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phy-testf08solutions - Fall 2,00(D RARLoN No\es Fob_ZooC1...

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Unformatted text preview: Fall 2,00% (D RARLoN No\es Fob_ZooC1 MULTIPLE CHOICE (’48 marks total) QHM ‘1’“ T25; ~ SOLMT ! 0N5 Constants: (D = did/Lit o. = dco/dt vf = (or ar = vz/r : (031' a[ = or st : V15 t 9:513? {of = (91+ (IN S[ : si + vest + seem): a = a + (1)1-At+ LearjAtf W52 = Visjt3W<AS (0f: = tof+2oA9 vzdF/dt a:dv/dt a = me P = (113/ at g = 980 13/51 G = 6.6734104] Mug/kg3 MEarth : 5-98X1024 kg REMl : 6.37xl06 m Question I The motion diagram for an object is shown to the right. It dispiays the position of the object at equally spaced time instants l, 2, 3, and 4. Which vector best represents the direction of the acceleration of the object at time instant 2? f \ + / {A} (B) (C ) (D) Question 2 The velocity-versus-time graph for an object is shown to the right. Which figure below best represents the object’s acceleration—versus-time graph? a a | Sat Obit {3:4 6 l (A) (B) (C) CD .L s L Question 3 In one of your Laboratory or Pilot Practical sessions, you were provided with a cart on a straight. flat track. To cause the cart to move at a constant velocity along the track, you IA) attached a fan accessory to the cart 5‘ 6/ Medfl“"‘5 Mm‘flie, 2, I raised one side of the track by putting spacers under the leg (C) eliminated friction by wiping the track down. PKG/k l‘ \l (4, 7 q ‘ (D) applied a constant forward force with your hand to the cart. (E) did nothing. The cart always moved at a constant velocity. Page 1. of 3 Question 4 A sphere with a diameter of O. 160 m rotates about an axis through its centre with constant angular acceleration. it goes from 0 rad/s to 33.0 rad/s in 12.8 s The magnitude ot‘the tangential acceleration at the equator of the sphere in 111/5" is (A) 0 (B) 0.440 (C) 0.220 (D) 87.1 (E) 174.2 Ott=0<r=erwo r :33‘0 0.\6 :O-ZzM/S’L Qeestion 5 At ‘1 2' Objects A and B have masses 111135.00 kg and mB=6.00 kg. The pulleys and strings shown in the figure have negligible mass. The coefficient of kinetic friction between Object A and the horizontai ta le is H150. l ()0. Nah. : ‘ 0'“: ‘ ‘e a: ‘ "Tie-e ‘ A: L V The magnitude 0t to accelei , ton 0t Liofet A in M18 1s MIAS'l aéswne (3)188 (C) 2.23 (D)3.06 (E) 4.52 ““047 “‘9 A ts +. tkz r" Questiort 6 A satellite orbits the Earth in a circle once every 12.0 hours. The Satellites orbital speed in m/s is (A) 3.87:<103 (B) 3.02m)3 ('C) 0 (o) 147 (E)5.‘93w104 Ll: Question 7 see 5“ Wh 0", Pay, 5 - Two wrestlers travel towards one another. Wrestler B exerts 1 force on Wrestler F (Ni ‘ A that varies with time as they collide. as shown in the figure. The magnitude of g TM!“ 5" the impulse on Wrestier A in his is u 6i}; f Fa.) (3)30 (CH) (mat) (Eta fiil /4 curve. Question 8 :60 pl K I 53 Alice and Bob are having a tug-of—war. They pull on the opposite ends of a long massiess rope. According to Newton’s Third Law, the force exerted by Alice on Bob Via the rope is equal and opposite / to the force exerted by Bob on Alice. Suppose Alice pulls Bob away from his original position. thus =40 M5 winning the tug—of-warz How is this possible? (A) Newton’s Third Law does not apply when Bob moves. (B) Bob has less mass than Alice. (C) Bob’s gravitational attraction towards the centre of the Earth is less than Alice‘s. JD) This is not possible. Bob Wili only move ifthe rope has mass. .Alice exerts more horizontai force on the gl‘fi‘tiiid than Bob. R COMM] ‘F/Om (4’0. Alva o! S'l'a‘l't'c ‘Oi‘ciibn oi iivw. jrowd alarm Nu‘u’s shoes , Moi m react/M '- Peru of [“1261 ShOeS o'x 44¢ SNHhoi. @is New-LON) 2114 Law ‘va GLEN} 5/ @ {5 Ngwhm’j 2M Law: TV 056%} A. 3 Unknowvxs '. T). ORA «M m. ,3 nnkAowM. .r, Mi“; 3 we maTI'MS Tb 501% 3 mud ‘1 MW. 0101“", '0“ ‘f , . XA 7A5 )CA Mommas, \13 (karma: by I‘ll} as mu ck . T’v‘x rdafi’ef N ,(Lccdzfah'm s «. a=~a B EAL?) +7 ifféflwmk." T M .flg ,' sow Tor on , TWuj® idv C993 VZT = mag .. MA L ,4 A 1- < , :ZI Z/MKmAg 2,14 q ZT H: 2T: El<mom:( T by 5.2, [:7 +ZMA “A M33 2 M4. ‘. °‘V€ ’FOP , Kunming}. and s @ G. N we can assume, 3 = 61.8%: 4+ 1%». aH’H'uaLt or 44le SA+CHI+€I WA we Can an. low-caf+k\orb.'+ eituaJfl'oflS or CAth/ 3". Eat/24$; T= zTrF =?T=~Z—IW 3 3 Edi/.g'll": V:W (”Nuke «two above (iua‘l’ionsz T: 21! V: 1:4 : 02 kr X géookir>(0l_gm/3> = G‘7K/04M/5 2“ zxr C ' “A‘P'Or+una"€\«// ~HM‘S I')’ 40+ 04C of ‘Hu. 5— Cite/<25] SO Hu‘; mus} be at Ckapfer (3 problem j [5 (cs; {92ch HUI. $a~+em+e I‘S Far -onm End—A's surface... Eq/AEZS’: T2 =(Lfn'2 >r3 :57 r=[TZGM]\/3 EoL.t3.ZZ- vleL/‘QTI 2 [411” 1‘4, {GMT’Z r‘ GMT1 \ . 1 Lx‘x V7- GZM? 411,114, :LG Mzzzfltlz 3 GMT‘ T1 \, V: ZTVGM>§ : Z-n' 641.40“ sum“)? ‘2 kr x 'ZQDOSA, R 01'“)ch A @ There are four parts to this question. Clearly show your reasoning and work as some part marks may he 3111211111: firite your final answers 111 1‘1e hoses provided. LONG ANSWER (52 marks total) A ball with mass 0.850 kg is suspended on a 1.31 m long massless string near the surface of the Earth. it moves at constant speed in a horizontal Circle as shown in the figure below. The string makes a constant angle 6=l0.0 degrees with the vertical. lgnere air resistance. PART A (7 marks) What is the 1ad1us 1 ot the circle? $1 Tomatrq- 511 9 =3, \.'3lm PART B (15 marks} D111 1311111 label the See-body iiegram for the all Only +1190 WCD’TLS'. N0 lL Cfih‘htfi'l’al “Para (‘5 l is; Ll: drawx on a ‘Hml. ; CQA+HPL'\'R[ 1‘; Pan/5124 61/ vllxk l 3 ngmg TEAS! on ’.I\ ““11! CA 526. (or konzon‘l'fil UOMPOM/ml o1C +£n$1on>( PART C (15 marks) Wéat is the net force 13net 0n the ‘5‘“ x m Em“ does nd’ “(Kile/n+6 vcr+t'ca//7 / 66am“ H‘ nlsz In Q Lor«'z°n’f&‘ Flam-2. C? R4130 = Tease ”“9 =9 T0059”?) 12mg— W 0059 FMX : _fsw : - Mama 2 -Mije 0058 \ijx: «(0.38 (afifim via/51) tan lo '5 ~ 1.4% 8% N | Wm r; Ndr - 'X‘a‘"’“+’° r ~ x'dir CHM r a PARTD(15marks) If“ 02/ Q o FNu—s -l."f7N 'C What is the angular velociig c3 of the ball? “Wm“ Srew‘ “5" Eva is m W‘i’fl'p—ehl form: Fuu : Marx/7' Flour " m’ufif 1" "' M wzw _..,,——————-—-—'1 SJVQ 4151‘ (’0‘. (A): E _: J l~4é$g m 0.9.; [0.227470 (A): 2754‘ Paul/S «from ol‘raléfo‘m/ Ol/‘raohbm [s C(oCk-W‘V/ (ZS VWANCI flow‘ alwx/C. 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