Physics 101
Exam 1
Spring 2013
Instructions
This is an open book open notes exam. You have fifty (50) minutes to complete it.
1. Use a #2 pencil. Do not use a mechanical pencil or pen. Darken each circle
completely, but stay within the boundary. If you de
171.101
Exam 1
Spring 2013
General Physics 171.101
Midterm Exam 1
February 21, 2013
Answer all five questions. VERY IMPORTANT: YOU MUST SHOW AND EXPLAIN
ALL YOUR WORKING TO GET FULL CREDIT. Partial credit will given for partiallycorrect answers. Pace your
Instructions
This is an open book open notes exam. You have fifty (50) minutes to complete it.
1. Use a #2 pencil. Do not use a mechanical pencil or pen. Darken each circle
completely, but stay within the boundary. If you decide to change an answer, erase
General Physics 171.101
Midterm Exam 3
April 18, 2012
Answer both questions. VERY IMPORTANT: YOU MUST SHOW AND EXPLAIN ALL
YOUR WORKING TO GET FULL CREDIT. Partial credit will given for partially-correct
answers
General Physics 171.101
Midterm Exam 3
April 18, 2012
Answer both questions. VERY IMPORTANT: YOU MUST SHOW AND EXPLAIN ALL
YOUR WORKING TO GET FULL CREDIT. Partial credit will given for partially-correct
answers
171.101
Exam 3
Spring 2013
General Physics 171.101
Midterm Exam 3
April 19, 2013
Answer all four questions.
VERY IMPORTANT: YOU MUST SHOW AND EXPLAIN ALL YOUR WORKING
TO GET FULL CREDIT. THIS INCLUDES MULTIPLE CHOICE QUESTIONS.
Partial credit will given f
171.101
Exam 3
Spring 2013
General Physics 171.101
Midterm Exam 3
April 19, 2013
Answer all four questions.
VERY IMPORTANT: YOU MUST SHOW AND EXPLAIN ALL YOUR WORKING
TO GET FULL CREDIT. THIS INCLUDES MULTIPLE CHOICE QUESTIONS.
Partial credit will given f
171.101 Midterm 1
Solu1ons
No horizontal accelera1on no horizontal force T1 cos 1 = T2 cos 2
T1 = T1 cos 2 /cos 1 > T2 if 2 < 1
The dashed line above represents her nal velocity. It is composed of her
Solutions for Midterm 2
1. This is an inelastic collision. Total momentum of the two-body system is conserved.
Initial momentum in the x direction is (70 kg)(5 m/s) (50 kg)(5 m/s)cos(30) = 133.5
kg*m/s. Initial momentum in the y direction is (50 kg)(5 m/s
Physics 101
Exam 1
Spring 2013
Instructions
This is an open book open notes exam. You have fifty (50) minutes to complete it.
1. Use a #2 pencil. Do not use a mechanical pencil or pen. Darken each circle
completely, but stay within the boundary. If you de
171.101
Exam 1
Spring 2013
General Physics 171.101
Midterm Exam 1
February 21, 2013
Answer all five questions. VERY IMPORTANT: YOU MUST SHOW AND EXPLAIN
ALL YOUR WORKING TO GET FULL CREDIT. Partial credit will given for partiallycorrect answers. Pace your
General Physics 171.101
Final Exam
May 16, 2012
Answer all five questions. VERY IMPORTANT: YOU MUST SHOW AND EXPLAIN ALL
YOUR WORKING TO GET FULL CREDIT. Partial credit will given for partially-correct
answers.
171.101
Final Exam
Spring 2013
General Physics 171.101
Final Exam
May 14th, 2013
Answer all six questions.
VERY IMPORTANT: YOU MUST SHOW AND EXPLAIN ALL YOUR WORKING
TO GET FULL CREDIT. THIS INCLUDES MULTIPLE CHOICE QUESTIONS.
Partial credit will given fo
1
1.1
Problem 1
Problem Statement
A square is constructed of 4 point masses, each of mass m, and 4 massless
rods of length L = 2d. About which axis (or axes) shown below is (are) the
moment(s) of inertia the greatest?
Figure 1: Four dierent axes for the s
1. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the tota
1. With speed v = 11200 m/s, we find
K= 1 2 1 mv = (2.9 105 ) (11200) 2 = 18 1013 J. . 2 2
2. (a) The change in kinetic energy for the meteorite would be
1 1 K = K f - K i = - K i = - mi vi2 = - 4 106 kg 15 103 m/s 2 2
(
)(
)
2
= -5 1014 J
1. (a) The center of mass is given by xcom = [0 + 0 + 0 + (m)(2.00) + (m)(2.00) + (m)(2.00)]/6.00m = 1.00 m. (b) Similarly, ycom = [0 + (m)(2.00) + (m)(4.00) + (m)(4.00) + (m)(2.00) + 0]/6m = 2.00 m. (c) Using Eq. 12-14 and noting that the gravitatio
1. Using the given conversion factors, we find (a) the distance d in rods to be d = 4.0 furlongs =
( 4.0 furlongs )( 201.168 m furlong )
5.0292 m rod
= 160 rods,
(b) and that distance in chains to be d =
( 4.0 furlongs )( 201.168 m furlong )
20.1
1. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one-fourth of a period. The period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period:
f = 1 1 = = 1.47 Hz. T 0.680 s
(c) A sinuso
1. (a) Since the gas is ideal, its pressure p is given in terms of the number of moles n, the volume V, and the temperature T by p = nRT/V. The work done by the gas during the isothermal expansion is
W=
V2 V1
p dV = n RT
V2 V1
dV V = n RT ln 2 . V
1. (a) With a understood to mean the magnitude of acceleration, Newton's second and third laws lead to m2 a2 = m1a1
c6.3 10 kghc7.0 m s h = 4.9 10 m =
-7 2 2
-7
9.0 m s
2
kg.
(b) The magnitude of the (only) force on particle 1 is
q q q F = m1
1. We note that the symbol q2 is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for q1 = q2 .
The following two sketches are for the cases q1 > q2 (left figu
1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is
= E A = EA cos = (1800 N C ) 3.2 10-3 m cos145 = -1.5 10-2 N m 2 C.
2
(
)
2. We u
1. Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery. The charge on the capacitor is then q = CV, and this is the same as the total charge that has passed through the battery.
1. (a) The charge that passes through any cross section is the product of the current and time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The number of electrons N is given by q = Ne, where e is the magnitu
1. (a) Eq. 28-3 leads to 6.50 10-17 N FB v= = = 4.00 105 m s . -19 -3 eB sin 160 10 C 2.60 10 T sin 23.0 .
c
hc
h
(b) The kinetic energy of the proton is
K=
2 1 2 1 mv = 167 10-27 kg 4.00 105 m s = 134 10-16 J. . . 2 2
c
hc
h
This is
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by
B=
0i
2r
.
With r = 20 ft = 6.10 m, we have
c4 10 B=
hb 2 b6.10 mg
-7
T m A 100 A
g = 3.3 10
-6
T = 3.3 T.
(b
1. The x and the y components of a vector a lying on the xy plane are given by
ax = a cos , a y = a sin
where a =| a | is the magnitude and is the angle between a and the positive x axis. (a) The x component of a is given by ax = 7.3 cos 250 = 2