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Unformatted text preview: First Name: Last Name: Section: I 1 February 14, 2003 Physics 207 EXAM 1 Print your name and section clearly on all 5 pages. (If you do not know your section
number, write your TA’s name.) Show all work in the space immediately below each problem. Your
final answer must be placed in the box provided. Problems will be graded on reasoning and
intermediate steps as well as on the ﬁnal answer. Be sure to include units wherever necessary, and the
direction of vectors. Each problem is worth 25 points. In doing the problems, try to be neat. Check
your answers to see that they have the correct dimensions (units) and are the right order of magnitude.
You are allowed only the equations and constantsin this exam booklet. The exam lasts exactly 50 minutes. (Do not write below) SCORE: 60LUT20 150$ Problem 1:
Problem 2:
Problem 3: Problem 4: TOTAL: Don't open the exam until you are instructed to start. Please relax! First Name: Last Name: Section: I 2 PROBLEM I
In order to hit more home runs, Mark McGwire is allowed to bat from an elevated platform as shown.
Because of the platform, the ball starts its trajectory 5.0 meters above the rest of the playing ﬁeld.
Suppose he hits a ball and gives it an initial speed V0. «$59 i
'Tlftfofm \ $1.0M ”t 3,0M £0406 __ I 60 M >
a.) The ball initially travels at an angle of 45° with respect to the horizontal direction. If the projectile
is to land on a spot on the ground that is at a range of 160 m from the base of the platform, what value of V0 is required? (9 pts.) /éo
X : (VD (”seabk :3? 7E: VDCoseg a = a. +®06M 60y“ i3}? l!
b.) How long is the .ball in the air before it hits the ground? (8 pts.) — [£4 :53
7k“ Vpéosaf c.) Now suppose McGwire hits the ball so that it travels initially at an angle of 0° with respect to the
horizontal direction. What value of V0 is required so the ball just clears the top of a fence that is 160
meters from the base of the platform and is 3.0 m above the ground? (8 pts.) X s \fofk' : IéDm Away" =2\ ‘
‘ __f_L ‘/2_ $0 v0: ﬂab Va
7% > a J (‘76?) <7 First Name: Last Name: Section: I 3 PROBLEM 2
A block with mass m1=12 kg is attached to another block with mass m2=31 kg by a massless rope
passed over a frictionless, massless pulley as shown. m2 slides with no friction on the surface of a table. a.) Which block has the larger acceleration? (5 pts.) b.) What is the acceleration of In]? (5 pts.) m1.
M “T: m‘ac .....———>p
T ‘3 T— 2 7 “ls:
0“ :T .
W‘ ‘Jm/ i, 4/ mac.
0»: it: 3 12x76” =
W'lwz BI'HZ.
c.) What is the tension in the rope ? (5 pts.)
l :: W10. ‘___, 39L) (1.) Consider the following unrelated problem. In the ﬁgure below, the spring in conﬁguration (a) is
stretched 0.10 m. How much will the same spring be stretched in conﬁguration (b)? (5 pts.) (a) (b) First Name: Last Name: Section: I 4 PROBLEM 3 A rocket is launched vertically from the earth (in the positive 2 direction). Shortly after launch there
are three constant forces acting on the rocket: i) Gravity. The mass of the rocket is 120 kg. 11) The rocket engines, which are badly misaligned with the rocket axis and produce a force of 5300 N
in the yz plane, 30° from z the axis. iii.) The wind, which exerts a force of 330 N in the x direction. a.)What are the three components of the resultant force vector F while the rocket is ﬁring‘7 ( 7 ts )
A _.., . p . . A “a? . — ‘\ _ k A
W :: ——w»& \C — ”120 $73!!  480?— ) Rani: 330k
_.., 4 .
F  5300 £06 30 V: —t 530056v~30 x : 4570 K
E“ T 6 " ’K 4. 4,
d ‘z ._., —~, .7 129505 530A. +2700 4 Sage K,
I” 3 L7 'i' rEV‘éL‘i PW“; . b.) What is the magnitude 0 F? (6 pts.) r—‘M / » 2 1 'Z 0 t 219152) + SLHO : 2L— 3 4300 M c.) The rocket engine ﬁres for 7.8 seconds. Where is the rocket ' 
shuts off? Assume the rocket starts at position (0,0,0). (6 pts.) (X, y, Z coordinates) when the engine
1 —
— U ”'7, iZi1:= F/m biz“L153
“Ce 9. )
53Dx*z (1.) Consider this unrelated problem: An object moves along the track shown in the topview diagram below. The track is curved ‘
near point B, straight near point C, and curved near point D. The object moves from point A to point E with constant speed. , _ i. ii. , /
Draw arrows on the diagram to [IL $65369"
represent the direction of the B acceleration of the object at points
B, C, and D. If the acceleration is zero at a point, then state that A
explicitly. Is the magnitude of the
acceleration at point B greater
than, less than, or equal to the
magnitude of the acceleration at point D? Top view First Name: Last Name: Section: I 5 PROBLEM 4
The ﬁrst two parts of this question are related to units and dimensional analysis, and require no
knowledge of Newton’s Law of Gravitation (which we’ll cover soon). a.) Newton’s Law of Gravitation determines the force of gravity between two objects with masses m1
and m2. F is measured 1n Newtons and is expressed as _ Gmlm2
_ r2
where r is the distance between the objects. In terms of kilograms, meters and seconds, what are the
units of G, the Universal Gravitational Constant? (6 pts.)
Z
I! v 3:; Jail
so [6?]: .
I b.) A neutron star has a radius of 15 km and a mass m = 1.4 x 1031 kg. What is the density of the star §
\
V‘ I“ in metric tons per cubic centimeter? (6 pts.) 7
A 5‘ m / [bl/oz W ’1‘
~(‘ .=‘ 15 #0 CM / ’ : __,,_. y
._ ——"’ I ‘S
2 37 fi— 17 ( 5 / aJXM C v»
M = l [1 We MHi 3 q 2L
7' fﬂ/ﬂ m. . Cw The last two parts of this problem are related to circular motion. 0.) Calculate the radial acceleration of a person standing on the Earth’s equator, due to the rotation of
the Earth about its axis. Radius of the Earth = 6.4 x 106 m. (7 pts.) 1
, \/ Z , LZlTT/TB T": 24 )LBLIaoo
V r 4v 9112‘? 3 i?
T d.) What tangential acceleration at the equator would be required to slow the Earth’s rotation to zero
over a period of 1 day with constant deceleration? (6 pts.) \l 5“. A we“ ”75‘ PHYSICS 207 Exam 1
Summary of Equations 2/14/03
CHAPTER 2 —— 1D Motion
Displacement Ax = x2 — x1
Ax
Average velocity UM = 
At
. lim Ax dx
Instantaneous velocu t = ——— ~ —
y 00 At —> 0 At dt
Relative velocity UPB = UM + 0A3
t t ld'
Average Speed __oa_1_sta_nce_ = E
total tlme At
AverageAcceleration at,v = 92
At
2
Instantaneous acceleration a — g—U — _i—)—C
dt alt2
Acceleration due to gravity g = 9.8lm/s2 = 32.2 ft/s2
. . . lim
Displacement and Veloc1ty as 1ntegrals Ax: At 020 At: [ 0 dt
90]
lim
a. At.= dt
=At —> 0 ,2. I a
ConstantAcceleration Equations
Velocity U = UO + at
Displacement in terms of vav Ax: x— xo— — U t— — £500 + 0))?
Displacement in terms of a Ax: x— x0 :01) t+ ; at2
vin terms ofaandAX ()2 = ug+2an CHAPTER 3 — 2 and 3D Motion Components Ax = A cos 6
Ay= A sin (9 Magnitude A = «A: + Ay2 Unit vectors A = Ax i + Ay j+ AZ k Adding vectors using components. if C = A + B then
CX = Ax + Bx
and Cy = Ay + By . a lim " ‘
Instantaneous  ve1001ty vector u = A: = d r
At —> 0 At d1
.  l A ” ‘
Instantaneous — acceleratlon vector a = 1m ——U— d—U
At —> 0 At dt
Relative Velocity OPE = 13!”, + 13 AB Projectile Motion Equation Uxt = ()0)r 1 . 1
y(t)= yo +00yt—Egt2 = yo + (00 51n0)t ”'5th
Uy(t)= 00y gt = u0 sine gt x(t) = x0 + ont = x0 + (00 cos 6)t g 2‘
Path x = tan6 x— ———x
y( > < o) [203 90
2
Range when initial and ﬁnal elevations are equal R = 2°— sin 260
g CHAPTER 4  Newton's Laws  F ~ ~
Newton's Laws :Second law a = ”3’ or 2F = Fm = mEi
m
Mass EL = 2‘
ml a2
Weight W = mg
Hooke's law F = —kAx CHAPTER 5  Applications of Newton's Laws Kinetic friction fk = ,ukE,
Static friction Q S ,uan
2
. . u 2
= _ = U ._ d0
Centnpetal acceleratlon a r a, g a ,
Speed and Period U = 2—2:; Mathematics Quadratic equation: the solutions of the quadratic equation ax2 +bx+c=0 are x=—2i:t31—\/b2 —4ac
a a Trigonometry sin(—0) = —sin 19 cos(—l9) = cos (9 sin(7z7’2  0) = cos0 cos(7z7’2  6) = sin0 sin(7t  0) = sin 19 cos(7r  0) = cos€ Constants Acceleration due to gravity at the earth’s surface: g = 9.81 rn/s2
Avogadro’s Number: NA = 6.02 x 1023 molecules/mole 1 metric ton = 1000 kg
Radius of the Earth = 6.4 x 106 m ...
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 Winnokur
 Physics

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