Name _______________________________________________________________________
Exam #2
Physics I
Fall 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.
Your name should be at the top of every page.
Section #
_____ 1
_____ 2
_____ 3
_____ 5
_____ 6
_____ 7
_____ 8
_____ 10
M/R 8-10 (Bedrosian)
M/R 10-12 (Wetzel)
M/R 12-2 (Wetzel)
T/F 10-12 (Bedrosian)
T/F 12-2 (Cummings)
T/F 2-4 (Cummings)
M/R 2-4 (Schroeder)
M/R 12-2 (Bedrosian)
Questions
Part A
Value
24
B-1
16
B-2
24
C-1
16
C-2
20
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 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 – 24 Points Total (6 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.
What principle of physics is expressed by equation 22 on the Formula Sheet?
Conservation of Momentum.
The Impulse-Momentum Theorem.
Conservation of Mechanical Energy.
The Work-Kinetic Energy Theorem.
Which of the following conditions are necessary to solve a problem correctly using
equations 45a and 45b on the Formula Sheet?
Put all that apply or “0” if none.
Two objects collide.
All velocities are in one dimension only (+ or –).
Total forces external to the system sum to zero or are negligible during the event.
The total kinetic energy of the system before the event is equal to the total kinetic
energy of the system after the event.
Two drivers with the same mass have a drag race in two electric cars, A and B.
Both cars experience the same net force as a function of time as shown in the graph
below and both start from rest at t = 0. The mass of car A is greater than the mass
of car B. Which car has greater kinetic energy at t = 20 seconds?
F (N)
Fmax
A.
B.
C.
D.
Car A.
t (sec)
0
Car B.
0
10
20
Both have the same kinetic energy at t = 20 seconds.
There is not enough information to determine which car has the greater kinetic
energy at t = 20 seconds.
2
Name _______________________________________________________________________
______ 4.
A.
B.
C.
D.
E.
F.
______ 5.
A.
B.
C.
D.
______ 6.
A.
B.
C.
D.
E.
F.
 
What is the direction of a  b for the vectors shown below?
To the right on the page.
To the left on the page.
Up on the page.
Down on the page.
Into the page.
Out of the page.
a
b
A satellite in space spinning on its axis uses its internal motors to unfold its solar
panels as shown in the figure below, so that in the final position its solar panels are
farther from the axis of rotation than in the initial position. During the unfolding of
the solar panels, which quantity below is conserved (stays constant)?
initial
Rotational inertia.
Angular velocity.
Angular momentum.
Rotational kinetic energy.
final
An automobile is moving north when the driver applies the brakes and stops. What
is the direction of the torque on the wheels of the car while slowing down?
North
South.
East.
West.
Up (toward the sky).
Down (toward the ground).
N
W
E
S
3
Name _______________________________________________________________________
B-1 – Graphing – 16 Points
A man went on a bicycle ride over a total horizontal distance of 10 km. Although he ended at
the same height he began, during the trip he went up and down over hills as shown in the graph
below. h is the height above his starting point and d is the horizontal distance ridden.
Being a professor of physics, he carefully pedaled the bicycle so that the work done pedaling
plus the work done by friction and air resistance equaled zero at every point along the way. He
was moving at 5 m/s at the highest point of the trip. The total mass of man + bicycle = 80 kg.
Plot the gravitational potential energy (U) and the kinetic energy (K) of the rider versus d.
Assume U = 0 at the starting point (d = 0) and use g = 9.8 N/kg.
h (m)
10.0
5.0
0
d (km)
2.0
4.0
6.0
8.0
10.0
-5.0
U (J)
d (km)
0
2.0
4.0
6.0
8.0
10.0
K (J)
d (km)
0
2.0
4.0
6.0
8.0
10.0
4
Name _______________________________________________________________________
B-2 – Graphing – 24 Points
The particle shown below begins at time t = 0.0 s at location (0.0,0.5) meters with zero speed. It
is subjected to a constant net force in the +X direction of 0.20 N for 10.0 s.
The Z axis (not shown) is positive out of the page. “+” marks the origin (0.0,0.0).
(0.0,0.5)
F
Y
X
origin
(0.0,0.0)
Part I –
Plot the linear momentum of the particle in the X direction (px) as a function of time.
Make sure to show the initial and final values clearly.
px (kg m/s)
0
t (s)
2.0
4.0
6.0
8.0
Part II is continued on the next page.
5
10.0
Name _______________________________________________________________________
Part II –
Plot the angular momentum of the particle (lx, ly, lz, ) as a function of time. We are asking for
all three (X,Y,Z) components of the angular momentum vector for this part of the problem.
Make sure to show the initial and final values clearly, and put the correct units on the l axes.
lx (
)
0
t (s)
2.0
ly (
4.0
6.0
8.0
10.0
)
t (s)
0
2.0
lz (
4.0
6.0
8.0
10.0
)
t (s)
0
2.0
4.0
6.0
8.0
10.0
6
Name _______________________________________________________________________
Problem C-1 (16 Points)
A hydrogen atom moving at +5.0 x 106 î m/s collided with a helium atom initially at rest. After
the collision, the velocity of the hydrogen atom was –3.0 x 106 î m/s. For this problem, we will
assume that the mass of a helium atom is exactly four times the mass of a hydrogen atom (to
make the calculations easier) and we will neglect all external forces during the collision.
Was the collision elastic? Explain your answer in a few sentences supported by calculations.
initial
+5.0 x 106 m/s
H
He
final
–3.0 x 106 m/s
?
H
7
He
Name _______________________________________________________________________
Problem C-2 (20 Points)
A 4.0 kg mass is attached to the end of a thin rod. The rod’s mass is 3.0 kg and its total length
is 0.50 meters. The rod is free to rotate about its center point without friction. The rod begins at
rest in a horizontal position, the initial position. Because of the unbalanced mass on the end, the
rod and attached mass will rotate when released. (“Down” is toward the bottom of the page.)
What is the angular speed of the rod with the attached mass when the mass reaches its lowest
point? Use g = 9.8 N/kg. You will find the following formula helpful:
I rod  121 M L2
L = 0.5 m
initial
final
Angular speed = _______________________________________ units ________
8
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
9
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
10