Phy 211: General Physics I
Chapter 9 Worksheet
1
Center of Mass:
1. a) Determine the center of mass for the Earth-Moon system.
rcom
rcom
rcom
Moon
vMoon
Ans. Using the center of mass for the earth as the origin (0,0) &
standard Cartesian oriented coordinates:
y
m
r
+ mearthrearth
mmoon rmoon
= moon moon
=
where rearth=0
mmoon + mearth
mmoon + mearth


7.36x1022 kg
8
ˆ
= 
 3.82x10 m j
22
24
7.36x10
kg
+
5.98x10
kg


6
ˆ
= 4.64x10 m j



vEarth

x
Earth
b) At a particular moment, the moon is moving with a linear
speed (vmoon) of 992 m/s (while in a “circular” orbit around the Earth) while the Earth is moving
at 2.99x104 m/s (vearth) at a 45o counter-clockwise angle to vmoon. Determine the velocity of the
center-of-mass for the Earth-Moon system.
Ans.
vcom =
vcom
vcom
mmoonvmoon + mearthvearth
mmoon + mearth





 7.36x1022kg - 992 m  ˆi + 5.98x1024kg 2.99x104 m sin 225o ˆi + cos 225o ˆj
s
s
= 

7.36x1022 kg + 5.98x1024 kg

= -2.09x104 m ˆi + -2.09x104 m ˆj

s


s

2. a) The average mass of an oxygen and hydrogen, respectively, is
15.999 amu and 1.0079 amu (1 amu = 1.661x10-27 kg). Calculate the
mass of the oxygen and hydrogen atom, respectively, in SI units.
Ans.
 15.999 amu   1.661 x 10
mO = 

1
1 amu


-27


The Water Molecule
O
H
-27
 1.0079 amu   1.661 x 10 kg 
-27
mH = 
 = 1.674 x 10 kg

1
1 amu



 
H
104.5o
kg 
-27
 = 26.574 x 10 kg

b) The center-to-center distance between the oxygen and either hydrogen atom is 0.957854 Å
(1 Å = 10-10 m). Calculate the distance between an oxygen atom and a hydrogen atom,
respectively, in SI units.
Ans. rH to O
o


-10

0.957854
Α

  1 x 10 m  = 9.57854 x 10-11 m
=
o



1
1 Α



c) Determine the center of mass for a water molecule.
Ans. Using the center of mass for the oxygen atom as the origin:
 mHrH1 + mOrO + mHrH2 
-12
ˆ
ˆ
rcom = 
 = 0 m i - 6.685x10 m j
2mH + mO




Phy 211: General Physics I
Chapter 9 Worksheet
2
Impulse and Momentum:
1.
a) What is the change in momentum for this object?
Ans.
p = 50kg 10 ms -30 ms  ˆi=-1000
kgm
s
ˆi
b) Calculate the impulse received by the object using the force equation above.
Ans.
J=
0.5s
0s
Fdt =
 -  4000  t - 1000N  dt ˆi
0.5s
N
s
0s

J= - 2000 Ns  t2 - 500N t

0.5s
0s
ˆi =-1000N  s ˆi
c) What is the average force exerted on the object?
Ans.
p -1000 kgsm ˆ
Favg= =
i = -2000 N ˆi
t
0.5 s
d) Calculate the impulse received by the object using the average force calculated in (c).
Ans.
J =Favgt =-2000 N0.5 s  ˆi =-1000N  s iˆ
e) If the deceleration from 30 m/s to 10 m/s had occurred over a time interval of 5 sec
(obviously subject to a different force equation), what is the impulse received by the object?
Ans.
Since the initial and final momentum values are the same as above,
J = p = 50kg 10 ms -30 ms  ˆi=-1000
kgm
s
ˆi
f) Calculate the average force exerted on the object in (e).
Ans.
Favg=
p -1000
=
t
5s
kgm
s
ˆi = -200 N ˆi
2. a) What is the initial and final linear momentum of the vehicle for this road test (in SI units)?
Ans.
pi = 0 kgsm ˆi & pf = 1581.8 kg 51.2 ms  ˆi=8.10x104 kgsm ˆi
b) What is the change in momentum for the Corvette during this trial?
Ans.
p = pf - pi = 8.10x104 kgsm ˆi
c) What is the rate of momentum change for the Corvette during this test? How does this value
compare the average force vector exerted by the road on the Corvette during this trial?
Ans.
8.09x104 kgsm ˆ
p
=
i = 6.33x103N ˆi = FNet
t
12.8 s
ˆi
F =f
+F = f
-F
.
Net
sby road
drag

sby road
drag

 fsby road =FNet + Fdrag  ˆi
Thus the rate of momentum change is less than the (static friction) force by the road exerted on
the vehicle.
d) How much impulse (change in momentum) does the earth receive during this trial?
Ans.
Jearth=-JCorvette=pCorvette= -8.10x104 kgsm ˆi
Phy 211: General Physics I
Chapter 9 Worksheet
3
e) Estimate the change in velocity ( vearth ) for the earth during this same road test. The mass
of the earth is 5.98x1024 kg.
Ans.
vearth=
Jearth  -8.10x104 kgsm  ˆ
-20 m ˆ
=
 i = -1.35x10 s i
24
mearth  5.98x10 kg 


Conservation of Linear Momentum:
1. An 100 kg object traveling at 50 m/s collides (perfectly inelastic) with a 50 kg object initially
at rest.
a. Determine the linear momentum vector for each object prior to the collision?
Ans.
p1i = 100 kg 50 ms  ˆi= 5000 kgsm ˆi & p2i = 0 kgsm ˆi
b. If the collision is perfectly inelastic, what is the total momentum vector for the 2-object
system following the collision?
Ans.
psystembefore = psystemafter = 5000 kgsm ˆi
c. What is the center-of-mass velocity vector for the two combined objects following the
collision?
Ans.
vsystem=
psystemafter
msystem
 5000 kgsm  ˆ
mˆ
= 
 i= 33.3 s i
150
kg


d. Calculate the impulse vector exerted on each object.
J1=p1=100 kg 33.3 ms -50 ms  ˆi=-1.67x103N  s
Ans.
and
J2=p2=50 kg 33.3 ms  ˆi=1.67x103N  s
e. What is the kinetic energy of each object following the collision?
K1= 12 m1v12= 12 100 kg 33.3 ms  =5.54x104J
2
Ans.
K2= 12 m2v22= 12 50 kg 33.3 ms  =2.77x104 J
2
f. Calculate the non-conservative work (Wnc) performed on the 2 object system.
Ans.
Since all of the forces associated with the collision are internal forces,
Ksystembefore = 12 100 kg 50 ms  =1.25x105J
2
Ksystembefore =5.54x104 J + 2.77x104 J =8.31x104J
WNC=Ksystem=Ksystemafter - K systembefore
WNC=8.31x104 J -1.25x105J = -4.90x104 J
Phy 211: General Physics I
Chapter 9 Worksheet
4
2. A 100 kg object traveling at 50 m/s collides head-on (perfectly elastic) with a 50 kg object
initially at rest.
a. What is the kinetic energy of each object prior to the collision?
K1= 12 m1v12= 12 100 kg 50 ms  =1.25x105J
2
Ans.
K2= 12 m2v22= 12 50 kg 0 ms  =0 J
2
b. What is the kinetic energy of the two object system immediately following the collision?
Ans.
Kbefore=Kafter =1.25x105J
By definition, KE is conserved:
c. Using the laws for conservation of linear momentum and conservation of mechanical energy,
determine the velocity vectors for each object following the collision?
Ans.
To begin, let’s establish that
psystem =m1v1 + m2v2 = 5000 kgsm
Ksystem = 12 m1v12 +
1
2
m2v12 = 1.25x105J
Solve for v1 in the momentum equation the substitute this relation into the energy
equation:
v1 =
5000 kgsm - 50 kg v2
5000 kgsm - m2v2
=
= 50 ms - 0.5v2
m1
100 kg
v12 = 50 ms - 0.5v2  = 2500 ms2 - 50 ms  v2+ 0.25v22
2
Ksystem = 12 m1v12 +
1
2

m2v22 = 1.25x105J

Ksystem = 12 100kg 2500 ms2 - 50 ms  v2+ 0.25v22 +
2

1
2
50kg v22
= 1.25x105J

Ksystem =1.25x105J - 2500 kgsm v2+ 12.5kg v22 + 25kg v22 = 1.25x105 J
 2500 kgsm  ˆ
mˆ
m
m
mˆ
ˆ
Solving for v2 and v1: v2=
 i=66.7 s i and v1= 50 s -0.5 66.7 s  i=16.7 s i
37.5kg


d. What is the impulse received by each object during the collision?
Ans.
J1 = p1 = 100kg 16.7 ms -50 ms  ˆi=-3330 N  s ˆi
J2 = p2 = 50kg 66.7 ms -0 ms  ˆi=3330 N  s ˆi
Phy 211: General Physics I
Chapter 9 Worksheet
5
2-D Collisions:
Two objects (m1=100 kg and m2 = 150 kg) collide as shown in the diagram (there are no
external forces acting on either object):
m1
30o
v1 = 3 m/s
m1
m2
60o
m2
vf = ??
v2 = 5 m/s

a) What is the magnitude and direction of the linear momentum vector (prior to the collision) for
each object, respectively?
Ans.
p1i = 100 kg 3 ms = 300 kgsm  θp1 = 300o
p2i = 150 kg 5 ms = 750 kgsm  θp2 = 210o
b) Express the linear momentum vector (prior to the collision) for each object in component
form.
Ans.


 -sin60 ˆi -cos60 ˆj= -650
p1i = 100 kg 3 ms  sin30o ˆi -cos30o ˆj = 150 kgsm iˆ -260 kgsm ˆj
p2i = 150 kg 5 ms
o
o
kgm
s
ˆi -375 kgm ˆj
s
c) Determine the linear momentum vector for the 2 object system (prior to the collision),
magnitude and direction.
Ans.

 

psystemi = p1i + p2i= 150 kgsm - 650 kgsm ˆi + -260 kgsm - 375 kgsm ˆj

 

psystemi = -500 kgsm ˆi + -635 kgsm ˆj
d) Considering the collision to be perfectly inelastic, what is the linear momentum vector (in
component form) for the 2 object system following the collision?
Ans.

 

psystemf = psystemi = -500 kgsm ˆi + -635 kgsm ˆj
e) Determine the velocity vector, in component form, for the 2 object system following the
collision?
vsystemf =
Ans.
vsystemf =
psystemf
msystem
-500
kgm
s
 ˆi +-635
kgm
s
250 kg
 ˆj
= -2 ms  ˆi + -2.54 ms  ˆj
f) What is the magnitude and direction for the velocity for the 2 object system following the
collision?
vsystemf =
Ans.
-2 ms 
2
+ -2.54 ms  = 3.23 ms
2
 2 ms 
θvsystem = tan-1 
= 38.2o (according to diagram)
m 
f
2.54
s 
