68 5. IMPULSE, MOMENTUM, AND IMPACT OF PARTICLES
Similarly, by considering particle B, one can show
e D
v
0
B
u
u v
B
: (5.13)
From Equations (5.12) and (5.13) the unknown velocity u is to be eliminated.
From Equation (5.12), one obtains
e
.
v
A
u
/
D u v
0
A
:
Rearranging terms to obtain u,
u D
ev
A
C v
0
A
1 C e
: (5.14)
From Equation (5.13), one has
u D
ev
B
C v
0
B
1 C e
: (5.15)
Equating (5.14) to (5.15), it gives
ev
A
C v
0
A
D ev
B
C v
0
B
:
Rearranging terms, one has
e D
v
0
B
v
0
A
v
A
v
B
: (5.16)
Equation (5.16) expresses the coefficient of restitution as a ratio of the relative velocity after
impact to that before impact of the two particles. Applying Equation (5.7) and (5.16) the two
unknown velocities after impact, v
0
A
and v
0
B
, can be solved.
Remarks:
(a) For perfectly plastic impact, e D 0:
(b) For perfectly elastic impact, e D 1: us, in this case, the total energy as well as the total
momentum of the two particles is conserved. at is, the kinetic energy of the two particles
is conserved
1
2
m
A
v
2
A
C
1
2
m
B
v
2
B
D
1
2
m
A
v
0
A
2
C
1
2
m
B
v
0
B
2
: (5.17)
5.3.2 OBLIQUE CENTRAL IMPACT
Now, consider two colliding particles whose velocities are not directed along the line of impact,
as shown in Figure 5.2b. is is the case known as oblique central impact. Suppose the velocities
*
v
0
A
and
*
v
0
B
after impact are in the directions shown in Figure 5.5. ese unknown vector quan-
tities constitute four unknowns, two for unknown magnitudes and two for unknown angles or
directions. eir solution requires four independent equations.