Page 109 - Schaum's Outline of Theory and Problems of Applied Physics

94 MOMENTUM [CHAP. 8

Fig. 8-2

and in the y direction we have
(1000 kg)(20 m/s)
m A v Ay + m B v By
v ABy = = = 8 m/s
m AB 2500 kg
The magnitude of the velocity v AB is therefore

2
2
v AB = v 2 + v 2 = (−7.2 m/s) + (8 m/s) = 10.8 m/s
ABx ABy
The direction of v AB may be specified by the angle θ between the +y direction, which is north, and v AB . From
Fig. 8-2(b), ignoring the sign of v ABx ,weget
v ABx 7.2 m/s
tan θ = = = 0.9 θ = 42 ◦
v ABy 8 m/s
◦
The wreckage begins to move at 10.8 m/s in a direction 42 west of north.

COEFFICIENT OF RESTITUTION
The coefficient of restitution e is defined as the ratio between the relative velocity of recession v − v after a

2 1
collision between two bodies and their relative velocity of approach v 1 − v 2 :
v − v 1

2
Coefficient of restitution = e =
v 1 − v 2
Values of e range from 0 to 1. In a perfectly elastic collision, e = 1 and the relative velocity after the collision is
the same as the relative velocity before it. In a perfectly inelastic collision, e = 0.
SOLVED PROBLEM 8.16
An object of mass m 1 and initial velocity v 1 undergoes an elastic head-on collision with another object
of mass m 2 that is at rest. How do the results of the collision depend on m 1 and m 2 ?
Let us call the final velocities of the objects v and v . From conservation of momentum,

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

1
The coefficient of restitution for an elastic collision is e = 1, so here, with v 2 = 0,

v 1 = v − v 1 v = v 1 + v 1
2
2