Name _______________________________________________________________________
Exam #2
Physics I
Spring 2007
If you would like to get credit for having taken this exam, we need
your name (printed clearly) at the top and section number below.
Your name should be at the top of every page.
Section #
_____ 1
_____ 2
_____ 4
_____ 5
_____ 7
_____ 9
_____ 10
_____ 11
_____ 12
_____ 14
_____ 15
M/R 8-10 (Bedrosian)
M/R 10-12 (Wilke)
M/R 12-2 (Yamaguchi)
M/R 2-4 (Yamaguchi)
M/R 4-6 (Wilke)
T/F 10-12 (Wetzel)
T/F 10-12 (Washington)
T/F 12-2 (Eah)
T/F 2-4 (Eah)
M/R 12-2 (Zhang)
M/R 2-4 (Bedrosian)
Questions
Part A
Value
40
Part B
24
Part C
36
Total
100
Score
You may not unstaple this exam.
Only work written on the same page as the question will be graded.
Cheating on this exam will result in an F in the course.
1
Name _______________________________________________________________________
On this exam, please neglect any relativistic and/or quantum mechanical effects. If you don’t
know what those are, don’t worry, we are neglecting them! On all multiple-choice questions,
choose the best answers in the context of what we have learned in Physics I.
On graphing and numerical questions, show all work to receive credit.
Part A – Multiple Choice – 40 Points Total (10 at 4 Points Each)
Write your choice(s) on the line to the left of the question number.
______ 1.
A)
B)
C)
D)
______ 2.
A)
B)
C)
D)
______ 3.
A)
B)
C)
D)
______ 4.
A)
B)
C)
D)
Equation 26 on the Formula Sheet expresses Conservation of Mechanical Energy
if and only if …
Non-conservative forces do no work.
The net external force is zero.
The change of kinetic energy equals the net work done.
Objects collide and bounce off each other without sticking together.
Comparing Equations 20 and 23 on the Formula Sheet, why was the subscript
"cons" added to the force vector in Equation 23?
Potential energy can only be defined for conservative forces.
Potential energy can only be defined when the force is constant.
Potential energy can only be defined when momentum is conserved.
Adding subscripts to equations makes them look more impressive.
Consider a coin glued to the outer edge of a rotating disk. We define the angle of
rotation of the coin about the center of the disk, , so that  = 0° at time t = 0 s.
In order to use Equation 31 correctly to predict the rotation angle of the coin at a
later time, it is necessary that
The angular speed of the coin is constant.
The linear speed of the coin is constant.
The centripetal acceleration of the coin is constant.
The tangential acceleration of the coin is constant.
Consider a system consisting of two objects that collide. We define an origin of
coordinates to measure their locations. In order for the angular momentum of the
system about the origin to be conserved, it is necessary that
The collision is elastic.
The net external force on the system is zero.
The net external torque on the system is zero.
The total work done on the system by all non-conservative forces equals zero.
2
Name _______________________________________________________________________
Questions 5-10 refer to Tarzan swinging from a vine
as shown in the figure to the right. In the initial
position, A, Tarzan is beginning the swing to Jane
(shown on the right) from rest at a height h above the
lowest point of the swing. Point B is the lowest point
of the swing. Assume the vine has a constant length,
d, and negligible mass. Ignore air resistance.
The illustration shows Tarzan as he starts.
Point A is his initial location starting at rest.
Point B is the lowest point of the swing.

The attachment point of the vine is the coordinate
origin for calculating angular momentum and torque.
______ 5.
A)
B)
C)
D)
During Tarzan's swing between A and B,
which quantities below are conserved?
Select all that apply or put "0" for none.
A
B
Kinetic Energy of Tarzan.
Potential Energy of Tarzan in gravity.
Total Mechanical Energy of Tarzan.
Angular Momentum of Tarzan.
For questions 6-10, select from the following directions.
A)
B)
C)
D)
To the right on the page.
To the left on the page.
Up on the page.
Down on the page.
E)
F)
G)
H)
Right and down on the page.
Out of the page.
Into the page.
No direction; quantity = 0.
______ 6.
When Tarzan is at point A (just starting from rest with his feet off the rock), what is
the direction of his angular momentum?
______ 7.
When Tarzan is at point A (just starting from rest with his feet off the rock), what is
the direction of the total torque on him?
______ 8.
When Tarzan is at point B (the lowest point), what is the direction of his angular
momentum?
______ 9.
When Tarzan is at point B (the lowest point), what is the direction of the total
torque on him?
______ 10. Suppose Tarzan let go of the vine after reaching point B. In the instant after letting
go, what would be the direction of Tarzan's angular momentum?
3
Name _______________________________________________________________________
B – Graphing – 24 Points
An object of unknown mass starting with 1.0 J of kinetic energy moving in the +X direction
moves in one dimension and is acted on by a net conservative force, from x = 0 to x = 10 cm.
The graph of total potential energy (U) is shown below. Note that the curved portions of the
potential energy graph are parabolas.
Graph the kinetic energy (K) and net force acting on the object as functions of displacement (x).
Make sure your plots clearly show:
A. Any minimum or maximum points.
B. Whether the graph segments are curved or straight.
C. The values of K and net force (Fnet) at x = 0, 2, 4, 5, 6, 8, and 10 cm.
Show all work, including what equations and/or principles of physics you are using.
Note: You may not assume an arbitrary value for the object’s mass to do this problem.
U (J)
2
1
0
x (cm)
-1
2
4
6
8
10
-2
K (J)
x (cm)
2
4
6
8
10
Fnet (N)
x (cm)
2
4
6
8
10
4
Name _______________________________________________________________________
Problem C (36 Points) – Ball on a Stick
Please note that this is a multi-step problem leading to the final answer. While each step will be
graded individually, you will want to double-check the earlier steps carefully. The final
numerical answer to the problem has been provided so that you can check your work. We will
be grading primarily on whether you are using correct equations and/or principles of physics.
A ball on a stick begins at rest in position A, with the stick horizontal. The stick is free to rotate
about a frictionless pivot at its center. In position B, the ball has moved to its lowest position
vertically below the pivot. We want to find the angular momentum of the ball and stick in
position B. The pivot point is the origin for calculating all angular quantities.
Position A
Position B
Y
g=
9.8 N/kg
5.0 kg
3.75 kg
stick
X
pivot
Y
X
+Z is out of the page
The mass of the stick is 3.75 kg and its total length is L = 4.0 m. Use I = 1/12 M L2 for the stick.
The mass of the ball is 5.0 kg. Treat it as a point mass on the end of the stick.
The system is the stick and the ball. Neglect air resistance and use g = 9.8 N/kg.
C-1: Can you correctly use Conservation of Angular Momentum to solve this problem? Explain
why or why not in one or a few complete sentences. (5 Points)
C-2: Can you correctly use Conservation of Mechanical Energy to solve this problem? Explain
why or why not in one or a few complete sentences. (5 Points)
5
Name _______________________________________________________________________
C-3: Calculate the rotational inertia of the system about the pivot point. (3 Points)
C-4: Calculate the angular acceleration of the system in Position A. (5 Points)
C-5: Calculate the angular acceleration of the system in Position B. (4 Points)
6
Name _______________________________________________________________________
C-6: Calculate the change in potential energy from Position A to Position B. (4 Points)
C-7: Calculate the change in kinetic energy from Position A to Position B. (4 Points)
C-8: Calculate the angular momentum of the system at Position B. (6 Points)
Answer: 70 kg m2/s in –Z direction.
Important:
Did you clearly show what equation(s) and/or principle(s) of physics you used?
Did you include units for all of your numerical answers?
For vector quantities, did you indicate both the magnitude and direction?
7
Formula Sheet for Homework and Exams – Page 1 of 2


U    Fcons  dx
1.
v f  v 0  a t f  t 0 
23.
2.
x f  x 0  v 0 ( t f  t 0 )  12 a ( t f  t 0 ) 2
24.
U g  m g (y  y 0 )
3.
x f  x 0  ( v 0  v f )( t f  t 0 )
25.
U s  12 k ( x  x 0 ) 2
4.
x f  x 0  v f ( t  t 0 )  12 a ( t f  t 0 ) 2
26.
27.
28.
 K   U  Wnoncons
s  r
v tangential   r
29.
a tangential   r
5.
6.
1
2
v f  v 02  2ax f  x 0 
 

 F  Fnet  m a
2
7.
2r
T
v
8.
a centripetal 
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
v2
 2 r
r
a radial  a centripetal


p  mv

 
dp
 F  Fnet  d t



J   Fnet dt   p


P   pi


dP
  Fext
dt
30.
  0  t  t 0 
31.
   0  0 ( t  t 0 )  12 ( t  t 0 ) 2
32.
   0  12 (0  )( t  t 0 )
33.
   0  ( t  t 0 )  12 ( t  t 0 ) 2
 2  02  2   0 
   
35a. a  b  a b sin( )
 
a  b  a y b z  a z b y î 
35b.
a z b x  a x b z  ĵ  a x b y  a y b x k̂
34.

36.
37.
M   mi
38.
1
1
x cm   m i x i y cm   m i y i
M
M


P  M v cm
   
a  b  a b cos()  a x b x  a y b y  a z b z
 
W  Fd
 
W   F  dx
21.
K  12 m v 2  12 m (v x  v y )
22.
K f  K i  Wnet
2
39.
40.
41.
42.
43.
2
I   m i ri


2
K rot  12 I  2
 
W     d
  
  r F

 dL

  I  d t
  
l  r p


L  l i


L  I
44x. m1 v1, x ,before  m 2 v 2, x ,before  m1 v1, x ,after  m 2 v 2, x ,after
44y. m1 v1, y ,before  m 2 v 2, y ,before  m1 v1, y ,after  m 2 v 2, y,after
44z. m1 v1,z ,before  m 2 v 2,z ,before  m1 v1,z ,after  m 2 v 2,z ,after
45a. v1,f 
m1  m 2
2 m2
v1,i 
v 2 ,i
m1  m 2
m1  m 2
45b.
8
v 2,f 
2 m1
m  m1
v1,i  2
v 2 ,i
m1  m 2
m1  m 2

Formula Sheet for Homework and Exams – Page 2 of 2
46a.
46b.
47a.
47b.
48a.
48b.
49.

m m
| F | G 1 2 2
r

m m
F  G 1 2 2 r̂
r

1 | q1 || q 2 |
| F |
4  0
r2

1 q1 q 2
F
(r̂ )
4  0 r 2

1 | qi |
| Ei |
4   0 ri 2

1 qi
E
(r̂i )
4   0 ri 2


F  qE
50.
51.
52.
1 qi
4   0 ri
U  qV
 
V    E  dx
V
V
x
V
53y. E y  
y
V
53z. E z  
z

 
54. F  q v  B
mv
55. r 
qB
53x. E x  
Useful Constants
(You can use the approximate values on exams.)
Universal Gravitation Constant
G  6.67310 11 N m 2 kg 2  6.67 10 11
Electrostatic Force Constant
1
 8.987551788 10 9 N m 2 C  2  9.0 10 9
4  0
Magnetic Constant
 0  4  10 7 H m 1  1.26 10 6
Speed of Light in Vacuum
c  2.99792458 10 8 m s 1  3.0 10 8
Charge of a Proton
e  1.602176462 10 19 C  1.6 10 19
Electron-Volt Conversion Constant
1eV  1.602176462 10 19 J  1.6 10 19
Mass of a Proton
m p  1.6726215810 27 kg  1.67 10 27
Mass of an Electron
m e  9.10938188 10 31 kg  9.110 31
9