Page 110 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 8] MOMENTUM 95

Substituting for v in the first equation gives

2

m 1 v 1 = m 1 v + m 2 v 1 + m 2 v 1
1

(m 1 + m 2 )(v ) = (m 1 − m 2 )(v 1 )
1

m 1 − m 2

v = (v 1 )
1
m 1 + m 2
In a similar way we find that

2m 1

v = (v 1 )
2
m 1 + m 2
From these results we draw the following conclusions, which are illustrated in Fig. 8-3.
(a)For m 1 < m 2 : Here v is opposite in direction to v 1 , so the incoming lighter object rebounds from the heavier

1
one. If m 1 m 2 , for instance, in the case of a ball striking a wall, then v ≈−v 1 .

1

(b)For m 1 = m 2 : Here v = 0 and v = v 1 , so the incoming object m 1 comes to a stop, and the struck object m 2
2
1
moves away with the same velocity as m 1 had initially. An example is a moving billiard ball striking a stationary
one head-on.
(c) For m 1 > m 2 : The incoming object m 1 continues with reduced velocity while the struck object m 2 moves away
ahead of m 1 at a higher velocity. An example is a tennis serve. If m 1 m 2 , then v ≈ 2v 1 .

2

Fig. 8-3

SOLVED PROBLEM 8.17

For the collisions considered in Prob. 8.16, find the ratio between the kinetic energy KE transferred
2
to the initially stationary object and the kinetic energy KE 1 of the initially moving object before the
collision.
From the formula
2m 1

v = v 1
2
m 1 + m 2
we find that
Transferred KE KE 2 1 2 m 2 v 2 4m 1 m 2 4 (m 2 /m 1 )
2
= = = =
Initial KE KE 1 1 m 1 v 2 (m 1 2 1 m 2 ) 2 (1 + m 2 /m 1 ) 2
2 1
This formula is plotted in Fig. 8-4. The transfer of energy is a maximum for m 1 = m 2 , when all of the energy of m 1
is transferred to m 2 . This is the situation shown in Fig. 8-3(b).