1/12/22, 9:33 PM
bonus assignment final week
bonus assignment final week
Due: 11:59pm on Saturday, January 15, 2022
You will receive no credit for items you complete after the assignment is due. Grading Policy
A Satellite in a Circular Orbit
Consider a satellite of mass m1 that orbits a planet of mass m2 in a circle a distance r from the center of the planet. The satellite's mass is
negligible compared with that of the planet. Indicate whether each of the statements in this problem is true or false.
Part A
The information given is sufficient to uniquely specify the speed, potential energy, and angular momentum of the satellite.
Hint 1. What constitutes sufficient initial conditions?
An initial position and velocity plus a knowledge of the forces at all points will suffice to specify the subsequent motion in Newtonian
mechanics. Are you able to determine the satellite's velocity from the information given?
ANSWER:
true
false
Correct
Part B
The total mechanical energy of the satellite is conserved.
Hint 1. When is mechanical energy conserved?
A system's total mechanical energy is conserved if no nonconservative forces act on the system. Is gravity a conservative force?
ANSWER:
true
false
Correct
Part C
The linear momentum vector of the satellite is conserved.
Hint 1. When is linear momentum conserved?
The linear momentum vector of a system is conserved if the net external force acting on it is zero.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
1/18
1/12/22, 9:33 PM
bonus assignment final week
ANSWER:
true
false
Correct
Part D
The angular momentum of the satellite about the center of the planet is conserved.
Hint 1. When is angular momentum conserved?
Angular momentum about a particular axis is conserved if there is no net torque about that axis.
ANSWER:
true
false
Correct
Part E
The equations that express the conservation laws of total mechanical energy and linear momentum are sufficient to solve for the speed
necessary to maintain a circular orbit at R without using F ⃗
= ma⃗
.
Hint 1. How are conservation laws used?
Conservation laws are generally applied to a system that has some "initial" and "final" conditions that are different. Motion in a
circular orbit, however, possesses no obvious initial and final points that are different.
ANSWER:
true
false
Correct
± Understanding Newton's Law of Universal Gravitation
Learning Goal:
To understand Newton's law of universal gravitation and be able to apply it in two-object situations and (collinear) three-object situations; to
distinguish between the use of G and g.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
2/18
1/12/22, 9:33 PM
bonus assignment final week
In the late 1600s, Isaac Newton proposed a rule to quantify the attractive force known as gravity between objects that have mass, such as
those shown in the figure. Newton's law of universal gravitation describes the
magnitude of the attractive gravitational force F g⃗ between two objects with masses
m 1 and m 2 as
Fg = G (
m 1m 2
r
2
)
,
where r is the distance between the centers of the two objects and G is the
gravitational constant.
The gravitational force is attractive, so in the figure it pulls to the right on m1 (toward
m 2 ) and toward the left on m 2 (toward m 1 ). The gravitational force acting on m 1 is
equal in magnitude to, but exactly opposite in direction from, the gravitational force
acting on m2, as required by Newton's third law. The magnitude of both forces is
calculated with the equation given above.
The gravitational constant G has the value
G = 6.67 × 10
−11
2
N ⋅ m /kg
2
and should not be confused with the magnitude of the gravitational free-fall acceleration constant, denoted by g, which equals 9.80 m/s2 near
the surface of the earth. The magnitude of G in SI units is tiny. This means that gravitational forces are sizeable only in the vicinity of very
massive objects, such as the earth. You are in fact gravitationally attracted toward all the objects around you, such as the computer you are
using, but the magnitude of that force is too small to be noticed without extremely sensitive equipment.
Consider the earth following its nearly circular orbit (dashed curve) about the sun.
The earth has mass mearth = 5.98 × 10 24 kg and the sun has mass
30
m sun = 1.99 × 10
kg . They are separated, center to center, by
r = 93 million miles = 150 million km .
Part A
What is the magnitude of the gravitational force acting on the earth due to the sun?
Express your answer in newtons.
Hint 1. What units to select
For the force to come out in newtons, all masses and distances used must be expressed in SI units. Note that neither
6
6
93 million miles (= 93 × 10 miles ) nor 150 million km (= 150 × 10 km) is in the SI unit of length, which is meters.
ANSWER:
3.53×1022
N
Correct
This force causes the earth to orbit the sun.
Part B
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
3/18
1/12/22, 9:33 PM
bonus assignment final week
At the moment shown in the figure of the earth and sun , what is the direction of
the gravitational force acting on the earth? The possible directions are displayed
in this figure .
ANSWER:
A
B
C
D
E
Correct
As the earth proceeds around its orbit, the direction of the gravitational force acting on it changes so that the force always points
directly toward the sun.
Part C
What is the magnitude of the gravitational force acting on the sun due to the earth?
Hint 1. Newton's third law
Newton's third law states that if object A exerts a force on object B, then object B must exert a force of the same type on object A.
These two forces are equal in magnitude and exactly opposite in direction. Since they act on different objects, these two forces
never cancel one another out.
ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
4/18
1/12/22, 9:33 PM
bonus assignment final week
The earth does not exert any gravitational force on the sun.
_______________
The earth exerts some force on the sun, but less than 3.53 × 10 22
much less massive than the sun.
_______________
The earth exerts 3.53 × 10 22
Part A.
_______________
N
N
because the earth, which is exerting the force, is so
of force on the sun, exactly the same amount of force the sun exerts on the earth found in
The earth exerts more than 3.53 × 10 22
more massive than the earth.
_______________
N
of force on the sun because the sun, which is experiencing the force, is so much
Correct
Also note that the force exerted on the sun by the earth is directed from the sun toward the earth, downward at the moment
shown in the figure. This is exactly the opposite direction of the force exerted on the earth by the sun found in Part B, in
accordance with Newton's third law.Recall that Newton's second law states that ΣF ⃗ = ma⃗ . The sun does not accelerate as
much as the earth due to this gravitational force since the sun is so much more massive. This force causes the earth to follow a
nearly circular path as it orbits the sun, but the same amount of force only causes the sun to wobble back and forth very slightly.
Part D
Which of the following changes to the earth-sun system would reduce the magnitude of the force between them to one-fourth the value
found in Part A?
Check all that apply.
ANSWER:
Reduce the mass of the earth to one-fourth its normal value.
Reduce the mass of the sun to one-fourth its normal value.
Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.
Increase the separation between the earth and the sun to four times its normal value.
Correct
The actions in choices A, B, and C would reduce the force to one-fourth its original value by modifying the product of the masses.
Increasing the separation distance by a factor of 4 would reduce the force by a factor of 42 = 16. To reduce the force by only a
factor of 4, increase the separation distance by a factor of 2 instead because 22 = 4 .
Part E
With the sun and the earth back in their regular positions, consider a space probe with mass mp = 125 kg launched from the earth
toward the sun. When the probe is exactly halfway between the earth and the sun along the line connecting them, what is the direction of
the net gravitational force acting on the probe? Ignore the effects of other massive objects in the solar system, such as the moon and other
planets.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
5/18
1/12/22, 9:33 PM
bonus assignment final week
Hint 1. How to approach the problem
The sun and the earth each exert a gravitational force on the probe. Use Newton's law of universal gravitation to find the magnitude
and direction of each of these forces. Then sum them (as vectors!) to find the net force acting on the probe.
Hint 2. Direction of the gravitational force due to the earth
What is the direction of the gravitational force acting on the probe due to the earth (refer to the figure)?
ANSWER:
A
B
C
D
E
Hint 3. Direction of the gravitational force due to the sun
What is the direction of the gravitational force acting on the probe due to the sun (refer to the figure)?
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
6/18
1/12/22, 9:33 PM
bonus assignment final week
ANSWER:
A
B
C
D
E
Hint 4. Comparing the gravitational forces acting on the probe
Which exerts the stronger gravitational force on the probe?
ANSWER:
The sun exerts a stronger force.
The earth exerts a stronger force.
The forces exerted on the probe by the sun and by the earth are the same magnitude.
ANSWER:
The force is toward the sun.
The force is toward the earth.
There is no net force because neither the sun nor the earth attracts the probe gravitationally at the midpoint.
There is no net force because the gravitational attractions on the probe due to the sun and the earth are equal in magnitude but
point in opposite directions, so they cancel each other out.
Correct
In calculating these forces, the mass of the probe is the same, as is the distance involved. However, the sun is over 300,000
times as massive as the earth, so at this location the probe is attracted much more strongly toward the sun than toward the earth.
Perhaps the most-common gravitational calculation is to determine the "weight" of an object of mass mo sitting on the surface of the earth (i.e.,
the magnitude of the gravitational attraction between the earth and the object). According to Newton's law of universal gravitation, the
gravitational force on the object is
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
7/18
1/12/22, 9:33 PM
bonus assignment final week
Fg = G (
me mo
2
re
) = (
Gm e
2
re
) mo
,
where me is the mass of the earth and re is the distance between the object and the center of the earth (i.e., re = radius of the earth =
3
6.38 × 10 km ).
Part F
What is the value of the composite constant (
Gm e
2
re
),
to be multiplied by the mass of the object mo in the equation above?
Express your answer numerically in meters per second per second.
ANSWER:
9.80
2
m/s
Correct
This particular combination of constants occurs so often that it is given its own name, g, and called the "gravitational free-fall
acceleration near the surface of the earth." Using g leads to the more familiar formula for calculating the gravitational weight Fg
of an object of mass m , namely
Fg = mg
.
This equation is limited to calculating the weight of objects near the surface of the earth. When calculating gravitational forces
between objects under other conditions, such as in the earth/probe calculation done earlier, use Newton's law of universal
gravitation instead.
Exercise 13.34 - Enhanced - with Solution
On October 15, 2001, a planet was discovered orbiting around the star HD68988. Its orbital distance was measured to be 10.5 million
kilometers from the center of the star, and its orbital period was estimated at 6.3 days.
For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Kepler's third law.
Part A
What is the mass of HD68988?
Express your answer in kilograms.
ANSWER:
M
= 2.3×1030
kg
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
8/18
1/12/22, 9:33 PM
bonus assignment final week
Correct
IDENTIFY: Knowing the orbital radius and orbital period of a satellite, we can calculate the mass of the object about which it is
revolving.
SET UP: The radius of the orbit is r
m s = 1.99 × 10
30
kg
EXECUTE: Solving T
2
mH D =
4π r
T
2
9
and its period is T
m
. The orbital period is given by T
=
2πr
3/2
−
−
−−
−
−
√Gm HD
9
4π (10.5 × 10
=
5
(5.443 × 10
=
2πr
3/2
−
−
−−
−
−
√Gm HD
= 6.3 days = 5.443 × 10
5
s
. The mass of the sun is
.
for the mass of the star gives
2
3
G
= 10.5 × 10
2
s)
3
m)
−11
(6.673 × 10
2
2
= 2.3 × 10
30
kg
N . m /kg )
Part B
What is the mass of HD68988?
Express your answer in terms of our sun's mass.
ANSWER:
M
= 1.2
Msun
Correct
m H D = 1.2 mS
EVALUATE: The mass of the star is only 20% greater than that of our sun, yet the orbital period of the planet is much shorter
than that of the earth, so the planet must be much closer to the star than the earth is.
Exercise 14.22
In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in
smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 28.0 nm long) with a laser, first
without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can
model such a process as a mass on a spring.
Part A
Find the ratio of the frequency with the virus attached ( f S+V ) to the frequency without the virus (f S ) in terms of mV and mS , where mV
is the mass of the virus and mS is the mass of the silicon sliver. Notice that it is not necessary to know or measure the force constant of
the spring.
Express your answer in terms of the variables mV and mS .
ANSWER:
fS+V
fS
=
1
−
−
−−−−
−
−
√(1+
m
V
m
S
)
Answer Requested
Part B
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
9/18
1/12/22, 9:33 PM
bonus assignment final week
In some data, the silicon sliver has a mass of
virus. What is the mass of the virus in grams?
2.10×10−16
g
and a frequency of 2.03×1015 Hz without the virus and 2.88×1014 Hz with the
Express your answer in grams.
ANSWER:
mV
= 1.02×10−14
g
Correct
Part C
What is the mass of the virus in femtograms?
Express your answer in femtograms.
ANSWER:
mV
= 10.2
fg
Correct
Exercise 14.35 - Enhanced - with Solution
A 2.00 kg frictionless block attached to an ideal spring with force constant 365 N/m is undergoing simple harmonic motion. When the block
has displacement +0.200 m, it is moving in the negative x-direction with a speed of 3.50 m/s .
For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Velocity, acceleration, and energy in shm.
Part A
Find the amplitude of the motion.
Express your answer with the appropriate units.
ANSWER:
A
= 0.327 m
Correct
IDENTIFY and SET UP: Velocity, position, and total energy are related by E
EXECUTE: E
=
1
2
2
mv x +
1
2
kx
2
E = 12.3 J + 7.30 J = 19.6 J
=
1
2
.E
2
(2.00 kg)(−3.50 m/s) +
=
1
2
kA
2
and A
−
−
−
=√
2E
k
1
2
=
1
2
kA
2
=
1
2
2
mv x +
2
(365 N/m)(+0.200 m)
1
2
kx
2
.
.
−−
−−
−
−
=√
2(19.6 J)
365 N/m
= 0.327 m
.
Part B
Find the block's maximum acceleration.
Express your answer with the appropriate units.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
10/18
1/12/22, 9:33 PM
bonus assignment final week
ANSWER:
= 59.7
amax
m
2
s
Correct
Acceleration and position are related by −kx
amax =
k
m
A =(
365 N/m
2.00 kg
= max
. The maximum magnitude of acceleration is at x
= ± A
.
2
)(0.327 m) = 59.7 m/s
Part C
Find the maximum force the spring exerts on the block.
Express your answer with the appropriate units.
ANSWER:
Fmax
= 119 N
Correct
2
Fmax = mamax = (2.00 kg)(59.7 m/s ) = 119 N
. Or, Fx
= −kx
gives Fmax
= kA = (365 N/m)(0.327 m) = 119 N
,
which checks.
EVALUATE: The maximum force and maximum acceleration occur when the displacement is maximum and the velocity is zero.
Problem 14.83
A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg that rests on a
frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses
the spring a maximum distance of 17.0 cm After the impact, the block moves in SHM.
Part A
Calculate the period of this motion.
Express your answer with the appropriate units.
ANSWER:
T
= 0.477 s
Correct
Problem 10.60
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the
free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
11/18
1/12/22, 9:33 PM
bonus assignment final week
pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.50 kg ) moves is frictionless. The system is released from
rest, and block B (mass 8.00 kg ) moves downward 1.80 m in 2.00 s .
Part A
What is the tension force that the rope exerts on block B ?
Express your answer with the appropriate units.
ANSWER:
TB
= 71.2 N
Correct
Part B
What is the tension force that the rope exerts on block A ?
Express your answer with the appropriate units.
ANSWER:
TA
= 2.25 N
Correct
Part C
What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?
Express your answer with the appropriate units.
ANSWER:
I
= 0.490 kg⋅m2
Correct
Problem 10.72
A thin-walled, hollow spherical shell of mass m and radius r starts from rest and rolls without slipping down the track shown in the figure .
Points A and B are on a circular part of the track having radius R . The diameter of the shell is very small compared to h0 and R , and the
work done by the rolling friction is negligible.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
12/18
1/12/22, 9:33 PM
bonus assignment final week
Part A
What is the minimum height h0 for which this shell will make a complete loop-the-loop on the circular part of the track?
Express your answer in terms of the variables m , R , and appropriate constants.
ANSWER:
h0
=
2.83R
Correct
Part B
How hard does the track push on the shell at point B , which is at the same level as the center of the circle?
Express your answer in terms of the variables m , R , and appropriate constants.
ANSWER:
N
=
2.2mg
Correct
Part C
Suppose that the track had no friction and the shell was released from the same height h0 you found in part (a). Would it make a complete
loop-the-loop?
ANSWER:
yes
no
Correct
Part D
In part (c), how hard does the track push on the shell at point A , the top of the circle?
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
13/18
1/12/22, 9:33 PM
bonus assignment final week
Express your answer in terms of the variables m , R , and appropriate constants.
ANSWER:
⃗∣
∣
N
∣
∣
=
2
3
mg
Correct
Part E
How hard did the track push on the shell at point A in part (a)?
Express your answer in terms of the variables m , R , and appropriate constants.
ANSWER:
⃗∣
∣
N
∣
∣
= 0
Correct
Exercise 10.46
A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg . Initially open and at rest, the door is
struck at its center by a handful of sticky mud with mass 0.600 kg kg , traveling perpendicular to the door at 12.0 m/s just before impact.
Part A
Find the final angular speed of the door.
Express your answer in radians per second.
ANSWER:
ω
= 0.267
rad/s
Correct
Part B
Does the mud make a significant contribution to the moment of inertia?
ANSWER:
yes
no
Correct
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
14/18
1/12/22, 9:33 PM
bonus assignment final week
Exercise 9.57 - Enhanced - with Solution
A slender rod with length L has a mass per unit length that varies with distance from the left-hand end, where x
2
dm/dx = γx , where γ has units of kg/m .
= 0
, according to
For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Uniform sphere with radius r, axis through
center.
Part A
Calculate the total mass of the rod in terms of γ and L.
Express your answer in terms of the variables γ and L.
ANSWER:
M
γL
=
2
2
Correct
IDENTIFY: Apply I
2
= ∫ r
SET UP: For this case, dm
EXECUTE: M
dm
and M
= ∫ dm .
.
= γx dx
= ∫ dm = ∫
L
0
γx dx = γ
x
2
2
L
|
=
0
γL
2
2
Part B
2
Use I = ∫ r dm to calculate the moment of inertia of the rod for an axis at the left-hand end, perpendicular to the rod. Use the
expression you derived in part A to express I in terms of M and L.
Express your answer in terms of the variables M and L.
ANSWER:
I
=
ML
2
2
Correct
I = ∫
L
0
2
x (γx)dx = γ
x
4
4
L
|
0
=
γL
4
4
=
M
2
L
2
. This is larger than the moment of inertia of a uniform rod of the same
mass and length, since the mass density is greater farther away from the axis than nearer the axis.
Part C
Repeat part B for an axis at the right-hand end of the rod.
Express your answer in terms of the variables M and L.
ANSWER:
I
=
1
6
ML
2
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
15/18
1/12/22, 9:33 PM
bonus assignment final week
Correct
I = ∫
L
0
2
(L − x) γxdx = γ ∫
L
0
2
(L x − 2Lx
2
3
+ x )dx = γ(L
2 x
2
2
− 2L
x
3
3
+
x
4
4
L
)|
0
= γ
L
4
12
=
M
6
2
L .
This is a third of the result of part b, reflecting the fact that more of the mass is concentrated at the right end.
EVALUATE: For a uniform rod with an axis at one end, I
part c
is smaller than this.
=
1
3
ML
2
. The result in part
b
is larger than this and the result in
Problem 7.70
A small block with mass 0.0475 kg slides in a vertical circle of radius 0.400 m on the inside of a circular track. During one of the revolutions of
the block, when the block is at the bottom of its path, point A , the magnitude of the normal force exerted on the block by the track has
magnitude 3.95 N . In this same revolution, when the block reaches the top of its path, point B , the magnitude of the normal force exerted on
the block has magnitude 0.665 N .
Part A
How much work was done on the block by friction during the motion of the block from point A to point B ?
Express your answer with the appropriate units.
ANSWER:
W f riction
= −9.84×10−2 J
Correct
Problem 5.94
You are riding in an elevator on the way to the 18th floor of your dormitory. The elevator accelerationg upward with a = 1.85 m/s2 . Beside you
is the box containing your new computer; box and contents have a total mass of 35.0 kg . While the elevator is accelerating upward, you push
horizontally on the box to slide it at constant speed toward the elevator door.
Part A
If the coefficient of kinetic friction between the box and elevator floor is μk = 0.32, what magnitude of force must you apply?
Express your answer with the appropriate units.
ANSWER:
F
= 130 N
Correct
Problem 3.53
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
16/18
1/12/22, 9:33 PM
bonus assignment final week
A toy rocket is launched with an initial velocity of 11.0 m/s in the horizontal direction from the roof of a 30.0-m-tall building. The rocket's engine
produces a horizontal acceleration of ( 1.60 m/s3 )t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g,
downward. Air resistance can be neglected.
Part A
What horizontal distance does the rocket travel before reaching the ground?
Express your answer with the appropriate units.
ANSWER:
l
= 31.3 m
Correct
Problem 1.82
Vector A ⃗ has magnitude 5.00 m and lies in the xy -plane in a direction 53.0∘ from the +x -axis measured toward the +y-axis. Vector B⃗ has
magnitude 8.00 m and a direction you can adjust.
Part A
You want the vector product A ⃗ × B⃗ to have a positive z-component of the largest possible magnitude. What direction angle should you
select for vector B⃗ measured from the +x -axis toward the +y-axis?
Express your answer in degrees.
ANSWER:
θ
= 143.0
∘
Correct
Part B
What is the direction angle of B⃗ for which A ⃗ × B⃗ has the most negative z-component? The angle is measured from the +x -axis toward
the +y-axis.
Express your answer in degrees.
ANSWER:
θ
= 323.0
∘
Correct
Part C
What are the two direction angles of B⃗ for which A ⃗ × B⃗ is zero? The angles are measured from the +x -axis toward the +y-axis.
Express your answers in degrees separated by a comma.
ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
17/18
1/12/22, 9:33 PM
θ1
, θ2 = 53.0,233.0
bonus assignment final week
∘
Correct
Score Summary:
Your score on this assignment is 0%.
You received 0 out of a possible total of 0 points, plus 12.95 points of extra credit.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9672969
18/18