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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 + VagiweﬁJ)“ 2 3(0960193
0
fo/Ving for V0: 1‘ “z '2. “‘1.
98 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=lZB ,
= 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 49185401. 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’tkll 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° 411201.: {.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? 1423) 2: ’l + 7.5'C23) —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+7S‘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 “Ag9‘ 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—23!) :: 30 IV f0!“ equate»: (3) 1 Tl: m} (oifa) == 2 (55’+234) 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—tﬂﬁ)
: 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. aM‘ka
.: —. (roXBSDUL)
:— 78903 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 poutF:
W 5 440 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
.. (IONS002 — ’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 too2x1)
_. fﬂ______.______.._.___~ ((0) (9—0 = 0"?“ 08.11.2007 :2 6 7 am Page 6/6 EP105 General Physics! ...
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