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1

Chapter 8

MomentumImpulse

and Collisions

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Analysis of motion: 2 key ideas

Newton’s laws of motion

Conservation of Energy

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3

Newton’s Laws

2nd Law:

1st Law: An object at rest or traveling in uniform motion will remain at rest or traveling in uniform motion unless and until an external force is applied

amFnetrr

=

zznetyynetxxnet maFmaFmaF === ,,, ,,

3rd Law: For every Action, there is an equal but opposite Reaction

4

Conservation of Mechanical EnergyThe mechanical energy of a system is the sum of its potential energy U and the kinetic energy K of the objects within it

UKE +=mech

In an isolated system where only conservative forces cause energy change, the kinetic energy and potential energy can change, but their sum, the mechanical energy of the system cannot change

ffii UKUK +=+

3

5

Conservation of EnergyThe total energy of a system can change only by amounts of energy that are transferred to or from that system

internalthermal EEUKE Δ+Δ+Δ+Δ=Δ

The total energy of an isolated system cannot change

0internalthermal =Δ+Δ+Δ+Δ EEUK

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Conservation of Mechanical Energy (cont.)

Examples:

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7

Problems with two (or more) objects

Three (or more) body problem is one of the most difficult in physics

Example

8

Part 1

Momentum and Impulse

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9

The linear momentum of a particle is a vector defined as

m is the mass of the particlev is the velocity of the particle

Newton’s 2nd Law is now

Linear Momentum

vmp =

dtpd

dtvmd

dtvdmamF

rrr ==== )( NOTE: Newton’s original

statement was in terms of linear momentum

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Impulse

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11

Impulse (cont)

tFdtFppJ ave

t

tif

f

i

Δ==−= ∫

average force

tpp

F ifave

Δ−

=

example: What the use of an air bag in a car?

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A car has abrupt stop by driving into a stone embankment: mass of the car is 1800 kg, mass of a driver is 60 kg, speed 27.8 m/s, (100 km/h), impact occurred over time interval 10−2 s.Estimate average size of force exerted on the driver:

F = 60.0 kg * 27.8 m/s/0.01s = 1.67*105 N

comparing to weight of the driver W = 60 kg * 9.8 m/s2

Stopping force is approximately 280 times weight

example

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13

Checkpoint

A student's life was saved in an automobile accident because an airbag expanded in front of his head. If the car had not been equipped with an airbag, the windshield would have stopped the motion of his head in a much shorter time. Compared to the windshield, the airbag:A) causes a much smaller change in momentumB) exerts a much smaller impulseC) causes a much smaller change in kinetic energyD) exerts a much smaller forceE) does much more work

tpp

F ifave

Δ−

=

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Part 2

Conservation of Momentum

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15

Conservation of momentum

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System of N particles

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17

Momentum conservation for a system ofparticles

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The law of conservation of linear momentum

If no net external force acts on a system of particles, the total momentum of the system can not change

for closed, isolated system

.constP = fi PP =

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

fi ttimelatersomeatmomentumlineartotal

ttimeinitialsomeatmomentumlineartotal

0==dtPdF net

momentum of a single particle is not necessarilyconserved, only momentum of a closed system

10

19

Checkpoint

A man, standing at rest on a horizontal frictionless floor, might get himself moving by: A) walking B) rolling C) exhaling vertically D) crawling slowly E) throwing a shoe horizontally

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Two masses on frictionless surface

example

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21

Part 3

System of particles

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Motion of system as a whole

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23

The center-of-mass

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Coordinates of the center-of-mass

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25

The center of mass (example)

dmm

mxCM21

2

+=

21

2211

mmxmxmxCM +

+=

∑=

=++++=

n

iiiCM rm

Mmmrmrmr

121

2211 1...

...

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System of three point particles in 2-D

example

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27

CM for Continuous Mass Distributions

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Mass distributed over a volume

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Mass distributed over a surface

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Mass distributed over a curve

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Constant linear density

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CM is in the middle point

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33

Constant surface density

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CM at center - again!

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CM of half-disk

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CM is Independent of Origin

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Importance of CM

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Example

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39

Newton’s Second Law for a system of particlesNet force on a system of particles

...21 ++= FFFnet

Definition for the center of mass

...2211 ++= rmrmrM CM

Differentiating twice with respect to time

...

...

2211

2211

++=

++=

amamaM

vmvmvM

CM

CM

CMnet aMF =

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Practical applications:

If the net force on a system of particles is zero

0...21 =++= FFFnet

then

and …

constva CMCM == ,0

CMnet aMF =

constrv CMCM == ,0

21

41

What is the change in velocity of the car if the man runs so that his speed relative to the car is vrel?

VMvmPi +== 0

relvVv −=

MVvVm rel +−= )(0 mMmvV rel

+=

example

42

The cat Tom, of mass 5 kg, and the mouse Jerry (see the cartoon "Tom and Jerry") are in 1 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 2.0 m apart and symmetrically located with respect to the canoe's center. The canoe moves 1.5 m relative to the shore during the exchange. What is Jerry's mass?

example

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43

Give a man a fish and he will eat for a day. Teach him how to fish, and he will sit in a boat and drink beer all day.

exampleFishing …

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Part 4

Collisions

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45

CollisionsA collision is an isolated event in which two or more bodies exert forces on each others for a relatively short time

force between bodies can be:– conservative (total energy conserved) =>

elastic collision

– dissipative (energy lost to thermal motion etc.) => inelastic collision

Either way, total momentum is conserved (as long as no external force is present)

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Collisions

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Types of collisions (again)

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Perfectly inelastic collisions

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One of the bodies is at rest

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Particular cases

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One-dimensional collisions

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Change in total kinetic energy

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53

Inelastic collision in one dimension

ffii vmvmvmvm 22112211 +=+

Because the motion is one dimension we can drop arrows for vector

Examples: traffic accidents

fi PP =

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Completely Inelastic collision in one dimension

Vmmvm i )( 2111 +=

After the collision two bodies move together

fi PP =

ivmm

mV 121

1

+=

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55

Example: ballistic pendulum

vMm

mV+

=

ghMmVMm )()(21 2 +=+

ghm

Mmv 2+=

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Two cars slide on an icy road as they attempt to stop at a traffic light. The mass of A is 100 kg, and the mass of B is 1400 kg. The coefficient of kinetic friction between the locked wheels of either car and the road is 0.13. Car A succeeds in stopping at the light, but car B cannot stop and rear-end car A. After the collision, A stops 8.2 m ahead of its position at impact, and B 6.1 m ahead. Both drivers had their brakes locked through the incident. From the distance each car moved after the collision, find the speed of cars A and B immediately after the impact. Use conservation of linear momentum to find speed at which car B struck car A

example

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57

There are two events in this problem:

1. the collision of moving car B with stationary car A

2. stopping due to frictional forces on cars

example

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example

find may we equation first the using then and find may we equations two last the from

:stopping

:collision

0

2

2

0

221

221

vvv

gdvgdmvm

gdvgdmvm

vmvmvm

ba

bkbbbkbb

akaaakaa

bbaab

μμ

μμ

==

==

+=

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59

Perfectly elastic collisions

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One-dimensional collisions

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One-dimensional collisions

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Equal masses

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Heavy body at rest

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Light body at rest

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65

Elastic collision in one dimension

ffi vmvmvm 221111 +=

fi

fi

EEPP

==

222

211

211 2

121

21

ffi vmvmvm +=

if

if

vmm

mv

vmmmmv

121

12

121

211

2+

=

+−=

Special situations

Equal masses

A massive target

A massive projectile

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Moving target

ffii vmvmvmvm 22112211 +=+

222

211

222

211 2

121

21

21

ffii vmvmvmvm +=+

iif

iif

vmmmmv

mmmv

vmm

mvmmmmv

221

211

21

12

221

21

21

211

2

2

+−+

+=

++

+−=

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67

Relative velocity

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Two-dimensional elastic collisions

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Two billiard balls

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Algebraic proof

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71

Velocities of balls