Name: _______________________________________________________________________
Exam #2
Physics I
Fall 2006
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
_____ 3
_____ 5
_____ 6
_____ 7
_____ 8
_____ 9
_____ 10
M/R 8-10 (Bedrosian)
M/R 10-12 (Bedrosian)
M/R 12-2 (Zhang)
M/R 2-4 (Schroeder)
T/F 10-12 (Wetzel)
T/F 12-2 (Wetzel)
T/F 2-4 (Eah)
M/R 4-6 (Bedrosian)
T/F 12-2 (Eah)
Questions
A
Value
32
B-1
20
B-2
20
C
28
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 answer(s) in the context of what we have learned in Physics I.
On graphing and numerical questions (Parts B and C), show all work to receive credit.
Part A – Multiple Choice – 32 Points Total
Write your choice(s) on the line to the left of the question number.
Questions on this page are worth 4 points each.
_______1.
_______2.
A)
B)
C)
D)

A force, F , has the following property: the right-hand side of Equation 20 on the
Formula Sheet always gives exactly zero whenever the initial and final points of
integration are the same point. What can you correctly conclude?
Pick all that apply or put "0" if none of them apply.

F is a conservative force.

Momentum is conserved for any system acted on by F only.

The direction of F is at a right angle to the motion of the system it acts upon.

F cannot change the kinetic energy of the system it acts upon.
_______3.
A)
B)
C)
D)
E)
Which equation on the Formula Sheet best expresses the definition of potential
energy? Write the number to the left.
An ice skater begins a spin with her arms out. As she brings her arms in closer to
her body, her angular speed increases. After a few seconds, she extends her arms
out again, her angular speed slows back to its original value, and then she breaks
out of the spin to complete her performance. Neglecting the friction of her skates
on the ice and air resistance during the spin, what principle of physics should you
use to explain the changes in her angular speed?
Conservation of linear kinetic energy.
Conservation of rotational kinetic energy.
Conservation of linear momentum.
Conservation of angular momentum.
None of the above.
_______4.
Sphere A has mass = M and radius = 2R. Sphere B has mass = 3M and radius = R.
Both spheres have uniform density. Which has the greater rotational inertia about
its center? [You do not need the formula to answer this question.]
A) Sphere A.
B) Sphere B.
C) Both have the same rotational inertia.
2
A
B
mass = M
radius = 2R
mass = 3M
radius = R
Name: _______________________________________________________________________
Questions 5-12 refer to the figure below. Each question on this page is worth 2 points.
Two objects of equal mass, A and B, are moving in uniform circular motion about the center
marked with an “X”. Object A is closer to the center. The speed is v = 1.0 m/s for both objects.
v = 1 m/s (both)
A
B
For the quantities given in the question below, pick the object that has the larger quantity from
the following choices:
A)
B)
C)
D)
The quantity is larger for object A.
The quantity is larger for object B.
The quantity is the same for both objects.
There is insufficient information to determine which object has the larger quantity.
For example, if the quantity is “speed,” the answer is C.
_______5.
Magnitude of angular velocity.
_______6.
Magnitude of linear acceleration.
_______7.
Magnitude of angular acceleration.
_______8.
Magnitude of angular momentum.
_______9.
Kinetic energy.
_______10. Period.
_______11. Magnitude of net force acting on it.
_______12. Magnitude of net torque acting on it.
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Name: _______________________________________________________________________
B-1 – Graphing – 20 Points
Carts A and B are moving in one dimension on a frictionless track. They undergo an elastic
collision. Cart A is moving at +4.00 m/s before the collision and –3.50 m/s after the collision as
shown in the graph below. Plot the velocity of Cart B.
Make sure to include the following features – they should be clearly shown on your graphs:
1. Clearly labeled vertical axis with values and units.
2. General shapes of the curves, noting any points where the curvature or slope changes.
Show all work, including what equations and/or principles of physics you are using.
Cart A
Cart B
0.50 kg
1.50 kg
VA (m/s)
4
3
2
1
t (msec)
0
-1
5
10
-2
-3
-4
VB (m/s)
0
t (msec)
5
10
4
Name: _______________________________________________________________________
B-2 – Graphing – 20 Points
A mass on a vertical spring begins its motion at rest at y = 0 cm. It reaches a maximum height
of y = 10 cm. The two forces acting on the mass are gravity and the spring force. The graph of
the total potential energy (PE) of the system versus position is given below.
Graph the kinetic energy of the mass and the total force on the mass versus y.
Be sure to include:
1. Shape of the curve(s).
2. Minimum and maximum points.
3. Clearly labeled scales on the axes.
Show all work, including what equations and/or principles of physics you are using.
PE (J)
2
1
y (cm)
0
-1
5
10
-2
KE (J)
y (cm)
0
5
10
0
Net Force (N)
0
y (cm)
5
10
5
Name: _______________________________________________________________________
Problem C (28 Points) – Unbalanced Ferris Wheel
A Ferris Wheel for children at an amusement park was designed for children up to 20.0 kg.
However, a poorly-trained attendant allowed a parent with a mass of 80.0 kg on the ride with
three 20.0 kg children, all sitting at 90° from each other on the wheel. The drive belt broke when
the parent was at the top. At that instant, the Ferris Wheel was rotating at +1.00 rad/s.
After the belt broke, the friction torque on the Ferris Wheel was a constant –339 N-m. The only
other torque was due to the unbalanced weights of the riders. The rotational inertia of the Ferris
Wheel around its center was 2.125 × 103 kg-m2 without the riders. Treat the riders as point
masses located 2.50 m from the center.
Find the rotational inertia of the system, the change in gravitational potential energy from the
initial position (parent on top) to the position rotated 180° (parent on bottom), the work done by
friction during 180° of rotation, and the rotation speed of the wheel after 180° of rotation.
80.0 kg
1.00 rad/s
20.0 kg
20.0 kg
20.0 kg
(Show additional work on the next page if you need more space.)
Rotational Inertia = _____________________________________________ units ____
Change in Gravitational PE = _____________________________________________ units ____
Work Done by Friction = _____________________________________________ units ____
Speed of Wheel = _____________________________________________ units ____
6
Name: _______________________________________________________________________
This page is for additional work on problem C if you need it. Don't put work for Part B here please.
7
Name: _______________________________________________________________________
Formula Sheet for Homework and Exams – Page 1 of 2


U    Fcons  dx
1.
v  v 0  a t  t 0 
23.
2.
x  x 0  v 0 ( t  t 0 )  12 a ( t  t 0 ) 2
24.
U g  m g (y  y 0 )
3.
x  x 0  12 ( v0  v)( t  t 0 )
25.
U s  12 k ( x  x 0 ) 2
4.
x  x 0  v( t  t 0 )  12 a ( t  t 0 ) 2
26.
27.
28.
 K   U  Wnoncons
s  r
v tangential   r
29.
a tangential   r
6.
v 2  v 02  2a x  x 0 
 

 F  Fnet  m a
7.
T
8.
a centripetal 
5.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
2r
v
v2
 2 r
r
a radial  a centripetal


p  mv

 
dp
F

F


net
dt



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