Physics I – Exam 2 – Fall 2003
Answer Key
Part A – 1. C, 2. B, 3. B, 4. C
4 points each, no partial credit.
Part B 16 Points
1.
Knowing to use (m g h) for the potential energy at the top.
2.
Knowing to use (K + U = –Fd) for the kinetic energy at the bottom.
3.
Straight line for PE from whatever at d = 0 to PE = 0 at d = 100 cm.
4.
Straight line for KE from KE = 0 at d = 0 to whatever at d = 100 cm.
5.
PE = (0.5) (9.8) (0.05) = 0.245 J at d = 0 cm.
6.
KE = 0.245 – (0.1) (1) = 0.145 J at d = 100 cm.
Part C
C-1 16 points
2
0  3600 
 377 rad / s
f  0
60
  0
 377
 f

 0.02618 rad / s 2
t
4  3600
  I   38.2  0.02618  1.00 N m
(The minus sign is dropped because the problem asks for magnitude, but OK if there.)
For # of revolutions, there are three methods:
A:    0 ( t )  12  ( t ) 2  377  14400  0.02618  14400 2  2  2.71 10 6 rad
B:    f ( t )  12  ( t ) 2  0  0.02618  14400 2  2  2.71 10 6 rad
C:    12 (0  f ) ( t )  (3600  0)( 4  60)  2  432000 rev
In method C, we can use units of rev and min.
C-2 16 points
1D work for variable force: W   F dx  Area  12  0.5  8000  2000 J
Work-Kinetic Energy Theorem: K  W  12 m v 2  W  2000 J
2W
2  2000

 1000 m / s
m
0.004
No points for trying to solve this with “Favg = m a” even if getting right answer!
v
C-3 8 points
Conservation of angular momentum: I1 1  I 2 2  2 
I1 1 1.25  12

 20 rad / s
I2
0.75
C-4 8 points
2
2
Work-Kinetic Energy Theorem: K  W  W  12 I 2 2  12 I1 1
W  0.5  0.75  20 2  0.5  1.25  12 2  150  90  60 J
C-5 12 points
1D elastic collision equations (for second object initially at rest):
m  m2
2 m1
0.5  1.5
2  0.5
v1f  1
v1i 
 4  2 m / s v 2f 
v1i 
 4  2 m / s
m1  m 2
0.5  1.5
m1  m 2
0.5  1.5
C-6 8 points
Conservation of mechanical energy: K  U  0  12 m v 2  12 k x 2
xv
m
1.5
 2
 0.0400 m  4.00 cm
k
3750