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 cir
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 cr
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
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
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
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
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
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 repre
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/
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 cir
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 cr
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
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
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 iner
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
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
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.
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
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 pe
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
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
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
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
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
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 elec
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 1
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
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