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0593_C09_fm Page 305 Monday, May 6, 2002 2:50 PM

Principles of Impulse and Momentum 305

FIGURE 9.10.4
Colliding particles with different
masses.
From the momentum conservation principle (see Eq. (9.7.1)), we have:

m v + mv = m v + mv ˆ (9.10.9)
ˆ
AA B B AA B B
Then, by solving Eqs. (9.10.8) and (9.10.9) for ˆ v and ˆ v we obtain:
A B
) ( [
e m v
ˆ v = (1 M m − em v ) +(1 + ) B B] (9.10.10)
A A B A
M m )[ e v ) em v ) B]
+
ˆ v = (1 (1 + A ( m − (9.10.11)
B A B A
where M is defined as:

M = m + m (9.10.12)
A B

Consider three special cases: (1) e = 0 , (2) e = 1 , and (3) m = m .
A B
Case 1: e = 0 (Plastic Collision)
In this case, Eqs. (9.10.10) and (9.10.11) simplify to the expressions:
ˆ v = ( m v + m v ) ( m + ) (9.10.13)
m
A A A B B A B
and

ˆ v = ( m v + m v ) ( m + ) (9.10.14)
m
B A A B B A B
Observe that ˆ v = ˆ v , as expected with a plastic collision.
A B
Case 2: e = 1 (Elastic Collision)
In this case, Eqs. (9.10.10) and (9.10.11) become:

B B]
ˆ v = ( [ m − ) m v ( m + )
A
A A m v + 2 A m B (9.10.15)
B
ˆ v =[ 2 m v +( m − m v ) B] ( m + m ) (9.10.16)
B A A B A A B

Case 3: m = m B
A
In this case, Eqs. (9.10.10) and (9.10.11) become:

12
ˆ v = ( ) ( [ 1 − e v ) +(1 + e v ) ] (9.10.17)
A A B