Page 108 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 8] MOMENTUM 93

2
1
1
(b) Initial KE = m 1 v + m 2 v 2
2 1 2 2
2
1
1
2
= ( )(40 kg)(4 m/s) + ( )(60 kg)(2 m/s) = 440 J
2 2
1
1
2
2
Final KE = (m 1 + m 2 )v = ( )(100 kg)(2.8 m/s) = 392 J
2 3 2
Therefore 48 J of energy is lost, 11 percent of the original amount.
SOLVED PROBLEM 8.14
The two skaters of Prob. 8.13 are moving in opposite directions and collide head-on. (a) If they remain
in contact, what is their final velocity? (b) How much kinetic energy is lost?
(a) We take into account the opposite directions of motion by letting v 1 =+4 m/s and v 2 =−2 m/s. Then
m 1 v 1 + m 2 v 2 (40 kg)(4 m/s) − (60 kg)(2 m/s)
v 3 = = =+0.4 m/s
m 1 + m 2 40 kg + 60 kg
Since v 3 is +0.4 m/s, the two skaters move off in the same direction as the 40-kg skater had originally, which
is to be expected since she had the greater initial momentum.
(b) Initial KE = 440 J (as in Prob. 8.13)
2
2
1
1
Final KE = (m 1 + m 2 )v = ( )(100 kg)(0.4 m/s) = 8J
2 3 2
Therefore 432 J of energy is lost, 98 percent of the original amount. This is the reason why head-on collisions
of automobiles produce much more damage than overtaking collisions.
COLLISIONS IN TWO AND THREE DIMENSIONS
The directional character of linear momentum must be kept in mind in problems that involve collisions between
bodies that do not move along the same straight line. In such cases, linear momentum must be conserved
separately in two perpendicular directions when the motions involve two dimensions, or separately in three
mutually perpendicular directions when the motions involve three dimensions. The following problem illustrates
the way such problems are treated.

SOLVED PROBLEM 8.15
A 1000-kg car moving north at 20 m/s collides with a 1500-kg car moving west at 12 m/s. If the cars stick
together after the collision, at what velocity and in what direction does the wreckage begin to move?

Linear momentum must be conserved separately in both the north-south and east-west directions, which we call
the y and x axes, respectively, as in Fig. 8-2(a). Thus we have
Momentum before crash = momentum after crash

x direction m A v Ax + m B v Bx = m AB v ABx

y direction m A v Ay + m B v By = m AB v ABy
Here
m A = 1000 kg m B = 1500 kg m AB = m A + m B = 2500 kg
v Ax = 0 v Bx =−12 m/s v ABx = ?
v Ay = 20 m/s v By = 0 v ABy = ?
In the x direction we have

m A v Ax + m B v Bx 0 + (1500 kg)(−12 m/s)
v ABx = = =−7.2 m/s
m AB 2500 kg