Transcript wooden toy cars

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)
The Center of Mass :
Consider a system of two particles of masses m1 and m2
at positions x1 and x2 , respectively. We define the
position of the center of mass (com) as follows:
m x  m2 x2
xcom  1 1
m1  m2
We can generalize the above definition for a system of n particles as follows:
xcom
m1 x1  m2 x2  m3 x3  ...  mn xn m1 x1  m2 x2  m3 x3  ...  mn xn
1



m1  m2  m3  ...  mn
M
M
n
m x
i 1
Here M is the total mass of all the particles M  m1  m2  m3  ...  mn .
We can further generalize the definition for the center of mass of a system of
particles in three-dimensional space. We assume that the ith particle
( mass mi ) has position vector ri .
rcom
1

M
n
m r
i 1
i i
(9-2)
i i
The position vector for the center of mass is given by the equation rcom
1

M
ˆ
The position vector can be written as rcom  xcom ˆi  ycom ˆj  zcom k.
The components of rcom are given by the equations
xcom
1

M
n
m x
i 1
i i
ycom
1

M
n
m y
i 1
i
i
zcom
1

M
n
m z
i 1
i i
The center of mass has been defined using the quations
given above so that it has the following property:
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
The above statement will be proved later. An example is
given in the figure. A baseball bat is flipped into the air
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)
n
m r.
i 1
i i
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.
x CM
m1x 1  m 2 x 2 (3.00 kg)( 5.00 m)  (4.00 kg)(3.00 m)


m1  m 2
(3.00 kg  4.00 kg)
 0.429 m.
The Center of Mass for Solid Bodies :
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
discrete distribution of mass become integrals:
1
1
1
xcom 
xdm
y

ydm
z

zdm
com
com



M
M
M
The integrals above are rather complicated. A simpler special case is that of
dm
M
uniform objects in which the mass density  
is constant and equal to
:
dV
V
1
1
1
xcom   xdV
ycom   ydV
zcom   zdV
V
V
V
In objects with symmetry elements (symmetry point, symmetry line, symmetry plane)
it is 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.
C
.
C
(9-4)
z
F1
m2
m1
m3
F2
x
Newton's Second Law for a System of Particles :
Consider a system of n particles of masses m1 , m2 , m3, ..., mn
F2
O
and position vectors r1 , r2 , r3 ,..., rn , respectively.
y
The position vector of the center of mass is given by
Mrcom  m1r1  m2 r2  m3 r3  ...  mn rn . We take the time derivative of both sides 
d
d
d
d
d
M rcom  m1 r1  m2 r2  m3 r3  ...  mn rn 
dt
dt
dt
dt
dt
Mvcom  m1v1  m2 v2  m3v3  ...  mn vn . Here vcom is the velocity of the com
and vi is the velocity of the ith particle. We take the time derivative once more 
d
d
d
d
d
vcom  m1 v1  m2 v2  m3 v3  ...  mn vn 
dt
dt
dt
dt
dt
Macom  m1a1  m2 a2  m3a3  ...  mn an . Here acom is the acceleration of the com
M
and ai is the acceleration of the ith particle.
(9-5)
Macom  m1a1  m2 a2  m3 a3  ...  mn an .
z
m1
F1
m2
F2
mi ai  Fi . Here Fi is the net force on the ith particle,
m3
F2
x
We apply Newton's second law for the ith particle:
O
Macom  F1  F2  F3  ...  Fn .
The force Fi can be decomposed into two components: applied and internal:
y
Fi  Fi app  Fi int . The above equation takes the form:

 F
 
 
 ...  F    F




Macom  F1app  F1int  F2app  F2int  F3app  F3int  ...  Fnapp  Fnint 
Macom
app
1
 F2app  F3app
app
n
int
1
 F2int  F3int  ...  Fnint
The sum in the first set of parentheses on the RHS of the equation above is
just Fnet .
The sum in the second set of parentheses on the RHS vanishes by virtue of
Newton's third law.
The equation of motion for the center of mass becomes Macom  Fnet .
In terms of components we have:
Fnet, x  Macom, x
Fnet, y  Macom, y
Fnet, z  Macom, z
(9-6)
Macom  Fnet
Fnet, x  Macom, x
Fnet, y  Macom, y
Fnet, z  Macom, z
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)
v
m
p
p  mv
Linear Momentum :
Linear momentum p of a particle of mass m and velocity v
is defined as p  mv .
The SI unit for linear momentum is the kg.m/s.
Below we will prove the following statement: The time rate of change of the
linear momentum of a particle is equal to the magnitude of net force acting on
the particle and has the direction of the force.
dp
In equation form: Fnet  . We will prove this equation using
dt
Newton's second law:
dp d
dv
p  mv 
  mv   m
 ma  Fnet
dt dt
dt
This equation is stating that the linear momentum of a particle can be changed
only by an external force. If the net external force is zero, the linear momentum
cannot change:
dp
(9-8)
Fnet 
dt
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 components
px = +9.0
and
py = −12.
The Linear Momentum of a System of Particles
z
m1
p1
m2
p3
linear momentum to a system of particles. The
m3
p2
x
In this section we will extend the definition of
O
y
ith particle has mass mi , velocity vi , and linear
momentum pi .
We define the linear momentum of a system of n particles as follows:
P  p1  p2  p3  ...  pn  m1v1  m2v2  m3v3  ...  mn vn  Mvcom .
The linear momentum of a system of particles is equal to the product of the
total mass M of the system and the velocity vcom of the center of mass.
dP d
  Mvcom   Macom  Fnet .
dt dt
The linear momentum P of a system of particles can be changed only by a
The time rate of change of P is
net external force Fnet . If the net external force Fnet is zero, P cannot change.
P  p1  p2  p3  ...  pn  Mvcom
dP
 Fnet
dt
(9-9)
Collision and Impulse :
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 collision 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 ti when the ball touches the bat
and ends at t f when the two objects separate.
The ball is acted upon by a force F (t ) during the collision.
The magnitude F (t ) of the force is plotted versus t in fig. a.
The force is nonzero only for the time interval ti  t  t f .
F (t ) 
dp
. Here p is the linear momentum of the ball,
dt
tf
tf
dp  F (t )dt   dp   F (t )dt
ti
ti
(9-10)
tf
tf
tf
 dp   F (t )dt  dp  p
ti
ti
f
 pi  p = change in momentum
ti
tf
 F (t )dt
is known as the impulse J of the collision.
ti
tf
J   F (t )dt The magnitude of J is equal to the area
ti
under the F versus t plot of fig. a  p  J .
In many situations we do not know how the force changes
with time but we know the average magnitude Fave of the
collision force. The magnitude of the impulse is given by
J  Fave t , where t  t f  ti .
p  J
J  Fave t
Geometrically this means that the area under the
F versus t plot (fig. a) is equal to the area under the
Fave versus t plot (fig. b).
(9-11)
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?
I
2.00
3
F 

 1.60  10 N
t 1/ 800
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).
Fave
Series of Collisions
Consider a target that collides with a steady stream of
identical particles of mass m and velocity v along the x-axis.
A number n of the particles collides with the target during a time interval t.
Each particle undergoes a change p in momentum due to the collision with
the target. During each collision a momentum change  p is imparted on the
target. The impulse on the target during the time interval t is J  np.
J np
n

  mv.
t
t
t
Here v is the change in the velocity of each particle along the x-axis due
m
to the collision with the target  Fave  
v.
t
m
Here
is the rate at which mass collides with the target.
t
If the particles stop after the collision, then v  0  v  v.
(9-12)
If the particles bounce backwards, then v  v  v  2v.
The average force on the target is Fave 
Conservation of Linear Momentum :
z
m1
p1
m2
p3
Consider a system of particles for which Fnet  0
dP
 Fnet  0  P  Constant
O
dt
y
x
If no net external force acts on a system of particles, the total linear momentum
p2
m3
P cannot change.
 total linear momentum   total linear momentum 

at some initial time t   at some later time t
f
i 



The conservation of linear momentum is an important principle in physics.
It also provides a powerful rule we can use to solve problems in mechanics
such as collisions.
Note 1: In systems in which Fnet  0 we can always apply conservation of
linear momentum even when the internal forces are very large as in the case
of colliding objects.
Note 2: We will encounter problems (e.g., inelastic collisions) in which the
(9-13)
energy is not conserved but the linear momentum is.
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?
Pi  Pf

m v
j
j
j,i
  m k v k,f
k
mGv G ,i  mbv b ,i  mGv G ,f  m bv b ,f
0  0  mGv G ,f  m bv b ,f
 mGv G ,f  mbv b ,f
v G ,f
(0.1)(150)

 3.3 m/s.
4
Momentum and Kinetic Energy in Collisions
Consider two colliding objects with masses m1 and m2 ,
initial velocities v1i and v2i , and final velocities v1 f and v2 f ,
respectively.
If the system is isolated, i.e., the net force Fnet  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 to determine the results of a collision
without knowing the details. Collisions are divided into two broad classes:
elastic and inelastic.
A collision is elastic if there is no loss of kinetic energy, i.e., K i  K f .
A collision is inelastic if kinetic energy is lost during the collision due to
conversion into other forms of energy. In this case we have K f  K i .
A special case of inelastic collisions are known as completely inelastic.
In these collisions the two colliding objects stick together and they move as a
single body. In these collisions the loss of kinetic energy is maximum.
(9-14)
One - Dimensional Inelastic Collisions :
In these collisions the linear momentum of the colliding
objects is conserved  p1i  p2i  p1 f  p2 f .
m1v1i  m1v2i  m1v1 f  m1v2 f
One - Dimensional Completely Inelastic Collisions :
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 v2i  0.  m1v1i  m1V  m1V 
m1
V
v1i
m1  m2
The velocity of the center of mass in this collision
is vcom
p1i  p2i
m1v1i
P



.
m1  m2 m1  m2 m1  m2
In the picture to the left we show some freeze-frames
of a totally inelastic 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:
M v  m 4v 4  m 8v 8  0  4 v 4  8  6iˆ
 v  12iˆ,
4
K 
 432 J
1
1
4
8
2
2
m 4 v 4   m 8 v 8   (12) 2  (6) 2
2
2
2
2
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
One - Dimensional Elastic Collisions
Consider two colliding objects with masses m1 and m2 ,
initial velocities v1i and v2i , and final velocities v1 f and v2 f ,
respectively.
Both linear momentum and kinetic energy are conserved.
Linear momentum conservation: m1v1i  m1v2i  m1v1 f  m1v2 f
2
2
m1v12i m1v22i m1v1 f m2v2 f
Kinetic energy conservation:



2
2
2
2
We have two equations and two unknowns, v1 f and v1 f .
(eq. 1)
(eq. 2)
If we solve equations 1 and 2 for v1 f and v1 f we get the following solutions:
m1  m2
2m2
v1 f 
v1i 
v2i
m1  m2
m1  m2
v2 f 
2m1
m  m1
v1i  2
v2i
m1  m2
m1  m2
(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?
Special Case of Elastic Collisions - Stationary Target  v 2 i = 0  :
We substitute v2i  0 in the two solutions for v1 f and v1 f :
v1 f 
v2 f
m1  m2
2m2
m  m2
v1i 
v2i  v1 f  1
v1i
m1  m2
m1  m2
m1  m2
2m1
m2  m1
2m1

v1i 
v2i  v2 f 
v1i
m1  m2
m1  m2
m1  m2
Below we examine several special cases for which we know the outcome
of the collision from experience.
v1i v2i = 0
1. Equal masses m1  m2  m
v1 f 
v2 f
m1  m2
mm
v1i 
v1i  0
m1  m2
mm
x
m
m
2m1
2m

v1i 
v1i  v1i
m1  m2
mm
v1f = 0
v2f
x
m
m
The two colliding objects have exchanged velocities.
(9-17)
v1i
2. A massive target
v2i = 0
x
m1
m2
v2f
v1f m
1
x
m2
m2
m1 
m1
m2
1
m1
1
m  m2
m
v1 f  1
v1i  2
v1i  v1i
m1
m1  m2
1
m2
v2 f
m 
2 1 
m2 
 m1 
2m1


v1i 
v1i  2   v1i
m
m1  m2
1
 m2 
1
m2
Body 1 (small mass) bounces back along the incoming path with its speed
practically unchanged.
m1
Body 2 (large mass) moves forward with a very small speed because
1.
m2
(9-18)
2. A massive projectile m1
v1i
m2 
v2i = 0
m1
m2
x
v1f
v2f
m1
m2
x
m2
m1
m2
m  m2
m1
v1 f  1
v1i 
v1i  v1i
m
m1  m2
1 2
m1
1
v2 f 
2m1
2
v1i 
v1i  2v1i
m
m1  m2
1 2
m1
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)
1
Collisions in 2-D
Collisions in Two Dimensions :
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 the xy -plane. Their masses are m1 and m2 .
The linear momentum of the system is conserved: p1i  p2i  p1 f  p2 f .
If the system is elastic the kinetic energy is also conserved: K1i  K 2i  K1 f  K 2 f .
We assume that m2 is stationary and that after the collision particle 1 and
particle 2 move at angles 1 and  2 with the initial direction of motion of m1.
In this case the conservation of momentum and kinetic energy take the form:
x  axis: m1v1i  m1v1 f cos 1  m2 v2 f cos  2 (eq. 1)
y  axis: 0  m1v1 f sin 1  m2v2 f sin  2 (eq. 2)
1
1
1
m1v12i  m1v22 f  m2v22 f (eq. 3) We have three equations and seven variables:
2
2
2
Two masses: m1 , m2 ; three speeds: v1i , v1 f , v2 f ; and two angles: 1 ,  2 . If we know
the values of four of these parameters we can calculate the remaining three. (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 conserved
in 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 )
dp
 p   Fdt  Area under the curve
dt
10  (6  2)

 40
2
p  m v f v i 
F
v f 
p
m
v i 
40
 4  8 m/s
10
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?
dp
F
 p   Fdt  Area under the curve
dt
5000  4 103

 10
2
p 10
m
v 
 2
m
5
s
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 )
x cm 
m1x 1  m 2 x 2  m 3x 3 1.0  0.0  1.0  2.0  2.0 1.0

 1.0 m
m1  m 2  m 3
4.0
y cm 
m1 y 1  m 2 y 2  m 3 y 3 1.0  0.0  2.0 1.0  1.0  0

 0.5 m
m1  m 2  m 3
4.0
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?
M v  m1v 1  m 2v 2
(1500  3000) v  1500  90 iˆ  3000  60 jˆ
km
 v   30  40   50
 13.9 m/s
hour
2
2
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)
0  m1v 1  m 2v 2
m
0  2 v  0.003  400  v  0.6 .
s
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, p  M v  m v f v i 
 (1)  2  (3)  5
The average force on the floor by the ball F 
p 5
  50 N (down)
t .1
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.
Consrevation of momentum  m bv 1  m bv f  m B V
m b v 1 v f
 V 
mB
  0.02  (10  5)  1.0 m/s
0.1
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 2.0kg 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 2.0kg 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?
Consrevation of momentum  mv  ( m  M )V
2 4
m
 2.67
(2  1)
s
1
1
Conservation of K.E. after collision 
( m  M )V 2  k x 2
2
2
(m  M )
3
 x V
 2.67
 0.33 m
k
200
 V 