Principles & Practice of
Physics 2nd Edition
ST
SOLUTIONS
UV
MANUAL
IA
_A
Volume 1: Chapters 1–21
Volume 2: Chapters 22–34
PP
Eric Mazur
RO
Comprehensive Solutions Manual for
VE
Instructors and Students
D?
© Eric Mazur. All rights reserved. Reproduction or distribution without permission is
prohibited.
©MedConnoisseur
, 1
ST
FOUNDATIONS
UV
Solutions to Developing a Feel Questions, Guided Problems,
and Questions and Problems
Developing a Feel
1. 10−2 m 2. 102 m 3. 102 m 4. 101 m 2 5. 103 6. 103 kg 7. 104 kg 8. 1013 kg 9. 105 kg 10. 1011
IA
11. 109 12. 105
Guided Problems
_A
1.2 Solar oxygen
1. Getting Started Much of the plan used in Worked Problem 1.1 can be used here. We wish to use the volume of
the sun and the percent of the Sun that is made up of oxygen to calculate the mass density and number density of
oxygen atoms. As before we assume the Sun is a perfect sphere.
PP
2. Devise Plan As before, we use the mass density ρ = m / V and the number density n = N / V . We will use
4
V = π R 3 for the volume of the Sun. We will use the total mass of the Sun and the fraction of that mass that is made
3
up of oxygen to calculate the mass of oxygen in the Sun. We will divide by volume to get the mass density. Then,
RO
one need only divide by the mass of a single oxygen atom to convert from mass density to number density.
3. Execute Plan First, the mass density is given by
m m (0.00970) M Sun
ρ = = oxygen =
V 4 π R3 4 3
π RSun
Sun
3 3
(0.00970)(1.99 × 1030 kg)
VE
ρ= = 13.7 kg/m3
4
π (6.96 × 10 m)
8 3
3
We convert from mass density to number density using
ρ 13.7 kg/m3
n= = = 5.14 × 1026 atoms/m3
matom 2.66 × 10−26 kg/atom
D?
4. Evaluate Result The calculated mass density of oxygen is approximately two orders of magnitude smaller than
the mass density of hydrogen calculated in Worked Problem 1.1. This is what we expect, because the oxygen
accounts for only about 1% of the mass of the Sun and hydrogen makes up approximately 70%. Since the oxygen
accounts for two orders of magnitude less mass, it is perfectly sensible that the mass density of oxygen would also be
two orders of magnitude smaller. Similarly, the number density of oxygen is about three orders of magnitude smaller
than the number density of hydrogen. If the only issue were the relative percentages of the Sun’s mass made up by
oxygen and hydrogen, we would expect a difference of only two orders of magnitude. However, oxygen is also much
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1-1
, 1-2 Chapter 1
more massive than hydrogen (around one order of magnitude more massive). Hence, mass densities that differ by two
orders of magnitude mean number densities that differ by three orders of magnitude. Our results are consistent with
those of Worked Problem 1.1.
1.4 Box volume
ST
1. Getting Started This problem is similar to Worked Problem 1.3 in that we are given a mixture of units which we must
convert to SI units. However, the expressions for volume here is completely different from that of Worked Problem 1.3.
2. Devise Plan We convert from feet to inches using the conversion factor 1 ft = 12 in, and convert from inches to
UV
meters using 1 in = 0.0254 m. We can convert millimeters and centimeters to meters by simply dividing by the
appropriate factor of ten. When all quantities are in units of meters, we proceed to find the volume of the box using
Vbox = A wh, where A, w, and h are the length, width, and height of the box, respectively.
3. Execute Plan We first convert the length, width and height to units of meters:
1m
A = 1420 mm × 3 = 1.420 m
IA
10 mm
12 in 0.0254 m
w = 2.75 ft × × = 0.838 m
1 ft 1 in
1m
_A
h = 87.8 cm × = 0.878 m
102 cm
Finally the volume is given by Vbox = A wh = (1.420 m)(0.838 m)(0.878 m) = 1.05 m3 .
4. Evaluate Result A volume of 1.05 m3 is reasonable. The box had a size that was on the order of 1.0 m in each
dimension. One dimension had a length greater than 1.0 m and the other two were smaller than 1.0 m. This is the
PP
approximate size we would expect.
1.6 Digits on your own
1. Getting Started The methods used in Worked Problem 1.5 are essentially the same that will be used here. The
RO
expressions involve quantities with varying numbers of significant digits. The number of significant digits in the
answer will depend on how many significant digits are in the given in the problem statement, and on the operation
carried out between quantities (subtraction, multiplication, etc).
2. Devise Plan As in Worked Problem 1.5, the number of significant digits in a product or quotient is the same as
the number of significant digits in the input quantity that has the fewest significant digits. The number of decimal
places in a sum or difference is the same as the number of decimal places in the input quantity that has the fewest
VE
decimal places. To express our answers as orders of magnitude, we write each in scientific notation, round the
coefficient either down to 1 (for coefficients ≤ 3) or up to 10 (for coefficients > 3). We then write the answer as a
power of ten without the coefficient.
3. Execute Plan (a) In (205)(0.0041)(489.62), the middle quantity (0.0041) has the fewest significant digits, with
D?
only two. Hence the answer must be given to only two significant digits: 4.1 × 102. Here the coefficient is greater
than 3, so we round it to 10 and obtain for the order of magnitude 10 × 102 = 103. (b) Here, the first factor is given to
four significant digits, which is the fewest of all factors (since π is known to many digits). Hence the answer must
be given to five significant digits: 2.475 × 101. Here the prefactor is less than 3, so we round it down to 1. This yields
an order of magnitude of 1× 10 = 101. (c) The first term is known to the thousandths place, whereas the second term is
known only to the tenths place. Hence the answer can be given only out to the tenths place: 6.9802 × 103. The
prefactor is greater than 3, so we round it to 10. This yields an order of magnitude of 10 × 103 = 104.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, Foundations 1-3
4. Evaluate Result We could check our answers by getting order of magnitude estimates for each number. (a) Using
orders of magnitude of (205)(0.0041)(489.62) yields (102 )(10−2 )(103 ) = 103 , which is consistent with our answer.
(b) Similarly, (190.8)(0.407 500)/π becomes (102 )(100 )/(101 ) = 101 , which is again consistent with our answer. (c)
Finally, to nearest order (6980.035) + (0.2) yields (104 ) − (10−1 ) = 104 , which is also consistent.
ST
1.8 Roof area
1. Getting Started We will approximate the United States as a rectangle 5,000 km wide and 3,000 km high. A very
small percentage of this area is occupied by buildings. Only in cities can we approximate the surface as being
UV
covered by structures.
2. Devise Plan The approximate area of the United States is (104 km)(103 km) = 107 km 2 . Depending on one’s
definition of a “city” the percentage of this surface area that is occupied by cities could be anywhere between 0.1%
and 1%. But since even cities are not completely filled with structures, we will use the smaller of these two
percentages for our estimate (0.1%). These two numbers can be used to find an order of magnitude estimate of the
IA
total roof area in the United States.
3. Execute Plan The roof area is given by Aroof = (0.1%) AUS = (0.001)(107 km 2 ) = 104 km 2 .
4. Evaluate Result We could check our result by estimating the roof surface area in a completely different way. Let
_A
us start with the population of the United States ( 3 × 108 people to one significant digit), and assume that there exists
a structure that houses every set of three or four people, and assume further that these structures have a footprint of
approximately 30 m 2 . Of course, many people live alone and many people live in apartment buildings that have
many floors (meaning less of a footprint per person). But these cases might cancel each other out, making our
estimate plausible. This yields a surface area of 2.3 × 109 m 2 . Let us double this number to account for the places of
PP
business that employ many of these people. This gives a total surface area of order 1010 m 2 . Using known
conversion factors, one can see that this is equivalent to 104 km 2 . This agrees with our previous estimate.
Questions and Problems
RO
1.1. The word “undetectable” prevents this from being a valid scientific hypothesis. A hypothesis must be
experimentally verifiable.
1.2. You assume that the competing product contains non-zero fat, and that the serving sizes of the two are equal.
Say the two foods contain the same amount of fat per ounce. The maker of the product being advertised could print
his label showing a recommended serving size 50 percent smaller than the recommended serving size of the
competing product. This makes the claim of 50 percent less fat per serving true but of course misleading.
VE
1.3. You assume the sequence is linear, meaning each entry is larger than the previous one by a constant amount. As
an alternative, the sequence could be formed by starting with 1, 2 and then setting the n th term cn equal to the sum
of the previous two terms: cn = cn −1 + cn − 2 . This would work for c3 = c2 + c1 = 2 + 1 = 3, and would yield 5 as the next
number in the sequence.
D?
1.4. If you assume that the coins are currently circulated U.S. currency, you would not be able to find a solution. If,
however, you consider that the word “cents” is also used to refer to hundredths of other currencies, then all that
would be required is that increments of 10 and 20 centers exist in some currency. As an example, “cents” may refer
to hundredths of a Euro. You would say the coins must be worth 10 and 20 Euro cents, respectively (which do exist).
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Physics 2nd Edition
ST
SOLUTIONS
UV
MANUAL
IA
_A
Volume 1: Chapters 1–21
Volume 2: Chapters 22–34
PP
Eric Mazur
RO
Comprehensive Solutions Manual for
VE
Instructors and Students
D?
© Eric Mazur. All rights reserved. Reproduction or distribution without permission is
prohibited.
©MedConnoisseur
, 1
ST
FOUNDATIONS
UV
Solutions to Developing a Feel Questions, Guided Problems,
and Questions and Problems
Developing a Feel
1. 10−2 m 2. 102 m 3. 102 m 4. 101 m 2 5. 103 6. 103 kg 7. 104 kg 8. 1013 kg 9. 105 kg 10. 1011
IA
11. 109 12. 105
Guided Problems
_A
1.2 Solar oxygen
1. Getting Started Much of the plan used in Worked Problem 1.1 can be used here. We wish to use the volume of
the sun and the percent of the Sun that is made up of oxygen to calculate the mass density and number density of
oxygen atoms. As before we assume the Sun is a perfect sphere.
PP
2. Devise Plan As before, we use the mass density ρ = m / V and the number density n = N / V . We will use
4
V = π R 3 for the volume of the Sun. We will use the total mass of the Sun and the fraction of that mass that is made
3
up of oxygen to calculate the mass of oxygen in the Sun. We will divide by volume to get the mass density. Then,
RO
one need only divide by the mass of a single oxygen atom to convert from mass density to number density.
3. Execute Plan First, the mass density is given by
m m (0.00970) M Sun
ρ = = oxygen =
V 4 π R3 4 3
π RSun
Sun
3 3
(0.00970)(1.99 × 1030 kg)
VE
ρ= = 13.7 kg/m3
4
π (6.96 × 10 m)
8 3
3
We convert from mass density to number density using
ρ 13.7 kg/m3
n= = = 5.14 × 1026 atoms/m3
matom 2.66 × 10−26 kg/atom
D?
4. Evaluate Result The calculated mass density of oxygen is approximately two orders of magnitude smaller than
the mass density of hydrogen calculated in Worked Problem 1.1. This is what we expect, because the oxygen
accounts for only about 1% of the mass of the Sun and hydrogen makes up approximately 70%. Since the oxygen
accounts for two orders of magnitude less mass, it is perfectly sensible that the mass density of oxygen would also be
two orders of magnitude smaller. Similarly, the number density of oxygen is about three orders of magnitude smaller
than the number density of hydrogen. If the only issue were the relative percentages of the Sun’s mass made up by
oxygen and hydrogen, we would expect a difference of only two orders of magnitude. However, oxygen is also much
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1-1
, 1-2 Chapter 1
more massive than hydrogen (around one order of magnitude more massive). Hence, mass densities that differ by two
orders of magnitude mean number densities that differ by three orders of magnitude. Our results are consistent with
those of Worked Problem 1.1.
1.4 Box volume
ST
1. Getting Started This problem is similar to Worked Problem 1.3 in that we are given a mixture of units which we must
convert to SI units. However, the expressions for volume here is completely different from that of Worked Problem 1.3.
2. Devise Plan We convert from feet to inches using the conversion factor 1 ft = 12 in, and convert from inches to
UV
meters using 1 in = 0.0254 m. We can convert millimeters and centimeters to meters by simply dividing by the
appropriate factor of ten. When all quantities are in units of meters, we proceed to find the volume of the box using
Vbox = A wh, where A, w, and h are the length, width, and height of the box, respectively.
3. Execute Plan We first convert the length, width and height to units of meters:
1m
A = 1420 mm × 3 = 1.420 m
IA
10 mm
12 in 0.0254 m
w = 2.75 ft × × = 0.838 m
1 ft 1 in
1m
_A
h = 87.8 cm × = 0.878 m
102 cm
Finally the volume is given by Vbox = A wh = (1.420 m)(0.838 m)(0.878 m) = 1.05 m3 .
4. Evaluate Result A volume of 1.05 m3 is reasonable. The box had a size that was on the order of 1.0 m in each
dimension. One dimension had a length greater than 1.0 m and the other two were smaller than 1.0 m. This is the
PP
approximate size we would expect.
1.6 Digits on your own
1. Getting Started The methods used in Worked Problem 1.5 are essentially the same that will be used here. The
RO
expressions involve quantities with varying numbers of significant digits. The number of significant digits in the
answer will depend on how many significant digits are in the given in the problem statement, and on the operation
carried out between quantities (subtraction, multiplication, etc).
2. Devise Plan As in Worked Problem 1.5, the number of significant digits in a product or quotient is the same as
the number of significant digits in the input quantity that has the fewest significant digits. The number of decimal
places in a sum or difference is the same as the number of decimal places in the input quantity that has the fewest
VE
decimal places. To express our answers as orders of magnitude, we write each in scientific notation, round the
coefficient either down to 1 (for coefficients ≤ 3) or up to 10 (for coefficients > 3). We then write the answer as a
power of ten without the coefficient.
3. Execute Plan (a) In (205)(0.0041)(489.62), the middle quantity (0.0041) has the fewest significant digits, with
D?
only two. Hence the answer must be given to only two significant digits: 4.1 × 102. Here the coefficient is greater
than 3, so we round it to 10 and obtain for the order of magnitude 10 × 102 = 103. (b) Here, the first factor is given to
four significant digits, which is the fewest of all factors (since π is known to many digits). Hence the answer must
be given to five significant digits: 2.475 × 101. Here the prefactor is less than 3, so we round it down to 1. This yields
an order of magnitude of 1× 10 = 101. (c) The first term is known to the thousandths place, whereas the second term is
known only to the tenths place. Hence the answer can be given only out to the tenths place: 6.9802 × 103. The
prefactor is greater than 3, so we round it to 10. This yields an order of magnitude of 10 × 103 = 104.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, Foundations 1-3
4. Evaluate Result We could check our answers by getting order of magnitude estimates for each number. (a) Using
orders of magnitude of (205)(0.0041)(489.62) yields (102 )(10−2 )(103 ) = 103 , which is consistent with our answer.
(b) Similarly, (190.8)(0.407 500)/π becomes (102 )(100 )/(101 ) = 101 , which is again consistent with our answer. (c)
Finally, to nearest order (6980.035) + (0.2) yields (104 ) − (10−1 ) = 104 , which is also consistent.
ST
1.8 Roof area
1. Getting Started We will approximate the United States as a rectangle 5,000 km wide and 3,000 km high. A very
small percentage of this area is occupied by buildings. Only in cities can we approximate the surface as being
UV
covered by structures.
2. Devise Plan The approximate area of the United States is (104 km)(103 km) = 107 km 2 . Depending on one’s
definition of a “city” the percentage of this surface area that is occupied by cities could be anywhere between 0.1%
and 1%. But since even cities are not completely filled with structures, we will use the smaller of these two
percentages for our estimate (0.1%). These two numbers can be used to find an order of magnitude estimate of the
IA
total roof area in the United States.
3. Execute Plan The roof area is given by Aroof = (0.1%) AUS = (0.001)(107 km 2 ) = 104 km 2 .
4. Evaluate Result We could check our result by estimating the roof surface area in a completely different way. Let
_A
us start with the population of the United States ( 3 × 108 people to one significant digit), and assume that there exists
a structure that houses every set of three or four people, and assume further that these structures have a footprint of
approximately 30 m 2 . Of course, many people live alone and many people live in apartment buildings that have
many floors (meaning less of a footprint per person). But these cases might cancel each other out, making our
estimate plausible. This yields a surface area of 2.3 × 109 m 2 . Let us double this number to account for the places of
PP
business that employ many of these people. This gives a total surface area of order 1010 m 2 . Using known
conversion factors, one can see that this is equivalent to 104 km 2 . This agrees with our previous estimate.
Questions and Problems
RO
1.1. The word “undetectable” prevents this from being a valid scientific hypothesis. A hypothesis must be
experimentally verifiable.
1.2. You assume that the competing product contains non-zero fat, and that the serving sizes of the two are equal.
Say the two foods contain the same amount of fat per ounce. The maker of the product being advertised could print
his label showing a recommended serving size 50 percent smaller than the recommended serving size of the
competing product. This makes the claim of 50 percent less fat per serving true but of course misleading.
VE
1.3. You assume the sequence is linear, meaning each entry is larger than the previous one by a constant amount. As
an alternative, the sequence could be formed by starting with 1, 2 and then setting the n th term cn equal to the sum
of the previous two terms: cn = cn −1 + cn − 2 . This would work for c3 = c2 + c1 = 2 + 1 = 3, and would yield 5 as the next
number in the sequence.
D?
1.4. If you assume that the coins are currently circulated U.S. currency, you would not be able to find a solution. If,
however, you consider that the word “cents” is also used to refer to hundredths of other currencies, then all that
would be required is that increments of 10 and 20 centers exist in some currency. As an example, “cents” may refer
to hundredths of a Euro. You would say the coins must be worth 10 and 20 Euro cents, respectively (which do exist).
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
