5.4. CONSIDERATION OF ENERGY AND MOMENTUM 71
direction of
*
v
0
B
. is means that one requires three independent equations to solve for the three
unknowns. To this end, one considers the following three situations.
First, since there is no friction between block A and ball B, the component along the t -axis
of the momentum of ball B is conserved. us, the t component of the velocity of ball B
remains unaltered. is enables one to write
*
v
B
t
D
*
v
0
B
t
: (5.21)
Second, the component along the x-axis of the total momentum of block A and ball B is
conserved. is means
m
A
v
A
C m
B
.
v
B
/
x
D m
A
v
0
A
C m
B
v
0
B
x
: (5.22)
ird, the component along the n-axis of the relative velocity of block A and ball B after
impact can be obtained by multiplying the coefficient of restitution and the n component
of the relative velocity before impact. at is
*
v
0
B
n
*
v
0
A
n
D e
*
v
A
n
*
v
B
n
: (5.23)
It may be appropriate to note that the validity of Equation (5.23) cannot be established via
a simple extension of the derivation for the direct central impact case. Without presenting
the proof it suffices to state that Equation (5.23) is valid to cases in which the motion of
one particle or both particles are constrained.
Equations (5.21), (5.22), and (5.23) can be applied to solve for the three unknowns (two
unknown magnitudes and one unknown phase or direction) of the velocities
*
v
0
A
and
*
v
0
B
after impact.
5.4 CONSIDERATION OF ENERGY AND MOMENTUM
Up to this stage three basic methods for the solution of problems of particles have been intro-
duced. ese methods are as follows.
Direct application of Newton’s second law of motion,
*
F D m
*
a :
In this method the principles of kinematics are applied such that one can find position,
velocity from
*
a at any time.
Method of work and energy which relates force, mass, velocity, and displacement.
72 5. IMPULSE, MOMENTUM, AND IMPACT OF PARTICLES
Method of impulse and momentum which has been presented in this chapter.
Of course, the choice of one or more of these methods is solely dependent of the type of
problem one is confronted with. For example, there are situations in which the solution of
the problem at hand may require the combined use of the foregoing methods. Specifically,
for example, in the case of two pendulums P and Q are hanging from the ceiling. When
mass m
P
of pendulum P is released from its horizontal position P
1
to hit mass m
Q
of
pendulum Q at rest, as illustrated in Figure 5.7a, the solution of this problem requires the
following three phases.
(a) In the first phase, pendulum P travels from position P
1
to P
2
, as shown in Figure 5.7a.
e principle of conservation of energy can be applied to find the velocity
*
v
P
2
of
pendulum P at position P
2
.
(b) In the second phase, pendulum P impacts pendulum Q, as shown in Figure 5.7b. Since
there is no externally applied impulse the total momentum of the two pendulums
is conserved. Further, the velocities
*
v
P
3
and
*
v
Q
3
of the two pendulums after
impact can be determined by applying the relation between their relative velocities.
(c) In the third and final phase, pendulum Q travels from position Q
3
to Q
4
, as shown in
Figure 5.7c. e principle of conservation of energy can be applied to find the max-
imum vertical distance y
4
that can be achieved by pendulum Q. e corresponding
angle can thus be found by trigonometry.
Example 5.1
A much simplified model of explosion of a grenade conceptually consists of two hemispheres
that are connected by an inextensible string holding the spring under compression, as shown
in Figure 5.8a in which m
A
D 2:0 kg, m
B
D 1:0 kg. Note that the spring is not attached to the
hemispheres that have unequal masses. It is known that the potential energy of the compressed
spring is 100 J or N.m and the whole system has an initial velocity
*
v
o
. When the string is cut,
which simulates the explosion event, the hemispheres fly apart and the angle between the axis of
the system and the horizontal is D 61
ı
. If the magnitude of the initial velocity is v
o
D 100 m/s,
determine the resulting velocity of each hemisphere.
Solution:
Let the frame Oxy moving with the mass center (z-axis being perpendicular to the Oxy plane)
while the x-axis and y-axis along the horizontal and vertical directions with the origin O at the
mass center, as indicated in Figure 5.8b. Note that since the frame is moving with the whole
system therefore, the velocity of each hemisphere is zero before the string is cut. is means that
the kinetic energy, immediately after the string is cut, is zero.
5.4. CONSIDERATION OF ENERGY AND MOMENTUM 73
Impact:
Conservation of Momentum
Relative Velocities
(a) (b)
(c) Conservation of Energy
Conservation of Energy
(
⃑
P
)
1
= 0
(
⃑
P
)
2
(
⃑
Q
)
2
= 0
(
⃑
Q
)
3
(
⃑
P
)
3
P
2
P
1
Q
1
Q
2
P
3
P
4
L
L
L
L
θ
y
4
Q
4
Q
3
Figure 5.7: ree phases in the solution of two impacting pendulums: (a) conservation of energy
in first phase, (b) impact of two pendulums, and (c) conservation of energy in third phase.
A
B
A
B
A
B
=
θ
θ
⃑
o
⃑
o
m
B
⃑'
B
m
A
⃑'
A
x
y
O
(a)
(b)
Figure 5.8: (a) Model of explosion of a grenade and (b) FBD with frame Oxy.
74 5. IMPULSE, MOMENTUM, AND IMPACT OF PARTICLES
Given data
m
A
D 2:0 kg, m
B
D 1:0 kg, V
1
D 100 N.m, and V
2
D 0, because there is no strain energy when
the string is cut.
Conservation of momentum
0 C 0 D m
A
*
v
0
A
m
B
*
v
0
B
:
Note that the negative sign on the rhs indicates that immediately after the string is cut the two
hemispheres travel in opposite directions. erefore,
*
v
0
A
D
m
B
m
A
*
v
0
B
:
Conservation of energy
T
1
C V
1
D T
2
C V
2
) 0 C 100 D
1
2
m
A
v
0
A
2
C
1
2
m
B
v
0
B
2
C 0
) 100 D
1
2
m
A
m
B
m
A
*
v
0
B
2
C
1
2
m
B
v
0
B
2
) 100 D
m
B
.
m
A
C m
B
/
2m
A
v
0
B
2
) v
0
B
D
s
200m
A
m
B
.
m
A
C m
B
/
:
Substituting for the given data, one obtains
v
0
B
D
s
200
.
2
/
1
.
2 C 1
/
m
s
D 20
r
1
3
m
s
; v
0
A
D 40
r
1
3
m
s
:
*
v
0
A
D 40
r
1
3
m
s
61
ı
;
*
v
0
B
D 20
r
1
3
m
s
61
ı
:
Velocities of A and B
e absolute velocities of A and B are
*
v
0
A
F
D
Œ
100 m/s !
C
"
40
r
1
3
m
s
61
ı
#
D 113:02
m
s
10:29
ı
;
*
v
0
B
F
D
Œ
100 m/s !
C
"
20
r
1
3
m
s
61
ı
#
D 94:94
m
s
6:07
ı
:
5.4. CONSIDERATION OF ENERGY AND MOMENTUM 75
e detailed steps of obtaining these absolute velocities are provided in the following.
Determination of
*
v
0
A
F
and
*
v
0
B
F
Recall the Oxy moves with the mass center. e unit vectors along the x-axis and y-axis are,
respectively,
*
{ and
*
| as shown in Figure 5.8b. us,
*
v
0
A
F
D
*
v
o
C
*
v
0
A
D 100
*
{ C 40
r
1
3
cos 61
ı
*
{ C sin 61
ı
*
|
m
s
)
*
v
0
A
F
D
100 C 40
r
1
3
cos 61
ı
!
*
{ C
40
r
1
3
sin 61
ı
!
*
|
m
s
)
*
v
0
A
F
D 111:20
*
{ C 20:20
*
| m/s or
*
v
0
A
F
D 113:02 m/s 10:29
ı
:
Similarly,
*
v
0
B
F
D
*
v
o
C
*
v
0
B
D 100
*
{ 20
r
1
3
cos 61
ı
*
{ C sin 61
ı
*
|
m
s
)
*
v
0
B
F
D
100 20
r
1
3
cos 61
ı
!
*
{
20
r
1
3
sin 61
ı
!
*
|
m
s
)
*
v
0
B
F
D 94:40
*
{ 10:10
*
| m/s or
*
v
0
B
F
D 94:94 m/s 6:07
ı
:
Example 5.2
A spherical ball B is attached at the free end of an inextensible cable which is fixed to the ceiling,
as shown in Figure 5.9a. Another spherical ball A having the same density and half of the radius
of ball B is released from rest when it is just touching the cable and acquires a velocity
*
v
o
before
striking ball B. Assuming the coefficient of restitution e D 1 and friction between the surfaces
of the balls being disregarded, find the velocity of each ball immediately after impact.
Solution:
Let the mass of ball B be m
B
D m and its radius r. en mass of ball A is m
A
D m=8 since
volume of ball B is V
B
D
4
3
r
3
and radius of ball A is r=2.
For ball A (Figure 5.9b)
Momentum is conserved along tangent to balls. By applying Equation (5.2), one has
m
8
*
v
A
C I
1!2
D
m
8
*
v
0
A
:
Components along t t ,
C &;
m
8
v
o
sin C 0 D
m
8
v
0
A
t
)
v
0
A
t
D v
o
sin
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