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Unformatted text preview: PHYS 101 — General Physicsl, First Midterm
Duration: 120 minutes October 17, 2009 NAME: .................................................................. .. Section: .................... .. (lug; 0.2 (25) 0.3 (25) l 0.4 (25) Total (100)] You must sign the Honor Code for your exam to be graded: “t pledge, on my Honor, not to tie, cheat, or steat in either my academic
or personal tife. t understand that such acts violate the Honor Code and
undermine the community of trust of which we are att stewards." I agree to abide by this Honor Code Signature:
during this exam. Don’t forget
to sign! Present your work in a legible and organized format, otherwise you may
lose signiﬁcant portion of your credit even if your solution is correct! Solutions will be posted to —> httpzf/wwwten.bilkent.edu.tr/~phys101 Some expressions from the Textbook which may or may NOT be useful for this exam (However, you are supposed to know what each symbol means) Straightline motion with constant acceleration: _ . 1 2 2 2 . _ 00x“):
UI—on+axt, x=AO+UDxI+—axt , or :00: +2ax(x—x0), X—lo— f 2 Straightline motion with varying acceleration: f l'
or=um+f0aI dr‘ x=x0+ 0L5r d: Projectile motion: 1 . {3 .
.‘c=(Uﬂ 005(10):. 1:2(Uusman)I——gr , ox :uucosag. U}. —t)0511‘1{1 —gr ‘ 2
Uniform and nonuniform circular motion:
3 ’J
. U" 4n“)?
_. j _ I_ =
U _ “ERET ‘ grad _ R ‘ grad T2 Newton’s second law: EF=mE Newton’s third law: .AlonB: Hons: Magnitude of kinetic friction force:
ﬂ=mn Magnitude of static friction force:
ﬁSM” Trigonometric table (degrees)
sin (O):cos(90]:0 sin (10)=cos(80}=0.17
sin (20)=cos(70)=0.34
sin (30)=cos(60)=0.5
sin(37):cos(53):0.6
sin (40)=cos(50):0.64
sin (50)=cos(40)=0.77
sm(53)=cos{37}=0.8
sin (60)=cos(30)=0.8?
sin (70) cos(20)=0.93
80) cos(10)=0.98
90) cos(0)=1 si n(
sin ( ll 0.1 (25 points) A block of mass M is released from a height H on an inclined plane with an angle 0t from the
horizontal surface. The block slides down the inclined plane and travels a distance of L on the horizontal surface and stops at the point B. The kinetic coefficient of friction for inclined surface
is til and the kinetic coefficient of friction for horizontal surface is [42. a) Draw the free body diagrams of the block when it is on the inclined surface and on the
~— horizontal surface. "‘9 i b) Find lhe distance traveled (L) on the horizontal surface in terms of the given parameter5_
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on intimate sun (3am; I an her. 25mm
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q 1 I"‘ 5 L : ﬂ ‘ a X
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m Q'V'Fthmgi f" Cale — W“ W
9' II  "L t NISWC‘JSW l
pl : PUT YOUR ANSWER IN THIS BOX
33nd eye», Cos cﬂ ﬂ”; and r_—_——— 1/
(1,3 (25 points) Two bullets are fired at the same time and position (As shown in the figure). The speed of
bullets is 100 m/sec and the angles with the horizontal are 37 degrees and 53 degrees. (Assume
g=10 m/secz) 2:52 v2 =100 m/sec v1=100 m/sec 3) Determine the instantaneous velocity vectors of the bullets 7 Seconds after they are fired. —t—b—> (using unit vectors i,j, k) b) What is the angle between the velocity vectors of the bullets 7" seconds after they are fired? c) What is the relative'velocity of the first bullet xvi 've the second bullet 7 seconds after they are fired? “Ll0" :“4 lo d) Determine the vector (cross) product of the initial velocity vectors of the bullets? o) 1‘ the; ’ .1 that _.. ICU. C9334; mg l VT
V01}: EGO31m: 50mg ) v”: mo. stir13:39 w; YotjZ. (— ’5
VI: ELM —l0ii. gJ'jPi Ill .2 L Mfg!
bl [19905 Jﬂ‘i P’OJM’L
i; A V U 5‘0 éﬁ—JUJO
I a C_ ‘— ll L 'H ’— _ U I J U t (3389 \ “s©_ n  r
/ V2 / I/‘ 3/ 10!:sz lKJEPJI/Ilgdh‘dlil 0.3 (25 points) Two blocks (the mass of the blocks is 2 kg) are sliding down an inclined plane with an angle of 60
degrees with the horizontal. The coefficient of kinetic friction between the upper block and the
surface is 0.2 and, the coefficient of kinetic friction between the lower block and the surface is
0.1. A string connects two blocks. The string makes an angle of 10 degrees with the inclined
surface. a) Draw the free body diagrams of the blocks b) Find the acceleration of the blocks. c)
Find the tension in the string. (Assume g=10 m/secz) 9*] MIC) iixz'mo“
i p}: O
\' 6’“) T— r ? ﬂ; : “OK (J)
ng L+ costL 1
0 ,— m? ”mt 30*” {"509 ’0 F *‘2 f m“ 9:) ’Nrﬂmc}mSgOH—r€;ﬂlo: Q :3; A4; M3Qs6J+ls?nlo Nag, # mg wsétj + T sin to r. O :9 Ni: m3 cos tieTatar} F“ 5: [M (memséo+ Tarn JO] (3)
Cl:ﬂ_z~ (mcﬁﬁogéo ——T 33h 10) Una; {mull ML 3*}? ETCJS 393;} "p1 0'“ ““W 2’4 “L 76% T;O,SN and 61353 WrL PUT YOUR ANSWER IN THIS BOX 8: Q.4 (25 points) Conceptual Questions Answer the following questions in the box for each part. To claim any credit for your discussions, base your reasoning on ghysics {laws}. (a) (7 pt.) We usually neglect the air drag for a projectile motion. Assume that there is a large
air drag on the object that is thrown with an initial velocity and at an angle with the
horizontai. it takes 3 sec. to reach its highest point during its projectile motion. After reaching the highest point. will it take longer or shorter than 3 sec. to fall back to the
ground? Why? ll will lake longer than 3 sec. As the objecl goes up the drag force is in the some dlrecllon
as the gravity and as the object goes down the drag force is in opposite direction. Thus ll
takes longer to go back to the ground. (7 pt.) The velocity of a particle as a function of time is given in the graph. Draw the position
and acceleration of the particle as a function of time. Initially the particle is at 0 position. Velocity
0
Position .v""‘”  “H V
—_—_——__— V “9&1”!
Projectile motion _ _ _ . . .
Uniform Circular motion Nonuniform Circular motion i—
(c) (5 pt.) What is the St unit for the following physical quantities  Centripetal acceleration : Po { 31  Angular velocity: (“9 A. /?Q C o Coefficient of kinetic friction: no (.1 ii i” L . FrequenCy: i‘II..l"S.~’C I? {.qu  Proportionality constant of fluid resistance lr: it: .W (f = in! Fluid resistance at low speed.) m " F’— ...
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This note was uploaded on 12/27/2009 for the course PHYS 101 taught by Professor Atillaaydinli during the Fall '09 term at Bilkent University.
 Fall '09
 AtillaAydinli
 mechanics

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