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Home Work Solutions PHYS111.02 SFSU Eradat [WH5 CHAPTERS 12-‐14] 1 +WHW5: Chapter 12: 7,43,57, Chapter 13: 55,81, Chapter 14: 23,25,37,50,51 Chapter 12 7. Picture the Problem: The spaceship is attracted gravitationally to both the Earth and the Moon.
Strategy: Use the Universal Law of Gravity (equation 12-‐1) to relate the attractive forces from the Earth and the Moon. Set the force due to the Earth equal to twice the force due to the Moon when the spaceship is at a distance r from the center of the Earth. Let R = 3.84 ×108 m, the distance between the centers of the Earth and Moon. Then solve the expression for the distance r.
Solution: 1. (a) Set FE = 2FM using equation 12-‐1 and solve for r:
GmsmE
r 2 = 2GmsmM
R − r( )2
mE R − r( )2= 2mMr 2
R − r = 2mM mE r
r = R1+ 2mM mE
= 3.84 ×108 m
1+ 2 × 7.35×1022 kg( ) 5.97 ×1024 kg( )= 3.32 ×108 m
2. (b) The answer to part (a) is independent of the mass of the spaceship because the spaceship’s mass is included in the force between it and both the Moon and the Earth, and so its value cancels out of the expression.
Insight: The distance in part (a) is the same for any mass, and corresponds to about 52 Earth radii or about 86% of the distance R between the Earth and the Moon. The two forces are equal at 3.46×108 m or about 90% of R.
43. Picture the Problem: An object is located at the surface of the Earth and later at an altitude of 350 km. Strategy: Use equation 12-8 to find the gravitational potential energy of the object as a function of its distance
r = RE + h from the center of the Earth. Then take the difference between the values at h = 0 and h = 350 km and compare it with the approximate change in potential, ΔU = mgh.
Solution: 1. (a) Calculate
U = −G
MEmRE + h
at h = 0: U0 = − 6.67 ×10−11 N ⋅m2 /kg2( ) 5.97 ×1024 kg( ) 8.8 kg( )
6.37 ×106 m= −5.5×108 J
2. (b) Calculate U at h = 350 km:
Uh = − 6.67 ×10−11 N ⋅m2 /kg2( ) 5.97 ×1024 kg( ) 8.8 kg( )
6.37 ×106 + 350 ×103 m= −5.2 ×108 J
3. (c) Take the difference ΔU : ΔU = Uh −U0 = −5.50 ×108 J( ) − −5.21×108 J( ) = 2.9 ×107 J
4. Compare with mgh: ΔU = mgh = 8.8 kg( ) 9.81 m/s2( ) 350 ×103 m( ) = 3.0 ×107 J
Insight: The two calculations of ΔU (without any rounding) differ by about 5%. The mgh calculation is an approximation because it assumes the value of g is constant over the 350 km, when in fact it gets smaller as the distance from the Earth’s center increases.
57. Picture the Problem: The planet is ten times more massive and has one-‐tenth the radius of Earth. A
projectile at its surface is given sufficient kinetic energy to escape the planet. Strategy: Use a ratio to compare the escape speed on the new planet with the escape speed on Earth. Use
equation 12-‐13 to form the ratio.
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Solution: Write a ratio of the escape speeds:
ve,new
ve,Earth
=2GMnew Rnew
2GME RE
=Mnew RE
ME Rnew
=10ME( )RE
ME1
10 RE( ) = 100 = 10
ve,new = 10 ve,Earth
Insight: Although the speed only needs to be increased by a factor of 10, the initial kinetic energy of a
projectile needs to be 100 times larger on this new planet in order for the projectile to escape that planet’s gravity.
Chapter 13 55. Picture the Problem: A block and spring are initially at rest as a bullet is fired at high speed directly
toward them. The bullet then embeds in the block and compresses the spring.
Strategy: The bullet and block first undergo an inelastic collision. Then they jointly compress the spring, converting their kinetic energy into potential energy of the spring. Use conservation of energy to relate the speed v of the block and bullet to the compression distance x. Finally, use conservation of momentum to find the initial speed of the bullet v0 from the combined speed of bullet and block. The time elapsed from impact to rest is one-‐quarter of a period.
Solution: 1. (a) Set the initial kinetic energy of the block and bullet to the final potential energy of the spring:
K i = U f
12
M + m( )v2 = 12
kA2
2. Solve for the speed of the bullet and block:
v = kA2
M + m=
785 N/m( ) 0.0588 m( )2
1.500 kg + 0.00225 kg= 1.344 m/s
3. Using conservation of momentum write the initial speed of the bullet in terms of the final speed of bullet and block:
mv0 = ( M + m)v
v0 =M + m
m⎛⎝⎜
⎞⎠⎟
v
4. Calculate initial speed of bullet:
v0 =
1.500 kg + 0.00225 kg0.00225 kg
⎛⎝⎜
⎞⎠⎟
1.344 m/s( ) = 897 m/s
5. (b) Calculate one-‐quarter period:
T4= π
2M + m
k= π
21.50225 kg
785 N/m= 0.0687 s
Insight: The initial kinetic energy of the bullet does not equal the final energy of the compressed spring. Some of the initial kinetic energy is lost due to the inelastic collision with the block.
81. Picture the Problem: A mass that is attached to a spring, displaced from equilibrium, and released, will
oscillate with simple harmonic motion. The maximum speed occurs as the mass passes through the equilibrium position. The maximum force occurs when the spring is stretched its maximum distance from equilibrium.
Strategy: Find the amplitude from the maximum speed and maximum acceleration. Then determine the maximum acceleration from the maximum force and Newton’s Second Law. Then obtain the angular speed ωfrom the maximum speed and acceleration, and finally calculate the force constant and frequency of oscillation fromω .
Solution: 1. (a) Combine the maximum velocity and maximum acceleration equations
vmax = Aω , amax = Aω 2
Home Work Solutions PHYS111.02 SFSU Eradat [WH5 CHAPTERS 12-‐14] 3
to solve for the amplitude:
vmax2
amax
=Aω( )2
Aω 2 = A
2. Using Newton’s Second Law write ampli-‐ tude as a function of the maximum force:
A =
vmax2
amax
=vmax
2
Fmax / m=
mvmax2
Fmax
3. Solve for the amplitude:
A =
3.1 kg( ) 0.68 m/s( )2
11 N= 0.13 m
4. (b) Combine maximum velocity and acceleration to find angular speed:
ω =
amax
vmax
=Fmax
mvmax
= 11 N3.1 kg( ) 0.68 m/s( ) = 5.2 rad/s
5. Solve equation 13-‐10 for k: k = mω 2 = 3.1 kg( ) 5.2 rad/s( )2
= 84 N/m
6. (c) Use the angular speed to determine f:
f = ω
2π= 5.2 rad/s
2π= 0.83 Hz
Insight: Another way to solve this problem is to first calculate the angular speed from the maximum force and velocity. The amplitude can then be found from the maximum speed divided by the angular speed.
Chapter 14 23. Picture the Problem: We are given the equation describing a wave and wish to determine the amplitude,
wavelength, period, speed, and direction of travel.
Strategy: The general form of a wave is given by
y = Acos 2π
λx − 2π
Tt
⎛⎝⎜
⎞⎠⎟
. Compare this equation to
y = 15 cm( )cos π
5.0 cmx − π
12 st
⎛⎝⎜
⎞⎠⎟
, the equation given in the problem, to identify the wave parameters.
Use equation 14-‐1 and the definition of frequency to calculate the wave speed.
Solution: 1. (a) Identify the amplitude as A: A = 15 cm
2. (b) Identify the wavelength as λ:
2πλ
= π5.0 cm
, so λ = 10 cm = 0.10 m
3. (c) Identify the period as T:
2πT
= π12 s
, so T = 24 s
4. (d) Use equation14-‐1 to calculate the speed:
v = λ f = λ
T= 10 cm
24 s= 0.42 cm/s
5. (e) The wave travels to the right, because the t-‐term and x-‐term have opposite signs.
Insight: The wave equation is a compact way of completely describing a wave, because it is possible to extract all of the wave properties from the equation.
25. Picture the Problem: Equations for four different waves are given. From these equations we need to
determine the directions the waves travel, which waves have the highest frequency, the largest wavelength, and the greatest speed.
Strategy: When x and t have opposite signs, the wave travels to the left. When they have the same sign, the wave travels to the right. Examine the equations to determine which waves travel in each direction. The coefficient of the t term is proportional to the frequency. Find the equation with the largest t-coefficient, and it
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will have the largest frequency. The coefficient of the x term is inversely proportional to the wavelength. Find the equation with the smallest x-coefficient and it will have the largest wavelength. The speed is the t-coefficient divided by the x-coefficient. Divide the coefficients and find the largest speed.
Solution: 1. (a) Find the waves for which x and t have opposite signs: Waves
yA and yC travel to the right.
2. (b) Find the waves for which x and t have the same sign:
Waves
yB and yD travel to the left.
3. (c) Find the wave with the largest t-‐coefficient: Wave
yC has the largest frequency.
4. (d) Find the wave with the smallest x-‐coefficient: Wave
yA has the largest wavelength.
5. (e) Find the wave for which the magnitude of the t-‐coefficient divided by the x-‐coefficient is greatest: Wave
yC has the greatest speed.
Insight: The wave speeds are: vA = 1.33 cm/s, vB = 0.8 cm/s, vC = 6.0 cm/s, and vD = 5.0 cm/s
37. Picture the Problem: The intensity level at a distance of 2.0 meters is given. We want to find the
intensity levels at 12 m and 21 meters. We also want to find the distance for which the intensity level is 0, the farthest point at which the siren can be heard.
Strategy: Insert equation 14-‐6 into equation 14-‐8 to create an equation for intensity level in relation to distance. Use this relationship to calculate the intensity levels at 12 m and 21 m. To calculate the farthest distance at which the siren can be heard, set the intensity level to zero, and solve the relation for distance.
Solution: 1. Insert equation 14-‐6 into equation 14-‐8:
β = 10 logI2
I0
⎛
⎝⎜⎞
⎠⎟= 10 log
r1
r2
⎛
⎝⎜⎞
⎠⎟
2I1
I0
⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢⎢
⎤
⎦⎥⎥
= 10 logI1
I0
⎛
⎝⎜⎞
⎠⎟+10 log
r1
r2
⎛
⎝⎜⎞
⎠⎟
2
2. (a) Insert β at r1 = 2.0 m and set r2=12 m:
β = 120 +10 log 2.0 m
12 m⎛⎝⎜
⎞⎠⎟
2
= 104 dB
3. (b) Repeat for r2=21 m:
β = 120 +10 log 2.0 m
21 m⎛⎝⎜
⎞⎠⎟
2
= 99.6 dB
4. (c) Set the intensity level equal to zero:
β = 10 logI1
I0
⎛
⎝⎜⎞
⎠⎟+10 log
r1
r2
⎛
⎝⎜⎞
⎠⎟
2
0 = 120 + 10 log 2.0 mr
⎛⎝⎜
⎞⎠⎟
2
5. Solve for r:
−10 log 2.0 mr
⎛⎝⎜
⎞⎠⎟
2
= 120
2.0 mr
⎛⎝⎜
⎞⎠⎟
2
= 10−12 ⇒ r = 2.0 m10−6 = 2.0 ×106 m
Insight: This is a theoretical limit that could be realized in an ideal case. In a more realistic scenario,
Home Work Solutions PHYS111.02 SFSU Eradat [WH5 CHAPTERS 12-‐14] 5
ambient noise, as well as energy losses when the sound waves are reflected or absorbed by surfaces, would prevent us from hearing the sound 2000 km away. Sometimes the real-‐world factors we ignore make a huge difference!
50. Picture the Problem: The image depicts
a person in a stationary vehicle listening to the siren of a fire engine as it passes. From the frequencies heard as the fire engine approaches and leaves we want to calculate the time it takes to reach a fire 5.0 km away.
Strategy: The frequency f emitted by the fire engine remains constant. Use equation 14-‐10 to write the observed frequencies in terms of the emitted frequency and speed, because the source (fire engine) is moving and the observer (you) is stationary. The negative sign is used when the engine is approaching and the positive sign when the engine is moving away. Combine the two equations to eliminate the emitted frequency and solve for the speed of the fire engine. Divide the distance by the speed to calculate the time to reach the fire.
Solution: 1. Write equation 14-‐10 for the approaching fire engine and solve for f:
′f1 =1
1− uv
⎛
⎝⎜⎞
⎠⎟f ⇒ f = ′f1 1− u
v⎛⎝⎜
⎞⎠⎟
2. Write equation 14-‐10 for the receding fire engine and solve for the emitted frequency:
′f2 =1
1+ uv
⎛
⎝⎜⎞
⎠⎟f ⇒ f = ′f2 1+ u
v⎛⎝⎜
⎞⎠⎟
3. Set the equations for the emitted frequencies equal and solve for the speed:
′f1 v − ′f2 v = ′f1 u + ′f2 u
u =′f1 − ′f2
′f1 + ′f2
⎛
⎝⎜⎞
⎠⎟v = 460 Hz − 410 Hz
460 Hz + 410 Hz⎛⎝⎜
⎞⎠⎟
343 m/s( ) = 19.7 m/s
4. Solve for the time of arrival: t = d
u= 5.0 ×103 m
19.7 m/s= 254 s = 4.2 min
Insight: The frequency of the stationary fire truck is 434 Hz. Note that the frequency of the approaching engine increases more than the frequency of the receding engine decreases.
51. Picture the Problem: As you approach a source of sound, the increase in observed frequency is related
to your speed. Strategy: We want to calculate the speed necessary for the observed frequency to be 1.15 times the
emitted frequency. This problem has a stationary source and moving observer, so employ equation 14-‐9 to calculate the observed frequency, using the plus sign because the observer is moving toward the source. Set the observed frequency equal to 1.15 times the emitted frequency and solve for the speed.
Solution: 1. Set the observed frequency in equation 14-‐9 equal to 1.15 f :
′f = 1+ u v( ) f
1.15 f = 1+ u v( ) f
2. Solve for the speed:
uv= 1.15−1= 0.15 ⇒ u = 0.15v = 0.15 343 m/s( ) = 51 m/s
Insight: Note that the fractional increase in frequency is equal to the ratio of your speed to the speed of
sound.