200 m rolling (without slipping and starting from rest) down a 5.00 m long 30° incline (angle measured with respect to the horizontal), followed by a 5.00 m long ﬂat stretch. [Assume that the sphere rolls without kinetic friction] a) Draw free-body diagrams of the sphere on the incline and on the ﬂat surface. b) What is the rotational kinetic energy ofthe sphere at the bottom ofthe incline? c) How long does it take the sphere to reach the end of the ﬂat stretch after starting to roll downhill from rest? ‘ d) What is the angular acceleration of the sphere and the torque acting on it as it rolls downhill? e) How long would it take if, instead, the sphere would roll down a 15° incline, 9.66 m long, connecting the start and ﬁnish points ofthe previous described incline plus flat stretch? Compare the result with that obtained in (c) and provide an explanation. at g in a“ is! © ,s a e 5 ms 3am? \$ng gnaw 89 Ill % {a ﬁwi‘ %} at} m it 3*: «W \W (at {age ﬁeﬂw emtaruaatﬁa tie eat/Meek tt ﬂKtﬁuse ago 9K§=éiantfaei w? ﬁkﬁﬁégwfiﬁmﬂﬁﬁafﬁ‘mﬁy « s; r , in *4}. A. L 41%(2, mg agglg ﬁgjég W [ﬂag %&M: 76% - “é, ff?” :1 Egg? g ,‘a A? an , ,2 C” azﬁsﬂ *- {nesz (5%,; éﬁhgxﬁaf amﬁéfm’égﬁk M Mr? gééfi/wwaf gig £2? W? a? - %%@ mm? m 5m 5%, M u‘ a, W'& ’ W7 w m M J? a My “3ng g W< '5 g “2% wﬁéﬁ ““ ‘ﬁéﬁ- ﬁuéwuggﬁ 3 am my” 37” i2 ﬁg}? {233 m i; 23533 & ’é’éﬁﬁiﬂyﬁ was {3:052} an??? g” g; @éﬂwx amgé‘ﬁi a3 %@ gawa 113.3% ml? ﬂﬁﬁé afﬁx g “if i ,1 QM %5 2 523% *3 éwwiéif"; 323:: g. M" @ﬁgmé ﬂ ﬁg K65 E? W ’gﬂmgﬁﬂ/gﬁﬁé 3555f gm”? % @Wwﬂggg, ,{3 53%. a WW5 gﬁwm£ x’iMﬁWm 69”; Q mévw" {C} ffegeééeéf The followino roblem is worth = 20 oints: P2. An object with mass 3.00 kg and moving frictionless with a speed of 18.0 m/s hits an identical object, which is initially at rest. After the collision, the first object moves at an angle 91:200; the second object moves off at an angle 63:400. Both angles are measured with respect to the original direction of motion ofthe first object (see illustration). [Neglect all effects of friction in the process] a) Find the speed of each mass after the collision. b) Calculate the magnitude and direction (with respect to the object’s original direction of motion) of the impulse received by the ﬁrst mass in the collision. c) Compare the original kinetic energy ol‘the ﬁrst object with the sum of the kinetic energies of both objects after the collision. How much kinetic energy was lost in the collision? Whenmsgﬁkineuaenergyuasaﬁmmmcmqbwnglemawatewbsehahmmwm WWW/mm object moves aFTEith”?C”6‘l”lisio g~~i’ti~tl~ieatirst;«cb}ectuscattg;s at the same fixed angle W WWWMWWW swwﬁaagoﬂasbeterewgerwwhiehhanglemegaiswthemeellisierrei e) How much kinetic energy would be lost i'fthe collision were completely inelastic? (eel [email protected] mwme‘mt' rest}: a target 2: gaff; v’e ﬁg «3% 23;, ﬁgsng «be A ea as; as g, , ass-“41%. %€:mazm. was “aggWgrgf ! Eta, Wgsemmatig: {xi}; fer Vi: @EE/We ,4ﬁeéﬁt‘? «at {K}: i geter {a} "lijg‘s V / I g e ljzlee‘lﬁﬁﬁ 55%an we @fgﬁg’} e lag (%Q+ws%} @ I / Law Wt t m gauges; were %=wL—azt.él§ (wager; The following problem is worth 7+7+6 = 20 points: P3. Consider the crane depicted in the illustration at right, which holds a mass m = 450 kg. The uniform boom AB, which is 12.0 m long and has a mass of 200 kg, is securedlby the steel cable CB. Assume that the system is in static equilibrium. 8) Calculate the tension Force in the steel cable in equilibrium. b) What are the horizontal and vertical components of the force with which the boom pushes on the pivot point A. c) lfthe cable were cut between B and C, but the mass remained secured to the end of the boom, how large would be the initial torque on the boom about point A? g g‘vw‘é gnu %‘K 2?; 7: CM ﬁg} 1 Wmawﬁ Lowe L ‘3 mom. 5 We 2ee% 65% mm Wﬁ‘g ﬁffptte airﬁeld if}! ya a: 17;.— gt. 5% 1500‘— mg L sill Keg w“ it? i: gel??? ii the} w we lie (Watt is or: test; if émgkgg‘g a a team it gm? \ W”; cm E: Egg 6% g; a} W gaze at” t w e W E Qfaﬂléﬁz’3ﬁlg ‘43? Qm‘lﬁ garage? “mgal‘ijéeggza M a 4" {9: w a. “3: 5“ .r #9 g” Willilvé‘l g ateﬁﬁ a {awt’l'l‘gldlg ,4: rm: tease M. A W n, “f: A g is". fay ’A._ I I»? :1 2L; 1? M ...
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