11. detailed slides to cover impulse and momentum635442187916733417

10/27/09

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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 Newtons 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 Newtons 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

Lets 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 + m2v2i = m1v1 f + m2v2 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(v1i2! v1 f

2 ) = m2 (v2 f2! 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 = !(v1 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 = !(v1 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

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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 = m1 + m2( )Vf

K +U conserved :

1

2m

1+ m

2( )Vf2= m

1+ m

2( )gh!Vf = 2gh

v1i = 1+

m2

m1

!

"#$

%&2gh

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Fraction of kinetic energy lost is ..

If m2m1: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 = !(v1 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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2008 by W.H. Freeman and Company

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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

! lnM

0

M0" Rt

#

$%&

'(" gt

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Variable mass

! Newtons 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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11. detailed slides to cover impulse and momentum635442187916733417

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