ep105-2007-mt1

# ep105-2007-mt1 - “guinea, UNIVERSITY or GAZIANI'EP t. l...

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Unformatted text preview: “guinea, UNIVERSITY or GAZIANI'EP t. l " DEPARTMENT or ENGINEERING PHYSICS 1%, ,1 EP 105 General physics I 9:. First Midterm Exam FALL SEMESTER Date: 08/11/2007 Time: 100 min. * * * SOLVE ONLY 5 OUT OF 6 QUESTIONS * * * ' The steps of solution of each problem should be shown clearly in the space given. ° Write your answers in boxes provided, otherwise your answer will not be considered. ' Constants: g = 9.8 m/s2 ; sin40°= 0.643 cos40° = 0.766 QUESTION 1 (20 %) A basketball player who is h = 2 m tall is standing on the ﬂoor R = 10 m from the basket, as in Figure. If he shoots the ball at a 6 = 40° angle with the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is H = 3.05 m. Let be be *3!“ éiM‘ 0i: {:ls‘gkt‘ ‘FOr Ha. lNIL“- . ’16:: VOCof9t ,Q. .- t ———«> '5 ""' W A .. v0 (1059' F 'F V0 C059- . I 1 . 44=ﬂa+vofm9fﬁzgé . ‘ / 7-. H:A +V95M9‘fp“??? Substihuﬁm} é; gi‘elols: #- L . R- J. H; \A + Vagiweﬁ-J)“ 2 3(0960193 0 fo/Ving for V0: 1‘ “z '2. “‘1. 9-8 x to Z ' ‘ = 40.5; M/ a. 9340' ( 2.4- 10 6mm; 4.0:] V0 5 08.11.2007 Page 1/6 EP105 General Physics I QUESTION 2 (20 "/o) _. Given the displacement vectors 3 = 3i — 4j + 4k (m) and B 2 2i + 3j - 7k (111) (a) FindavectorE such that 2K—3—E=0 -v "' 4 --) ZA—-8—c::o-—>Z=lZ-B , = 2(3?— 44+“) "(7"4'314? (b) Using the deﬁnition of dot product, ﬁnd the angle in degrees between X and B Iii/=- A = J3‘+<—¢)‘+ 4“ == 4! m [57:8 :V21+?1+(‘7)i = “‘2’” —+ a " -a (gt—45+4k)'(17+35"”)- “3" C 59:: 22W" 0 A; ‘f‘w‘ fa 50.4.1 ’2 3 u. a» 3 --4 4:: 167+293+r7a (M1) Axe: 7. 3 ---7- 4» ~ ~ (1 , A a A“? —~ W: 0,431+o.7223+ MAL [5 LL 2-: .. [ZxBi ‘11614- 291' 4471' A A h A u=o.u’si 4-9-1854-0-1. 08.11.2007 page 2/6 EP105 General Physics I QUESTION 3 (20 °/o) Table giVes data on the position (measured in meter) of a particle moving along a straight line at various times (measured in seconds). Assume that the particle acceleration is constant. HMU‘Ix‘lQQO’tkl-l 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 (a) Find an expression as for the particle position as a function of time. we. need 60 . . d i 7' general Cfuaé’O/I, x65) .- Xo+Vo£+1 “‘5 Find x.) Una. p... . a4:- ‘i:=:O;—-) 1(a): 2.9+ 049° 4112-01.: {.90—3 25.3 "‘ {00 M «11- f=1;—5 7(0):." 4+ 00.5.9470: ~> 4+l’afq/13LS‘0 (4) at {-225 a 9th.) :3 4' + 2", +2“ _’ "1' 2a 2 8'00 Solving. V, Mei at Fr.“ c,qu Hon (0 and (2) o 1. ‘- x(t)= 1+ ?.S‘t~2¢- (M) (b) What is the position of the particle at t = 2.3 s? 142-3) 2: ’l + 7.5'C2-3) -—2(2‘3)1' “ 7'5?“ x: 7.57m (c) What is the average velocity of the particle between t = 3.5 s and t = 1.0 5? 0mm :3 ~ ~ 3 . 5“ f. o 2 - f (d) What is the instantaneous velocity of the particle at t = 2.0 s? at d ’1}: 7‘ ::_.—-—-[ 4+7-S‘t—Zé7'] all- ou— : 7-? - (‘f 74‘s.. 402—) - Mo My page 3,6 EP105 General Physics I 0(1) H H 08.11.2007 QUESTION 4 (20 %) Three blocks are connected on the table, as shown in Figure. The table surface is rough and has a coefﬁcient of kinetic friction of 0.350. The blocks have masses of M1: 4 kg, M2 = 1 kg, and M3 = 2 kg, and the pulleys are frictionless. 1'4 ‘1. N T7. 7‘: _ N 1; T; T“ k #I‘ a L ﬂ. = "Ma 2» {My “Ag-9‘ M33 (b) Determine the acceleration of the system and its direction. Newfon'i in” {nu 'For' ego/I é/ocL 1'! a! fo//0w5.‘ M43— 'T( :1 ’“ca (1) [MI—'H‘MI—‘ME A .'. a r: T‘ "Tm -48“ _ M1“ (2‘) ml+m1+mx 3— Tx- M33 2 MM (3) :(4 — 0.3s‘(¢)-2)7‘8 New} (0 + (2) +0) ¢+ (+1 .. = 2.34 mg" Mig- H F]: ML: _ n3? 3 * (Ml‘ﬂnx—f’z)“ I'd flu all‘rctél‘on 0‘ --.X o. (0) Determine the tensions in the two cords ‘1 = 2 ' 3 ( Ml S imp» equuHM U) 2 T4: M‘ (3.42.) z: 4( 94—2-3!) :: 30 IV f0!“ equate»: (3) 1 Tl: m} (oi-fa) == 2 (5-5’+2-34) 2: 24.22.” 08.11.2007 Page 4/6 EP105 General Physics I QUESTION 5 (20 %) A block of mass m = 50 kg is pulled up at a constant acceleration of a = 2 m/s2 by means of a small winch as shown in Figure. The block being initially at rest rises to a height of h = 16 m. Ignore the mass of the pulley and all possible frictions in the system. (21) Find the work done by the winch at the end of the rise “g [wwéM’J 2nd law: W“, =: TA Coxo' Tamg :ma =(59°)((6)CO T:M(a+g) = 94403 —_- 50(z—tﬂ-ﬁ) : 5m?” m=04¢oJ (b) Find the work done by the gravity (i.e. weight of the block) at the end of the rise Wﬁ :: (1051.80. :2. a-M‘ka -.: —-. (roXB-SDUL) :—- 7-8903 Wg=_?2¢oJ. (c) Find the average power delivered by the winch during the period of the total rise "Ti/M reqw'neol {a com poi/n4 B I}: lq‘VU‘AJX pout-F: W 5 44-0 3‘ W 92...... 1—- 1: 4 I 08.11.2007 Page 5/6 EP105 General Physics I QUESTION 6 (20 %) A m = 10~kg block is released from point A whose height is h = 3 m, as in Figure given below. The track is frictionless except for the portion between B and C, which has a length of d = 2 m. The block travels down the track, hits a spring of force constant k = 2250 N/m, and compresses the spring x = 0.4 m from its equilibrium position before coming to rest momentarily. Then the block goes back and gains a new height between A and B, after passing through the points C and B respectively. Mao mu... (a) Detemiine the coefﬁcient of kinetic friction between the block and rough surface work (“AL I; the. ﬁr’célbﬂ 1'; change In chMI'cd/ ener”. — .— WF AE o 0 .. {3nd z: “1; 114‘) "" ( a; “limoqd: jkxm _ mg}; . F I 50/ \$1111} or [Ii M (“7‘ = I”: d I .. (IONS-002 —- ’17.. (zzcoﬂwc) ((0)01) (2.) (b) Find the height to which the block rebounds back after passing through the points C and B. —- E {-13 -llkmgd: mgH- flex?- W U9. kim— FILM?“ a! "’3' m. rammed“ .. (9. \$7M too-2x1) _. fﬂ______.______.._.___~ ((0) (9—0 = 0"?“ 08.11.2007 :2 6 7 am Page 6/6 EP105 General Physics! ...
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ep105-2007-mt1 - “guinea, UNIVERSITY or GAZIANI'EP t. l...

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