P131 Practice Final
WARNING: This is NOT the real exam. Do this exam at your own risk! The real final exam will be different. The midterm will have 20 multiple choice questions (100 points total) and 4 show work problems (100 points total). As exampl
Physics 133 Sample Quiz Recitation Instructor (circle one): Able
h = 6.63 x 10 J s = 4.14 x 10 c = 3.00 x 108 m/s me = 9.11 x 10-31 kg
-34 -15
Name_
eV s
Baker Charlie Easy Fox George h = 1.05 x 10-34 J s = 6.58 x 10-16 eV s
hc = 1240 eV nm me c2 = 511 ke
Physics 131
Spring 2008
Dongping Zhong
Lecture 4 (Monday, Mar. 31)
Vectors
Key concept: Scalar and Vector Vector (direction and magnitude)
Page 1
v y ( j)
Ay
A Ax
v v v v A = Ax i + Ay j + Az k v v v v aA = aAx i + aAy j + aAz k v v aA = aA
Physics 131
Spring 2008
Dongping Zhong
Lecture 3 (Friday, March 28)
Complex Kinematics
Key concept: Free-Fall Acceleration (g=9.8m/s2)
Page 1
Motion with constant acceleration:
v 2v v dv d x a= = 2 = const. dt dt
a 2 x = x(t ) = x0 + v0t + t
Physics 131
Spring 2008
Dongping Zhong
Lecture 2 (Wednesday, March 26)
Kinematics
Key concepts: constant acceleration motion uniform motion
Page 1
Instantaneous velocity:
v x v v v = v (t ) = lim t v dx ds = = ( = slope of position graph) dt dt
1. We apply Eq. 4-35 to solve for speed v and Eq. 4-34 to find acceleration a. (a) Since the radius of Earth is 6.37 106 m, the radius of the satellite orbit is r = (6.37 106 + 640 103 ) m = 7.01 106 m. Therefore, the speed of the satellite is
2
1. (a) The height is h = d sin, where d = 12.5 m and = 20.0. Therefore, h = 4.28 m. (b) The horizontal distance is d cos = 11.7 m.
2. (a) With r = 15 m and = 30, the x component of r is given by rx = rcos = (15 m) cos 30 = 13 m. (b) Similarly, the
1. The metric prefixes (micro (), pico, nano, .) are given for ready reference on the inside front cover of the textbook (also, Table 12). (a) 1 century = (10-6 century )
100 y 1 century 365 day 1y 24 h 1 day 60 min = 52.6 min . 1h
(b) The percent
Physics 131
Spring 2008
Dongping Zhong
Lecture 5 (Wednesday, April 2)
Motion in 2D & 3D
Relative motion
Page 1
2-Diemension motion
v y ( j)
P
v r
`
v X (i )
v v v v r = r (t ) = x(t )i + y (t ) j v v dr dx v dy v v v V= j = Vx i + V y j = i
Physics 131
Spring 2008
Dongping Zhong
Lecture 6 (Friday, April 4 )
Projectile Motion
Page 1
Projectile Motion
y
v ( j)
v V0
x-motion no acceleration
v x (i )
y-motion at constant acceleration ay=-g=-9.8 m/s2, ax=0
OSU Physics Department
Physics 131
Spring 2008
Dongping Zhong
Lecture 7 (Monday, Apr. 7)
Circular Motion
Angular variables
Page 1
s (radians ) , s = arc length r 2r = 2 (rad ) full circle = r 3600 = 57.30 1 rad = 1 rad x 2 rad average angular velocity t d = dt
P
Physics 133 Sample Quiz Recitation Instructor (circle one): Able Problem 1. A positive lens is placed a distance D to the right of the negative lens. The final system image is 620 mm to the right of the positive lens, the same size as the arrow and pointi
Physics 133 Sample Quiz Recitation Instructor (circle one): Able
Name_ Baker Charlie QUIZ #5 Easy Fox George
Problem 1. In a two-slit interference experiment, the m=2 bright spot is found to be 3.0 mm away from the m = 0 bright spot when 750 nm light is u
Physics 133 Sample Quiz Recitation Instructor (circle one): Able
Name_ Baker Charlie Easy Fox George
QUIZ #3 Problem 1. A green laser beam travels inside a glass block (nglass = 1.3) along the path shown. It hits the left side at the critical angle and to
Physics 133 Sample Quiz Recitation Instructor (circle one): Able
Name_ Baker Charlie QUIZ #2 Easy Fox George
(1) A transverse, string wave is described by: y = (20 mm) sin[(2.0 rad/m) x + (16 rad/s) t] where x and y are in meters and t is in seconds. (a)
Physics 133 Sample Quiz Recitation Instructor (circle one): Able
Name_ Baker Charlie QUIZ #6 Easy Fox George
A tiny rock lies motionless with respect to the earth a distance 2.0 ly away. A rocket, rest length 100 m, travels from the earth to the asteroid
Lecture 21 Summary
Impulse = change in momentum!
J = p = p f - pi =
I Physics131
tf
ti
F dt
Inelastic Collision: Momentum is conserved but not KE
(e.g. trains stick together after collision)
m1v1 I + m 2 v 2 I = ( m1 + m 2 ) v f v2 I = 0 v f
P131 Practice Final
WARNING: This is NOT the real exam. Do this exam at your own risk! The real final exam will be different. The midterm will have 20 multiple choice questions (100 points total) and 4 show work problems (100 points total). As exampl
Physics 131
Spring 2008
Dongping Zhong
Lecture 11 (Wednesday, Apr. 16)
Acceleration constraints (Rope and Pulley) Solving Complex System
Page 1
All three systems are stationary. Rank the rope tensions.
A B C
What do the pictures mean?
Page 2
Physics 131
Spring 2008
Dongping Zhong
Lecture 9 (Friday, Apr 11)
Force and Motion
Equilibrium, Dynamics, Kinematics
Page 1
Equilibrium (1st law) Object at rest or moving with constant velocity
v v Fnet = Fi = 0
i
Dynamics (2nd law) Object a
Physics 131
Spring 2008
Dongping Zhong
Lecture 8 (Wednesday, Apr 9)
Force and Motion
Key concept: Newton's first and second laws
Page 1
Newtonian Mechanics Force (a push/pull, vector, N or lb, a contact or long-rang force) Newton's First Law,
Summary for Lecture 19 on October 31, 2007 Center of Mass (COM)
rCOM x COM y COM z COM x COM 1 M 1 M 1 M 1 M
n i 1 n
m i ri mi x i
i 1 n
mi yi
i 1 n
mi zi
i 1
xdm for a solid object
.and similar exp ressions for y and z.
Center of Mass #2:
Summary for Lecture 29 on November 28, 2007. Angular momentum
P mv LrP
momentum , sin gle body angular momentum , sin gle
CCW is .
po int like body
L r mv sin L I
L ri m i v i sin i dL or , dt f dt L f Li
i
( 2 po int like b
Summary for Lecture 28 on November 26, 2007: There were no new concepts; we just demonstrated and solved one statics problem:
r
rF
F
r F sin
Statics : Fi 0 i 0
CCW is positive equations
Two
CCW rotation is .
A ladder of length