Multiple choice questions [00 points] Answe all of the following questions. Read each question caefully. Fill the coect ule on you scanton sheet. Each coect answe is woth 4 points. Each question has exactly one coect answe.. As shown in the figue, a mass M is hanging y thee massless stings fom the ceiling of a oom. Let T, T, and T 3 denote the tensions in the 3 stings. 30 60 T T 3 T M Choose one among the following A. T > T 3 T < T B. 3 T cos30=t 3 cos60 C. T = T 3 Wite that the node (whee the 3 stings meet) is in equiliium:. Still efeing to the situation descied in question, choose one among the following: A. T > T + T3 T < T + T In a ectangle tiangle, the hypothenuse is less than the sum of 3 the sides of the ight angle. T 3 B. C. T = T + T3 T T
3. M L L/ As shown in the figue, a small sphee of mass M attached to a massless sting of length L is eleased at height L/ and allowed to swing ack and foth. Ignoe ai esistance. Let you system consist of the sphee at the end of the sting. Fo this system: A. only kinetic enegy is conseved B. only total mechanical enegy is conseved C. only total mechanical enegy and linea momentum is conseved D. each of total mechanical enegy, linea momentum, and angula momentum is conseved. E. each of total mechanical enegy and the sum of linea plus angula momentum is conseved. 4. A man tuns with an angula velocity on a otation tale, holding two equal masses at ams' length. If he dops the two masses without moving his ams, his angula velocity A. deceases B. emains the same C. inceases 5. A wheel is otating feely with an angula speed of 0 ad/s on a shaft whose moment of inetia is negligile. A second identical wheel, initially at est, is suddenly coupled to the same shaft. The angula speed of the coupled wheels is A. 0 ad/s(angula momentum is conseved: 0I = Iω) B. 4 ad/s C. 0 ad/s D. 8 ad/s E. 40 ad/s
Questions 6 though 3 all efe to the same polem. A small all of mass m=00 g hangs y a massless inextensile sting fom the ceiling of a ailway ca. You may teat the all as a point paticle. At the instant shown in figue, the tain is at est in the station. Figue 6. The tension in the sting is m A. 0.98 N B. 9.8 N C. 0 N D. Othe E. Can't tell. Thee is not enough infomation. Sometime late, the tain is moving with deceasing speed. At the instant shown in figue, the speed is v T and the all is oseved to hang as shown. 7. The velocity of the tain is diected: 60 Figue A. To the ight B. To the left (The acceleation a is to the ight. Thus, the change of velocity v = a t is to the ight. If the speed is deceasing, v and v have opposite diections). C. Can't tell. Thee is not enough infomation. m 8. The tension in the sting is A. 0.98 N B..3 N C. 0.849 N D..96 N E. Othe 60 W BE T BS T BS sin 60 = mg
The tain continues to move with constant acceleation. These sketches ae fo questions 9, 0 and. A. B. C. D. E. θ>60 60 θ<60 60 9. Which of the sketches aove est epesents the all when the speed of the tain is 0.5v T? g Use T BS cos θ = ma and T BS sin θ = mg to find tan θ = a If a doesn't change, θ emains the same. C 0. Which of the sketches aove est epesents the all when the tain is evesing diection and has 0 speed? a emains the same: C Elsewhee on the tain, a second all of mass m hangs fom a sting of the same length as that of the fist all.. Which of the sketches aove est epesents all at the instant depicted in figue? g tan θ = doesn t' depend on the mass: C a Suppose that at the instant depicted y figue, the sting holding the all of mass m eaks.. Which of the following est epesents the tajectoy of the all as seen y an oseve on the gound (i.e. in the efeence fame of the gound)? A. B. C. D. E. Fee fall with a hoizontal initial velocity diected to the left.
3. The point of the floo that is diectly elow the all when the sting eaks is maked with an 'X'. When the all falls, it lands A. on the X B. to the left of the X (The tain slows downs as the all falls). C. to the ight of the X D. Can't tell. Thee is not enough infomation. Questions 4 though 8 all efe to the same polem. A ace ca stats at est on a flat cicula tack with a adius of 00m. With unifomly d v acceleated motion ( a// = = constant), the ca completes one lap aound the tack in dt 60 seconds. The ca has a mass of 500 kg. 4. As viewed fom the cente of the tack, what is the otational acceleation of the ace ca? A..7 x 0-3 ad/s B. 3.5 x 0-3 ad/s 4π ( α = a// = constant, thus θ = αt. α = ) 60 C. 0.0 ad/s D. 0.7 ad/s E. 0. ad/s 5. As viewed fom the cente of the tack, what angle has the ace ca taveled afte 0 seconds? (The answes ae expessed in adians) A. π/0 B. π/9 C. π/9 (use θ = αt fo t=0s) D. π/3 E. π/3 6. What is the aveage speed of the ca duing the fist lap aound the tack? A. 0 B..67 m/s C. 3.77 m/s D. 5.4 m/s π 00π E. 0.5 m/s ( v avg = ) t 60 = lap
7. Let f denote the magnitude of the fictional foce of the oad on the ca, and let W denote the magnitude of the weight of the ca. As the ca dives aound the tack, what can you say aout the atio f/w? A. f/w inceases B. f/w deceases C. f/w stays the same D. Can't say anything without any moe infomation. Top view Race tack f CT Ca (moving in counteclockwise diection) Side view f CT N CT W CE f f ma = m ma ( α ) ma CT = = // + CT v + As the ca dives aound the tack, v inceases. Thus, f CT inceases 8. Late, duing a ace, the ace ca will tavel at constant speed aound the tack. The dive needs to know the asolute maximum speed that he might e ale to dive aound the tack without sliding. Fotunately, he hies you to calculate this. You answe should e: A. The speed is 3x0 8 m/s. Wahoo! B. As you ties get moe and moe fictional, you can dive faste and faste, until you ae limited y the pefomance of you ca's engine. C. Even if you have the est ties, you cannot possily dive faste than 6 m/s on that tack, and you might not even e ale to go that fast without sliding. D. Even if you have the est ties, you cannot possily dive faste than 3 m/s on that tack, and you might not even e ale to go that fast without sliding. f µ N = µ W CT s CT S CE
Since Also, µ S, f CT WCE v v m. It follows that m WCE f CT. That is v g = 3.3m/s E. Even if you have the est ties, you cannot possily dive faste than 5 m/s on that tack, and you might not even e ale to go that fast without sliding.
Questions 9 though all efe to the same polem. You ae standing in the middle of a oad, and a tuck is diving diectly towad you at a constant speed of 0m/s. To get the dive's attention, you decide to thow a pefectly elastic ue all with a mass of 0. kg diectly at the font of the tuck. The font of the tuck is fictionless, pefectly vetical, and flat. The all has a velocity of 5.0 xˆ m/s just efoe it stikes the tuck ( xˆ is a hoizontal unit vecto). Assume that the collision is pefectly elastic. 9. Immediately afte the collision, the hoizontal velocity of the all is appoximately A. 5.0 xˆ m/s B. 0.0 xˆ m/s C. 5.0 xˆ m/s D. 0.0 xˆ m/s E. 5.0 xˆ m/s f f i i The collision is elastic thus: v v = ( v v ) The velocity of the tuck will not change significantly efoe and afte the collision since M t >> M. f f i i i i Thus: v = v + v v v v = ( 0xˆ) 5xˆ 0. What is the impulse eceived y the all duing the collision? t t t t t A. 3.0 xˆ kg m/s B..5 xˆ kg m/s C..0 xˆ kg m/s D..5 xˆ kg m/s E..0 xˆ kg m/s The impulse is equal to the change of momentum: f i I = p = M v M v = 0.( 5xˆ 5xˆ). As viewed in the tuck dive's efeence fame, the asolute magnitude of the hoizontal velocity of the all afte the collision is The collision is elastic thus: A. the same as efoe the collision f f i i v vt = ( v vt ) B. smalle than efoe the collision C. geate than efoe the collision D. cannot e detemined fom the infomation given.
Questions though 5 all efe to the same polem. A unifom oad has length 8x, width x (x is an unknown distance), and unknown mass m. A piece of clay of mass M is placed on the oad as shown. The oad with clay attached is found to alance when placed on a fictionless pivot as shown.. How does the mass of the oad m compae to the mass of the piece of clay M? clay of mass M A. m = M B. m = M/ The cente of mass of the oad and the clay is located just aove the pivot Take an axis (laeled s) with its oigin at the pivot: Ms clay + ms oad = ( M + m) s oad + clay Mx + mx = 0 C. m = M D. Can't say anything without any moe infomation. 3. Which of the following est descies the oad? A. At est in stale equiliium B. At est in unstale equiliium (The cente of mass is aove the pivot) C. Not in equiliium. It will tip down to the ight. D. Not in equiliium. It will tip down to the left. E. Thee is not enough infomation to detemine. 4. The otational moment of inetia of the oad alone aout the pivot is I. What is the otational moment of inetia of the oad and the piece of clay aout the pivot? A. I + Mx (The moment of inetia of the piece of clay is Mx ) B. I + Mx C. I + Mx D. I + 3 Mx E. I Mx 5. Late, you ae told that M=0 g and the dimensions of the oad ae 40cm y 0cm. Fo fun, you stat spinning the system aound the pivot with a constant otational velocity of ev/sec. The otational kinetic enegy of the piece of clay aound the pivot is appoximately: A..5 x0-5 J B. 5.0 x0-4 J C..0 x0-3 J ( KE = I clay ω = (Mx ) ω = 0.0 (0.05) (π ) ) D. 50 J E. 9870 J pivot Cente of mass of the oad and the piece of clay s Cente of mass of the oad
PROBLEM [35 points] a y m x M θ As shown in the figue, a toy tain is pulling a wedge of mass M acoss a hoizontal taletop. A lock of mass m is esting on the inclined suface of the wedge; let θ e the angle that the incline makes with the hoizontal. Thee is no fiction etween the lock and the wedge and thee is no fiction etween the wedge and the tale. Assume that the massless sting that connects the tain to the wedge is exactly hoizontal. As you watch the expeiment, you notice that the tain is acceleating and you notice that the lock emains at the same height on the incline (in othe wods the acceleation of the lock is hoizontal). ). [5 pts] The magnitude of the net acceleation vecto of the lock is (no explanation necessay) a. geate than the magnitude of the acceleation vecto of the wedge.. less than the magnitude of the acceleation vecto of the wedge. c. equal to the magnitude of the acceleation vecto of the wedge. (if not, the lock would slide up o down the wedge). ). [5 pts] In the oxes elow, daw fee ody diagams fo the lock and the wedge. Lael each foce vecto. N WT θ N BW T WS W BE N WB θ W WE Block Wedge
3). [0 pts] In the space elow, wite the x-component and y component equations fom Newton's second law fo the lock. Clealy indicate if any acceleation component is zeo. Use notation consistent with the laeling of the foces in you fee ody diagams. W + N = ma BE BW a is along xˆ Along x: N BW sin θ = ma Along y: N BW cos θ mg = 0 4). [0 pts] In the space elow, wite the x-component and y component equations fom Newton's second law fo the wedge. Clealy indicate if any acceleation component is zeo. Use notation consistent with the laeling of the foces in you fee ody diagams. TWS + NWT + WWE + NWB = Ma Along x: TWS NWB sin θ = Ma Along y: N Mg cos θ = 0 WT N WB 5). [5 pts]solve fo the magnitude of the tension in the sting in tems of M, m, θ, and g. Show you wok. N = (3 d law) WB N BW Fom 3): a = g tanθ and And fom 4) TWS = Mg tanθ + mg tanθ T = ( M + m) g tanθ WS N BW = mg cosθ
TOP VIEW PROBLEM [5 points] A hexagonal lock is at est upon a level fictionless suface, as shown in the top view diagam at ight. Note: The lock is not attached to the suface and is fee to move. A [6 pts] Two foces F and F, equal in magnitude and opposite in F diection, ae exeted on the lock as indicated. a. [8 pts] Find the diection of α, the lock's angula acceleation vecto aout its cente of mass (). If α =0, then indicate that explicitly. Explain. Toque fo F o : F sin(50 ) into the page Toque fo F : F out of the page The net toque τ is out of the page Since τ = Iα, α is out of the page (the lock stats otating counteclockwise). X 60 30 30 30 F. [8 pts] Find the diection of a, the acceleation vecto of the lock's cente of mass. If a =0, then indicate that explicitly. Explain. F F ma + = = a = 0 0 B [9 pts] In this pat of the polem conside the toque poduced y F (taken aout the cente of mass) when it is exeted at each of the points a,, c, and d laeled in the diagam elow. X X X d X c a Rank the magnitudes of these toques fom lagest to smallest. Explain the easoning you
used to detemine you anking. Use τ = F, τ = F sinθ τ a τ a = F sin(50) = F = F sin(90) = F τ c = F sin(90) = F τ d = F sin(30) = F τ = τ = τ < τ c d