Physics I – Exam 2 – Spring 2004
Answer Key
Part A – 1. B, 2. C, 3. 41, 4. A 4 points each, no partial credit.
B-1 8 Points
Potential energy at the top =
potential energy at the bottom:
m g h = ½ k (-h)2
k = 2 m g / h = 98. N/m
Net Force (N)
+4.9
B-2 22 Points
Straight line for force as shown.
Parabolas for energy as shown.
Max/min energy at 5 cm
Features
Force:
straight line
slope = -98 N/m
F = 0 at y = 5 cm
Total PE:
parabola
0.49 J at y = 0 or y = 10.
min value at 5 cm = 0.3675 J
same value at y = 10 and y = 0.
KE:
inverted shape of PE curve
0 J at y = 0 or y = 10.
max value at 5 cm = PEmax – PEmin
same value at y = 10 and y = 0.
0
y (cm)
5
10
5
10
5
10
-4.9
Total PE (J)
0.49
0.3675
y (cm)
0
KE (J)
0.1225
y (cm)
0
0
Part C
Must show work to receive credit.
C-1 16 points
Can solve this problem in angle units of rev, rev/s, and rev/s2.
Conversion from rev to rad is OK but unnecessary.
t 2  4 t1
Equation 31: 1  12 1 t 1  0.4 t 1
2
2
Equation 33:  2   12  2 t 2  0.1 t 2  1.6 t 1
2
2
2
  1   2  2.0 rev  2.0 t 1  t 1  1s
t1  t 2  5 t1  5s
2
C-2 18 points
Equation 26: (0  12 m v 2 )  (m g h  0)  F d
1
(80)(8) 2  (66)  2352.655 J
2
2352.655
h
 3.0 m
(80)(9.8)
mgh 
C-3 20 points
Use conservation of angular momentum. It is in the +Z direction.
  


The bug: l  r  p  (0.1)(0.003)( 2.0) k  6.0 10 4 k kg m 2 / s
(y directed velocity does not contribute to the cross product.)
OK to consider instantaneous 0 = vtang/r of bug before landing and then L0 = Ibug 0.
I total  I turntable  I bug  5.97 10 3  (0.003)(0.1) 2  6.00 10 3 kg m 2


L 6.0  10 4
L  I total     
 0.1 rad / s
I 6.0  10 3