Download Chapter 5 Center of Mass and Linear Momentum

Chapter 5
Chapter 5
Center of Mass and Linear Momentum
Center of Mass and Linear Momentum
Basic Requirements：
1. Understand the concept of the center of mass
2. Master Newton's second law for a system of particles
3. Master the law of conservation of linear momentum
4. Understand the motion of a system with varying mass
5. Understand the external forces and internal energy changes
Review and Summary
Center of Mass The center of mass of a system of n particles defined to be the point
whose coordinates are given by
xcom 
1
M
n
 mi xi
i 1

1
rcom 
M
or
ycom 
,
n
1 n
 mi yi
M i 1
, zcom 
1 n
 mi zi
M i 1

m r
i 1
(5-5)
(5-8)
i i
where M is the total mass of the system. If the mass is continuously distributed, the
center of mass is given by
xcom 
1
xdm
M
,
ycom 
1
M
 ydm
, zcom 
1
zdm
M
(5-9)
If the density (mass per unit volume) is uniform, the Eq.5-9 can be written as
xcom 
1
xdV
V
,
ycom 
1
ydV
V
, zcom 
1
zdV
V
(5-11)
where V is the volume occupied by M.
Newton’s Second Law for a System of Particles The motion of the center mass of
any system of particles is governed by Newton’s second law for a system of particles,
which is


Fnet  Macom
- 111 -
(5-14)
Chapter 5
Center of Mass and Linear Momentum

Here Fnet is the net force of all the external force acting on the system, M is

the total mass of the system, and acom is the acceleration of the system’s center of
mass.
Linear Momentum and Newton’s Second Law For a single particle, we define a

quantity p called its linear momentum as


p  mv
(5-22)
and can write Newton’s second law in terms of this momentum:


dp
Fnet 
dt
(5-23)
For a system of particles these relations become




dP
P  Mvc o m and Fnet 
dt
Conservation of Linear Momentum
(5-25, 5-27)
If a system is isolated so that no net external

force acts on the system, the linear momentum P of the system remains constant:

P =constant
(closed, isolated system)
(5-42)
This can also be written as
 
Pi  Pf
(closed, isolated system)
(5-43)

where the subscripts refer to the value of P at some initial time and at a later time.
Eqs. 5-42 and 5-43 are equivalent statements of the law of conservation of linear
momentum.
Collisions In a collision, two bodies exert strong forces on each other for a relatively
short time. These forces are internal to the two-body system and are significantly
larger than any external force during the collision.
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Chapter 5
Center of Mass and Linear Momentum
Impulse and Linear Momentum Applying Newton’s second law in momentum form
to a particle-like body involved in a collision leads to the impulse-linear momentum
theorem:


 
p f  pi  p  J




where p f  pi  p is the change in the body’s linear momentum, and J is the

impulse due to the force F t  exerted on the body by the other body in the collision:

tf 
J   F t dt
ti

If Favg is the average magnitude of F t  during the collision and t is the
duration of the collision, the for one-dimensional motion
J  Favg t
When a steady stream of bodies, each with mass m and speed v, collides with a body
whose position is fixed, the average force on the fixed body is
Favg  
n
n
p   mv
t
t
where n / t is the rate at which the bodies collide with the fixed body, and v is
the change in velocity of each colliding body. This average force can also be written as
Favg  
m
v
t
where m / t is the rate at which mass collides with the fixed body. In Eqs.5-15 and
5-16, v  v if the bodies stop upon impact, or v  2v if they bounce
directly backward with no change in their speed.
Inelastic Collision—One Dimension In an inelastic collision of two bodies, the
kinetic energy of the two-body system is not conserved. If the system is closed and
isolated, then the total linear momentum of the system must be conserved, which we
can write in vector form as




p1i  p2i  p1 f  p2 f
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(5-77)
Chapter 5
Center of Mass and Linear Momentum
where subscripts i and f refer to values just before and just after the collision,
respectively.
If the motion of the bodies is along a single axis, the collision is one-dimensional
and we can write Eq.5-77 in terms of velocity components along the axis:
mi v1i  m2v2i  m1v1 f  m2v2 f
If the bodies stick together, the collision is a completely inelastic collision and the
bodies have the same final velocity V (because they are stuck together).
Motion of the Center Mass The center of mass of a closed, isolated system of two

colliding bodies is not affected by the collision. In particular, the velocity vcom of the
center of mass cannot be changed by the collision and its related to the constant total

momentum P of the system by

vcom 



P
p1i  p2i

m1  m2 m1  m2
Elastic Collisions—One Dimension An elastic collision is a special type of collision
in which the kinetic energy of the system of colliding bodies is conserved. Some
collisions in the everyday world can be approximated as being elastic collisions. If the
system is closed and isolated, its linear momentum is also conserved. For a
one-dimensional collision in which body 2 is a target and body 1 is an incoming
projectile, conservation of kinetic energy and linear momentum yield the following
expressions for the velocities immediately after the collisions:
and
v1 f 
m1  m2
v1i
m1  m2
v2 f 
2m1
v1i
m1  m2
If both bodies are moving prior to the collision, their velocities immediately after the
collision are given by
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Chapter 5
Center of Mass and Linear Momentum
v1 f 
and
v2 f 
m1  m2
2m2
v1i 
v2i
m1  m2
m1  m2
2m1
m  m1
v1i  2
v2i
m1  m2
m1  m2
Note the symmetry of subscripts 1 and 2 in Eqs.5-22 and 5-23.
Collisions in Two Dimensions If two bodies collide and their motion is not along a
single axis (the collision is not head-on), then the collision is two-dimensional. If the
two-body system is closed and isolated, then the law of conservation of momentum
applies to the collision and can be written as




P1i  P2i  P1 f  P2 f
In component form, the law gives two equations that describe the collision (one
equation for each of the two dimensions). If the collision is also elastic (a special case),
then the conservation of kinetic energy during the collision gives a third equation:
K1i  K 2i  K1 f  K 2 f
Examples
Example 1 Assume that a cannon-shot with mass of 2m is projected out from the
ground, when it reaches the maximum height it explodes into two fragments with
equal masses (Fig.5-1), among them one piece falls vertically freely, and the other
piece horizontally projects out, they reach the ground at the same time. Where does the
second fragment land?
Solution: Consider the cannon-shot as a system
and ignore the air resistance. Before and after the
explosion the center of mass of the cannon-shot
moves on the same parabola. That is to say, after
the explosion the trajectory of the center of mass of
the two fragments still follows the parabolic
trajectory of the cannon-shot before the explosion.
Fig.5-1 Example 1
If we choose where the first fragment lands as the origin O of the coordinates system
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Chapter 5
Center of Mass and Linear Momentum
and the rightward horizontal axis as the positive direction of the Ox axis, and
we assume m1 and m2 as the masses of the two fragments, we have as given by the
question m1  m2  m ; assume x1 and x 2 are the distances between where the
fragments land and the origin O. From the figure we know that x1  0 , and we
have
xC 
m1 x1  m2 x2
m1  m2
Since m1  m2  m , we have
x2  2 xC
(Answer)
i.e., the horizontal distance between the landing point of the second fragment and the
landing point of the first fragment is twice the horizontal distance between the center
of mass of the fragments and the landing point of the first fragment. Although this
problem can also be solved by the kinetics method, it would be much more
complicated than the above method.
Example 2 There is soft chain with length l and uniform mass density, the mass
per unit is  , we put it in a pile on the ground, if we hold on one end of the chain and
pull it up with a constant speed v , when the hand-holding end of the chain reaches a
height y above the ground, what is the pulling force of the hand?
Solution: From the figure we can see that the coordinate ycom
of the mass center of the part of the chain already above the
ground changes as the chain is pulled up, select the coordinates
system as shown in the figure and the center of mass is at
ycom
my

m
i
i
i

y
y
  l  y   0
y2
2

l
2l
(1)
Fig.5-2 Example 2
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Chapter 5
Where
Center of Mass and Linear Momentum
 is the mass per unit length of the chain. Since the combined



d 2 ycom
external force on the chain if F  yg  l
dt 2
i.e.,
F  yg  ˆj  l d
2
ycom ˆ
j
dt 2
(2)
Taking the second derivative with respect to time t on Eq.(1), we have
2
d 2 ycom 1  dy 
d2y


y




dt 2
l  dt 
dt 2 
Considering v=dy/dt, v=0, we have yd y / d t  0 . The above becomes
2
2
d 2 y com v 2

l
dt 2
Substituting the above into Eq.(2), we have
F  yg  ˆj  v 2 ˆj
F  v 2  yg
and
Example 3
(Answer)
Assume that a nucleus at rest decays and emits an electron and a
neutrino and becomes a new nucleus. It is known that the directions of motion of the
electron and the neutrino are perpendicular to each other, the momentum of the
1.2  10 22 kg  m / s and the momentum of the neutrino is
6.4  10 23 kg  m / s , what is the magnitude and direction of the momentum of the
electron
is
new nucleus?



Solution: Let pe , pv and p N denote the momentum of the electron, the neutrino,


and the new nucleus respectively, and pe and pv are perpendicular as shown in the
figure. In the short time period of the nuclear decay, the internal force between the
particles are much greater than the external forces acting on the particle system,
therefore the momentum of the particle system is conserved before and after the decay
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Chapter 5
Center of Mass and Linear Momentum
process. Consider that the nucleus was at rest before the decay, then after the
decay the sum of the momentum of the electron, the neutrino, and the new nucleus is
also equal to 0, i.e.,



pe  pv  pN  0


Since pe and pv are perpendicular, we have

pN  pe2  pv2

1/ 2
Substituting the known data in it we, have

 
2
pN  1.2  10 22  6.4  10 23

2 1/ 2
kg  m  s 1  1.36  10 22 kg  m  s 1
The angle α in the figure is:
  arctan
pe
1.2 1022
 arctan
 61.9
pv
6.4 1023

Or the angle between the momentum p N of the new nucleus and the momentum

pv of the neutrino is
  180  61.9  118.1
(Answer)
Example 4 A returning rocket flies at 2.5  10 m / s along the horizontal direction
3
with respect to the ground, ignore the air resistance. The control system separates the
rocket into two parts, the front part is the instrumental module with mass of
100kg and the back part is the rocket container with mass of 200kg If the speed of
the instrumental module with respect to the rocket container is 1.0103 m/s, what are
the speeds of the instrumental module and the rocket container with respect to the
ground?
Solution: As shown in the figure, take the ground as an inertial frame S (O xyz ) , let

v be the velocity of the rocket with respect to the inertial frame S before the


separation, v1 and v2 be the velocities of the instrumental module and the rocket
- 118 -
Chapter 5
Center of Mass and Linear Momentum
container respectively, with respect to the inertial frame S after the separation,

v ' be the velocity of the instrumental module with respect to the rocket container
after the separation, and take the rocket container as another inertial frame

S (O x y z ) , the S  system moves with velocity v2 along the xx’ axis respect to
the S frame. With the velocity formula of the relative motions, we have
  
v1  v2  v '



Since v1 , v2 and v ' are all in the same horizontal direction, the above equation
becomes
v1  v2  v'
Before and after the rocket separation, it is only subject to the gravity in the vertical
direction, so the momentum component in the horizontal direction is conserved, we
have
m1  m2 v  m1v1  m2v2
Solving the above two equations, we have
v2  v 
m1
v'
m1  m2
Substituting the data in it, we have
1


v2   2.5  103   1.0  103 m / s  2.17  103 m / s
3


3
3
v1  2.17  10  1.0  10 m / s  3.17  103 m / s
(Answer)
Both v1 and v2 are positive numbers, indicating that their velocities are in the same

direction as v , only that the instrumental module increased its speed by the thrust of
the rocket, while the speed of the rocket container got slowed down, therefore the
momentum transfer is realized.
Example 5 Assume there are dusts with density ρ in the universe, such dusts are at
rest with respect to the inertial reference frame. A spaceship with mass m0 penetrates
- 119 -
Chapter 5
Center of Mass and Linear Momentum
through the dusts with the initial velocity v0 , since the dusts get stuck on the
spaceship, it causes the speed of the spaceship to change, what is the relationship
between the speed of the spaceship and its time of flight in the dusts?
For
convenience, assume the external shape of the spaceship is a cylinder with the
cross-sectional area S.
Solution: According to the conditions given in the problem, we may consider the
collisions between the dusts and the spaceship are complete inelastic collisions, we
treat the dusts and the spaceship as a single system, also consider the spaceship flies in
free space and there are no external forces on this system, therefore the momentum of
the system in conserved. Before entering the dusts (i.e.,
t=0) the mass and the speed of the spaceship are m0 and
v0 respectively, and when the spaceship is in the dusts
(at time t) the mass and the speed are m and v
respectively. Therefore, according to momentum
conservation, we have
Fig.5-3 Example 5
m0 v0  mv
(1)
Also, in the time interval t  t  dt , since the collision between the spaceship and
the dusts are complete inelastic collisions, the mass of the dusts stuck on the spaceship,
i.e., the increase of the mass of the spaceship, is
dm  Svdt
(2)
From (1) we have
dm  
m0 v0
dv
v2
Therefore, we have
Svdt  
m0 v 0
dv
v2
From the given conditions, integrating the above, we have
dv
S

3
v0 v
m0 v 0

v
We obtain
- 120 -

dt
0
dt
Chapter 5
Center of Mass and Linear Momentum
1 1
1 
S
 2  2
t


2v
v0  m0 v0
We have
v
m0
v0
2Sv0t  m0
(Answer)
Obviously, the longer the spaceship flies in the dusts, the lower its speed.
Example 6 As shown in the figure, assume two elastic small balls with masses m1


and m2 and velocities v10 and v 20 collide in such a way that their centers are aligned
in a straight line, the directions of the two velocities are the same. If the collision is






m1v10  m2 v20  m1v1  m2 v2
(1)
completely elastic, what are the velocities v1 and v2 after the collision?
Solution: From the theorem of momentum conservation,
we have
From the law of conservation of mechanical energy we
have
1
1
1
1
2
2
m1v10
 m2 v20
 m1v12  m2 v22
2
2
2
2
(2)
Fig.5-4 Example 6
Eq.(1) can be written as
m1 v10  v1   m2 v2  v20 
Eq.(2) can be written as



m1 v10  v1  m2 v2  v20
2
2
2
2
(3)

(4)
From Eqs.(3) and (4) we have
v10  v1  v2  v20
Or
v10  v20  v2  v1
- 121 -
(5)
Chapter 5
Center of Mass and Linear Momentum
Eq.(5) indicates that before the collision the relative velocity v10  v20  of
the two balls approaching each other is equal to the relative velocity v2  v1  of the
two balls separating each other after the collision.
From Eq.(3) and Eq.(5), we obtain:
v1 
v2
m1  m2 v10  2m2 v20
m1  m2
(Answer)
m  m1 v20  2m1v10
 2
m1  m2
Problem Solving
1 (2) In the ammonia ( NH 3 ) molecule of Fig.5-5, three hydrogen ( H ) atoms form
an equilateral triangle, with the center of the triangle at distance d  9.40  10
11
m
from each hydrogen atom. The nitrogen ( N ) atom is at the apex of a pyramid, with
the three hydrogen atoms forming the base. The nitrogen-to-hydrogen atomic mass
ratio is 13.9 , and the nitrogen-to-hydrogen distance is L  10.14  10
are the (a) x and (b) y coordinates of the molecule’s center of mass?
Solution: The position of the molecule’s center of mass
can be found out by the center of mass equations of a
system of n particles. But here we can first get the
11
m . What
position of the center of mass of the three hydrogen atoms
on the base, then the center of mass of the molecule can be
determined by the coordinates of the center of mass of the
base and the nitrogen atom.
Fig.5-5 Problem 1
Obviously, the coordinates of the center of mass of equilateral triangle of hydrogen
atoms has
xbase  d ,
ybase  0
The x and y coordinates of the nitrogen atom are
x  0,
y  L2  d 2 
10.14 10   9.40 10 
11 2
11 2
 3.80  1011 m
Then, the x and y coordinates of center of mass of the molecule are
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Chapter 5
Center of Mass and Linear Momentum
xcom  0
y com 
3m H ybase  m N y N
13.9m H y N

3m H  m N
3m H  13.9m H
(Answer)
13.9

3.80  10 11  3.13  10 11 m
16.9
2 (4) A metal soda can of uniform composition has a mass of 0.140kg and is
12.0cm tall (Fig.5-6). The can is filled with 1.31kg of soda. Then small holes are
drilled in the top and bottom (with negligible loss of metal) to drain the soda. What is
the height h of the center of mass of the can and the contents (a) initially and (b)
after the can loss all the soda? (c) What happens to h as the soda drains out? (d) If
x is the height of the remaining soda at any given instant, find x when the center of
mass reaches its lowest point.
Solution:
(a) Since the can is uniform and filled with soda, the center of mass initially is right at
the center of the can:
hi 
12.0
 6.0cm
2
(Answer)
(b) After the can loss all the soda, the center of mass the system is
that of the can:
hf 
12.0
 6.0cm
2
(Answer)
Fig.5-6
Problem 2
(c,d) The system consists of the can and the remained soda, so
0.14kg6cm   1.31kg  x  x cm 
h
 12
 2
1.31kg
0.14kg 
x
12

dh
 0 , from the calculus, x has the minimum value that means
dx
the center of mass reaches its lowest point is x  2.84cm
So h decreases first, after the center of mass reaches its lowest point when
We know when
- 123 -
Chapter 5
Center of Mass and Linear Momentum
x  2.84cm , it begins to increase until the can loss all the soda.
3 (5) In Figure 5-7, two particles are launched from the origin of the coordinate
system at time t  0 . Particle 1 of mass m1  5.00g is shot directly along the x
axis (on a frictionless floor), where it moves with a constant speed of 10.0m/ s .
Particle 2 of mass m2  3.00 g is shot with a velocity of magnitude 20.0m/ s at
an upward angle such that it always stays directly above particle 1 during its flight. (a)
What is the maximum height H max reached by the com of the two-particle system?
In unit-vector notation, what are the (b) velocity
and (c) acceleration of the com when the com
reaches H max ?
Solution:
(a) Suppose the launching angle of Particle 2 is
 . To make Particle 2 always stay directly
above particle 1
v2 x  v2 cos  v1
(1)
Fig.5-7 Problem 3
Substituting v1  10.0m / s, v2  20m / s into Eq.(1) yields
1


m2  v2 sin   t  gt 2 
m2 y2
2

 
So ycom 
m1  m2
m1  m2
  60
(2)
Substituting m1  5.00 g , m2  3.00 g , g  9.8m / s into Eq.(2) and suppose
2
dycom
0
dt
yields
t  3s
so H max  ycom (t  3s)  5.625m
(Answer)
(b) As the com reaches Hmax, v2 y  0 , so
vcom, x  10m / s, vcom, y  0
- 124 -
(3)
Chapter 5
And
Center of Mass and Linear Momentum

vcom  vcom, xiˆ  vcom, y ˆj  10iˆm / s

(Answer)

(c) From the equation Fnet  Ma com , we get

m2 g ˆ
a  a y ˆj 
j  3.75 ˆjm / s 2
m1  m2
(Answer)
4 (6) Ricardo, of mass 80kg , and Carmelita, who is lighter, are enjoying Lake
Merced at dusk in a 30kg canoe. When the canoe is at rest in the placid water, they
exchange seats, which are 3.0m apart and symmetrically located with respect to the
canoe’s center. Ricardo notices that the canoe moves 40cm horizontally relative to a
pier post during the exchange and calculates Carmelita’s mss. What is it?
Solution: The key idea here is that during the exchange, there is no external force on
the two-people and the canoe system, and the total linear momentum is conserved. So
the center of mass of the system will keep
Ricardo
Carmelita
unchanged.
Choose the initial position of the center of
mass of the canoe as origin and construct x
x
axis as shown. As the diagrammatic sketch
O
shown, Ricardo notices that the canoe Carmelita
Ricardo
moves 40cm relative to a pier post, that
means the canoe moves 20cm relative to
the unchanged center of mass of the system.
40 cm
(The initial and final position of the canoe is
symmetric about the center of mass of the
Center of Mass
system.)
Since
Fig. 5-8 Problem 4
xcom 
mRicardox Ricardo  mCarmelitaxCarmelita  mcanoe xcanoe
mRicardo  mCarmelita  mcanoe
we have
- 125 -
Chapter 5
Center of Mass and Linear Momentum
80kg   3m   mCarmelita  3m   30kg   0

2 
 2 
 3m

 3m

 80kg 
 20cm   mCarmelita  
 20cm   30kg    20cm 
 2

 2

yields mCarmelita  68.125kg
(Answer)

5 (9) A 0.30kg softball has velocity of 15m / s at an angle of 35 below the
horizontal just before making contact with the bat. What is the magnitude of the
change in momentum of the ball while it is in contact with the bat if the ball leaves the
bat with a velocity of (a) 20m/ s , vertically downward, and (b) 20m/ s ,
horizontally back toward the pitcher?
Solution:
(a) Construct the coordinates as shown in Fig. 5-9. The change in momentum


p  mv f  mvi
  
O

 0.30kg 20 ˆj  15  cos 35iˆ  15  sin 35 ˆj m / s 
  3.69iˆ  3.42 ˆj kg  m / s 
Magnitude

p 
 3.692  3.422
 5.03kg  m / s
(Answer)
(b) The change in momentum of the ball is

 
35º

vi
x
y
Fig. 5-9 Problem 5


p  0.30kg  20iˆ  15  cos 35iˆ  15  sin 35 ˆj m / s 
 9.69iˆ  2.58 ˆj kg  m / s 

p 
 9.692   2.582
 10.03kg  m / s
(Answer)
6 (13) Figure 5-10 shows a two-ended “rocket” that is initially stationary on a
frictionless floor, with its center at the origin of an x axis. The rocket consists of a
central block C (of mass M  6.00kg ) and blocks L and R (each of mass
m  2.00kg ) on the left and right sides. Small explosions can shoot either of the side
blocks away from block C and along the x axis. Here is the sequence: (1) At
time t  0 , block L is shot to the left with a speed of 3.00m/ s relative to the
- 126 -
Chapter 5
Center of Mass and Linear Momentum
velocity that the explosion gives the rest of the rocket. (2) Next, at
time t  0.80s , block R is shot to the right with a speed of 3.00m/ s relative to the
velocity that block C then has. At t  2.80s , what are (a) the velocity of block C and
(b) the position of its center?
Solution: (a) From time t  0s to t  0.80s ,
the velocity of block C is v C 1 , as the linear
momentum is conserved:
Fig.5-10 Problem 6
2.00kgvC1  3.00m / s   6.00kg  2.00kgvC1  0
yields
vC1  0.6m / s
(Answer)
From time t  0.80s , the velocity of block C is vC 2 , and
2.00kgvC 2  3.00m / s   6.00kgvC 2  6.00kg  2.00kgvC1
yields
vC 2  0.15m / s
At t  2.80s , the velocity of block C is vC 2  0.15m / s .
(Answer)
(b) The position of center of block C:
x  vC1  0.80s  vC 2 2.80s  0.80s   0.18m
(Answer)
7 (19) In Fig.5-11, a 10 g bullet moving directly upward at 1000m/ s strikes and
passes through the center of mass of a 5.0kg block initially at rest. The bullet
emerges from the block moving directly upward at
400m/ s . To what maximum height does the block then
rise above its initial position?
Solution: Choose the initial position of the block as origin
and the upward direction as positive. The linear momentum
of the bullet-block system is conserved during the
interaction:
mbulletvbullet,i  mbulletvbullet, f  mblock vblock
- 127 -
Fig.5-11 Problem 7
Chapter 5
Center of Mass and Linear Momentum
10 g 1000m / s   10 g 400m / s   5.0kgvblock
vb l o ck 1.2m / s
yields
The maximum height of the block:
vblock
1.2m / s 2  0.073m  7.3cm

2g
2  9.8m / s 2
2
y max 
(Answer)
8 (20) A completely inelastic collision occurs between two balls of wet putty that
move directly toward each other along a vertical axis. Just before the collision, one
ball, of mass 3.0kg , is moving upward at 20m/ s and the other ball, of mass
2.0kg , is moving downward at 12m / s . How high do the combined two balls of
putty rise above the collision point? (Neglect air drag)
Solution: Choose the collision point as origin and the upward direction as positive.
The velocity of the combined balls right after the collision is v . The linear
momentmum of the system is conservedin the collision, so we get
3.0kg20m / s  2.0kg 12m / s  3.0kg  2.0kgv
Solving for v , yields
v  7.2m / s
Then the maximum height that the combined balls rise is
7.2m / s   2.64m
v2

2 g 2  9.8m / s 2
2
ymax 
(Answer)
9 (21) In Fig.5-12, block 1 (mass 2.0kg ) is moving rightward at 10 m/s and block 2
(mass 5.0kg ) is moving rightward at 3.0m/ s . The surface is frictionless, and a
spring with a spring constant of 1120N / m is fixed to block 2. When the blocks
collide, the compression of the spring is maximum at the instant the blocks have the
same velocity. Find the maximum compression.
Fig.5-12 Problem 9
- 128 -
Chapter 5
Center of Mass and Linear Momentum
Solution: The linear momentum of the two blocks and the spring system is
conserved, and the mass of the spring is negligible. Choose rightward as positive
direction and suppose when the compression of the spring is maximum, the velocity of
the blocks is v , we have
2.0kg10m / s  5.0kg3.0m / s  2.0kg  5.0kgv
Solving for v , yields
v  5m / s
Because of the compression of the spring, the total kinetic energy of the system
decreased, gives us
1
1
1
1
2
2
2
2
kxmax  2.0kg 10m / s   5.0kg 3m / s    2.0kg  5.0kg 5m / s  
2
2
2
2

Substituting k=1120N/m into the above equation yields the maximum compression of
the spring:
xmax  0.25m
(Answer)
10 (25) In Fig.5-13, puck 1 of mass m1  0.20kg is sent sliding across a
frictionless lab bench, to undergo a one-dimensional elastic collision with stationary
puck 2. Puck 2 then slides off the bench and lands a distance d from the base of the
bench. Puck 1 rebounds from the collision and slides off the opposite edge of the
bench, landing a distance 2d from the base of the bench. What is the mass of puck 2?
(Hint: Be careful with signs.)
Fig.5-13 Problem 10
Solution: Choose rightward as positive direction. The linear momentum of the
two-puck system is conserved during the collision, so
m1v1i  m1v1 f  m2v2
With the landing distance of the two pucks, we have
- 129 -
(1)
Chapter 5
Center of Mass and Linear Momentum
v1 f
v2

 2d
 2
d
(2)
During one-dimensional elastic collision, the total kinetic energy is conserved:
1
1
1
2
2
2
m1v1i  m1v1 f  m2v2
2
2
2
(3)
From Eqs.(1)~(3), we get the mass of puck 2:
m2  5m1  5  0.20kg  1.00kg
(Answer)
11 (33) Figure 5-14 shows a uniform square plate of edge length
6d  6.0m from which a square piece of edge length 2d has
been removed. What are (a) the x coordinate and (b) the y
coordinate of the center of mass of the remaining piece?
Solution: The square plate of edge length 6d consists of the
square piece of edge length 2d and the remaining piece, and
obviously the center of mass of the uniform square plate is at the
origin. So we have
xcom, plate 
ycom, plate 
m piece xcom, piece  mremainxcom, remian
m plate
m piece ycom, piece  mremain ycom, remian
m plate
Fig.5-14 Problem 11
0
(1)
0
where xcom, piece  2d , ycom, piece  0
(2)
The uniform square gives us
m p i e ce
2d 2 m  1 m ,
6d 2 p l a t e 9 p l a t e
8
mr e m a i nm p l a 
m p l a t e (3)
t e m p i e ce
9
Eqs.(1)~(3) give us the coordinates of the center of mass of the remaining piece:
1
xcom, remian   d  0.25m ,
4
12 (44)
ycom, remain  0
(Answer)
Two identical coins are initially held at height h  11.0m . Coin 1 is
- 130 -
Chapter 5
Center of Mass and Linear Momentum
dropped at time t  0 and then lands on a muddy field where it sticks. Coin
2 is dropped at t  0.500s and then lands on the field. What is the acceleration

acom of the center of mass (com) of the two-coin system (a) between t  0 and
t  0.500s , (b) between t  0.500s and time t1 when coin hits and sticks, and (c)
between t1 and time t 2 when coin 2 hits and sticks? What is the speed of the center
of mass when t is (d) 0.250s , (e) 0.750s , (f) 1.75s ?
Solution: Choose the dropping point as origin and downward as positive
direction,construct x coordinate.
t1 
2h
2 11m
2h

 1.50s , t 2  0.500s 
 2.00s
2
g
9.8m / s
g
(a) between t  0 and t  0.500s , coin 1 is free falling and coin 2 is at rest, so

m g ˆ 1 ˆ
acom 
i  gi  4.9iˆ m / s 2
2m
2


(Answer)
(b) between t  0.500s and time t1  1.50s , coin 1 and 2 are both free falling,
and

m g  m g ˆ
ac o m
i  giˆ  9.8iˆ m / s 2
2m


(Answer)
(c) between t1  1.50s and time t 2  2.00s , coin 1 is at rest and coin 2 is free
falling, and

m g ˆ 1 ˆ
acom 
i  gi  4.9iˆ m / s 2
(Answer)
2m
2
(d) When t  0.250s , coin 1 is free falling and coin 2 is at rest, and the speed of com


is
vcom 


mv1 1
1
  gt  9.8m / s 2 0.250s   1.225m / s
2m 2
2
(Answer)
(e) When t  0.750s , 0.500s  t  t1 , coin 1 and 2 are both free falling, and
mv1  mv2 1
1
 gt  g t  0.500s   g 2t  0.500s 
2m
2
2
1
 9.8m / s 2 2  0.750s  0.500s 
2
 4.9m / s
vcom 


- 131 -
(Answer)
Chapter 5
Center of Mass and Linear Momentum
t1  t  t 2 , coin 1 is at rest and coin 2 is free falling,
(f) When t  1.75s ,
and
vcom 


mv2 1
1
 g t  0.500s   9.8m / s 2 1.75s  0.500s   6.125m / s
2m 2
2
(Answer)
13 (47) In Fig.5-15, a 3.2kg box of running shoes slides on a horizontal frictionless
table and collides with a 2.0kg box of ballet slippers
initially at rest on the edge of the table, at height
h  0.40m . The speed of the 3.2kg box is
3.0m/ s just before the collision. If the two boxes
stick together because of packing tape on their sides,
what is their kinetic energy just before they strike the
floor?
Fig.5-15 Problem 13
Solution: Choose the edge of the table as origin, and construct x  y coordinates as
shown. Just before the boxes strike the floor, their velocity

v  vxiˆ  v y ˆj
During the collision, the linear momentum of the two-box system is conserved:
3.2kg3.0m / s   3.2kg  2.0kgvx
vx  1.85m / s
Solving for v x , yields

So

vy  2 gh  2 9.8m / s 2 0.40m  2.8m / s
Along the y axis
v 2  vx  v y  1.85m / s   2.80m / s   11.2625m / s 
2
2
2
2
2
The kinetic energy of the two boxes just before they strike the floor is
K


1
3.2kg  2.0kg v 2  1 5.2kg  11.2625m2 / s 2  29.28 J
2
2
(Answer)
14 (49) In Fig.5-16, block 1 slides along an x axis on a frictionless floor with a
speed of 0.75m/ s . When it reaches stationary block 2, the two blocks undergo an
elastic collision. The following table gives the mass and length of the (uniform) blocks
- 132 -
Chapter 5
Center of Mass and Linear Momentum
and also the locations of their centers at time t  0 . Where is the center of
mass of the two-block system located (a) at t  0 , (b) when the two blocks first
touch, and (c) at t  4.0s ?
Block
Mass (kg)
Length (cm)
Center at t=0
1
2
0.25
0.50
5.0
6.0
x=-1.50m
x=0
Fig.5-16 Problem 14
Solution:
(a) At t  0 , the coordinate of the center of mass:
xcom 
0.25kg 1.50m  0.50kg0  0.5m
0.25kg  0.50kg
(Answer)
(b) When the two blocks first touch, the coordinate of com:
0.25kg  5.0cm   0.50kg0
xcom 
2 

0.25kg  0.50kg
 0.83cm
(c) Suppose block 1reaches stationary block 2 at time t1 , then
 5.0cm 6.0cm 


  1.50m
2
2 

t1 
 1.93s
0.75m / s
After collision, the velocity of com is that of the two boxes:
v
0.25kg0.75m / s   0.25m / s
0.25kg  0.50kg
At t  4.0s , the position of the com
- 133 -
(Answer)
Chapter 5
Center of Mass and Linear Momentum
xcom  0.83cm  vt  t1   0.83cm  0.25m / s 4.0s  1.93s   50.92cm
(Answer)
15 (52) Two particles P and Q are released from rest 1.0m apart, P has a
mass of 0.10kg , and Q a mass of 0.30kg . P and Q attract each other with a
2
constant force of 1.0  10 N . No external forces act on the system. (a) What is the
speed of the center of mass of P and Q when the separation is 0.50m ? (b) At
what distance from P' s original position do the particles collide?
Solution: (a) Choose P' s original position as origin and P' s velocity as positive.
Since no external forces act on the system, the linear momentum of the system is
conserved, and the speed of the center mass will keep unchanged as zero.
vcom  0
(Answer)
(b) As the speed of the com is zero, the position of the com is at rest, and the two
particles will collide at that point, that is
xcom 
0.10kg0  0.30kg1.0m  0.75m
0.10kg  0.30kg
So the particles collide 0.75m from P' s original position.
- 134 -
(Answer)