Tenth Edition
CHAPTER
13
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. Self
California Polytechnic State University
Kinetics of Particles: Energy
and Momentum Methods
PART 2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
The potential energy stored at
the top of the roller coaster is
transferred to kinetic energy
as the cars descend.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
The elastic potential energy
stored in the trampoline is
transferred to kinetic energy
and
gravitational
potential
energy as the girl flies upwards.
2-2
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Potential Energy
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
Yes
No
Friction
Yes
No
Normal force
Yes
No
Spring force
Yes
No
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2-3
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Potential Energy

• Work of the force of gravity W,
U12  W y1  W y2
• Work is independent of path followed; depends
only on the initial and final values of Wy.
V g  Wy
 potential energy of the body with respect
to force of gravity.
U12  V g   Vg 
1
2
• Choice of datum from which the elevation y is
measured is arbitrary.
• Units of work and potential energy are the same:
Vg  Wy  N  m  J
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 4
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Potential Energy
• 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,
GMm GMm
U12 

r2
r1
• Potential energy Vg when the variation in the
force of gravity can not be neglected,
GMm
WR 2
Vg  

r
r
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 5
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Potential Energy
• Work of the force exerted by a spring depends
only on the initial and final deflections of the
spring,
U12  12 kx12  12 kx22
• The potential energy of the body with respect
to the elastic force,
Ve  12 kx 2
U12  Ve 1  Ve 2
• Note that the preceding expression for Ve is
valid only if the deflection of the spring is
measured from its undeformed position.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 6
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Conservation of Energy
• Work of a conservative force,
U12  V1  V2
• Concept of work and energy,
U12  T2  T1
• Follows that
T1  V1  T2  V2
E  T  V  constant
T1  0 V1  W
T1  V1  W
T2  12 mv22 
1W
2 g   W V2  0
2g
T2  V2  W
• When a particle moves under the action of
conservative forces, the total mechanical
energy is constant.
• 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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 7
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Motion Under a Conservative Central Force
• When a particle moves under a conservative central
force, both the principle of conservation of angular
momentum
r0 mv0 sin 0  rmv sin 
and the principle of conservation of energy
T0  V0  T  V
1 mv 2
0
2

GMm 1 2 GMm
 2 mv 
r0
r
may be applied.
• Given r, the equations may be solved for v and j.
• At minimum and maximum r, j  90o. Given the
launch conditions, the equations may be solved for
rmin, rmax, vmin, and vmax.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 8
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.6
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.
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.
• Solve for the kinetic energy and velocity
at 2.
If the collar is released from rest at
position 1, determine its velocity after
it has moved 6 in. to position 2.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 9
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of energy between
positions 1 and 2.
Position 1: Ve  12 kx12  12 3 lb in.8 in.  4 in.2  24 in.  lb
V1  Ve  Vg  24 in.  lb  0  2 ft  lb
T1  0
Position 2: Ve  12 kx22  12 3 lb in.10 in.  4 in.2  54 in.  lb
Vg  Wy  20 lb  6 in.  120 in.  lb
V2  Ve  Vg  54  120  66 in.  lb  5.5 ft  lb
T2  12 mv22 
1 20 2
v2  0.311v22
2 32.2
Conservation of Energy:
T1  V1  T2  V2
0  2 ft  lb  0.311v22  5.5 ft  lb
v2  4.91ft s 
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 10
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.7
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.
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
• 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.
13 - 11
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.7
SOLUTION:
• Setting the force exerted by the loop to zero, solve for the
minimum velocity at D.
2
   Fn  man : W  man
mg  m vD
r
2
vD
 rg  2 ft 32.2 ft s   64.4 ft 2 s 2
• Apply the principle of conservation of energy between
points A and D.
V1  Ve  Vg  12 kx2  0  12 36 lb ft x 2  18 x 2
T1  0
V2  Ve  Vg  0  Wy  0.5 lb 4 ft   2 ft  lb
2
T2  12 mvD



1 0.5 lb
2 2
64
.
4
ft
s  0.5 ft  lb
2 32.2 ft s 2
T1  V1  T2  V2
0  18 x 2  0.5  2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
x  0.3727 ft  4.47 in.
13 - 12
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.9
SOLUTION:
• For motion under a conservative central
force, the principles of conservation of
energy and conservation of angular
momentum may be applied simultaneously.
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.
• 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
Determine (a) the maximum altitude
insertion angle error.
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
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 13
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.9
• 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:
TA  VA  TA  VA
1 mv 2
0
2

GMm 1 2 GMm
 2 mv1 
r0
r1
Conservation of angular momentum:
r
r0mv0  r1mv1
v1  v0 0
r1
Combining,
2

r0 2GM
1 v 2 1  r0   GM 1  r0 
1




2 0
2
r0  r1 
r1 r0v02
 r1 
r0  6370 km  500 km  6870 km
v0  36900 km h  10.25  106 m s


2
GM  gR 2  9.81m s 2 6.37  106 m  398  1012 m3 s 2
r1  60.4  106 m  60400 km
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 14
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.9
• 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:
GMm
1 mv 2  GMm  1 mv 2
T0  V0  TA  VA

0
max
2
2
r0
rmin
Conservation of angular momentum:
r
r0mv0 sin 0  rmin mvmax
vmax  v0 sin 0 0
rmin
Combining and solving for sin j0,
sin 0  0.9801
j 0  90  11.5
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
allowable error  11.5
13 - 15
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
ASSIGNMENT: Problem 1
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.
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.
©
2013 The McGraw-Hill Companies, Inc. All rights reserved.
ANSWER:
N D  1260 lbs
• 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.
2 - 16
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Impulsive Motion
The thrust of a rocket acts
over a specific time period
to give the rocket linear
momentum.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
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.
2 - 17
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Principle of Impulse and Momentum
• From Newton’s second law,
 d 

F  mv 
mv  linear momentum
dt


Fdt  d mv 
t2



F
dt

m
v

m
v

2
1
t1
• Dimensions of the impulse of
a force are
force*time.
• Units for the impulse of a
force are


N  s  kg  m s  s  kg  m s
2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
t2


 Fdt  Imp12  impulse of the force F
t1


mv1  Imp12  mv2
• 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.
13 - 18
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Impulsive Motion
• 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,



mv1   F t  mv2
• 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

Ft is small and therefore, may be
neglected – an example of this is the weight
of the baseball.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 19
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.10
SOLUTION:
• Apply the principle of impulse and
momentum. The impulse is equal to the
product of the constant forces and the
time interval.
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 20
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.10
SOLUTION:
• Apply the principle of impulse and
momentum.


mv1   Imp12  mv2
Taking components parallel to the
incline,
mv1  W sin 5t  Ft  0
 4000 

88 ft s   4000sin 5t  1500t  0
 32.2 
t  9.49 s
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 21
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.11
SOLUTION:
• Apply the principle of impulse and
momentum in terms of horizontal and
vertical component equations.
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 22
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.11
SOLUTION:
• Apply the principle of impulse and momentum in
terms of horizontal and vertical component equations.


mv1  Imp12  mv2
x component equation:
 mv1  Fx t  mv2 cos 40
4 16
80  Fx 0.15  4 16 120 cos 40
32.2
32.2
Fx  89 lb

y component equation:
y
0  Fy t  mv2 sin 40
x
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
4 16
120 cos 40
32.2
Fy  39.9 lb



F  89 lb i  39.9 lb  j , F  97.5 lb
Fy 0.15 
13 - 23
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.12
SOLUTION:
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
• Apply the principle of impulse and
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.
13 - 24
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.12
SOLUTION:
• Apply the principle of impulse and momentum to the package-cart
system to determine the final velocity.
y
x




m pv1   Imp12  m p  mc v2
x components:


m p v1 cos 30  0  m p  mc v2
10 kg 3 m/s cos 30  10 kg  25 kg v2
v2  0.742 m/s
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 25
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.12
• Apply the same principle to the package alone to determine the impulse
exerted on it from the change in its momentum.
y
x


m pv1   Imp12  m pv2
x components:
m p v1 cos 30  Fx t  m p v2
10 kg 3 m/s cos 30  Fx t  10 kg v2
y components:
Fx t  18.56 N  s
 m p v1 sin 30  Fy t  0
 10 kg 3 m/s sin 30  Fy t  0
 Imp12



 Ft   18.56 N  s i  15 N  s  j
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Fy t  15 N  s
Ft  23.9 N  s
13 - 26
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.12
To determine the fraction of energy lost,
T1  12 m p v12 
T2 
1
2
1
2
10 kg  3m s   45 J
 mp  mc  v22 
2
1
2
10 kg  25 kg  0.742 m s   9.63 J
2
T1  T2 45 J  9.63 J

 0.786
T1
45 J
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 27
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
ASSIGNMENT: Problem 2
SOLUTION:
• Draw impulse and momentum diagrams
of the jumper.
• Apply the principle of impulse and
momentum to the jumper to determine
the force exerted on the foot.
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
ANSWER:
Favg  1.295W i  6.21W j
2 - 28
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
ASSIGNMENT: Problem 2 (Concept question)
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
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 29
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
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
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 30
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Impact
• 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.
Direct Central 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.
Oblique Central Impact
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 31
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Direct Central Impact
• 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,
m Av A  mB v B  mB vB  mB vB
• A second relation between the final
velocities is required.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 32
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Direct Central Impact
e  coefficient of restitution
• Period of deformation: m Av A   Pdt  m Au
Rdt u  vA


 Pdt v A  u
0  e 1
• Period of restitution:
m Au   Rdt  m AvA
• A similar analysis of particle B yields
vB  u
e
u  vB
• Combining the relations leads to the desired
second relation between the final velocities.
vB  vA  ev A  v B 
• Perfectly plastic impact, e = 0: vB  vA  v
m Av A  mB v B  m A  mB v
• Perfectly elastic impact, e = 1:
Total energy and total momentum conserved.
vB  vA  v A  v B
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 33
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Oblique Central Impact
• 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.
• Normal component of total
momentum of the two particles is
conserved.
• Normal components of relative
velocities before and after impact
are related by the coefficient of
restitution.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
v A t  vA t
v B t  vB t
m A v A n  mB v B n  m A vA n  mB vB n
vB n  vA n  ev A n  v B n 
13 - 34
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Oblique Central Impact
• Block constrained to move along horizontal
surface.


• Impulses from internal forces F and  F
along the n axis and from external force Fext
exerted by horizontal surface and directed
along the vertical to the surface.
• Final velocity of ball unknown in direction and
magnitude and unknown final block velocity
magnitude. Three equations required.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 35
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Oblique Central Impact
• Tangential momentum of ball is
conserved.
v B t  vB t
• Total horizontal momentum of block
and ball is conserved.
m A v A   mB v B  x  m A vA   mB vB  x
• Normal component of relative
velocities of block and ball are related
by coefficient of restitution.
vB n  vA n  ev A n  v B n 
• Note: Validity of last expression does not follow from previous relation for
the coefficient of restitution. A similar but separate derivation is required.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 36
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Problems Involving Energy and Momentum
• 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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 37
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.14
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.
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
• 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.
13 - 38
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.14
SOLUTION:
• Resolve ball velocity into components parallel and
perpendicular to wall.
vn  v cos 30  0.866v
vt  v sin 30  0.500v
• Component of ball momentum tangential to wall is conserved.
vt  vt  0.500v
t
n
• Apply coefficient of restitution relation with zero wall
velocity.
0  vn  evn  0
vn  0.90.866v   0.779v



v   0.779v n  0.500v t
 0.779 
v  0.926v tan 1
  32.7
 0.500 
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 39
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.15
SOLUTION:
• Resolve the ball velocities into components
normal and tangential to the contact plane.
• Tangential component of momentum for
each ball is conserved.
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
• 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.
13 - 40
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.15
SOLUTION:
• Resolve the ball velocities into components normal and
tangential to the contact plane.
v A n  v A cos 30  26.0 ft
s
vB n  vB cos 60  20.0 ft
s
v A t  v A sin 30  15.0 ft
s
vB t  vB sin 60  34.6 ft
s
• Tangential component of momentum for each ball is
conserved.
vA t  v A t  15.0 ft
s
vB t  vB t
 34.6 ft s
• Total normal component of the momentum of the two
ball system is conserved.
mA v A n  mB vB n  mA vA n  mB vB n
m26.0  m 20.0  mvA n  mvB n
vA n  vB n  6.0
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 41
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.15
• The normal relative velocities of the balls are related by the
coefficient of restitution.
vA n  vB n  ev A n  vB n 
 0.9026.0   20.0  41.4
• Solve the last two equations simultaneously for the normal
velocities of the balls after the impact.
vA n  17.7 ft s
vB n  23.7 ft s



v A  17.7t  15.0n
n
 15.0 
vA  23.2 ft s tan 1
  40.3
 17.7 



vB  23.7t  34.6n
t
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
 34.6 
vB  41.9 ft s tan 1
  55.6
23
.
7


13 - 42
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.16
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.
Ball B is hanging from an inextensible
• The relative velocities along the line of
cord. An identical ball A is released
action before and after the impact are
from rest when it is just touching the
related by the coefficient of restitution.
cord and acquires a velocity v0 before
striking ball B. Assuming perfectly
• Solve the last two expressions for the
elastic impact (e = 1) and no friction,
velocity of ball A along the line of action
determine the velocity of each ball
and the velocity of ball B which is
immediately after impact.
horizontal.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 43
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.16
r
 0.5
2r
  30
sin  
SOLUTION:
• Determine orientation of impact line of action.
• The momentum component of ball A
tangential to the contact plane is
conserved.



mv A  Ft  mv A
mv0 sin 30  0  mvA t
vA t  0.5v0
• The total horizontal (x component)
momentum of the two ball system is
conserved.




mv A  Tt  mv A  mvB
0  mvA t cos 30  mvA n sin 30  mvB
0  0.5v0 cos 30  vA n sin 30  vB
0.5vA n  vB  0.433v0
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 44
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.16
• The relative velocities along the line of action before
and after the impact are related by the coefficient of
restitution.
vB n  vA n  ev A n  vB n 
vB sin 30  vA n  v0 cos 30  0
0.5vB  vA n  0.866v0
• 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.
vA n  0.520v0
vB  0.693v0



v A  0.5v0t  0.520v0n
vA  0.721v0
  tan 1
0.52 
  46.1
 0.5 
  46.1  30  16.1
vB  0.693v0 
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 45
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.17
SOLUTION:
• Apply the principle of conservation of
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.
A 30 kg block is dropped from a height
of 2 m onto the the 10 kg pan of a
• Apply the principle of conservation of
spring scale. Assuming the impact to be
energy to determine the maximum
perfectly plastic, determine the
deflection of the spring.
maximum deflection of the pan. The
constant of the spring is k = 20 kN/m.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 46
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.17
SOLUTION:
• Apply principle of conservation of energy to
determine velocity of the block at instant of impact.
T1  0
V1  WA y  30 9.812   588 J
T2  12 m A v A 22  12 30 v A 22
V2  0
T1  V1  T2  V2
0  588 J  12 30 v A 22  0
v A 2  6.26 m s
• Determine velocity after impact from requirement that
total momentum of the block and pan is conserved.
mA v A 2  mB vB 2  mA  mB v3
306.26  0  30  10v3
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
v3  4.70 m s
13 - 47
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.17
• Apply the principle of conservation of energy to
determine the maximum deflection of the spring.
T3  12 m A  mB v32  12 30  10 4.7 2  442 J
V3  Vg  Ve
0
1 kx 2
2 3

1
2
20  10 4.91 10 
3
3 2
 0.241 J
T4  0
Initial spring deflection due to
pan weight:
x3 
WB 10 9.81
3


4
.
91

10
m
3
k
20  10
V4  Vg  Ve  WA  WB  h   12 kx42


 392x4  4.91  103   12 20  103 x42
 392 x4  x3   12 20  103 x42
T3  V3  T4  V4

 

442  0.241  0  392 x4  4.91  103  12 20  103 x42
x4  0.230 m
h  x4  x3  0.230 m  4.91  103 m
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
h  0.225 m
13 - 48
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
ASSIGNMENT: Problem 3
SOLUTION:
• This is a multiple step problem.
Formulate your overall approach.
• Use work-energy to find the
velocity of the block just
before impact
A 2-kg block A is pushed up against a spring • Use conservation of
compressing it a distance x= 0.1 m. The block is
momentum to determine
then released from rest and slides down the 20º
the speed of ball B after
incline until it strikes a 1-kg sphere B, which is
the impact
suspended from a 1 m inextensible rope. The
spring constant k=800 N/m, the coefficient of • Use work energy to find
the velocity at 
friction between A and the ground is 0.2, the
distance A slides from the unstretched length of
nd
the spring d=1.5 m, and the coefficient of • Use Newton’s 2 Law to
find tension in the rope
restitution between A and B is 0.8. When =40o,
find (a) the speed of B (b) the tension in the rope.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 49
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
ASSIGNMENT: Problem 3 (Concept question)
Compare the following statement to the problem you just solved.
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
If the mass of A is smaller
than the 2 kg given in the
problem, the tension T will
be…
Smaller
Bigger
Bigger
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 50
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Summary
Approaches to Kinetics Problems
Forces and
Accelerations
Velocities and
Displacements
Velocities and
Time
Newton’s Second
Law (last chapter)
Work-Energy
ImpulseMomentum
 F  ma
G
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
T1  U12  T2
t2
mv1   F dt  mv2
t1
2 - 51