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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 ] 1inch piece of paper with
whatever you wish written on both sides. All questions (equally weighted) are multiplechoice — 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) lbftfsec2 {1 _, / E,  2 .0 “r G. 'U " F, ' 3
(2) lbsec (ft c.  _ ‘ (3) lbft—secz 32‘ .I 1. 3,4,. '49:: 3
(4) ftseczflb 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)106N = O_ 25 m (6) None of the above (Note: Newton‘s Universal Gravitational Constant = 6.673(10'”) n13,'(kg52). 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%: ‘: 41 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 " " — labsL?) ' "t" )0 / \ _' *“ iﬁi '. 4_—:_ f) ._ .f)‘ IEE— 7:) ‘ Z Wt \. A3 6. Determine the xcoordinate 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 REEElms; 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')": Shaw33:" x=¢<ot 'Ka' ‘ K” 0 ‘3 \Z ‘3 4‘ £0 3
f 2 (ma) 5m Q0 C05 0 ~ .. _ — 29, r
32.1 ___1; 78. As it passes point A of the horizontalplane track, the car P has a s eed of 10 ﬁfsec and it immediately
begins to have a constant timerateof—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 rF‘) _ 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 910. 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) COS3’3
” ‘ " \z .0
t *2 r914? locosaos = {socOoszs + Uses33
‘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 1112. 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'rfrn. 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. ﬂaws. ‘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 ,
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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.
 Fall '08
 LGKraige

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