Chapter 8 Conservation of Linear Momentum · PDF fileChapter 8 Conservation of Linear Momentum...

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Chapter 8

Conservation of Linear Momentum

Physics 201

October 22, 2009

Conservation of Linear Momentum

•! Definition of linear momentum,

! p

! p = m

! v

Linear momentum is a vector.

Units of linear momentum are kg-m/s.

Can write Newton’s second law in terms of momentum:

d! p

dt=

d(m! v )

dt= m

d! v

dt= m! a

!d! p

dt=! F net

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Momentum of a system of particles

•! The total momentum of a system of

particles is the vector sum of the momenta of

the individual particles:

From Newton’s second law, we obtain

Psys

! "!!= m

i

"vi

i

! ="pi

i

!

!Fext

i

! =!Fnetext =

d!Psys

dti

!

Conservation of Momentum

•! Law of conservation of momentum:

–! If the sum of the external forces on a system is

zero, the total momentum of the system does not

change.

If then

Momentum is always conserved (even if forces are nonconservative).

!Psys = mi

!vi

i

! = M!vCM = const

" !""""

!Fext

i

! = 0

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Collisions

“before” m1 m2

“after” m1 m2

momentum before collision = momentum after collision

Always -

But only if

! F

external= 0

Explosion - I

“before” M

“after” m1 m2

v1 v2

Example: m1 = M/3 m2 = 2M/3

After explosion, which block has larger momentum? (left, right, same)

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“before” M

“after” m1 m2

v1 v2



Each has the same momentum

Explosion - I

“before” M

“after” m1 m2

v1 v2




Which block has larger velocity?

Explosion - I

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“before” M

“after” m1 m2

v1 v2





mv is the same for each block, so smaller mass has larger velocity

Explosion - I

“before” M

“after” m1 m2

v1 v2






Is kinetic energy conserved?

Explosion - I

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Explosion - I

“before” M

“after” m1 m2

v1 v2






Is kinetic energy conserved? NO! K was 0 before, it is greater after the explosion.

Explosion - I

“before” M

“after” m1 m2

v1 v2






Is kinetic energy conserved? (green=yes, red=no) NO!

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This is like a microscopic explosion ….

Momentum and Impulse

!Fave!t " I definition of impulse

!F = m

!a = m

d!v

dt=d!p

dt# !!p =!F!t

!p ! m

!v

!! For single object….

"! If F = 0, then momentum conserved (p = 0)

!psys

=!pi

i

!

Internal forces " forces between objects in system

External forces " any other forces

#!psys

=!Fext#t

Thus, if !Fext

= 0, then #!psys

= 0

i.e. total momentum is conserved!

•!For “system” of objects …

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Momentum and Impulse


!F = m

!a = m

d!v

dt=d!p

dt# !!p =!F!t

!p ! m

!v

!! For single object….

"! If F = 0, then momentum conserved (p = 0)

!psys

=!pi

i

!

Internal forces " forces between objects in system

External forces " any other forces

#!psys

=!Fext#t

Thus, if !Fext

= 0, then #!psys

= 0

i.e. total momentum is conserved!

•!For “system” of objects …

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!F = m

!a = m

d!v

dt=d!p

dt# !!p =!F!t

Let’s estimate the average force during the collision

Club speed: 50 m/s

Assume that impulse is given after 5 cm

--> whiteboard

!Fave =

I

!t=1

!t

!Fdt

ti

t f

"

Some Terminology

•! Elastic Collisions:

collisions that conserve kinetic energy

•! Inelastic Collisions:

collisions that do not conserve kinetic energy

*! Completely Inelastic Collisons:

objects stick together

n.b. ALL CONSERVE MOMENTUM!!

If external forces = 0

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Elastic Collision in 1-Dimension

m1v1i + m2

v2i = m1

v1 f + m2

v2 f

1

2m1v1i

2+1

2m2v2i

2=1

2m1v1 f

2+1

2m2v2 f

2

Energy conserved (for elastic

collision only)

Linear momentum is conserved

Initial Final

Elastic Collision Conservation of Momentum

m1v1i + m2v2i = m1v1 f + m2v2 f

m1(v1i ! v1 f ) = m2 (v2 f ! v2i )

Conservation of Kinetic Energy

1

2m1v1i

2+

1

2m2v2i

2=

1

2m1v1 f

2+

1

2m2v2 f

2

m1(v1i

2! v1 f

2 ) = m2 (v2 f

2! v2i

2 )

m1(v1i ! v1 f )(v1i + v1 f ) = m2 (v2 f ! v2i )(v2 f + v2i )

Combining the above two equations

v1i + v1 f = v2i + v2 f

v1i ! v2i = !(v1 f ! v2 f )

Magnitude of relative velocity is conserved.

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Is this an elastic collision?

v1i ! v2i = !(v

1 f ! v2 f )For elastic collision only:

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What is the speed of the golf ball, in

case of an elastic collision

Club speed: 50 m/s

Mass of clubhead: 0.5kg

Mass of golfball: 0.05kg

Two unknowns:

speed of club and

speed of golfball after impact

Problem solving strategy:

-! Momentum conservation

-! Energy conservation (or

use the derived equation

for relative velocities)

--> whiteboard

Is this an elastic collision?

v1i ! v2i = !(v

1 f ! v2 f )Yes, the relative speeds

are approximately the same

before and after collision

For elastic collision only:

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v1 f =

m1! m

2

m1+ m

2

v1i

v2 f =

2m1

m1+ m

2

v1i

Result:

Special cases: 1)! Golf shot: m1>>m2

Club speed almost unchanged

Ball speed almost 2 x club speed

2) Neutron scatters on heavy nucleus: m1<<m2

neutron scatters back with almost same speed speed of nucleus almost unchanged

Completely inelastic collision

•! Two objects stick together and move with the center

of mass:

•! If pAi=0:

!PAi +

!PBi =

!PAf +

!PBf =

!PCM

!PBi =

!PCM

mBv!

Bi = mA + mB( )v!

f

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!! Two stage process:

1. m collides with M, inelastically. Both M and m then move together with a velocity V

f

(before having risen significantly).

2. Both (m1 + m2) rise a height h, conserving energy E. (no non-conservative forces acting after collision)

Ballistic

Pendulum

What is the initial

velocity vli of the

projectile? Known quantities:

m1, m2, h

•! Stage 1: Momentum is conserved

Energy is not conserved

in x-direction:

!! Stage 2 (after the collision): Energy is conserved

Substituting for V gives:

Ballistic

Pendulum

m1v1i = m

1+ m

2( )Vf

K +U conserved :

1

2m

1+ m

2( )Vf

2= m

1+ m

2( )gh!Vf = 2gh

v1i = 1+

m2

m1

!

"#$

%&2gh

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Fraction of kinetic energy lost is …………..

If m2<<m1: not much energy is lost

Of m2>>m1:almost all energy is lost

Ballistic

Pendulum

Wthermal =U ! Ki

= m1+ m

2( )gh !1

2m1v1i

2

= m1+ m

2( )gh !1

2m1

m1+ m

2

m1

"

#$%

&'

2

2gh

= m1+ m

2( )ghm1

m2

•! How much energy is dissipated?

Wthermal

Ki

=m1

m1+ m

2

Coefficient of restitution e=1/2

Inelastic collision

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Coefficient of restitution

v1i ! v2i = !(v

1 f ! v2 f )

Perfectly elastic collision:

The coefficient of restitution is a

measure of the “inelasticity:

Elastic collision: e=1

Perfectly inelastic collision: e=0

e =v1 f ! v2 f

v1i ! v2i

Collisions or Explosions in Two Dimensions

y

x

before after

•! Ptotal,x and Ptotal,y independently conserved

*!Ptotal,x,before = Ptotal,x,after

*!Ptotal,y,before = Ptotal,y,after

Ptotal ,before

! "!!!!!!!= Ptotal ,after

! "!!!!!!

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Explosions

“before” M

A

Which of these is possible?

A

B

both

B

“after”

Explosions

“before” M

A


A (p appears conserved)

B (p not conserved in y direction)

both

B

“after”

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Explosions

“before” M

A B


A

B

neither

“after”

Explosions

“before” M

A B


A (p not conserved in y direction)

B

neither

“after”

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(Inelastic) Car – truck collision •! Knowns: m1, m2, v1, v2,

Mcm=m1+m2

•! Unknowns: final velocity vector

(in x and y)

•! Equation(s): Momentum

conservation (in x and y)

•! Strategy: write out conservation

of momentum equation.

( 2 unknowns, 2 equations, ! piece of cake)

Elastic collision in 2 dimensions

•! Assume we know all initial conditions, mass and

momentum.

•! 4 Unknown quantities:

Equations:

Momentum conservation: 2 (x and y)

Energy conservation: 1

Need one more piece of information to solve the

problem: often a measurement.

What is unspecified above is the impact parameter

(and the precise nature of the interaction)

v1 f

! "!,v2 f

! "!!

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•! Assuming –! Collision is elastic (KE is conserved)

–! No spin is imparted –! Balls have the same mass

–! One ball starts out at rest

Shooting Pool...

Shooting Pool •! Elastic collision means conservation of kinetic energy

if m1 = m2:

•! Conservation of momentum:

if

1

2m1v1i

2=p1i

2

2m1

=p1 f

2

2m1

+p2 f

2

2m2

!P1i +!P2i =!P1 f +!P2 f

!P2i= 0

!P1i =!P1 f +!P2 f

p1i

2= p

1 f

2+ p

2 f

2

P1 and p2

Form a right

angle!

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•! Tip: If you shoot a ball spotted on the

“dot”, you have a good chance of

scratching !

Shooting Pool...

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v

!

cm =m1v1

"!+ m

2v2

"!"

m1+ m

2

Subtract vcm from

all velocities

Collisions in the CM frame

The transformation to the cm frame is not

necessary, but it is often convenient to switch to

the CM frame

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Rocket equation

•! The mass is changing

•! Thrust is generated by impulse of exhaust of mass

with velocity v: vdm

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The Saturn V

The Saturn V rocket:

•! 111 m tall

•! 10m diameter

•! 3000 tons at start

•! Thrust: 34 MN

Rocket equation

•! Mass change

•! Thrust

•! Weight: F = M g

•! Rocket equation:

•! Integration yields:

dM

dt= R = const

M (t) = M0! Rt

dM

M

dM

dt!u

ex=d(M !u

ex)

dt=dP

dt= "F

thrust

Mdv

dt= R !uex " Mg

dv

dt=

R !uex

M0" Rt

" Mg

v = uex! ln

M0

M0" Rt

#

$%&

'(" gt

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Variable mass

•! Newton’s second law for continuously variable

mass:

•! Where

is the velocity of impacting material relative to

object with mass M at a given time.

!Fnet ,ext

+dM

dt

!vrel= M

dv

!

dt

v

!

rel = u

!! v

!

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