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346 CHAPTER 11 Collisions

or

2m 1
final target velocity in elastic collision v v (11.14)
2 1
m m
1 2
Equations (11.13) and (11.14) give us the final velocities v and v in terms of the
1 2
initial velocity v .
1

An empty boxcar of mass m 20 metric tons rolling on a
EXAMPLE 4 1
straight track at 5.0 m s collides with a loaded stationary boxcar
of mass m 65 metric tons (see Fig. 11.5). Assuming that the cars bounce off
2
each other elastically, find the velocities after the collision.
SOLUTION: With m 20 tons and m 65 tons, Eqs. (11.13) and (11.14) yield
1 2
20 tons 65 tons
v 5.0 m s 2.6 m s
1
20 tons 65 tons
2 20 tons
v 5.0 m s 2.4 m s
2
20 tons 65 tons
Thus, boxcar 2 acquires a speed of 2.4 m s, and boxcar 1 recoils with a speed of
2.6 m s (note the negative sign of v ).
1

(a)
Boxcar 1 is Boxcar 2 is
moving “projectile.” stationary “target.”
v 1

1 2
x

(b)

v'
v' 1 2
1 2
x
Here, boxcar 1
recoils backward.

FIGURE 11.5 (a) Initially, boxcar 1 is moving toward the right, and boxcar 2 is stationary.
(b) After the collision, boxcar 1 is moving toward the left, and boxcar 2 is moving toward
the right.

Note that if the mass of the target is much larger than the mass of the projectile,
then m can be neglected compared with m . Equation (11.13) then becomes
1
2
m 2
v v v (11.15)
1 m 1 1
2