chapter 9 center of mass and linear momentum in this chapter we will introduce the following new...
TRANSCRIPT
Chapter 9
Center of Mass and Linear Momentum
In this chapter we will introduce the following new concepts:
-Center of mass (com) for a system of particles -The velocity and acceleration of the center of mass-Linear momentum for a single particle and a system of particles
We will derive the equation of motion for the center of mass, and discuss the principle of conservation of linear momentum.
Finally, we will use the conservation of linear momentum to study collisions in one and two dimensions and derive the equation of motion for rockets. (9-1)
1 2
1 2
1 1 2 2com
1 2
Consider a system of two particles of masses and
at positions and , respectively. We define the
position of the center of mass (com) as follows:
m x m xx
m m
x x
m m
The Center of Mass :
1 1 2 2 3 3 1 1 2 2 3 3com
11 2 3
...
We can generalize the above definition for a system of particles as follows:
Here is the total mass of
.
all th
.. 1
...
e partic
nn n n n
i iin
m x m x m x m x m x m x m x m xx m x
m m
M
m M
n
m M
1 2 3
les ... .
We can further generalize the definition for the center of mass of a system of
particles in three-dimensional space. We assume that the th particle
( mass ) has position
n
i
M m m m m
i
m
com1
vector
1
.n
i ii
i
r m rM
r
(9-2)
com1
com com com com
com
1The position vector for the center of mass is given by the equation .
ˆ ˆ ˆThe position vector can be written as i j k.
The components of are given by the e
n
i ii
r m rM
r x y z
r
com com com1 1 1
1 1 1
quat
ions
n n n
i i i i i ii i i
x m x y m y z m zM M M
The center of mass has been defined using the quations
given above so that it has the following prope
The center of mass of a system of particles moves as though
all the system's mass were conc
rty:
entr
The above statement will be proved la
ated there, and that the
vector sum o
ter. An example is
given in the figur
f all the
e. A bas
external forces were appli
eball bat is flipped into t
ed ther
he a
e
ir
and moves under the influence of the gravitation force. The
center of mass is indicated by the black dot. It follows a
parabolic path as discussed in Chapter 4 (projectile motion).
All the other points of the bat follow more complicated paths.(9-3)
Example: A 3.00 kg particle is located on the x axis at x = −5.00 m and a 4.00 kg particle is on the x axis at x = 3.00 m. Find the center of mass of this two–particle system.
Example: A 3.00 kg particle is located on the x axis at x = −5.00 m and a 4.00 kg particle is on the x axis at x = 3.00 m. Find the center of mass of this two–particle system.
1 1 2 2CM
1 2
(3.00 kg)( 5.00 m) (4.00 kg)(3.00 m)
(3.00 kg 4.00 kg)
0.429 m.
m x m xx
m m
Solid bodies can be considered as systems with continuous distribution of matter.
The sums that are used for the calculation of the center of mass of systems with
dis
The Center of Mass for Solid Bodies :
com com com
1 1 1
crete distribution of mass become integrals:
The integrals above are rather complicated. A simpler special case is that of
uniform objects in w
h
x xdm y ydm z zdmM M M
com com com
1 1 1
ich the mass density is constant and equal to :
In objects with symmetry elements (symmetry point, symmetry line, symmetry plane)
it is
x xdV y ydV z zdVV V
dm M
dV V
V
not necessary to evaluate the integrals. The center of mass lies on the symmetry
element. For example, the com of a uniform sphere coincides with the sphere center.
In a uniform rectangular object the com lies at the intersection of the diagonals.
(9-4).CC
O
m1
m3
m2
F1
F2
F2
x y
z
1 2 3,
1 2 3
Consider a system of particles of masses , , ...,
and position vectors , , ,..., , respectively.
The position vector of the center of mas
n
n
Newton's Second Law for a System
n
of Particles :
m m m m
r r r r
s is given by
com 1 1 2 2 3 3
com 1 1 2 2 3 3
com 1 1 2 2 3 3 com
... . We take the time derivative of both sides
...
... . Here is the velocity of the
n n
n n
n n
Mr m r m r m r m r
d d d d dM r m r m r m r m r
dt dt dt dt dtMv m v m v m v m v v
com 1 1 2 2 3 3
com 1 1 2 2 3 3 com
com
and is the velocity of the th particle. We take the time derivative once more
...
... . Here is the accelera
i
n n
n n
v i
d d d d dM v m v m v m v m v
dt dt dt dt dtMa m a m a m a m a a
tion of the com
and is the acceleration of the th particle.ia i
(9-5)
O
m1
m3
m2
F1
F2
F2
x y
zcom 1 1 2 2 3 3
com 1 2 3
... .
We apply Newton's second law for the th particle:
. Here is the net force on the th particle,
... .
n n
i i i i
n
Ma m a m a m a m a
i
m a F F i
Ma F F F F
app int
app int app int app int app intcom 1 1 2 2 3 3
co
The force can be decomposed into two components: applied and internal:
. The above equation takes the form:
...
i
i i i
n n
F
F F F
Ma F F F F F F F F
Ma
app app app app int int int intm 1 2 3 1 2 3
net
... ...
The sum in the first set of parentheses on the RHS of the equation above is
just .
The sum in the second set of parentheses
n nF F F F F F F F
F
com net
net, com, net , com,
on the RHS vanishes by virtue of
Newton's third law.
The equation of motion for the center of mass becomes
In terms of components we have
.
:
x x y y
Ma F
F Ma F Ma
net, com,
z zF Ma (9-6)
com net Ma F
net, com,
net, com,
net, com,
x x
y y
z z
F Ma
F Ma
F Ma
The equations above show that the center of mass of a system of particles
moves as though all the system's mass were concentrated there, and that the
vector sum of all the external forces were applied there. A dramatic example is
given in the figure. In a fireworks display a rocket is launched and moves under
the influence of gravity on a parabolic path (projectile motion). At a certain point
the rocket explodes into fragments. If the explosion had not occurred, the rocket
would have continued to move on the parabolic trajectory (dashed line). The forces
of the explosion, even though large, are all internal and as such cancel out. The
only external force is that of gravity and this remains the same before and after the
explosion. This means that the center of mass of the fragments follows the same
parabolic trajectory that the rocket would have followed had it not exploded. (9-7)
mv
p Linear momentum of a particle of mass and velocity
The SI unit for linear momentum is the kg.m
is defined a
/s
s .
.
p mv
p m v
Linear Momentum :
p mv
The time rate of change of the
linear momentum of a particle is equal to the magnitude of net force acting on
t
Below we will prove the fol
he particle and has the dir
lowing statem
ection of the
ent:
for
net
net
In equation form: . We will prove this equation using
Newton's second law:
This equation is stating that the linear momentum of a particle can be c
c
a
.
h
e
n
dpF
dt
dp d dvp mv mv m ma F
dt dt dt
ged
only by an external force. If the net external force is zero, the linear momentum
cannot change:net
dpF
dt
(9-8)
Example: A 3.00 kg particle has a velocity of (3.0 i − 4.0 j) m/s. Find its x and y components of momentum and the magnitude of its total momentum.
Using the definition of momentum and the given values of m and v we have:p = mv = (3.00 kg)(3.0 i − 4.0 j) = (9.0 i − 12. j)So the particle has momentum componentspx = +9.0 and py = −12.
O
m1
m3
m2
p1
p2
p3
x y
z
In this section we will extend the definition of
linear momentum to a system of particles. The
th particle has mass , velocity , and linear
momentum i i
i
i m v
p
The Linear Momentum of a System of Particles
.
1 2 3 1 1 2 2 3 3 com
We define the linear momentum of a system of particles as follows:
... ... .
The linear momentum of a system of particles is equal to the product of tn n n
n
P p p p p m v m v m v m v Mv
com
com com net
he
total mass of the system and the velocity of the center of mass.
The time rate of change of is .
The linear momentum of a system of particles can be change
M v
dP dP Mv Ma F
dt dt
P
net net
d only by a
net external force . If the net external force is zero, cannot change.F F P
1 2 3 com... nP p p p p Mv
net
dPF
dt
(9-9)
We have seen in the previous discussion that the momentum of an object can
change if there is a nonzero external force acting on the object. Such forces
exist during the collisio
Collision and Impulse :
n of two objects. These forces act for a brief time
interval, they are large, and they are responsible for the changes in the linear
momentum of the colliding objects.
Consider the collision of a baseball with a baseball bat.
The collision starts at time when the ball touches the bat
and ends at when the two objects separate.
The ball is acted upon by a force
i
f
t
t
( ) during the collision.
The magnitude ( ) of the force is plotted versus in fig. a.
The force is nonzero only for the time interval .
( ) . Here is the linear momentum of the bal
i f
F t
F t t
t t t
dpF t p
dt
l,
( ) ( )f f
i i
t t
t t
dp F t dt dp F t dt
(9-10)
( ) = change in momentum
( ) is known as the impulse of the collision.
( ) The magnitude of is equal to the area
under the ver
f f f
i i i
f
i
f
i
t t t
f i
t t t
t
t
t
t
dp F t dt dp p p p
F t dt J
J F t dt J
F
ave
sus plot of fig. a .
In many situations we do not know how the force changes
with time but we know the average magnitude of the
collision force. The magnitude of the impulse is given by
t p J
F
J F
ave
ave
, where .
Geometrically this means that the area under the
versus plot (fig. a) is equal to the area under the
versus plot (fig. b).
f it t t t
F t
F t
p J
(9-11)aveJ F t
Example: A child bounces a ball on the sidewalk. The linear impulse delivered by the sidewalk is 2.00 N· s during the 1/800 s of contact. What is the magnitude of the average force exerted on the ball by the sidewalk?
32.001.60 10 N
1/800t
I
F
Example: A 3.0 kg steel ball strikes a wall with a speed of 10 m/s at an angle of 60o with the surface. It bounces off with the same speed and angle, as shown in the figure. If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the wall by the ball?
Since has no x component, the average force has magnitude 2.6 ×102 N and points in the y direction (away from the wall).
Consider a target that collides with a steady stream of
identical particles of mass and velocity along the -axis.m v x
Series of Collisions
A number of the particles collides with the target during a time interval .
Each particle undergoes a change in momentum due to the collision with
the target. During each collision a momentum
n t
p
ave
change is imparted on the
target. The impulse on the target during the time interval is .
The average force on the target is .
Here is the change in the ve
p
t J n p
J n p nF m v
t t tv
ave
locity of each particle along the -axis due
to the collision with the target .
Here is the rate at which mass collides with the target.
If the particles stop after the collision
x
mF v
tm
t
, then 0 .
If the particles bounce backwards, then 2 .
v v v
v v v v
(9-12)
Fave
O
m1
m3
m2
p1
p2
p3
x y
z
net
net
Consider a system of particles for which 0
0 Constant
F
dPF P
dt
Conservation of Linear Momentum :
total linear momentumtotal linear momentum
at some later time at some initial time
If no net external force acts on a system of particles, the total linear momentum
cannot change.
Th
fi tt
P
e conservation of linear momentum is an important principle in physics.
It also provides a powerful rule we can use to solve problems in mechanics
suc
Note 1:
h as collisions.
In systems in which F
net 0 we can always apply conservation of
linear momentum even when the internal forces are very large as in the case
of colliding objects.
We will encounter problems (e.g., inelasticNote 2 co s: lli
ions) in which the
energy is not conserved but the linear momentum is. (9-13)
Example: A 4.5 kg gun fires a 0.1 kg bullet with a muzzle velocity of +150 m/s. What is the recoil velocity of the gun?
j kj k
m m i f j,i k,fP P v v
, , , ,
, ,
, ,
,
0 0
(0.1)(150)3.3 m/s.
4
G G i b b i G G f b b f
G G f b b f
G G f b b f
G f
m v m v m v m v
m v m v
m v m v
v
1 2
1 2 1 2
Consider two colliding objects with masses and ,
initial velocities and , and final velocities and ,
respectively.
i i f f
m m
v v v v
Momentum and Kinetic Energy in Collisions
netIf the system is isolated, i.e., the net force 0, linear momentum is conserved.
The conservation of linear momentum is true regardless of the collision type.
This is a powerful rule that allows us
F
to determine the results of a collision
without knowing the details. Collisions are divided into two broad classes:
and .
A collision is if there is
elastic in
no loss o
elastic
ela f kinetic stic energy, i.e.,
A collision is if kinetic energy is lost during the collision due to
conversion into other forms of energy. In this case we have
A special case
.
i
of inelastic collisi
nelas
tic
.
i f
f i
K K
K K
ons are known as .
In these collisions the two colliding objects stick together and they move as a
single body. In these collisions the loss of kinetic
co
e
mpletely in
nergy is ma
elastic
ximum.(9-14)
1 2 1 2
1 1 1 2 1 1 1 2
In these collisions the linear momentum of the colliding
objects is conserved .i i f f
i i f f
p p p p
m v m v m v m v
One - Dimensional Inelastic Collisions :
2
In these collisions the two colliding objects stick together
and move as a single body. In the figure to the left we show
a special case in which 0iv
One - Dimensional Completely Inelastic Collisions :
1 1 1 1
11
1 2
1 2 1 1com
1 2 1 2 1 2
.
The velocity of the center of mass in this collision
is .
In the picture to the left we show some freeze-frames
of a totally inelasti
i
i
i i i
m v mV mV
mV v
m m
p p m vPv
m m m m m m
c collision.(9-15)
Example: An object of 12.0 kg at rest explodes into two pieces of masses 4.00 kg and 8.00 kg. The velocity of the 8.00 kg mass is 6.00 m/s in the +ve x-direction. The change in the kinetic is:
Example: An object of 12.0 kg at rest explodes into two pieces of masses 4.00 kg and 8.00 kg. The velocity of the 8.00 kg mass is 6.00 m/s in the +ve x-direction. The change in the kinetic is:
4 4 8 8 4
4
2 2 2 24 4 8 8
ˆ0 4 8 6
ˆ12 ,
1 1 4 8(12) (6)
2 2 2 2432 J
M v m v m v v i
v i
K m v m v
Inelastic collisions in 1D
Example: A 10.0 g bullet is stopped in a block of wood (m = 5.00 kg). The speed of the bullet–plus–wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet?
Elastic collisions in 1D
1 2
1 2 1 2
Consider two colliding objects with masses and ,
initial velocities and , and final velocities and ,
respectively.
i i f f
m m
v v v v
One - Dimensional Elastic Collisions
1 1 1 2 1 1 1 2
2 22 21 1 2 21 1 1 2
Both linear momentum and kinetic energy are conserved.
Linear momentum conservation: (eq. 1)
Kinetic energy conservation: (eq. 2)2 2 2 2
We have tw
i i f f
f fi i
m v m v m v m v
m v m vm v m v
1 2 21 1 2
1 2 1 2
1 2 12 1 2
1 2 2
1 1
1 1
1
o equations and two unknowns, and .
If we solve equations 1 and 2 for and we get the following solution
2
2
s:
f f
f i i
f i i
f f
m m mv v v
m m m m
m m mv v v
m m
v v
m
v v
m
(9-16)
Example: A 10 kg ball (m1) with a velocity of +10 m/s collides head on in an elastic manner with a 5 kg ball (m2) at rest. What are the velocities after the collision?
Example: A 10 kg ball (m1) with a velocity of +10 m/s collides head on in an elastic manner with a 5 kg ball (m2) at rest. What are the velocities after the collision?
2 1 1We substitute 0 in the two solutions for and :i f fv v v
2Special Case of Elastic Collisions - Stationary Target v = 0 :i
1 2 21 1 2
1 2 1 2
1 2
1 2 12 1 2
1 2 1
1 11 2
12 1
1 22
2
2
Below we examine several special cases for which we know the outcome
of the collision from exp
1. E
erience.
q
2
u
f i i
f i
f i
f ii
m mv v
m m
m
m m mv v v
m m m m
m m mv v v
m m m mv v
m m
1 21 1
1
11 2
12 1 1 1
1 2
2
0
2 2
The two colliding objects have exchanged velocit
al
ie
masses
s.
f i i
f i i i
m m m mv v v
m m m m
m mv v
m m
v vm m m m
m
m
m
m
m
v1i
v1f = 0 v2f
v2i = 0
x
x
(9-17)
12 1
2
1
1 2 21 1 1 1
11 2
2
1
21 12 1 1 1
11 2 2
2
1
1
1
22
2. A massive targ
2
et
1
f i i i
f i i i
mm m
m
mm m m
v v v vmm mm
mmm m
v v v vmm m mm
1
2
Body 1 (small mass) bounces back along the incoming path with its speed
practically unchanged.
Body 2 (large mass) moves forward with a very small speed because 1. m
m
(9-18)
v2i = 0
m1
m1
m2
m2
v1i
v1f
v2f
x
x
m1
m1
m2
m2
v1i
v1f v2f
v2i = 0
x
x
21 2
1
2
1 2 11 1 1 1
21 2
1
12 1 1 1
21 2
1
12. A massive projectil e
1
1
2 22
1
f i i i
f i i i
mm m
m
mm m m
v v v vmm mm
mv v v v
mm mm
Body 1 (large mass) keeps on going, scarcely slowed by the collision.
Body 2 (small mass) charges ahead at twice the speed of body 1.
(9-19)
Collisions in 2-D
In this section we will remove the restriction that the
colliding objects move along one axis. Instead we assume
that the two bodies that participate in the collision
move in
Collisions in Two Dimensions :
1 2 the -plane. Their masses are and .xy m m
1 2 1 2
1 2 1 2
2
The linear momentum of the system is conserved: .
If the system is elastic the kinetic energy is also conserved: .
We assume that is stationary and that after the
i i f f
i i f f
p p p p
K K K K
m
1 2 1
1 1 1
collision particle 1 and
particle 2 move at angles and with the initial direction of motion of .
In this case the conservation of momentum and kinetic energy take the form:
axis: i
m
x m v m
1 1 2 2 2
1 1 1 2 2 2
2 2 21 1 1 2 2 2
1 2 1 1
cos cos (eq. 1)
axis: 0 sin sin (eq. 2)
1 1 1 (eq. 3) We have three equations and seven variables:
2 2 2Two masses: , ; three speeds: , ,
f f
f f
i f f
i f
v m v
y m v m v
m v m v m v
m m v v v
2 1 2; and two angles: , . If we know
the values of four of these parameters we can calculate the remaining three.
f
(9-20)
Example: An unstable nucleus of mass 17 × 10**27 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.0×10**27 kg, moves along the y axis with a speed of 6.0 × 10**6 m/s . Another particle of mass 8.4 × 10**27 kg, moves along the x axis with a speed of 4.0 × 10**6 m/s . Find(a) the velocity of the third particle and (b) the total energy given off in the process.
The parent nucleus is at rest , so that the total momentum was (and remains) zero: Pi = 0.
Example: A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33 m/s at an angle of 30.0degree with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball’s velocity.
Then the condition that the total x momentum be conserved gives us:
We can similarly find vy by using the condition that the total y momentum be conservedin the collision. This gives us:
A 10.0 kg toy car is moving along the x axis. The only force Fx acting on the car is shown in Fig. 5 as a function of time (t). At time t = 0 s the car has a speed of 4.0 m/s. What is its speed at time t = 6.0 s? (Ans: 8.0 m/s )
A 10.0 kg toy car is moving along the x axis. The only force Fx acting on the car is shown in Fig. 5 as a function of time (t). At time t = 0 s the car has a speed of 4.0 m/s. What is its speed at time t = 6.0 s? (Ans: 8.0 m/s )
Area under the curve
10 (6 2)40
2
404 8 m/s
10
f i
f i
dpF p Fdt
dt
p m v v
pv v
m
An impulsive force Fx as a function of time (in ms) is shown in the figure as applied to an object (m = 5.0 kg) at rest. What will be its final
speed?
An impulsive force Fx as a function of time (in ms) is shown in the figure as applied to an object (m = 5.0 kg) at rest. What will be its final
speed?
3
Area under the curve
5000 4 1010
2
10 m2
5 s
dpF p Fdt
dt
pv
m
If the masses of m1 and m3 in Fig. 5 are 1.0 kg each and m2 is 2.0 kg, what are the coordinates of the center of mass? (Ans: (1.00, 0.50) m )
If the masses of m1 and m3 in Fig. 5 are 1.0 kg each and m2 is 2.0 kg, what are the coordinates of the center of mass? (Ans: (1.00, 0.50) m )
1 1 2 2 3 3
1 2 3
1 1 2 2 3 3
1 2 3
1.0 0.0 1.0 2.0 2.0 1.01.0 m
4.0
1.0 0.0 2.0 1.0 1.0 00.5 m
4.0
cm
cm
m x m x m xx
m m m
m y m y m yy
m m m
A 1500 kg car traveling at 90.0 km/h east collides with a 3000 kg car traveling at 60.0 km/h south. The two cars stick together after the collision (see Fig 2). What is the speed of the cars after collision?
A 1500 kg car traveling at 90.0 km/h east collides with a 3000 kg car traveling at 60.0 km/h south. The two cars stick together after the collision (see Fig 2). What is the speed of the cars after collision?
1 1 2 2
2 2
ˆ ˆ(1500 3000) 1500 90 3000 60
km30 40 50 13.9 m/s
hour
M v m v m v
v i j
v
A 2.00 kg pistol is loaded with a bullet of mass 3.00 g. The pistol fires the bullet at a speed of 400 m/s. The recoil speed of the pistol when the bullet was fired is: (Ans: 0.600 m/s)
A 2.00 kg pistol is loaded with a bullet of mass 3.00 g. The pistol fires the bullet at a speed of 400 m/s. The recoil speed of the pistol when the bullet was fired is: (Ans: 0.600 m/s)
1 1 2 20
m0 2 0.003 400 0.6 .
s
m v m v
v v
A 1.0 kg ball falling vertically hits a floor with a velocity of 3.0 m/s and bounces vertically up with a velocity of 2.0 m/s . If the ball is in contact with the floor for 0.10 s, the average force on the floor by the ball is:
A 1.0 kg ball falling vertically hits a floor with a velocity of 3.0 m/s and bounces vertically up with a velocity of 2.0 m/s . If the ball is in contact with the floor for 0.10 s, the average force on the floor by the ball is:
The change of momentum of the particle,
(1) 2 ( 3) 5
5The average force on the floor by the ball 50 N (down)
.1
f ip M v m v v
pF
t
A 20-g bullet is fired into a 100-g wooden block initially at rest on a horizontal frictionless surface. If the initial speed of the bullet is 10 m/s and it comes out of the block with a speed of 5.0 m/s, find the speed of the block immediately after the collision.
A 20-g bullet is fired into a 100-g wooden block initially at rest on a horizontal frictionless surface. If the initial speed of the bullet is 10 m/s and it comes out of the block with a speed of 5.0 m/s, find the speed of the block immediately after the collision.
1
1
Consrevation of momentum
0.02 (10 5)1.0 m/s
0.1
b b f B
b f
B
m v m v m V
m v vV
m
A 1.0-kg block at rest on a horizontal frictionless surface is connected to a spring (k = 200 N/m) whose other end is fixed (see figure). A 2.0-kg block moving at 4.0 m/s collides with the 1.0-kg block. If the two blocks stick together after the one-dimensional collision, what maximum compression of the spring does occur when the blocks momentarily stop?
A 1.0-kg block at rest on a horizontal frictionless surface is connected to a spring (k = 200 N/m) whose other end is fixed (see figure). A 2.0-kg block moving at 4.0 m/s collides with the 1.0-kg block. If the two blocks stick together after the one-dimensional collision, what maximum compression of the spring does occur when the blocks momentarily stop?
2 2
Consrevation of momentum ( )
2 4 m2.67
(2 1) s
1 1Conservation of K.E. after collision ( )
2 2
( ) 32.67 0.33 m
200
mv m M V
V
m M V k x
m Mx V
k