Test1Sol - ESM 2304 Name Lil l 3'1” l‘l Test 1(Form A 9...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ESM 2304 Name: Lil l 3'1”) l‘l Test 1 (Form A) 9!] HO? Rules: Closed notes, closed book, no personal aid. You may use one 8.5 x ] 1-inch piece of paper with whatever you wish written on both sides. All questions (equally weighted) are multiple-choice — no partial credit will likely be awarded. Show all work on these test pages and/or attached sheets and then mark the best answer on the provided opscan form. Place your name on this test and your name and ID number on the opscan sheet. Turn in the opscan form in one stack and this test plus any attached pages in a second stack. In the unlikely case that your Opscan form cannot be graded, your worked—out solutions will be graded instead, so you must be neat. Please sign the standard academic pledge below. Further notes: oVector quantities are denoted by boldface Roman type, as in i (as opposed to i). «Scalar quantities are denoted by lightface italic type, as in v (as opposed to v). oThe usual rules (as clearly established in all mechanics classes) for significant figures apply. Further opscan issues: 1. On your opscan sheet, fill in the Form Letter printed near the top left of this page. 2. On your opscan sheet, fill in the Group Number Dr. Kraige MWF: Group 1 according to your section: Dr. Kraige TR: Group 2 I pledge that I have neither given nor received unauthorized aid on this test. (Signature) 1. The base units of the US. derived unit of mass, the slug, are as follows: 1:: i‘"f"-L) lb-ftfsec2 {1 _, / E, - 2 .0 “r G. 'U " F, ' 3 (2) lb-sec (ft c. - _ ‘- (3) lb-ft—secz 32‘ .I 1. 3,4,. '49:: 3 (4) ft-seczflb a _____ ___F (5) The slug is a base unit in US. units. (6) None of the above 2. The gravitational force F which steel sphere A exerts on steel sphere B is (1) (?.59i+4.38j)10'6N _, g a 2) (-2.31i—1.336j)10"°N . a \ (3) (-?.S9i—4.38j)10"‘l\‘ 3M, m, f i, = 0.5 m (4) (2.31i+ 1.336j)10'9 N (5) (-1.333i—2j)10-6N = O_ 25 m (6) None of the above (Note: Newton‘s Universal Gravitational Constant = 6.673(10'”) n13,'(kg-52). densin of steel is 7830 kgmf). AZ 3. The acceleration of a particle that is moving along a straight line is given by a = kt3, where the constant k = 0.5 ms" and I is the time in s. lfthe particle velocity at time I = 0 is vo =10 mfs as it passes the position so = 2 m, determine its velocity v at time I =45. ,5 :H 51..) kt * (l) v = 22 mfs ' ‘ (2) v = 13.33 mfs g, '; + -_ . (3) v = 18.63 mt's k L “i r t" (4) v = 138 mfs . r 4‘ J a . r L/(S) v = 42 mfs K. L," "1"" "i w: 4 ’i (6) None of the above 0 to I 4' k -. .": it " \ ah T i 5') 1: (0+ 0.5%:- ‘: 4-1 mfg 4. The acceleration of a particle that is moving along a straight line is given by a = 41:9, where the constant k = 0.5 5'2 and s is the position in m. Ifthe particle velocity at time t = 0 is vo = 10 mfs as it passes the position so = 2 m, determine its velocity v when the position is s = 5 m. /(1) 1.»: 9.46 mfs * k5 : "'1 To (2) v = 8.49 mfs _ H __ i (3 v = 9.62 mfs * k5: “fl-— "‘ i "i- i (4) v = 6.66 mfs I, 1 __ t (5) v = 5.33 mfs w ‘1; _,. t; —_ I (1.} (6) None of the above : ' F] a; x I.) - v 1 — i ___ (K 5 C .— 5: D J i "1 r F; , .- z: 2‘3 _ "E {- ':._ \ 3 "- 1M '- C .3 ,1 : 15¢ - it» "2 "Mrs ml: 5. The acceleration of a particle that is moving along a straight line is given by a = -kv, where the constant k = 0.5 s" and v is the velocity in mfs. lfthe particle velocity at time t = 0 is vo =10 mfs as it passes the position so = 2 m, determine the position 3 when the velocity has been reduced to 40% of its initial value. wit) (1) s = 19 m n Re — a! {2) 5:17.33 m H k A; a: \/('3) s = 12.86 m r_ r .1 S = m .— Ik’ 'I.‘ u .3 1 :L I (5) s=4m ‘46 "' (6) None of the above “'3 " " — labs-L?) ' "t" )0 / \ _' *“ ifii '. 4_—:_- f) ._ .f)‘ IEE— 7:) ‘ Z Wt \. A3 6. Determine the x-coordinate of the point of impact of the projectile which is launched from point 0 with the initial conditions shown. B )C ‘ .i. ‘ . l /(1) f=269fi REE-Elms; 1.th l'rnpalc? 3' ‘00 fiisec' *3 F333): is “thaw: Vi V‘ECIMKI x: I L \I!I ___ (4) x=359fi ""““ ‘ (5) x=401ft ___ ,X' (6) None of the above I 0 ( . 3 _t L -. x , _ : ‘1]. ‘e F J Lo“ :2“. in 31" (Dzo'i \J 5‘“ 5 fl; :3: ’1" Emmi: w w it: ~ 13H - l’ a z 3 ° 2min . "i J o A; A— i n “\3 go“ 6’ LEW-(3310.6) : __QK')": Shaw-33:" x=¢<ot 'Ka' ‘ K” 0 ‘3 \Z ‘3 4‘ £0 3 f 2 (ma) 5m Q0 C05 0 ~ .. _ — 29, r 32.1 ___1; 7-8. As it passes point A of the horizontal-plane track, the car P has a s eed of 10 fifsec and it immediately begins to have a constant time-rate-of—change of speed equal to 5 fifsec . The car travels 200 feet between points A and B. At point B, the radius of curvature of the track is 120 ft. "f. Determine the magnitude of its total acceleration just as the car passes point B. 1) 5 fifscc2 i/iz) 13.20 reset? (3) 17.5 fifsecz (4) 32.2 ftr'sec2 (5) 19.97 fusec2 (6) None of the above l l "L r-F‘) _ q- r. r-. 3 :- i a 4%; \f‘e “P 8. Determine the angle (positive CCW) which the total acceleration of Problem 7 makes with the positive r- axis (tan em to the ath at B). i _ —\ an 8- 13 Y1 l a H IQ 6 : ,tCm .___Q+I (I) 15.95° i 'I MR ’75 %) 105.9° an it : Um .. fl _ 3) 74.10 ‘ J t D (4) 240.30 1.63 J“ “’ :— 5 (5) 343° qt __ (6) None of the above H4 9-10. A particle P moves in a vertical straight line and has the velocity and acceleration as indicated on the figure at the instant under consideration. 30" r ._ 30m ’% "Ur :: 2' — '21) 95:33” “I ' l731t‘c; ‘* n 0 ,_ # 1;. 39 git—0.0625 “’9 a (a n “2.0 ‘5m 3:) l0 MJS lo J {Gal-:3 " "3.. _ a QK “5 re Ya : if) COS-3’3 ” ‘ " \z .0 t *2 r914? locosaos = {soc-O-oszs + Uses-33 ‘3 (129 mist 9. The instantaneous value of the range rate 3‘ is (1) ll)me (2) -10mfs (3) ~20mr’s /(4) -1132 mfs (5) -l 1.55 mfs (6) None of the above 10. The instantaneous value of i’ is (1) (2) (3) 4) (5) (6) 0.625 mfsz -0525 mfsz -1 .333 ms2 —9.81 m2 9.29 mfsz None of the above PIS 11-12. On a football kickoff, the football B is launched from ground level at O with the initial conditions shown in the figure. The receiver/1 makes a “shoestring” catch — that is, he catches it when it is basically back to ground level. The receiver is running at a speed of 15 fifsec toward the kicking team at the instant of the catch. Assume that all action occurs in the plane of the paper. Also make the classic projectile assumptions of no aerodynamic drag, constant acceleration due to gravity, and a flat earth. 1 1. The velocity of the ball B with respect to the receiver A at the instant of the catch is _,-(1) vB/A = 84.3i - 40] ft/sec (2) V’ng = - fil/SCC VB/A = ‘l‘ WSCC (4) V3,“ = lSi - 40j fifsec V3“ = WSCC (6) None of the above CU ‘r'r-frn. Q J fin. I"? " '1 ‘73" Kt Q C J L,- ‘ - ‘I :_n .I--_‘_ {I I' ""3 C... :3, ‘3" ‘l‘me C ‘1": fin. flaw-s. ‘W l [5 T: C: \ C353 3:3 g — C 3 *‘V v3 3,- Qi. 3, - we”) lid-.drget n -—_ J — - l ' 1 F ~43 - r ‘3'“ I’l’glfg. ’ fill it D — : ' L r- A L I) 12. The acceleration of the ball B with respect to the receiver A at the instant of the catch is (1) am = 32.2j ftr’sec2 «(2) am = —32.2j WSec2 E"? 8 m. - :9 ,9 (3) a3,” = 0 (zero vector) 7 ,7 r, - (4) a3,” = -80i + 16.1 j arses2 ' “ ~':- A - a 3 (5) 33.,4 = -80i -l6.1j fU’secz ‘- (6) None of the above .. _ .,_.. ,, - r __ ' as 13. Sliders A and B move in fixed slots as shown, and a bar of constant length L = 5 m connects the two sliders. Determine the velocity of slider A when the position of slider B is 4 m from the origin 0 and the cerresponding velocity of slider B is 10 mfs up. (l) 10 mfs left / (2) 7.5 mfs left i (3) 7.5 mfs right i 4) l2 mfs left / (5) 13.33 mfs left if (6) None ofthe above , f O L 7— _H ’2. fits 4 QB — e , r z ‘ ‘3 e“ “F f\ + C3 13>: -~ g T _ Q1338 _‘ fl- [loj : __ [333 H'" \“ ti) ll UK] ( N fl l l l“ | l l I or all r _2_ t. 3 'L (l. I‘ ‘ .._ 'L “‘1- r- I n \ 3 F-\\ H J $53 Kig‘DHEFfiE eel—Ts LJBJ \m23)_ _ 4-2 7 8.7-5 \_ "u .-.--‘ ’ - ‘ -.' ) :— _— 5‘— {KF-Qfl; 53-3) "e-“W “\\H\ ...
View Full Document

This note was uploaded on 04/16/2008 for the course ESM 2304 taught by Professor Lgkraige during the Fall '08 term at Virginia Tech.

Page1 / 6

Test1Sol - ESM 2304 Name Lil l 3'1” l‘l Test 1(Form A 9...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online