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Physics 11a Practice Problem Solutions

1

Practice Problem - Finals Prep Solutions 1. This is a kinematics problem with an energy twist. We can use the kinematics we learned in Chapter 3 first to calculate the velocity (and thus the speed) of the arrow at any position for part (a) and then to calculate the position at any speed for part (b). In addition, because we now know the connection between speed and kinetic energy, we can connect position and kinetic energy. The arrow is neither an isolated nor a closed system, so we draw no system diagram. Instead, taking the kinematics approach, we draw a motion diagram (see Figure).

The definition of kinetic energy should be sufficient to tackle part (a) because both inertia and initial velocity are known. Parts (b) and (c) are more involved. In free fall, acceleration is downward and constant. With our choice in Figure 5.04 to have the positive x axis upward, we have ax = –g. The relationship between velocity and position that does not explicitly involve time is Eq. 3.13:

vx ,f

2 ! vx ,i2 = 2a(xf ! xi )

This equation may be applied between any two points in the motion, and so we can use it to find either the arrow's height at a specified speed or its speed at a specified height. This plus the definition of kinetic energy should take care of parts (b) and (c).

(a) Kstart =12 mvstart

2

= 12 (0.12 kg) (40 m s)2 = 96 J


2

(b) Before we can determine the value of K at the instant the arrow reaches half its maximum height,, we must find the distance to the top of the flight. We know the initial position (0), initial velocity (+40 m/s), final velocity (0; why?), and acceleration (–9.8 m/s2; why minus?) and so can write

vx ,top2 = vx ,start

2 + 2ax (xtop ! xstart )

xtop = xstart +vx ,top

2 ! vx ,start2

2ax

= 0 +(0)2 ! (40 m/s)2

2(!9.8 m/s2 )

= 82 m

We can now find the squared velocity at half this maximum height, xhalf = xtop/2:

vx ,half2 = vx ,start

2 + 2ax (xhalf ! xstart )

= (+40 m s)2 + 2(!9.8 m s2 ) (41 m ! 0) = 796 m2 /s2

The arrow's speed at half its maximum height is therefore

vhalf = vx ,half

2 = 796 m2 /s2 = 28 m/s

(Notice that this speed is not half the initial speed.) The kinetic energy at this point is thus

2 2 21 1half half2 2 (0.12 kg) (796 m s ) 48 JK mv= = =

This is half the initial kinetic energy. Having gained half its maximum height has apparently converted half the arrow's kinetic energy to internal energy, or transferred it from the arrow to something else. (c) We know from (b) that at half its maximum height the arrow's speed is 28 m/s. This means that the position where the speed is 20 m/s, half the initial value, must be above the halfway point. Let’s make a guess. Because the velocity in K = 1

2 mv2 is squared, reducing v to half its initial value reduces K to a quarter of its initial value. We know that v = 0 at the top of the arrow’s trajectory (Section 3.3). From (b) we know that when the arrow is at half its maximum height, half of its kinetic energy had been converted to internal energy. Perhaps at three-quarters of the distance to the top of the flight, three-quarters of the kinetic energy has been converted to internal energy, as shown in the energy bars of Figure 5.05.


3

Figure 5.05 Thus, we estimate that the height at which the velocity is +20 m/s, half the original velocity, is about

(0.75)(82 m) = 61.5 m ! 62 m above the starting position. Comparing the answers to (a) and (b) shows that the arrow's speed decreases as the arrow rises, as expected. A maximum distance of 82 m is a reasonable height for an arrow, consistent with the 40 m/s initial speed. We assumed the arrow was in free fall so that we could use the free-fall value for acceleration, and this is a reasonable assumption for an arrow on its upward flight. The rest of the solution follows from kinematics and the definition of kinetic energy. In (c), we were told to use energy arguments to determine a position. If we had not been restricted that way, we could have used our kinematics equation to find the position at which the arrow was moving at +20 m/s, and so let's do that now as a check, using P to denote that position:

vx ,P2 = vx ,start

2 + 2a(xP ! xstart )

xP = xstart +vx ,P

2 ! vx ,start2

2a= 0 +

(20 m/s)2 ! (40 m/s)2

2(!9.8 m/s2 )= 61.2 m " 61 m

in nice agreement with our energy-based guess, the difference being in the third significant digit.


4

2 (a) Because the collision is elastic, their relative speed does not change, so it remains 5.0 m/s. (b) Because your brother has slightly more inertia, he continues to travel in the same direction he had before the collision, but more slowly. You move in that same direction after the collision, too. Because their relative velocity hasn't changed, your final velocity is –5 m/s + (–0.36 m/s) = –5.36 m/s. (c) Yes; it must be, because we used the definition of relative velocity, together with the values given and the answer to (a), to calculate the answer to (b). (d) It must be zero, because the collision was elastic. To confirm

KEinitial

= 12 60 kg( ) 5.0 m/s( )2

= 750 J ! 7.5"102 J

KEfinal

= 12 52 kg( ) #5.36 m/s( )2

+ 12 60 kg( ) #0.36 m/s( )2

= 750 J ! 7.5"102 J

3. Because at the peak of its flight the shell has an instantaneous velocity of zero, its momentum at that instant is zero. Conservation of momentum implies that the pieces fly

off in opposite directions and, further,

4. Both ratios are equal to one. Since identical particles have the same inertia, they each have equal speed both before and after the collision in the zero-momentum frame. Since the collision is elastic, their relative velocity remains the same after the collision as it was before. In the collision, the direction of their velocities changes, but not the magnitude. This is the case regardless of how they collide, provided it’s elastically. 5. The elastic potential energy in the spring was converted into kinetic energy in the moving carts. We know the speed of one cart, and can calculate the speed of the other from conservation of momentum. Then, we can calculate the kinetic energy of the two-cart system.

(0.36 kg)(1.1 m/s) + (0.12 kg)vf ,x = 0vf ,x = !3.3 m/s

K = 12 (0.36 kg) 1.1 m/s( )2 + 1

2 (0.12 kg) 3.3 m/s( )2 = 0.87 J

6. (a) While I don’t know when, in its flight, a plane actually reaches its cruising speed, it does get moving pretty fast well before it reaches its cruising altitude. This suggests that it takes more energy to reach cruising altitude than it does to reach cruising speed, because the engines are providing energy for a longer period of time. (b) Ignoring drag, it takes the amount of kinetic energy the plane has at cruising speed.

K = 12 mv

2 = 12 (2.1!105 kg) 270 m/s( )2 = 7.7 !109 J

0 = m1v1 ! m2v2m1(3v2 ) = m2v2m1

m2

=13

KE1KE2

=12 m1v1

2

12 m2v2

2 =1 " 32

3 "12= 3


5

(c) If the plane’s velocity doesn’t change, then neither does its kinetic energy, so the only thing that changes is its gravitational potential energy.

!U = mgh = (2.1"105 kg)(9.8 m/s2 )(10400 m) = 2.1"1010 J 7. The speed will be greater for the configuration on the right, where the smaller block moves horizontally. That’s because, when the smaller block moves horizontally, its gravitational potential energy does not change. When the smaller block moves vertically, its gravitational potential energy increases as it is lifted by the larger block’s falling. This offsets some of the larger block’s loss of gravitational potential energy. Since less gravitational potential energy is lost, less kinetic energy is gained. Since both systems have the same inertia, less kinetic energy means less speed. In equations, when the larger block falls a distance D, we have

!Uleft = "mgD + 12 mgD = " 1

2 mgD!Uright = "mgD!K = "!UKleft = 1

2 (32 m)vleft

2 = 12 mgD

Kright = 12 (

32 m)vright

2 = mgD

vleft = 23 gD

vright = 43 gD = 2vleft

8. (a) After the block has fallen a distance D, we can calculate its speed from conservation of energy.

12 mblockvblock

2 = mblockgD

vblock = 2gD

We can then calculate the speed of the block with embedded pellet immediately after the collision from conservation of momentum.

mpelletvpellet ! mblockvblock = (mpellet + mblock )v

v =mpelletvpellet ! mblockvblock

mpellet + mblock

We can then calculate Hmax from conservation of energy. (mpellet + mblock )gHmax = (mpellet + mblock )g(H ! D) + 1

2 (mpellet + mblock )v2

Hmax = H ! D +v2

2g

Combining the preceding, we get

Hmax = H ! D +12g

mpelletvpellet ! mblock 2gDmpellet + mblock

"

#$

%

&'

2


6

(b) The energy dissipated is just the difference in object energies immediately before and immediately after the collision. 9. (a) The forces acting on the piano are just the force of gravity and the tension in the rope, which are responsible for its acceleration. The person has to pull with enough force to provide the tension.

mg ! FT = m g8

FT = 78

mg = 78

(3000 N) = 2625 N

(b) Here the forces acting on the piano are the force of gravity and twice the tension in the rope.

mg ! 2FT = m g8

FT = 716

mg = 716

(3000 N) = 1313 N

10. Because, when he first hits the scale, he has a lot of momentum. That is, besides just supporting his weight, the springs in the scale first have to provide a sufficient impulse to bring him to rest with respect to the floor, and they do so very quickly. This is not a good thing to try at home, because such a force might easily exceed the elastic limit for the spring scale and thus ruin it for future use. 11. (a) Since the child is not accelerating, the vector sum of the forces acting on her must total zero. Those forces are the force of gravity, pulling her down, and the force of the trampoline pushing her up.

mg = k!x

k = mg!x

= (20 kg)(9.8 m/s2 )0.11 m

= 1.8 "103 N/m

(b) The change in length of a spring is proportional to the force acting on. Here, the force is the weight of the person standing on the trampoline, which is proportional to the person’s inertia.

!x = 75 kg

20 kg0.11 m = 0.41 m

12. The forces acting on the block on the left are the force of gravity, the force exerted by the spring, and the tension in the rope. The forces acting on the pair of blocks on the right are the force of gravity, and the tension in the rope. The rope ensures that both sides accelerate at the same rate. Combining these, we get:


7

FT ! kD ! mg = ma(2m)g ! FT = (2m)a2FT ! 2kD ! 2mg = 2mg ! F T

FT = 2kD + 4mg3

a = g ! FT

2m= g

3! kD

3m

13. The work required to stretch a spring from its equilibrium length equals the potential energy stored in the spring, which is proportional to the square of the distance it’s stretched.

W =U = 12 kd 2

12 k(0.10 m)2 = 18 J

k = 3600 N/m!U = 1

2 (3600 N/m)(0.20 m)2 "18 J = 54 J

14. (a) Since the skier is not accelerating, the tension in the rope must equal the magnitude of the component of the force of gravity parallel to the slope.

T = mg sin(40°) = (55 kg)(9.8 m/s2 )sin(40°) = 3.5!102 N (b) The work done by the rope equals the product of its force (the tension) times the displacement of its point of application.

W = F!xF = (3.5"102 N)(100 m) = 3.5"104 J

15. Since the sign is not accelerating, the vector sum of the forces acting on it must be zero. Each component must also be zero, in particular, the horizontal component, which is the sum of the horizontal components of the cables’ tensions.

Tx + (80 N)cos(30°) = 0

Tx = !69 N

T =Tx

cos(135°)= 98 N

For the vertical component of the forces acting on the sign to be zero, the sum of the vertical components of the cables’ tensions must equal the force of gravity on the sign, which gives us its inertia.

mg = Ty + (80 N)sin(30°) = T sin(135°) + (80 N)sin(30°) = 109 N

m = 11 kg

16. It is always best to draw a sketch of the situation showing the given information.


8

Here we have two kinds of motion, the rotational motion of the disk about the axle and the linear motion of the block falling under the influence of gravity. We might try a conserved quantity such as energy or angular momentum to find the speed, but will likely need Newton’s Second Law or kinematical equations to get acceleration. If we include the disk, block and string in our system, then the tension forces (which are tangential) should cancel in pairs. The gravitational force on the block may also be tangential, though. Clearly the angular momentum of the system (about the axle) does not remain constant, because both the block and disk contribute ncreasing angular momentum in the same direction. We have no experience as yet in dealing with possible changes in the angular momentum of the Earth, so perhaps we should try energy. If we include the disk, block, string and Earth, we form a closed system. We can determine the speed of the block from its kinetic energy 1 2( )mv2 . It gains kinetic energy through the conversion of gravitational potential energy as it falls. Some of the initial potential energy must also go into the kinetic energy of the disk, however, because the rotational and linear motions are connected by the string. Fortunately, we know the relationship: because of the connecting string, a point on the outer rim of the disk must always have the same linear speed as the block. Thus vblock = ωdiskR. Because there is no source of energy dissipation apparent, we assume that the total mechanical energy K + U of the system is constant:

K f +U f = Ki +Ui

We can use our energy accounting tools to determine the block's linear speed after it has fallen a distance d. We can then use kinematical equations to get the block's linear acceleration. The “disk” is a very short solid cylinder, so we can use the solid cylinder rotational inertia from Table 11.3. Because we want to use the formula for gravitational potential energy (mgy), we must choose our positive y axis upward because that direction was assumed in the derivation of this equation. We know there are going to be many things to keep track of, and so we had better draw a robust diagram that we can add elements to as we go along


9

For convenience, we have chosen the origin at the initial level of the block, making yi = 0 m (and thus Ui = mgyi = 0 J). Because the system starts at rest, Ki = 0 J. After the block has fallen a distance d, it is at yf = –d, information that we put in our diagram. We can get the ball's final linear speed vf by using this final position –d and the relationship between the block's linear speed and the disk’s rotation: vblock = vf = R!disk = R! f

K f +U f = Ki +Ui

12 mv f

2 + 12 I! f

2 + mgy f = 0 + 0 + mgyi

12 mv f

2 + 12 ( 1

2 MR2 )v f

R

"

#$

%

&'

2

+ mg((d) = mg(0)

12 mv f

2 + 14 Mv f

2 = mgd

v f2 =

2mm + 1

2 Mgd

v f =2m

m + 12 M

gd (1)

Because we do not know how much time it takes the block to travel the distance d but do know the initial and final speeds, we can use the equation

vf2 = vi

2 + 2a(xf – xi) for the block’s motion along the y axis to get its acceleration:


10

v f , y2 = vi, y

2 + 2ay ( y f ! yi )

2mm + 1

2 Mgd = (0)2 + 2ay (!d ! 0) = !2ayd

ay = !m

m + 12 M

g

If the block were not connected to the disk by the string, the block would be in free fall and its speed after it had fallen a distance d would be the familiar free-fall result v = 2gd . In our case, the speed is less because of the presence of the term 1 2( )M in the denominator of Eq. 1. We expect this lower speed because some of the gravitational potential energy the system loses as the block falls must go into increasing the rotational kinetic energy of the disk rather than all of it going into increasing the kinetic energy of the block. We also expect that if the disk’s inertia is much smaller than the block’s (M << m), the disk should have little effect. If we set M = 0 kg to get the limiting case, we indeed get the expected free-fall speed. As for the acceleration, the negative sign indicates the acceleration is downward because the positive y axis points upward, as we expect for a falling block that starts from rest. Because not all of the system's potential energy goes into translational kinetic energy, the block cannot fall as fast as it would if there were no rotational diversion. An acceleration magnitude less than g makes sense. 17. As the ball falls, it converts some of its initial gravitational potential energy (strictly speaking, the gravitational potential energy of the Earth-ball system) into kinetic energy, so its speed increases.

12 mv2 = mg!h!h = Lsin"

v = 2g!h = 2gLsin"

In order for the ball to travel in a circle, the radial component of the net force acting on it must provide the centripetal acceleration necessary at that speed. The forces acting on the ball are the force of gravity and the tension in the string.

ar =1m

(T ! mg sin") = v2

LTm! g sin" = 2gLsin"

L= 2g sin"

T = 3mg sin"

(b) The tension in the string is greatest when sin! = 1, that is, when the ball is at the bottom of its trajectory, at which point the tension equals three times the weight of the ball, that is, 3mg.


11

18. The problem is asking for the details of an orbit whose period is T = 24.0 h. If the satellite must always appear at the same location as seen from the ground, then its orbital speed should be constant. This implies a circular orbit. We draw a sketch of the system, using symbolic values for the known gravitational mass of Earth me and the unknown gravitational mass of the satellite ms (Figure W13.1-1).

Figure W13.1-1(new-WP_13.1a.jpg) Kepler’s third law says the square of the orbital period is proportional to the cube of the distance from the object exerting the gravitational force. Thus a specified period implies a certain radius of orbit. We know the gravitational mass of the object around which the satellite orbits (Earth, me=5.97×1024 kg) and the period in which it must complete its orbit: T = 24.0 h = 8.64 !104 s . We need to find an expression for the orbital radius r in terms of these parameters. Because the satellite’s velocity changes direction, we know there is a force exerted on it, and this force is obviously a gravitational force. So we should employ Newton’s second law:

!Fs = ms

!a! It is beneficial to draw a free-body diagram with the pertinent information (see Figure below). As usual, it is wise to choose one axis (an x axis in this case) to be in the direction of the acceleration, which we know to be directed toward the center of the orbit.


12

Because the orbit is circular, we can use the relationship between centripetal acceleration and rotational speed ω, which for the satellite is 2π radians in 24 h, making ω = 2π/T.

ac =!2r

The centripetal acceleration must be supplied by the only force available: gravity. However, because the satellite may not be near Earth’s surface, we cannot set ac equal to g. Instead, we use Newton’s expression for the magnitude of the universal gravitational force:

FEs

g = GmEms

r 2

This should allow us to solve for r. Using Newton’s second law, we substitute known formulae:

FEs,xg = msax

+FEsg = ms (+ac )

GmEms

r 2 = ms (!2r)

GmE =! 2r3

We want to solve for r in terms of T:

GmE =2!T

"#$

%&'

2

r3

r3 = GmE

T2!

"#$

%&'

2


13

This is Kepler’s third law in mathematical form. Putting in numbers, we get

r3 = (6.67 !10"11 Nim2 kg2 )(5.97 1024 kg)8.64 !104 s

2#$

%&'

()

2

r = 4.22 !107 m * 42 !103 km

This looks pretty far up, considering that the space shuttle orbits only a couple of hundred km above the ground. This is the distance from Earth’s center to the satellite, though, and so we need to subtract 6.4×103 km to get the distance from the ground. Also, our value for r is much less than the Earth-Moon distance, and so the answer is plausible. The use of a circular orbit seems reasonable, because we want the satellite to always remain at about the same height. There is another restriction, though: the satellite can only remain directly above a point on the Earth’s equator! The circular orbit must be equatorial, because an orbit designed to pass directly above, say, Kansas City, would only pass directly above that location once in each 24 hours. This is because the center of the circular orbit must be the center of the Earth so that the gravitational force is central. Points on the Earth, though, move in circles parallel to the equator as Earth spins on its axis. Only by making the satellite orbit the equator can it be seen as essentially stationary from points on the ground The r3 expression tells us that the relationship between period T and orbital radius r depends, not surprisingly, on G and on the gravitational mass of the body around which the object orbits. Notice, however, that the T - r relationship does not depend on the gravitational mass of the satellite. Any satellite, regardless of gravitational mass, has the same geosynchronous orbit. Because a geosynchronous orbit means the same equatorial altitude for all satellites, a practical problem arises: how do you get all the satellites people want up there in the limited space available? 19. We can relate the height above the surface of the Earth to the acceleration due to Earth’s gravity at that height.

a(h) =GmE

RE + h( )2 =GmE

RE2 1+ h

RE

!

"#$

%&

2 = g

1+ hRE

!

"#$

%&

2

h = ga(h)

'1!

"#

$

%& RE

So

a(h) = 0.999g

h = 10.999

!1"

#$

%

&' (6.38 (106 m) = (5.00 (10!4 )(6.38 (106 m) = 3.2 (103 m


14

a(h) = 0.99g

h = 10.99

!1"

#$

%

&' (6.38 (106 m) = (5.04 (10!3)(6.38 (106 m) = 3.2 (104 m

a(h) = 0.9g

h = 10.9

!1"

#$

%

&' (6.38 (106 m) = (5.41(10!2 )(6.38 (106 m) = 3.5(105 m

20. [Note: Checkpoint 13.23 works through the details of showing that the force of gravity inside a spherical shell is zero.] Following the hint, let’s imagine that we’re in a very deep mine shaft, and consider the gravitational force due to that portion of the Earth that’s farther from the center than we are, and that portion that’s closer, separately. Since the force of gravity inside a spherical shell is zero, it’s also zero within any number of them. Since we can consider that portion of the Earth that’s farther from the center than we are as a collection of thin spherical shells, all of which exert zero force on us, the force due to that portion is zero, too. All that’s left is the force of gravity due to the portion of the Earth that’s closer to the center than we are. If we assume the Earth has uniform density

a(r) =Gm<r

r 2

m<r =43!r3" = 4

3!r3 mE43!RE

3

#

$%

&

'( =

mEr3

RE3

a(r) =GmEr3

r 2 RE3 =

GmErRE

3 = gRE

r

a(r) = 9.8 m/s2

6.378 )106 mr = (1.5)10*6 s-2 )r

So, below the surface of the Earth, the acceleration of gravity is proportional to the distance from the Earth’s center. We might notice that this has the same form as Hooke’s law, for the force due to a spring. So, if we could drill a hole all the way through the Earth and dropped a mass down the hole, we’d expect it to travel up and down through the hole, just as if it were bouncing on a spring. 21. The object reaches an altitude at which all of its initial kinetic energy has been converted into gravitational potential energy in the Earth-object system.


15

!U G = "!K"GmEmRE + h

""GmEm

RE

= 12 mvi

2

"GmE

RE + h""GmE

RE

= 12 vi

2

"GmE

RE 1+ hRE

#

$%&

'(

""GmE

RE

= 12 vi

2

"GmE

RE

1

1+ hRE

"1

#

$

%%%%

&

'

((((

= 12 vi

2

1

1+ hRE

"1="REvi

2

2GmE

1+ hRE

= 1

1"REvi

2

2GmE

h = 1

1"REvi

2

2GmE

"1

#

$

%%%%

&

'

((((

RE

h = 1

1" (6.378 )106 m)(4.0 )103 m/s)2

2(6.67 )10"11 N *m2 /kg2 )(5.97 )1024 kg)

"1

#

$

%%%%

&

'

((((

RE

h = 0.15RE = 0.15(6.378 )106 m) = 9.4 )105 m

This is a great enough height that, if we had used the near-Earth approximation for the gravitational potential energy, we’d have gotten a significantly smaller answer.

mgh = 12 mvi

2

h =vi

2

2g

h = (4.0 !103 m/s)2

2(9.8 m/s2 )= 8.2 !105 m


16

The approximation gives a smaller value because it overestimates the change in gravitational potential energy at a given height or, equivalently, it overestimates the force of gravity at a given height. 22.

.

Call the period in the frame of the stationary observer TO and the period in the reference frame of the clock TC.

T ! TO,i =1

1"vi

2

c2

TB,i

TO,f = 1

1"4vi

2

c2

TB,f =1

1"4vi

2

c2

TB,i =1"

vi2

c2

1"4vi

2

c2

T = 3T

(1"vi

2

c2 ) = 9(1"4vi

2

c2 ) # "8 = "35vi

2

c2 # vi =835

c = 0.478c

vf = 2vi = 0.956c

82.87 10 m/s! 23. (a)

(b)


17

24. The proper length is measured in the rest frame of the mile markers: 5 miles = 8047 m. b) The proper time is measured by the clock in the car, where the events occur at the same position. The distance between mile markers in the frame of the car is reduced by

length contraction: !L =L"

, ! = 1" vc0

#$%

&'(

2#

$%

&

'(

"12

= 1" 0.42( )"12 = 1.091, so

!L =8047 m1.091

= 7376 m .

so the proper time is 7376 m

0.4( ) 3.00 !108 m/s( )= 61.5 µs .

25. Let frame A be your reference frame at rest with the ground. For two events 1 and 2 we have, from Eq. 14.29, the time difference between events as measured by some other

frame B: !tB = tB2 " tB1 = # !tA "

vABx

c02 !xA

$

%&

'

() . We want to find the x-velocity vABx of

frame B of your friend with respect to A such that ΔtB = 0, i.e. events 1 and 2 occur at the same time in B. There is a similar Lorentz transformation equation involving the y-velocity vABy but because ΔyA = 0 it cannot contribute to ΔtB. Setting ΔtB = 0 we have

!tA "

vABx

c02 !xA = 0 ,

vABx =

!tAc02

!xA

=10"5 s( )3#108 m/s( )

2 #104 mc0 = +0.15 c0

. 26. The total energy after collision is the total energy of both protons, or 14,000 GeV. Since the momentum is zero, all of this energy can be converted into the rest energy of the new particle:


18

E0 = mc02 = 14,000 GeV , m =

14 !1012 eV( )1.60 !10"19 J/eV( )3.00 !108 m/s( )2

= 2.49 !10"23 kg.

27. We can find the “spring constant” of Earth from the relationship of its mass and angular frequency.

km

! =

2k m!= We could substitute the known mass of Earth and its angular frequency of 2 rad y! to find a numeric value for the “spring constant.” 28. The general form of the equation of simple harmonic motion is

( ) ( )sin ix t A t! "= + . In this case,

( ) 2sin 82

x t t!!" #= +$ %& '

(a) The amplitude of this motion is 2 cm 0.020 mA = = .

(b) The frequency of oscillation is 18 4 s

2 2f ! "

" "#= = =

(c) The period of this motion is

1

1 1 0.25 s4 s

Tf != = = .

(d) Since 2i!" = , x takes the value 0x = at 4t T= , 3 4t T= , and 5 4t T= .

29. For an object in simple harmonic motion,

2 212E m A!= .

Maximum acceleration occurs at maximum amplitude, and is 2 2

xa x A! != " = " . We notice that acceleration has the opposite sign of A, being ! out of phase with displacement.

2 xaA

! "=

221 1

2 2x

xa

E m A ma AA!" #= = !$ %& '


19

( )( )( )212 1.0 kg 5.0m s 0.12 m 0.30 JE = ! ! =

30. ▲Focus problem

The block provides the tension in the wire. We want to find what tension is required to create, along the horizontal section of the wire, a three-antinode standing wave vibrating at 550 Hz. The number of antinodes is related to the number of wavelengths of the traveling waves that fit into the horizontal region of the wire. We assume that the block moves negligibly when the string is vibrating. We need to relate the frequency f, wavelength λ, and tension force F = mg. The can be done using Equation 16.25,

c = F µ = ! f . (1) For the third harmonic, there are three antinodes and four nodes. The fourth node (the one for which n = 3 in Eq. 16.21) occurs at x = L, where L is the length of the wire containing the waves. Substituting L for x and 3 for n in Eq. 16.22, x, = 0, λ/2, λ, 3λ/2, . . . , we have

L = (3)

!2

"#$

%&'= 3

2 ! . (2)

Knowing this expression for λ, we can then solve for the tension F in terms of the known quantities. From Eq. 2, we know that the third harmonic has a wavelength λ = 2L/3. Solving Eq. 1 for the tension force F = mg gives us


20

mg = µ(! f )2 = µ 23 Lf( )2

m =4µ(Lf )2

9g

=4(0.0013 kg m) (2.3 m)(550 Hz)"# $%

2

9(9.8 m s2 )= 94.3 kg

That is a pretty massive block. A large tension is needed to excite high frequencies, especially in long strings, however, and so the answer is not unreasonable. The units on the left side of our numeric equation yield an inertia unit, which is what we need. We know you have to make a string tighter to get a higher pitch (higher frequency) from it. Our algebraic expression says this in that the block’s inertia must increase if the frequency f is to increase. We would also expect the inertia to be inversely proportional to g because if the gravitational pull were stronger, a block of less inertia would be required. Our assumption that the block does not move is likely to be accurate with this much inertia. We neglected the inertia of the hanging portion of the wire in our calculation, but it would have to be more than 1000 m in order to become a noticeable contributor relative to the block. 31. Only the reflected pulse in (c) is depicted correctly. When a pulse reflects off a fixed end, it’s inverted, so (b) and (d) are incorrect (although (d) does correctly depict a pulse reflected off a free end). When a pulse is reflected, the leading edge of the incident pulse becomes the leading edge of the reflected pulse, so (a) and (b) are incorrect.

32. Approaching ambulance:

! f =f

1" vS v( )

Departing ambulance:

! ! f =f

1" " vS v( )( )

Since

! f = 560 Hz and

! ! f = 480 Hz

560 1 !vSv

" # $

% & ' = 480 1 +

vSv

" # $

% & '

33. (b)

sin! =vvS

=1

3.00 ;

! = 19.5°

tan! =hx

;

x =h

tan!

x =20 000 mtan19.5°

= 5.66! 104 m = 56.6 km


21

(a) It takes the plane

t =xvS

=5.66 ! 104 m

3.00 335 m s( ) = 56.3 s to travel this distance.

t = 0

a.

!

h

Observer b.

!

h

Observer hears the boom

x

1 040 vSv

= 80.0

vS =80.0 343( )

1 040 m s = 26.4 m s

34. To find the separation of adjacent molecules, use a model where each molecule occupies a sphere of radius r given by

!air =average mass per molecule

43" r3

or

1.20 kg m3 =4.82 ! 10"26 kg

43# r3 ,

r =3 4.82 ! 10"26 kg( )4# 1.20 kg m3( )

$

% & &

'

( ) )

1 3

= 2.12 ! 10"9 m .

Intermolecular separation is

2r = 4.25! 10"9 m , so the highest possible frequency sound wave is

fmax =v

!min=

v2r =

343 m s4.25 " 10#9 m

= 8.03" 1010 Hz ~ 1011 Hz .

35. At equilibrium

F! = 0 or

Fapp + mg = B

where B is the buoyant force.

The applied force,

Fapp = B ! mg

where

B = Vol !water( )g

and

m = Vol( )! ball .

So,

Fapp = Vol( )g !water " ! ball( ) =43#r3g !water " ! ball( )

Fapp =43 ! 1.90" 10#2 m( )3

9.80 m s2( ) 103 kg m3 # 84.0 kg m3( ) = 0.258 N


22

36. Assuming the top is open to the atmosphere, then

P1 = P0.

Note

P2 = P0 . Flow rate

= 2.50! 10"3 m3 min = 4.17 ! 10"5 m3 s .

(a)

A1 >> A2 so

v1 << v2

Assuming

v1 = 0 ,

P1 +!v1

2

2 + !gy1 = P2 +!v2

2

2 + !gy2

v2 = 2gy1( )1 2 = 2 9.80( ) 16.0( )[ ]1 2 = 17.7 m s

(b) Flow rate

= A2v2 =!d2

4"

# $

%

& ' 17.7( ) = 4.17 ( 10)5 m3 s

d = 1.73 ! 10"3 m = 1.73 mm 37. c 38. It would not maintain its regular motion, but instead exhibit erratic pseudo-Brownian motion.

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