chapter 9 linear momentum and collisionsnsmn1.uh.edu/rbellwied/classes/spring2013/ch9-notes.pdf ·...

Copyright © 2010 Pearson Education, Inc.

Chapter 9

Linear Momentum andCollisions


Units of Chapter 9

• Linear Momentum

• Momentum and Newton’s Second Law

• Impulse

• Conservation of Linear Momentum

• Inelastic Collisions

• Elastic Collisions


9-1 Linear Momentum

Momentum is a vector; its direction is thesame as the direction of the velocity.


9-1 Linear Momentum

Change in momentum:

(a) mv

(b) 2mv


Momentum in collisions

Consider two interacting bodies with m2 > m1:

If we know the net force on each body then tm

t !=!=! netF

av

The velocity change for each mass will be different if themasses are different.

F21 F12m1 m2


tm !=!2111Fv

Rewrite the previous result for each body as:

ttm !"=!=!211222FFv

Combine the two results:

( ) ( )ifif mm

mm

222111

2211

vvvv

vv

!!=!

"!="


21

2211

pp

vv

!"=!

!"=! mm

The change in momentum of the two bodies is “equal andopposite”. Total momentum is conserved during theinteraction; the momentum lost by one body is gained bythe other.


9-2 Momentum and Newton’s Second LawNewton’s second law, as we wrote it before:

is only valid for objects that have constantmass. Here is a more general form, alsouseful when the mass is changing:


9-3 Impulse

Impulse is a vector, in the same directionas the average force.


9-3 Impulse

We can rewrite

as

So we see that

The impulse is equal to the change inmomentum.


9-3 Impulse

Therefore, the samechange in momentummay be produced by alarge force acting for ashort time, or by asmaller force acting for alonger time.


9-4 Conservation of Linear MomentumThe net force acting on an object is the rateof change of its momentum:

If the net force is zero, the momentum does notchange:


Example: A force of 30 N is applied for 5 sec to each of twobodies of different masses.

30 N

m1 or m2

Take m1 < m2

(a) Which mass has the greater momentum change?

t!=! FpSince the same force is applied toeach mass for the same interval, Δpis the same for both masses.


(b) Which mass has the greatest velocity change?

Example continued:

m

pv

!=!

Since both masses have the same Δp,the smaller mass (mass 1) will have thelarger change in velocity.

(c) Which mass has the greatest acceleration?

t!

!=v

aSince a∝Δv the mass with the greatervelocity change will have the greatestacceleration (mass 1).


Example: An object of mass 3.0 kg is allowed to fall fromrest under the force of gravity for 3.4 seconds. What is thechange in momentum? Ignore air resistance.

Want Δp = mΔv.

(downward) m/s kg 100

m/sec 3.33

=!=!

"=!"=!

!=!

vp

v

av

m

tg

t


Example: What average force is necessary to bring a 50.0-kg sled from rest to 3.0 m/s in a period of 20.0 seconds?Assume frictionless ice.

( )( )N 5.7

s 0.20

m/s 0.3kg 0.50av

av

av

==

!

!=

!

!=

!=!

F

t

m

t

t

vpF

Fp

The force will bein the directionof motion.


9-4 Conservation of Linear MomentumInternal Versus External Forces:

Internal forces act between objects within thesystem.

As with all forces, they occur in action-reactionpairs. As all pairs act between objects in thesystem, the internal forces always sum to zero:

Therefore, the net force acting on a system isthe sum of the external forces acting on it.


9-4 Conservation of Linear Momentum

Furthermore, internal forces cannot change themomentum of a system.

However, the momenta of components of thesystem may change.


9-4 Conservation of Linear Momentum

An example of internal forces movingcomponents of a system:


Example: A rifle has a mass of 4.5 kg and it fires a bullet of10.0 grams at a muzzle speed of 820 m/s. What is the recoilspeed of the rifle as the bullet leaves the barrel?

As long as the rifle is horizontal, there will be no netexternal force acting on the rifle-bullet system andmomentum will be conserved.

m/s 1.82m/s 820kg 4.5

kg 01.0

0

!=""#

$%%&

'!=!=

+=

=

b

r

br

rrbb

fi

vm

mv

vmvm

pp


9-5 Inelastic Collisions

Collision: two objects striking one another

Time of collision is short enough that externalforces may be ignored

Inelastic collision: momentum is conserved butkinetic energy is not

Completely inelastic collision: objects sticktogether afterwards



A completely inelastic collision:



Solving for the final momentum in terms of theinitial momenta and masses:



Ballistic pendulum: the height h can be foundusing conservation of mechanical energy afterthe object is embedded in the block.



For collisions in two dimensions, conservationof momentum is applied separately along eachaxis:


9-6 Elastic CollisionsIn elastic collisions, both kinetic energy andmomentum are conserved.

One-dimensional elastic collision:


9-6 Elastic Collisions

We have two equations (conservation ofmomentum and conservation of kinetic energy)and two unknowns (the final speeds). Solving forthe final speeds:


9-6 Elastic CollisionsTwo-dimensional collisions can only be solved ifsome of the final information is known, such asthe final velocity of one object:


Example: In a railroad freight yard, an empty freight car ofmass m rolls along a straight level track at 1.0 m/s andcollides with an initially stationary, fully loaded, boxcar ofmass 4.0m. The two cars couple together upon collision.

(a) What is the speed of the two cars after the collision?

( )

m/s 2.0

0

1

21

1

212111

2121

=!!"

#$$%

&

+=

+=+=+

+=+

=

vmm

mv

vmmvmvmvm

pppp

pp

ffii

fi


(b) Suppose instead that both cars are at rest after thecollision. With what speed was the loaded boxcar movingbefore the collision if the empty one had v1i = 1.0 m/s.

m/s 25.0

00

1

2

1

2

2211

2121

!=""#

$%%&

'!=

+=+

+=+

=

ii

ii

ffii

fi

vm

mv

vmvm

pppp

pp

Example continued:


Example: A projectile of 1.0 kg mass approaches a stationarybody of 5.0 kg mass at 10.0 m/s and, after colliding, reboundsin the reverse direction along the same line with a speed of 5.0m/s. What is the speed of the 5.0 kg mass after the collision?

( )

( )( ) m/s0.3m/s 5.0m/s 10kg 0.5

kg 0.1

0

11

2

12

221111

2121

=!!=

!=

+=+

+=+

=

fif

ffi

ffii

fi

vvm

mv

vmvmvm

pppp

pp


Example (text problem 7.58): Body A of mass M has anoriginal velocity of 6.0 m/s in the +x-direction toward astationary body (B) of the same mass. After the collision,body A has vx = +1.0 m/s and vy = +2.0 m/s. What is themagnitude of body B’s velocity after the collision?

A

B

vAi

Initial

B

AFinal


fxfxix

fxfxixix

fxix

vmvmvm

pppp

pp

221111

2121

0 +=+

+=+

=

fyfy

fyfyiyiy

fyiy

vmvm

pppp

pp

2211

2121

00 +=+

+=+

=

x momentum:

Solve for v2fx: Solve for v2fy:

m/s 00.5

11

2

1111

2

=

!=

!=

fxix

fxix

fx

vv

m

vmvmv

y momentum:

m/s 00.2

1

2

11

2

!=

!=!

= fy

fy

fy vm

vmv

m/s 40.522

22

2=+= fxfyf vvvThe mag. of v2 is

Example continued:


9-7 Center of Mass

The center of mass of a system is the point wherethe system can be balanced in a uniformgravitational field.


9-7 Center of Mass

For two objects:

The center of mass is closer to the moremassive object.


9-7 Center of Mass

The center of mass need not be within the object:


9-7 Center of MassMotion of the center of mass:


Example: Particle A is at the origin and has a mass of 30.0grams. Particle B has a mass of10.0 grams. Where must particle B be located so that thecenter of mass (marked with a red x) is located at the point(2.0 cm, 5.0 cm)?

x

y

A

x

ba

bb

ba

bbaa

cm

mm

xm

mm

xmxmx

+=

+

+=

ba

bb

ba

bbaacm

mm

ym

mm

ymymy

+=

+

+=


( )

cm 8

cm 2g30g 10

g 10

=

=+

=

b

b

cm

x

xx

( )

cm 20

cm 5g 10g 30

g 10

=

=+

=

b

bcm

y

yy

Example continued:


Example: The positions of three particles are (4.0 m, 0.0 m),(2.0 m, 4.0 m), and (−1.0 m, −2.0 m). The masses are 4.0 kg,6.0 kg, and 3.0 kg respectively. What is the location of thecenter of mass?

x

y

3

2

1


( )( ) ( )( ) ( )( )( )

m 92.1

kg 364

m 1kg 3m 2kg 6m 4kg 4

321

332211cm

=

++

!++=

++

++=

mmm

xmxmxmx

( )( ) ( )( ) ( )( )( )

m 38.1

kg 364

m 2kg 3m 4kg 6m 0kg 4

321

332211cm

=

++

!++=

++

++=

mmm

ymymymy

Example continued:


9-7 Center of Mass

The total mass multiplied by the acceleration ofthe center of mass is equal to the net externalforce:

The center of massaccelerates just asthough it were a pointparticle of mass Macted on by


Example: Body A has a mass of 3.0 kg and vx = +14.0 m/s.Body B has a mass of 4.0 kg and has vy = −7.0 m/s. What isthe velocity of the center of mass of the two bodies?

Consider a body made up of many different masseseach with a mass mi.

The position of each mass is ri and the displacement ofeach mass is Δri = viΔt.


For the center of mass:!

! "

="="

i

i

i

ii

m

m

t

r

vrcmcm

Solving for the velocity of the center of mass:

!

!

!

!=

"

"

=

i

i

i

ii

i

i

i

i

i

m

m

m

tm v

r

vcm

Or in component form:

!

!=

i

i

i

xii

m

vm,

xcm,v

!

!=

i

i

i

yii

m

vm,

ycm,v

Example continued:


Applying the previous formulas to the example,

( )( ) ( )( )( )

m/s 6kg 43

m/s 0kg 4m/s 14kg 3

,,

,

=+

+=

+

+=

ba

xbbxaa

xcm

mm

vmvmv

( )( ) ( )( )( )

m/s 4kg 43

m/s 7kg 4m/s 0kg 3

,,

,

!=+

!+=

+

+=

ba

ybbyaa

ycmmm

vmvmv

Example continued:


9-8 Systems with Changing Mass:Rocket Propulsion

If a mass of fuel Δm is ejected from a rocketwith speed v, the change in momentum of therocket is:

The force, or thrust, is

chapter 9 linear momentum and collisionsnsmn1.uh.edu/rbellwied/classes/spring2013/ch9-notes.pdf ·...

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