Physics I – Exam 2 – Spring 2003
Answer Key
Part A –
1. C, 2. B, 3. B,
4. D, 5. B, 6. A
B-1
 0
 (3600) 2

 648 rev / min 2
2  2 (10000)
2

rev 2 rad 1 min 2
 1.13 rad / s 2
2
2
min 1 rev 3600 sec
  I   50 (1.13)  56.55 N m (question asks for magnitude, don’t need – sign)
Convert to rad/s2:   648
B-2
Both pucks have the same mass, so we can divide m from the momentum equations and
just deal with the components of velocity. First, find the X and Y components of final
velocity for puck A:
v AfX  2.5 cos(20)  2.349 v AfY  2.5 sin( 20)  0.855
Now find the X and Y components of final velocity of puck B using cons. of mom.
v BfX  v AiX  v AfX  4.000  2.349  1.651 v BfY  v AiY  v AfY  0.000  0.855  0.855
Finally, get magnitude and angle
v Bf  v BfX  v BfY  (1.651) 2  (0.855) 2  1.859 m / s
2
v
 B  arctan  BfY
 v BfX
2

  0.855 
  arctan 
  27.38
 1.651 

B-3
K  U  (K f  K i )  ( U f  U i )  Wnoncons
K f  K i  U f  U i  Wnoncons
K f  0  0  12 k x 2  Ffriction d  0.5 *125 * (0.02) 2  0.0475 * 0.10  0.02025 J
1
2
m v2  Kf
v
2 Kf
2 * 0.02025

 9m/s
m
0.0005
B-4
I1 1  L  (I1  I book ) 2
I book  I1
1  2
2
 3  2.5 
3
2
I book  0.010 * 
  2.0  10 kg m
2
.
5


C-1
A. I TOTAL  I student  I stool  I ball  1.4  0.1  m ball r 2
I TOTAL  1.4  0.1  m ball r 2  1.5  0.1 * (0.8) 2  1.564 kg m 2
B. Positive direction for angular momentum is UP. The angular momentum of the ball
(before it is caught) is in the negative (DOWN) direction.
L before  L studentstool  L ball  (I student  I stool ) before  r m ball v ball
L before  (1.4  0.1) * 4  0.8 * 0.1 *16.35  4.692 kg m 2 / s
L
4.692
after 

 3 rad / s
L before  L after  I TOTAL after
I TOTAL 1.564
C-2
The warning in the problem is clear that it is incorrect to do this problem with the concept
of “average acceleration” or “average force”. The correct solution of this problem
requires the Impulse-Momentum Theorem and the Work-Energy Theorem.
Let the +X direction be east and the +Y direction be north. The initial momentum is
p Xi  m v  10 kg m / s.
p Yi  0
The impulse changes the momentum:
p Xf  p Xi  J X  10  18  8 kg m / s.
p Yf  p Yi  J Y  0  6  6 kg m / s
Divide by mass to get velocity and get the final speed.
v Xf  p Xf / m  8 / 2  4 m / s.
v Yf  p Yf / m  6 / 2  3 m / s
v f  v Xf  v Yf  (4) 2  (3) 2  5 m / s
Since the speed did not have a net change, only the direction of velocity, the final and
initial kinetic energies are equal. Therefore, by the Work-Energy Theorem:
The force did no (zero) net work over the specified time interval. Answer = 0 J.
2
2