Download Linear Momentum and Collisions

small car moving with the same speed. The more momentum an object has, the
harder it is to stop it, and the greater effect it will have if it is brought to rest by
bnpact or collision. A football player is more likely to be stunned if tackled by a
beavy opponent running at top speed than by a lighter or slower-moving tackler. A
beavy, fast-moving truck can do more damage than a slow-moving motorcycle.
A force is required to change the momentum of an object, whether it is to
crease the momentum, to decrease it (such as to bring a moving object to rest), or
to change its direction. Newton originally stated his second law in terms of momen
tum (although he called the product my the “quantity of motion”). Newton’s state
ment of the second law of motion, translated into modern language, is as follows:
servation of linear momentum is another of the great
ervation laws of physics. Collisions, as between billiard or
I balls, illustrate this vector law very nicely: the total vector
) before the collision equals the total vector
1
nentum (m v
nentum just after the collision. In this photo, the moving
ball strikes the 11 ball at rest. Both balls move after the
isbn, at angles, but the sum of their vector momenta equals
initial momentum of the incoming cue ball. We will
ider both elastic collisions (where kinetic energy is also
served) and inelastic collisions. We also will examine the
ept of center of mass, and how it can make the study of
plex motion more readily analyzed and understood.
The rate of change of momentum of an object is equal to the net force
applied to it.
We can write this as an equation,
2F
CHAPTER
dp
=
(9-2)
—‘
NEWTON’S SECOND LAW
where F is the net force applied to the object (the vector sum of all forces acting
ma, from
on it). We can readily derive the familiar form of the second law, IF
0 is the initial velocity of an object and v
Eq. 9—2 for the case of constant mass. If v
Is its velocity after a time di’ has elapsed, then
“1
Linear Momentum and Collisions
2F
he law of conservation of energy, which we discussed in the previous chap
ter, is one of several great conservation laws in physics. Among the other
quantities found to be conserved are linear momentum, angular momen
tum, and electric charge. We will eventually discuss all of these because the conser
vation laws are among the most important in all of science. In this chapter, we
discuss linear momentum and its conservation. We then make use of the laws of
conservation of linear momentum and of energy to analyze collisions. Indeed, the
law of conservation of momentum is particularly useful when dealing with two or
more bodies that interact with each other, as in collisions.
Our focus up to now has been mainly on the motion of a single object. In this
chapter we will deal with systems of two or more objects. An important concept
for this study is that of center of mass, which we discuss later in the chapter.
=
dp
di’
—
=
d(mv)
di’
=
m
dv
dt
—
=
[constant mass]
ma,
because, by definition, a = dv/dt. Newton’s statement, Eq. 9—2, is actually more gen
eral than the more familiar one because it includes the situation in which the mass
may change. This is important in certain circumstances, such as for rockets which
lose mass as they burn fuel (Section 9—10) and in relativity theory (Chapter 37).
T
Washing a car: momentum change and force. Water
at
rate
of 1.5 kg/s with a speed of 20 rn/s and is aimed at the side
leaves a hose
a
a
car,
which
stops
it,
Fig. 9—1. (That is, we ignore any splashing back.) What is
of
the force exerted by the water on the car?
[ExAMPLE 9-1
SOLUTION
Momentum and Its Relation to Force
We take the x direction positive to the right. In each second, water
.
iiioineiitiini
—
Pfinal
0
Pinitial
—
—st—
t
—
30kg
1.Os
.
rn/s
—30N
‘I’he minus sign indicates that the force on the water is opposite to the water’s
original velocity. The car exerts a force of 30 N to the left to stop the water, so by
Newton’s third law, the water exerts a force of 30 N on the car.
FIGURE 9—1
(9—1)
I Water splashes back. What if the water splashes
back from the car in Example 9—1? Would the force on the car be greater or less?
Since velocity is a vector, momentum is a vector. The direction of the momentum
is the direction of the velocity, and the magnitude of the momentum is p = my.
Since v depends on the reference frame, this frame must be specified. The unit of
momentum is simply that of mass X velocity, which in SI units is kg rn/s. There is
no special name for this unit.
Everyday usage of the term momentum is in accord with the definition
above. For according to Eq. 9—1, a fast-moving car has more momentum than a
slow-moving car of the same mass, and a heavy truck has more momentum than
RESPONSE If the water splashes back toward the hose, the change in momen
turn will be greater in magnitude, and so the force on the car will be greater in
magnitude. Note that Pfinal will now point in the negative x direction, as shown
in Fig. 9—2 (instead of being zero as in Example 9—1). So the result for F (see dis
played equation in Example 9—1), will be minus something of magnitude more
than 30 N (i.e., —35 to —40 N, depending on the water’s rebound speed). To put it
simply, the car exerts not only a force to stop the water, but also an additional
force to give it momentum in the opposite direction.
FIGURE 9—2 Example 9—2.
Momentum of water before and after
splashing back, and p.
The linear momentum (or “momentum” for short) of an object is defined as the
product of its mass and its velocity. Momentum (plural is momenta) is usually rep
resented by the symbol p. If we let m represent the mass of an object and v repre
sent its velocity, then its momentum p is
Linear
20 m/s
with a momentum of Px
mv = (1.5 kg)(20 m/s) = 30 kg rn/s is brought to
rest at the moment it hits the car. The magnitude of the force (assumed constant)
that the car must exert to change the momentum of the water by this amount is
F
9—1
v
p
=
my.
.
CONCEPTUAL EXAMPLE 9—2
Example 9—1.
—
P initial
Pfinal
=
P final — P initial
SECTION 9—1
1
207
9—2 Conservation of Momentum
1
m
2v
m
2
1
V
0—
Ia
/
x
FIGURE 9—3 Momentum is
conserved in a collision of two balls.
The concept of momentum is particularly important because, under certain circum
stances, momentum is a conserved quantity. In the mid-seventeenth century, shortly
before Newton’s time, it had been observed that the vector sum of the momenta of
two colliding objects remains constant. Consider, for example, the head-on collision
of two billiard balls, as shown in Fig. 9—3. We assume the net external force on this
system of two balls is zero—that is, the only significant forces are those that each
ball exerts on the other during the collision. Although the momentum of each of
the two balls changes as a result of the collision, the sum of their momenta is found
1 is the momentum of ball num
1v
to be the same before as after the collision. If m
ber 1 and m
v the momentum of ball 2, both measured before the collision, then
2
2v
. After the
2
1 + m
1v
the total momentum of the two balls before the collision is m
collision, the balls each have a different velocity and momentum, which we will
1 v and m
2 v. The total momentum after
designate by a “prime” on the velocity: m
2 v. No matter what the velocities and masses involved
1v + m
the collision is m
are, it is found that the total momentum before the collision is the same as after
wards, whether the collision is head-on or not, as long as no net external force acts:
momentum after
momentum before
P
or
L
v
2
v + m
1
m
(‘O,VSLRVAiION Of
LI\fAR MOMENTUM
(1110 bodies (oI1i(gJ
FIGURE 9—4 Collision of two
objects. Their momenta before
collision are Pi and P2’ and after
collision are p and p. At any
moment during the collision each
exerts a force on the other of equal
magnitude but opposite direction.
Before
collision
Pi
P2
v
2
l + m
m
v
1
=
P1P2
That is, the total vector momentum of the system of two balls is conserved: it stays
constant.
Although the law of conservation of momentum was discovered experimen
tally, it is closely connected to Newton’s laws of motion and they can be shown to
be equivalent, which we now do.
2 that have momenta Pi and P2
1 and m
Let us consider two bodies of mass m
before they collide and p and p after they collide, as in Fig. 9—4. During the colli
sion, suppose the force exerted by body 1 on body 2 at any instant is F. Then, by
Newton’s third law, the force exerted by body 2 on body 1 is —F. During the brief
collision time, we assume no other (external) forces are acting (or that F is much
greater than any other external forces acting). We rewrite Eq. 9—2 as dp = F dt
and integrate both sides of this equation over the brief time interval of the colli
sion, from t1 to tf.
(tf
dp=
We apply this first to body 2, on which the force F acts (Fig. 9—4),
=
P
—
P2
=
Fdt.
j
(9—4a)
On body 1, the force is —F, so
(tf
After
collision
P1
‘Pi
—
Pi
=
Fdt.
—
.Jti
We compare these two equations and see that
P’i/
p1
Pi
=
P2
ii
=
(p
—
—
P2).
Rearranging, we have
208
CHAPTER 9
+ rn,,v,,
=
.
1
2p
dP
cit
—
1
dp
dt
—
(9—5)
F,
th
1
object. The forces can be of two types:
where F
1 represents the net force on the
(1) external forces on objects of the system, exerted by objects outside the system,
and (2) internal forces that objects within the system exert on other objects in the
system. By Newton’s third law, the internal forces occur in pairs: if one object
exerts a force on a second object, the second exerts an equal and opposite force on
(he first object. Thus, in the sum over all the forces in Eq. 9—5, all the internal
iorces cancel each other in pairs. Thus we have
(9—6)
=
Pi + P2 = Pi + p,
which is Eq. 9—3, the law of conservation of momentum.
(9—4b)
\f: IO\.s SI ( O\I) I in
(hr a s (/11 o/ )IJ’ (
where 2
Fext is the sum of all external forces acting on our system. If the net exter
0, so zP = 0 or P = constant. Thus we have
nal force is zero, then dP/dt
This is the law of conservation of momentum. It can also be stated as
ftf
collision
v + m-)v
1
m
2 +
when the net external force on a system is zero, the total momentum remains
constant.
Fdt.
.Jt
Jti
=
We differentiate with respect to time:
(9—3)
f
t
(
F
We have put this derivation in the context of a collision. As long as no exter
flal forces act, Eqs. 9—4 are valid over any time interval, and conservation of
momentum is always valid as long as no external forces act. In the real world, exter
nal forces do act: friction on billiard balls, gravity acting on a baseball, and so on. So
it may seem that conservation of momentum cannot be applied. Or can it? In a col
lision, the force each body exerts on the other acts only over a very brief time inter
val, and is very strong. During the brief time, F in Eqs. 9—4 is very much larger than
bc other forces acting (gravity, friction). If we measure the momenta just before,
md just after, the collision, momentum will be very nearly conserved. We cannot
Wait for the external forces to produce their effect before measuring p and p.
For example, when a racket hits a tennis ball or a bat hits a baseball, both
before and after the “collision” the ball moves as a projectile under the action of
gravity and air resistance. However, when the bat or racket hits the ball, during this
brief time of the collision, these external forces are insignificant compared to the
llision force the bat exerts on the ball (and vice versa). From Eqs. 9—4 we see
then that momentum is conserved (or very nearly so) as long as we measure
p and P2 just before the collision and p and p immediately after the collision.
Our derivation of the conservation of momentum via Eqs. 9—4 can be
extended to include any number of interacting bodies. Let P represent the total
momentum of a system of n interacting objects:
the total momentum of an isolated system of bodies remains constant.
I 1W Of
( O\SI 1?i lIIO\
I)! LIV! 1/?
1IO’ilI \ Ill!
13y an isolated system, we mean one on which no external forces act—the only
forces acting are those between objects of the system.
If a net external force acts on a system, then the law of conservation of
momentum will not apply. However, if the “system” can be redefined so as to
include the other objects exerting these forces, then the conservation of momen
tum principle can apply. For example, if we take as our system a falling rock, it
does not conserve momentum since an external force, the force of gravity exerted
by the Earth, is acting on it and its momentum changes. However, if we include the
Farth in the system, the total momentum of rock plus Earth is conserved. (This of
course means that the Earth comes up to meet the ball. Since the Earth’s mass is
so great, its upward velocity is very tiny.)
SECTION 9—2
Conservation of Momentum
209
_______________
As with energy, the importance of the momentum concept is that momentum
is conserved under quite general conditions. Although the law of conservation of
momentum follows from Newton’s second law, as we have seen, it is in fact more
general than Newton’s laws. In the tiny world of the atom, Newton’s laws fail, but
the great conservation laws—those of energy, momentum, angular momentum.
and electric charge—have been found to hold in every experimental situation
tested. It is for this reason that the conservation laws are considered more basic
than Newton’s laws.
EXAMPLE 9—4 Rifle recoil. Calculate the recoil velocity of a 5.0-kg rifle that
shoots a 0.050-kg bullet. at a speed of 120 m/s, Fig. 9—7.
SOLUTION The total momentum of the system is conserved. We let the sub
scripts B represent the bullet and R the rifle; the final velocities are indicated by
primes. Then conservation of momentum in the x direction gives
mBvB
+
mRvR
(a) Before shooting
v
(b) After shooting
= mBvB + mRvR
FIGURE 9—7 Example 9—4.
0 + 0 = (0.O5Okg)(120m/s) + (5.0kg)(v)
v = —1.2 rn/s.
EXAMPLE 9—3 Railroad cars collide: momentum conserved. A 10,000-kg
railroad car traveling at a speed of 24.0 rn/s strikes an identical car at rest. If the
cars lock together as a result of the collision, what is their common speed after
ward? See Fig. 9—5.
Since the rifle has a much larger mass, its (recoil) velocity is much less than that
of the bullet. The minus sign indicates that the velocity (and momentum) of the
rifle is in the negative x direction, opposite to that of the bullet. Notice that it is
the vector sum of the momenta that is conserved.
SOLUTION
1v
1 (since v
2 = 0), to the
The initial total momentum is simply m
right in the +x direction. After the collision, the total momentum will be the same,
and it will be shared by both cars. Since the two cars become attached, they will
have the same speed, call it v’. Then:
)v’ =
2
1 + m
(m
v
1
m
1
m
V
—
1
m
+ m
2
=V
1
V
= 12.0 rn/s
2
to the right. Their mutual speed after collision is half the initial speed of car 1.
EXAMPLE 9-5 Billiard ball collision in 2-D. A billiard ball moving with
1 = 3.0 rn/s in the +x direction (Fig. 9—8) strikes an equal-mass ball ini
speed v
tially at rest. The two balls are observed to move off at 45°, ball 1 above the x axis
and ball 2 below. That is, O = 45° and O = —45° in Fig. 9—8. What are the speeds
of the two balls after the collision?
SOLUTION We set up an xy coordinate systern as shown in Fig. 9—8. From sym
metry, we might guess that the two balls have the same speed. But let us not
assume that now. Instead, let us apply conservation of momentum. We are given
(= m) so
2
= m
FIGURE 9—8 Example 9—5.
‘Cl,)
vi
•
= —45°
‘initial = ‘finaI
1 = mv + mv.
my
iRE 9—5 Example 9—3.
(a) Before collision
tELAPJEiETL
=?
Vector momentum is conserved, which means that each component is conserved.
We can write the x and y components of this vector equation as
1 = mvcos(45°) + mvcos(—45°)
my
and
0 = mv sin(45°) + mv sin(—45°)
The m’s cancel out in both equations. The second equation yields [recall
sin(—O) = —sin 0]:
sin(45°)
sin 45°
=
sin(—45°)
\—sin45°)
so they do have equal speeds as we guessed at first. The x component equation
gives [recall cos(—0) = cos0]:
= vcos(45°) + vcos(45°) = 2vcos(45°)
so
3.0 rn/s
1
V
=
=
= 2.lm/s.
2(0.707)
2 cos(45°)
—
—
(b) After collision
iRE 9—6 (a) A rocket,
aining fuel, at rest in some
rence frame. (b) In the same
rence frame, the rocket fires and
s are expelled at high speed out
ear. The total vector momentum,
+ Procket’ remains zero.
p=o
(a)
rocket
gas
(b)
CHAPTER 9
The law of conservation of momentum is particularly useful when we are
dealing with fairly simple systems such as collisions and certain types of explo
sions. For example, rocket propulsion, which we saw in Chapter 4 can be understood on the basis of action and reaction, can also be explained on the basis of the
conservation of momentum. Before a rocket is fired, the total momentum of
rocket plus fuel is zero. As the fuel burns, the total momentum remains unchanged:
the backward momentum of the expelled gases is just balanced by the forward
momentum gained by the rocket itself (see Fig. 9—6). Thus, a rocket can accelerate
in empty space. There is no need for the expelled gases to push against the Earth
or the air (as is sometimes erroneously thought), as we already discussed in
Chapter 4. Similar examples are the recoil of a gun and the throwing of a package
from a boat.
Linear Momentum and Collisions
Zi] Collisions and Impulse
F9
Conservation of momentum is a very useful tool for dealing with collision
processes, as we already saw in the Examples of the previous Section. Collisions
are a common occurrence in everyday life: a tennis racket or a baseball bat strik
ing a ball, two billiard balls colliding, one railroad car stridng another, a hammer
hitting a nail. At the subatomic level, scientists learn abou”the structure of nuclei
and their constituents, and about the nature of the forces involved, by careful study
of collisions between nuclei and/or elementary particles.
SECTION 9—3
Collisions and Impulse
211
In a collision of two ordinary objects, both objects are deformed, often con
siderably, because of the large forces involved (Fig. 9—9). When the collision
occurs, the force usually jumps from zero at the moment of contact to a very large
value within a very short time, and then abruptly returns to zero again. A graph of
the magnitude of the force one object exerts on the other during a collision, as a
function of time, is something like that shown by the red curve in Fig. 9—10. The
time interval zt is usually very distinct and usually very small.
From Newton’s second law, the net force on an object is equal to the rate of
change of its momentum:
dt
FIGURE 9—9 Tennis racket striking
a ball. Note the deformation of both
ball and racket due to the large force
each exerts on the other.
=
F dt.
f
ii
SOLUTION (a) We don’t know F and can’t calculate the impulse J = fF dt
directly, but we can use the fact that the impulse equals the change in momentum
of the object. We need to determine the velocity of the person just before striking
the ground, which we can do using conservation of energy (Eq. 8—9), iK =
v
J
=
( h)
=
Fdt.
=
=
=
p
—
p, is equal to the impulse
,
(9—7)
1
it
The units for impulse are the same as for momentum, kg m/s (or N s) in SI. Since
J = fF di’, we can state that the impulse J of a force is equal to the area under the
F versus t curve, as indicated by the shading in Fig. 9—10.
Equation 9—7 is true only if F is the net force on the object. It is valid for any
net force F where p
1 and p correspond precisely to the times t1 and tf. But the
impulse concept is really most useful for so-called impulsive forces—that is, for a
force like that shown in Fig. 9—10, which has a very large magnitude over a very
short time interval and is essentially zero outside this time interval. For most colli
sion processes, the impulsive force is much larger than any other force acting, and
the others can be neglected. Then the impulsive force is essentially the net force,
and the change in momentum of an object during a collision is due almost entirely
to the impulsive force. For such an impulsive force, the time interval over which we
take the integral in Eq. 9—7 is not critical as long as we start before t1 and end after
. (Of course, if
1
t
tf, since F is essentially zero outside this time interval zXt = tf
the chosen time interval is too large, the effect of the other forces does become sig
nificant—such as the flight of a tennis ball which, after the impulsive force admin
istered by the racket, begins to slowly fall under gravity.)
It is sometimes useful to speak of the average force, F, during a collision. It
is defined as that constant force which, if acting over the same time interval
= tf
1 as the actual force, would produce the same impulse and change in
t
momentum. Thus
.
Time,
t
FIGURE 9—10 Force as a function
of time during a typical collision.
FIGURE 9—11 The average force F
over t gives the same impulse
(F t) as the actual force.
F
0
V2g(y
\/i(9.8m/s2)(3.Om)
=
y)
—
—
=
Ft
p
=
p
0
=
.
—
—
—
p
(7Okg)(7.7rn/s)
=
*v
7.7 rn/s.
—540 N
FIGURE 9—12 Period during which
impulse acts (Example 9—6).
s.
(7.7 rn/s
Om/s)/2
+
=
3.8rn/s,
o the collision lasts a time
1
it
Fdt=J.
=0
In coming to rest, the body decelerates from 7.7 rn/s to zero in a distance
1.0cm = 1.0 X 102 rn. The average speed during this brief period is
1
it
0
—mg(y
=
The negative sign tells us that the force is opposed to the original momentum—that
is, the force acts upward.
d
Thus the change in momentum of an object, zp
acting on it:
—
=
tf
J=
0
v7.7 rn/s
Yo)’
where we assume he started from rest (vo = 0), and Yo = 3.0 m and y = 0. Thus,
after falling 3.0 m, the person’s velocity just before hitting the ground will be
IFdt,
_lti
where p
1 and p are the momenta of the object just before and just after the colli
sion. The integral of the force over the time interval during which it acts is called
the impulse, J:
I111/)11/’
height of 3.0 m. Then estimate the average force exerted on the person’s feet by
the ground, if the landing is (b) stiff-legged, and (c) with bent legs. In the former
case, assume the body moves 1.0 cm during impact, and in the second case, when
the legs are bent, about 50 cm.
tf
=
1
dp=p—p
(a) Calculate the impulse
As the person strikes the ground, the momentum is quickly brought to zero,
Fig. 9—12. The impulse on the person is
If we integrate this over the duration of a collision, we have
f
Bend.your knees when landing.
2
mv
(We have written F instead of ZF for the net force, which we assume is entirely
due to the brief but large force that acts during the collision.) This equation
applies, of course, to each of the objects in a collision. During the infinitesimal time
interval dt, the momentum changes by
dp
EXAMPLE 9-6
experienced when a 70-kg person lands on firm ground after jumping from a
d
(1.0 x 10
m)
2
v
(3.8m/s)
Since the magnitude of the impulse is
average net force F has magnitude
s.
3
=2.6X10
F iXt
=
540 N
s, and
it
=
2.6
3,
X
the
F=L= 540Ns =2.lxlO
N.
5
tt
2.6
iOs
x
The force F is the net force upward on the person (we calculated it from Newton’s second law). F is the surn of the average force upward on the legs exerted
by the ground, Fgrd ‘ which we take as positive, plus the downward force of gray—mg (see Fig. 9—13):
ity,
F
Since
mg
=
=
)
2
(7Okg)(9.8rn/s
Fgrd
F+ mg
=
=
=
540N .
0.13s
5
=
5 the
force the ground exerts upward on
690N, then
2.1
(c) This is like part (b), except d
F
net force during impact is
F = Fgrd mg, where Fgrd
mg
Fgrd
=
X
N
5
10
+
0.50 m, so
FIGURE 9—13 When the person
lands on the ground, the average
0.690
iXt
=
X
N
3
10
2.1 x 10
N.
5
(0.50 rn)/(3.8 m/s)
=
the person.
0.13 s, and
4.2 X 10
N
3
[he upward force exerted on the person’s feet by the ground is, as in part (b):
Fgrd
F
+ mg
=
4.2
x
N
3
10
+
0.69 X 10
N
3
=
4.9
X
i0 N.
—
itf
0
212
ti
CHAPTER 9
tf
t
t=j
Fdt.
Figure 9—11 shows the magnitude of the average force, F, for the impulsive force of
Fig. 9—10. The rectangular area F zXt equals the area under the impulsive force curve.
Clearly, the force on the feet and legs is much less when he knees are bent. In
fact, the ultimate strength of the leg bone (see Chapter 2, Table 12—2) is not
great enough to support the force calculated in part (b), so the leg would likely
break in such a stiff landing, whereas it probably wouldn’t in part (c).
SECTION 9—3
Collisions and Impulse
213
I 9—4
L
Liastu
(01115101?
LcL
ZE%
(a) Approach
(b) Collision
(c) If elastic
(d) If inelastic
FIGURE 9—14 Two equal mass
objects (a) approach each other with
equal speeds, (b) collide, and then
(c) bounce off with equal speeds in
the opposite directions if the
collision is elastic, or (d) bounce
back much less or not at all, if the
collision is inelastic.
i’IoJ;?entll,?? (o1iscJ?’u!ioI?
the collision, vj and v. We will do this in a moment in some Examples, but first we
derive a useful result. To do so we rewrite the momentum equation as
Conservation of Energy and Momentum
in Collisions
1
mi(v
(v
2
),
2
v
v) = m
we
rewrite the kinetic energy equation as
and
During most collisions, we usually don’t know how the collision force varies over
time, and so analysis using Newton’s second law becomes difficult or impossible.
But we can still determine a lot about the motion after a collision, given the initial
motion, by making use of the conservation laws for momentum and energy. We saw
in Section 9—2 that in the collision of two objects such as billiard balls, the total
momentum is conserved, as long as any other “external” forces can be ignored. If
the two objects are very hard and no heat is produced in the collision, then kinetic
energy is conserved as well. By this we mean that the sum of the kinetic energies of
the two objects is the same after the collision as before. Of course, for the brief
moment during which the two objects are in contact, some (or all) of the energy is
stored momentarily in the form of elastic potential energy. But if we compare the
total kinetic energy before the collision with the total after the collision, they are
found to be the same. Such a collision, in which the total kinetic energy is con
served, is called an elastic collision. If we use the subscripts 1 and 2 to represent the
two objects, we can write the equation for conservation of total kinetic energy as
1 v + mv
m
=
1v
m
2 +
m v.
[elastic collisionl
(9—8)
Here, primed quantities (‘) mean after the collision and unprimed mean before the
collision, just as in Eq. 9—3 for conservation of momentum.
At the atomic level the collisions of atoms and molecules are often elastic.
But in the “macroscopic” world of ordinary objects, an elastic collision is an ideal
that is never quite reached, since at least a little thermal energy (and perhaps
sound and other forms of energy) is always produced during a collision. Collisions
of two air table pucks or two hard elastic balls, such as billiard balls, are very close
to perfectly elastic, and we often treat them as such. Even when the kinetic energy
is not conserved, the total energy is, of course, always conserved.
Collisions in which kinetic energy is not conserved are said to be inelastic
collisions. The kinetic energy that is lost is changed into other forms of energy,
often thermal energy, so that the total energy (as always) is conserved. In this case,
we can write that
2
1 + K
K
=
K’1 + K + thermal and other forms of energy.
(2_
(vi
1
m
—
b)(a + b)
=
2
a
+ v’1)
—
—
V + 1
1
V’
=
—
1
V
[-5 Elastic Collisions in One Dimension
We now apply the conservation laws for momentum and kinetic energy to an elas
tic collision between two small objects that collide head-on, so all the motion is
along a line. We deal here only with translational motion, but our analysis works
well even for hard billiard balls because at the instant of collision the rolling has
little effect. Let us assume that both objects are initially moving with velocities v
1
and v
2 along the x axis, Fig. 9—15a. After the collision, their velocities are v’1 and v,
Fig. 9—15b. For any v > 0, the object is moving to the right (increasing x), whereas
for v < 0, the object is moving to the left (toward decreasing values of x).
From conservation of momentum, we have
V + m
1
m
V
2
+
energy coiiser ?‘ahiol?
CHAPTER 9
2
V
x
=
—
)(v + v
2
v
).
2
2
v’1 and v
(ii)
v)t obtain
V + V
.
2
1
m
We can rewrite this equation as
2
m
vi
=
—
=
2
V
—
—(v’1
—
[head-on elastic collisionj
v).
(9—9)
This is an interesting result: it tells us that for any elastic head-on collision, the rel
ative speed of the two objects after the collision has the same magnitude as before
(hut opposite direction), no matter what the masses are.
Now let us look at some special cases of elastic head-on collisions. We assume
,m
2
, and m
1
2 are known, and we wish to solve for v and v, the velocities of
.v
1
v
the two objects after the collision.
x
(b)
FIGURE 9—15 Two small objects, of
masses m
1 and m
, (a) before the
2
collision, and (b) after the collision.
EXAMPLE 9—7 Equal masses. A billiard ball of mass m moving with speed
v collides head-on with a second ball of equal mass. What are the speeds of the
two balls after the collision, assuming it is elastic? Assume (a) both balls are mov
ing, (b) ball 2 is initially at rest (v
2 = 0).
SOLUTION
(a) Since m
1
1 + v
v
2
=
2
m
=
=
m, then conservation of momentum gives
v + v.
We need a second equation, since there are two unknowns. We could use the con
servation of kinetic energy equation, or the simpler Eq. 9—9 derived above:
Vt
—
2
V
=
V
—
.
1
V’
Now we add these two equations and get
=
1
V
and then subtract the two equations to get
=
V
1
m
v’1
=
.
2
v
‘[hat is, the balls exchange velocities as a result of the collision: ball 2 acquires the
velocity that ball 1 had before the collision, and vice versa.
(h) If ball 2 is at rest initially, so that v
2
=
1
V
and
V’1
=
=
0, we have
0.
That is, ball 1 is brought to rest by the collision, wheres ball 2 acquires the orig
inal velocity of ball 1. This result is often observed by billiard and pool players,
and is valid only if the two balls have equal masses (and no spin is given to the
halls). See Fig. 9—16.
+ m
V.
2
V
2
m
=
Linear Momentum and Collisions
FIGURE 9—16
In this multi-flash
photo of a head-on collision
between two balls of equal mass, the
white cue ball is accelerated from
rest by the cue stick and then strikes
the red ball, initially at rest. The
white ball stops in its tracks and the
(equal mass) red ball moves off with
the same speed as the white ball had
before the collision. See Example 9—7.
V.
2
V’1 + m
1
m
2
We have two equations, so we can solve for two unknowns. If we know the masses
and initial velocities, then we can solve these two equations for the velocities after
214
—
(a)
Because the collision is assumed to be elastic, kinetic energy is also conserved:
Kinetic
2
m
} we write this as
2
b
We divide Eq. (ii) by Eq. (i), and (assuming v
1
2
V
See Fig. 9—14.
—
1
m
2
I ‘2_ 2
v
2
m
V
‘2’_
11
V
v
1
m
or [noting that (a
(i)
—
—
y
Note that Eqs. (i) and (ii), which are the conservation laws for momentum and kinetic energy, are both
satisfied by the solution v = v
1 and v = v
. This is a valid solution, but not very interesting. It corre
2
ponds to no collision at all—when the two objects miss each other.
SECTION 9—5
215
________________
ring ol)/r( I ((il/ides ii,tli
%C( (md buI immmn Ills (ii res I
Wo
EXAMPLE 9-8 Unequal masses, target at rest. A very common practical
, the “target”)
2
) to strike a second object (m
1
situation is for a moving object (m
2 = 0). Assume the objects have unequal masses and that the collision
at rest (v
occurs along a line (head-on). (a) Derive equations for v and v in terms of the
. (b) Determine the final
2
1 and m
1 and the masses m
1 of mass m
initial velocity v
1 >> m
).
2
velocities if the moving object is much more massive than the target (m
(c) Determine the final velocities if the moving object is much less massive than
).
2
1 << m
the target (m
SOLUTION
2
(a) We combine the momentum equation, which (with v
mi(vi
v
2
m
=
(
0) is
+
v
=
v, to obtain
2m
2
1 + m
\m
mV +
lit iii V ,,iissile, I/c/lI 1(11
t
I
v’,
=
1
m
=
,
2
m
and
1
v
and we obtain
v
0
=
=
Ve
—
0.
V,
=
Ve
—
tflV
Solving for
V[je,
=
mpVe
.
1
—v
The massive object remains essentially at rest and the very light incoming object
rebounds with essentially its same speed but in the opposite direction. For exam
ple, a tennis ball colliding head-on with a stationary bowling ball will hardly affect
the bowling ball, but will rebound with nearly the same speed it had initially, just
as if it had struck a hard wall.
It can readily be shown (it is given as Problem 40) for any elastic head-on
collision that
/1-n,
+v
(
2
1
2rn
—V
2
V
1
11l(11lO11,,
ihese
((f iluliuli
J
2
1 + m
m
—
)
2
+m
and
1
(m
1
V
=
—
2
m
V
1
2
1 + m
\,m
/
+ V
1
2
1
\,m
2
2m
+ m
2
These general equations, however, should not be memorized. They can always be
216
CHAPTER 9
Linear Momentum and Collisions
mV
+ mHeVFle.
we obtain
+
1
V
Thus the velocity of the heavy incoming object is practically unchanged, whereas
the light object, originally at rest, takes off with twice the velocity of the heavy
one. The velocity of a heavy bowling ball, for example, is hardly affected by strik
ing the much lighter bowling pins.
1 << m
. A moving light object strikes a very
2
2 = 0 and m
(c) Finally, we let v
massive object at rest. In this case, using the equations in part (a)
11(11
—
2mv
0
Do
V,.
and substituting this into the momentum equation we get
VF1e
ic/mi 1,1/551/C. limo er 1(11cm-i
+ mHeVIe.
=
0
—
2v
/
\ucleur (0/1151011
Thus
This is the same case treated in Example 9—7, and we get the same result: for
objects of equal mass, one of which is initially at rest, the velocity of the one mov
ing initially is completely transferred to the object originally at rest.
. A very heavy moving object strikes a light object
2
1 >> m
(b) Let v
2 = 0 and m
at rest, and we have, using the relations for v and v above,
1
V’
APPLIED
Because the collision is elastic, kinetic energy is conserved and we can use
Eq. 9—9, which becomes
—
As a check, we let
PHYSICS
Call the initial direction of motion the +x direction. We have
1 = v, = 3.60 x 1O m/s. We want to find the velocities v, and
= VHe = 0 and v
Vje after the collision. From conservation of momentum we have
1
V
7
m
1
(m
=v
1
V
I
m
1 + 2
\m
1’ (/11(1/ 111(15 SI’S hull
OH/li/n’s speed iii litill /
EXAMPLE 9—9 A nuclear collision. A proton of mass 1.01 u (unified atomic
mass units) traveling with a speed of 3.60 X iO m/s has an elastic head-on colli
sion with a helium (He) nucleus (mH = 4.00 u) initially at rest. What are the
velocities of the proton and helium nucleus after the collision? (As mentioned in
Chapter 1, 1 u = 1.66 X 10-27 kg, but we won’t need this fact.)
SOLUTION
—
1
with Eq. 9—9, rewritten as v
2
V
=
derived quickly from the conservation laws. For many problems, it is simplest
just to start from scratch, as we did in the special cases above and as shown in the
next Example.
2(1.01 u)(3.60 X 10
m/s)
4
—
5.01
mHC
=
U
1.45 X i0 rn/s.
VP
The other unknown is v, which we can now obtain from
V,
Vjj
=
—
He
(a)
Vp
1.45 X 10 rn/s
—
3.60 >< 10 rn/s
=
—2.15 x 10 rn/s.
I The minus sign tells us that the proton reverses direction upon collision, and we
its speed is less than its initial speed (see Fig. 9—17). This makes sense
/ see that
ordinary
from
experience: the lighter proton would be expected to “bounce back”
/
not with its full original
somewhat from the more massive helium nucleus, but
velocity as it would from a rigid wall (which would correspond to extremely large,
[nfinite mass).
VHe
(b)
FIGURE 9—17 Example 9—9:
(a) before collision, (b) after
collision.
[] Inelastic Collisions
Collisions in which kinetic energy is not conserved are called inelastic collisions.
Some of the initial kinetic energy in such collisions is transformed into other types
of energy, such as thermal or potential energy, so the total final kinetic energy is
less than the total initial kinetic energy. The inverse can also happen when poten
tial energy (such as chemical or nuclear) is released, in which case the total final
kinetic energy can be greater than the initial kinetic energy. Explosives are exam
ples of this type. Typical macroscopic collisions are inelastic, at least to some
extent, and often to a large extent. If two objects stick together as a result of a col
lision, the collision is said to be completely inelastic. Two colliding balls of putty
(
‘iiip/elr/s ujelustu
SECTION 9—6
eullisioii
Inelastic Collisions
217
that stick together or two railroad cars that couple together when they collide are
examples of completely inelastic collisions. The kinetic energy in some cases is all I
transformed to other forms of energy in an inelastic collision, but in other cases
only part of it is. In Example 9—3, for instance, we saw that when a traveling rail- I
road car collided with a stationary one, the coupled cars traveled off with some
kinetic energy.
I3allistic J)C11(I1tIuh1Z
VM
EXAMPLE 9—10 Ballistic pendulum The ballisticpendulum is a device used
to measure the speed of a projectile, such as a bullet. The projectile, of mass m, is I
fired into a large block (of wood or other material) of mass M, which is suspended
like a pendulum. (Usually, M is somewhat greater than m.) As a result of the col
lision, the pendulum-projectile system swings up to a maximum height h,
Fig. 9—18. Determine the relationship between the initial horizontal speed of the
projectile, v, and the height h.
SOLUTION We analyze this process by dividing it into two parts: (1) the colli
sion itself, and (2) the subsequent motion of the pendulum from the vertical
hanging position to height h. In part (1), Fig. 9—18a, we assume the collision time
is very short, and so the projectile comes to rest in the block before the block has
moved significantly from its position directly below its support. Thus there is no
net external force and momentum is conserved:
0
(a)
my
‘
1Ev’J”
M+m
(b)
FIGURE 9—18 Ballistic pendulum
=
(m + M)v’,
where v’ is the speed of the block and embedded projectile just after the collision,
before they have moved significantly. Once the pendulum begins to move (part 2,
Fig. 9—18b), there will be a net external force (gravity, tending to pull it back to
the vertical position). So, for part (2), we cannot use conservation of momentum,
But we can use conservation of mechanical energy since the kinetic energy imme
diately after the collision is changed entirely to gravitational potential energy
when the pendulum reaches its maximum height, h. Therefore (letting y = 0 for
the pendulum in the vertical position):
(Example 9—10).
1 + U
K
1
=
2 + U
K
2
f] Collisions in Two or Three Dimensions
Conservation of momentum and energy can also be applied to collisions in two or
three dimensions, and the vector nature of momentum is especially important. One
common type of non-head-on collision is that in which a moving object (called the
“projectile”) strikes a second object initially at rest (the “target”) object. This is
the common situation in games such as billiards, and for experiments in atomic and
fluclear physics (the projectiles, from radioactive decay or a high-energy accelera
br, strike a stationary target nucleus).
Figure 9—19 shows the incoming projectile, m
, heading along the x axis
1
, which is initially at rest. If these are billiard balls, m
2
1 strikes
toward the target, m
and they go off at the angles O and 9, respectively, which are measured rela
’s initial direction (the x axis). The objects may begin to deflect even
1
tive to m
they
touch if electric, magnetic, or nuclear forces act between them.
before
Let us apply the law of conservation of momentum to a collision like that
of Fig. 9—19. We choose the xy plane to be the plane in which the initial and
final momenta lie. Because momentum is a vector, and is conserved, its compo
nents in the x and y directions remain constant. In the x direction, we have
+ P2x = Px + Px’ or:
1v
m
1
=
1 v cos 0 + m
m
2 y cos Q.
(9—lOa)
I)
(O’l(’rl’(’(I
Because there is no motion in the y direction initially, the y component of the total
momentum is zero:
0
=
1 v sinO + 2
m
m sinO.
v
(9—lOb)
p. o,i.sered
When we have two independent equations, we can solve for, at most, two
Unknowns.
If we know that a collision is elastic, we can then apply conservation of
1netic energy and obtain a third equation:
1 + 1<2
K
K + K
or, for the collision shown in Fig. 9—19,
or
mv
+0=0+(m+M)gh,
2
(m+M)v’
so v’
=
(ii)
We combine equations (i) and (ii) to obtain
v=
m+M
v
m
=
m+M ,—
V2gh,
m
=
1 v1
m
2 +
mv.
[elastic collisioni
(9—lOc)
Kiiu’iic
(‘1l(’iL
(o1zserled
Jf the collision is elastic, we have three independent equations and we can solve
for three unknowns. If we are given m
,m
1
,v
2
1 (and v
, if it is not zero), we cannot,
2
for example, predict the final variables, v, v, O, and O&, because there are four of
Ihem. However, if we measure one of these variables, say O, then the other three
ariab1es (vi, v, and o) are uniquely determined, and we can determine them
using Eqs. 9—lOa, b, c.
which is the final result. To obtain this result, we had to be opportunistic, in that
we used whichever conservation laws we could: in (1) we could use only conser
vation of momentum, since the collision is inelastic and conservation of mechani
cal energy is not validt; and in (2), conservation of mechanical energy is valid, but
not conservation of momentum. In part (1), if there were significant motion of
the pendulum during the deceleration of the projectile in the block, then there
would be an external force during the collision—so conservation of momentum
would not be valid, and this would have to be taken into account.
FIGURE 9—19 Object 1,the
projectile, collides with object 2, the
target. They move off, after the
collision, with momenta p and p at
angles O’ and O.
,
/\\
p
.
/
x
Pi
Total energy is conserved, of course.
218
CHAPTER 9
Linear Momentum and Collisions
SECTION 9—7
Collisions in Two or Three Dimensions
219
EXAMPLE 9-11 Proton—proton collision. A proton traveling with speed
8.2 X lu m/s collides elastically with a stationary proton in a hydrogen target as
in Fig. 9—19. One of the protons is observed to be scattered at a 60° angle. At
what angle will the second proton be observed, and what will be the velocities of
the two protons after the collision?
SOLUTION We saw a two-dimensional collision in Example 9—5, where we I
needed to use only conservation of momentum. Now we are given less information:
we have three unknowns instead of two, so we need the kinetic energy equation as
, Eqs. 9—lOa, b, and c become
2
1 = m
well as the two momentum equations. Since m
v
1
v
=
=
0
=
2
v
+
v
vi cos oç
v sin 0
+
+
(ii)
v cos O
v sin O ,
(iii)
8.2 x i0 m/s and O
= 60° are given. In the second and third equa
1 =
where v
tions, we move the v terms to the left side and square both sides of the equations:
2O
2 O’j = v cos
2 cos
1 v cos 0 + v
v — 2v
sin
2
v
O
=
.
vsin
O
2
20
2 0 + cos
We add these two equations and use sin
1 v cos O + ;2
2v
vi2
Into this equation we substitute 2
v?
—
=
=
1 to get:
.
.
.
.
=
VT
“
1 ,
V
from equation
.
(i)
above, and get
1 v cos 0
2v
2
2v
or
v;
=
1 cos O
v
=
(8.2 X i0 m/s)(cos 60°)
=
4.1 x iO m/s.
To obtain v, we use equation (i) above (conservation of kinetic energy):
10
m
/s.
\/v — v’
2 = 7.1 X 5
1
Finally, from equation (iii), we have
)
0866
=
(4.1 X 10 m/s(
= —0.50,
sin
sin O = —
\7.1 x 10 rn/s/
so O = —30°. (The minus sign means particle 2 moves at an angle below the x
axis if particle 1 is above the axis, as in Fig. 9—19.) An example of such a collision
is shown in the bubble chamber photo of Fig. 9—20. Notice that the two trajecto- I
ries are at right angles to each other after the collision. This can be shown to be
true in general for non-head-on elastic collisions of two particles of equal mass,
one of which was at rest initially (see Problem 57).
=
FIGURE 9—20 Photo of a
proton—proton collision in a
hydrogen bubble chamber (a device
that makes visible the paths of
elementary particles). The many
lines represent incoming protons
which can strike the protons of the
hydrogen in the chamber.
PROBLEM
SOLVING
Momentum Conservation and Collisions
1. Be sure no significant external force
acts on your chosen system. That is, the forces that act
between the interacting bodies must be the only sig
nificant ones if momentum conservation is to be used.
[If this is valid for a portion of the problem, you can
use momentum conservation for that portion only.]
2. Draw a diagram of the initial situation, just before
the interaction (collision, explosion) takes place, and
represent the momentum of each object with an
arrow and label. Do the same for the final situation,
just after the interaction.
3. Choose a coordinate system and “+“ and “—“ direc
tions. (For a head-on collision, you will need only an
220
CHAPTER 9
Linear Momentum and Collisions
x axis.) It is often convenient to choose the +x axis in
the direction of one object’s initial velocity.
4. Write momentum conservation equation(s):
total initial momentum = total final momentum.
You have one equation for each component (x, y, z);
only one equation for a head-on collision.
5. If the collision is elastic, you can also write down a
conservation of kinetic energy equation:
total initial kinetic energy = total final kinetic energy.
2 = v
1 — v
[Alternately, you could use Eq. 9—9: v
if the collision is one dimensional (head-on).]
6. Solve algebraically for the unknown(s).
—
Center of Mass (CM)
Until now, when we have dealt with an extended body (that is, a body that has size),
We have assumed that it could be approximated as a point particle or that it undergoes only translational motion. Real extended bodies, however, can undergo rotational and other types of motion as well. For example, the diver in Fig. 9—21a
undergoes only translational motion (all parts of the body follow the same path),
whereas the diver in Fig. 9—21b undergoes both translational and rotational motion.
We will refer to motion that is not pure translation as general motion.
Observations indicate that even if a body rotates, or there are several bodies
that move relative to one another, there is one point that moves in the same path
(hat a particle would if subjected to the same net force. This point is called the
Center of mass (abbreviated CM). The general motion of an extended body (or sys
tern of bodies) can be considered as the sum of the translational motion of the CM,
plus rotational, vibrational, or other types of motion about the CM.
As an example, consider the motion of the center of mass of the diver in
Fig. 9—21; the CM follows a parabolic path even when the diver rotates, as shown
In Fig. 9—21b. This is the same parabolic path that a projected particle follows
when acted on only by the force of gravity (that is, projectile motion). Other points
in the rotating diver’s body follow more complicated paths.
Figure 9—22 shows a wrench translating and rotating along a horizontal sur
face—note that its CM, marked by a red +, moves in a straight line, as shown by the
dashed white line.
We show below that the important properties of the CM follow from Newton’s
laws if the CM is defined in the following way. We can consider any extended body
us being made up of many tiny particles. But first we consider a system made
up of only two particles (or small objects), of mass m
1 and m
. We choose a coordinate
2
system so that both particles lie on the x axis at positions x
1 and x
, Fig. 9—23. The
2
center of mass of this system is defined to be at the position, XCM, given by
XCM
m
x
1
1
m
+
+
m
x
2
m,
=
(a)
x + 2
1
m
m
x
M
where M = m
2 is the total mass of the system.The center of mass lies on the line
1 + m
joining m
1 and m
. If the two masses are equal (m = m
2
2 = m), X(M is midway
between them, since in this case X(M = m(xi + x
)/2m = (x
2
1 + x,)/2. If one mass is
greater than the other, say, m
1 > m
, then the CM is closer to the larger mass. If all the
2
mass is concentrated at x
, say, so m
2
1 = 0, then XcM = (Ox
1 + m
)/(0 + m
x
2
) =
2
as we would expect.
FIGURE 9—22 Translation plus rotation: a wrench moving over
a horizontal surface. The CM, marked with a red +, moves in a
straight line.
(b)
FIGURE 9—21 The motion of the
diver is pure translation in (a), but is
translation plus rotation in (b).
FIGURE 9—23 The center of mass
of a two-particle system lies on the
line joining the two masses. Here
1 > m so the CM is closer to m
m
1
than to
.
2
m
y
xiH
x
,nl
v,
In-,
XCM
SECTION 9—8
Center of Mass (CM)
221
Now let us consider a system consisting of n particles, where n could be very
large. This system could be an extended body which we consider as being made up
of n tiny particles. If these n particles are all along a straight line (call it the x axis),
we define the CM of the system to be located at
XCM
=
x +
2
m + m
x
1
2 +
1 + m
m
XCM
M
=
At
I=
o
5.Om 6.Om
FIGURE 9—24 Example 9—12.
1.Om
X
1
EXAMPLE 9—12 CM of three guys on a raft. Three people of roughly equivalent mass m on a lightweight (air-filled) banana boat sit along the x axis at posi
3 = 6.0 m (Fig. 9—24). Find the position of the CM.
2 = 5.0 m, and x
1 = 1.0 m, x
tions x
SOLUTION We assume each x is the horizontal location of each person’s CM.
Treating them as particles, the overall CM is:
)
3
2 + x
1 + x
m(x
3
2 + mx
1 + mx
mx
x CM =
=
3m
m+m+m
(1.Om + 5.0m + 6.Om)
3
=
12.Om
=4.Om.
If the particles are spread out in two or three dimensions, as for a typical
extended body, then we define the coordinates of the CM as
mjyj
2m,x
—
XCM
‘
YCM
(9—11)
_
‘
ZCM
‘
where x , y , z are the coordinates of the particle of mass m, and again M
=
is the total mass.
Although from a practical point of view we usually calculate the components of
the CM (Eq. 9—i 1), it is sometimes convenient (for example, for derivations) to write
th parti
k is the position vector of the
1
1 = xi + y
j + z
1
Eq. 9—11 in vector form. If r
then
cle, and rCM
1 + YcMJ + ZCM k is the position vector of the center of mass,
XCM
rCM
m3
¶
CM
l.50m
m2
2.00 m
FIGURE 9—25
x
=
Example 9—13.
(9—12)
M
EXAMPLE 9—13 Three particles in 2-D. Three particles, each of mass
2.50 kg, are located at the corners of a right triangle whose sides are 2.00 m and
1.50 m long, as shown in Fig. 9—25. Locate the center of mass.
SOLUTION We choose our coordinate system as shown (to simplify calcula
1 has coordinates
2 on the x axis. Then m
1 at the origin and m
tions) with m
3 has coordinates
2 = 2.0 m, Y2 = 0; and m
2 has coordinates x
0; m
xl = Yi
3 = 1.5m.Then, from Eq. 9—11,
3 = 2.Om,y
x
(2.SOkg)(0) + (2.SOkg)(2.OOm) + (2.5Okg)(2.OOm)
3(2.50 kg)
—‘
XCM
YCM
=
YcM
dm,
=
ZCM
=
where the sum over all the mass elements is fdm
Object. In vector notation, this becomes
fz dm,
(9—13)
M, the total mass of the
(9—14)
frdm.
=
rCM
=
j
A concept similar to center of mass is center of gravity (Co). The CO of a
body is that point at which the force of gravity can be considered to act. Of course,
the force of gravity actually acts on all the different parts or particles of a body, but
for purposes of determining the translational motion of a body as a whole, we can
assume that the entire weight of the body (which is the sum of the weights of all its
parts) acts at the CO. Strictly speaking, there is a conceptual difference between the
center of gravity and the center of mass, but for practical purposes, they are gener
ally the same point. [There would be a difference between the two only if a body
Were large enough so that the acceleration due to gravity was different at different
parts of the body.j
y
x
FIGURE 9—26 An extended body,
here shown in only two dimensions,
can be considered to be made up of
many tiny particles (n), each having
a mass zeimj. One such particle is
shown located at a point
1 = x
r
i + y,j + z
1
k.We take the
1
limit of n —* cc so im becomes
the infinitesimal dm.
=
_
(eiiter of tnas.s
Jx dm,
=
‘
,x
1
,... x,, are their posi
2
,m
1
,... m are the masses of each particle and x
2
where m
is the summation sign meaning to sum over all the parti
tions. The symbol
x,
des, where i takes on integer values from 1 to n. (Often we simply write
.
1
leaving out the i = I to n.) The total mass of the system is M = Ym
I
(oiitiniioit.s (Iistribuhiofl of inass
x
1
m
+ mx
+ m
It is often convenient to think of an extended body as made up of a continm
us distribution of matter. In other words, we consider the body to be made up of
$ particles, each of mass im in a tiny volume around a point x,, y, z, and we take
the limit of n approaching infinity (Fig. 9—26). Then IXm becomes the infinitesimal
mass dm at points x, y, z. The summations in Eqs. 9—li and 9—12 become integrals:
(2.SOkg)(0) + (2.5Okg)(0) + (2.5Okg)(1.50m)
7.50 kg
EXAMPLE 9—14 CM of a thin rod. (a) Show that the CM of a uniform thin
rod of length L and mass M is at its center. (b) Determine the CM of the rod
assuming its linear mass density A (its mass per unit length) varies linearly from
0 at the left end to double that value, A = 2A
A
, at the right end.
0
A
SOLUTION
We choose a coordinate system so that the rod lies on the x axis
with the left end at x = 0, Fig. 9—27. Then
= 0 and ZCM = 0.
The
rod
is
uniform,
its
mass
a)
so
per
unit
length
(linear mass density A) is constant
(
and we write it as A = M/L. We now imagine the rod as divided into infinitesimal
elements of length dx, each of which has mass dm = A dx. We use Eq. 9—13:
1
A
0.50 m.
The CM and the position vector rQM are shown in Fig. 9—25.
=
which satisfies A
other words, a
XCM
M
L
2
Ax
Axdx
AL
2
iE
=
L
=
I
Example
9—14.
y
/
0
4
4
x
dx
dm
=
?dx
x
-
L
(1 + ax)
0
A
0 at x = 0 and gives A = 2A
A
0 at x
ilL. Again we use Eq. 9—13:
=
L
Ax dx
=
ix=()
L
1
A
0
M Jo \
—
I(
x
dx
=
L if (1 + aL)
=
(
=
Ao(x +
A
2
x
=
2. In
\\\ L
x
3
+
3L)
=
6M
2
L
But what is M in terms of A
0 and L? We can write
L
1
M=
dm=
Jx=O
=
L
1
=
=
j-j 0
j
where we used A = M/L. This result, X(M at the center, is what we expected.
A at x = 0 and we are told that A increases linearly to
( b) Now we have A = 0
0 at x = L. So we write
A = 2A
1.33 m
=
L
1
;: jxdm
xCM
FIGURE 9—27
ILAdxAOIL(l
JO
Jo \
Then
0
5A
ML
6
xCM
2
x
)dx
x2)IL
I
lo
=
3
—A
L
0
.
2
5
=L,
which is more than half way along the rod, as we would expect since there is
Note that the numerical values of xCM and YcM depend on the coordinate system
chosen, but the physical position of the CM (relative to the three particles) does not
222
CHAPTER 9
depend on the reference frame.
more mass to the right.
SECTION 9—8
Center of Mass (CM)
223
_____
___________
\J li))
St 1)/Ik117af
i)l)/(
(ji
z
x
FIGURE 9—28 Cylindrical disk with
origin of coordinates at geometric
center.
iii
I ‘xtend’d hüd,es
FIGURE 9—29 Example 9—15. This
L-shaped object has thickness t (not
shown on diagram).
EXAMPLE 9—15 CM of L-shaped flat object. Determine the CM of the un
form thin L-shaped construction brace shown in Fig. 9—29.
SOLUTION Consider the object as two rectangles: rectangle A, which
2.06 m X 0.20 m, and rectangle B, which is 1.48 m X 0.20 m. With the origin at’
as shown, the CM of A is at
V
2.06m
0
H
O.20m
A
For symmetrically shaped bodies of uniform composition, such as spheres.
cylinders, and rectangular solids, the CM is located at the geometric center of the
body. Consider a uniform circular cylinder, such as a solid circular disk. We expect
the CM to be at the center of the circle. To show that it is, we first choose a coordi
perpendicular
nate system whose origin is at the center of the circle with the z axis 1
in Eq. 9—11, there is as much.
to the disk (Fig. 9—28). When we take the sum
mass at any +x, as there is at —xi. So all terms cancel out in pairs and XCM = 0. The I
same is true for YcM In the vertical (z) direction, the CM must lie halfway betweet
the circular faces: if we choose our origin of coordinates at that point, there is
much mass at any +z as at —z, so ZCM = 0. For other uniform, symmetricallvj
shaped bodies, we can make similar arguments to show that the CM must lie on
line of symmetry. If a symmetric body is not uniform, then these arguments do not
hold. For example, the CM of a wheel or disk weighted on one side is not at
geometric center but closer to the weighted side.
To locate the center of mass of a group of extended bodies, we can u
1 are the
Eq. 9—11, where the m are the masses of these bodies and x,, y, and z
coordinates of the CM of each of the bodies.
.
CMB
xA =
The
1.48 m
CM
1.03m,
YA
=
0.lOm.
=
1.96m,
YB
where p is the density. The mass of B is
MB = (1.48 m)(0.20 m)(pt)
and the total mass is M
Thus
=
MAXA + MBXB
XCM
=
m
(pt),
2
(0.412
)
)(pt),
2
(0.296 m
m
(pt).
2
(0.708
)
)(1.96 m)
2
)(1.03 m) + (0.296 m
2
(0.412 m
=
)
2
(0.708 m
where pt was canceled out in numerator and denominator. Similarly,
)(—0.74 m)
2
)(0.10 m) + (0.296 m
2
(0.412 m
point
1.42
M
YcM =
=
—0.25 m,
)
2
(0.708 m
point so labeled in Fig. 9—29. For
the
which puts the CNapproximately at
object is assumed to be uniform.
the
since
object not perfectly thin, ZCM = t/2,
FIGURE 9—30 Determining the
of a flat uniform body.
FIGURE 9—31
Finding the
CM
CG.
-
I
224
CHAPTER 9
I
.
Note in this last Example that the CM can actually lie outside the boa
Another example is a doughnut whose CM is at the center of the hole.
It is often easier to determine the CM or CO of an extended body experimen
tally rather than analytically. If a body is suspended from any point, it will
(Fig. 9—30) unless it is placed so its CO lies on a vertical line directly below
point from which it is suspended. If the object is two-dimensional, or has a plane o
symmetry, it need only be hung from two different pivot points and the respecti
vertical (plumb) lines drawn. Then the CO will be at the intersection of the tv
lines, as in Fig. 9—31. If the object doesn’t have a plane of symmetry, the CO Wi
respect to the third dimension is found by suspending the object from at le
three points whose plumb lines do not lie in the same plane.
Linear Momentum and Collisions
As mentioned in Section 9—8, a major reason for the importance of the concept of
enter of mass is that the translational motion of the CM for a system of particles (or
extended bodies) is directly related to the net force on the system as a whole. We
flow show this, by examining the motion of a system of n particles of total mass M,
which we assume remains constant. We begin by rewriting Eq. 9—12 as
MrCM
=
.
r
1
m
We differentiate this equation with respect to time:
1
dr
drCM
M-j----
=
MVCM
=
or
Where v,
of the
(9—15)
/dt is the velocity of the jth particle of mass m
1
dr
1 and VCM is the velocity
We take the derivative with respect to time again and obtain
=
CM.
dv
M—f
=
Where a,
/dt is the acceleration of the th particle. Now dvQM/dt is the acceler
1
dv
ation of the CM, aCM. By Newton’s second law, m,a
1 = F
1 where F
1 is the net force
on the jth particle. Therefore
(9—16)
Thai is, the vector sum of all the forces acting on the system is equal to the total
—0.74m.
The mass of A, whose thickness is t, is
MA = (2.06m)(0.20m)(t)(p)
0.20m
Center of Mass and Translational Motion
F.
l+F
fl=2
M=F
...+F
2
MaC
+
of B is at
XB
[9_i]
mass of the system times the acceleration of its center of mass. Note that our sys
(cm of n particles could be the n particles that make up one or more extended
hodies.
The forces F
1 exerted on the particles of the system can be divided into two
lypes: (1) external forces exerted by objects outside the system and (2) internal
forces that particles within the system exert on one another. By Newton’s third
law. the internal forces occur in pairs: if one particle exerts a force on a second partide in our system, the second must exert an equal and opposite force on the first.
Thus, in the sum over all the forces in Eq. 9—16, these internal forces cancel each
øiher in pairs. We are left, then, with only the external forces on the right side of
Lq. 9—16:
MaCM =
Fext,
[constant M}
(9—17)
I
ii IO\ .S SJ( ()\l) /111
ii t
where Fext is the sum of all the external forces acting on our system, which is the
0ff force acting on the system. Thus
the sum of all the forces acting on the system is equal to the total mass of the
system times the acceleration of its center of mass.
\/•tt•I)\. I( ()\I) I
W
1
(1(11 (1 SVSlt’iii)
ibis is Newton’s second law for a system of particles. It also applies to an
vtended body (which can be thought of as a collection of particles), and to a
4y%tem of bodies. Thus we conclude that
the center of mass of a system of particles (or bodies) with total
mass M moves like a single particle of mass M acted upon by the same net
external force.
I,i’I’.t/iIll;,’,’/
1iiOfiOii o/ \I
Thai is, the system moves as if all its mass were concentrated at the CM and all the
xtcrnal forces acted at that point. We can thus treat the translational motion of
any body or system of bodies as the motion of a particle (see Figs. 9—21 and 9—22).
SECTION 9—9
Center of Mass and Translational Motion
225
This theorem clearly simplifies our analysis of the motion of complex systems and
extended bodies. Although the motion of various parts of the system may be com
plicated, we may often be satisfied with knowing the motion of the CM. This theo
rem also allows us to solve certain types of problems very easily, as illustrated by
the following Example.
CONCEPTUAL EXAMPLE 9—jj A two-stage rocket. A rocket is shot into
the air as shown in Fig. 9—32. At the moment it reaches its highest point, a hori
zontal distance d from its starting point, a prearranged explosion separates it into
two parts of equal mass. Part I is stopped in midair by the explosion and falls ver
tically to Earth. Where does part II land? Assume g = constant.
After the rocket is fired, the path of the CM of the system continues
to follow the parabolic trajectory of a projectile acted on only by a constant grav
itational force. The CM will thus arrive at a point 2d from the starting point. Since
the masses of I and II are equal, the CM must be midway between them. There
fore, II lands a distance 3d from the starting point. (If part I had been given a
kick up or down, instead of merely falling, the solution would have been some
what more complicated.)
An interesting example is the discovery since 1996 of nearby stars that seem
to “wobble.” What could cause such a wobble? Very likely a planet
that orbits about the star, each exerting a gravitational force on the other. The
planets are too small and too far away to have been observed directly by existing
telescopes. But the slight wobble in the motion of the star suggests that both the
planet and the star (its sun) orbit about their mutual center of mass, and hence
the star appears to have a wobble. Irregularities in the star’s motion can be
øhtained to an accuracy of 3 m/s, and from the data, the size of the planets’ orbits
Can be obtained, as well as their masses (if as large as Jupiter’s). See Fig. 6—17 in
Chapter 6.
P H Y S I C S
A P P L I E D
Distant planets discovered
SOLUTION
(ETEI
—
,
,
‘I
FIGURE 9-32
Example 9-16.
I
0
\4
d
d
We can write Eq. 9—17, MaCM = Fext, in terms of the total momentum P of
a system of particles. P is defined, as we saw in Section 9—2 as
P
From Eq. 9—15
iOt(ll lflOllI(’lltllllI
P
v
1
m
=
(MvCM =
+
v
2
m
m
v
1
)
+
+
we
mv
=
.
1
p
have
(9—18)
= Pv.tvCM
Thus, the total linear momentum of a system of particles is equal to the product of
the total mass M and the velocity of the center of mass of the system. Or, the linear
momentum of an extended body is the product of the body’s mass and the velocity
of its CM.
If we differentiate Eq. 9—18 with respect to time, we obtain (assuming the
total mass M is constant)
dP
=
M
We now treat systems whose mass varies. Such systems could be treated as a type
= Fext where P is the
total momentum of the system and Fext S the net external force exerted on it.
Great care must be taken to define the system, and to include all changes in
momentum. An important application is to rockets, which propel themselves forward by the ejection of burnt gases: the force exerted by the gases on the rocket
accelerates the rocket. The mass M of the rocket decreases during this process, so
dM/dt < 0. Another application is the dropping of material (gravel, packaged
*oods) onto a conveyor belt. In this situation, the mass M of the loaded conveyor
hclt increases, so dM/dt > 0.
To treat the general case of a system of variable mass, let us consider the sys
tem shown in Fig. 9—33. At some time t, we have a system of mass M and momen
tum Mv; we also have a tiny (infinitesimal) mass dM traveling with velocity u
which is about to enter our system. An infinitesimal time dt later, the mass dM
combines with the system. For simplicity we will refer to this as a “collision.” So
Our system has changed in mass from M to M + dM in the time dt. Note that dM
can be less than zero, as for a rocket propelled by ejected gases.
In order to apply Eq. 96, dP/dt = IFext, we must consider a definite fixed
system of particles. That is, in considering the change in momentum, dP, we must
Consider the momentum of the same particles initially and finally. We will define
our total system as including M plus dM. Then initially, at time t, the total
momentum is Mv + udM (Fig. 9—33). At time t + di’, after dM has combined
with M, the velocity of the whole is now v + dv and the total momentum is
(Al + dM)(v + dv). So the change in momentum dP is
of inelastic collision, but it is simpler to use Eq. 9—6, dP/dt
dP=(M+dM)(v+dv)—(Mv+udM)
=
= MaCM.
dP
dt
—
Fext,
(9—6)
where Fext is the net external force on the system. This is just Eq. 9—6 obtained
earlier: Newton’s second law for a system of objects. It is valid for any definite
we can determine how the total
fixed system of particles or objects. If we know
momentum changes.
226
CHAPTER 9
Linear Momentum and Collisions
M dv
+
dP
From Eq. 9—17, we see that
WTONS SECOND LAW
(jr a svteiJ
,
vdM
+
—
dM dv
dM
(a)
(b)
FIGURE 9—33
(a)
At time
t,
a mass
dM is about to be added to our
system M. (b) At time t + dt, the
mass dM has been added to our
system.
udM
then, by Eq. 9—6, we have
dvCM
-a;-
•—1o Systems of Variable Mass; Rocket Propulsion
Mdv
+
vdM
di’
—
udM
where we have dropped the term dM dv/dt, since in the limit of infinitesimals it is
Zero (dv/dt may not approach zero but dM surely does). Thus we get
‘ext
dv
M —-
—
(u
—
v)
dM
(9—19a)
.
*SECTION 9—10
Systems of Variable Mass; Rocket Propulsion
227
Note that the quantity (u
M. That is,
—
v) is the relative velocity,
Vrej,
of dM with respect
tol
VrejUV
is the velocity of the entering mass dM as seen by an observer on M. So we can
rearrange Eq. 9—19a:
M
I /1HI,S1
0/?
dv
=
EXAMPLE 9—18 Rocket propulsion. A fully fueled rocket has a mass of
kg, of which 15,000 kg is fuel. The burned fuel is spewed out the rear at a
rate of 190 kg/s with a speed of 2800 m/s relative to the rocket. If the rocket is
fired vertically upward (Fig. 9—35) calculate: (a) the thrust of the rocket; (b) the
net force on the rocket at blastoff and just before burnout (when all the fuel has
been used up); (c) the rocket’s velocity as a function of time, and (d) its final
velocity at burnout. Ignore air resistance and assume the acceleration due to
gravity is constant at g
9.80 rn/s
2
Fext + Vrel
dM
(9—19b)
We can interpret this equation as follows. Mdv/dt is the mass times the accelera-i
tion of M. The first term on the right, IFext, refers to the external force on the
mass M (for a rocket, it would include the force of gravity and air resistance). It
does not include the force that dM exerts on M as a result of their collision. This i
taken care of by the second term on the right, Vrei(dM/dt), which represents
rate at which momentum is being transferred to (or from) the mass M because c
the mass that is added to (or leaves) it. It can thus be interpreted as the force
exerted on the mass M due to the addition (or ejection) of mass. For a rocket thir
term is called the thrust, since it represents the force exerted on the rocket by
expelled gases.
APPLIED
‘!() /iI1L H)ii
/evor belt
EXAMPLE 9—17 Conveor belt. You are designing a conveyor system for
gravel yard. A hopper drops gravel at a rate of 75.0 kg/s onto a conveyor belt that I
2.20m/s (Fig. 9—34). (a) Determine the force
moves at a constant speed v
needed to keep the conveyor belt moving. (b) What power output must the
motor have that drives the conveyor belt.
.
V
I
Fext
=
M
dv
(u
—
—
v)
dM
=
Vrel
=
5.3 X iO N.
i/mist
FG
where we have taken upward as positive 50 Vrel is negative because it is down
ward, and dM/dt is negative because the rocket’s mass is diminishing.
(b) Fext = Mg = (2.1 X i0 kg)(9.80 m/s
)
2
2.1 X
N initially, and
(6.0
i0 kg)(9.80 m/s
)
2
5.9 X iO N at burnout. Hence, the net force on the
rocket at blastoff is
=
Mg
‘-dM
x
=
5.3 X 10
N
5
—
2.1 X 10
N
5
=
3.2 X 10
N,
5
[blastoff]
Vgases
and just before burnout it is
FIGURE 9—35
Fnet
=
5.3
N
5
10
X
—
5.9
N
4
10
X
=
4.7 X 10 N.
After burnout, of course, the net force is that of gravity, —5.9
(c) From Eq. 9—19b we have
Fext
dv
=
dt
(
(V
=
(—2800 m/s)(—190 kg/s)
=
+ Vrel
[burnout]
X
iO N.
Example 9—18;
M is the mass
of the rocket at any instant and is
decreasing until burnout.
VreI
=
Vgases
—
Vrocket.
dM
where Fext = —Mg, and M is the mass of the rocket and is a function of time.
Since vrel is constant, we can integrate this easily:
dM
—-
FIGURE 9—34 Gravel dropped
from hopper onto conveyor belt.
rocket
(a) The thrust (see discussion after Eq. 9—19b) is:
Fthrust
SOLUTION We assume that the hopper is at rest so u = 0 and that the hopper
has just begun dropping gravel, so dM/dt = 75.0 kg/s. Since the belt moves at?’
constant speed (dv/dt = 0), we have from Eq. 9—19a:
// —7Trr—7—rT7r7T7—>
APPLIED
RocIct propulsion
Fthmst
SOLUTION
Fnet
PHYSICS
PHYSICS
000
LI,
0— (0—
j
dv
=
—
0
V
Jo
rMdM
g dt
+ Vrel
I
0
JM
M
or
=
v-
=
(2.20 m/s)(75.0 kg/s)
=
165 N.
v(t)
=
0
v
—
gt
+ VreI
ln
M
1’ (i: /1iIlt1101l
uf
t
(b) From Eq. 8—22, this force does work at the rate
dW
d
dM
2
=
Fext
V
=
V
=
363 W,
which is the power output required of the motor.
[But note that this work does not all go into kinetic energy of the gravel, since
where v(t) is the rocket’s velocity and M its mass at any time t. Note that Vrel 5
negative (—2800 m/s in our case) because it is opposite to the motion, and that
In (M/M
) is also negative because M
0
0 > M. Hence, the last term—which repre
sents the thrust—is positive and acts to increase the velocity.
(d) The time required to reach burnout is the time needed to use up all the fuel
(15,000kg) at a rate of 190 kg/s; so at burnout,
1.50
1dM 2
dK
d(1
—j_=j-Mv) =-—v,
which is only half the work done by text. The other half of the external work
done goes into thermal energy produced by friction between the gravel and the
belt (the same friction force that accelerates the gravel).]
228
CHAPTER 9
Linear Momentum and Collisions
=
If we take v
0
=
=
X iO kg
190 kg/s
=
79.0 s.
0, then using the result of part (c):
—(9.80 m/s
)(79 s)
2
+
(—2800 m/s)(ln
6000 kg \
21,000 kg)
*SECTION 9—10
=
2730
rn/s.
i’
at biertiout
Systems of Variable Mass; Rocket Propulsion
229
fl Summary
The linear momentum, p, of an object is defined as the
product of its mass times its velocity,
p
my.
=
mv + mv
In terms of momentum, Newton’s second law can be
written as
=
dp
—.
dt
That is, the rate of change of momentum of an object
equals the net force exerted on it.
When the net external force on a system of objects is
zero, the total momentum remains constant. This is the
law of conservation of momentum. Stated another way,
the total momentum of an isolated system of objects
remains constant.
The law of conservation of momentum is very useful in
dealing with the class of events known as collisions. In a
collision two (or more) bodies interact with each other
for a very short time and the force between them during
this time is very large compared to any other forces act
ing. The impulse of such a force on a body is defined as
J
=
fFdt
and is equal to the change in momentum of the body as
long as F is the net force on the body:
ftf
P=PfPi=
Fdt=J.
1
it
Total momentum is conserved in any collision:
Pi + P2
=
Pi + P.
The total energy is also conserved, but this may not be
useful unless the only type of energy transformation
=
v.
2
v + m
1
m
If kinetic energy is not conserved, the collision is called
inelastic. If two colliding objects stick together as the
result of a collision, the collision is said to be completely
inelastic.
For a system of particles, or for an extended body that
can be considered as having a continuous distribution of
matter, the center of mass (CM) is efined as
or
XCM
=
jIxdm,
ydm, ZCM
YcM =
=
jzdm,
where M is the total mass of the system.
The center of mass (cM) of a system is important
because this point moves like a single particle of mass M
acted on by the same net external force, IFext. In equa
tion form, this is just Newton’s second law for a system of
particles (or extended bodies):
MaCM =
=
Fext.
fl Problems
Section 9—1
1. (I) Calculate the force exerted on a rocket at takeoff given
that the propelling gases are expelled at a rate of 1200 kg/s
with a speed of 50,000 rn/s.
2. (I) The momentum of a particle, in SI units, is given by
i
2
8.Oj
p = 4.8t
8.9tk. What is the force as a function
of time?
3. (II) (a) What is the momentum of a 30.0-g sparrow with a
speed of 12 rn/s? (b) What will be its momentum 12 s later if
a constant 2.0 X 102 N force due to air resistance acts on it?
4. (II) A 145-g baseball, moving along the x axis with speed
30.0 rn/s, strikes a fence at a 45° angle and rebounds along
the y axis with unchanged speed. Give its change in momen
turn using unit vector notation.
5. (II) The force on a particle of mass m is given by
F = 26i
j where F is in N and t in s. What will be the
2
12t
change in the particle’s mornenturn between t = 1.0 s and
t = 2.Os?
(. (II) A 4200-kg rocket is traveling in outer space with a
velocity of 120 rn/s toward the Sun. It needs to alter its
course by 23.0°, which can be done by shooting its rockets
briefly in a direction perpendicular to its original rnotion. If
the rocket gases are expelled at a speed of 2200 rn/s relative
to the rocket, what mass of gas must be expelled?
—
H Questions
1. We claim that momentum is conserved. Yet most moving
objects eventually slow down and stop. Explain.
1 and m
2 rest on a frictionless table
2. Two blocks of mass m
and are connected by a spring. The blocks are pulled apart,
stretching the spring, and then released. Describe the subse
quent motion of the two blocks.
3. A light body and a heavy body have the same kinetic
energy. Which has the greater momentum?
4. When a person jumps from a tree to the ground, what hap
pens to the momentum of the person upon landing?
5. Explain, on the basis of conservation of momentum, how a
fish propels itself forward by swishing its tail back and forth.
6. Why, when you release an inflated, untied balloon, does it
fly across the room?
19. Show on a diagram how your CM shifts when you move
from a lying position to a sitting position.
20. Describe an analytic way of determining the CM of any thin,
triangular-shaped, uniforrn plate.
21. Place yourself facing the edge of an open door. Position
your feet astride the door with your nose and abdomen
touching the door’s edge. Try to rise on your tiptoes. Why
can’t this be done?
22. If only an external force can change the momentum of the
center of mass of an object, how can the internal force of
the engine accelerate a car?
23. How can a rocket change direction when it is far out in
space and essentially in a vacuum?
24. In observations of nuclear a-decay, the electron and recoil
nucleus often do not separate along the same line. Use conservation of momentum in two dimensions to explain why
this implies the emission of at least one other particle in the
disintegration.
25. Bob and Jim decide to play tug-of-war on a frictionless lake.
Jim is considerably stronger than Bob, but Bob weighs
160 lbs while Jirn weighs 145 lbs. Who loses by crossing over
the rnidline first?
26. At a carnival game you try to knock over a heavy cylinder
by throwing a small ball at it. You have a choice of throwing either a ball that will stick to the cylinder, or a second
ball of equal rnass and speed that will bounce backward
off the cylinder. Which ball is more likely to make the
cylinder move?
IFext,
where M is the total mass of the system, aCM is
the acceleration of the CM of the system, and IFext is the
total (net) external force acting on all parts of the system.
For a system of particles of total linear momentum
P =
= MvCM, Newton’s second law is
T
11. The speed of a tennis ball on the return of a serve can be
just as fast as the serve, even though the racket isn’t swung
very fast. How can this be?
12. Is it possible for a body to receive a larger impulse from a
small force than from a large force?
13. How could a force give zero impulse over a nonzero time
interval even though the force is not zero for at least a part
of that time interval?
14. In a collision between two cars, which would you expect to
be more damaging to the occupants: if the cars collide and
remain together, or if the two cars collide and rebound
backward? Explain.
15. A superball is dropped from a height h onto a hard steel
plate (fixed to the Earth), from which it rebounds at very
nearly its original speed. (a) Is the momentum of the ball
conserved during any part of this process? (b) If we con
sider the ball and the Earth as our system, during what parts
of the process is momentum conserved? (c) Answer part
(b) for a piece of putty that falls and sticks to the steel plate.
16. It used to be common wisdom to build cars to be as rigid as
possible to withstand collisions. Today, though, cars are
designed to have “crumple zones” that collapse upon
impact. What advantage does this have?
17. A rocket following a parabolic path through the air sud
denly explodes into many pieces. What can you say about
the motion of this system of pieces?
18. Why is the CM of a 1-rn length of pipe at the pipe’s mid
point, whereas this is not true for your arm or leg?
7. It is said that in ancient times a rich man with a bag of gold
coins was stranded on the surface of a frozen lake and froze
to death. Because the ice was frictionless, he could not push
himself to shore. What could he have done to save himself
had he not been so miserly?
8. If a falling ball were to make a perfectly elastic collision with
the floor, would it rebound to its original height? Explain.
9. (a) An empty sled is sliding on friqtionless ice when Susan
drops vertically from a tree above onto the sled. When she
lands, does the sled speed up, slow down, or keep the same
speed? (b) Later, Susan falls off the sled. When she drops off.
does the sled speed up, slow down, or keep the same speed?
10. Is it easier to hit a home run from a pitched ball than
from one tossed in the air by the batter? Explain.
—
—
7. (II) A ball of mass m falls freely a distance h from rest. It
hits the floor and rebounds to its starting height. (a) What
total time is required for the motion? (Ignore the time of
contact with the floor.) (b) What is the speed of the ball as
it strikes the floor? (c) What is the change in mornenturn of
the ball as it strikes the floor? (d) Calculate the average
force the ball exerts on the floor averaged over the tirne
interval of part (a). Is this result surprising?
8. (III) Air in a 100-krn/h wind strikes head-on the face of a
12. (I) A 10,500-kg railroad car travels alone on a level fric
tionless track with a constant speed of 15.0 rn/s. An addi
tional 6350 kg load is dropped from a tower onto the car.
What then will be its speed?
13. (II) A child in a boat throws a 5.40-kg package out horizon
tally with a speed of 10.0 rn/s, Fig. 9—36. Calculate the veloc
ity of the boat immediately after, assuming it was initially at
rest. The mass of the child is 26.0 kg and that of the boat is
55.0 kg.
14. (II) A 12-gram bullet traveling 190 rn/s penetrates a 2.0-kg
block of wood and emerges going 150 rn/s. If the block is
stationary on a frictionless surface when hit, how fast does it
move after the bullet emerges?
15. (II) An explosion breaks an object, originally at rest, into
two fragments. One fragment acquires twice the kinetic
energy of the other. What is the ratio of their masses?
Which is greater?
16. (II) A ball moving with a speed of 17 rn/s strikes an identi
cal ball that is initially at rest. After the collision, the incom
ing ball has been deviated by 45° from its original direction,
and the struck ball moves off at 30° from the original direc
tion (Fig. 9—37). What are the speeds of the two balls after
the collision?
(l7mIs)’
FIGURE 9—37
19. (II) A radioactive nucleus at rest decays into a second
nucleus, an electron, and a neutrino. The electron and neu
trino are emitted at right angles and have momenta of
8.6 x 10_23 kg rn/s and 6.2 X 1023 kg m/s, respectively. I
What is the magnitude and the direction of the momentum
of the recoiling nucleus?
20. (II) A 900-kg two-stage rocket is traveling at a speed of
6.50 x io rn/s away from Earth when a predesigned explo
sion separates the rocket into two sections of ei/al mass that
then move with a relative speed (relative to ,‘ach other) of
2.80 x io rn/s along the original line of motion. (a) What is
the speed and direction of each section (relative to Earth)
after the explosion’ (fr) How much energy was supplied by
the explosion? {Hii: Vhat is the change in kinetic energy as
a result of the explosion?]
21. (III) A 200-kg projectile, fired with a speed of 100 rn/sat a
60° angle, breaks into three pieces of equal mass at the high
est point of its arc. Two of the fragments move with the
same equal speed right after the explosion as the entire pro
jectile had just before the explosion; one of these moves
vertically downward and the other horizontally. Determine
(a) the velocity of the third fragment immediately after the
explosion and (b) the energy released in the explosion.
.
.
(II) (a) A molecule of mass m and speed v strikes a wall at
right angles and rebounds with the same speed. If the colli
sion time is t, what is the average force on the wall during
the collision? (b) If molecules, all of this type, strike the wall
at intervals a time t apart (on the average) what is the aver
age force on the wall averaged ov’ra long time?
/
(II) Suppose the force acting on a tennis ball (mass
0.060 kg) as a function of time is given by the graph of
Fig: 9—39. Use the graph to estimate (a) the total impulse
given the ball, and (b) the speed of the ball after being
struck, assuming the ball is being served so it is nearly at
rest initially.
Section 9—3
22. (I) A 0.145-kg baseball pitched at 35.0 rn/s is hit on a hori
zontal line drive straight back at the pitcher at 56.0 rn/s. If
the contact time between bat and ball is 5.00 X io s, calcu
late the force (assumed to be constant) between the ball
and bat.
23. (I) The tennis ball may leave the racket of a top player on
the serve with a speed of 65.0 m/s. If the ball’s mass is
0.0600 kg and is in contact with the racket for 0.0300 s, what
is the average force on the ball? Would this force be large
enough to lift a 60-kg person?
24. (II) A tennis ball of mass m = 0.060 kg and speed
v = 28 rn/s strikes a wall at a 45° angle and rebounds with
the same speed at 45° (Fig. 9—38). What is the impulse given
the wall?
j5°
(II) The force on a bullet is given by the formula
26 >< 10
F = (780
t) N over the time interval t = 0 to
5
t = 3.0 x io s. In this formula t is in seconds and F is
in newtons. (a) Plot a graph of F versus t for t = 0 to
= 3.0 x i0 s. (b) Estimate, using graphs, the impulse
given the bullet. (c) Determine the impulse by integration.
(d) If the bullet achieves a speed of 300 rn/s as a result of
this impulse, given to it in the barrel of a gun, what must the
bullet’s mass be?
—
(III) From what maximum height can a 75-kg person jump
without breaking the lower leg bone on either leg? Ignore
air resistance and assume the body of the person moves a
distance of 0.60 rn from the standing to the seated position
(that is, in breaking the fall). Assume the breaking strength
, and its
2
(force per unit area) of bone is 170 X 106 N/rn
2
smallest cross-sectional area is 2.5 X i0 rn
Problem 16.
17. (II) A rocket of mass m traveling with speed v
0 along the x
axis suddenly shoots out one-third its mass parallel to the y
. Express
0
axis (as seen by an observer at rest) with speed 2v
the velocity of the remainder of the object in I, j, k notation.
18. (II) The decay of a neutron into a proton, an electron, and
a neutrino is an example of a three-particle decay process.
Use the vector nature of momentum to show that if the neu
tron is initially at rest, the velocity vectors of the three must
be coplanar (that is, all in the same plane). The result is not
true for numbers greater than three.
(II) A 140-kg astronaut (including space suit) acquires a
speed of 2.50 rn/s by pushing off with his legs from a
1800-kg space capsule. (a) What is the change in speed of
the space capsule? (b) If the push lasts 0.500 s, what is the
average force exerted by each on the other? As the refer
ence frame, use the position of the capsule before the push.
(c) What is the kinetic energy of each after the push?
FIGURE 9—38
Problem 24.
25. (II) Water strikes the turbine blades of a generator so that
its rebounding velocity is 0.75 of its original magnitude and
reversed in direction. If the flow rate is 60 kg/s and the
original water speed is 10 rn/s. what is the average force on
the blades?
(III) A scale is adjusted so that when a large, flat pan is
placed on it, it reads zero. A water faucet at height
h = 2.5 m above is turned on and water falls into the pan at
a rate R = 0.12 kg/s. Determine (a) a formula for the scale
reading as a function of time t and (b) the reading for
= 15 s. (c) Repeat (a) and (b), but replace the flat pan
2
with a tall, narrow cylindrical container of area A = 20 cm
(the level rises in this case).
Sections 9—4 and 9—5
(II) A ball of mass 0.540 kg moving east (+x direction) with
a speed of 3.90 m/s collides head-on with a 0.320-kg ball at
rest. If the collision is perfectly elastic, what will be the
speed and direction of each ball after the collision?
(II) A 0.450-kg hockey puck, moving east with a speed of
4.20 m/s, has a head-on collision with a 0.900-kg puck mitially at rest. Assuming a perfectly elastic collision, what will
be the speed and direction of each object after the collision?
(II) A 0.060-kg tennis ball, moving with a speed of 7.50 rn/s,
has a head-on collision with a 0.090-kg ball initially moving
away from it at a speed of 3.00 rn/s. Assuming a perfectly
elastic collision, what is the speed and direction of each ball
after the collision?
(II) A softball of mass 0.220 kg that is moving with a speed
of 6.5 rn/s collides head-on and elastically with another ball
initially at rest. Afterward it is found that the incoming
ball has bounced backward with a speed of 3.8 rn/s. Calcu
late (a) the velocity of the target ball after the collision, and
(b) the mass of the target ball.
(II) A 0.280-kg croquet ball makes an elastic head-on colli
sion with a second ball initially at rest. The second ball moves
off with half the original speed of the first ball. (a) What is
the mass of the second ball? (b) What fraction of the original
kinetic energy gets transferred to the second ball?
(II) An atomic nucleus initially moving at 500 m/s emits an
alpha particle in the direction of its velocity, and the new
nucleus slows to 450 m/s. If the alpha particle has a mass of
4.0 u and the original nucleus has a mass of 222 u, what
speed does the alpha particle have when it is emitted?
(II) A ball of mass m makes a head-on-elastic collision with a
second ball (at rest) and rebounds with a speed equal to onefourth its original speed. What is the mass of the second ball?
(II) Determine the fraction of kinetic energy lost by a neutron
1 = 1.01 u) when it collides head-on and elastically with
(m
(a) H (m = 1.01u);(b)H(heavyhydrogen,m = 2.Olu);
(c) 12 C (m = 12.00 u); (d) 208 Pb (lead, m = 208 u).
(II) Show that, in general, for any head-on one-dimensional
elastic collision, the speeds after collision are
2
I’m
m
1
2m
(\m
v = 1
1 + m
\m
1 + 1112)
)
7
1
/m
1
v=v
—
2
m
2
+v
7
2m
),
2
(mi+m
)
2
mi+m
where v
2 are the initial speeds of the two objects of
1 and v
mass m
.
2
1 and m
(III) A 2.0-kg
Section 9—6
42. (II) An 18-g rifle bullet traveling 180 rn/s buries itself in a
3.6-kg pendulum hanging on a 2.8-rn-long string, which
makes the pendulum swing upward in an arc. Determine the
horizontal component of the pendulum’s displacement.
43. (II) (a) Derive a formula for the fraction of kinetic energy
lost, AK/K, for the ballistic pendulum collision of Example
380 g.
14.0 g and M
9—10. (b) Evaluate for m
pieces, one of
two
into
object
an
breaks
explosion
44. (II) An
which has 1.5 times the mass of the other. If 17,500 J were
released in the explosion, how much kinetic energy did each
piece acquire?
45. (II) After a completely inelastic collision between two
objects of equal mass, each having initial speed, v, the two
move off with speed v/3. What was the angle between their
initial directions?
-kg Toyota collides into the rear end of a
3
46. (II) A 0.95 X 10
-kg Cadillac stopped at a red light. The bumpers
3
2.2 x 10
lock, the brakes are locked, and the two cars skid forward
4.8 m before stopping. The police officer, knowing that the
coefficient of kinetic friction between tires and road is
0.40, calculates the speed of the Toyota at impact. What
was that speed?
47. (II) A measure of inelasticity in a head-on collision of two
bodies is the coefficient of restitution, e, defined as
—
e=
2
V
—
1
V
v is the relative velocity of the two bodies after
1
where v’
1 is their relative velocity before.
v
2
the collision and v
(a) Show that for a perfectly elastic collision, e = 1, and for
0. (b) A simple method
a completely inelastic collision, e
for a body collid
restitution
of
for measuring the coefficient
drop the body
to
is
steel
like
surface
hard
very
ing with a
onto a heavy steel plate, as shown in Fig. 9—41. Determine a
formula for e in terms of the original height h and the max
imum height reached after collision h’.
—
—
0
Ii
r jv’i
1
v
49. (III) A 5.0-kg body moving in the +x direction at 5.5 rn/s
collides head-on with a 3.0-kg body moving in the —x direction at 4.0 m/s. Find the final velocity of each mass if: (a) the
bodies stick together; (b) the collision is elastic; (c) the 5.0-kg
body is at rest after the collision; (d) the 3.0-kg body is at rest
after the collision; (e) the 5.0-kg body has a velocity of’
4.0 rn/s in the —x direction after the collision. Are the results
in (c), (d), and (e) “reasonable”? Explain.
50. (III) For the ballistic pendulum, let us estimate the ccuracv
of our approximation that the block (mass M) doeski’t move
during the collision. Use the symbols of Example 9—10 and
Fig. 9—18, and assume the projectile is decelerated uniformly
within the block in a distance d. Estimate: (a) the collisionl
time t; (b) by how much is momentum not conserved, zip,
because there is a net external force acting during this time;
and (c) by what fraction the calculated speed of the bullet I
will be in error if we use the equation given in Example 9—10,
$5, (II) A neutron collides elastically with a helium nucleus (at
rest initially) whose mass is four times that of the neutron.
The helium nucleus is observed to rebound at an angle
= 45°. Determine the angle of the neutron, 0, and the
speeds of the two particles, v and
after the collision.
5 rn/s.
The neutron’s initial speed is 6.2 X i0
16. (III) A neon atom (m
= 20 u) makes a perfectly elastic col
lision with another atom. After the impact, the neon atom
travels away at a 55.6° angle from its original direction and
the unknown atom travels away at a 50.0° angle. What is the
mass (in u) of the unknown atom?
$7. (III) Prove that in the elastic collision of two objects of
identical mass, with one being a target initially at rest, the
angle between their final velocity vectors is always 90°.
(III) For an elastic collision between a projectile particle of
1 and a target particle (at rest) of mass m
, show that
2
mass m
the scattering angle, O, of the projectile (a) can take any
value, 0 to 180°, for m
, but (b) has a maximum angle
2
1 < m
.
2
given by cos
2 = 1
1 > m
/m form
(m
2
)
1
Section 9—7
1 = 7.8 rn/s
1 = 3.3 kg) moving with speed v
51. (II) An eagle (m
= 4.6 kg
2
(m
eagle
second
a
with
course
is on a collision
10.2 rn/s in a direction at right angles to the
2
moving at V
first. After they collide, they hold onto one another. In what
direction, and with what speed, are they moving after th&
collision?
52. (II) A billiard ball of mass mA = 0.400 kg moving with speed
1.80 rn/s strikes a second ball, initially at rest, of mass
VA
0.500 kg. As a result of the collision, the first ball is
mB
deflected off at an angle of 30.0° with a speed V = 1.10 m/s.
(a) Taking the x axis to be the original direction of motion of
ball A, write down the equations expressing the conservation
of momentum for the components in the x and y directions
separately. (b) Solve these equations for the speed, V and
angle, 0’, of ball B. Do not assume the collision is elastic.
53. (II) An atomic nucleus of mass m traveling with speed ‘v col
lides elastically with a target particle of mass 2m (initially at
rest) and is scattered at 90°. (a) At what angle does the tar
get particle move after the collision? (b) What are the final
speeds of the two particles? (c) What fraction of the initial
kinetic energy is transferred to the target particle?
54. (II) Two billiard balls of equal mass move at right angles I
and meet at the origin of an xy coordinate system. One is’
moving upward along the y axis at 2.0 m/s, and the other
is moving to the right along the x axis with speed 3.7 rn/s.
After the collision (assumed elastic), the second ball is mov
ing along the positive y axis (Fig. 9—42). What is the final
direction of the first ball, and what are their two speeds?
,
—
FIGURE 9—44
Section 9—8
(I) The distance between a carbon atom (m = 12 u) and
an oxygen atom (m = 16 u) in the CO molecule is
1.13 X 10_b m. How far from the carbon atom is the center
of mass of the molecule?
61).
(I) An empty 1150-kg car has its CM 2.50 m behind the front
of the car. How far from the front of the car will the CM be
when two people sit in the front seat 2.80 m from the front
of the car, and three people sit in the back seat 3.90 m from
the front? Assume that each person has a mass of 70.0 kg.
61. (II) Find the center of mass of the ammonia molecule. The
. The hydrogens are at the corners
3
chemical formula is NH
of an equilateral triangle (with sides 0.16 nm) that forms the
base of a pyramid, with nitrogen at the apex (0.037 nm ver
tically above the plane of the triangle).
62. (II) Three cubes, of side 1, 21, and 31, are placed next to
one another (in contact) with their centers along a straight
line and the I = 2l cube in the center (Fig. 9—43). What is
the position, along this line, of the CM of this system?
Assume the cubes are made of the same uniform material.
Problem 47. Measurement of coefficient
48. (II) A wooden block is cut into two pieces, one with three
times the mass of the other. A depression is made in both
faces of the cut, so that a firecracker can be placed in it with
the block reassembled. The reassembled block is set on a
rough-surfaced table and the fuse is lit. When the fire
cracker explodes, the two blocks separate and slide apart.
What is the ratio of distances each block travels?
2lo 3lo
2)
(‘2
V23.7 mIs
+x
0
FIGURE 9—42
Problem 54. (Ball 1
2.Om/s
v
=
1
(:)
Problem 64.
65. (II) A uniform thin wire is bent into a semicircle of radius r.
Determine the coordinates of its center of mass with respect
to an origin of coordinates at the center of the “full” circle.
66. (III) Determine the
CM
of a thin, uniform, semicircular plate.
67. (III) Determine the CM of a machine part that is a uniform
cone of height h and radius R, Fig. 9—45. [Hint: Divide the
cone into an infinite number of disks of thickness dz, one of
which is shown.]
I
I
I
y
x
FIGURE 9—45
Problem 67.
68. (III) Determine the CM of a uniform pyramid that has four
triangular faces and a square base with equal sides all of
length s. [Hint: See Problem 67.]
‘-7-’
FIGURE 9—41
of restitution.
64. (II) A uniform thin machine part is a flat circular plate of
radius 2R that has a circular hole of radius R cut out of it.
The center of the hole is a distance 0.80R from the center of
the plate, Fig. 9—44. What is the position of the center
of mass of the plate? [Hint: Think of a solid plate as being
the sum of the given plate plus a small one of radius R.j
71. (II) A 50-kg woman and a 70-kg man stand 11.0 m apart on
frictionless ice. (a) How far from the man is their CM? (b) If
they hold onto the two ends of a rope, and the man pulls on
the rope so that he moves 2.8 m, how far from the woman
will he now be? (c) How far will the man have moved when
he collides with the woman?
Section 9—10
*
76. (II) A 2500-kg rocket is to be accelerated at 3.0 g at take
off from the Earth. If the gases can be ejected at a rate of
30 kg/s, what must be their exhaust speed?
*
77. (II) Suppose the conveyor belt of Example 9—17 is retarded
by a friction force of 140 N. Determine the required output
power (hp) of the motor as a function of time from the
moment gravel first starts falling (t = 0) until 3.0 s after the
gravel begins to be dumped off the end of the 20-m-long
conveyor belt.
72. (II) Suppose that in Example 9—16 (Fig. 9—32) m
11 = 3m
.
1
(a) Where then would m
11 land? (b) What if m
1 = 3m
?
1
73. (II) A helium balloon and its gondola, of mass M, are in the
air and stationary with respect to the ground. A passenger,
of mass m, then climbs out and slides down a rope with
speed v, measured with respect to the balloon. With what
speed and direction (relative to Earth) does the balloon
then move? What happens if the passenger stops?
74. (II) Two people, one of mass 75 kg and the other of mass
55 kg, sit in a rowboat of mass 80 kg. With the boat ini
tially at rest, the two people, who have been sitting at
opposite ends of the boat, 3.0 m apart from each other,
exchange seats. How far and in what direction will the boat
move?
75. (III) A 200-kg flatcar 25 m long is moving with a speed of
5.0 m/s along horizontal frictionless rails. A 90-kg man then
starts walking from one end of the car to the other in the
direction of motion, with speed 2.0 rn/s with respect to
the car. In the time it takes for him to reach the other end,
how far has the flatcar moved?
/
78. (II) A rocket traveling 1850 rn/s away from the Earth at an
altitude of 6400 km fires its rockets, which eject gas at a
speed of 1200 rn/s (relative to the rocket). If the mass of the
rocket at this moment is 25,000 kg and an acceleration of
1.7 m/s
2 is desired, at what rate must the gases be ejected?
* 79.
(II) The jet engine of an airplane takes in 100 kg of air per
second, which is burned with 4.2 kg of fuel per second. The
burned gases leave the plane at a speed of 550 rn/s (relative
to the plane). If the plane is traveling 270 rn/s (600 mi/h),
determine (a) the thrust due to ejected fuel; (b) the thrust
due to accelerated air passing through the engine; and
(c) the power (hp) delivered.
*
*
80. (III) A sled filled with sand slides without friction down a 30°
slope. Sand leaks out a hole in the sled at a rate of 2.0 kg/s. If
the sled starts from rest with an initial total mass of 40.0 kg,
how long does it take the sled to travel 120 m along the slope?
83. A 0.145-kg pitched baseball moving horizontally at 35.0 rn/s
strikes a bat and is popped straight up to a height of 55.6 m
before turning around. If the contact time is 0.50 ms, calcu
late the average force on the ball during the contact.
83. In order to convert a tough split in bowling, it is necessary to
strike the pin a glancing blow as shown in Fig. 9—47. Assume
that the bowling ball, initially traveling at 12.0 m/s, has five
times the mass of a pin and that the pin goes off at 80° from
the original direction of the ball. Calculate the speed (a) of
the pin and (b) of the ball just after collision, and (c) calcu
late the angle through which the ball was deflected. Assume
the collision is elastic and ignore any spin of the ball.
80°
Cuebafl
FIGURE 9—46
Problem 81.
82. During a Chicago storm, winds can whip horizontally at
speeds of 100 km/h. If the air strikes a person at the rate of
40 kg/s per square meter and is brought to rest, calculate
the force of the wind on a person. Assume the person’s area
to be 1.50 m high and 0.50 m wide. Compare to the typical
maximum force of friction (
1.0) between the person
and the ground, if the person has a mass of 70 kg.
236
CHAPTER 9
Linear Momentum and Collisions
FIGURE 9—47
Problem 84.
92. A (light) pallet has a load of identical cases of tomato
paste (see Fig. 9—49), each of which is a cube of length 1.
Find the center of gravity in the horizontal plane, so that a
crane operator can pick up the load without tipping it.
FIGURE 9—49
Uv=3lOm/s
i1
FIGURE 9—48
*
Problem 86.
$7. A ball of mass m makes a head-on elastic collision with a
second ball (at rest) and rebounds with a speed equal to
0.600 of its original speed. What is the mass of the second
ball?
fl General Problems
81. A novice pool player is faced with the corner pocket shot
shown in Fig. 9—46. The relative size of some of the dimensions
(the units aren’t important, only their ratios) are also shown.
Should the player be worried about this being a “scratch shot,”
one where the cue ball will also fall into a pocket? Give
details. Assume equal mass balls and an elastic collision.
$5. A 15-g bullet strikes and becomes embedded in a 1.10-kg
block of wood placed on a horizontal surface just in front of
the gun. If the coefficient of kinetic friction between the
block and the surface is 0.25, and the impact drives the block
a distance of 9.5 rn before it comes to rest, what was the
muzzle speed of the bullet?
*6 A gun is fired vertically into a 1.40-kg block of wood at rest
on a thin horizontal sheet directly above it, Fig. 9—48. If the
bullet has a mass of 21.0 g and a speed of 310 m/s, how high
will the block rise into the air after the bullet becomes
embedded in it?
$8. You have been hired as an expert witness in a court case
involving an automobile accident. The accident involved a
car of mass 2000 kg (car A) which approached a stationary
car of mass 1000 kg (car B). The driver of car A applied his
brakes 15 m before he crashed into car B. After the colli
sion, car A slid 15 m while car B slid 30 m. The coefficient of
kinetic friction between the locked wheels and the road was
measured to be 0.60. Prove to the court that the driver of
car A was exceeding the 55-mph speed limit before applying
the brakes.
$9 For the completely inelastic collision of two railroad cars
that we considered in Example 9—3, calculate how much of
the initial kinetic energy is transformed to thermal or other
forms of energy.
90 A hockey puck of mass 4m has been rigged to explode, as
part of a practical joke. Initially the puck is at rest on a fric
tionless ice rink. Then it bursts into three pieces. One chunk,
of mass m, slides across the ice at speed v. Another chunk, of
mass 2m, slides across the ice at speed 2v, in a direction at
right angles to the direction of the first piece. From this
information, find the velocity of the final chunk.
91 Two blocks of mass m
1 and m
, resting on a frictionless
2
table, are connected by a stretched spring and then released.
(a) Is there a net external force on the system? (b) Deter
mine the ratio of their velocities, 2
/v (c) What is the ratio
1
v
.
of their kinetic energies? (d) Describe the motion of the CM
of this system. (e) How would the presence of friction alter
the above results?
Problem 92.
93. A 5800-kg open railroad car coasts along with a constant
speed of 8.60 m/s on a level track. Snow begins to fall verti
cally and fills the car at a rate of 3.50 kg/mm. Ignoring fric
tion with the tracks, determine the speed of the car after
60.0 mm using momentum conservation.
94. Consider the railroad car of Problem 93, which is slowly fill
ing with snow. (a) Determine the speed of the car as a func
tion of time using Eq. 9—19. (b) What is the speed of the car
after 60.0 mm. Does this agree with the simpler calculation
(Problem 93) based on Section 9—2?
95. A meteor whose mass was about 108 kg struck the Earth
(m = 6.0 x 1024 kg) with a speed of about 15 km/s and
came to rest in the Earth. (a) What was the Earth’s recoil
speed? (b) What fraction of the meteor’s kinetic energy was
transformed to kinetic energy of the Earth? (c) By how
much did the Earth’s kinetic energy change as a result of
this collision?
96. A block of mass m = 2.20 kg slides down a 30.0° incline
which is 3.60 m high. At the bottorn, it strikes a block of
mass M = 7.00 kg which is at rest on a horizontal surface,
Fig. 9—50. (Assume a smooth transition at the bottom of the
incline.) If the collision is elastic, and friction can be
ignored, determine (a) the speeds of the two blocks after
the collision, and (b) how far back up the incline the srnaller
mass will go.
3.6m
FIGURE 9—50
Problems 96 and 97.
97. In Problem 96 (Fig. 9—50), what is the upper limit on the
mass m if it is to rebound from M, slide up the incline, stop,
slide down the incline, and collide with M again?
General Problems
237
r
—T---..
Skeet
----:-------y
4 v=200m/s
h
t
Pellet
0 = 30 rn/s
Skeet _p’-V
c2 30°
4
.
\
5
\
FIGURE 9—51
A 0.25-kg skeet (clay target) is fired at an angle of 30° to
the horizon with a speed of 30 rn/s (Fig. 9—51). When it
reaches the maximum height, it is hit from below by a 15-g
pellet traveling vertically upward at a speed of 200 rn/s.
The pellet is embedded in the skeet. (a) How much higher
did the skeet go up? (b) How much extra distance, zx,
does the skeet travel because of the collision?
). The gravitational slingshot effect in Fig. 9—52, shows the
planet Saturn moving in the negative x direction at its
orbital speed (with respect to the Sun) of 9.6 km/s. The
mass of Saturn is 5.69 X 1026 kg. A spacecraft with mass
825 kg approaches Saturn, moving initially in the +x direc
tion at 10.4 km/s. The gravitational attraction of Saturn (a
conservative force) causes the spacecraft to swing around it
(orbit shown as dashed line) and head off in the opposite
direction. Estimate the final speed of the spacecraft after it
is far enough away to be nearly free of Saturn’s gravita
tional pull.
.
km/s
4
VplO.
—
x
= —9.6 km/s
=?
FIGURE 9—52
Problem 99.
Problem 98.
101. A golf ball of mass 0.045 kg is hit off the tee at a speed of
50 rn/s. The golf club was in contact with the ball for
5.0 x io- s. Find (a) the impulse imparted to the golf ball,
and (b) the average force exerted on the ball by the
golf club.
102. The space shuttle launches an 800-kg satellite by ejecting it
from the cargo bay. The ejection mechanism is
activated and is in contact with the satellite for 4.0 s to give
it a velocity of 0.30 rn/s in the z-direction relative to the
shuttle. The mass of the shuttle is 90,000 kg. (a) Determine
the component of velocity Vf of the shuttle in the minus
z-direction resulting from the ejection. (b) Find the aver
age force that the shuttle exerts on the satellite during the
ejection.
103. Car manufacturers conduct crash tests on new vehicles to
ensure the safety of the passengers. In one such test a car
of mass 1000 kg traveling at 50 km/h strikes a brick wall.
(a) Is this an elastic or inelastic collision? (b) Assume that
the front of the vehicle collapsed by 0.70 m. What is the
impact time? (c) What is the average impulsive force?
104. A massless spring with spring constant k is placed between
a block of mass m and a block of mass 3m. Initially the
blocks are at rest on a frictionless surface and they are
held together so that the spring between them is cornpressed by an amount D from its equilibrium length. The
blocks are then released and the spring pushes them off in
opposite directions. Find the final speeds of the two blocks.
105. In a physics lab, a small cube slides down a frictionless
incline as shown in Fig. 9—54, and elastically strikes a cube
that is only one-half its mass. If the incline is 20cm high and
the table is 90 cm off the floor, where does each cube land?
. A mallet consists of a uniform cylindrical head of mass
M
2.00 kg and a diameter 0.0800 m mounted on a uniform
cylindrical handle of mass 0.500 kg and length 0.240 m, as
shown in Fig. 9—53. If this mallet is tossed, spinning, into
the air, how far above the bottom of the handle is the point
that will follow a parabolic trajectory?
m
I
8.00cm
3
CHAPTER 9
Problem 100.
Linear Momentum and Collisions
CHAPTER
111
.
Rotational Motion About
a Fixed Axis
ntil now, we have been concerned with translational motion. In this chap
ter and the next, we will deal with rotational motion. The material has
been divided so that this Chapter 10 covers all the basic aspects of a rigid
body rotating about a fixed axis. Chapter 11, on the other hand, presents a more
general approach and some slightly more advanced (optional) topics.
Although we will sometimes deal with systems of many particles, we will
mainly be concerned with rigid bodies. By a rigid body we mean a body with a def
inite shape that doesn’t change, so that the particles that compose it stay in fixed
positions relative to one another. Of course, any real body is capable of vibrating
or deforming when a force is exerted on it. But these effects are often small, so the
concept of an ideal rigid body is very useful as a good approximation.
U
I
24.0 cm
FIGURE 9—53
When you want to go faster on a
bicycle, as Lance Armstrong does
in this 1999 Tour de France photo,
you exert a strong force on the pedals.
The speed of a bike depends on the
rotation speed of the wheels and the
pedals. The angular velocity of the
rotating pedals increases not simply
in proportion to the force applied,
but rather in proportion to the net
torque. Torque is the product of the
force times the lever arm
(= perpendicular distance from the
axis of rotation to the line of action of
the force). That is: the angular
acceleration of a rotating body is
proportional to the net applied torque.
When an object is rotating, it has
rotational kinetic energy and angular
momentum. Angular momentum, and
its conservation, play an interesting
and often unexpected role in the real
world, as we will see.
FIGURE 9—54
Problem 105.
239