Cutnell_9th_problems_ch_1_thru_10_converted

push the disks outward. Consider the situation where disk 1 comes to a momentary halt shortly after
the spring is released. Assuming that

,

, and

, find the velocity of disk 2 at that

moment.

19.

A lumberjack

is standing at rest on one end of a floating

that is also at

rest. The lumberjack runs to the other end of the log, attaining a velocity of

relative to the shore, and then

hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between
the logs and the water.

(a) What is the velocity of the first log just before the lumberjack jumps off?

Answer:

REASONING The system consists of the lumberjack and the log. For this system, the sum of the
external forces is zero. This is because the weight of the system is balanced by the corresponding
normal force (provided by the buoyant force of the water) and the water is assumed to be frictionless.
The lumberjack and the log, then, constitute an isolated system, and the principle of conservation of
linear momentum holds.

SOLUTION

(a) The total linear momentum of the system before the lumberjack begins to move is zero, since all
parts of the system are at rest. Momentum conservation requires that the total momentum remains
zero

during the motion of the lumberjack.

Here the subscripts ”1” and ”2” refer to the first log and lumberjack, respectively. Let the direction of
motion of the lumberjack be the positive direction. Then, solving for

gives

The minus sign indicates that the first log recoils as the lumberjack jumps off.

(b) Now the system is composed of the lumberjack, just before he lands on the second log, and the
second log. Gravity acts on the system, but for the short time under consideration while the

lumberjack lands, the effects of gravity in changing the linear momentum of the system are

negligible. Therefore, to a very good approximation, we can say that the linear momentum of the

system is very nearly conserved. In this case, the initial momentum is not zero as it was in part (a);
rather the initial momentum of the system is the momentum of the lumberjack just before he lands on
the second log. Therefore,

In this expression, the subscripts ”1” and ”2” now represent the second log and lumberjack,

respectively. Since the second log is initially at rest,

. Furthermore, since the lumberjack and

the second log move with a common velocity,

. The statement of momentum

conservation then becomes

Solving for , we have

The positive sign indicates that the system moves in the same direction as the original direction of the
lumberjack's motion.

(b) Determine the velocity of the second log if the lumberjack comes to rest on it.

Answer:

REASONING The system consists of the lumberjack and the log. For this system, the sum of the
external forces is zero. This is because the weight of the system is balanced by the corresponding
normal force (provided by the buoyant force of the water) and the water is assumed to be frictionless.
The lumberjack and the log, then, constitute an isolated system, and the principle of conservation of
linear momentum holds.

SOLUTION

(a) The total linear momentum of the system before the lumberjack begins to move is zero, since all
parts of the system are at rest. Momentum conservation requires that the total momentum remains
zero

during the motion of the lumberjack.

Here the subscripts ”1” and ”2” refer to the first log and lumberjack, respectively. Let the direction of
motion of the lumberjack be the positive direction. Then, solving for

gives

The minus sign indicates that the first log recoils as the lumberjack jumps off.

(b) Now the system is composed of the lumberjack, just before he lands on the second log, and the
second log. Gravity acts on the system, but for the short time under consideration while the

lumberjack lands, the effects of gravity in changing the linear momentum of the system are

negligible. Therefore, to a very good approximation, we can say that the linear momentum of the

system is very nearly conserved. In this case, the initial momentum is not zero as it was in part (a);
rather the initial momentum of the system is the momentum of the lumberjack just before he lands on
the second log. Therefore,

In this expression, the subscripts ”1” and ”2” now represent the second log and lumberjack,

respectively. Since the second log is initially at rest,

. Furthermore, since the lumberjack and

the second log move with a common velocity,

. The statement of momentum

conservation then becomes

Solving for , we have

The positive sign indicates that the system moves in the same direction as the original direction of the
lumberjack's motion.

REASONING The system consists of the lumberjack and the log. For this system, the sum of the
external forces is zero. This is because the weight of the system is balanced by the corresponding
normal force (provided by the buoyant force of the water) and the water is assumed to be frictionless.
The lumberjack and the log, then, constitute an isolated system, and the principle of conservation of
linear momentum holds.

SOLUTION

(a) The total linear momentum of the system before the lumberjack begins to move is zero, since all
parts of the system are at rest. Momentum conservation requires that the total momentum remains
zero during the motion of the lumberjack.

Here the subscripts ”1” and ”2” refer to the first log and lumberjack, respectively. Let the direction of
motion of the lumberjack be the positive direction. Then, solving for

gives

The minus sign indicates that the first log recoils as the lumberjack jumps off.

(b) Now the system is composed of the lumberjack, just before he lands on the second log, and the
second log.

Gravity acts on the system, but for the short time under consideration while the lumberjack lands, the
effects of gravity in changing the linear momentum of the system are negligible. Therefore, to a very
good

approximation, we can say that the linear momentum of the system is very nearly conserved. In this
case, the initial momentum is not zero as it was in part (a); rather the initial momentum of the system
is the momentum of the lumberjack just before he lands on the second log. Therefore,

In this expression, the subscripts ”1” and ”2” now represent the second log and lumberjack,
respectively. Since the second log is initially at rest,

. Furthermore, since the lumberjack and the second log move with a

common velocity,

. The statement of momentum conservation then becomes

Solving for , we have

The positive sign indicates that the system moves in the same direction as the original direction of the
lumberjack's motion.

20.

An astronaut in his space suit and with a propulsion unit (empty of its gas propellant) strapped to his
back has a mass of 146 kg. The astronaut begins a space walk at rest, with a completely filled
propulsion unit. During the space walk, the unit ejects some gas with a velocity of

. As a result, the astronaut recoils with a velocity of

. After the gas is ejected, the mass of the astronaut (now wearing a partially empty propulsion unit) is
165 kg. What percentage of the gas was ejected from the completely filled propulsion unit?

21.

A two-stage rocket moves in space at a constant velocity of

. The two stages are then separated by a

small explosive charge placed between them. Immediately after the explosion the velocity of the
1200-kg upper stage is

in the same direction as before the explosion. What is the velocity (magnitude and direction) of the
2400-kg lower stage after the explosion?

Answer:

, in the same direction the rocket had before the explosion

REASONING The two-stage rocket constitutes the system. The forces that act to cause the separation
during the explosion are, therefore, forces that are internal to the system. Since no external forces act
on this system, it is isolated and the principle of conservation of linear momentum applies:

where the subscripts ”1” and ”2” refer to the lower and upper stages, respectively. This expression
can be solved for

.

SOLUTION Solving for
gives
Since
is positive,
.
*22.
A 40.0-kg boy, riding a 2.50-kg skateboard at a velocity of
across a level sidewalk, jumps

forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the

sidewalk is

,

above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the

skateboard's velocity relative to the sidewalk at this instant? Be sure to include the correct algebraic
sign with your answer.

*23.

The lead female character in the movie Diamonds Are Forever is standing at the edge of an offshore
oil rig. As she fires a gun, she is driven back over the edge and into the sea. Suppose the mass of a
bullet is 0.010 kg and its velocity is

. Her mass (including the gun) is 51 kg.

(a) What recoil velocity does she acquire in response to a single shot from a stationary position,
assuming that no external force keeps her in place?

Answer:

(b) Under the same assumption, what would be her recoil velocity if, instead, she shoots a blank
cartridge that ejects a mass of

at a velocity of

?

Answer:

*24.

A 0.015-kg bullet is fired straight up at a falling wooden block that has a mass of 1.8 kg. The bullet
has a speed of

when it strikes the block. The block originally was dropped from rest from the top of a building and
has been falling for a time t when the collision with the bullet occurs. As a result of the collision, the
block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the
building. Find the time t.

*25.

By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the
floor. As the plate falls, its momentum has only a vertical component and no component parallel to
the floor. After the collision, the component of the total momentum parallel to the floor must remain
zero, since the net external force acting on the plate has no component parallel to the floor. Using the
data shown in the drawing, find the masses of pieces 1 and 2.

Problem 25

Answer:

,

REASONING No net external force acts on the plate parallel to the floor; therefore, the component of
the momentum of the plate that is parallel to the floor is conserved as the plate breaks and flies apart.
Initially, the total momentum parallel to the floor is zero. After the collision with the floor, the
component of the total momentum parallel to the floor must remain zero. The drawing in the text
shows the pieces in the plane parallel to the floor just after the collision. Clearly, the linear
momentum in the plane parallel to the floor has two components; therefore the linear momentum of
the plate must be conserved in each of these two mutually perpendicular directions. Using the drawing
in the text, with the positive directions taken to be up and to the right, we have

(1)

(2)

These equations can be solved simultaneously for the masses

and

.

SOLUTION Using the values given in the drawing for the velocities after the plate breaks, we have,
(1)

(2)

Subtracting 2 from 1, and noting that

, gives

. Substituting this value into

either 1 or 2 then yields

.

**26. Adolf and Ed are wearing harnesses and are hanging at rest from the ceiling by means of ropes
attached to them. Face to face, they push off against one another. Adolf has a mass of 120 kg, and Ed
has a mass of 78 kg. Following the push, Adolf swings upward to a height of 0.65 m above his
starting point. To what height above his own starting point does Ed rise?

**27.

A cannon of mass

is rigidly bolted to the earth so it can recoil only by a negligible amount. The

cannon fires an 85.0-kg shell horizontally with an initial velocity of

. Suppose the cannon is then

unbolted from the earth, and no external force hinders its recoil. What would be the velocity of an
identical shell fired by this loose cannon?

(Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the
system.) Answer:

REASONING The cannon and the shell constitute the system. Since no external force hinders the
motion of the system after the cannon is unbolted, conservation of linear momentum applies in that
case. If we assume that the burning gun powder imparts the same kinetic energy to the system in each
case, we have sufficient information to develop a mathematical description of this situation, and solve

it for the velocity of the shell fired by the loose cannon.

SOLUTION For the case where the cannon is unbolted, momentum conservation gives

(1)

where the subscripts ”1” and ”2” refer to the cannon and shell, respectively. In both cases, the burning
gun power imparts the same kinetic energy to the system. When the cannon is bolted to the ground,
only the shell moves and the kinetic energy imparted to the system is

The kinetic energy imparted to the system when the cannon is unbolted has the same value and can be
written using the same notation as in equation 1:

(2)

Solving equation 1 for

, the velocity of the cannon after the shell is fired, and substituting the resulting expression

into Equation 2 gives

(3)

Solving equation 3 for

gives

Section 7.3 Collisions in One Dimension

Section 7.4 Collisions in Two Dimensions

28. After sliding down a snow-covered hill on an inner tube, Ashley is coasting across a level
snowfield at a constant

velocity of

. Miranda runs after her at a velocity of

and hops on the inner tube. How fast do

the two slide across the snow together on the inner tube? Ashley's mass is 71 kg and Miranda's is 58
kg. Ignore the mass of the inner tube and any friction between the inner tube and the snow.

29. Consult Multiple-Concept Example 8 for background pertinent to this problem. A 2.50-g bullet,
traveling at a speed of

, strikes the wooden block of a ballistic pendulum, such as that in Figure 7.12. The block has a mass
of 215

g.

(a) Find the speed of the bullet–block combination immediately after the collision.

Answer:

(b) How high does the combination rise above its initial position?

Answer:

1.22 m

30.

One object is at rest, and another is moving. The two collide in a one-dimensional, completely
inelastic collision.

In other words, they stick together after the collision and move off with a common velocity.
Momentum is conserved.

The speed of the object that is moving initially is

. The masses of the two objects are 3.0 and 8.0 kg.

Determine the final speed of the two-object system after the collision for the case when the large-
mass object is the one moving initially and the case when the small-mass object is the one moving
initially.

31.

Batman

jumps straight down from a bridge into a boat

in which a criminal is

fleeing. The velocity of the boat is initially

. What is the velocity of the boat after Batman lands in it?

Answer:

REASONING Batman and the boat with the criminal constitute the system. Gravity acts on this system
as an external force; however, gravity acts vertically, and we are concerned only with the horizontal
motion of the system.

If we neglect air resistance and friction, there are no external forces that act horizontally; therefore,
the total linear momentum in the horizontal direction is conserved. When Batman collides with the
boat, the horizontal component of his velocity is zero, so the statement of conservation of linear
momentum in the horizontal direction can be written as Here,

is the mass of the boat, and
is the mass of Batman. This expression can be solved for , the velocity of
the boat after Batman lands in it.
SOLUTION Solving for gives
The plus sign indicates that the boat continues to move in its initial direction of motion.
32. A car
is traveling at
when it collides head-on with a sport utility vehicle
traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was
the sport utility vehicle traveling?
33.
A 5.00-kg ball, moving to the right at a velocity of
on a frictionless table, collides head-on with a
stationary 7.50-kg ball. Find the final velocities of the balls if the collision is
(a) elastic and
Answer:
,
REASONING The system consists of the two balls. The total linear momentum of the two-ball system
is conserved because the net external force acting on it is zero. The principle of conservation of
linear momentum applies whether or not the collision is elastic.
When the collision is elastic, the kinetic energy is also conserved during the collision

SOLUTION

(a) The final velocities for an elastic collision are determined by simultaneously solving the above
equations for the final velocities. The procedure is discussed in Example in the text, and leads to
Equations 7.8a and 7.8b. According to Equation 7.8a:

Let the initial direction of motion of the 5.00-kg ball define the positive direction. Substituting the
values given in the text, these equations give

The signs indicate that, after the collision, the 5.00-kg ball reverses its direction of motion, while the
7.50-kg ball moves in the direction in which the 5.00-kg ball was initially moving.

(b) When the collision is completely inelastic, the balls stick together, giving a composite body of
mass that moves with a velocity . The statement of conservation of linear momentum then

becomes

The final velocity of the two balls after the collision is, therefore,

(b) completely inelastic.

Answer:

REASONING The system consists of the two balls. The total linear momentum of the two-ball system
is conserved because the net external force acting on it is zero. The principle of conservation of
linear momentum applies whether or not the collision is elastic.

When the collision is elastic, the kinetic energy is also conserved during the collision

SOLUTION

(a) The final velocities for an elastic collision are determined by simultaneously solving the above
equations for the final velocities. The procedure is discussed in Example in the text, and leads to
Equations 7.8a and 7.8b. According to Equation 7.8a:

Let the initial direction of motion of the 5.00-kg ball define the positive direction. Substituting the

values given in the text, these equations give
The signs indicate that, after the collision, the 5.00-kg ball reverses its direction of motion, while the
7.50-kg ball moves in the direction in which the 5.00-kg ball was initially moving.
(b) When the collision is completely inelastic, the balls stick together, giving a composite body of
mass that moves with a velocity . The statement of conservation of linear momentum then
becomes
The final velocity of the two balls after the collision is, therefore,
REASONING The system consists of the two balls. The total linear momentum of the two-ball system
is conserved because the net external force acting on it is zero. The principle of conservation of
linear momentum applies whether or not the collision is elastic.
When the collision is elastic, the kinetic energy is also conserved during the collision
SOLUTION
(a) The final velocities for an elastic collision are determined by simultaneously solving the above
equations for the final velocities. The procedure is discussed in Example in the text, and leads to

Equations 7.8a and 7.8b.
According to Equation 7.8a:
Let the initial direction of motion of the 5.00-kg ball define the positive direction. Substituting the
values given in the text, these equations give
The signs indicate that, after the collision, the 5.00-kg ball reverses its direction of motion, while the
7.50-kg ball moves in the direction in which the 5.00-kg ball was initially moving.
(b) When the collision is completely inelastic, the balls stick together, giving a composite body of
mass that moves with a velocity . The statement of conservation of linear momentum then becomes
The final velocity of the two balls after the collision is, therefore,

34. Multiple-Concept Example 9 reviews the concepts that play roles in this problem. The drawing
shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.025 kg and is
moving along the x axis with a velocity of
. It makes a collision with puck B, which has a mass of 0.050 kg and is initially at rest. The
collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the
drawing. Find the final speeds of

(a) puck A and

(b) puck B.

35.

A projectile

is fired at and embeds itself in a stationary target

. With what

percentage of the projectile's incident kinetic energy does the target (with the projectile in it) fly off
after being struck?

Answer:

7.4%

REASONING We obtain the desired percentage in the usual way, as the kinetic energy of the target
(with the projectile in it) divided by the projectile's incident kinetic energy, multiplied by a factor of
100. Each kinetic energy is given by Equation 6.2 as

, where and are mass and speed, respectively. Data for the masses are given, but

the speeds are not provided. However, information about the speeds can be obtained by using the
principle of conservation of linear momentum.

SOLUTION We define the following quantities:

The desired percentage is

(1)

According to the momentum-conservation principle, we have

Note that the target is stationary before being struck and, hence, has zero initial momentum. Solving
for the ratio

, we find that

Substituting this result into Equation 1 gives

36.

Object A is moving due east, while object B is moving due north. They collide and stick together in a
completely inelastic collision. Momentum is conserved. Object A has a mass of

and an initial velocity of

, due east. Object B, however, has a mass of

and an initial velocity of

, due north. Find the magnitude and direction of the total momentum of the two-object system

after the collision.

37. Multiple-Concept Example 7 deals with some of the concepts that are used to solve this problem.
A cue ball is at rest on a frictionless pool table. The ball is hit dead center by a pool stick, which

applies an impulse of

to the ball. The ball then slides along the table and makes an elastic head-on collision with a

second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is
struck.

Answer:

*38.

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held
horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a
block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the
collision is elastic. The masses of the ball and block are, respectively, 1.60 kg and 2.40 kg, and the
length of the wire is 1.20 m. Find the velocity (magnitude and direction) of the ball

(a) just before the collision, and

(b) just after the collision.

*39.

A girl is skipping stones across a lake. One of the stones accidentally ricochets off a toy boat that is
initially at rest in the water (see the drawing). The 0.072-kg stone strikes the boat at a velocity of

,

below due east,

and ricochets off at a velocity of

,

above due east. After being struck by the stone, the boat's velocity is

, due east. What is the mass of the boat? Assume the water offers no resistance to the boat's motion.

Answer:

0.062 kg
*40.
A mine car
rolls at a speed of
on a horizontal track, as the drawing shows. A 150-kg
chunk of coal has a speed of

when it leaves the chute. Determine the speed of the car–coal system after the
coal has come to rest in the car.
Problem 40
*41.
A 50.0-kg skater is traveling due east at a speed of
. A 70.0-kg skater is moving due south at a speed
of
. They collide and hold on to each other after the collision, managing to move off at an angle south of
east, with a speed of . Find
(a) the angle and

Answer:
REASONING The two skaters constitute the system. Since the net external force acting on the system
is zero, the total linear momentum of the system is conserved. In the direction (the east/west
direction), conservation of linear momentum gives
, or
Note that since the skaters hold onto each other, they move away with a common velocity . In the
direction,
, or
These equations can be solved simultaneously to obtain both the angle and the velocity .

SOLUTION
(a) Division of the equations above gives

(b) Solution of the first of the momentum equations gives

(b) the speed , assuming that friction can be ignored.

Answer:

REASONING The two skaters constitute the system. Since the net external force acting on the system
is zero, the total linear momentum of the system is conserved. In the direction (the east/west
direction), conservation of linear momentum gives

, or

Note that since the skaters hold onto each other, they move away with a common velocity . In the
direction,

, or

These equations can be solved simultaneously to obtain both the angle and the velocity .

SOLUTION

(a) Division of the equations above gives

(b) Solution of the first of the momentum equations gives

REASONING The two skaters constitute the system. Since the net external force acting on the system

is zero, the total linear momentum of the system is conserved. In the direction (the east/west
direction), conservation of linear momentum gives
, or

Note that since the skaters hold onto each other, they move away with a common velocity . In the
direction,
, or
These equations can be solved simultaneously to obtain both the angle and the velocity .
SOLUTION
(a) Division of the equations above gives
(b) Solution of the first of the momentum equations gives

*42.

A 4.00-g bullet is moving horizontally with a velocity of

, where the

sign indicates that it is

moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a
horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the
first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that
both blocks are moving after the collision with the bullet. The mass of the first block is 1150 g, and its
velocity is

after the bullet passes

through it. The mass of the second block is 1530 g.

(a) What is the velocity of the second block after

the bullet embeds itself?

(b) Find the ratio of the total kinetic energy after the collisions to that before the collisions.
*43.
An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is
1837 times that of the electron. Assume that all motion, before and after the collision, occurs along
the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision
to that of the electron before the collision?
Answer:
*44. A 60.0-kg person, running horizontally with a velocity of
, jumps onto a 12.0-kg sled that is initially at
rest.
(a) Ignoring theeffects of friction during the collision, find the velocity of the sled and person as they
move away.
(b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of
kinetic friction between the sled and the snow?
**45.
Starting with an initial speed of
at a height of 0.300 m, a 1.50-kg ball swings downward and strikes a
4.60-kg ball that is at rest, as the drawing shows.
(a) Using the principle of conservation of mechanical energy, find the speed of the
1.50-kg ball just before impact.
Answer:
REASONING The two balls constitute the system. The tension in the
wire is the only nonconservative force that acts on the ball. The tension does no
work since it is perpendicular to the displacement of the ball. Since
,
the principle of conservation of mechanical energy holds and can be used to find
the speed of the 1.50-kg ball just before the collision. Momentum is conserved

during the collision, so the principle of conservation of momentum can be used
to find the velocities of both balls just after the collision. Once the collision has
occurred, energy conservation can be used to determine how high each ball rises.
SOLUTION
(a Applying the principle of energy conservation to the 1.50-kg ball, we have
)
If we measure the heights from the lowest point in the swing,
,
and the expression above simplifies to
Solving for , we have
(b If we assume that the collision is elastic, then the velocities of both balls
) just after the collision can be obtained from Equations 7.8a and 7.8b:

Since
corresponds to the speed of the 1.50-kg ball just before the
collision, it is equal to the quantity calculated in part (a). With the given
values of
and
, and the value of
obtained in part (a), Equations 7.8a and 7.8b yield the
following values:
The minus sign in
indicates that the first ball reverses its direction as a
result of the collision.
(c If we apply the conservation of mechanical energy to either ball after the
) collision we have
where
is the speed of the ball just after the collision, and is the final
height to which the ball rises. For either ball,
, and when either
ball has reached its maximum height,

. Therefore, the

expression of energy conservation reduces to

Thus, the heights to which each ball rises after the collision are

(b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls
just after the collision.

Answer:

,

REASONING The two balls constitute the system. The tension in the wire is the only nonconservative
force that acts on the ball. The tension does no work since it is perpendicular to the displacement of
the ball. Since

, the principle of conservation of mechanical energy holds and can

be used to find the speed of the 1.50-kg ball just before the collision. Momentum is conserved during
the collision, so the principle of conservation of momentum can be used to find the velocities of both
balls just after the collision. Once the collision has occurred, energy conservation can be used to
determine how high each ball rises.

SOLUTION

(a) Applying the principle of energy conservation to the 1.50-kg ball, we have

If we measure the heights from the lowest point in the swing,

, and the expression above

simplifies to

Solving for , we have
(b) If we assume that the collision is elastic, then the velocities of both balls just after the collision can
be obtained from Equations 7.8a and 7.8b:
Since
corresponds to the speed of the 1.50-kg ball just before the collision, it is equal to the
quantity calculated in part (a). With the given values of
and
, and the
value of
obtained in part (a), Equations 7.8a and 7.8b yield the following values:
The minus sign in
indicates that the first ball reverses its direction as a result of the collision.
(c) If we apply the conservation of mechanical energy to either ball after the collision we have where
is the speed of the ball just after the collision, and is the final height to which the ball rises.
For either ball,

, and when either ball has reached its maximum height,

.

Therefore, the expression of energy conservation reduces to

Thus, the heights to which each ball rises after the collision are

(c) How high does each ball swing after the collision, ignoring air resistance?

Answer:

0.409 m (1.50-kg ball), 0.380 m (4.60-kg ball)

REASONING The two balls constitute the system. The tension in the wire is the only nonconservative
force that acts on the ball. The tension does no work since it is perpendicular to the displacement of
the ball. Since

, the principle of conservation of mechanical energy holds and can

be used to find the speed of the 1.50-kg ball just before the collision. Momentum is conserved during
the collision, so the principle of conservation of momentum can be used to find the velocities of both
balls just after the collision. Once the collision has occurred, energy conservation can be used to
determine how high each ball rises.

SOLUTION

(a) Applying the principle of energy conservation to the 1.50-kg ball, we have

If we measure the heights from the lowest point in the swing,
, and the expression above
simplifies to
Solving for , we have
(b) If we assume that the collision is elastic, then the velocities of both balls just after the collision can
be obtained from Equations 7.8a and 7.8b:
Since
corresponds to the speed of the 1.50-kg ball just before the collision, it is equal to the
quantity calculated in part (a). With the given values of
and
, and the
value of
obtained in part (a), Equations 7.8a and 7.8b yield the following values:
The minus sign in
indicates that the first ball reverses its direction as a result of the collision.

(c) If we apply the conservation of mechanical energy to either ball after the collision we have where

is the speed of the ball just after the collision, and is the final height to which the ball rises.

For either ball,

, and when either ball has reached its maximum height,

.

Therefore, the expression of energy conservation reduces to

Thus, the heights to which each ball rises after the collision are

REASONING The two balls constitute the system. The tension in the wire is the only nonconservative
force that acts on the ball. The tension does no work since it is perpendicular to the displacement of
the ball. Since

, the principle of conservation of mechanical energy holds and can be used to find the speed of the
1.50-kg ball just before the collision. Momentum is conserved during the collision, so the principle of
conservation of momentum can be used to find the velocities of both balls just after the collision.
Once the collision has occurred, energy conservation can be used to determine how high each ball
rises.

SOLUTION

(a) Applying the principle of energy conservation to the 1.50-kg ball, we have

If we measure the heights from the lowest point in the swing,
, and the expression above simplifies to
Solving for , we have
(b) If we assume that the collision is elastic, then the velocities of both balls just after the collision can
be obtained from Equations 7.8a and 7.8b:
Since
corresponds to the speed of the 1.50-kg ball just before the collision, it is equal to the quantity
calculated in part (a). With the given values of
and
, and the value of
obtained in part (a), Equations 7.8a and 7.8b yield the following values:
The minus sign in
indicates that the first ball reverses its direction as a result of the collision.
(c) If we apply the conservation of mechanical energy to either ball after the collision we have where
is the speed of the ball just after the collision, and is the final height to which the ball rises. For either
ball,

, and when either ball has reached its maximum height,

. Therefore, the expression of

energy conservation reduces to

Thus, the heights to which each ball rises after the collision are

**46.

Multiple-Concept Example 7 outlines the general approach to problems like this one. Two identical
balls are traveling toward each other with velocities of

and

, and they experience an elastic head-on

collision. Obtain the velocities (magnitude and direction) of each ball after the collision.

**47. A ball is dropped from rest from the top of a 6.10-m-tall building, falls straight downward,
collides inelastically with the ground, and bounces back. The ball loses 10.0% of its kinetic energy
every time it collides with the ground. How many bounces can the ball make and still reach a
windowsill that is 2.44 m above the ground?

Answer:

8 bounces

Section

7.5

Center
of
Mass
48. Two particles are moving along the x axis. Particle 1 has a mass
and a velocity
. Particle 2 has a
mass
and a velocity
. The velocity of the center of mass of these two particles is zero. In other
words, the center of mass of the particles remains stationary, even though each particle is moving.
Find the ratio of the masses of the particles.
49.
Consider the two moving boxcars in Example 5. Determine the velocity of their center of mass (a)
before and (b) after the collision. (c) Should your answer in part (b) be less than, greater than, or
equal to the common velocity of the two coupled cars after the collision? Justify your answer.
Answer:
(a)
(b)
(c) equal to
REASONING AND SOLUTION The velocity of the center of mass of a system is given by Equation
7.11.
Using the data and the results obtained in Example 5, we obtain the following:
(a) The velocity of the center of mass of the two-car system before the collision is
(b) The velocity of the center of mass of the two-car system after the collision is
(c) The answer in part (b)

as the common velocity . Since the cars are coupled together,
every point of the two-car system, including the center of mass, must move with the same velocity.
50.
John's mass is 86 kg, and Barbara's is 55 kg. He is standing on the x axis at
, while she is standing
on the x axis at
. They switch positions. How far and in which direction does their center of mass
move as a result of the switch?
51.
Two stars in a binary system orbit around their center of mass. The centers of the two stars are
apart. The larger of the two stars has a mass of
and its center is
from the system's center
of mass. What is the mass of the smaller star?
Answer:
*52. The drawing shows a sulfur dioxide molecule. It consists of two oxygen atoms and a sulfur
atom. A sulfur atom is twice as massive as an oxygen atom. Using this information and the data
provided in the drawing, find (a) the x coordinate and
(b) the y coordinate of the center of mass of the sulfur dioxide molecule. Express your answers in
nanometers
.

Copyright © 2012 John Wiley & Sons, Inc. Al rights reserved.
Problems
Section 8.1 Rotational Motion and Angular
Displacement
Section 8.2 Angular Velocity and Angular Acceleration
1.
A pitcher throws a curveball that reaches the catcher in 0.60 s. The ball curves because it is spinning at
an average angular velocity of
(assumed constant) on its way to the catcher's mitt. What is the angular
displacement of the baseball (in radians) as it travels from the pitcher to the catcher?
Answer:

21 rad

REASONING The average angular velocity is equal to the angular displacement divided by the
elapsed time (Equation 8.2). Thus, the angular displacement of the baseball is equal to the product of
the average angular velocity and the elapsed time. However, the problem gives the travel time in
seconds and asks for the displacement in radians, while the angular velocity is given in revolutions
per minute. Thus, we will begin by converting the angular velocity into radians per second.

SOLUTION Since

and

, the average angular velocity (in rad/s) of the baseball is

Since the average angular velocity of the baseball is equal to the angular displacement

divided by the elapsed time

, the angular displacement is

(8.2)

2.

The table that follows lists four pairs of initial and final angles of a wheel on a moving car. The
elapsed time for each pair of angles is 2.0 s. For each of the four pairs, determine the average angular
velocity (magnitude and direction as given by the algebraic sign of your answer).

Initial angle Final angle

(a)

0.45 rad

0.75 rad

(b)

0.94 rad

0.54 rad

(c)

5.4 rad

4.2 rad

(d)

3.0 rad
3.8 rad
3. The earth spins on its axis once a day and orbits the sun once a year
. Determine the average angular
velocity (in
) of the earth as it
(a) spins on its axis and
Answer:
(b) orbits the sun. In each case, take the positive direction for the angular displacement to be the
direction of the earth's motion.
Answer:

4. Our sun rotates in a circular orbit about the center of the Milky Way galaxy. The radius of the orbit
is

,

and the angular speed of the sun is

. How long (in years) does it take for the sun to make one

revolution around the center?

5.

In Europe, surveyors often measure angles in grads. There are 100 grads in one-quarter of a circle.
How many grads are in one radian?

Answer:

63.7 grad

REASONING AND SOLUTION Since there are

radians per revolution and it is stated in the problem that

there are 100 grads in one-quarter of a circle, we find that the number of grads in one radian is 6.

The initial angular velocity and the angular acceleration of four rotating objects at the same instant in
time are listed in the table that follows. For each of the objects (a), (b), (c), and (d), determine the final
angular speed after an elapsed time of 2.0 s.

Initial angular velocity

Angular acceleration

(a)

(b)

(c)

(d)

7.

The table that follows lists four pairs of initial and final angular velocities for a rotating fan blade.
The elapsed time for each of the four pairs of angular velocities is 4.0 s. For each of the four pairs,
find the average angular acceleration (magnitude and direction as given by the algebraic sign of your
answer).

Initial angular velocity

Final angular velocity

(a)

(b)
(c)
(d)
Answer:
(a)
(b)
(c)
(d)
8. Conceptual Example 2 provides some relevant background for this problem. A jet is circling an
airport control tower at a distance of 18.0 km. An observer in the tower watches the jet cross in front
of the moon. As seen from the tower, the moon subtends an angle of
. Find the distance traveled (in meters) by the jet as the observer
watches the nose of the jet cross from one side of the moon to the other.
9.
A Ferris wheel rotates at an angular velocity of
. Starting from rest, it reaches its operating speed
with an average angular acceleration of
. How long does it take the wheel to come up to operating
speed?
Answer:
8.0 s
REASONING Equation 8.4
indicates that the average angular acceleration is equal to the

change in the angular velocity divided by the elapsed time. Since the wheel starts from rest, its initial
angular velocity is

. Its final angular velocity is given as

. Since the average angular acceleration is given

as

, Equation 8.4 can be solved to determine the elapsed time .

SOLUTION Solving Equation 8.4 for the elapsed time gives

10.

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular
velocity of and is covered with a soft material that does the polishing. An operator holds the polisher
in one place for 45 s, in order to buff an especially scuffed area of the floor. How far (in meters) does
a spot on the outer edge of the disk move during this time?

*11. The sun appears to move across the sky, because the earth spins on its axis. To a person standing
on the earth, the sun subtends an angle of

(see Conceptual Example 2). How much time (in seconds) does it take

for the sun to move a distance equal to its own diameter?

Answer:

128 s

*12.

A propeller is rotating about an axis perpendicular to its center, as the drawing shows. The axis is
parallel to the ground. An arrow is fired at the propeller, travels parallel to the axis, and passes
through one of the open spaces between the propeller blades. The angular open spaces between the
three propeller blades are each

. The vertical drop of the arrow may be ignored. There is a maximum value for the angular speed of
the propeller, beyond which the arrow cannot pass through an open space without being struck by one
of the blades. Find this maximum value when the arrow has the lengths L and speeds v shown in the
following table.

L

v

(a) 0.71 m

(b) 0.71 m

(c) 0.81 m

*13. Two people start at the same place and walk around a circular lake in opposite directions. One
walks with an angular speed of

, while the other has an angular speed of

. How long will it be before

they meet?

Answer:

1200 s

*14. A space station consists of two donut-shaped living chambers, A and B, that have the radii shown
in the drawing. As the station rotates, an astronaut in chamber A is moved
along a circular arc. How far along a circular
arc is an astronaut in chamber B moved during the same time?

Problem 14
*15.
The drawing shows a device that can be used to measure the speed of a bullet. The device consists of
two rotating disks, separated by a distance of
, and rotating with an angular speed of
. The bullet
first passes through the left disk and then through the right disk. It is found that the angular
displacement between the two bullet holes is
. From these data, determine the speed of the bullet.
Answer:
*16.
An automatic dryer spins wet clothes at an angular speed of
. Starting from rest, the dryer reaches its
operating speed with an average angular acceleration of
. How long does it take the dryer to come up to
speed?
*17.
A stroboscope is a light that flashes on and off at a constant rate. It can be used to illuminate a rotating
object, and if the flashing rate is adjusted properly, the object can be made to appear stationary.
(a) What is the shortest time between flashes of light that will make a three-bladed propeller appear
stationary when it is rotating with an angular speed of
?

Answer:
REASONING AND SOLUTION
(a) If the propeller is to appear stationary, each blade must move through an angle of
or
between flashes. The time required is
(b) The next shortest time occurs when each blade moves through an angle of
, or
rad,
between successive flashes. This time is twice the value that we found in part a, or
.
(b) What is the next shortest time?
Answer:
REASONING AND SOLUTION
(a) If the propeller is to appear stationary, each blade must move through an angle of
or
between flashes. The time required is