To do so, we rewrite Eq. 9-63 as
m 1 (v 1 iv 1 f)m 2 v 2 f (9-65)
and Eq. 9-64 as*
(9-66)
After dividing Eq. 9-66 by Eq. 9-65 and doing some more algebra, we obtain(9-67)
and (9-68)Note that v 2 fis always positive (the initially stationary target body with mass m 2
always moves forward). From Eq. 9-67 we see that v 1 fmay be of either sign (the
projectile body with mass m 1 moves forward if m 1 m 2 but rebounds if m 1 m 2 ).
Let us look at a few special situations.
1.Equal masses Ifm 1 m 2 , Eqs. 9-67 and 9-68 reduce to
v 1 f 0 and v 2 fv 1 i,
which we might call a pool player’s result. It predicts that after a head-on colli-
sion of bodies with equal masses, body 1 (initially moving) stops dead in its
tracks and body 2 (initially at rest) takes off with the initial speed of body 1. In
head-on collisions, bodies of equal mass simply exchange velocities. This is
true even if body 2 is not initially at rest.
2.A massive target In Fig. 9-18, a massive target means that m 2 m 1 .For
example, we might fire a golf ball at a stationary cannonball. Equations 9-67
and 9-68 then reduce to(9-69)This tells us that body 1 (the golf ball) simply bounces back along its incom-
ing path, its speed essentially unchanged. Initially stationary body 2 (the
cannonball) moves forward at a low speed, because the quantity in paren-
theses in Eq. 9-69 is much less than unity. All this is what we should expect.
3.A massive projectile This is the opposite case; that is,m 1 m 2. This time, we
fire a cannonball at a stationary golf ball. Equations 9-67 and 9-68 reduce to
v 1 fv 1 i and v 2 f 2 v 1 i. (9-70)
Equation 9-70 tells us that body 1 (the cannonball) simply keeps on going,
scarcely slowed by the collision. Body 2 (the golf ball) charges ahead at twice
the speed of the cannonball. Why twice the speed? Recall the collision de-
scribed by Eq. 9-69, in which the velocity of the incident light body (the golf
ball) changed from vtov, a velocity changeof 2v. The same changein ve-
locity (but now from zero to 2v) occurs in this example also.Moving Target
Now that we have examined the elastic collision of a projectile and a stationary
target, let us examine the situation in which both bodies are moving before they
undergo an elastic collision.
For the situation of Fig. 9-19, the conservation of linear momentum is written as
m 1 v 1 im 2 v 2 im 1 v 1 fm 2 v 2 f, (9-71)v 1 f v 1 i and v 2 f
2 m 1
m 2 v 1 i.v 2 f2 m 1
m 1 m 2v 1 i.v 1 fm 1 m 2
m 1 m 2v 1 im 1 (v 1 iv 1 f)(v 1 iv 1 f)m 2 v 22 f.238 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
*In this step, we use the identity a^2 b^2 (ab)(ab). It reduces the amount of algebra needed to
solve the simultaneous equations Eqs. 9-65 and 9-66.