L2.3: Chapter 2
Equations of Motion
Motion in One Dimension
Kinematic Equations
summary
Kinematic Equations
The kinematic equations can be used with
any particle under uniform acceleration.
The kinematic equations may be used to
solve any problem involv
L2.2: Chapter 2 1D
Accelerated Motion
Motion in One Dimension
Vocabulary Note
Velocity and speed will indicate
instantaneous values or as explained.
Average will be used when the average
velocity or average speed is indicated
Analysis Models
Based on four
L5.1- Newtons Laws
& FBDs
Introduction to Laws of Motion
Sir Isaac Newton
1642 1727
Formulated basic laws
of mechanics
Discovered Law of
Universal Gravitation
Invented form of
calculus
Many observations
dealing with light and
optics
Forces
Forces are what
L3.1 - Chapter 3
Vectors and Motion
Examples:
At t = 0, a particle is located at x = 25 m and has a
velocity of 15 m/s in the positive x direction. The
acceleration of the particle varies with time as shown in
the diagram. What is the velocity of the part
Chapter 2
Motion in One Dimension
Kinematics
Describes motion while ignoring the agents
that caused the motion
For now, will consider motion in one
dimension
Along a straight line
Will use the particle model
A particle is a point-like object, has mass but
L4.1Chapter 4
Motion in Two & Three
Dimensions
Components of a Vector,
Introduction
A component is a projection of a vector
along an axis.
Any vector can be completely
described by its components.
It is useful to use rectangular components.
These are the
L1.3 Using Uncertainties
Fall 2015
PHYS 191, Arts & Science Program, The Petroleum Institute
Mass per unit volume is called density
Examples Calculate . . .
Density of material: (18 kg) / (0.032 m3) = 560 kg/m3
o
o
o
Mass of object: (380 kg/m3) x (0.0040
Chapter 4
Motion in Two & Three
Dimensions
More About the Range of a
Projectile
Example: If Initial velocity is 11.00 m/s, and the angle the long jumper
is having with the x-axis is 20. Where will the jumper lands?
Non-Symmetric Projectile
Motion
Follow t
The Petroleum Institute
141 n_._a.;1Jl
THE PETIDLEUM "CSTIYUTE
lvuun Abuu-lu-u . .4, -.;A 4.
PHYS 241 - Physics II - Test 1
Electromagnetism and Optics
Fall Semester 2015
Time allowed: 50 minutes
Direction to Students:
All questions are to be attempted.
The Petroleum Institute
PHYS 241 Physics II
Electromagnetism and Optics
Final Exam
Fall Semester 2015
Time allowed: 150 minutes
Direction to Students:
All questions are to be attempted.
All answers are to be written in the space provided below each part o
The Petroleum Institute
PHYS 241 Physics II
Electromagnetism and Optics
Final Exam
Spring Semester 2015
Time allowed: 150 minutes
Direction to Students:
All questions are to be attempted.
All answers are to be written in the space provided below each part
TECHNICAL REPORT CORROSION PROTECTION
Co-Extruded 3-ply Tape Systems are
Advanced Technology Long-Term
Experiences are the Best Proof
In April 2013, the author had the pleasure to be the chairman of the Pipeline Panel at the 7th African Energy Forum
in Ha
93. (a) Centimeters are to be understood as the length unit and seconds as the time unit. Making sure our (graphing) calculator is in radians mode, we find
(b) The previous graph is at t = 0, and this next one is at t = 0.050 s.
And the final one, shown b
92. (a) For visible light f min = and f max = (b) For radio waves min = and max = (c) For X rays f min = and f max = c min = 3.0 108 m s = 3.0 1019 Hz. 1.0 1011 m c max = 3.0 108 m s = 6.0 1016 Hz 9 5.0 10 m c min = 3.0 108 m s = 2.0 102 m. 1.5 106 Hz c m
91. Using Eq. 16-50, we have y ' = 0.60cos sin 5x 200t + 6 6
with length in meters and time in seconds (see Eq. 16-55 for comparison). (a) The amplitude is seen to be 0.60cos = 0.3 3 = 0.52 m. 6
(b) Since k = 5 and = 200, then (using Eq. 16-12) v = (c) k
90. (a) The wave number for each wave is k = 25.1/m, which means = 2/k = 250.3 mm. The angular frequency is = 440/s; therefore, the period is T = 2/ = 14.3 ms. We plot the superposition of the two waves y = y1 + y2 over the time interval 0 t 15 ms. The fi
89. (a) The wave speed is v= (b) The time required is
t= m 2 ( + ) 2 ( + ) = = 2 1+ . v k k ( + ) / m
F
=
k k ( + ) = . m /( + ) m
Thus t
if
/
1,
then
t
/ 1/ ;
and
if
/
1,
then
2 m / k = const.
88. (a) The frequency is f = 1/T = 1/4 Hz, so v = f = 5.0 cm/s. (b) We refer to the graph to see that the maximum transverse speed (which we will refer to as um) is 5.0 cm/s. Recalling from Ch. 11 the simple harmonic motion relation um = ym = ym2f, we hav
87. (a) From the frequency information, we find = 2f = 10 rad/s. A point on the rope undergoing simple harmonic motion (discussed in Chapter 15) has maximum speed as it passes through its "middle" point, which is equal to ym. Thus, 5.0 m/s = ym ym = 0.16
86. Repeating the steps of Eq. 16-47 Eq. 16-53, but applying cos + cos = 2 cos
+
2
cos
2
(see Appendix E) instead of Eq. 16-50, we obtain y = [0.10cos x ]cos 4t , with SI units understood. (a) For non-negative x, the smallest value to produce cos x = 0 is
85. (a) With length in centimeters and time in seconds, we have u= dy x = 60 cos 4 t . dt 8
Thus, when x = 6 and t = 1 , we obtain 4 u = 60 cos so that the speed there is 1.33 m/s. (b) The numerical coefficient of the cosine in the expression for u is 60.
84. (a) Let the displacements of the wave at (y,t) be z(y,t). Then z(y,t) = zm sin(ky t), where zm = 3.0 mm, k = 60 cm1, and = 2/T = 2/0.20 s = 10 s1. Thus z ( y , t ) = (3.0 mm)sin ( 60cm 1 ) y (10 s 1 ) t . (b) The maximum transverse speed is um = zm =
83. (a) Let the cross-sectional area of the wire be A and the density of steel be . The tensile stress is given by /A where is the tension in the wire. Also, = A. Thus, vmax =
max A = max =
7.00 108 N m 2 = 3.00 102 m s 7800 kg m3
(b) The result does no
82. (a) This distance is determined by the longitudinal speed:
d = t = ( 2000 m/s ) 40 106 s = 8.0 102 m. (b) Assuming the acceleration is constant (justified by the near-straightness of the curve a = 300/40 106) we find the stopping distance d:
(
)
= +
81. To oscillate in four loops means n = 4 in Eq. 16-65 (treating both ends of the string as effectively fixed). Thus, = 2(0.90 m)/4 = 0.45 m. Therefore, the speed of the wave is v = f = 27 m/s. The mass-per-unit-length is
= m/L = (0.044 kg)/(0.90 m) = 0
80. (a) Since the string has four loops its length must be two wavelengths. That is, = L/2, where is the wavelength and L is the length of the string. The wavelength is related to the frequency f and wave speed v by = v/f, so L/2 = v/f and
L = 2v/f = 2(40
79. We use Eq. 16-2, Eq. 16-5, Eq. 16-9, Eq. 16-13, and take the derivative to obtain the transverse speed u. (a) The amplitude is ym = 2.0 mm. (b) Since = 600 rad/s, the frequency is found to be f = 600/2 95 Hz. (c) Since k = 20 rad/m, the velocity of th
2 78. We use P = 1 2 ym vf 2 f 2 . 2
(a) If the tension is quadrupled, then P2 = P 1
2 4 1 =P = 2P . 1 1 1 1
f2 f1
2
(b) If the frequency is halved, then P2 = P 1
=P 1
f1 / 2 f1
2
=
1 P. 1 4
77. (a) The wave speed is
v=
120 N = = 144 m/s. 8.70 103 kg /1.50 m
(b) For the one-loop standing wave we have 1 = 2L = 2(1.50 m) = 3.00 m. (c) For the two-loop standing wave 2 = L = 1.50 m. (d) The frequency for the one-loop wave is f1 = v/1 = (144 m/s)
76. (a) At x = 2.3 m and t = 0.16 s the displacement is
y ( x, t ) = 0.15sin [( 0.79 ) ( 2.3) 13 ( 0.16 )] m = 0.039 m.
(b) We choose ym = 0.15 m, so that there would be nodes (where the wave amplitude is zero) in the string as a result. (c) The second wa