Download Ch 8 HW Day 4: p 254 – 265, #`s 5, 11 – 15, 18, 21, 67, 71 – 74

Ch 8 HW Day 4 (8.5 – 8.6): p 254 – 265, #’s 5, 11 – 15, 18, 21, 67, 71 –
74
21. Determine the Concept. We can find the loss of kinetic
energy in these two collisions by finding the initial and
final kinetic energies. We’ll use conservation of
momentum to find the final velocities of the two masses in
each perfectly elastic collision.
(a) Letting V
represent the
velocity of the
masses after their
perfectly inelastic
collision, use
conservation of
momentum to
determine V:
Express the loss of
kinetic energy for
the case in which the
two objects have
oppositely directed
velocities of
magnitude v/2:
Letting V represent
the velocity of the
masses after their
perfectly inelastic
collision, use
conservation of
momentum to
determine V:
pbefore  pafter
or
mv  mv  2mV  V  0
  v 2 
K  K f  K i  0  2 12 m  
 2 


mv2

4
pbefore  pafter
or
mv  2mV  V  12 v
Express the loss of
kinetic energy for
the case in which the
one object is initially
at rest and the other
has an initial
velocity v:
K  K f  K i
2
mv
v
 12 2m    12 mv2  
4
2
2
The loss of kinetic energy
is the same in both cases.
(b) Express the
percentage loss for
the case in which the
two objects have
oppositely directed
velocities of
magnitude v/2:
2
mv
K

 100%
2
K before
mv
Express the
percentage loss for
the case in which the
one object is initially
at rest and the other
has an initial
velocity v:
2
1
mv
K
 14 2  50%
K before 2 mv
1
4
1
4
The percentage loss is greatest for
the case in which t he two objects
have oppositely directed velocitie s
of magnitude v/2.
73 ••
Picture the Problem. We can find the velocity of the center
of mass from the definition of the total momentum of the
system. We’ll use conservation of energy to find the
maximum compression of the spring and express the initial
(i.e., before collision) and final (i.e., at separation) velocities.
Finally, we’ll transform the velocities from the center of
mass frame of reference to the table frame of reference.
(a) Use the definition
of the total
momentum of a
system to relate the
initial momenta to
the velocity of the
center of mass:



P   mi vi  Mv cm
i
or
m1v1i  m2 v2i  m1  m2 vcm
Note: The RHS is true because of
the definition of the velocity of the
com of a system, not because it is a
PIC.
Solve for vcm:
vcm 
Substitute numerical
values and evaluate
vcm:
vcm 
m1v1i  m2 v2i
m1  m2
2 kg 10 m/s   5 kg 3 m/s 
2 kg  5 kg
 5.00 m/s
(b) Find the kinetic
energy of the system
at maximum
compression
(u1 = u2 = 0):
Use conservation of
energy to relate the
kinetic energy of the
system to the
potential energy
stored in the spring
at maximum
compression:
2
K  K cm  12 Mvcm

7 kg 5 m/s 2  87.5 J
K  U s  0
or
K f  K i  U sf  U si  0
K cm  K i  k x  0
Because Kf = Kcm and
Usi = 0:
Solve for x:
1
2
1
2
x 
2
2K i  K cm 
k


2 12 m1v12i  12 m2 v22i  K cm
k

m1v12i  m2v22i  2 K cm
k
Substitute numerical values and evaluate x:
x 
2 kg 10 m/s 2  5 kg 3 m/s 2  287.5 J   
1120 N/m
1120 N/m 
0.250 m

(c) Find u1i, u2i, and
u1f for this elastic
collision:
Use conservation of
mechanical energy
to set the velocity of
recession equal to
the negative of the
velocity of approach
and solve for u2f:
Transform u1f and
u2f to the table frame
of reference:
u1i  v1i  vcm  10 m/s  5 m/s  5 m/s,
u 2i  v 2i  vcm  3 m/s  5 m/s  2 m/s,
and
u1f  v1f  vcm  0  5 m/s  5 m/s
u2f  u1f  u2i  u1i 
and
u2f  u2i  u1i  u1f
  2 m/s   5 m/s  5 m/s
 2 m/s
v1f  u1f  vcm  5 m/s  5 m/s  0
and
v 2f  u 2f  vcm
 2 m/s  5 m/s  7.00 m/s
*74
••
Picture the Problem. Let the system include the earth, the bullet,
and the sheet of plywood. Then Wext = 0. Choose the zero of
gravitational potential energy to be where the bullet enters the
plywood. We can apply both conservation of energy and
conservation of momentum to obtain the various physical quantities
called for in this problem.
(a) Use
conservation of
mechanical
energy after the
bullet exits the
sheet of
plywood to
relate its exit
speed to the
height to which
it rises:
Solve for vm:
Proceed
similarly to
relate the initial
velocity of the
plywood to the
height to which
it rises:
(b) Apply
conservation of
momentum to
K  U  0
or, because Kf = Ui = 0,
2
1
m
2
 mv  mgh  0
vm 
vM 
2 gh
2 gH


pi  pf
or
the collision of
the bullet and
the sheet of
plywood:
Substitute for
vm and vM and
solve for vmi:
mvmi  0  mvm  MvM
mvmi  m 2 gh  M 2 gH
vmi 
M
2 gh 
2 gH
m
(c) Express the
initial
mechanical
energy of the
system (i.e., just
before the
collision):
Express the
final
mechanical
energy of the
system (i.e.,
when the
bullet and
block have
reached their
maximum
heights):
KEi  12 mvm2 i
(d) Use the
work-energy
theorem with
Wext = 0 to find
the energy
dissipated by
KEf  KEi  Wfriction  0

2M
 mg h 
m

2
M  
hH    H 
 m  
KEf  mgh  MgH  g mh  MH 
and
friction in the
inelastic
collision:
Wfriction  KEi  KEf
 h M 
 gMH 2

 1
 H m 