9781118230725.pdf

momentum of the grapefruit changes, but since no horizontal external force acts on
the grapefruit, the horizontal component of the linear momentum cannot change.
Note that we focus on the external forces acting on a closed system.
Although internal forces can change the linear momentum of portions of the sys-
tem, they cannot change the total linear momentum of the entire system. For ex-
ample, there are plenty of forces acting between the organs of your body, but they
do not propel you across the room (thankfully).
The sample problems in this module involve explosions that are either one-
dimensional (meaning that the motions before and after the explosion are along
a single axis) or two-dimensional (meaning that they are in a plane containing
two axes). In the following modules we consider collisions.

9-5 CONSERVATION OF LINEAR MOMENTUM 231

We can relate the vMSto the known velocities with

.

In symbols, this gives us

vHSvrelvMS (9-47)

or vMSvHSvrel.

Substituting this expression for vMSinto Eq. 9-46, and then

substituting Eqs. 9-45 and 9-46 into Eq. 9-44, we find

Mvi0.20M(vHSvrel)0.80MvHS,

which gives us

vHSvi0.20vrel,

or vHS2100 km/h(0.20)(500 km/h)

2200 km/h. (Answer)



velocity of

hauler relative

to Sun 





velocity of

hauler relative

to module 





velocity of

module relative

to Sun 

Sample Problem 9.05 One-dimensional explosion, relative velocity, space hauler

One-dimensional explosion:Figure 9-12ashows a space hauler

and cargo module, of total mass M, traveling along an xaxis in

deep space. They have an initial velocity of magnitude 2100

km/h relative to the Sun. With a small explosion, the hauler

ejects the cargo module, of mass 0.20M(Fig. 9-12b). The hauler

then travels 500 km/h faster than the module along the xaxis;

that is, the relative speed vrelbetween the hauler and the mod-

ule is 500 km/h. What then is the velocity of the hauler rela-

tive to the Sun?

KEY IDEA

Because the hauler – module system is closed and isolated,

its total linear momentum is conserved; that is,

, (9-44)

where the subscripts iandfrefer to values before and after

the ejection, respectively. (We need to be careful here:

Although the momentum of the systemdoes not change, the

momenta of the hauler and module certainly do.)

Calculations:Because the motion is along a single axis, we can

write momenta and velocities in terms of their xcomponents,

using a sign to indicate direction. Before the ejection, we have

PiMvi. (9-45)

LetvMSbe the velocity of the ejected module relative to the

Sun. The total linear momentum of the system after the ejec-

tion is then

Pf(0.20M)vMS(0.80M)vHS, (9-46)

where the first term on the right is the linear momentum of

the module and the second term is that of the hauler.

P

:

iP

:

f

v:HS

v:i

Figure 9-12(a) A space hauler, with a cargo module, moving at initial

velocity (b) The hauler has ejected the cargo module. Now the

velocities relative to the Sun are for the module and for the

hauler.

:vMS :vHS

:vi.

(a)(b)

Cargo module

Hauler

0.20M

vi vMS vHS

0.80M

x x

The explosive separation can change the momentum

of the parts but not the momentum of the system.

Additional examples, video, and practice available at WileyPLUS

Checkpoint 6

An initially stationary device lying on a frictionless floor explodes into two pieces, which

then slide across the floor, one of them in the positive xdirection. (a) What is the sum of

the momenta of the two pieces after the explosion? (b) Can the second piece move at an

angle to the xaxis? (c) What is the direction of the momentum of the second piece?