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Chapter 9
Center of Mass
and
Linear Momentum
9.2 The Center of Mass
The center of mass of a
system of particles is the
point that moves as
though (1) all of the
system’s mass were
concentrated there and
(2) all external forces
were applied there.
The center of mass (black
dot) of a baseball bat
flipped into the air follows a
parabolic path, but all other
points of the bat follow
more complicated curved
paths.
9.2 The Center of Mass: A System of Particles
Consider a situation in which n particles are strung out along
the x axis. Let the mass of the particles are m1, m2, ….mn, and
let them be located at x1, x2, …xn respectively. Then if the
total mass is M = m1+ m2 + . . . + mn, then the location of the
center of mass, xcom, is
9.2 The Center of Mass: A System of Particles
In 3-D, the locations of the center of mass are given by:
The position of the center of mass can be expressed as:
9.2 The Center of Mass: Solid Body
In the case of a solid body, the “particles” become
differential mass elements dm, the sums become integrals,
and the coordinates of the center of mass are defined as
where M is the mass of the object.
If the object has uniform density, r, defined as:
Then
Where V is the volume of the object.
Sample problem, COM
Calculations: First, put the stamped-out disk (call it
disk S) back into place to form the original composite
plate (call it plate C). Because of its circular symmetry,
the center of mass comS for disk S is at the center of S,
at x =-R. Similarly, the center of mass comC for
composite plate C is at the center of C, at the origin.
Assume that mass mS of disk S is concentrated in a
particle at xS =-R, and mass mP is concentrated in a
particle at xP. Next treat these two particles as a two
particle system, and find their center of mass xS+P.
Next note that the combination of disk S and plate P is
composite plate C. Thus, the position xS+P of comS+P
must coincide with the position xC of comC, which is at
the origin; so xS+P =xC = 0.
But,
and xS=-R
Sample problem, COM of 3 particles
The total mass M of the system is 7.1 kg.
The coordinates of the center of mass are
therefore:
We are given the following data:
Note that the zcom = 0.
9.3: Newton’s 2nd Law for a System of Particles
The vector equation that governs the motion of the center
of mass of such a system of particles is:
Note that:
1. Fnet is the net force of all external
forces that act on the system. Forces
on one part of the system from
another part of the system (internal
forces) are not included
2. M is the total mass of the system.
M remains constant, and the system
is said to be closed.
3. acom is the acceleration of the
center of mass of the system.
9.3: Newton’s 2nd Law for a System of Particles: Proof of final result
For a system of n particles,
where M is the total mass, and ri are the position vectors of the masses mi.
Differentiating,
where the v vectors are velocity vectors.
This leads to
Finally,
What remains on the right hand side is the vector sum of all the external forces that
act on the system, while the internal forces cancel out by Newton’s 3rd Law.
Sample problem: motion of the com of 3 particles
Calculations: Applying Newton’s second law
to the center of mass,
Linear Momentum
I) One particle system
Definition of linear momentum:
The linear momentum of an object is the product of its mass and
velocity.
Note that momentum is a vector—it has both a magnitude and a
direction.
SI unit of momentum: kg • m/s. This unit has no special name.
Linear momentum
(Newton’s 2nd Law - acceleration)
.
dv dmv dp
 F  ma  m


dt
dt
dt
The time rate of change of the momentum of a particle is equal to
the net force acting on the particle and is in the direction of that
force.
(Newton’s 2nd Law - momentum)
Collision and Impulse
The change in linear momentum is related to the force by
Newton’s second law written in the form
 
dp  Fnet dt

p2
t
 2 
 dp   Fnet dt

p1
t1
The let side of the equation is the change of the momentum of the

particle:
p

p 

d
p

2

p1
The right side of the equation is a measure of both the magnitude and
the duration of the net force, and is called the impulse of the (net)
force, J.
 t2 
J   Fnet dt
t1
Impulse
In this case, the impact is brief, and the ball
experiences a force that is great enough to slow,
stop, or even reverse its motion.
The figure depicts the force of impact at one
instant. The ball experiences a force F(t) that
varies during the collision and changes the
linear momentum of the ball.
Graphical Representation of Impulse
 

J  Favg t  p
Instead of the ball, one can focus on the
bat. At any instant, Newton’s third law
says that the force on the bat has the same
magnitude but the opposite direction as the
force on the ball.
That means that the impulse on the bat has
the same magnitude but the opposite
direction as the impulse on the ball.
Impulse-Momentum
Example: Bouncing vs. Stopping
Actual contact times may be very short.
Example Tennis
• A tennis ball weighing 50 g and moving horizontally with a velocity
of 12 m/s, is hit by a racket such that its velocity becomes 20 m/s in
exactly the opposite direction.
a) What is the change in the momentum of the ball?
p  p f  pi
p  mv f mvi  m(v f  vi )
p  0.05kg[20m / s  (12m / s)]  1.6 kg  m / s
b) What impulse is delivered by the racket to the ball?
J = Δp = 1.6 Ns
c) What force is exerted by the racket on the ball if the time of impact
is 5 ms?
F Δt = Δp therefore F x 0.005 s= =1.6 Ns, F = 320 N
Example Tennis
• The tennis ball (50 g) moving with a velocity of 12 m/s, hits a wall
at an angle of 30o with the normal and bounces with the velocity of
10 m/s at an angle of 45o with the normal.
a) What is the change in the magnitude of the momentum of the ball?
 

pi
p  p f  pi
30



 
p  mv f mvi  m(v f  vi ) p45
f
p x  m(v fx  vix )
p y  m(v fy  viy )
p x  0.05kg[(10m / s) cos 45o  (12m / s) cos 30o ]  0.873kgm / s
p y  0.05kg[(10m / s) sin 45o  (12m / s) sin 30o ]  0.053kgm / s
p  0.8732  0.0532  0.875kgm / s
b) What impulse is delivered by the wall to the ball?
J = Δp = 0.875 Ns
c) What force is exerted by the wall on the ball if the time of impact is 5 ms?
F Δt = Δp therefore F x 0.005 s= 0.875 Ns, F = 174.9 N
Force & Impulse
•
Two boxes, one heavier than the other, are initially at rest on a horizontal
frictionless surface. The same constant force F acts on each box for exactly
1 second.
– Which box has the most momentum after the force acts?
(a) heavier
F
light
(b)
lighter
F
(c) same
heavy
Solution
We know Fav 
p
t
so
p  Fav t
In this problem F and t are the same for both boxes!
The boxes will have the same final momentum.
F
light
F
heavy
II) Systems of Particles
For a system of objects, the total momentum
is the vector sum of each.
System of Two Particles
• Suppose we have a system of two particles as shown. The particles interact with
each other through equal and opposite internal forces F, and in addition there are
external forces pushing on particles 1 and 2.
• We will apply the impulse momentum theorem separately for each particle:
1) (F1, EXT + F12)Δt = Δp1
2) (F2, EXT + F21)Δt = Δp2
F2,EXT
Adding the two equations yields:
(F1, EXT + F2, EXT+ F12 + F21)Δt = Δp1+ Δp2
But because of Newton’s 3rd Law:
m1 F12
F12 = -F21, so F12 + F21 =0
(F1, EXT + F2, EXT)Δt = Δ(p1+ p2)
F21
F1,EXT
If we denote:
ΣFEXT= F 1, EXT + F 2, EXT
and
Ptot = p1+ p2
m2
System of Two Particles
we get:
(ΣFEXT) Δt = ΔPtot
or (ΣFEXT) = ΔPtot/ Δt
Impulse-Momentum Theorem for Systems of Particles
All the “internal” forces cancel !!
Only the “external” forces matter !!
Note: Internal forces cannot change the
total momentum of the system.
If (ΣFEXT) =0 then ΔPtot=0,
and Ptot=const.
Conservation of Linear
Momentum-System
If (ΣFEXT) =0 than ΔPtot=0, and Ptot=const.
If there is no net external force acting on a
system, its total momentum cannot change,
and stays constant.
This is the law of conservation of momentum.
If there are internal forces, the momenta of
individual parts of the system can change, but
the overall momentum stays the same.
Example: A rifle has a mass of 4.5 kg and it fires a bullet of
10.0 grams at a muzzle speed of 820 m/s. What is the recoil
speed of the rifle as the bullet leaves the barrel?
As long as the rifle is horizontal, there will be no net
external force acting on the rifle-bullet system and
momentum will be conserved.
pi  p f
0  mb vb  mr vr
 0.01 kg 
mb
820 m/s  1.82 m/s
vr  
vb  
mr
 4.5 kg 
25
Examples: Conservation of Linear
Momentum
In this example, there is no external force, but
the individual components of the system do
change their momenta:
9.6: Collision and Impulse: Series of Collisions
Let n be the number of projectiles that collide in a
time interval t.
Each projectile has initial momentum mv and
undergoes a change p in linear momentum
because of the collision.
The total change in linear momentum for n
projectiles during interval t is np. The
resulting impulse on the target during t is along
the x axis and has the same
magnitude of np but is in the opposite direction.
In time interval t, an amount of mass m =nm
collides with the target.
Sample problem: 2-D
impulse, cont.
Sample problem: 1-D explosion
Sample problem: 2-D explosion
9.8: Momentum and Kinetic Energy in Collisions
In a closed and isolated system, if there are two colliding bodies, and
the total kinetic energy is unchanged by the collision, then the
kinetic energy of the system is conserved (it is the same before and
after the collision). Such a collision is called an elastic collision.
If during the collision, some energy is always transferred from
kinetic energy to other forms of energy, such as thermal energy or
energy of sound, then the kinetic energy of the system is not
conserved. Such a collision is called an inelastic collision.
Sample problem: two pendulums
Collisions
Collisions: an interaction that lasts a very
short time
Elastic vs. Inelastic Collisions
• A collision is said to be elastic when kinetic energy as well as
momentum is conserved before and after the collision.
Kbefore = Kafter
– Carts colliding with a spring in between, billiard balls, etc.
vi
• A collision is said to be inelastic when kinetic energy is not
conserved before and after the collision, but momentum is
conserved.
Kbefore  Kafter
– Car crashes, collisions where objects stick together, etc.
Completely Inelastic Collisions
Objects stick together after collision.
m1
m1+m2
m2
v2
v1
P before
=
v’
P after
p1 + p2 = P’
m1v1 + m2v2
= (m1 + m2) v’


 m1v1  m2 v2
v'
m1  m2
Energy
Total kinetic energy is not conserved in an inelastic
collision.
K1  K 2  K '
2D Inelastic Collisions
Again, in a completely
inelastic collision the
objects stick together
afterwards.
vy’
vx’
p + P = P’
P =MV
P’=(m+M)v’
p = mv



mv  MV  (m  M )v '


 mv  MV
v'
mM
vx ' 
mv
mM
vy ' 
MV
mM
Inelastic collisions in 1-D: Velocity of Center of Mass
Fig. 9-16 Some freeze frames
of a two-body system, which
undergoes a completely inelastic
collision. The system’s center of
mass is shown in each freezeframe. The velocity vcom of the
center of mass is unaffected by the
collision.
Because the bodies stick together
after the collision, their common
velocity V must be equal to vcom.
Explosion (inelastic un-collision)
Before the explosion:
M
After the explosion:
v1
v2
m1
m2
Explosion...
• No external forces, so P is conserved.
• Initially: P = 0
• Finally: P = m1v1 + m2v2 = 0
m1v1 = - m2v2
M
v1
v2
m1
m2
Explosion 2
• A bomb explodes into 3 identical pieces. Which of the
following configurations of velocities is possible?
(a) 1
(b) 2
v
v
m
m
v
V
m
m
(1)
(c) both
v
v
m
m
(2)
Explosion 2
• No external forces, so P must be conserved.
• Initially: P = 0
• In explosion (1) there is nothing to balance the upward
momentum of the top piece so Pfinal  0.
v
mv
m
mv
mv
v
v
m
m
(1)
Explosion 2
• No external forces, so P must be conserved.
• All the momenta cancel out.
• Pfinal = 0.
mv
v
mv
mv
m
v
v
m
m
(2)
Ballistic Pendulum
L
L
L
m v
H
M+m
M

L
V
A projectile of mass m moving horizontally with speed v
strikes a stationary mass M suspended by strings of length
L. Subsequently, m + M rise to a height of H.
Given H, what is the initial speed v of the projectile?
V=0
Ballistic Pendulum...

Two stage process:
1. m collides with M, inelastically. Both M and
m then move together with a velocity V’
(before having risen significantly).
2. M and m rise a height H, conserving K+U energy E.
(no non-conservative forces acting after collision)
Ballistic Pendulum...
• Stage 1: Momentum is conserved
in x-direction:

mv  (m  M )V '
 m
V ' 
mM

v

Stage 2: K+U Energy is conserved
( EI  EF )
1
2
(m  M )V '2  (m  M ) gH
Eliminating V’ gives:
V '2  2 gH
 M
v  1   2 gH
m

Sample problem: conservation of
momentum
The collision within the bullet– block system is so
brief. Therefore:
(1) During the collision, the gravitational force on
the block and the force on the block from the
cords are still balanced. Thus, during the
collision, the net external impulse on the
bullet–block system is zero. Therefore, the
system is isolated and its total linear
momentum is conserved.
(2) The collision is one-dimensional in the sense
that the direction of the bullet and block just
after the collision is in the bullet’s original
direction of motion.
As the bullet and block now swing up
together, the mechanical energy of the bullet–
block–Earth system is conserved:
Combining steps:
Elastic Collisions
For an elastic collision, both the kinetic
energy and the momentum are conserved:
Before
v1
m1
After
v1’
v2
m2
m1
v2’
m2


'
'
m1v1  m2v2  m1v1  m2v2
1
1
1
1
2
2
'2
'2
m1v1  m2 v2  m1v1  m2 v2
2
2
2
2
2D Elastic Collisions (off Center)

'
'
m1v1  m1v1  m2v2
'2
1 1
'2
2 2
m1v1 / 2  m v / 2  m v / 2
2
If m1 = m2 = m
'2
1
' 2
2
v1  v  v
2
Pythagorean Theorem, angle 90o
Billiards.
• The final directions are separated by 90o.
pf
pi
vcm
Pf
F
initial
final
Billiards.
• So, we can sink the red ball without sinking
the white ball.
Billiards.
• So, we can sink the red ball without sinking the white ball.
• However, we can also scratch. All we know is that the
angle between the balls is 90o.
Billiards.
• Tip: If you shoot a ball spotted on the
“dot”, you will sink both balls !
Elastic Collisions in 2-D
• A moving ball initially traveling in the direction shown hits
an identical but stationary ball. The collision is elastic.
– Describe one possible direction of both balls just after
the collision.
(a)
(b)
(c)
Billiards Solution
• In the first solution, the angle between the balls is not 90o.
• In the second solution, there are no downward y
components to balance out the upward y components.
Billiards Solution
• The third choice both balances the y components and has
90o between the final direction vectors of the two balls.
• As a result, the third choice is the only one of the three that
fits all necessary criteria.
Collision “timescales”
• Collisions typically involve interactions that
happen quickly.
vf
vi
Vf
F
initial
final
The balls are in contact
for a very short time.
Ex. 1
• 1) Jennifer hits a stationary 200 gram ball and it leaves the
racket at 40 m/s. If time lapse photography shows that the
ball was in contact with the racket for 40 ms:
• (a) What average force was exerted on the racket?
• (b) What is the ratio of this force to the weight of the
ball?
a) Δp = mv-0=0.2 kg x 40 m/s= 8 kg x m/s
Δp = F x Δt
F = Δp/Δt = 8/0.04 = 200 N
b) Fw = mg = 0.2 x 10 = 2 N
F/Fw = 200/2 = 100
Ex. 2
• 2) A 0.32 kg ball is moving horizontally 30 m/s just before bouncing off
a wall, thereafter moving 25 m/s in the opposite direction.
• (a) What is the magnitude of its change in momentum?
• (b) What percentage of the kinetic energy was lost in the collision?
a) Δp = mvf – mvi =
= 0.32kg x 25 m/s – 0.32 kg x (-30 m/s) =
17.6 Ns
b)
0.32kg  (25m / s) 2


 0.69
2
2
K i mvi / 2 0.32kg  (30m / s)
Kf
mv 2f / 2
31% lost
Ex. 3
• 3) In space, a 4.0 kg metal ball moving 30. m/s has a head-on collision
with a stationary 1.0 kg second ball. After the elastic collision, what are the
velocities of the balls?
m1v1  0  m v  m v
'
1 1
'
2 2
'2
1 1
'2
2 2
m1v1 / 2  0  m v / 2  m v / 2
2
4 kg x 30m/s = 4kg v1’ + 1 kg x v2’
4 kg x (30m/s)2 = 4kg v1’2 + 1 kg x v2’2
120 = 4v1’ + v2’
3600 = 4v1’2 + v2’2
120 - 4v1’ = v2’
3600 = 4v1’2 + (120 - 4v1’)2
3600 = 4v1’2 + 14400 – 960 v1’ + 16 v1’2
20 v1’2 - 960 v1’ +10800= 0
V1’ = 18 m/s, 30 m/s
v1’2 - 48 v1’ +540= 0
V2’ = 48 m/s
Ex. 5
•
•
•
5) A 2200 kg auto moving at 12 m/s runs into a 3800 kg truck which is moving
in the same direction at 5 m/s. The collision is completely inelastic.
(a) What is the speed of the auto immediately after the collision?
(b) What percentage kinetic energy did the truck gain in the collision?
m1v1  m2v2  (m1  m2 )v'
2200kg x 12 m/s + 3800 kg x 5 m/s =
=(2200 kg + 3800 kg)v’
26400 + 19000 = 6000v’
3800  (7.57) 2
K / Kt 
 2.29
2
3800  (5)
'
t
129% gain
V’ = 7.57 m/s
Ex. 6
6) A handball of mass 0.10 kg, traveling
horizontally at 25m/s, strikes a wall and
rebounds at 19 m/s. What is the change in
the momentum of the ball?
Δp = mvf – mvi =
= 0.1kg x 19 m/s – 0.1 kg x (-25 m/s)= 4.4 Ns