Page 111 - Fisika Terapan for Engineers and Scientists

10.1 Momentum 311

(a)
v = 0.
2
m 1 m 2
v 1

x

(b) Cars locked together,
=
so v' 1 v' 2 = v' .
m 1 m 2

v'
x

FIGURE 10.4 (a) Initially, the red automobile has a speed of 24 m/s, and the blue automobile is at rest.
(b) After the collision, both automobiles are in motion with velocity v .

By momentum conservation, the momenta P and P before and after the collision
x x
must be equal:
m v (m m )v (10.6)
1 1 1 2
,
When we solve this for the velocity of the wreck v we find
m v
1 1
v
m m
1 2
(10.7)
1500 kg 24 m s
12 m s
1500 kg 1500 kg
The forces acting during the firing of the gun or the collision of the automobiles
are quite complicated, but momentum conservation permits us to bypass these com-
plications and directly obtain the answer for the final velocities. Incidentally: It is easy
to check that kinetic energy is not conserved in these examples. During the firing of
the gun, kinetic energy is supplied to the shot and the gun by the explosive combus-
tion of the gunpowder, and during the collision of the automobiles, some kinetic energy
is used up to produce changes in the shapes of the automobiles.
The conservation law for momentum depends on the absence of “extra” forces. If
the particles are not isolated from the rest of the Universe, then besides the mutual
forces exerted by one particle on the other, there are also forces exerted by other bodies
not belonging to the particle system.The former forces are called internal forces of the internal forces and external forces
system and the latter external forces. For instance, for the colliding automobiles of
Example 3 the gravity of the Earth, the normal force of the road, and the friction of
the road are external forces. In Example 3 we ignored these external forces, because
gravity and the normal force cancel each other, and the friction force can be neglected
in comparison with the much larger impact force that the automobiles exert on each
other. But if the external forces are significant, we must take them into account, and
we must modify Eq. (10.5). If the internal force on particle 1 is F and the external
1,int
force is F , then the total force on particle 1 is F F and its equation of
1,ext 1,int 1,ext
motion will be
dp
1
F 1,int F 1,ext (10.8)
dt