Note - Chapter 08

8/11/2019 Note - Chapter 08

1/15

1

Chapter 8

Linear Momentum, Impulse, and Collisions

8.1 Linear Momentum and Impulse

The linear momentum pr

of a particle of mass mmoving

with velocity vr

is defined as:

r

p " mr

v

Note that pr

is a vector that points in the same direction as

the velocity vector vr

.

One can show that Newtons second law of motion,

! = amF r

r

, may be written in terms of linear momentum.


2/15

2

dt

pd

dt

vmd

dt

vdmamF

rrr

r

r

====!

That is,

! =dt

pdF

r

r

The net force!Fr

acting on a particle is equal to the timerate of change of its linear momentum.

Law of Conservation of Linear Momentum

When the net external forceacting on a system iszero,

then the total linear momentum of the system remainsconstant (or conserved). This is so because if ! = 0F

r

,

then 0=dt

pdr

! pr

= constant.

The Impulse - Momentum Theorem

The impulser

J of the net external force !Fr

acting on a

particle during a time interval t! is equal to the change is

linear momentum pr

! of the particle during that interval.


3/15

3

r

J " #r

p =r

F$( )t0

tf

% dt=r

Fave #t

that is,

r

J =

"r

p

orr

Fave"t

#

$%

&%

Impulse equals the areaunder the force versus time

curve.


4/15

4

8.2 Conservation of Linear Momentum

When the net external forceacting on a system iszero,

then the total linear momentum of the system remains

constant (or conserved). This is so because if ! = 0Fr

,

then 0=dtpdr

! pr

= constant.

8.3 and 8.4 Momentum Conservation and Collisions

A collisionis an event during which two objects come

close to each other and interact by means of forces.


5/15

5

A. Elastic Collisions

An elastic collision is one in which the total kinetic

energy of the system is the same before and after the

collision. That is, the total kinetic energy is conserved.

Consider the following elasticcollision in one dimension:

(i) conservation of total linear momentum (if !Fr

=0)means that

BfBAfABiBAiA vmvmvmvm rrrr

+=+

which in this case yields:

)()()()( BfBAfABiBAiA vmvmvmvm +!=!+

(ii) conservation of total kinetic energy gives:


6/15

6

2222

2

1

2

1

2

1

2

1

BfBAfABiBAiA vmvmvmvm +=+

which may be re-written for 1 dimensional collisions(using conservation of linear momentum) as

( )BiAiBfAf vvvv rrrr

!!=!

B. Inelastic CollisionsAn inelastic collisionis one for which the total kinetic

energy of the system is notconserved.

C. Completely (Perfectly) Inelastic Collisions

A completely inelastic collisionis one for which the

colliding bodies stick together and move as a unit after the

collision.


7/15

7

Collisions in Two Dimensions

If 0=! extFr

, then the two components of the total linear

momentum of the system are conserved.

!"

!#$

=

=

%constP

constP

total

y

total

x

Often times, it is easier to solve for the components of the

final velocities than to solve directly for the magnitudes

of the final velocities.

8.5 The Center of Mass

The center of mass (CM)of a system of particles moves

as if all the mass of the system were concentrated at thatpoint.


8/15

8

A.System consisting of discrete particles

The x-coordinatecm

x of the CM of a system of N discrete

particles is given by:

!

!

=

=

=

+++

+++

=N

i

i

N

i

ii

cm

m

xm

mmm

xmxmxmx

1

1

321

332211

...

...

wherexiis the x-coordinate of the ith

particle. Also, the y-

coordinate of the center of mass of the system of discreteparticles is equal to

!

!

=

=

=

+++

+++

=N

ii

N

iii

cm

m

ym

mmm

ymymymy

1

1

321

332211

...

...

The position vector cmr

r

of the center of mass hascoordinated (

cmcm yx , and equals

jyixr cmcmcm +=r

or

!

!

=

=

=N

i

i

N

i

ii

cm

m

rm

r

1

1

r

r


9/15

9

B.Center of Mass of an Extended Object

M

dmx

xdm

dmx

x cmcm!

!

!=

"=

M

dmyy

dm

dmyy cmcm

!

!

!="=

whereMis the total mass.


10/15

10

C.Motion of a System of Particles

A. Position Vectorcmr

r

of a system of N discrete particles:

M

rm

m

rm

r

N

i

ii

N

i

i

N

i

ii

cm

!

!

!=

=

=

==1

1

1

rr

r

B. Velocity cmV

r

of the CM of a system of N particles:

i

N

i

i

i

N

i

i

cm

cm vm

Mdt

rdm

Mdt

rdV

r

rr

r

!!==

===

11

11

i

N

i

icm vm

M

V r

r

!=

=

1

1

C. Total Linear Momentumtotal

Pr

of system of N

particles:

cmtotal VMP

rr

=

!!==

="=N

i

ii

N

i

iitotal vmvm

M

MP

11

1 rrr

!=

=

N

iitotal pP

1

r

r


11/15

11

D. Linear Accelerationcma

r

of system of N particles:

dt

Vd

a

cm

cm

r

r

=

!=

=

N

i

i

icmdt

vdm

Ma

1

1 r

r

!==

N

i

iicm am

M

a

1

1 rr

E. Net External Force on system ofNparticles:

Re-arrange last formula to obtain:

!! ! ="==

cmext

N

i

iiext aMFamF r

r

r

r

1

Note that if 0=! extFr

, then 00 =!=dt

Vda

cm

cm

r

r

"

r

Vcm= const "

r

Ptotal

=Mr

Vcm= const

That is, the net linear momentum of a system of particles

remains constant (is conserved) if no net external force

acts on the system.

If an isolated system consisting of two or more particles is

initially at rest, then its center of mass CM remains at rest

unless acted upon by a net external force.


12/15

12


13/15

13

8.6 Rocket Propulsion

The operation of a rocket depends on the law of

conservation of linear momentum applied to a system,

where the system is the rocket and the ejected fuel.

As a rocket moves in free space (a vacuum) its linear

momentum changes when some of its mass is released in

the form of ejected gases. The rocket is thus accelerated

as a result of the push or thrust from the exhaust

gases.

The process represents the inverseof a completely

inelastic collision; that is, linear momentum is conserved,

but the kinetic energy of the system is increasedat the

expense of the energy stored in the fuel of the rocket. Let

m= mass of the rocket plus fuel(-dm)= mass of ejected fuel

vex= exhaust speed of the ejected fuel relative to the

rocket. Thus vfuel= v - vex= speed of ejected fuel.

v= speed of rocket before ejecting the fuel


14/15

14

After the rocket ejects fuel of mass (dm), the speed of

the rocket increases to v + dv.

Applying the law of conservation of linear momentum tothis system,

r

Pbeforetotal

=

r

Paftertotal

m v = m+ dm( ) v+ dv( )+ "dm( ) v " vex( )

0 = mdv+ dmdv+ vexdm

neglect term dmdvbecause it is a product of two small

quantities and thus it is much smaller than the other terms,

dv = "vex

dm

m

integrate both sides of the equations (note the change to

dummy variables) so that

d "vvo

vf

# =$vexd "m

"

mmo

mf

#

vf" vo =vex ln mo

mf

#

$%

&

'(


15/15

15

Comments:

1.The increase in the speed of the rocket is

proportional to the exhaust speed vex.

2.

Thrust on the rocket is ma = mdvdt

= vexdm

dt. Note

that the thrust increases as the rate of change of mass

increases (i.e., as the burn rate increases).

3.We have assumed throughout this analysis that the

rocket is in gravity-free outer space. However,

gravity must be taken into account when a rocket islaunched from the surface of a planet. (see problem

8.112).

Documents

Note - Chapter 08