Name ________________________________________________________________________
Exam #2
Physics I
Spring 2004
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.
Section #
_____ 1
_____ 2
_____ 3
_____ 4
_____ 5
_____ 6
_____ 7
_____ 9
_____ 10
_____ 11
_____ 12
_____ 14
_____ 15
M/R 8-10 (Schowalter)
M/R 10-12 (Schowalter)
M/R 10-12 (Stoler)
M/R 12-2 (Bedrosian)
M/R 2-4 (Bedrosian)
M/R 2-4 (Schroeder)
M/R 4-6 (Bedrosian)
T/F 10-12 (Adams)
T/F 10-12 (Washington)
T/F 12-2 (Wilke)
T/F 2-4 (Wilke)
M/R 12-2 (Stoler)
T/F 12-2 (Adams)
Questions
Part A
Value
16
B-1
8
B-2
22
C-1
16
C-2
18
C-3
20
Total
100
Score
Cheating on this exam will result in an F in the course.
Sharing information about this exam with people who have not yet
taken it is cheating on the exam for both parties involved.
The Formula Sheets are the last two pages. Detach carefully for easier reference if you wish.
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 answer in the context of what we have learned in Physics I.
On numerical questions, show all work to receive credit.
Part A – Warm-Ups – 16 Points Total (4 at 4 Points Each)
Write your choice on the line to the left of the question number.
______ 1.
A.
B.
C.
D.
______ 2.
A.
B.
C.
D.
An asteroid in space initially at rest explodes into two pieces, A and B, which then
move in opposite directions. Piece A has greater mass than piece B. Which piece
has greater kinetic energy? Ignore all external forces.
Piece B
Piece A
Piece A.
Piece B.
The pieces have equal kinetic energy.
There is not enough information to determine which has greater kinetic energy.
A particle begins with kinetic energy K1 at point A. It moves along path 1 to point
B, then along path 2 back to point A. It ends with kinetic energy K2 where K2 > K1.
Which statement below can one correctly conclude?
Path 2
The net force on the particle was always centripetal.
A
The potential energy of the system decreased.
At least one non-conservative force acted on the particle.
The kinetic energy of the particle at point B is > K1 and < K2.
B
Path 1
______ 3.
Which equation on the Formula Sheet defines the angular momentum of a particle?
(Write the equation number on the line to the left.)
______ 4.
Two spherical asteroids, A and B, have equal masses and volumes. A is made of
uniform rock. B has a spherical iron core surrounded by a spherical shell of ice.
The density of rock is greater than ice but less than iron. Which asteroid has the
Asteroid A
Asteroid B
greater rotational inertia about its central axis?
Rock
A.
B.
C.
D.
Ir on
Ice
Asteroid A.
Asteroid B.
The asteroids have equal rotational inertia.
There is not enough information to determine which has greater rotational inertia.
2
Name ________________________________________________________________________
B – Graphing – 30 Total Points
zero spring force
lowest point
y = 10 cm
y = 0 cm
In the illustration above, a mass of 0.50 kg is attached to a hanging spring that is initially at its
equilibrium point (before attaching the mass). The mass is released from rest and moves
downward 10.0 cm to its lowest point. Use g = 9.8 N/kg as the acceleration constant of gravity.
Assume the spring is massless and follows Hooke’s Law: F = –k (y–y0).
B-1 – Determining the Spring Constant (8 Points)
Find the spring constant, k, from the information given above.
Reminder: To get credit, you must show what equations and/or principles of physics you used.
98.
Spring Constant k =
3
units
N/m
Name ________________________________________________________________________
B-2 – Force and Energy Plots (22 Points)
(Continued from the previous page.)
Plot the net force on the mass, total potential energy (PE) of the system, and kinetic energy (KE)
of the mass versus height (y) from y = 0 (the bottom) to y = 10 cm (the top).
Make sure to include the following features:
1. General shapes of the curves, noting any points where the curvature or slope changes.
2. The values at any minimum or maximum points.
3. The values at y = 0, y = 5, and y = 10 cm.
Net Force (N)
0
y (cm)
5
10
Total PE (J)
y (cm)
0
KE (J)
5
10
y (cm)
0
0
5
10
4
Name ________________________________________________________________________
Part C – Problems – 54 Points Total
C-1 (16 points)
A TV game show contestant spins a prize wheel starting at rest with a constant angular
acceleration of 0.8 rev/s2 for t1 seconds. She releases the wheel and it slows to a stop at a
constant angular acceleration of –0.2 rev/s2 for t2 seconds. The wheel turns a total of 2.0
revolutions during this process. What is the total time the wheel is turning (t1+t2)?
Total time = ________________________________________________ seconds
5
Name ________________________________________________________________________
C-2 (18 points)
A skateboarder begins rolling up a curved ramp at 8.0 m/s at the bottom. While he rolls on the
ramp, there is a 66 N force of friction (constant magnitude) directly opposite his motion (always
in the opposite direction the skateboard is moving). When he reaches the end of the ramp, he is
moving upward (vertically). The shape of the ramp is a circular arc covering ¼ of a circle with
radius = 2.00 m. What is the maximum height of the center of mass of the skateboarder and his
skateboard measured from its initial height before going up the ramp? The total mass of
skateboarder and skateboard = 80. kg. Ignore air resistance and use g = 9.8 N/kg.
Notes: The skateboarder is not rotating in the air. There is no friction except on the curved part
of the ramp. Treat the skateboarder as a point mass for this problem.
2 m radius
8 m/s
Maximum height of center of mass: _________________________ units _______
6
Name ________________________________________________________________________
C-3 (20 points)


A large insect with mass 3.0 g is flying with a velocity of (2.0 i  1.0 j) m/s when it lands
without slipping on a turntable at location (0.0,–10.0) cm with respect to the center of the
turntable at (0,0). The rotational inertia of the turntable is 5.97  10 3 kg m2. The turntable is
initially at rest and you can ignore the friction of its bearings. Treat the insect as a point mass.
What is the angular speed of the turntable + insect after the insect lands?

j (Y)

i (X)
Angular speed: ___________________________ units ________
7
Name ________________________________________________________________________
Formula Sheet for Homework and Exams – Page 1 of 2
1.
v  v 0  a t  t 0 
21.
2.
x  x 0  v 0 ( t  t 0 )  12 a ( t  t 0 ) 2
K  12 m v 2  12 m (v x  v y )
22.
3.
x  x 0  12 ( v0  v)( t  t 0 )
23.
K f  K i  Wnet


U    Fcons  dx
4.
x  x 0  v( t  t 0 )  12 a ( t  t 0 ) 2
24.
U g  m g (y  y 0 )
25.
U s  12 k ( x  x 0 ) 2
26.
27.
28.
 K   U  Wnoncons
s  r
v tangential   r
a tangential   r
2
2
6.
v 2  v 02  2a x  x 0 
 

 F  Fnet  m a
7.
T
8.
a centripetal 
29.
30.
  0  t  t 0 
Fcentripetal


p  mv

 
dp
 F  Fnet  d t



J   Fnet dt   p


P   pi


dP
  Fext
dt
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
M   mi
38.
5.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
2r
v
v2
 2 r
r
v2
m
 m 2 r
r
35.
 2  02  2   0 
   
a  b  a b sin( )
36.
I   m i ri
34.
37.
39.
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
 
W  Fd
 
W   F  dx
40.
41.
42.
43.
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
m1  m 2
2 m2
v1,i 
v 2 ,i
m1  m 2
m1  m 2
2 m1
m  m1

v1,i  2
v 2 ,i
m1  m 2
m1  m 2
45a. v1,f 
45b. v 2,f
8
2
K rot  12 I  2
 
W     d
  
  r F

 dL

  I  d t
  
l  r p


L  l i


L  I
Name ________________________________________________________________________
Formula Sheet for Homework and Exams – Page 2 of 2

m m
46a. | F |  G 1 2 2
r

m m
46b. F  G 1 2 2 r̂
r

1 | q1 || q 2 |
47a. | F | 
4  0
r2

1 q1 q 2
47b. F 
(r̂ )
4  0 r 2

1 | qi |
48a. | E i | 
4   0 ri 2

1 qi
(r̂i )
48b. E  
4   0 ri 2


49. F  q E
50.
51.
52.
1 qi
4   0 ri
U  qV
 
V    E  dx
V
V
x
V
53y. E y  
y

 
54. F  q v  B
mv
55. r 
qB
53x. E x  
Useful Constants
(You can use the approximate values on tests.)
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