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Chapter 13
Kinetics of Particles:
Energy and Momentum
Methods
1
Kinetics of Particles
 Newton’s Second Law
(บทที่ 12)
 Work and Energy Methods (บทที่ 13)
 Impulse and Momentum Methods (บทที่ 13)
2
Work and Energy Methods
3
Introduction
• Previously, problems dealing with the motion of particles
 were

solved through the fundamental equation of motion, F  ma.
Current chapter introduces two additional methods of analysis.
• Method of work and energy: directly relates force, mass,
velocity and displacement.
• Method of impulse and momentum: directly relates force,
mass, velocity, and time.
4
Work of a Force

• Differential vector dr is the particle displacement.
• Work of the force is
 
dU  F  dr
 F ds cos 
 Fx dx  Fy dy  Fz dz
• Work is a scalar quantity, i.e., it has magnitude and
sign but not direction.
• Dimensions of work are length  force. Units are
1 J  joule   1 N 1 m 
งาน = แรงในทิศทางการเคลื่อนทีข่ องวัตถุ x ระยะทางทีว่ ัตถุนั้นเคลื่อนทีไ่ ด้
แรงทีม่ ีทศิ ทางสวนกับทิศการเคลือ่ นทีข่ องวัตถุ จะทาให้ งานมีค่าเป็ นลบ
5
Work of a general force
• Work of a force during a finite displacement,
U12 
A2 

F

d
r

A1


s2
s2
s1
s1
 F cos ds   Ft ds
A2
 Fx dx  Fy dy  Fz dz 
A1
• Work is represented by the area under the
curve of Ft plotted against s.
6
Work of the force of gravity
• Work of a constant force in rectilinear motion,
U12  F cos  x
• Work of the force of gravity,
dU  Fx dx  Fy dy  Fz dz
 W dy
y2
U12    W dy
y1
 W  y 2  y1   W y
• Work of the weight is equal to product of
weight W and vertical displacement y.
• Work of the weight is positive when y < 0,
i.e., when the weight moves down.
7
Work of the force exerted by spring
• Magnitude of the force exerted by a spring is
proportional to deflection,
F  kx
k  spring constant N/m or lb/in.
• Work of the force exerted by spring,
dU   F dx  kx dx
x2
U12    kx dx  12 kx12  12 kx22
x1
• 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,
U12   12  F1  F2  x
8
Work of a gravitational force
Work of a gravitational force (assume particle M
occupies fixed position O while particle m follows path
shown),
Mm
dU   Fdr  G 2 dr
r
r2
Mm
r1
r2
U12    G
dr  G
Mm
Mm
G
r2
r1
9
Principle of Work & Energy

• Consider a particle of mass m acted upon by force F
dv
Ft  mat  m
dt
dv ds
dv
m
 mv
ds dt
ds
F t ds  mv dv
• Integrating from A1 to A2 ,
s2
v2
s1
v1
 Ft ds  m  v dv  12 mv2  12 mv1
U12  T2  T1
2
2
T  12 mv 2  kinetic energy
T1  U12  T2
• Units of work and kinetic energy are the same:
2
m
 m


2
T  12 mv  kg    kg 2 m  N  m  J
s
 s 
10
Advantage of work and energy methods
• Velocity 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.
11
Applications of the Principle of Work
• Wish to determine velocity of pendulum bob
and Energy
at A . Consider work & kinetic energy.
2

• Force P acts normal to path and does no
work.
T1  U12  T2
0  Wl 
1W 2
v2
2g
v2  2 gl
12
Applications of the Principle of Work
• Principle of work and energy cannot be
and Energy
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 ,
 Fn  m an
v2  2 gl
W v22
P W 
g l
W 2 gl
P W 
 3W
g l
13
Power and Efficiency
• Power  rate at which work is done.
 
dU F  dr


dt
dt
 
 F v
• Dimensions of power are work/time or force*velocity.
Units for power are
J
m
ft  lb
1 W (watt)  1  1 N 
or 1 hp  550
 746 W
s
s
s
•   efficiency
output work

input work
power output

power input
14
Sample Problem 13.1
SOLUTION:
• Evaluate the change in kinetic energy.
• Determine the distance required for the
work to equal the kinetic energy change.
An automobile weighing 4000 N is
driven down a 5o incline at a speed of
88 m /s when the brakes are applied
causing a constant total breaking force
of 1500 N.
Determine the distance traveled by the
automobile as it comes to a stop.
15
Sample Problem 13.1
SOLUTION:
• Evaluate the change in kinetic energy.
v1  88 m s
T1  12 mv12  12 4000 / 9.8188  1578797 N  m
2
v2  0
T2  0
• Determine the distance required for the work
to equal the kinetic energy change.
U12   1500 N x  4000 N sin 5x
 1151N x
T1  U12  T2
1578797 N  m  1151 N x  0
x  1371.67 m
16
Sample Problem 13.2
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.
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 mk = 0.25 and that the
pulley is weightless and frictionless.
17
Sample Problem 13.2
SOLUTION:
• Apply the principle of work and energy separately
to blocks A and B.


W A  200 kg  9.81m s 2  1962 N
FA  m k N A  m k W A  0.251962 N   490 N
T1  U12  T2 :
0  FC 2 m   FA 2 m   12 m Av 2
FC 2 m   490 N 2 m   12 200 kg v 2


WB  300 kg  9.81m s 2  2940 N
T1  U12  T2 :
0  Fc 2 m   WB 2 m   12 m B v 2
 Fc 2 m   2940 N 2 m   12 300 kg v 2
18
Sample Problem 13.2
• When the two relations are combined, the work of the
cable forces cancel. Solve for the velocity.
FC 2 m   490 N 2 m   12 200 kg v 2
 Fc 2 m   2940 N 2 m   12 300 kg v 2
2940 N 2 m   490 N 2 m   12 200 kg  300 kg v 2
4900 J  12 500 kg v 2
v  4.43 m s
19
Sample Problem 13.3
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.
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 • Apply the principle of work and energy
for the rebound of the package. The
compressed 120 mm. The package has a
only unknown in the relation is the
velocity of 2.5 m/s in the position shown
and the maximum deflection of the spring velocity at the final position.
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.
20
Sample Problem 13.3
SOLUTION:
• Apply principle of work and energy between initial
position and the point at which spring is fully compressed.
T1  12 mv12  12 60 kg 2.5 m s 2  187.5 J
U12  f
  m kW x

T2  0

  m k 60 kg  9.81m s 2 0.640 m   377 J m k
Pmin  kx0  20 kN m 0.120 m   2400 N
Pmax  k  x0  x   20 kN m 0.160 m   3200 N
U12 e   12 Pmin  Pmax x
  12 2400 N  3200 N 0.040 m   112.0 J
U12  U12  f  U12 e  377 J m k  112 J
T1  U12  T2 :
187.5 J - 377 J m k  112 J  0
mk  0.20
21
Sample Problem 13.3
• Apply the principle of work and energy for the rebound
of the package.
T2  0
T 3 12 mv32  12 60kg v32
U 23  U 23  f  U 23 e  377 J m k  112 J
 36.5 J
T2  U 23  T3 :
0  36.5 J  12 60 kg v32
v3  1.103 m s
22
Sample Problem 13.4
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.
A 2000 N car starts from rest at point 1
• Apply principle of work and energy to
and moves without friction down the
determine velocity at point 3.
track shown.
• Apply Newton’s second law to find
Determine:
minimum radius of curvature at point 3
such that a positive normal force is
a) the force exerted by the track on
exerted by the track.
the car at point 2, and
b) the minimum safe value of the
radius of curvature at point 3.
23
Sample Problem 13.4
SOLUTION:
• Apply principle of work and energy to determine
velocity at point 2.
T1  0
T2  12 mv22 
U12  W 40 m 
T1  U12  T2 :
1W 2
v2
2 g
0  W 40 m  

v22  240 m g  240 m  9.81 m s 2
1W 2
v2
2 g

v2  28.01 m s
• Apply Newton’s second law to find normal force by
the track at point 2.
   Fn  m an :
W v22 W 240 m g
 W  N  m an 

g  2 g 20 m
N  5W
N  10000N
24
Sample Problem 13.4
• Apply principle of work and energy to determine
velocity at point 3.
T1  U13  T3
0  W 25 m  

1W 2
v3
2 g
v32  225 m g  225 m  9.81 m s 2

v3  22.15 m s
• 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.
   Fn  m an :
W  m an
W v32 W 225 m g


g 3 g
3
3  50 m
25
Sample Problem 13.5
D
SOLUTION:
Force exerted by the motor
cable has same direction as
the dumbwaiter velocity.
Power delivered by motor is
equal to FvD, vD = 8 m/s.
The dumbwaiter D and its load have a
combined weight of 600 N, while the
counterweight C weighs 800 N.
• In the first case, bodies are in uniform
motion. Determine force exerted by
motor cable from conditions for static
equilibrium.
Determine the power delivered by the
electric motor M when the dumbwaiter
(a) is moving up at a constant speed of
8 m/s and (b) has an instantaneous
velocity of 8 m/s and an acceleration of
2.5 m/s2, both directed upwards.
• In the second case, both bodies are
accelerating. Apply Newton’s
second law to each body to
determine the required motor cable
force.
26
Sample Problem 13.5
• In the first case, bodies are in uniform motion (a = 0).
Determine force exerted by motor cable from
conditions for static equilibrium.
Free-body C:
   Fy  0 :
2T  800 N  0
T  400 N
Free-body D:
   Fy  0 : F  T  600 N  0
F  600 N  T  600 N  400 N  200 N
Power  Fv D  200 N 8 m s 
 1600 N  m s
Power  1600 N  m s 
27
Sample Problem 13.5
• In the second case, both bodies are accelerating. Apply
Newton’s second law to each body to determine the required
motor cable force.
a D  2.5 m s 2 
aC   12 a D  1.25 m s 2 
Free-body C:
   Fy  mC aC :
800  2T 
800
1.25
10
T  100 N
Free-body D:
   Fy  m D a D :
600
2.5
10
F  100  600  150
F  650 N
F  T  600 
Power  Fv D  650 N 8 m s   5200 N  m s
Power  5200 N  m s 
28
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.
 1  Vg 2
U12  V g
• 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
29
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,
Vg  
GMm
r
30
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.
31
Conservative Forces
• Concept of potential energy can be applied if the
work of the force is independent of the path
followed by its point of application.
U12  V  x1, y1, z1   V  x2 , y2 , z2 
Such forces are described as conservative forces.
• For any conservative force applied on a closed path,
 
 F  dr  0
• Elementary work corresponding to displacement
between two neighboring points,
dU  V  x, y, z   V  x  dx, y  dy, z  dz 
 dV  x, y, z 
 V
V
V 
Fx dx  Fy dy  Fz dz  
dx 
dy 
dz 
y
z 
 x

 V  V  V  
F  
i
j
k   grad V

x

y

z


32
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 
T2  V2  W
1W
2 g   W V2  0
2g
• 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.
33
Motion Under a Conservative Central
• When a particle moves under a conservative central
Force
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.
34
Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of
energy between positions 1 and 2.
A 20 N collar slides without friction
along a vertical rod as shown. The
spring attached to the collar has an
undeflected length of 4 cm and a
constant of 3 N/cm.
• 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.
If the collar is released from rest at
position 1, determine its velocity after
it has moved 6 cm. to position 2.
35
Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of energy between
positions 1 and 2.
Position 1: Ve  12 kx12  12 3N / cm 8 cm  4 cm   24 N  cm
2
V1  Ve  Vg  24 N  cm  0
T1  0
Position 2: Ve  1 kx22  1 3N / cm 10 cm  4 cm 2  54 N  cm
2
2
Vg  Wy  20 N  6 cm   120 N  cm
V2  Ve  Vg  54  120  66 N  cm
T2 
1 mv 2
2
2
1 20 2

v2
2 10
Conservation of Energy:
T1  V1  T2  V2
0  24 N  cm  v22  66 N  cm
v2  9.5 m s 
36
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 N 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.
• 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.
37
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 v D
r


2
vD
 rg  2 m  10 m s 2  20 m 2 s 2
• Apply the principle of conservation of energy between
points A and D.
V1  Ve  V g  12 kx 2  0  12 3N / cm x 2  1.5 x 2
T1  0
V2  Ve  Vg  0  Wy  0.5 N 4 m   2 N  m
2
T2  12 mvD



1 0.5 N
2 2
20
m
s  0.5 N  m
2
2 10 m s
T1  V1  T2  V2
0  1.5 x 2  0.5  2
x  1.3 m
38
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
39
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
40
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
j0  90  11.5
allowable error  11.5
41
Impulse and Momentum Methods
42
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 2  s  kg  m s
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.
43
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.
44
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 N is
driven down a 5o incline at a speed of
88 m/s when the brakes are applied,
causing a constant total braking force of
1500 N.
Determine the time required for the
automobile to come to a stop.
45
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 m s   4000sin 5t  1500t  0
 10 
t  30.57 s
46
Sample Problem 13.11
SOLUTION:
• Apply the principle of impulse and
momentum in terms of horizontal and
vertical component equations.
A 0.5 kg baseball is pitched with a
velocity of 80 m/s. After the ball is hit
by the bat, it has a velocity of 120 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.
47
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
0.5
80  Fx 0.15  0.5 120 cos 40
10
10
Fx  57.3 N

y component equation:
0  Fy t  mv2 sin 40
y
x
0.5
120 cos 40
10
Fy  30.64 N



F  57.3 N  i  30.64 N  j , F  64.9 N
Fy 0.15 
48
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.
• 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.
49
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
50
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
Fy t  15 N  s
Ft  23.9 N  s
51
Sample Problem 13.12
To determine the fraction of energy lost,
T1  12 m p v12  12 10 kg 3 m s 2  45 J


T1  12 m p  mc v22  12 10 kg  25 kg 0.742 m s 2  9.63 J
T1  T2 45 J  9.63 J

 0.786
T1
45 J
52
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
53
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.
54
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
• Combining the relations leads to the desired
second relation between the final velocities.
• Perfectly plastic impact, e = 0: vB  vA  v
• Perfectly elastic impact, e = 1:
Total kinetic energy and total momentum
conserved.
vB  u
u  vB
vB  vA  ev A  v B 
e
m Av A  mB v B  m A  mB v
vB  vA  v A  vB
1
1
1
1
mAv A2  mB vB2  mAv'A2  mB v'B2
2
2
2
2
55
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.
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 
56
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.
57
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.
58
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.
• 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.
59
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 
60
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.
• 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.
61
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 m s v A t  v A sin 30  15.0 m
vB n  vB cos 60  20.0 m s vB t  vB sin 60  34.6 m
• Tangential component of momentum for each ball is
conserved.
vA t  v A t  15.0 m s
vB t  vB t  34.6 m 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
62
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 m s
vB n  23.7 m s



v A  17.7t  15.0n
 15.0 
vA  23.2 m s tan 1 
  40.3
 17.7 




v B  23.7t  34.6n
n
t
 34.6 
vB  41.9 m s tan 1 
  55.6
23
.
7


63
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
action before and after the impact are
cord. An identical ball A is released
related by the coefficient of restitution.
from rest when it is just touching the
cord and acquires a velocity v0 before • Solve the last two expressions for the
striking ball B. Assuming perfectly
velocity of ball A along the line of action
elastic impact (e = 1) and no friction,
and the velocity of ball B which is
determine the velocity of each ball
horizontal.
immediately after impact.
64
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
65
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 
66