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

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