test 1 solutions

test 1 solutions - EXAM 1 Physics 122 Thursday, September...

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Unformatted text preview: EXAM 1 Physics 122 Thursday, September 20th IMPORTANT: This test has 4 multiple-choice questions and 4 problems to be worked out. To assist in grading — please place the answer to each question parting the box provided. Show your calculations, methods, and approach clearly where possible. Showing your work will increase your chances of receiving partial credit. You have one hour to work the exam. Use g = 10.0 m/s2 for all problems. Assume all problems have three significant figures. Talking or looking at a neighbor’s exam will be considered academic dishonesty. N0 scratch sheets or additional paper is allowed on this exam. Use the space provided on this exam only. DO NOT OPEN THIS TEST UNTIL TOLD TO DO SO! NO CALCULATORS! For Instructors Use only: Problem Score Grader MC 5 6 7 8 PHYS 122 Equation Sheet —— Knight st : Vis + aSAt 1 sf 2 si + VisAt + —aS 2 2 2 st : Vis + 2asAs _. Fet=ma n sz (F net )r=mar= 1‘ st/r (0sz /dt Vt=(1)1‘ 27: r_2n T:— V 0) 0f =9i +(nAt f = 6i + (mm + EL(At)2 2r a V: rg 11:12 Lf = Li 13sz 1X: fo(t)dt Apx = x L=rmvt Kf+Uf =Ki +UI Ug 22gmy Us 2%k(As)2 (A02 = I‘l’lCOZI' AEsys 2 AK + AU + AEth = Wext Kf + Uf +AEth = Ki + Ui + Wext AK : Wnet : We + Wdiss + Wext WC 2 ~AU Wdiss : "'AEth F = —dU/ds P : dESys dt P = 1:“ V = FvcosG W = _stds =F-Ai‘ GmM Fm on M I 1.2 GM V = —————— r T2 2 47:2 r3 GM g _ GM surface r2 2GM Vescape : * R _ GM 2 % rgeo — 47:2 T U I _ GMm G = 6.67x10‘11Nm2 /kg2 Multiple Choice: select the best answer; no partial credit. 1)LetE=2i+3j' and F=2i—2}.Findthemagnitudeofé=E+2F. A) «H? B) 45—1}- C) J37 D) 6§+1j :(2i‘+3<\)r2('2'{"23‘) " 20 r33" +43'Llf‘ 1&3ng MHWYHZ‘: V5"? 2) Mary needs to row her boat across a 200—m-wide river that is flowing to the west at a speed of 1.0 m/s. Mary can row the boat with a speed of 4.0 m/s relative to the water. If Mary rows straight north, how far downstream will she land? 9 1:? X 200 /\ m . 333?: 311323 N Ali/LT” b (fizgfrl—Q/ob-E‘r VMO X32ézi—Uojc1-VMO Root/name (a : urn/«sea E; 505 ” “som 3) If the acceleration of a particle is zero, its velocity must also be zero. 3; A Particle mow“ n Lgin an 7 non— acre 5 ve\oc ‘4 has '26er CKCCC kamiicm ( Qé’nélqfll ) 3‘1? W 4) A child slides 5 meters down the surface of an icy slide. The top of the slide is 3 meters above the surface of the level ground. What is the component of the child’s acceleration along the slide? a. 3/5m/s2 b. 4/5m/s2 3 (9“ 3% c. 6m/52 5‘“ ’ a m '2. d. 8m/32 be Is Worked out problems. Please show all work. Place your answers in the box provided. 5. A woman stands at the edge of a cliff, holding one ball in each hand. Simultaneously, she throws one ball straight up with speed 10.0 m/s and the other straight down, with speed10.0 rn/s. Use g = 10.0 m/sz. a) Which ball has the greater acceleration at the instant of release? b) Which ball has the greater speed at the instant of release? both ba\\s have ONLFC’TVEF‘JF Velocities bur hoof WC] 579666; 0) Which ball hits the ground with greater speed? both ba\\$ hit w‘xl’h 'H/lfl Spccpd (1) Assuming the ball that is thrown downward strikes the ground 2 seconds after it is thrown, at what height was the ball when released? m/ if} L3: ~30 evoth + lgcujjg?’ (3)0 5 \ m 2 :2 0: 5o» IBM/3(Zs\~z[\0 @2305) e 5 “:BOZQLOMJrZOmrqor“ “1% HOW» 6. A hot air balloon is traveling vertically upward at a constant speed of 4.00 m/s. When Ll mi: it is 1953\m above the ground, apackage is released from the balloon. Use g = 10.0 m/sz. AR a. For how long after it is released is the package in the air? 0 \ 03K V‘s“? 1 “Bi + 0 3J5 V3102: v51?“ + 2a 053.32%, V I i Z _ m 2 flaw x 2 V5; -VSL V j; : GYM/S) +2( )0 BY ® ,. 0 t .M 5 \ a5 : W+38Llrlémo x/ m i M 2 20 "Vs m iOm/SZ fl: ‘3 ‘ t 9 H95 l» as b. What is the velocity of the package just before impact with the ground? 36,: work aha UC 6&0 “v5 7. A lamp hangs vertically from a cord in a descending elevator. In coming to a stop, the elevator decreases its speed at a rate of 2.00 m/s2 and the tension in the cord is 108 N. Use g = 10.0 III/$2. a) Draw the fiee body diagram of the system: b) Using Newton’s second law, find the mass of the lamp. 8. A 6.00 kg weight is connected by a string over a pulley to a 2.00 kg block that is sliding”? a at table as shown. The direction of the motion ofl‘block is downward and the coefficient of fiiction between the 2.00 kg block and the table is 0.200. Use g = 10.0 III/$2. a) Draw a free body diagram for each mass. “T’~MN Z mcx i’ (Goff? 0‘ c) Find the magnitude of acceleration for the blocks. 8:75 wrlrb ...
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This note was uploaded on 04/17/2008 for the course PHYSICS 122 taught by Professor Pope during the Spring '08 term at Clemson.

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test 1 solutions - EXAM 1 Physics 122 Thursday, September...

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