tenth edition - apl100.wdfiles.comapl100.wdfiles.com/local--files/notes/13_lecture_ppt mod.pdf ·...

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Tenth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Phillip J. Cornwell

Lecture Notes:

Brian P. Self California Polytechnic State University

CHAPTER

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

13 Kinetics of Particles:

Energy and Momentum

Methods


Vector Mechanics for Engineers: Dynamics

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Contents

13 - 2

Introduction

Work of a Force

Principle of Work & Energy

Applications of the Principle of

Work & Energy

Power and Efficiency

Sample Problem 13.1

Sample Problem 13.2

Sample Problem 13.3

Sample Problem 13.4

Sample Problem 13.5

Potential Energy

Conservative Forces

Conservation of Energy

Motion Under a Conservative

Central Force

Sample Problem 13.6

Sample Problem 13.7

Sample Problem 13.9

Principle of Impulse and Momentum

Impulsive Motion

Sample Problem 13.10



Impact

Direct Central Impact

Oblique Central Impact

Problems Involving Energy and

Momentum



Sample Problems 13.16




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Energy and Momentum Methods

2 - 3

The pogo stick allows the boy

to change between kinetic

energy, potential energy from

gravity, and potential energy

in the spring.

Accidents are often analyzed

by using momentum methods.



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Introduction

13 - 4

• Previously, problems dealing with the motion of

particles were solved through the fundamental

equation of motion,

• The current chapter introduces two additional

methods of analysis.

.F ma

• Method of work and energy: directly relates force,

mass, velocity and displacement.

• Method of impulse and momentum: directly

relates force, mass, velocity, and time.



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Introduction

2 - 5

Forces and

Accelerations

Velocities and

Displacements

Velocities and

Time

Approaches to Kinetics Problems

Newton’s Second

Law (last chapter)

Work-Energy Impulse-

Momentum

2211 TUTGF ma

2

11 2

t

tmv F dt mv



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Work of a Force

13 - 6

• Differential vector is the particle displacement. rd

• Work of the force is

dzFdyFdxF

dsF

rdFdU

zyx

cos

• Work is a scalar quantity, i.e., it has magnitude and

sign but not direction.

force. length • Dimensions of work are Units are

J 1.356lb1ftm 1N 1 J 1 joule



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Work of a Force

13 - 7

• Work of a force during a finite displacement,

2

1

2

1

2

1

2

1

cos

21

A

Azyx

s

st

s

s

A

A

dzFdyFdxF

dsFdsF

rdFU

• Work is represented by the area under the

curve of Ft plotted against s.

• Ft is the force in the direction of the

displacement ds



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2 - 8



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Work of a Force

13 - 9

What is the work of a constant force in

rectilinear motion?

1 2 cosU F x

1 2 sinU F x

1 2U F x

1 2 0U

a)

b)

c)

d)



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Work of a Force

13 - 10

• Work of the force of gravity,

yWyyW

dyWU

dyW

dzFdyFdxFdU

y

y

zyx

12

21

2

1

• Work of the weight is equal to product of

weight W and vertical displacement y.

• In the figure above, when is the work done by the weight positive?

a) Moving from y1 to y2 b) Moving from y2 to y1 c) Never



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Work of a Force

13 - 11

• Magnitude of the force exerted by a spring is

proportional to deflection,

lb/in.or N/mconstant spring k

kxF

• Work of the force exerted by spring,

222

1212

121

2

1

kxkxdxkxU

dxkxdxFdU

x

x

• Work of the force exerted by spring is positive

when x2 < x1, i.e., when the spring is returning to

its undeformed position.

• Work of the force exerted by the spring is equal to

negative of area under curve of F plotted against x,

xFFU 2121

21



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Work of a Force

13 - 12

As the block moves from A0 to A1, is

the work positive or negative?

Positive Negative

As the block moves from A2 to Ao, is

the work positive or negative?

Positive Negative

Displacement is

in the opposite

direction of the

force

Displacement is

in the same

direction of the

force



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Work of a Force

13 - 13

Work of a gravitational force (assume particle M

occupies fixed position O while particle m follows path

shown),

12221

2

2

1r

MmG

r

MmGdr

r

MmGU

drr

MmGFdrdU

r

r



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2 - 14

Does the normal force do work as the

block slides from B to A?

YES NO

Does the weight do work as

the block slides from B to A?

YES NO

Positive or

Negative work?



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Work of a Force

13 - 15

Forces which do not do work (ds = 0 or cos :

• Weight of a body when its center of gravity

moves horizontally.

• Reaction at a roller moving along its track, and

• Reaction at frictionless surface when body

in contact moves along surface,

• Reaction at frictionless pin supporting rotating body,



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Particle Kinetic Energy: Principle of Work & Energy

13 - 16

dvmvdsF

ds

dvmv

dt

ds

ds

dvm

dt

dvmmaF

t

tt

• Consider a particle of mass m acted upon by force F

• Integrating from A1 to A2 ,

energykineticmvTTTU

mvmvdvvmdsFv

v

s

st

221

1221

212

1222

12

1

2

1

• The work of the force is equal to the change in

kinetic energy of the particle.

F

• Units of work and kinetic energy are the same:

JmNms

mkg

s

mkg

2

22

21 mvT



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2 - 17



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Th

2 - 18

The work-kinetic energy is a scalar equation. Although in Cartesian coordinates

we find change in kinetic energy in x direction is equal to work done in that

direction. This is not true for cylindrical coordinates.



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Applications of the Principle of Work and Energy

13 - 19

• The bob is released

from rest at position A1.

Determine the velocity

of the pendulum bob at

A2 using work & kinetic

energy.

• Force acts normal to path and does no

work.

P

glv

vg

WWl

TUT

2

2

10

2

22

2211

• Velocity is found without determining

expression for acceleration and integrating.

• All quantities are scalars and can be added

directly.

• Forces which do no work are eliminated from

the problem.



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Applications of the Principle of Work and Energy

13 - 20

• Principle of work and energy cannot be

applied to directly determine the acceleration

of the pendulum bob.

• Calculating the tension in the cord requires

supplementing the method of work and energy

with an application of Newton’s second law.

• As the bob passes through A2 ,

Wl

gl

g

WWP

l

v

g

WWP

amF nn

32

22

glv 22

If you designed the rope to hold twice the weight of the bob, what would happen?



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Power and Efficiency

13 - 21

• rate at which work is done.

vF

dt

rdF

dt

dU

Power

• Dimensions of power are work/time or force*velocity.

Units for power are

W746s

lbft550 hp 1or

s

mN 1

s

J1 (watt) W 1

•

inputpower

outputpower

input work

koutput wor

efficiency



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Sample Problem 13.3

13 - 22

A spring is used to stop a 60 kg package

which is sliding on a horizontal surface.

The spring has a constant k = 20 kN/m

and is held by cables so that it is initially

compressed 120 mm. The package has a

velocity of 2.5 m/s in the position shown

and the maximum deflection of the spring

is 40 mm.

Determine (a) the coefficient of kinetic

friction between the package and surface

and (b) the velocity of the package as it

passes again through the position shown.

SOLUTION:

• Apply the principle of work and energy

between the initial position and the

point at which the spring is fully

compressed and the velocity is zero.

The only unknown in the relation is the

friction coefficient.

• Apply the principle of work and energy

for the rebound of the package. The

only unknown in the relation is the

velocity at the final position.



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Sample Problem 13.3

13 - 23

SOLUTION:

• Apply principle of work and energy between initial

position and the point at which spring is fully compressed.

0J5.187sm5.2kg60 22

212

121

1 TmvT

kk

kfxWU

J377m640.0sm81.9kg60 2

21

J0.112m040.0N3200N2400

N3200m160.0mkN20

N2400m120.0mkN20

21

maxmin21

21

0max

0min

xPPU

xxkP

kxP

e

J112J377212121 kefUUU

0J112J 377-J5.187

:2211

k

TUT

20.0k



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Sample Problem 13.3

13 - 24

• Apply the principle of work and energy for the rebound

of the package.

232

1232

132 kg600 vmvTT

J36.5

J112J377323232 kefUUU

232

1

3322

kg60J5.360

:

v

TUT

sm103.13v



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Group Problem Solving

2 - 25

Packages are thrown down an

incline at A with a velocity of 1

m/s. The packages slide along

the surface ABC to a conveyor

belt which moves with a

velocity of 2 m/s. Knowing that

k= 0.25 between the packages

and the surface ABC, determine

the distance d if the packages

are to arrive at C with a velocity

of 2 m/s.

SOLUTION:

The problem deals with a change in

position and different velocities, so use

work-energy.

• Find the kinetic energy at points A

and C.

• Determine the work done between

points A and C as a function of d.

• Use the work-energy relationship

and solve for d.

• Draw FBD of the box to help us

determine the forces that do work.



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SOLUTION:

Given: vA= 1 m/s, vC= 2 m/s, k= 0.25

Find: distance d

Will use:

Draw the FBD of the

block at points A and C

A A B B C CT U U T

Determine work done A → B

cos30

0.25 cos 30

sin 30

(sin 30 cos 30 )

AB

AB k AB

A B AB

k

N mg

F N mg

U mg d F d

mg d

7 m

BC BC

BC k

B C k BC

N mg x

F mg

U mg x

Determine work done B → C



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2 - 27

A A B B C CT U U T

Determine kinetic energy at A and at C

Substitute values into

21and 1 m/s

2A A AT mv v 21

and 2 m/s2

C C CT mv v

2 20

1 1(sin30 cos30 )

2 2A k k BCmv mg d mg x mv

2 2

2 2

/2 /2

(sin30 cos30 )

(2) /(2)(9.81) (0.25)(7) (1) /(2)(9.81)

sin30 0.25cos30

C k BC A

k

v g x v gd

6.71 md

Divide by m and solve for d



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2 - 28

If you wanted to bring the package to a complete stop at the

bottom of the ramp, would it work to place a spring as shown?

Would the package ever come to a stop?

k= 0.25

No, because the potential energy of the spring would turn

into kinetic energy and push the block back up the ramp

Yes, eventually enough energy would be dissipated

through the friction between the package and ramp.



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2 - 29

The potential energy stored at

the top of the roller coaster is

transferred to kinetic energy

as the cars descend.

The elastic potential energy

stored in the trampoline is

transferred to kinetic energy

and gravitational potential

energy as the girl flies upwards.



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Potential Energy

2 - 30

If the work of a force only depends on differences in

position, we can express this work as potential energy.

Can the work done by the following forces be expressed as

potential energy?

Weight

Friction

Spring force

Normal force

Yes

No

Yes

No

No

Yes

Yes

No



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Potential Energy

13 - 31

2121 yWyWU

• Work of the force of gravity , W

• Work is independent of path followed; depends

only on the initial and final values of Wy.

WyVg

potential energy of the body with respect

to force of gravity.

2121 gg VVU

• Units of work and potential energy are the same:

JmNWyVg

• Choice of datum from which the elevation y is

measured is arbitrary.



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Potential Energy

13 - 32

• Previous expression for potential energy of a body

with respect to gravity is only valid when the

weight of the body can be assumed constant.

• For a space vehicle, the variation of the force of

gravity with distance from the center of the earth

should be considered.

• Work of a gravitational force,

1221

r

GMm

r

GMmU

• Potential energy Vg when the variation in the

force of gravity can not be neglected,

r

WR

r

GMmVg

2



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Potential Energy

13 - 33

• Work of the force exerted by a spring depends

only on the initial and final deflections of the

spring,

222

1212

121 kxkxU

• The potential energy of the body with respect

to the elastic force,

2121

221

ee

e

VVU

kxV

• Note that the preceding expression for Ve is

valid only if the deflection of the spring is

measured from its undeformed position.



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Conservative Forces

13 - 34

• Concept of potential energy can be applied if the

work of the force is independent of the path

followed by its point of application.

22211121 ,,,, zyxVzyxVU

Such forces are described as conservative forces.

• For any conservative force applied on a closed path,

0rdF

• Elementary work corresponding to displacement

between two neighboring points,

zyxdV

dzzdyydxxVzyxVdU

,,

,,,,

Vz

V

y

V

x

VF

dzz

Vdy

y

Vdx

x

VdzFdyFdxF zyx

grad



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2 - 35



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Conservation of Energy

13 - 40

• Work of a conservative force,

2121 VVU

• Concept of work and energy,

1221 TTU

• Follows that

constant

2211

VTE

VTVT

• When a particle moves under the action of

conservative forces, the total mechanical

energy is constant.

WVT

WVT

11

11 0

WVT

VWgg

WmvT

22

2222

12 02

2

1• Friction forces are not conservative. Total

mechanical energy of a system involving

friction decreases.

• Mechanical energy is dissipated by friction

into thermal energy. Total energy is constant.



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Motion Under a Conservative Central Force

13 - 41

• When a particle moves under a conservative central

force, both the principle of conservation of angular

momentum

and the principle of conservation of energy

may be applied.

sinsin 000 rmvmvr

r

GMmmv

r

GMmmv

VTVT

221

0

202

1

00

• Given r, the equations may be solved for v and

• At minimum and maximum r, 90o. Given the

launch conditions, the equations may be solved for

rmin, rmax, vmin, and vmax.



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Sample Problem 13.7

13 - 42

The 250 g pellet is pushed against the

spring and released from rest at A.

Neglecting friction, determine the

smallest deflection of the spring for

which the pellet will travel around the

loop and remain in contact with the

loop at all times.

SOLUTION:

• Since the pellet must remain in contact

with the loop, the force exerted on the

pellet must be greater than or equal to

zero. Setting the force exerted by the

loop to zero, solve for the minimum

velocity at D.

• Apply the principle of conservation of

energy between points A and D. Solve

for the spring deflection required to

produce the required velocity and

kinetic energy at D.



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Sample Problem 13.7

13 - 43

SOLUTION:

• Setting the force exerted by the loop to zero, solve for the

minimum velocity at D.

:nn maF2

22 2 20.5 m 9.81m s 4.905m s

n D

D

W ma mg mv r

v rg

• Apply the principle of conservation of energy between

points A and D.

2 2 21 11 2 2

1

0 600N/m 300

0

e gV V V kx x x

T

22

2 2 212 2

0 0.25kg 9.81m/s (1 m) 2.45J

10.25 kg 4.905m s 0.613J

2

e g

D

V V V Wy

T mv

1 1 2 2

20 300 2.45J 0.613J

T V T V

x0.101 m 10/mmx



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2 - 44



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2 - 45



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Sample Problem 13.9

13 - 47

A satellite is launched in a direction

parallel to the surface of the earth

with a velocity of 36900 km/h from

an altitude of 500 km.

Determine (a) the maximum altitude

reached by the satellite, and (b) the

maximum allowable error in the

direction of launching if the satellite

is to come no closer than 200 km to

the surface of the earth

SOLUTION:

• For motion under a conservative central

force, the principles of conservation of

energy and conservation of angular

momentum may be applied simultaneously.

• Apply the principles to the points of

minimum and maximum altitude to

determine the maximum altitude.

• Apply the principles to the orbit insertion

point and the point of minimum altitude to

determine maximum allowable orbit

insertion angle error.



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Sample Problem 13.9

13 - 48

• Apply the principles of conservation of energy and

conservation of angular momentum to the points of minimum

and maximum altitude to determine the maximum altitude.

Conservation of energy:

1

212

1

0

202

1

r

GMmmv

r

GMmmvVTVT AAAA

Conservation of angular momentum:

1

0011100

r

rvvmvrmvr

Combining,

2001

0

1

0

02

1

202

021 2

111vr

GM

r

r

r

r

r

GM

r

rv

23122622

60

0

sm10398m1037.6sm81.9

sm1025.10hkm36900

km6870km500km6370

gRGM

v

r

km 60400m104.60 61r



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Sample Problem 13.9

13 - 49

• Apply the principles to the orbit insertion point and the point

of minimum altitude to determine maximum allowable orbit

insertion angle error.

Conservation of energy:

min

2max2

1

0

202

100

r

GMmmv

r

GMmmvVTVT AA

Conservation of angular momentum:

min

000maxmaxmin000 sinsin

r

rvvmvrmvr

Combining and solving for sin ,

5.1190

9801.0sin

0

0

5.11error allowable



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2 - 50

A section of track for a roller coaster

consists of two circular arcs AB and CD

joined by a straight portion BC. The radius

of CD is 80 m. The car and its occupants,

of total mass 300 kg, reach Point A with

practically no velocity and then drop freely

along the track. Determine the normal

force exerted by the track on the car at

point D. Neglect air resistance and rolling

resistance.

SOLUTION:

• This is two part problem – you

will need to find the velocity of

the car using work-energy, and

then use Newton’s second law

to find the normal force.

• Draw a diagram with the car

at points A and D, and

define your datum. Use

conservation of energy to

solve for vD

• Draw FBD and KD of the car

at point D, and determine the

normal force using Newton’s

second law.



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2 - 51

SOLUTION:

Given: vA= 0 m/s, rCD= 800 m, W=300 kg

Find: ND

Define your datum, sketch the

situation at points of interest

Datum

Use conservation of energy to find vD A A D DT V T V

0 0A Av T

2(300 kg)(9.81m/s )(30 m 20 m) =147150 JA AV mgy

2 2 21 1(300) 150

2 2D D D DT mv v v

Find TA

Find VA

Find TD

Find VD 0 0D Dy V

Solve for vD 2

2

150 147150

981

D

D

v

v

13.321 m/sDv



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2 - 52

Draw FBD and KD at point D

Use Newton’s second law in the normal direction

n nF ma

W

ND

man

mat

2D

D

vN mg m

R

981300 9.81

80DN

6620 NDN

en

et

2D

D

vN m g

R



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2 - 53

What happens to the normal

force at D if….

a) ND gets larger

b) ND gets smaller

c) ND stays the same

…we move point A higher? …the radius is smaller?

…we include friction?

a) ND gets larger

b) ND gets smaller


a) ND gets larger

b) ND gets smaller




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Impulsive Motion

2 - 54

The impulse applied to the railcar

by the wall brings its momentum

to zero. Crash tests are often

performed to help improve safety

in different vehicles.

The thrust of a rocket acts

over a specific time period

to give the rocket linear

momentum.



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Principle of Impulse and Momentum

13 - 55

• From Newton’s second law,

vmvmdt

dF

linear momentum

2211

21 force theof impulse 2

1

vmvm

FdtFt

t

Imp

Imp

• The final momentum of the particle can be

obtained by adding vectorially its initial

momentum and the impulse of the force during

the time interval.

12

2

1

vmvmdtF

vmddtF

t

t

• Dimensions of the impulse of

a force are

force*time.

• Units for the impulse of a

force are

smkgssmkgsN 2



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Impulsive Motion

13 - 56

• Force acting on a particle during a very short

time interval that is large enough to cause a

significant change in momentum is called an

impulsive force.

• When impulsive forces act on a particle,

21 vmtFvm

• When a baseball is struck by a bat, contact

occurs over a short time interval but force is

large enough to change sense of ball motion.

• Nonimpulsive forces are forces for which

is small and therefore, may be

neglected – an example of this is the weight

of the baseball.

tF



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13 - 57

An automobile of mass 1800 kg is

driven down a 5o incline at a speed of

100 km/h when the brakes are applied,

causing a constant total braking force of

7000 N.

Determine the time required for the

automobile to come to a stop.

SOLUTION:

• Apply the principle of impulse and

momentum. The impulse is equal to the

product of the constant forces and the

time interval.



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13 - 58

SOLUTION:


momentum.

2211 vmvm

Imp

Taking components parallel to the

incline,

1

2

sin5 0

km 1000 m 1h100 km/h 100 27.78 m/s

h 1 km 3600 s

1800 kg 2778 m/s 1800 kg 9.81m/s (sin 5 ) (7000 N) 0

mv mg t Ft

t t

9.16st



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13 - 59

A 120-g baseball is pitched with a

velocity of 24 m/s. After the ball is hit

by the bat, it has a velocity of 36 m/s in

the direction shown. If the bat and ball

are in contact for 0.015 s, determine the

average impulsive force exerted on the

ball during the impact.

SOLUTION:


momentum in terms of horizontal and

vertical component equations.



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13 - 60

SOLUTION:

• Apply the principle of impulse and momentum in

terms of horizontal and vertical component equations.

2211 vmvm

Imp

x

y

x component equation:

1 2 cos40

(0.12 kg)(24 m/s) (0.015 s) (0.12 kg)(36 m/s) cos 40

412.6 N

x

x

x

mv F t mv

F

F

y component equation:

20 sin 40

0.015 (0.12 kg)(36 m/s) sm40

185.1N

y

y

y

F t mv

F

F

(413 N) (185.1N) , 452NF i j F



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13 - 61

A 10 kg package drops from a chute

into a 24 kg cart with a velocity of 3

m/s. Knowing that the cart is initially at

rest and can roll freely, determine (a)

the final velocity of the cart, (b) the

impulse exerted by the cart on the

package, and (c) the fraction of the

initial energy lost in the impact.

SOLUTION:


momentum to the package-cart system

to determine the final velocity.

• Apply the same principle to the package

alone to determine the impulse exerted

on it from the change in its momentum.



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13 - 62

SOLUTION:

• Apply the principle of impulse and momentum to the package-cart

system to determine the final velocity.

2211 vmmvm cpp

Imp

x

y

x components:

2

21

kg 25kg 1030cosm/s 3kg 10

030cos

v

vmmvm cpp

m/s 742.02v



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13 - 63

• Apply the same principle to the package alone to determine the impulse

exerted on it from the change in its momentum.

x

y

2211 vmvm pp

Imp

x components:

2

21

kg 1030cosm/s 3kg 10

30cos

vtF

vmtFvm

x

pxp

sN56.18tFx

y components:

030sinm/s 3kg 10

030sin1

tF

tFvm

y

yp

sN15tFy

sN 9.23sN 51sN 56.1821 tFjitF

Imp

The t is of order 10-4 or 10-5 s, F is much larger than weight, i.e. weight is

neglected. Impulse due to weight is 100*10-4 Ns



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13 - 64

To determine the fraction of energy lost,

221 11 12 2

221 12 22 2

10 kg 3m s 45 J

10 kg 25 kg 0.742m s 9.63 J

p

p c

T m v

T m m v

786.0J 45

J 9.63J 45

1

21

T

TT



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2 - 65

The jumper approaches the takeoff

line from the left with a horizontal

velocity of 10 m/s, remains in contact

with the ground for 0.18 s, and takes

off at a 50o angle with a velocity of 12

m/s. Determine the average impulsive

force exerted by the ground on his

foot. Give your answer in terms of the

weight W of the athlete.

SOLUTION:

• Draw impulse and momentum diagrams

of the jumper.


momentum to the jumper to determine

the force exerted on the foot.

Here, dt is quite large and weight

should not be neglected



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2 - 66

Draw impulse and momentum diagrams of the jumper

1mv2mv

+ = 50º

Use the impulse momentum equation in y to find Favg

W t

Given: v1 = 10 m/s, v2= 12 m/s at 50º,

Δt= 0.18 s

Find: Favg in terms of W

avgF t

1 2( ) 0.18 sm t m tv P W v

x

y



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2 - 67

1mv2mv

+ = 50º

Use the impulse momentum equation in x and y to find Favg

W tavgF t

1 2( ) 0.18 savgm t m tv F W v

0 ( )(0.18) (12)(sin 50 )

(12)(sin 50 )

(9.81)(0.18)

avg y

avg y

WF W

g

F W W

(10) ( )(0.18) (12)(cos 50 )

10 (12)(cos 50 )

(9.81)(0.18)

avg x

avg x

W WF

g g

F W

1.295 6.21avg W WF i j

x

y

Favg-x is positive, which means we

guessed correctly (acts to the left)



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2 - 68

Car A and B crash into

one another. Looking

only at the impact, which

of the following

statements are true?

The total mechanical energy is the

same before and after the impact

If car A weighs twice as much as car B,

the force A exerts on car B is bigger

than the force B exerts on car A.

The total linear momentum is the same

immediately before and after the impact

True False

True False

True False



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2 - 69

The coefficient of restitution is used

to characterize the “bounciness” of

different sports equipment. The U.S.

Golf Association limits the COR of

golf balls at 0.83

Civil engineers use the

coefficient of restitution to

model rocks falling from

hillsides



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Impact

13 - 70

• Impact: Collision between two bodies which

occurs during a small time interval and during

which the bodies exert large forces on each other.

• Line of Impact: Common normal to the surfaces

in contact during impact.

• Central Impact: Impact for which the mass

centers of the two bodies lie on the line of impact;

otherwise, it is an eccentric impact..


• Direct Impact: Impact for which the velocities of

the two bodies are directed along the line of

impact.


• Oblique Impact: Impact for which one or both of

the bodies move along a line other than the line of

impact.



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13 - 71

• Bodies moving in the same straight line,

vA > vB .

• Upon impact the bodies undergo a

period of deformation, at the end of which,

they are in contact and moving at a

common velocity.

• A period of restitution follows during

which the bodies either regain their

original shape or remain permanently

deformed.

• Wish to determine the final velocities of the

two bodies. The total momentum of the

two body system is preserved,

BBBBBBAA vmvmvmvm

• A second relation between the final

velocities is required.



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13 - 72

• Period of deformation: umPdtvm AAA

• Period of restitution: AAA vmRdtum

10 e

uv

vu

Pdt

Rdt

nrestitutio of tcoefficien e

A

A



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13 - 73

• A similar analysis of particle B yields B

B

vu

uv e

• Combining the relations leads to the desired

second relation between the final velocities. BAAB vvevv

• Perfectly plastic impact, e = 0: vvv AB vmmvmvm BABBAA

• Perfectly elastic impact, e = 1:

Total energy and total momentum conserved. BAAB vvvv



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13 - 74

• Final velocities are

unknown in magnitude

and direction. Four

equations are required.

tBtBtAtA vvvv

nBBnAAnBBnAA vmvmvmvm

nBnAnAnB vvevv



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2 - 75

In deriving above equations the balls nothing was said as to

how they were supported. They could be supported in z-

direction (lying on a table) but impact was in x-y plane, so no

external forces due to reactions were there in x-y plane.

In the next problem, reaction forces are in the plane of impact.



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13 - 76

• Block constrained to move along horizontal

surface.

• Impulses from internal forces

along the n axis and from external force

exerted by horizontal surface and directed

along the vertical to the surface.

FF

and

extF

• Final velocity of ball unknown in direction and

magnitude and unknown final block velocity

magnitude. Three equations required.



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13 - 77

• Tangential momentum of ball is

conserved. tBtB vv

• Total horizontal momentum of block

and ball is conserved. xBBAAxBBAA vmvmvmvm

• Normal component of relative

velocities of block and ball are related

by coefficient of restitution.

nBnAnAnB vvevv

• Note: Validity of last expression does not follow from previous relation for

the coefficient of restitution. A similar but separate derivation is required.



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2 - 78



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2 - 79



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2 - 80



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2 - 81



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Problems Involving Energy and Momentum

13 - 82

• Three methods for the analysis of kinetics problems:

- Direct application of Newton’s second law

- Method of work and energy

- Method of impulse and momentum

• Select the method best suited for the problem or part of a problem

under consideration.



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13 - 83

Ball B is hanging from an inextensible cord. An identical

ball A is released from rest when it is just touching the

cord and acquires a velocity v0 before striking ball B.

Assuming perfectly elastic impact (e = 1) and no friction,

determine the velocity of each ball immediately after

impact.



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13 - 84

SOLUTION:

• Determine orientation of impact line of

action.

• The momentum component of ball A

tangential to the contact plane is

conserved.

• The total horizontal momentum of the

two ball system is conserved.

• The relative velocities along the line of

action before and after the impact are

related by the coefficient of restitution.

• Solve the last two expressions for the

velocity of ball A along the line of action

and the velocity of ball B which is

horizontal.



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13 - 85

SOLUTION:

• Determine orientation of impact line of action.

30

5.02

sinr

r

• The momentum component of ball A

tangential to the contact plane is

conserved.

0

0

5.0

030sin

vv

vmmv

vmtFvm

tA

tA

AA



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13 - 86

• The total horizontal (x component)

momentum of the two ball system is

conserved.

0

0

433.05.0

30sin30cos5.00

30sin30cos0

vvv

vvv

vmvmvm

vmvmtTvm

BnA

BnA

BnAtA

BAA



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13 - 87

• The relative velocities along the line of action before

and after the impact are related by the coefficient of

restitution.

0

0

866.05.0

030cos30sin

vvv

vvv

vvevv

nAB

nAB

nBnAnAnB



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13 - 88

• Solve the last two expressions for the velocity

of ball A along the line of action and the

velocity of ball B which is horizontal.

00 693.0520.0 vvvv BnA

0

10

00

693.0

1.16301.46

1.465.0

52.0tan721.0

520.05.0

vv

vv

vvv

B

A

ntA



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13 - 89

A 30 kg block is dropped from a height of 2 m onto the the

10 kg pan of a spring scale. Assuming the impact to be

perfectly plastic, determine the maximum deflection of the

pan. The constant of the spring is k = 20 kN/m.



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13 - 90

SOLUTION:


energy to determine the velocity of the

block at the instant of impact.

• Since the impact is perfectly plastic, the

block and pan move together at the same

velocity after impact. Determine that

velocity from the requirement that the

total momentum of the block and pan is

conserved.


energy to determine the maximum

deflection of the spring.



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13 - 91

SOLUTION:

• Apply principle of conservation of energy to determine

velocity of the block at instant of impact.

sm26.6030 J 5880

030

J 588281.9300

2222

1

2211

2222

1222

12

11

AA

AAA

A

vv

VTVT

VvvmT

yWVT



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13 - 92

• Determine velocity after impact from requirement that total

momentum of the block and pan is conserved.

sm70.41030026.630 33

322

vv

vmmvmvm BABBAA



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13 - 93

Initial spring deflection due to

pan weight:

m1091.41020

81.910 3

33k

Wx B

• Apply the principle of conservation of energy to determine the

maximum deflection of the spring.

2 21 13 32 2

3

22 3 31 132 2

4

30 10 4.7 442 J

0 20 10 4.91 10 0.241 J

0

A B

g e

T m m v

V V V

kx

T



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13 - 94

214 42

3 214 3 42

3 3 214 42

392 20 10

392 4.91 10 20 10

g e A BV V V W W h kx

x x x

x x

m 1091.4m 230.0 334 xxh m 225.0h



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13 - 95

m 230.0

10201091.43920241.0442

4

24

3

213

4

4433

x

xx

VTVT



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2 - 96



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2 - 97

A 2-kg block A is pushed up against a spring

compressing it a distance x= 0.1 m. The block is

then released from rest and slides down the 20º

incline until it strikes a 1-kg sphere B, which is

suspended from a 1 m inextensible rope. The

spring constant k=800 N/m, the coefficient of

friction between A and the ground is 0.2, the

distance A slides from the unstretched length of

the spring d=1.5 m, and the coefficient of

restitution between A and B is 0.8. When =40o,

find (a) the speed of B (b) the tension in the rope.



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2 - 98

SOLUTION:

• This is a multiple step problem.

Formulate your overall approach.

• Use work-energy to find the

velocity of the block just

before impact



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2 - 99

• Use conservation of momentum to

determine the speed of ball B after the

impact

• Use work energy to find the velocity at

• Use Newton’s 2nd Law to find tension

in the rope



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2 - 100

Given: mA= 2-kg mB= 1-kg,

k= 800 N/m, A =0.2, e= 0.8

Find (a) vB (b) Trope



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2 - 101

• Use work-energy to find the velocity of

the block just before impact

Determine the friction force acting on the

block A



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2 - 102

cos 0AN m g cos

(2)(9.81)cos 20

18.4368 N

(0.2)(18.4368)

3.6874 N

A

f k

N m g

F N

0:yF

Sum forces in

the y-direction

Solve for N



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2 - 103

Set your datum, use work-energy

to determine vA at impact.

1 1 1 1 2 2 2 2( ) ( ) ( ) ( )e g e gT V V U T V V

Determine values for each term.

2 21 1 1

1 10, ( ) (800)(0.1) 4.00 J

2 2eT V k x

1 1( ) ( )sin (2)(9.81)(1.6)sin 20 10.7367 Jg A AV m gh m g x d

1 2 ( ) (3.6874)(1.6) 5.8998 JfU F x d

2 2 22 2

1 1(1)( ) 1.000 0

2 2A A A AT m v v v V

2

x d

1

Datum



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2 - 104


2 21 1 1

1 10, ( ) (800)(0.1) 4.00 J

2 2eT V k x

1 1( ) ( )sin (2)(9.81)(1.6)sin 20 10.7367 Jg A AV m gh m g x d

1 2 ( ) (3.6874)(1.6) 5.8998 JfU F x d

2 2 22 2

1 1(1)( ) 1.000 0

2 2A A A AT m v v v V



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2 - 105

• Use conservation of

momentum to determine the

speed of ball B after the

impact

• Draw the

impulse

diagram

Note that the ball is

constrained to move only

horizontally immediately

after the impact.



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2 - 106

Note that the ball is constrained to move only

horizontally immediately after the impact.

Apply

conservation of

momentum in

the x direction

(2)(2.9727)cos20 2 cos20 (1.00)A Bv v

cos 0 cosA A A A B Bm v m v m v

(1)



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2 - 107

Note that the ball is constrained to move only

horizontally immediately after the impact.

( ) ( ) [( ) ( ) ]B n A n B n A nv v e v v

cos [ 0]

cos20 (0.8)(2.9727)

B A A

B A

v v e v

v v

Use the relative

velocity/coefficient

of restitution

equation

(2)

Solve (1) and (2) simultaneously

1.0382 m/s 3.6356 m/sA Bv v



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2 - 108

• Use work energy to find

the velocity at

Set datum, use Work-Energy

to determine vB at = 40o 1

2

Datum

1 1 1 1 2 2 2 2( ) ( ) ( ) ( )e g e gT V V U T V V


21 1

22 2 2 2

1( ) 0

2

1(1 cos )

2

B B

B B B

T m v V

T m v V m gh m gl



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2 - 109

1

2

Datum

2 21 1 2 2 2

2 22

2

2 2

1 1: ( ) 0 (1 cos )

2 2

( ) 2 (1 cos )

(3.6356) (2)(9.81)(1 cos 40 )

8.6274 m /s

B B B B

B

T V T V m v m v m g

v v gl

2 2.94 m/sv

Substitute into the Work-

Energy equation and solve

for vA



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2 - 110

• Use Newton’s 2nd Law to

find tension in the rope

• Draw your free-

body and kinetic

diagrams

:n B nF m acos

( cos )

B B n

B n

T m g m a

T m a g

• Sum forces in the

normal direction

et

en



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2 - 111

222

1.00 m

8.62748.6274 m/s

1.00n

va

(1.0)(8.6274 9.81cos40 )T

• Determine normal

acceleration

16.14 NT

• Substitute and solve

cos

( cos )

B B n

B n

T m g m a

T m a g



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Concept Question

2 - 112

If the coefficient of

restitution is smaller

than the 0.8 in the

problem, the tension

T will be…

Smaller Bigger

Compare the following statement to the problem

you just solved.



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Concept Question

2 - 113

If the rope length is smaller than the

1 m in the problem, the tension T

will be…

Smaller Bigger



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Concept Question

2 - 114

If the coefficient of

friction is smaller

than 0.2 given in

the problem, the

tension T will be…

Smaller Bigger



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Concept Question

2 - 115

If the mass of A is

smaller than the 2

kg given in the

problem, the

tension T will be…

Smaller Bigger



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Summary

2 - 116

Forces and

Accelerations

Velocities and

Displacements Velocities

and Time


Newton’s

Second Law

(last chapter)

Work-

Energy

Impulse-

Momentum

2211 TUT

GF ma

2

11 2

t

tmv F dt mv



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Summary

2 - 117

Forces and

Accelerations

Velocities and

Displacements

Velocities and

Time


Newton’s Second

Law (last chapter)

Work-Energy Impulse-

Momentum

2211 TUTGF ma

2

11 2

t

tmv F dt mv



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2 - 118



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2 - 119



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