Name:_______KEY__________________
Phys.115 Exam II
30 October 2006
Please do not turn the page until you are told to do so. Make sure that you have all three problems on your
copy of the test. In order to get credit on a problem, you must show your work. If you only write down an answer
without the work leading up to it, you will get no credit for it, even if it is the right answer.
1) A bowling ball encounters a 0.760 m vertical rise on the way back to the ball rack, as the drawing illustrates.
Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of
the ball is 3.50 m/s at the bottom of the rise.
0.760 m
a) (7 points) Please find the translational speed at the top.
Ei  E f
KETi  KERi  PEi  KET f  KER f  PE f
1 2 1 2
1
1
mvi  Ii  0  mv 2f  I 2f  mgh
2
2
2
2
2
1 2 1  2 2  vi 
1
12
 v f 
mvi   mr    mv 2f   mr 2    mgh
2
25
2
25
 r 
 r 
7 2 7 2
vi  v f  gh
10
10
10
10
2
v f  vi2  gh  3.50 m/s  
9.8 m/s 2 0.760 m   1.3 m/s
7
7
b) (6 points) Assume that after the scenario described above, for whatever reason, the ball stops spinning and it
slides with a constant speed of 1.0 m/s. At that speed, it collides inelastically with a rubber stopper at the end of
the rack and sticks to it. If the mass of the stopper is twice that of the ball, what is the speed of the stopper after
the collision? (Ignore friction between the ball/stopper and floor from now on.)
One-dimensional inelastic collision
pi  p f
2


mballvi  0  mball  mstopper v f
1
1
mballvi  mball  2mball v f ;  v f  vi  1.0 m/s   0.33 m/s
3
3
c) (7 points) The stopper is connected to a spring which will compress by 0.10 m before coming to rest. With
what angular frequency will the ball and stopper oscillate around the equilibrium point?
Conservation of energy (KE at equilibrium converted to PE elastic at maximum spring compression)
Ei  E f
KEi  PEel f
1
mball  mstopper v 2f  1 kx2
2
2

v 2f v f 0.33 m/s
k



 3.3 rad/s (or Hz)
m
x2
x
0.10 m
Problem1
Problem2
Problem3
Total/60
2) A 22-kg girl is bouncing on a pogo stick. Each bounce takes her 31 cm into the air, above the release point
of the pogo stick. Assume that the pogo stick is very light (almost massless).
a) (6 points) Using conservation of energy, what is the vertical component of her velocity at the start of each
bounce (just when she leaves the ground)?
Ei  E f
KEi  PE f
1 2
mv  mgh
2 i


v  2 gh  2 9.8 m/s 2 0.31 m   2.5 m/s
b) (7 points) Given that the spring in the pogo stick has a spring constant of k =106 N/m and the spring
compresses a distance of 1.0 cm at the bottom of the bounce, how much energy is lost to friction in the
mechanism of the pogo stick during the compression of the spring (you can ignore the gravitational potential
energy of the girl, as it is negligible compared to the net energy)?
1
1
1
1
2
2
W f  E f  Ei  kx2  mv 2  10 6 N/m 0.010 m   22 kg 2.5 m/s   18.75 J
2
2
2
2
19 J of energy were lost.
c) (7 points) Remembering that the girl must provide twice that energy at the bottom of each bounce (to
overcome the energy lost both while it is compressing and uncompressing), how high would she be able to
jump if she jumped with the same amount of energy on solid ground?
38 J  mgh
38 J
h
 18 cm
22 kg 9.8 m/s 2 
3) (20 points) The drawing shows an outstretched arm (0.61 m in length) that is parallel to the floor. The arm
is pulling downward against the ring attached to the pulley system in order to hold the 98 N weight stationary.

To pull the arm downward, the latissimus dorsi muscle applies the force M in the drawing, at a point that is
0.069 m from the shoulder joint and oriented at an angle of 29 . The arm has a weight of 47 N and a center of

gravity (cg) that is located 0.28 m from the shoulder joint. Find the magnitude of M .
T=98 N
mg

axis at
shoulder
jo int
0
 (0.069 m) M sin 29  (0.28 m)mg  (0.61 m)T  0
M 
 (0.28 m)mg  (0.61 m)T  (0.28 m)( 47 N)  (0.61 m)(98 N)

 1394 N
(0.069 m) sin 29
(0.069 m) sin 29
2R
v
1
KE  mv 2
2
T
W  ( F cos ) s
PE  mgh
Wnc  E f  Ei
E  KE  PE


p  mv


pi  p f
s  r

t


t

  0  t
1
2
2
2
  0  2
   0  0t  t 2
vt  r
at  r
ar 
T
v2
  2r
r
2

KE 
1 2
I
2
  Fl
 net  I
1
1 2
I CM  2  mvCM
2
2
L  I
WR  
Fel   kx
KE 
x  A cos t
v  A sin t
a   2 A cos t
2
 2f
T
k

m
1
PEel  kx2
2


mgL
I

g
L
;
f 
1
T
 L 
 A
F  Y 
 L0 
F
P
A
 V 

P   B 
V
0


.......... other ..........
f  FN
Ax  Bx 2  C  0
x
 B  B 2  4 AC
2A
.........constants .........
g  9.8 m/s 2