Dynamics13lecture

VECTOR MECHANICS FOR ENGINEERS:
DYNAMICSDYNAMICS
TenthTenth
EditionEdition
Ferdinand P. BeerFerdinand P. Beer
E. Russell Johnston, Jr.E. Russell Johnston, Jr.
Phillip J. CornwellPhillip J. Cornwell
Lecture Notes:Lecture Notes:
Brian P. SelfBrian P. Self
California Polytechnic State UniversityCalifornia 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: DynamicsVector Mechanics for Engineers: DynamicsTenth
Edition
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

Edition
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.

Edition
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 maΣ =
r r
• Method of work and energy: directly relates force,
mass, velocity and displacement.
• Method of impulse and momentum: directly
relates force, mass, velocity, and time.

Edition
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 TUT =+ →GF ma=∑
r r 2
1
1 2
t
t
mv F dt mv+ =∫
vv v

Edition
Work of a Force
13 - 6
• Differential vector is the particle displacement.rd
r
• Work of the force is
dzFdyFdxF
dsF
rdFdU
zyx ++=
=
•=
αcos
rr
• Work is a scalar quantity, i.e., it has magnitude and
sign but not direction.
force.length ×• Dimensions of work are Units are
( ) ( )( ) J1.356lb1ftm1N1J1 =⋅=joule

Edition
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
A
zyx
s
s
t
s
s
A
A
dzFdyFdxF
dsFdsF
rdFU
α
rr
• 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

Edition
Work of a Force
13 - 8
What is the work of a constant force in
rectilinear motion?
( )1 2 cosU F xα→ = ∆
( )1 2 sinU F xα→ = ∆
1 2U F x→ = ∆
1 2 0U → =
a)
b)
c)
d)

Edition
Work of a Force
13 - 9
• 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

Edition
Work of a Force
13 - 10
• Magnitude of the force exerted by a spring is
proportional to deflection,
( )lb/in.orN/mconstantspring=
=
k
kxF
• Work of the force exerted by spring,
2
22
12
12
1
21
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 ∆+−=→ 212
1
21

Edition
Work of a Force
13 - 11
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

Edition
Work of a Force
13 - 12
Work of a gravitational force (assume particle M
occupies fixed position O while particle m follows path
shown),
12
221
2
2
1
r
Mm
G
r
Mm
Gdr
r
Mm
GU
dr
r
Mm
GFdrdU
r
r
−=−=
−=−=
∫→

Edition
2 - 13
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?

Edition
Work of a Force
13 - 14
Forces which do not do work (ds = 0 or cos α = 0):
• 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,

Edition
Particle Kinetic Energy: Principle of Work & Energy
13 - 15
dvmvdsF
ds
dv
mv
dt
ds
ds
dv
m
dt
dv
mmaF
t
tt
=
==
==
• Consider a particle of mass m acted upon by force F
r
• Integrating from A1 to A2 ,
energykineticmvTTTU
mvmvdvvmdsF
v
v
s
s
t
==−=
−==
→
∫∫
2
2
1
1221
2
12
12
22
1
2
1
2
1
• The work of the force is equal to the change in
kinetic energy of the particle.
F
r
• Units of work and kinetic energy are the same:
JmNm
s
m
kg
s
m
kg 2
2
2
2
1 =⋅=





=





== mvT

Edition
Applications of the Principle of Work and Energy
13 - 16
• 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
r
glv
v
g
W
Wl
TUT
2
2
1
0
2
2
2
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.

Edition
Applications of the Principle of Work and Energy
13 - 17
• 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 ,
W
l
gl
g
W
WP
l
v
g
W
WP
amF nn
3
2
2
2
=+=
=−
=∑
glv 22 =
If you designed the rope to hold twice the weight of the bob, what would happen?

Edition
Power and Efficiency
13 - 18
• rate at which work is done.
vF
dt
rdF
dt
dU
Power
rr
rr
•=
•
==
=
• Dimensions of power are work/time or force*velocity.
Units for power are
W746
s
lbft
550hp1or
s
m
N1
s
J
1(watt)W1 =
⋅
=⋅==
•
inputpower
outputpower
input work
koutput wor
efficiency
=
=
=η

Edition
Sample Problem 13.1
13 - 19
An automobile weighing 4000 lb is
driven down a 5o
incline at a speed of
60 mi/h when the brakes are applied
causing a constant total breaking force
of 1500 lb.
Determine the distance traveled by the
automobile as it comes to a stop.
SOLUTION:
• Evaluate the change in kinetic energy.
• Determine the distance required for the
work to equal the kinetic energy
change.

Edition
Sample Problem 13.1
13 - 20
SOLUTION:
• Evaluate the change in kinetic energy.
( )( ) lbft481000882.324000
sft88
s3600
h
mi
ft5280
h
mi
60
2
2
12
12
1
1
1
⋅===
=

















=
mvT
v
( ) 0lb1151lbft481000
2211
=−⋅
=+ →
x
TUT
ft418=x
• Determine the distance required for the work
to equal the kinetic energy change.
( ) ( )( )
( )x
xxU
lb1151
5sinlb4000lb150021
−=
°+−=→
00 22 == Tv

Edition
Sample Problem 13.2
13 - 21
Two blocks are joined by an inextensible
cable as shown. If the system is released
from rest, determine the velocity of block
A after it has moved 2 m. Assume that
the coefficient of friction between block
A and the plane is µk = 0.25 and that the
pulley is weightless and frictionless.
SOLUTION:
• Apply the principle of work and
energy separately to blocks A and B.
• When the two relations are combined,
the work of the cable forces cancel.
Solve for the velocity.

Edition
Sample Problem 13.2
13 - 22
SOLUTION:
• Apply the principle of work and energy separately
to blocks A and B.
( )( )
( )
( ) ( )
( ) ( )( ) ( ) 2
2
1
2
2
1
2211
2
kg200m2N490m2
m2m20
:
N490N196225.0
N1962sm81.9kg200
vF
vmFF
TUT
WNF
W
C
AAC
AkAkA
A
=−
=−+
=+
====
==
→
µµ
( )( )
( ) ( )
( ) ( )( ) ( ) 2
2
1
2
2
1
2211
2
kg300m2N2940m2
m2m20
:
N2940sm81.9kg300
vF
vmWF
TUT
W
c
BBc
B
=+−
=+−
=+
==
→

Edition
Sample Problem 13.2
13 - 23
• When the two relations are combined, the work of the
cable forces cancel. Solve for the velocity.
( ) ( )( ) ( ) 2
2
1 kg200m2N490m2 vFC =−
( ) ( )( ) ( ) 2
2
1 kg300m2N2940m2 vFc =+−
( )( ) ( )( ) ( )
( ) 2
2
1
2
2
1
kg500J4900
kg300kg200m2N490m2N2940
v
v
=
+=−
sm43.4=v

Edition
13.2 – Alternate Solution, Group Problem Solving
2 - 24
Could you apply work-energy to the combined
system of blocks?
Given: v1= 0, distance = 2 m, µk = 0.25
( ) 21
24900 J 500kg v=
Note that vA = vB
( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 0.25 200 9.81 2m 300 9.81 2m 4900 JU → = − + =
What is T1 of the system?
1 0T =
What is the total work done between points 1 and 2?
2 m
2 m
( ) ( )2 2 2 21 1 1 1
2 2 2 2 2200kg 300kgA BT m v m v v v= + = +
What is T2 of the system?
Solve for v
sm43.4=v
1
1
2
2

Edition
Sample Problem 13.3
13 - 25
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:
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.
energy for the rebound of the package.
The only unknown in the relation is the
velocity at the final position.

Edition
Sample Problem 13.3
13 - 26
SOLUTION:
• Apply principle of work and energy between initial
position and the point at which spring is fully compressed.
( )( ) 0J5.187sm5.2kg60 2
2
2
12
12
1
1 ==== TmvT
( )
( )( )( ) ( ) kk
kf xWU
µµ
µ
J377m640.0sm81.9kg60 2
21
−=−=
−=→
( )( )
( ) ( )( )
( ) ( )
( )( ) J0.112m040.0N3200N2400
N3200m160.0mkN20
N2400m120.0mkN20
2
1
maxmin2
1
21
0max
0min
−=+−=
∆+−=
==∆+=
===
→ xPPU
xxkP
kxP
e
( ) ( ) ( ) J112J377212121 −−=+= →→→ kef UUU µ
( ) 0J112J377-J5.187
:2211
=−
=+ →
k
TUT
µ 20.0=kµ

Edition
Sample Problem 13.3
13 - 27
• Apply the principle of work and energy for the rebound
of the package.
( ) 2
32
12
32
1
32 kg600 vmvTT ===
( ) ( ) ( )
J36.5
J112J377323232
+=
+−=+= →→→ kef UUU µ
( ) 2
32
1
3322
kg60J5.360
:
v
TUT
=+
=+ →
sm103.13 =v

Edition
Sample Problem 13.4
13 - 28
A 2000 lb car starts from rest at point 1
and moves without friction down the
track shown.
Determine:
a) the force exerted by the track on
the car at point 2, and
b) the minimum safe value of the
radius of curvature at point 3.
SOLUTION:
• Apply principle of work and energy to
determine velocity at point 2.
• Apply Newton’s second law to find
normal force by the track at point 2.
• Apply principle of work and energy to
determine velocity at point 3.
• Apply Newton’s second law to find
minimum radius of curvature at point 3
such that a positive normal force is
exerted by the track.

Edition
Sample Problem 13.4
13 - 29
SOLUTION:
• Apply principle of work and energy to determine
velocity at point 2.
( )
( )
( ) ( )( ) sft8.50ft2.32ft402ft402
2
1
ft400:
ft40
2
1
0
2
22
2
2
22211
21
2
2
2
22
1
21
===
=+=+
+=
===
→
→
vsgv
v
g
W
WTUT
WU
v
g
W
mvTT
• Apply Newton’s second law to find normal force by
the track at point 2.
:nn amF =↑+ ∑
( )
WN
g
g
Wv
g
W
amNW n
5
ft20
ft402
2
2
2
=
===+−
ρ
lb10000=N

Edition
Sample Problem 13.4
13 - 30
• Apply principle of work and energy to determine
velocity at point 3.
( )
( ) ( )( ) sft1.40sft2.32ft252ft252
2
1
ft250
3
2
3
2
33311
===
=+=+ →
vgv
v
g
W
WTUT
• Apply Newton’s second law to find minimum radius of
curvature at point 3 such that a positive normal force is
exerted by the track.
:nn amF =↓+ ∑
( )
33
2
3 ft252
ρρ
g
g
Wv
g
W
amW n
==
=
ft503 =ρ

Edition
Sample Problem 13.5
13 - 31
The dumbwaiter D and its load have a
combined weight of 600 lb, while the
counterweight C weighs 800 lb.
Determine the power delivered by the
electric motor M when the dumbwaiter
(a) is moving up at a constant speed of
8 ft/s and (b) has an instantaneous
velocity of 8 ft/s and an acceleration of
2.5 ft/s2
, both directed upwards.
SOLUTION:
Force exerted by the motor
cable has same direction as
the dumbwaiter velocity.
Power delivered by motor is
equal to FvD, vD = 8 ft/s.
• In the first case, bodies are in uniform
motion. Determine force exerted by
motor cable from conditions for static
equilibrium.
• In the second case, both bodies are
accelerating. Apply Newton’s
second law to each body to
determine the required motor cable
force.

Edition
Sample Problem 13.5
13 - 32
• In the first case, bodies are in uniform motion.
Determine force exerted by motor cable from
conditions for static equilibrium.
( )( )
slbft1600
sft8lb200
⋅=
== DFvPower
( ) hp91.2
slbft550
hp1
slbft1600 =
⋅
⋅=Power
Free-body C:
:0=↑+ ∑ yF lb4000lb8002 ==− TT
Free-body D:
:0=↑+ ∑ yF
lb200lb400lb600lb600
0lb600
=−=−=
=−+
TF
TF

Edition
Sample Problem 13.5
13 - 33
• In the second case, both bodies are accelerating. Apply
Newton’s second law to each body to determine the required
motor cable force.
↓=−=↑= 2
2
12
sft25.1sft5.2 DCD aaa
Free-body C:
:CCy amF =↓+ ∑ ( ) lb5.38425.1
2.32
800
2800 ==− TT
Free-body D:
:DDy amF =↑+ ∑ ( )
lb1.2626.466005.384
5.2
2.32
600
600
==−+
=−+
FF
TF
( )( ) slbft2097sft8lb1.262 ⋅=== DFvPower
( ) hp81.3
slbft550
hp1
slbft2097 =
⋅
⋅=Power

Edition
Group Problem Solving
2 - 34
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
µk= 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.

Edition
2 - 35
SOLUTION:
Given: vA= 1 m/s, vC= 2 m/s, µk= 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
sin30
(sin30 cos30 )
AB
AB k AB
A B AB
k
N mg
F N mg
U mg d F d
mg d
µ
µ
→
= °
= = °
= ° −
= ° − °
7 mBC BC
BC k
B C k BC
N mg x
F mg
U mg x
µ
µ→
= =
=
= −
Determine work done B → C

Edition
2 - 36
A A B B C CT U U T→ →+ + =
Determine kinetic energy at A and at C
Substitute values into
21
and 1m/s
2
A A AT mv v= = 21
and 2 m/s
2
C C CT mv v= =
2 2
0
1 1
(sin30 cos30 )
2 2
A k k BCmv mg d mg x mvµ µ+ ° − ° − =
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 g
d
µ
µ
 + − =
° − °
+ −
=
° − °
6.71 md =
Divide by m and solve for d

Edition
2 - 37
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?
µk= 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.

Edition
2 - 38
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.

Edition
Potential Energy
2 - 39
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

Edition
Potential Energy
13 - 40
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:
JmN =⋅== WyVg
• Choice of datum from which the elevation y is
measured is arbitrary.

Edition
Potential Energy
13 - 41
• 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,
12
21
r
GMm
r
GMm
U −=→
• Potential energy Vg when the variation in the
force of gravity can not be neglected,
r
WR
r
GMm
Vg
2
−=−=

Edition
Potential Energy
13 - 42
• Work of the force exerted by a spring depends
only on the initial and final deflections of the
spring,
2
22
12
12
1
21 kxkxU −=→
• The potential energy of the body with respect
to the elastic force,
( ) ( )2121
2
2
1
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.

Edition
Conservative Forces
13 - 43
• 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,
0=•∫ rdF

• Elementary work corresponding to displacement
between two neighboring points,
( ) ( )
( )zyxdV
dzzdyydxxVzyxVdU
,,
,,,,
−=
+++−=
V
z
V
y
V
x
V
F
dz
z
V
dy
y
V
dx
x
V
dzFdyFdxF zyx
grad−=





∂
∂
+
∂
∂
+
∂
∂
−=






∂
∂
+
∂
∂
+
∂
∂
−=++


Edition
Conservation of Energy
13 - 44
• 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
VWg
g
W
mvT
=+
====
22
2
2
22
1
2 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.

Edition
Motion Under a Conservative Central Force
13 - 45
• 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.
φφ sinsin 000 rmvmvr =
r
GMm
mv
r
GMm
mv
VTVT
−=−
+=+
2
2
1
0
2
02
1
00
• Given r, the equations may be solved for v and ϕ.
• At minimum and maximum r, ϕ = 90o
. Given the
launch conditions, the equations may be solved for
rmin, rmax, vmin, and vmax.

Edition
Sample Problem 13.6
13 - 46
A 20 lb collar slides without friction
along a vertical rod as shown. The
spring attached to the collar has an
undeflected length of 4 in. and a
constant of 3 lb/in.
If the collar is released from rest at
position 1, determine its velocity after
it has moved 6 in. to position 2.
SOLUTION:
• Apply the principle of conservation of
energy between positions 1 and 2.
• The elastic and gravitational potential
energies at 1 and 2 are evaluated from
the given information. The initial
kinetic energy is zero.
• Solve for the kinetic energy and
velocity at 2.

Edition
Sample Problem 13.6
13 - 47
SOLUTION:
• Apply the principle of conservation of energy between
positions 1 and 2.
Position 1: ( )( )
0
lbft20lbin.24
lbin.24in.4in.8in.lb3
1
1
2
2
12
12
1
=
⋅=+⋅=+=
⋅=−==
T
VVV
kxV
ge
e
Position 2: ( )( )
( )( )
2
2
2
2
2
22
1
2
2
2
2
12
22
1
311.0
2.32
20
2
1
lbft5.5lbin.6612054
lbin.120in.6lb20
lbin.54in.4in.01in.lb3
vvmvT
VVV
WyV
kxV
ge
g
e
===
⋅−=⋅−=−=+=
⋅−=−==
⋅=−==
Conservation of Energy:
lbft5.50.311lbft20 2
2
2211
⋅−=⋅+
+=+
v
VTVT
↓= sft91.42v

Edition
Sample Problem 13.7
13 - 48
The 0.5 lb 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.
energy between points A and D. Solve
for the spring deflection required to
produce the required velocity and
kinetic energy at D.

Edition
Sample Problem 13.7
13 - 49
SOLUTION:
• Setting the force exerted by the loop to zero, solve for the
minimum velocity at D.
:nn maF =↓+ ∑
( )( ) 222
2
sft4.64sft32.2ft2 ===
==
rgv
rvmmgmaW
D
Dn
• Apply the principle of conservation of energy between
points A and D.
( )
0
18ftlb360
1
22
2
12
2
1
1
=
==+=+=
T
xxkxVVV ge
( )( )
( ) lbft5.0sft4.64
sft2.32
lb5.0
2
1
lbft2ft4lb5.00
22
2
2
2
1
2
2
⋅===
⋅==+=+=
D
ge
mvT
WyVVV
25.0180 2
2211
+=+
+=+
x
VTVT
in.47.4ft3727.0 ==x

Edition
Sample Problem 13.9
13 - 50
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.

Edition
Sample Problem 13.9
13 - 51
• 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
2
12
1
0
2
02
1
r
GMm
mv
r
GMm
mvVTVT AAAA −=−+=+ ′′
Conservation of angular momentum:
1
0
011100
r
r
vvmvrmvr ==
Combining,
2
001
0
1
0
0
2
1
2
02
02
1 2
111
vr
GM
r
r
r
r
r
GM
r
r
v =+





−=








−
( )( ) 23122622
6
0
0
sm10398m1037.6sm81.9
sm1025.10hkm36900
km6870km500km6370
×=×==
×==
=+=
gRGM
v
r
km60400m104.60 6
1 =×=r

Edition
Sample Problem 13.9
13 - 52
• 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
2
max2
1
0
2
02
1
00
r
GMm
mv
r
GMm
mvVTVT AA −=−+=+
Conservation of angular momentum:
min
0
00maxmaxmin000 sinsin
r
r
vvmvrmvr φφ ==
Combining and solving for sin ϕ0,
°±°=
=
5.1190
9801.0sin
0
0
ϕ
φ
°±= 5.11errorallowable

Edition
2 - 53
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 240 ft. The car and its occupants,
of total weight 560 lb, 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.

Edition
2 - 54
SOLUTION:
Given: vA= 0 ft/s, rCD= 240 ft, W=560 lbs
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= =
(560 lb)(90 60)=84,000 ft lbsA AV Wy= = + g
2 2 21 1 560
8.6957
2 2 32.2
D D D DT mv v v
 
= = = ÷
 
Find TA
Find VA
Find TD
Find VD
0 0D Dy V= =
Solve for vD
2
8.6957 84000Dv =
98.285 ft/sDv =

Edition
2 - 55
Draw FBD and KD at point D
Use Newton’s second law in the normal direction
n nF ma=∑
240 ft
W
ND
man
mat
2
D
D
v
N W m
R
 
− =  ÷ ÷
 
2
560 98.285
560
32.2 240
DN
 
= +  ÷ ÷
 
1260 lbsDN =
en
et

Edition
2 - 56
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

Edition
Impulsive Motion
2 - 57
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.

Edition
Principle of Impulse and Momentum
13 - 58
• From Newton’s second law,
( ) == vmvm
dt
d
F

linear momentum
2211
21 forcetheofimpulse
2
1
vmvm
FdtF
t
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
⋅=⋅⋅=⋅

Edition
Impulsive Motion
13 - 59
• 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∆


Edition
13 - 60
An automobile weighing 4000 lb is
driven down a 5o
incline at a speed of
60 mi/h when the brakes are applied,
causing a constant total braking force
of 1500 lb.
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.

Edition
13 - 61
SOLUTION:
momentum.
2211 vmvm

=+ ∑ →Imp
Taking components parallel to the
incline,
( )
( ) ( ) 015005sin4000sft88
2.32
4000
05sin1
=−°+





=−°+
tt
FttWmv
s49.9=t

Edition
13 - 62
A 4 oz baseball is pitched with a
velocity of 80 ft/s. After the ball is hit
by the bat, it has a velocity of 120 ft/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.

Edition
13 - 63
SOLUTION:
• Apply the principle of impulse and momentum in
terms of horizontal and vertical component equations.
2211 vmvm

=+ →Imp
x
y
x component equation:
( ) ( ) ( )
lb89
40cos120
2.32
164
15.080
2.32
164
40cos21
=
°=+−
°=∆+−
x
x
x
F
F
mvtFmv
y component equation:
( ) ( )
lb9.39
40cos120
2.32
164
15.0
40sin0 2
=
°=
°=∆+
y
y
y
F
F
mvtF
( ) ( ) lb5.97,lb9.39lb89 =+= FjiF


Edition
13 - 64
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.

Edition
13 - 65
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
kg25kg1030cosm/s3kg10
030cos
v
vmmvm cpp
+=°
+=+°
m/s742.02 =v

Edition
13 - 66
• 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
kg1030cosm/s3kg10
30cos
vtF
vmtFvm
x
pxp
=∆+°
=∆+°
sN56.18 ⋅−=∆tFx
y components:
( )( ) 030sinm/s3kg10
030sin1
=∆+°−
=∆+°−
tF
tFvm
y
yp
sN15 ⋅=∆tFy
( ) ( ) sN9.23sN51sN56.1821 ⋅=∆⋅+⋅−=∆=∑ → tFjitF

Imp

Edition
13 - 67
To determine the fraction of energy lost,
( ) ( )
( ) ( ) ( )
221 1
1 12 2
221 1
2 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
= = =
= + = + =
786.0
J45
J9.63J45
1
21 =
−
=
−
T
TT

Edition
2 - 68
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.

Edition
2 - 69
Draw impulse and momentum diagrams of the jumper
1mv
v 2mv
v
+ =
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 t+ − ∆ = ∆ =v P W v
x
y

Edition
2 - 70
1mv
v 2mv
v
+ =
50º
Use the impulse momentum equation in x and y to find Favg
W t∆
avgF t∆
1 2( ) 0.18 savgm t m t+ − ∆ = ∆ =v F W v
0 ( )(0.18) (12)(sin 50 )
(12)(sin 50 )
(9.81)(0.18)
avg y
avg y
W
F W
g
F W W
−
−
+ − = °
°
= +
(10) ( )(0.18) (12)(cos 50 )
10 (12)(cos 50 )
(9.81)(0.18)
avg x
avg x
W W
F
g g
F W
−
−
+ − = °
− °
=
1.295 6.21avg W W= − +F i j
x
y
Favg-x is positive, which means we
guessed correctly (acts to the left)

Edition
2 - 71
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

Edition
2 - 72
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

Edition
Impact
13 - 73
• 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.

Edition
13 - 74
• 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.

Edition
13 - 75
• Period of deformation: umPdtvm AAA =− ∫
• Period of restitution: AAA vmRdtum ′=− ∫
10 ≤≤
−
′−
==
=
∫
∫
e
uv
vu
Pdt
Rdt
nrestitutiooftcoefficiene
A
A
• 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 −=′−′

Edition
13 - 76
• Final velocities are
unknown in magnitude
and direction. Four
equations are required.
• No tangential impulse component;
tangential component of momentum
for each particle is conserved.
( ) ( ) ( ) ( )tBtBtAtA vvvv ′=′=
• Normal component of total
momentum of the two particles is
conserved.
( ) ( ) ( ) ( )nBBnAAnBBnAA vmvmvmvm ′+′=+
• Normal components of relative
velocities before and after impact
are related by the coefficient of
restitution.
( ) ( ) ( ) ( )[ ]nBnAnAnB vvevv −=′−′

Edition
13 - 77
• 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.

Edition
13 - 78
• 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.

Edition
Problems Involving Energy and Momentum
13 - 79
• 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.

Edition
13 - 80
A ball is thrown against a frictionless,
vertical wall. Immediately before the
ball strikes the wall, its velocity has a
magnitude v and forms angle of 30o
with the horizontal. Knowing that
e = 0.90, determine the magnitude
and direction of the velocity of the
ball as it rebounds from the wall.
SOLUTION:
• Resolve ball velocity into components
normal and tangential to wall.
• Impulse exerted by the wall is normal
to the wall. Component of ball
momentum tangential to wall is
conserved.
• Assume that the wall has infinite mass
so that wall velocity before and after
impact is zero. Apply coefficient of
restitution relation to find change in
normal relative velocity between wall
and ball, i.e., the normal ball velocity.

Edition
13 - 81
• Component of ball momentum tangential to wall is conserved.
vvv tt 500.0==′
• Apply coefficient of restitution relation with zero wall
velocity.
( )
( ) vvv
vev
n
nn
779.0866.09.0
00
−=−=′
−=′−
SOLUTION:
• Resolve ball velocity into components parallel and
perpendicular to wall.
vvvvvv tn 500.030sin866.030cos =°==°=
n
t
°=





=′
+−=′
−
7.32
500.0
779.0
tan926.0
500.0779.0
1
vv
vvv tn λλ


Edition
13 - 82
The magnitude and direction of the
velocities of two identical
frictionless balls before they strike
each other are as shown. Assuming
e = 0.9, determine the magnitude
and direction of the velocity of each
ball after the impact.
SOLUTION:
• Resolve the ball velocities into components
normal and tangential to the contact plane.
• Tangential component of momentum for
each ball is conserved.
• Total normal component of the momentum
of the two ball system is conserved.
• The normal relative velocities of the
balls are related by the coefficient of
restitution.
• Solve the last two equations
simultaneously for the normal velocities of
the balls after the impact.

Edition
13 - 83
SOLUTION:
• Resolve the ball velocities into components normal and
tangential to the contact plane.
( ) sft0.2630cos =°= AnA vv ( ) sft0.1530sin =°= AtA vv
( ) sft0.2060cos −=°−= BnB vv ( ) sft6.3460sin =°= BtB vv
• Tangential component of momentum for each ball is
conserved.
( ) ( ) sft0.15==′ tAtA vv ( ) ( ) sft6.34==′ tBtB vv
• Total normal component of the momentum of the two
ball system is conserved.
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) 0.6
0.200.26
=′+′
′+′=−+
′+′=+
nBnA
nBnA
nBBnAAnBBnAA
vv
vmvmmm
vmvmvmvm

Edition
13 - 84
°=





=′
+=′
°=





=′
+−=′
−
−
6.55
7.23
6.34
tansft9.41
6.347.23
3.40
7.17
0.15
tansft2.23
0.157.17
1
1
B
ntB
A
ntA
v
v
v
v
λλ
λλ


t
n
• The normal relative velocities of the balls are related by the
coefficient of restitution.
( ) ( ) ( ) ( )[ ]
( )[ ] 4.410.200.2690.0 =−−=
−=′−′ nBnAnBnA vvevv
• Solve the last two equations simultaneously for the normal
velocities of the balls after the impact.
( ) sft7.17−=′ nAv ( ) sft7.23=′ nBv

Edition
13 - 85
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.
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.

Edition
13 - 86
SOLUTION:
• Determine orientation of impact line of action.
°=
==
30
5.0
2
sin
θ
θ
r
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
=′
′=+°
′=∆+

• 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
=′+′
′−°′−°=
′−°′−°′=
′+′=∆+


Edition
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
=′−′
−°=′−°′
−=′−′
• 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
1
0
00
693.0
1.16301.46
1.46
5.0
52.0
tan721.0
520.05.0
vv
vv
vvv
B
A
ntA
α
β
λλ


Edition
13 - 88
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.
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.

Edition
13 - 89
SOLUTION:
• Apply principle of conservation of energy to
determine velocity of the block at instant of impact.
( )( )( )
( ) ( )( )
( )( ) ( ) sm26.6030J5880
030
J588281.9300
2
2
22
1
2211
2
2
22
12
22
1
2
11
=+=+
+=+
===
====
AA
AAA
A
vv
VTVT
VvvmT
yWVT
• Determine velocity after impact from requirement that
total momentum of the block and pan is conserved.
( ) ( ) ( )
( )( ) ( ) sm70.41030026.630 33
322
=+=+
+=+
vv
vmmvmvm BABBAA

Edition
13 - 90
Initial spring deflection due to
pan weight:
( )( ) m1091.4
1020
81.910 3
33
−
×=
×
==
k
W
x B
• Apply the principle of conservation of energy to
determine the maximum deflection of the spring.
( ) ( )( )
( )( )
( )( )
( ) ( )
( ) ( ) 2
4
3
2
13
4
2
4
3
2
1
34
2
42
1
4
4
233
2
12
32
1
3
2
2
12
32
1
3
10201091.4392
1020392
0
J241.01091.410200
J4427.41030
xx
xxx
kxhWWVVV
T
kx
VVV
vmmT
BAeg
eg
BA
×+×−−=
×+−−=
+−+=+=
=
=××=+=
+=
=+=+=
−
−
( ) ( )
m230.0
10201091.43920241.0442
4
2
4
3
2
13
4
4433
=
×+×−−=+
+=+
−
x
xx
VTVT
m1091.4m230.0 3
34
−
×−=−= xxh m225.0=h

Edition
2 - 91
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 α=40o
,
find (a) the speed of B (b) the tension in the rope.
SOLUTION:
• This is a multiple step problem.
Formulate your overall approach.
• Use work-energy to find the
velocity of the block just
before impact
• Use conservation of
momentum to determine
the speed of ball B after
the impact
• Use work energy to find
the velocity at α
• Use Newton’s 2nd
Law to
find tension in the rope

Edition
2 - 92
Given: mA= 2-kg mB= 1-kg,
k= 800 N/m, µA =0.2, e= 0.8
Find (a) vB (b) Trope
• Use work-energy to find the velocity of
the block just before impact
Determine the friction force acting on the block A
cos 0AN m g θ− =
cos
(2)(9.81)cos20
18.4368 N
(0.2)(18.4368)
3.6874 N
A
f k
N m g
F N
θ
µ
=
= °
=
= =
=
0:yF∑ =
Sum forces in the y-direction
Solve for N

Edition
2 - 93
Set your datum, use work-energy to determine vA at impact.
1
2
Datumx
d
1 1 1 1 2 2 2 2( ) ( ) ( ) ( )e g e gT V V U T V V→+ + + = + +
Determine values for each term.
2 2
1 1 1
1 1
0, ( ) (800)(0.1) 4.00 J
2 2
eT V k x= = = =
1 1( ) ( )sin (2)(9.81)(1.6)sin 20 10.7367 Jg A AV m gh m g x d θ= = + = ° =
1 2 ( ) (3.6874)(1.6) 5.8998 JfU F x d→ = − + = − = −
2 2 2
2 2
1 1
(1)( ) 1.000 0
2 2
A A A AT m v v v V= = = =
2
1 1 1 2 2 2: 0 4.00 10.7367 5.8998 1.000 0AT V U T V v→+ + = + + + − = +
2 2 2
8.8369 m /sAv = 2.9727 m/sA =v
Substitute into the Work-Energy equation and solve for vA

Edition
2 - 94
• 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.
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 vθ θ′+ = +
( ) ( ) [( ) ( ) ]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
θ′ ′− = −
′ ′° − =
Use the relative velocity/coefficient
of restitution equation
(1) (2)
Solve (1) and (2) simultaneously
1.0382 m/s 3.6356 m/sA Bv v′ ′= =

Edition
2 - 95
• Use work energy to find the velocity at α
Set datum, use Work-Energy to determine vB at α= 40o
1
2
Datum
1 1 1 1 2 2 2 2( ) ( ) ( ) ( )e g e gT V V U T V V→+ + + = + +
Determine values for each term.
2
1 1
2
2 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 α
′= =
= = = −
2 2
1 1 2 2 2
2 2
2
2
2 2
1 1
: ( ) 0 (1 cos )
2 2
( ) 2 (1 cos )
(3.6356) (2)(9.81)(1 cos40 )
8.6274 m /s
B B B B
B
T V T V m v m v m g
v v gl
α
α
′+ = + + = + −
′= − −
= − − °
=
2 2.94 m/sv =
Substitute into the Work-Energy equation and solve for vA

Edition
2 - 96
• Use Newton’s 2nd
Law to find tension in the rope
• Draw your free-body and
kinetic diagrams
:n B nF m a∑ =
cos
( cos )
B B n
B n
T m g m a
T m a g
α
α
− =
= +
et
en
2
22
1.00 m
8.6274
8.6274 m/s
1.00
n
v
a
ρ
ρ
=
= = =
(1.0)(8.6274 9.81cos40 )T = + °
• Sum forces in the normal direction • Determine normal acceleration
16.14 NT =
• Substitute and solve

Edition
Concept Question
2 - 97
If the coefficient of restitution is
smaller than the 0.8 in the
problem, the tension T will be…
Smaller Bigger
If the rope length is smaller than
the 1 m in the problem, the
tension T will be…
Smaller Bigger
If the coefficient of friction is
smaller than 0.2 given in the
problem, the tension T will be…
Smaller Bigger
If the mass of A is smaller
than the 2 kg given in the
problem, the tension T will
be…
Smaller Bigger
Compare the following statement to the problem you just solved.

Edition
Summary
2 - 98
Forces and
Accelerations
Velocities and
Displacements
Velocities and
Time
Approaches to Kinetics Problems
Newton’s Second
Law (last chapter)
Work-Energy Impulse-
Momentum
2211 TUT =+ →GF ma=∑
  2
1
1 2
t
t
mv F dt mv+ =∫
vv v

Dynamics13lecture

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