CK-12 Physics Intermediated3r1z7wkgqhj9d.cloudfront.net/Physics Rhetoric Resources... · 2018-06-01 · 1. What Is Science? Problem Sets Professor A in the previous problem. 9.True

CK-12 Physics IntermediateProblem Sets

Karson Bader, Ph.D.

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Printed: August 23, 2014

AUTHORKarson Bader, Ph.D.

TECHNICALREVIEWERSteven Lai, Peter Lai, ConnieCheng, Benjamin Fischer, AllenCheng, Sam Smith, VincentPolitzer

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Contents www.ck12.org

Contents

1 What Is Science? Problem Sets 11.1 What Is Science? Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 One-Dimensional Motion Problem Sets 72.1 One-Dimensional Motion Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Two-Dimensional Motion Problem Sets 233.1 Two-Dimensional Motion Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

4 Newton’s Three Laws Problem Sets 494.1 Newton’s Three Laws Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5 Forces in Two Dimensions Problem Sets 575.1 Forces in Two Dimensions Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

6 Work and Energy Problem Sets 716.1 Work and Energy Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 726.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

7 Momentum Problem Sets 957.1 Momentum Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 967.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

8 Angular Motion and Statics Problem Sets 1158.1 Angular Motion and Statics Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1168.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

9 Newton’s Universal Law of Gravity Problem Sets 1279.1 Newton’s Universal Law of Gravity Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . 1289.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

10 Periodic Motion Problem Sets 13510.1 Periodic Motion Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13610.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

11 Vibrations and Sound Problem Sets 14311.1 Chapter 11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

12 Fluid Mechanics Problem Sets 153

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www.ck12.org Contents

12.1 Chapter 12 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

13 Heat Problem Sets 16713.1 Heat Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

14 Thermodynamics Problem Sets 17214.1 Chapter 14 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

15 Electrostatics Problem Sets 17715.1 Chapter 15 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

16 Electric Potential Problem Sets 18316.1 Chapter 16 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18416.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

17 Circuits Problem Sets 19017.1 Chapter 17 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

18 Magnetism Problem Sets 20318.1 Chapter 18 Problems Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

19 Electromagnetism Problem Sets 20819.1 Chapter 19 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

20 Geometric Optics Problem Sets 21420.1 Chapter 20 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21520.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

21 Physical Optics Problem Sets 22421.1 Chapter 21 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225

22 The Special Theory of Relativity Problem Sets 23022.1 Chapter 22 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

23 Quantum Physics Problem Sets 23723.1 Chapter 23 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

24 Atomic Physics Problem Sets 24424.1 Chapter 24 Problems Atomic Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

25 Nuclear Physics Problem Sets 24825.1 Chapter 25 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

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www.ck12.org Chapter 1. What Is Science? Problem Sets

CHAPTER 1 What Is Science? ProblemSets

Chapter Outline1.1 WHAT IS SCIENCE? PROBLEM SETS

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1.1. What Is Science? Problem Sets www.ck12.org

1.1 What Is Science? Problem Sets

TABLE 1.1: Currency Exchange Rates

Currency Country Units Units/USD USD/UnitAUD Australia Dollars 0.969065814 1.031922BRL Brazil Reais 1.715000096 0.58309CAD Canada Dollars 0.985016592 1.015211COP Colombia Pesos 1,825.00 0.000548EGP Egypt Pounds 5.954000156 0.167954XAU Gold Ounces 0.000562237 1,778.61HKD Hong Kong Dollars 7.791262076 0.128349INR India Rupees 47.5599822 0.021026IQD Iraq Dinars 1,167.50 0.000857JPY Japan Yen 76.70421393 0.013037MXN Mexico Pesos 12.95827175 0.077171XPD Palladium Ounces 0.001382698 723.224PHP Philippines Pesos 43.30000244 0.023095RUB Russia Rubles 30.41087231 0.032883KRW South Korea Won 1,111.93 0.000899THB Thailand Baht 30.35499378 0.032944GBP United Kingdom Pounds 0.632646737 1.580661VND Vietnam Dong 20,890.00 4.79E-05

1. Billy decides to take 400 U.S. dollars with him on his vacation to Rio de Janeiro. After converting his money,how much will Billy have during his stay in Brazil? Use the given currency exchange rates ( Table 1.1) toanswer this question.

2. While shopping in London, Julia decides to buy her grandmother a necklace on sale for 33 pounds. Howmuch does she need in U.S. dollars to purchase the gift? Use the given currency exchange rates ( Table 1.1)to answer this question.

3. On his way back from a six-month trip to South Korea, Albert decides to convert the remainder of his moneyinto U.S. dollars. If all Albert has left is 56 won, how much does he have in dollars? Use the given currencyexchange rates ( Table 1.1) to answer this question.

4. Tara is driving down from Toronto, Ontario to visit her cousins in Buffalo, New York. If she has 330 Canadiandollars on her, how much is that worth across the border? Use the given currency exchange rates ( Table 1.1)to answer this question.

5. Returning to the U.S. from a trip to Japan, Nicole has a layover in Hong Kong. During her wait, she grabslunch at an airport cafe. If all Nicole has on her is 5,000 yen, does she have enough money to pay for a mealthat is worth HK$116.85 (Hong Kong dollars)? Use the given currency exchange rates ( Table 1.1) to answerthis question.

6. Three college students are planning a month-long trip to Japan. According to their calculations, the trip willcost a total of $9,000. They find part-time jobs to save up money, working 20 hours per week for 3 months.Will they make enough to cover the trip if they each earn $8.00 per hour? Assume there are 4 full weeks in amonth and use the given currency exchange rates ( Table 1.1) to answer this question.

7. Professor A gets paid $1,800 per month for teaching a course that meets 3 times a week. If each class is 3hours long, how much is the professor being paid per hour? Assume there are 4 full weeks in a month.

8. Professor B is paid $3,000 for every 24 classes she teaches. Determine whether she earns more or less than

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Professor A in the previous problem.9. True of False: The technique of dimensional analysis can help you determine numerical constants in a

calculation.10. The driving distance between San Francisco, California and Sacramento, California is 86.9 miles. Given that

1 meter = 3.28 feet, what is the distance between the cities in kilometers?11. A car is driving through a 25-mph school zone at 21.0 m/s. Is the driver speeding?12. In the equation x = v0t + 1

2 at2, x represents distance, v represents velocity, t is time, and a represents acceler-ation. Show that the dimensions of both sides of the equation match up.

13. In the equation mgh= 12 mv2, h represents distance, g represents acceleration due to gravity, m represents mass,

and v represents velocity. Show that the dimensions of both sides of the equation match up.14. Convert 90 km/h to mph, given that 1 inch = 2.54 cm.15. Convert 2 light-years to meters.16. Convert between the following units:

a. 1 second to hoursb. 1 m2 to cm2

17. Convert between the following units:

a. 1 km/h to m/sb. 1 mi/h to ft/s

18. Convert 1 g/cm3 to kg/m3.19. In some parts of the world, the speed limit is 100 km/h. What is this in mph?20. In 23.0 g of sodium there are 6.02×1023 atoms. Calculate the time it would take you to individually count the

atoms in that sodium sample if you had the technology to count 5 atoms per second.21. Convert 1 L to cm3 and m3.22. While preparing for a physics exam, one of your friends states that newtons are a fundamental unit in physics

and that this can be understood from Newton’s Laws. Is your friend correct?23. Show that dividing velocity by acceleration will yield a quantity measured in seconds (time).24. Kinetic energy is measured in joules. If the equation for kinetic energy is KE = 1

2 mv2, how can joules beexpressed in fundamental units?

25. Convert between the following units:

a. 1 m to cmb. 1 m2 to cm2

c. 1 m3 to cm3

26. Given that 1 inch = 2.54 cm, convert 50 mph to km/h.27. Calculate the number of kilometers in a mile.28. A deep-sea probe descends to its destination at a speed of 2.3 m/s. Given that 1 fathom = 6 feet, what is its

speed in fathoms per minute?29. How many square meters are in an area of 2.0 km2?30. Convert 65 mph to km/h.31. Which distance is greater: 100 yards or 100 meters?32. How many micrometers are in a meter?33. Jupiter has a surface area of 23.71 billion mi2. What is the gas giant’s surface area in square kilometers?34. Your new apartment has 3 rooms with the following dimensions: 10 f t×10 f t, 8 f t×13 f t, and 9 f t×10 f t.

What is the average area of the three rooms in your apartment?35. A small square block of mass 2.5 g has a volume of 113 cm3. What is the density of the block in kg/m3?36. If each step a person takes while walking is approximately 2 feet, how many steps will be taken in a mile?37. (a) How many breaths does a person take in one day if the average number of breaths taken is 500 per hour?

(b) How many more breaths are taken in a day if the average is 550 per hour?38. Using the correct number of significant digits, calculate the area of a room that is 13.73 by 4.33 m.39. Calculate the density of a 2 kg cylinder that is 3.0 cm tall with a radius of 5.0 cm.

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40. The dimensions of your dorm room is 10 ft by 10 ft, calculate the area in meters.41. If a pump at a gas station takes approximately 4.5 minutes to fill a 12 gallon gas tank, calculate the rate at

which the gas is flowing out of the pump.42. A small titanium disc has a radius of 2.5 cm and a thickness of 0.1 cm. What is the mass of the disc? (Assume

the density is 4.5 g/cm3)43. The length of a rectangle is 8 m more than twice the width. Determine the length and the width if the perimeter

is 112 m.44. Calculate the volume of a sphere with a diameter of 2.7 cm and a surface area of 22.9 cm2.45. One level of a parking lot structure measures 845 ft by 900 ft. Calculate the area in m2.46. How many square miles are in 3200 acres?47. Determine the units of the variable ’a’ in the following equation, if x is in meters, v is in meters per second

and t is in seconds: x = vt + 0.5at2.48. How many degrees are in 1π radians?49. Calculate the average of the following numbers: 2.5, 3.6, 7.1 and 3.3.50. If you are driving 120 m/s, how fast are you driving in km/h?51. Determine the speed, in m/s, if you are traveling at 70 mph.52. Determine the units of d in the following equation if P is in units of kg / (ms2) and v is in units of m/s: P +

dv2.53. How many seconds are in one year?54. Convert 1 L of water to cm3.55. Sum the following numbers keeping the correct number of significant digits: 2.030, 0.11352.56. In the following equation, what are the units of β : γ = 1√

1−β2.

57. In the following equation, what are the units of x, if A is in meters and t is in seconds: y = Ae−xt ?

58. If you could count 1 penny per second, how long would it take you to count 1 million dollars?59. When multiplying two numbers, do the exponents add or multiply?60. Is it possible for the left and right side of an equation to have different units; if so, how?61. How many seconds are in a femtosecond?

Solutions

1. (400 dollars)(1.715000096 reais

1 dollar

)= 686 reais

2. (33 pounds)(

0.632646737 dollars1 pound

)= 20.88 dollars

3. (56 won)(0.000899 dollars

1 won

)= 0.05 dollars

4. (330 Canadian dollars)( 1.015211 dollars

1 Canadian dollars

)= 335.02 dollars

5. (5,000 yen)(

0.013037 dollars1 yen

)(7.791262076 Hong Kong dollars

1 dollar

)= 507.87 Hong Kong dollars; yes, Nicole has enough

money to pay for her meal.6. In 3 months, each student will have earned (3 months)

( 4 weeks1 month

)(20 hours1 week

)(8 dollars1 hour

)= 1,920 dollars, which

adds up to a total of $5,760. No, the students will not make enough to cover their trip.7.(

1,800 dollars1 month

)(1 month4 weeks

)( 1 week3 classes

)( 1 class3 hours

)= 50 dollars/hour

8. Professor A teaches 12 classes per month. For 24 classes, Professor A would teach for 2 months and earn atotal of (2 months)

($1,800

1 month

)= $3,600, which is $600 more than what Professor B earns teaching the same

number of classes. Therefore, Professor B earns less than Professor A.9. False. Dimensional analysis can only help you determine the units of an answer or of a variable in question.

10. (86.9 mi)(

5,280 f t1 mi

)(1 m

3.28 f t

)(1 km

1,000 m

)= 140 km

11. (21.0 m/s)(

3.28 f t1 m

)(1 mi

5,280 f t

)( 60 s1 min

)(60 min1 h

)= 47.0 mi/h; yes, the car is driving well above the speed limit

of 25 mph (mi/h).

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12.

x = v0t +12

at2

[m] =[m]

[s][s]+

[m]

[s2][s]2

[m] = [m]+ [m]

13.

mgh =12

mv2

[kg][m

s2

][m] = [kg]

[ms

]2

[kg][

m2

s2

]= [kg]

[m2

s2

]14. 90 km

h

∣∣∣ 1 cm10−5 km

∣∣∣ 1 in2.54 cm

∣∣∣ 1 f t12 inch

∣∣∣ 1 mi5280 f t = 53.6 mph

15. 2 lightyear = 2 c · yr∣∣∣3.0E8 m/s

1 c

∣∣∣365 days1 yr

∣∣∣ 24 h1 day

∣∣∣3600 s1 h = 1.89E16 m

16. (a) 1 s∣∣∣1 min

60 s

∣∣∣ 1 h60 min = 2.78E−8 h

(b) 1 m2∣∣∣1000 cm2

1 m2 = 10000 cm2

17. (a) 1 kmh

∣∣∣ 1 h3600 s

∣∣∣1000 m1 km = 0.278 m/s

(b) 1 mih

∣∣∣ 1 h3600 s

∣∣∣5280 f t1 mi = 1.47 f t/s

18. 1 gcm3

∣∣∣ 1 kg1000 g

∣∣∣1E6 cm3

1 m3 = 1000 kg/m

19. 100 kmh

∣∣∣1 miles1.6 km = 6.25 mph

20. 6.02E23 atoms∣∣∣ 1 s

5 atoms = 1.204E23 s = 3.8E15 yrs

21. 1 L = length3 = (10 cm)3 = 10−3 m3

22. Newtons are not fundamental units since they can be derived from other units, namely kilograms, meters, andseconds.

23. va = m

ss2

m = s

24. 12 mv2⇒ [kg]

[m2

s2

]25. (a) 1 m

∣∣∣1E2 cm1 m = 100 cm

(b) 1 m2∣∣∣1E4 cm2

1 m2 = 10,000 cm2

(c) 1 m3∣∣∣1E6 cm3

1 m3 = 1,000,000 cm3

26. 50 milehour = 5280 f t

mi = 12 inf t = 2.54 cm

in = 1E−5 kmcm = 80.5 km

h

27. 1 mile = 63360 inch1 mile = 2.54 cm

1 inch = 1E−5 kmcm = 1.6 km

28. 2.3 ms = 3.28 f t

1 m = 1 f athom6 f t = 60 s

1 min = 75.4 f athommin

29. 2.0 km2 = 1000 mkm = 1000 m

km = 2×106 m2

30. 65 mileh = 1.61 km

1 mile = 104.7 km/h31. 100 yard = 91.44 m. Therefore 100 meters is a longer distance.32. 106 micrometers = 1 meter33. Surface area = 4πr2 = 4π(7.15×104 km)2 = 6.42×1010 km2

34. The average area is 98 ft2

35. The mass of the block is 0.0025 kg, while volume in meter3 is 1.13×10−4 m3. Therefore, ρ = 22.12 kg/m3.36. 5280 f t

mi

∣∣∣1 step2 f t = 2640 steps

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37. (a) 500h

24 h1 day = 12000 breaths

(b) 550h

24 h1 day = 13200 breaths

13200−12000 = 1200 more breaths taken38. 59.5 m2

39. ρ = MV = 2 kg

(πr2h) =2 kg

(π((.05)2)m2)3.0) = 8.49×104 kg/m3

40. 30.5 m2

41. 12 gal4.5 min = 2.67 gal/min

42. M = ρV =(

4.5 gcm3

)(π(2.5)20.1 cm3) = 8.84 g

43.

let w = width

2(2w+8)+2w = 112

6w = 96

w = 16

→ l = 2w+8 = 40

44. V = 43 πr3 = 10.3 cm3

45.

800 f t = 243.8 m

900 f t = 274.3 m

A = 6.69×104 m2

46.

1 mile2 = 640 acre

5 mile2 = 3200 acre

47. [a] = m/s2

48. 1π radians = 180

49. 4.1350. 120 m

s

( 1 km1000 m

)(3600 s1 h

)= 432 km/h

51. 70 mih

(1609 m1 mi

)( 1 h3600 s

)= 31.3 m/s

52. kg/m3

53. 1 yr(

365 days1 yr

)(24 h1 day

)(3600 s1 h

)= 3.15×107 s

54. 1 L = 103 cm3

55. 2.14456. β, has no units.57. Since the exponential must have no units, the units of x is seconds.58. $1×106

( $0.01s )

= 1×108 s = 76 yrs

59. The exponents add together.60. No, it is not possible for an equation to have different units on each side of the equal sign.61. 1×10−15 s = 1 f s

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www.ck12.org Chapter 2. One-Dimensional Motion Problem Sets

CHAPTER 2 One-Dimensional MotionProblem Sets

Chapter Outline2.1 ONE-DIMENSIONAL MOTION PROBLEM SETS

2.2 REFERENCES

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2.1. One-Dimensional Motion Problem Sets www.ck12.org

2.1 One-Dimensional Motion Problem Sets

Solutions

1. B. The displacement is equal to zero.2. True. By simply throwing a ball straight up, assuming you define up as positive and down as negative, the

velocity would be positive while the acceleration due to gravity would be negative.3. C. s = d

t =200 m40 s = 5 m

s4. A. v = 10 m

st = 30 sd = |v|t = (5 m

s )(30 s) = 150 m5. False. ~vSara = 5 m

s , ~vBilly = 5 ms~vSara−~vBilly = 0 m

s

6. D. t = xv =

1.5×1011 m3.0×108 m/s = 5.0×102 s

7. A. ∆~x = x f − xi = 0−1.5×1011 =−1.5×1011 m i8. Let your work be the origin: xi = 0

∆x = x f − xi = 10 km−0 = 10 km9. vavg =

∆x∆t =

10 kmtdrive+twalk

= 10 km8.0 km

70 km/h+( 40 min60 min

h)= 12.8 km/h

10.

vavg =∆x∆t

=x2− x1

t=

2.0×1010 m −4.0×1010 m12 months

=−2.0×1010

12m

month=−1.6×109 m

month

11. If the car’s velocity were constant, its position would be changing at a constant rate. However, because theequation for position has a degree greater than 1, the rate of change cannot be constant.

12.

v =∆x∆t

t =xv=

20 m120 km/h

=20 m

(120×103) m/h= 1.66×10−4h = 0.6 s

13. (a) vavg =∆x∆t =

(3 m/s)(60 s)+(2 m/s)(60 s)120 s = 2.5 m/s

(b) vavg =∆x∆t =

80 m40 m5 m/s+

40 m2 m/s

= 2.86 m/s

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(c)14. vavg =

x2(t2)−x1(t1)t2−t1

= 12 a(t1 + t2)

a. 24.5 m/sb. 14.7 m/sc. 7.35 m/s

15. (a) x = 42.7 m(b) x = 177.4 m(c) x = 1154.2 m(d) x(8)− x(0) = 1154.2 m−7 m = 1147.2 m

16. (a) vavg =x(5)−x(0)

∆t = 5(5)+5(25)5 = 30 m/s

(b) v = x(10)−x(5)∆t = 5(10)+5(100)−5(5)+5(25)

5 = 80 m/s17. (a) vavg =

∆x∆t =

100 km50 km

40 km/h+50 km

60 km/h= 48 km/h

(b) Because the direction is the same throughout the motorcycle’s journey, its average speed is the same asits average velocity: 48 km

h .18. a. Since Jimmy runs faster, find out how long it takes for him to run 100 m:

vJimmy = (.20)vBilly + vBilly = 6.0ms

t =x

vJimmy=

100 m6.0 m/s

= 16.67 s

Then, determine the distance Billy runs in that time:x = vBillyt = (5.0 m/s)(16.67 s) = 83.35 mJimmy won by x = 100 m−83.35 m = 16.65 m

b.

tBilly =x

vBilly=

100 m5.0 m/s

= 20 s

|tJimmy− tBilly|= |20 s−16.67 s|= 3.3 s

19. Let the distance to the park be x:vavg =

2xx

4.0 m/s+x

3.0 m/s= 3.43 m/s

20. v = xt =

1800 miles30 hours = 60 mph

21. (a) v2(t = 6)− v1(t = 2) = (2+5(6)2)− (2+5(2)2) = 160m/s(b) [a] = m

s , [b] =ms3

(c) aavg =∆v∆t =

v2−v1t2−t1

= 182−224 = 40 m/s2

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x = x0 + v0t +12

at2 ,

v = v0 +at = 3+4(3) = 15 m/s2

x = 5+(3)(3)+12(4)(3)2 = 32 m

22.

v = at

aavg =∆v∆t

=v(t2)− v(t1)(t2− t1)

=(a)(t2)− (a)(t1)

(t2− t1)= a

23. For the first 2 seconds:

v = v0 +at

v = (3)(2) = 6 m/s

Position reached in that time:x = v0t + 1

2 at2 = 6 mFor the 18 remaining seconds:x = x0 + v0t + 1

2 at2 = 6+(6 m/s)(18 s)+0 = 114 m24.

a =∆v∆t

=60 mph

3 s=

26.82 ms

3 s= 8.94

ms2

25. 1 mph = 0.447 m/s

(a)

v2 = v20 +2a∆x

a =v2− v2

02∆x

=0−44.7 m2/s2

2(200 m)=−0.11 m/s2

(b)

v = v0 +at

⇒ t =v− v0

a= 400 s

26. v2 = v20 +2a∆x

(a) a =v2−v2

02∆x = (25−4) m2/s2

2((125−5) m) = 0.088 m/s2

(b) v = v0 +at⇒ t = 34.3 s

27. To determine the time, set the final position of the cars equals to one another:

(a)

xp = xop + vopt +12

apt2; xs = xos + vost +12

ast2

vop = xop = xos = 0

xp = xs

⇒ 12

apt2 = vost

⇒ t =2vos

ap= 44.7 s

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(b) xp =12 apt2 = 2000 m

28. (a) Use these two equations to solve for the acceleration:x = v0t + 1

2 at2 (1)v = v0 +at (2)Solve for v0 from equation 2 and substitute into equation 1 to determine a:v0 = v−atx = (v−at)t + 1

2 at2⇒ a = 2(vt−x)t2 = 2.0 m/s2

(b) Use equation 2 to solve for the initial velocity:v0 = v−at = 10.0 m/s

29. First, solve for the object’s acceleration:

v2 = v20 +2a∆x⇒ a =

v2− v20

2∆x

v = v0 +at⇒ t =v− v0

a

t =2∆x

v+ v0= 31.6 s

30. To determine the time, set the final position of the vehicles equal to one another:

xt = xot + vott +12

att2; xc = xoc + voct +12

act2

xp = xs

⇒ xot + vot = voct +12

act2

Since the vehicles experience a near miss⇒ vt = vc, use v = v0 +at to solve for time:t = 2xotvoc−vot

= 30.0 sThus, a = vot−voc

t =−0.33 m/s2

31. Coming to a stop implies v0 = 0 m/s.

⇒ ∆x = −v20

−2|a|

(a) ∆x = 50 m(b) ∆x = 128 m

32. Since the car has a negative accel. of 4 m/s2, we can calculate the distance traveled using the kinematicequation:

v2 = v2o +2a∆x

∆x =v2− v2

o

2a∆x = 2 meters

33. aavg =∆v∆t =

v f−v0t2−t1

= (20−(−16) m/s)(4×10−3 s) = 9.00×103 m/s2

34.

x f = xi + vavgt

x f = xi +vi + v f

2t

⇒ vi =2(x f − xi)

t− v f = 28.64 m/s

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35. (a) x f = xi + v0t + 12 at2

Rewrite as the following:12 at2 + v0t−∆x = 0Use the quadratic formula to solve for time:

t = −b−√

b2−4ac2a where a =−2.50 m/s2, b = v0 = 49.17 m

s , and c =−∆x = 100 m⇒ t = 2.30, 17.33The answer must be the smaller of the two choices. The later event occurs if the car continues its negativeacceleration backwards!

(b) v = v0 +at = 49.1 m/s+(−5.0 m/s2)(2.30 s) = 37.7 m/s36. Compare the equations:

x = 3+3t−5t2

x = x0 + v0t +12

at2

⇒ x0 = 3

v0 = 3

a =−10

The particle turns around when the final velocity reaches zero:

⇒ v f = v0 +at

t =−v0

a= 0.3 s

37. (a)

v = v0 +at; v = 0

→ t =v0

|a|=

100 m/s4.8 m/s2 = 20.8 s

(b)

v2 = v20−2|a|∆x; v = 0

→ ∆x =v2

0a

= 2083 m

No, the pilot would not be able to stop the plane on the runway without a stronger acceleration.38. (a) Ttotal = t1 + t2 + t3

For the acceleration phase:t1 = v

|a| =10 m/s

1.5 m/s2 = 6.67 sThe cat runs for an additional 10 seconds:⇒ t2 = 10 sFinally, solve for the time it takes for the cat to come to a stop:t3 = 7 s⇒ Ttotal = 23.7 s

(b) vavg =xtotal

T =v2

2a1+vt2+

vt32

T = 7.11 m/s39. Choose your locations as the origin. To check for a collision look for a time that the vehicles are at the same

position:

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vo,cart +12

acart2 = xo,van + vo,vant

12

acart2 +(vo,car− vo,van)t− xo,van = 0

t =−(vo,car− vo,van)±

√(vo,car− vo,van)2 +(−2acarx)acar

⇒ t = 10 s,13.3 s

Because we have two real answers, a collision does in fact occur. The point at which the cars meet is thus thesooner of the two times. Where does the collision happen?x = xo,van + vo,vant = 200 m+(15 m

s )10 s = 350 mThe collision occurs 350 m from the point at which you apply the brakes.

40. aavg =∆v∆t =

v f−v0∆t =

44.7 ms

8 s = 5.58 ms2

The distance the car travels is just ∆x = 12 at2 = 178.8m

41. (a) Keeping in mind the final velocity will be downward (negative):

v2y = v2

oy +2a∆y

vy =−√

2a∆y =√

2(−9.8)(−200) =−62.6ms

(b)

vy = voy +at

⇒ t =vy

a= 6.39 s

42. (a)

v2y = v2

oy +2a∆y

∆y =v2

y− v2oy

2a= 31.9 m above the launching point

(b)

v = v0 +at

t =v− v0

a= 2.55 s

(c) When the ball has returned to its launch point, ∆y = 0:

v0t +12

at2 = 0

⇒ t = 5.10 s

(d)

v f = v0 +at

v f = 25ms+(−9.8

ms2 )(5.1 s)

⇒ v f =−25ms

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43. Assume up is positive.

y = y0 + v0t +12

at2;v = v0 +at

⇒ y =12

at2;v = at

(a) t = 1 s:y =−4.9 m,v =−9.8 m/s

(b) t = 2 s:y =−19.6 m,v−19.6 m/s

(c) t = 3 s:y =−44.1 m,v =−29.4 m/s

44. Assume up is positive. The first two equations are used for a and b; the second pair of equations are used forc and d.

y = y0 + v0t +12

at2

⇒ y = v0t +12

at2

v2 = v20 +2a∆y,v0 = 15

ms

⇒ v =√

v20 +2gy, where g =−9.8

ms2

(a) t = 2 sy = 10.4 m

(b) t = 4 sy =−18.4 m

(c) y = 2 m⇒ v = 13.6 m/s

(d) y = 5 m⇒ v = 11.3 m/s

45. Assume up is positive and the maximum height v f inal,y = 0.

(a)

v2 = v20 +2g∆y,v = 0

⇒ ∆y =−v2

02g

= 29.4 m

(b)

∆y = v0t +12

gt2

Using the quadratic formula:

t =−v0±

√v2

0−4(12 g)(0)

2 12 g

t =−v0± v0

g

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The time when the ball returns must be the nonzero solution:

t =−2v0

g⇒ t = 4.90 s

46. Write the height as a function of time:

t1 = 0.8 s, t2 = 2.8 s

h = v0t1 +12

at21 on the way up

h = v0t2 +12

at22 on the way down

v0t2 +12

at22 = v0t1 +

12

at21

v0 =−12

g(

t21 − t2

2t2− t1

)=

g2(t1 + t2) = 17.64 m/s

h = v0t1 + 12 at2

1 = 10.96 m47. Assume up is positive.

ytool = y0 +12

att2,y f loor =12

a f t2

y0 +12

att2 =12

a f t2

⇒ t =

√2y0

(a f +g)= 0.40 s

48. Position:

y =12

at2

⇒ y(t = 0.05) =12(−9.8)(0.05)2 =−0.01 m

y(t = 0.1) =12(−9.8)(0.1)2 =−0.05 m

Velocity:

v = v0 +at = at

⇒ v(t = 0.05) = (−9.8)(0.05) =−0.5ms

v(t = 0.1) = (−9.8)(0.1) =−0.98ms

49. Assume up is positive.

y = y0 + v0t +12

at2

⇒ y = v0t +12

at2

(a) At the peak, v f = 0 = v0 +gt.

v0 =−gt = (−9.8 m/s2)2.0 s = 19.6 m/s

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(b)

v2 = v20 +2g∆y

⇒ ∆y =−v2

02g

= 19.6 m

50. Assume down is positive.

12

gt2 + v0t−∆y = 0

12(−9.8)t2 +(−5.0)t− (−20) = 0

t =5.0±

√(−5.0)2−4(−4.9)(20)−9.8

t = 1.57 s,−2.63 s

51. 220 mph = 98.35 m/s

(a)

v2 = v20 +2g∆y

→ ∆y =−v2

02g

= 493 m

(b)

v = v0 +at

t =v0

a= 10 s to reach max height

Double for round trip:⇒ Ttotal = 2t = 20 s

52. Yes, it is possible to have negative instantaneous velocity, but positive average velocity. Remember thataverage velocity can be calculated by vavg =

∆x∆t . Thus, all that matters in calculating average velocity are the

endpoints of the voyage and the time it took in moving from start to finish. If at any point during travel theinstantaneous velocity was negative, it would have the effect of lowering the average velocity, which wouldremain positive.

53. Yes, it is possible for instantaneous velocity to be greater than average velocity. Imagine an object whichtravels from point A to point B, at some times traveling very fast and at other times traveling very slow.On average, its velocity would lie somewhere between these extremes. Conversely, it is NOT possible foran object’s average velocity to exceed instantaneous velocity. The average velocity is constrained by themaximum and minimum instantaneous velocities within the time period of interest or, written mathematically,vmin ≤ vavg ≤ vmax.

54. Yes. A nonzero velocity means that the object is moving. If the acceleration equals zero, it simply impliesthat the object is neither speeding up nor slowing down.

55. Consider a ball being thrown vertically upward. The v vs. t graph below describes its velocity during a portionof its ascent. Velocity is positive throughout the interval because the ball is moving. Meanwhile, accelerationis negative since the ball slows down (decelerates) as it nears its maximum height, due to the effect of gravity.

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56. Upon being released from your hand, both objects will experience the same acceleration due to gravity:g/approx−9.8 m

s2 .57. x = vt = (30 m/s)(0.45 s) = 13.5 m58. vavg =

∆x∆t =

(50 m+45 m)50 m

.80 m/s+45 m

.90 m/s= 95 m

62.5 s+50 s = 0.84 m/s

59. Let the distance to the store be defined as x:vavg =

x+xx

45 mph+x

60 mph= 2x

105x2700

= 51.4 mph

60. The distance between the walls is shrinking at a rate of 5 m/s. If the walls are 15 m apart, the total time beforethe walls touch is 3.0 s. Using the robot’s velocity, he travels a total distance of (10 m/s)(3.0 s) = 30.0 m.

61. Looking at the velocity function, the particle stops when the velocity is 0. Thus:

0 = 2−2t

⇒ t = 1 s

From the position function, we have:

x = 2(1)− (1)2

→ x = 1 m

62. aavg =∆v∆t =

v2−v1t2−t1

=(−2.2 m

s )−(1ms )

3.0 s =−1.07 ms2

63. Since the problem mentions constant acceleration, you can use the kinematic equations.

v2f = v2

0 +2a∆x

⇒ a =v2

f − v20

2∆x=

(75 mph)2− (25 mph)2

2(2.50 miles)= 1000 miles/h2

64. (a) Solve for the time in two phases:1. When speeding up to 30 m/s:

v = v0 +at

→ t =va=

30 m/s3.3 m/s2 = 9.09 s

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2. When slowing to 0 m/s:

v = v0 +at

→ t =−v0

a=−30 m/s−1.5 m/s2 = 20.0 s

⇒ ttotal = 20.0+9.09 = 29.09 s

(b) Solve for the distance in two phases:1. When speeding up to 30 m/s:

v2f = v2

0 +2a∆x

→ ∆x =v2

f − v20

2a=

(302−0)2(3.3)

= 136.36 m

2. When slowing to 0 m/s:

v2f = v2

0 +2a∆x

→ ∆x =v2

f − v20

2a=

(0−302)

2(−3.3)= 136.36 m

xtotal = 272.7 m

65. (a) v = v0 +atSince the car is being brought to a stop, v = 0.t = −v0

a = −15.5 m/s−5.13 m/s2 = 3.02 s

(b)

t = 3.02 s

x = v0t +12

at2

x = (15.5 m/s)(3.02 s)+12(−5.13 m/s2)(3.02)2

x = (46.81−23.39)m = 23.42 m

66. The best way to approach this is to calculate how much distance is needed to slow down the line from themaximum speed of 3.2 m/s.

v2 = v20 +2a∆x

∆x =−v2

02a

=− (3.2)2

2(−3.2)= 1.6 m

Distance traveled before having to slow down:x = 230 m−∆x = 228.4 m

67.

90 mph = 40.23ms

55 mph = 24.59ms

v = v0 +at

→ t =v− v0

a=

(24.59−40.23) ms

−4.7 ms2

= 3.33 s

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68. (a) Assuming that the acceleration is constant: 60 mph = 26.8 ms

x = v0t +12

at2

a =2(x− v0t)

t2 =2(22.5 m− (26.8 m

s )(1.0 s))(1.0 s)2 =−8.6 m/s2

(b)

v = v0 +at

⇒ v = 26.8 m/s−8.6 m/s(1 s) = 18.2 m/s

69. The equation gives you the time and height:

y = v0t +12

at2

⇒ v0 =yt− 1

2at = v0 =

0.41 m0.32 s

− 12(−9.8

ms

2)(0.32)

v0 = 2.85ms

70. (a)

v2 = v20 +2a∆y

v0 =√−2a∆y =

√−2(−9.8 m/s2)(25 m) = 22.1

ms

(b)

v = v0 +at

→ t =−v0

a=

22.1 m/s−9.8 m/s2 = 2.26 s

Total time is then:

2t = 4.52 s

71. (a)

v2 = v20 +2a∆y

→ ∆y =v2− v2

02g

=(30 m

s −0)2

2(−9.8 ms2 )

= 45.92 m

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(b)

72. v2 = v20 +2a∆y =

√(15

ms)2 +2(−9.8

ms)(−25 m) = 26.74 m/s

73. (a) Total time:

tT =

√2∆y

a=

√2(−100)(−9.8)

= 4.52 s

(b) The last 15% of the fall would be a distance of 15 m.Time for the screw to fall through the first 85%:

t85% =

√2∆y

a=

√2(−85)(−9.8)

= 4.16 s

Therefore the time to fall through the last 15% is:tT − t85% = 0.35 s

74. To determine your teacher’s speed, you need to determine the relationship between the time it takes the

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snowball to drop and the distance your teacher is from the building.Time it takes the snowball to drop:

t =

√2∆y

a=

√2(h′−h)

aKnowing that the snowball makes contact with your teacher’s head in this time, you can calculate his speed:

v0 =xt = x

√a

2(h′−h)75. To determine the average acceleration, you need to look at the bottle’s velocity before hitting the ground and

after hitting the ground.

v0 =−√

2(−9.8 m/s2)(−1.5 m) =−5.42 m/s

v f =√

2(−9.8 m/s2)(−0.5 m) = 3.13 m/s

aavg =∆vt

=3.13− (−5.42)

0.01= 855 m/s2

76. (a)

x f = xi +12(vi + v f )t

→ t =2∆x

vi + v f= 4.28×10−6 s

(b)

x f = xi +12(vi + v f )t

→ t =2∆x

vi + v f= 4.28×10−6 s

t ′ = 8.56×10−6 s

Using the information found above, determine the acceleration:

a =v2− v2

02∆x

=(5 km

s )2− (2 kms )2

2(.015)= 7.00×108 m

s2

Therefore: ∆x = v0t + 12 at2 = .0428 m

Thus, the particle travels 0.28 cm further.77. Yes, it is possible. This could be accomplished by the particle traveling in one direction and then turning

around and going back in the opposite direction.78. (a) t = 2 min = 120 s

x = vt = 4.0 m/s(120 s) = 480 m(b) x = 2v(t) = (8.0 m/s)(120 s) = 960 m

xtotal = 1440 m79. The average displacement is zero.80. If the acceleration is zero, then the velocity of the object is constant.81. Yes, it is possible for an object’s velocity and acceleration to have opposite signs. This implies that the object’s

speed is decreasing.82. The kinematic equations can be used for any object that is undergoing constant acceleration, regardless of

other environmental factors.

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2.2. References www.ck12.org

2.2 References

1. Laura Guerin. CK-12 Foundation .2. Laura Guerin. CK-12 Foundation .3. Laura Guerin. CK-12 Foundation .4. Laura Guerin. CK-12 Foundation .

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www.ck12.org Chapter 3. Two-Dimensional Motion Problem Sets

CHAPTER 3 Two-Dimensional MotionProblem Sets

Chapter Outline3.1 TWO-DIMENSIONAL MOTION PROBLEM SETS

3.2 REFERENCES

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3.1. Two-Dimensional Motion Problem Sets www.ck12.org

3.1 Two-Dimensional Motion Problem Sets

Solutions

1. Define the direction east as +x and north as +y. Thus, ~A = 5x and ~B = 8y.Draw the vector diagram for ~A+~B :

~A+~B = 5x+8yMagnitude: |~A+~B|=

√52 +82 = 9.4

Direction: θ = tan−1(8

5

)= 58 north of east

2.

C2 = A2 +B2

C =√

(2)2 +(5)2 = 5.4 m

θ = tan−1(

52

)= 68 south of east

3.√

(7−0)2 +(3−1)2 =√

49+4 =√

53

4.√(3−8)2 +(25)2 =

√(−5)2 +(25)2 =

√25+625 =

√650 = 5

√26

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5.

~A = 10 m, ~B = 12 m at θ = 60 south of east~Cx = ~Ax +~Bx

~Cx = 10+12cos60 = 16~Cy = 12sin60 = 10.4

C =√

Cx2 +Cy

2 = 19.1 m

θ = tan−1(

Cy

Cx

)= 33

6. Define the position vector as:~r =~r0 +~v0t + 1

2~at2 = (3.00t)x+ 12(2.50t2)y

Define the velocity vector as:~v =~v0 +at = (3.00)x+(2.50t)y

7. Determine how far the boy walks in the x-direction:x = (2 m/s)(5 min)

( 60 s1 min

)= 600 m

Determine how far the boy walks in the y-direction:y = (3 m/s)(2 min)

( 60 s1 min

)= 360 m

Displacement:√

x2 + y2 = 700 mAverage speed: total distance

total time = 600 m + 360 m420 s = 2.29 m/s

8. Find the car’s total displacement in both the north-south and east-west directions:

x1 = (50 mph)(40 min)(

1 h60 min

)= 33.3 miles east

x2x = (50cos20 mph)(20 min)(

1 h60 min

)= 15.7 miles east

x2y = (50sin20 mph)(20 min)(

1 h60 min

)= 5.70 miles north

x3 = (50 mph)(30 min)(

1 h60 min

)= 25 miles east

Find the car’s average velocity:

vavg =total displacement

total time

vavg =74.0 miles

90 minx+

5.70 miles90 min

y

|vavg|= 0.82 miles/min = 49.5 mph

9. Define down as positive.

(a)

y =12

at2

t =

√2ya

=

√2(0.02)

9.81= 0.06 s

(b)

v =xt= x

√a2y

= 30

√9.81

2(0.02)= 470.0 m/s

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10.

x = v0xt +12

axt2

Since ax = 0, the equation can be simplified to:

x = v0xt = (v0 cosθ)t = (25cos45)t

(a) x(t = 1.0 s) = 17.7 m(b) x(t = 3.0 s) = 53.0 m

11. Determine how long it takes for the bullet to reach the target:

vx =xt

t =xvx

=50500

= 0.1 s

Then, solve for the vertical displacement between the fired shot and the bullseye:

y =12

ayt2 =12(9.8)(0.1)2 = 0.049 m

Therefore, the barrel should be pointed 0.049 m above the bullseye.12.

v2y = v2

0y+2a∆y

y =v2

y− v20y

2a

At a projectile’s maximum height, vy = 0.

⇒ y =0− v2

0y

2a=−v2

0y

2(−g)=

(v0 sinθ)2

2g

13.

v0x =xt=

405

= 8 m/s

v0y =y+(1

2

)|a|t2

t=−3.0+

(12

)(9.8)(5)2

5= 23.9 m/s

Magnitude: |v0|= 25.2 m/sDirection: θ = tan−1

(23.98

)= 71.5

14. Define down as positive. Note that the package’s initial velocity is equivalent to the velocity of the plane.

(a)

v0y = v0 sinθ =y− 1

2 at2

t

v0 =y− 1

2 at2

t sinθ=

700− 12(9.8)(6)

2

6sin30= 175 m/s

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(b)

v0x = v0 cosθ =xt

x = (v0 cosθ)t = 907 m

(c)

vx = v0x +at = v0 cosθ = 175cos30 = 151 m/s

vy = v0y +at = v0 sinθ+gt = 175sin30+(9.8)(6) = 146 m/s

15. Assume at the start of the ball’s trajectory x0 = 0 and y0 = 0. At the end of the trajectory, x = 107 m and y =−1.5 m.Determine how long it takes for the ball to reach the fence in the x-direction:

v0x = v0 cosθ =xt

t =x

v0 cosθ

Now, consider the y-direction:

y = (v0 sinθ)t +12

at2

Substitute the expression for time derived above and solve for v0:

y = (v0 sinθ)

(x

v0 cosθ

)+

12

a(

xv0 cosθ

)2

y = x tanθ+ax2

2v20 cos2 θ

y− x tanθ =ax2

2v20 cos2 θ

v20 =

ax2

2(y− x tanθ)cos2 θ

v0 =

√ax2

2(y− x tanθ)cos2 θ

=

√(−9.8)(107)2

2(−1.5−107tan45)cos2 45= 32.16 m/s

Determine the height of the ball at the fence (the top of the fence has x = 97 m and y = 7−1.5 = 5.5 m):

y = x tanθ+ax2

2v20 cos2 θ

= 97tan45+−9.8(97)2

2(32.16)2 cos2 45= 7.8 m

The ball will pass 7.8−5.5 = 2.3 m above the fence.16. Define up as positive.

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(a) The cops leaps with v0x = 5 m/s and v0y = 0.Determine how long it takes the cops to drop 2 m in the y-direction:

y =12

at2

t =

√2ya

=

√2(−2)−9.8

= 0.639 s

Determine how far the cops travel in the x-direction during that time:

x = v0xt = (5 m/s)(0.639 s) = 3.2 m < 4 m

No, the cops do not clear the gap.(b) The caped crusader leaps with v0 = 5 m/s at an angle of θ = 45.

Determine how long it takes the caped crusader to drop 2 m in the y-direction, using the quadraticequation to solve for t:

y = v0yt +12

at2

−2 = (5sin45)t− 12

9.8t2

0 = 2+5√2

t−4.9t2

t =− 5√

2±√( 5√

2)2−4(−4.9)(2)

−9.8=−0.373,1.094

Determine how far the caped crusader travels in the x-direction during that time:

x = v0xt = (5cos45)(1.094 s) = 3.87 m < 4 m

No, the caped crusader does not clear the gap either. He misses by 4−3.87 = 0.13 m.

17. Define up as positive.

(a)

y = v0yt +12

at2

0 = v0yt +12

at2

t =−2v0y

a=−2(20.0sin50)

−9.8= 3.13 s

(b)

x = v0xt = (20cos50)(3.13) = 40.2 m


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(a) Find the x-component of the initial velocity:

v0x =xt=

12 m0.80 s

= 15 m/s

Find the y-component of the initial velocity:

y = v0yt +12

at2

v0y =y− 1

2 at2

t=

3.2− 12(−9.8)(0.8)2

0.8= 7.92 m/s

Put the components together to find the initial speed and direction of the ball:

v0 =√

v20x+ v2

0y= 17.0 m/s

θ = tan−1(

v0y

v0x

)= 27.8

(b) At the ball’s maximum height, vy = 0.

vy = v0y +at

t =v0y

|a|=

7.929.8

= 0.81 s

19. Define up as positive.Find how long the package is in the air:

y = v0yt +12

at2

−100 = (35sin30)t +12(−9.8)t2

0 =−4.9t2 +17.5t +100

t =−3.07,6.64

Find how far the package travels in the x-direction during that time to determine where it lands:

x = v0xt = (35cos30)(6.64) = 201 m

20. Define up as positive.To find out when the mountain lion reaches its maximum height, use the fact that at the trajectory’s peak vy = 0:

vy = v0y +at = 0

t =−v0y

a=

v0y

g

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Substitute t into the equation below to find v0y :

y = v0yt +12

at2

y = v0y

(v0y

g

)− 1

2(g)(

v0y

g

)2

y =v2

0y

2g

v0y =√

2gy =√

2(9.8)(4) = 8.85 m/s

Derive v0 from v0y :

v0 =v0y

sin45= 12.5 m/s

21. Start by using the range equation:

R =v2

0 sin2θ

g

To achieve the maximum range, θ = 45:

g =v2

0 sin90

R=

v20

R=

52

15= 1.66 m/s2

22. Use the range equation to find the angle needed:

R =v2 sin2θ

g

3000 =(1500)2 sin2θ

9.8sin2θ = 0.0131

2θ = 0.749

θ = 0.374


(a) x = v0xt = (7.5 m/s)(cos30)(4 s) = 26.0 m(b) y = v0yt +

12 at2 = (7.5 m/s)(sin30)(4 s)+ 1

2(9.8 m/s2)(16 s) = 63.4 m(c)

y = 8 m

y = v0yt +12

at2

0 =12

at2 +(vsinθ)t− y

t =−7.5sin30±

√(7.5sin30)2−4(4.9)(−8)

9.8t = 0.95 s

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24. Define up as positive.When the rock is kicked, no velocity is imparted to the vertical component. Thus, the rock is under free fall.

y =12

at2

t =

√2ya

=

√2(−50)−9.8

= 3.19 s

The remainder of the time is the time it takes for the sound to travel back.

ts = 3.5−3.19 = 0.31 s

The sound waves cover the distance:

(343 m/s)(0.31 s) = 106.33 m

Therefore, the horizontal distance traveled by the rock is:

x =√

106.332−502 = 93.8 m

From that, the initial velocity of the rock can be calculated:

v0x =xt=

93.8 m3.19 s

= 29.4 m/s

25. Assume the starting point is the origin.

v2fy= v2

0y+2a∆y

⇒ v0y =√

2gh

At 13 of the ball’s maximum height:

v2y( 1

3 h) = v20y+2a

(13

y f − yi

)= 2gh−2|a|

(13

h− yi

)⇒ v0y =

√(13

)gh

At the ball’s maximum height:

vx =12

√v2

x + v2y( 1

3 )h=

12

√v2

x +

(13

)gh

⇒ vx =

√gh9

θ = tan−1 v0y

vx= tan−1

√2gh√

gh9

= 86.8

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26. Define up as positive.Use vector notation to view the displacement from the ball to the hoop:5 m at 60 = (5cosθ)x+(5sinθ)yIf the ball drops 0.10 m, then the y-displacement is:5sinθ+0.10 = 4.43yThus, the new displacement is:2.5x+4.42y = 5.09 at 60.6

The time it takes to travel 5 m in the x-direction:

∆x = v0xt +12

at2

⇒ t =∆xv0x

=5cos60

v0 cos60.6

The drop in vertical motion is described by:

/∆y = v0yt +12

at2⇒ ∆y =12

at2

∆y =12

a(

5cos60

v0 cos60.6

)2

v0 =5cos60

cos60.6

√a

2∆y=

5cos60

cos60.6

√−9.8

2(−0.10)= 35.6 m/s

27. Use the range equation:

R =v2

0 sin2θ

g

R =(10 m/s)2

9.8 m/s2 sin114 = 9.32 m < 15 m

No, competitor #2 falls short.28. Break the ball’s velocity into components:

vx = v0 cosθ2vy = v0 sinθ2Solve for t by looking at the motion in the x-direction along the incline:

∆x = vxot

d cosθ1 = v0 cosθ2t

⇒ t =d cosθ1

v0 cosθ2

Next, consider motion in the y-direction, substituting in the expression for t derived above:

∆y = v0yt +12

at2

d sinθ1 = v0 sinθ2t +12

at2

d sinθ1 = v0 sinθ2

(d cosθ1

v0 cosθ2

)+

12

a(

d cosθ1

v0 cosθ2

)2

d sinθ1 = d tanθ2 cosθ1 +12

a(

d cosθ1

v0 cosθ2

)2

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Divide by d to simplify:

sinθ1 = tanθ2 cosθ1 +12

ad cos2 θ1

v20 cos2 θ2

Solve for d:

d =2v2

0 cos2 θ2 sinθ1

acos2 θ1−

2v20 cos2 θ2

(sinθ2cosθ2

)cosθ1

acos2 θ1

d =2v2

0 cos2 θ2 sinθ1

acos2 θ1−

2v20 cosθ2 sinθ2 cosθ1

acos2 θ1

d =2v2

0 cosθ2 (cosθ2 sinθ1− sinθ2 cosθ1)

acos2 θ1=

2v20 cosθ2 sin(θ1−θ2)

acos2 θ1

29. Define up as positive.It’s easy to first consider part b since we know how far away the wall is.b.

x = v0xt = (v0 cosθ)t

⇒ v0 =x

t cosθ

∆y = v0yt +12

at2 =( x

t cosθ

)(sinθ)t +

12

at2 = x tanθ+12

at2

⇒ t =

√2∆y−2x tanθ

a=

√2(15 m)−2(100 m) tan40

−9.8 m/s2

t = 4.22 s

a.

x = v0xt = (v0 cosθ)t

v0 =x

(cosθ)t=

100 m(cos40)(4.22 s)

= 30.9 m/s


(a)

∆y = v0yt +12

at2

0 =12

at2 +(v0 sinθ)t−∆y

t =−v0 sinθ±

√(v0 sinθ)2 +2a∆ya

t =−20sin35±

√(20sin35)2 +2(−9.8)(−50)−9.8

t =−2.2 s,4.6 s

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(b)

v fy = v0y +at = 20sin35+(−9.8)(4.6) =−33.6 m/s

v =√

v2x + v2

y =√(20cos35)2 +(−33.6)2 = 37 m/s

31. (a)

vavgx =∆x∆t

=6.4−1.0

5.0= 1.08 m/s

vavgy =∆y∆t

=1.5−3.9

5.0=−0.48 m/s

(b)

|v|=√

v2x + v2

y = 1.18 m/s

θ = tan−1(

vy

vx

)= 24

32. (a)

v =xt

x = vxt = (2.0 m/s)(6.4 s) = 12.8 m

y = vyt = (−2.0 m/s)(6.4 s) =−12.8 m

r =√

x2 + y2 =√(12.8)2 +(−12.8)2 = 18.1 m

After 6.4 s, the mouse is located at (12.8, -12.8), which is 18.1 m from the origin.(b)

v =xt

x = vxt = (−2.0 m/s)(6.4 s) =−12.8 m

y = vyt = (2.0 m/s)(6.4 s) = 12.8 m

r =√

x2 + y2 =√(−12.8)2 +(12.8)2 = 18.1 m

After 6.4 s, the mouse is located at (-12.8, 12.8), which is 18.1 m from the origin.33.

~vavg =∆r∆t

=[0.03+2.0(2.0)2]− [0.03+0]

2x+

[−0.5(2.0)]− [0]2

y = 4.0x−0.5y

|v|=√

(vx)2 +(vy)2 = 4.03 m/s

θ = tan−1(

vy

vx

)=−7.13

34.

aavgx =∆vx

∆t=

(8.0−0) m/s105 s

= 0.076 m/s2

aavgy =∆vy

∆t=

(5.5−0) m/s105 s

= 0.052 m/s2


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(a)

∆y = v0yt +12

at2

∆y =12

at2 =12(9.8 m/s2)(0.40 s)2 = 0.78 m

(b)

x = v0xt = (2.5 m/s)(0.40 s) = 1 m

(c)

36. Define down as positive.Determine how long it takes for the roses to hit the ground:

y =12

at2

t =

√2ya

=

√2(200 m)

(9.8 m/s2)= 196 s

∆x = v0xt = (15.0 m/s)(196 s) = 2940 m

37. Use the time it takes Ivan to hit the water to determine how far Carlos travels:

x = vt = (2.5 m/s)(4.0 s) = 10 m

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38. Define down as positive.It will take the backpackers a time t to fall 8.0 m.

t =

√2ya

=

√2(8.0 m)

(9.8 m/s2)= 1.28 s

Therefore, the velocity they need to clear the ledge is:

v =xt=

2.3 m1.28 s

= 1.80 m/s


(a) Determine the time it will take for your car to drop 19.0 m:

t =

√2ya

=

√2(19.0 m)

(9.8 m/s2)= 1.96 s

Therefore, the velocity you need to clear the river is:

v =xt=

50 m1.96 s

= 25.5 m/s

(b)

vx = 25.5 m/s

vy =√

2a∆y = 19.3 m/s

v =√

v2x + v2

y = 32.0 m/s

40. Determine how long it takes the package to drop the vertical distance:

t =

√2ya

= 1.43 s

Thus, the plane should be flying at a speed of:

v =xt=

5.0 m1.43 s

= 3.5 m/s

41. x = vtSolve for t from the vertical displacement:

t ′ =

√2ya

If a is 14 of what it is on Earth, time, and therefore distance traveled, would be doubled.


(a)

v2y = v2

0y+2a∆y

∆y =v2

y− v20y

2a=

0−15.22 m/s2(−9.8 m/s2)

= 11.8 m

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(b)

y =12

at2

t =

√2ya

=2(11.8 m)

(9.8 m/s2)= 15.5 s


(a)

v0x = v0 cosθ = (60.0 m/s)cos65 = 25.4 m/s

v0y = v0 sinθ = (60.0 m/s)sin65 = 54.4 m/s

(b) At maximum height, vy = 0.

vy = v0y +at

t =0− v0y

a=−54.4 m/s−9.8 m/s2 = 5.55 s

(c) y = v0yt +12 at2 = (54.4 m/s)(5.55 s)+ 1

2(−9.8 m/s2)(5.55 s)2 = 151 m(d) vx = v0x = 25.4 m/s

vy = 0ax = 0ay =−9.8 m/s2

44. Define up as positive.At maximum height, vy = 0.

v2y = v2

0y+2a∆y

0 = (v0 sinθ)2 +2a∆y

∆y =−(v0 sinθ)2

2a=−(40sin50)2

2(−9.8)= 48 m

Find the time it takes to reach the maximum height:

vy = v0y +at

t =−v0y

a=−v0 sinθ

a=−40sin50

−9.8= 3.127 s

Doubling t would represent the total time the bolt is in the air:

ttotal = 2t = 6.25 s

The total distance traveled is then:

x = (v0 cosθ)ttotal = (40.0cos50)(6.25 s) = 161 m

45.

v0x = v0 cosθ = 24cos37.8 = 18.96 m/s

vx =xt⇒ t =

xvx

=x

v0x

=20

18.96= 1.05 s

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46. (a)

vx = v0 cosθ = 25cos25 = 22.7 m/s

vy =√

v20y+2a∆y =

√(25sin25)2 +2(−9.8)(−30) = 26.5 m/s

v =√

v2x + v2

y = 35 m/s

(b) Find the time it takes for the baseball to hit the ground:∆y = v0 sinθt + 1

2 at2

Use the quadratic equation to solve for t:t = 1.62 sThe total horizontal distance traveled is then: x = vxt = (v0 cosθ)t = (22.7 m/s)(1.62 s) = 36.7 m

47.

x = v0xt⇒ t =x

v0 cosθ

y = (v0 sinθ)t +12

at2

Combine the two equations:

y =(v0 sinθ)x

v0 cosθ+

12

a(

xv0 cosθ

)2

= x tanθ+ax2

2v20 cos2 θ

⇒ v0 =

√ax2

2cos2 θ(y− x tanθ)=

√(−9.8 m/s2)(75.0 m)2

2cos2 45(30.0 m−75.0tan45)= 5.2 m/s

48. Since your speed is slower than the rate at which you are being pushed back, you will never reach the otherend.

49.

v2 = 2a∆x = 2(1.0 m/s2)(150.0 m)

v = 17.32 m/s

v2y = v2

0y+2ag∆y

∆y =0− v2

0y

2ag=−(17.32 m/s)2

2(−9.8 m/s2)= 15.3 m

50. Determine how long it takes for the rope to reach the water:y = v0yt +

12 at2

Use the quadratic equation to solve for t:t = 3.26 sAt this time, the rope will land a distance xfrom the cliff:x = v0 cosθt = (16.0 m/s)(cos50)(3.26 s) = 33.53 mIn the time determined above, the boat will travel a distance x′:x′ = (0.6 m/s)(3.26 s) = 1.96 mTherefore, the rope should be thrown when the boat is 35.5 m from the cliff.

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51. At the minimum velocity, y = 2d and x = 6d.

x = (v0 cosθ)t

6d = (v0 cosθ)t

⇒ t =6d√

2v0

y = (v0 sinθ)t +12

at2

Substitute in the expression for t derived above and solve for v0:

v0 = 3√

gd

52. Determine the time it will take to hit the professor’s head (∆y = 15.0 m):∆y = (v0 sinθ)t + 1

2 at2

Use the quadratic equation to solve for t:t = 1.37 s

(a) Determine the distance traveled in the x-direction:

v0 cosθ =xt

x = (v0 cosθ)t = (6.5cos40)(1.37) = 6.82 m

(b) No, the balloon misses.

53.

R = (v0 cosθ)

(2v0 sinθ

g

)v0 =

√gR

2sinθcosθ=

√(9.8 m/s2)(5.68 m)

2sin(21)cos(21)= 9.1 m/s

54.

R = (v0 cosθ)

(2v0 sinθ

g

)gRv2

0= 2sinθcosθ = sin2θ

θ =12

sin−1(

gRv2

0

)=

12

sin−1((9.8 m/s2)(6.01 m)

(8.17 m/s)2

)= 31

55. sin2(2θ) = sin2(180−2θ)56. The velocity of the ball when it strikes the ground can be determined from one of the kinematic equa-

tions: v2 = v20 + 2a∆y. The ball’s final velocity therefore can be expressed as v =

√v2

0 +2a∆y, which isclearly independent of θ.

57. You would not be able to determine the object’s instantaneous velocity. The given points may not be closeenough together to give you an accurate measurement.

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58.

59.60. To calculate the highest point in the trajectory of a thrown object, the velocity first has to be broken into

components and then the time it takes to reach the peak position can be calculated. Once the time is known,the position of the object at that time can be determined. (This is one method to calculate the position of theobject’s highest point.)

61. The path of the object would follow a parabolic shape.62. Acceleration and horizontal velocity remain constant during parabolic motion.63. A passenger outside of the train would see the ball following a parabolic path, while a passenger inside the

train would see the ball travel a linear path.64. Use the range equation for changes in elevation:

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65.

R =v0 cosθ

g

(vsinθ+

√v2 sin2

θ+2gy)

Rgcosθ

= v2 sinθ+

√v2 sin2

θ+2gy(Rg

cosθ

)2

= v4 sin2θ+ v2 sin2

θ+2gy

0 = v4 sin2θ+ v2 sin2

θ+[2gy−(

Rgcosθ

)2

]

Use the quadratic equation to solve for v:

v2 = 994.3 m/s

v = 31.5 m/s

v f = v0 +at

v fx = 5.0+2.0 t

v fy =−10.0

66.

~v = (5.0−2.0(2.0))i−8 j = 1.0i−8 j

|v|=√

1.02 +(−8.0)2 = 8.1 m/s

67. The velocity and acceleration vectors will be perpendicular at the highest point of the projectile’s parabolicpath.

68. The velocity vector and the acceleration vector are never parallel during projectile motion.69. Throwing the ball straight upward (θ = 90) will achieve the maximum height possible.70. The given angle pairs have the same displacement from 45°. Therefore, each angle would give the same range

when a particle is fired in projectile motion.71. Define up as positive.

y = v0yt +12

at2

−50 = (15sin35)t +12(−9.8)t2

Use the quadratic equation to solve for t:

t = 4.19 s

72. Define up as positive.First, determine the time it takes the object to hit the ground:

y = v0yt +12

at2

−30 = (25sin40)t +12(−9.8)t2

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t = 4.6 s

Plug in t to find the magnitude of the final velocity:

|v|=√

v2x + v2

y

=√

v20x+(v0y +at)2

=

√(v0 cosθ)2 +[v0 sinθ+at]2

=

√(25cos40)2 +[25sin40+(−9.8)(4.6)]2

= 35 m/s


y =12

at2

t =

√2ya

=

√2(−400 m)

(−9.8 m/s2)= 9.04 s

74.

~vavg =∆r∆t

=[(1.0(3.0)+1.0)− (1.0(1.0)+1.0)]i+[(0.5(3.0)2 +0.2)− (0.5(1.0)2 +0.2)] j

2.0

=(2i+4 j)

2.0= 1.0i+2.0 j

75. aavg =∆v∆t =

42−2510 = 1.7 m/s2

76. θ = tan−1(

ayax

)= tan−1

(0.50.7

)= 35.5

77.

(a) Start by using the range equation:

R =v2 sin2θ

g

To achieve the maximum range, θ = 45:

g =v2

R=

(8 m/s)2

25 m= 2.56 m/s2

(b) R = v2

g = (8 m/s)2

9.8 m/s2 = 6.53 m

78. x = vxt = (5.0 m/s)(4.0 s) = 20.0 m79. x = vxt = (8.0 m/s)(cos35)(4.0 s) = 26.2 m

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80. (a) y = 12 at2 = 1

2(9.8 m/s2)(4.0 s)2 = 78.4 m

(b) Use the quadratic equation to find the length of time t for which Meta’s pebble is in the air:

y = v0yt +12

at2

78.4 = (5cos35)t +12(9.8)t2

t = 4.3 s

Therefore, Meta’s pebble is in the air for 0.3 seconds longer than Aliya’s.

81. ~r = (3 mi)x+(5 mi)y82. ~A−2~B = (5−2)i+(4−4) j = 3i+0 j = 3i83.

2~B = 2i−3 j

~B = 1i− 32

j

~A+~B = 4 j

~A = 4 j−~B =−1i+(

4+32

)j =−1i+

112

j

84. False; this would only be true if both vectors pointed in the same direction, which isn’t always the case.85. Since the wind is blowing east, the plane must be directed west of south.

θ = sin−1(~vwind~vplane

)= sin−1 ( 60

300

)= 11.5 west of south

86. ~v(t) = d~r(t)dt = 3.0i− (3)(4.8t2) j = 3.0i− (14.4t2) j

87.

~v(t) =d~r(t)

dt= (3.0−2t)i− (10.0t) j

~a(t) =d~v(t)

dt=−2i−10.0 j


y = v0t +12

at2

10.0 = 16.3t +12(−9.8)t2

0 =−4.9t2 +16.3t−10.0


t = 0.81 s,2.5 s


v = v0 +at

At the ball’s maximum height, v = 0:

0 = v0 +at

t =−v0

a=− 8.3

(−9.8)= 0.85 s

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90.

Ax = Acosθ = (5 m)cos50 = 3.2 m

Ay = Asinθ = (5 m)sin50 = 3.8 m

91. ~B =−2.4~A92. |v|=

√v2

x + v2y =

√(1.1 m/s)2 +(0 m/s+(2.0 m/s2)(3.0 s))2 = 6.1 m/s

93. |v|=√

v2x + v2

y =√(0 m/s)2 +(0 m/s+(2.0 m/s2)(3.0 s))2 = 6.0 m/s

94. True; because both objects have the same downward acceleration, they will strike the ground at the same time.95. The magnitude of the resultant vector can only be written as A+B if the two vectors, ~Aand~B, are parallel and

point in the same direction.96. Let ~A = Axx+Ayy.

If the magnitude of the vector is A, then its components in x and y would be given by:Ax = Acosθ and Ay = Asinθ

In the case of a two-dimensional vector, 0 < θ < 90, which means both sinθ and cosθ are less than 1.Therefore, Ax = Acosθ < A and Ay = Asinθ < A.

97. Let ~A = Axx+0y, where Ax is any nonzero number.

|A|=√

A2x +02 = |Ax|> 0

Therefore, as long as Ax is nonzero,then the magnitude will always be positive.98.

Ax = 2cos20, Ay = 2sin20

Bx = 3cos35, By = 3sin35

~A+~B = (2cos20+3cos35)x+(2sin20+3sin35)y

= 4.33x+2.40y

|~A+~B|= 5.0

99. No, by definition a vector has both a direction and a magnitude.100. For simplicity, let’s look at two vectors, ~A and ~B, with only x-components. Let’s say A = 2i and therefore B =

−2i.

~A+~B = (2−2)i = 0i

|~A+~B|= 0

101.

Ax = 5cos20, Ay = 5sin20

Bx = 3cos65, By = 3sin65

(a)

~A+~B = (5cos20+3cos65)x+(5sin20+3sin65)y

= 5.96x+4.43y

|~A+~B|= 7.43

(b)

θ = tan−1(

4.435.96

)= 36.6

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102. (a)

~A = 4y~B =−4x

~A+~B =−4x+4y

|~A+~B|= 5.66

(b)103.

tan35 =h

70.0 mh = 49.0 m

104. The two vectors are perpendicular to one another.105. There is no method that can perform such an operation. Vectors and scalars cannot be added.106.

~A = 3cos20x+3sin20y~B = 3cos90x+3sin90y

(a) ~A+~B = 2.81x+4.02y

(b) ~A−~B = 2.81x−1.97y

(c) 2~A−2~B = 5.63x−3.95y

107. (a) tan−1(4.02

2.81

)= 55

(b) tan−1(−1.97

2.81

)= 325

(c) tan−1(−3.95

5.63

)= 325

108. θ = tan−1(

AyAx

)= tan−1

(−2.0−3.0

)= 33.7

Since the vector is in the third quadrant, add 180:⇒ θ = 213.7

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109.

A =−20.0x+55.0y

|~A|=√

(−20.0)2 +(55.0)2 = 58.5

θ = tan−1(

55.0−20.0

)=−70

⇒ θ = 110 with respect to the positive x-axis

110. |~A|=√

(3)2 +(5)2 +(7)2 = 9.11111.

112. Let ~A = Axx+Ayy and ~B = Bxx+Byy.

~A−~B = ~A− (~B)

(Ax−Bx)x+(Ayx−By)y = (Axx+Ayy)+(−1)(Bxx+Byy)

→ ~A− (~B) = (−Bxx−Byy)

⇒ ~A−~B = ~A− (~B)

113. Let ~A = Axx+Ayy and ~B = Bxx+Byy.

2(~A+~B) = 2(AxBx)x+2(Ay +By)y

2~A+2~B = 2Axx+2Ayy+2Bxx+2Byy = 2(Ax +Bx)x+2(Ay +By)y

⇒ 2(~A+~B) = 2~A+2~B

114. The statement ~A = ~B is true only when all of the corresponding components of each vector are equal.115.

~A = (200,−100,50), ~B = (150,−75,160)~A−~B = (200−150,−100+75,50−160) = (50,−25,−110)

|~A−~B|=√

502 +(−25)2 +(−110)2 = 123 m

116.

~A+~B = (2+0,−1+3,5+3) = (2,2,8)

|~A+~B|=√

22 +22 +82 = 8.5

117.

~A+~B = 6x+3y+8z

118.

~A+~B−~C = (1+0+1,5+7+2,0+2−2)

= (2,14,0)

|~A+~B−~C|=√

22 +142 +02 = 14.14

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119.

−~A−~B+~C = (−1−0−1,−5−7−2,0−2+2)

= (−2,−14,0)

|−~A−~B+~C|=√(−2)2 +(−14)2 +02 = 14.14

120.

~A = 30 m at 20, ~B = 20 m at 75

Ax = 30cos20 = 28.2

Ay = 30sin20 = 10.26

Bx = 20cos75 = 5.18

By = 20sin75 = 19.32

121.

~A = 30 m at 20, ~B = 20 m at 75

Ax = 30cos160 =−28.2

Ay = 30sin160 = 10.26

Bx = 20cos255 =−5.17

By = 20sin255 =−19.32

122.

~C = (Ax +Bx)x+(Ay +By)y

Cx = Ax +Bx = 2cos20+1cos90 = 1.88

Cy = Ay +By = 2sin20+1sin90 = 1.68~C = 1.88x+1.68y

C =√

C2x +C2

y = 2.52

θ = tan−1(

Cy

Cx

)= 42

123.

~D = (Ax−Bx)x+(Ay−By)y

D = Ax−Bx = 2cos20−1cos90 = 1.88

D = Ay−By = 2sin20−1sin90 =−0.32~D = 1.88x−0.32y

124. The angle between ~A and ~B is 90 since the vectors are perpendicular to one another.

47

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3.2 References

1. Haven Giguere. CK-12 Foundation .2. Haven Giguere and Laura Guerin. CK-12 Foundation .3. Haven Giguere. CK-12 Foundation .4. Haven Giguere. CK-12 Foundation .5. Laura Guerin. CK-12 Foundation .6. Haven Giguere. CK-12 Foundation .7. Haven Giguere. CK-12 Foundation .8. Laura Guerin. CK-12 Foundation .9. Haven Giguere. CK-12 Foundation .

48

http://www.ck12.org

www.ck12.org Chapter 4. Newton’s Three Laws Problem Sets

CHAPTER 4 Newton’s Three LawsProblem Sets

Chapter Outline4.1 NEWTON’S THREE LAWS PROBLEM SETS

4.2 REFERENCES

49

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4.1. Newton’s Three Laws Problem Sets www.ck12.org

4.1 Newton’s Three Laws Problem Sets

Solutions

1. Since the objects are attached, the tension is the same. Assume clockwise is positive and counter-clockwise isnegative.

∑F1 = T −m1g = m1a

∑F2 =−T +m2g = m2a

→ (m2−m1)g = (m1 +m2)a

a =(m2−m1)g(m1 +m2)

⇒ T −m1g = m1

((m2−m1)g

m1 +m2

)T = m1

((m2−m1)g

m1 +m2

)+m1g

T = m1g(

m2−m1

m1 +m2

)+m1g

(m1 +m2

m1 +m2

)T =

(2m1m2

m1 +m2

)g

2.

∑F = ~F = (m1 +m2)~a

~a =~F

(m1 +m2)

3. The statement is incorrect. The fact that the jar is motionless only implies that the sum of forces is 0. Inthis problem, gravity pulls the jar towards the Earth and the table pushes back with an equal magnitude in theopposite direction.

4.

F = m1a1;F = m2a2;m1a1 = m2a2

m1

m2=

a2

a1=

0.5 m/s2

4.0 m/s2 = 0.125

5. Since the cart is not accelerating the total force on the cart is equal to 0.6. (a)

1 lb.= 4.4 N

→ 185 lb.= 814 N

(b)

185 lb.= 83.25 kg

F = (83.25 kg)(1.62 m/s2) = 135 N

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7.

∑F1 = T −m1g = m1a

∑F2 = m2g−T = m2a

Adding the equations together

(a) a = (m2−m1)gm1+m2

= (6.0 kg)(9.8 m/s2)(10.0 kg) = 5.9 m/s2

(b) T = m1(g+a) = (2.0 kg)(9.8+5.9) = 31.4 N(c) The acceleration is positive, so the pulley will rotate clockwise. Mass 1 will move up and mass 2

down. y = 12 at2 = (0.5)(5.9 m/s2)(0.5 s)2 = 0.7 m

(d)

a = 5.9 m/s2

v2 = v2o +2a4y

4y =− v2o

2a=−(−2.4 m/s)2

2(5.9 m/s2)= 0.5 m

8. (a) ∑F = F−T = mtruckax = (1000 kg)(3.0 m/s2) = 3 kNx(b) The net force on the car ∑F = T = mcarax = (500 kg)(3.0 m/s2) = 1.5 kNx

9.

v2 = v2o +2a4x

a =v2− v2

o

24x=

(1202−752) mi2/h2

2(2 mi)= 2193.75 mi/h2

a = 0.272 m/s2

F = ma = (750 kg)(0.272 m/s2) = 204 N

10. 20 lb = 9.1 kg

(a)

∑F = N−mg = ma = 0

N = mg = (9.1 kg)(9.8 m/s2) = 89.2 N

(b)

∑F = N +15 N−mg = ma = 0

N = mg−15 N = 89.2−15 = 74.2 N

11. a = Fm = 140 N

53 kg = 2.64 m/s2

12. F = ma = (125 kg)(1.53 m/s2) = 191 N13. m = F

a = 400 N(2.20 m/s2)

= 182 kg14. (a)

4x =12

at2

→ a =24x

t2 =2(12.0 m)

36.0 s2 = 0.67 m/s2

m =Fa=

75.0 N(0.67 m/s2)

= 112 kg

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(b) v = at = (0.67 m/s2)(6.0 s) = 4.0 m/s The frictional force will now give the block an acceleration.

a =Fm

=(−35 N)

112 kg=−0.31 m/s2

⇒4x =v2− v2

o

2a=

0− (4.0 m/s)2

2(−0.31 m/s2)= 26.0 m

15. (a) a = Fm = (200 N)

75.0 kg = 2.67 m/s2

(b) 4x = 12 at2 = 33.4 m

(c) v = at = (2.67 m/s2)(5.0 s) = 13.4 m/s16.

m =(500 N)

9.8 m/s2 = 51.0 kg

→ F = ma = (51.0 kg)(15.0 m/s2) = 765 N

17.

m =Fg=

(71.2 N)

(9.8 m/s2)= 7.27 kg

a =Fm

=(150 N)

(7.27 kg)= 20.6 m/s2

18. m = Feg = (50.0 N)

9.8 m/s2 = 5.1 kg

a. m = 5.1 kgb. aup =

Fupm = (20.5 N)

(5.1 kg) = 4.02 m/s2

19.

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20.

21.22. (a)

a =v2− v2

o

24x=

0− (100 m/s)2

2(0.05 m)=−1×105 m/s2

v = vo +at

→ t =v− vo

a=

0−100 m/s−1×105 m/s2 = 1×10−3 s

(b) F = ma = (0.02 kg)(−1×105 m/s2) =−2 kN

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23.

∑F = T −mg = ma

a =T −mg

m=

25000 N− (1500 kg)(9.8 m/s2)

(1500 kg)= 6.9 m/s2

24. Force due to gravity, normal force.25. Force due to gravity, force due to air resistance.26. The student’s statement is wrong because gravity and the normal force from the table act on it. It should be

restated: “Because the forces on the marker are equal and opposite, the marker is in equilibrium.”27. The cup falls because the friction between the car and the cup is not great enough to hold the cup in place as

the car moves out from under it.28. The normal force from the floor pushes up on the cup as is hits the ground. The cup’s natural springiness

compresses and this stored potential energy is released as the cup bounces up from the ground.29. When a ball is being thrown, the force due to gravity and the force due to air resistance are acting on it.30. The astronaut can throw his jet pack in the opposite direction from the ship, causing him to drift towards the

ship.31.

∑F = T −mg = ma = 0

mg = T

m =Tg= 0.36 kg

32.

500 lb = 226.8 kg

F = mg = (226.8 kg)(9.8 m/s2) = 2222.64 N

2222.6−2000 = 222.6 N downward

33.

a =v− vo

t=

66 m/s1.3 s

= 50.8 m/s2

F = (90 kg)(50.8 m/s2) = 4.57 kN

34. F = ma = (5.0 kg)(5.0 m/s2) = 25 N Therefore, the student pushing the block to the left is pushing with only5 N of force.

v = vo +at

→ a =−vo

t=

(−3.0 m/s)(0.3 s)

= 10 m/s2

35. The mass of the robber is: m= Frg = (800 N)

(9.8 m/s2)= 81.6 kg Therefore, the force needed is: F =ma=(81.6 kg)(10 m/s2)=

816 N36.

a =v2− v2

o

24x=

0− (29.1 m/s)2

2(20 m)= 21.2 m/s2

1000 lb = 453.6 kg

F = (453.6 kg)(21.2 m/s2) = 9.62 kN

54

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37. Since it is descending at a rate less than 9.8 m/s2 there must be a force pushing up on it.F = (1.0 kg)(9.8−2.5) m/s = 7.3 N

38. m = 25 N9.8 m/s2 = 2.55 kg

39.

∑F = T −mg = ma

T = m(a+g) = (2.55 kg)(1.22+9.8) = 28.1 N

a =Fm

=200 N90 kg

= 2.22 m/s2

40. F = ma = (200 kg)(9.8 m/s2) = 1.96 kN41. No, 0 acceleration implies there are either no forces or all the forces cancel each other.42. 1 N of gold on Mercury has more mass because the acceleration from gravity is less, making it worth more.43. (a) The tension is 2000 N, as this is the force needed to keep the system in equilibrium according to Newton’s

first law.(b) The tension is still 2000 N. This is the force needed to ensure that no net force is present in the system.

44. The net force on the mass is 0 N, as the mass is not moving.45. The corresponding force is the force of gravity that the book exerts upon the earth. It is not the force that the

table exerts on the book.46. The net force on the dumbbell is 0 N, as the dumbbell is not moving during the 5 minutes.47. There is a non-zero net force on the race car.48. F = ma; aman = 2m/s; agirl = 4m/s

The man travels at 2 m/s, while the girl travels at 4 m/s.49. The skydiver weighs 750 N. Since he is traveling at a constant velocity, his net force must be 0 N. Thus, he

experiences a 750 N force in the upwards direction.50. According to Newton’s first law, objects will only travel in uniform motion if they experience no net force.

However, objects will experience a net force if they are acted on by only one force.

55

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4.2 References

1. Haven Giguere. CK-12 Foundation .2. Haven Giguere. CK-12 Foundation .3. Haven Giguere. CK-12 Foundation .4. Laura Guerin. CK-12 Foundation .5. Haven Giguere. CK-12 Foundation .6. Haven Giguere. CK-12 Foundation .7. Haven Giguere and Laura Guerin. CK-12 Foundation .8. Haven Giguere. CK-12 Foundation .9. Haven Giguere. CK-12 Foundation .

10. Haven Giguere. CK-12 Foundation .

56

http://www.ck12.org

www.ck12.org Chapter 5. Forces in Two Dimensions Problem Sets

CHAPTER 5 Forces in Two DimensionsProblem Sets

Chapter Outline5.1 FORCES IN TWO DIMENSIONS PROBLEM SETS

5.2 REFERENCES

57

http://www.ck12.org

5.1. Forces in Two Dimensions Problem Sets www.ck12.org

5.1 Forces in Two Dimensions Problem Sets

Solutions

1. T = mv2

r = (1.0 kg)(2 m/s)2

(1.5 m) = 2.67 N2.

T =mv2

r

→ v =

√Trm

=

√(100 N)(1.2 m)

2.5 kg= 6.9 m/s

3. f = mv2

r = (35 kg)(4 m/s)2

1.5 m = 373.3 N4. If there were no forces on it, then there would be no possible way to have the car move in a circular path.5. To determine the tension in each of the ropes, break the forces into components Assume right and up are

positive.∑Fx = T2 cosθ2−T1 cosθ1 = 0 . . . . . .(1)∑Fx = T1 sinθ1 +T2 sinθ2−200 = 0 . . . . . .(2)Solving for T2 in equation (1):

T2 = T1

(cosθ1

cosθ2

)

Solving into equation (2):

T1 sinθ1 +T1

(cosθ1

cosθ2

)sinθ2−200 = 0

T1 sinθ1 +T1 tanθ2 cosθ1−200 = 0

T1(sinθ1 + tanθ2 cosθ1) = 200

→ T1 =200

(sinθ1 + tanθ2 cosθ1)= 84.85 N

⇒ T2 = 173.9 N

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6.

(a)

∑Fx = mgsinθ = ma

→ a = gsinθ

(b)

4x = vot +12

axt2

⇒ t =

√2dax

7. v2f = v2

o +2ax4xSince the object is released from rest, vo = 0 .v f =

√2axL =

√2(gsinθ)L

8.

∑Fx = 3.0cos45+7.0cos330 = max

ax = 3.72 m/s2

∑Fy = 3.0sin45+7.0sin330 = may

ax =−0.63 m/s2

a =√

a2x +a2

y = 3.77 m/s2

9.

~F = m~a~F = (5.0 kg)(1.0 m/s2x−2.5 m/s2y) = 5 Nx−12.5 Ny

|F |=√

(5)2 +(−12.5)2 = 13.5 N

10.

~F = ~F1 + ~F2 + ~F3 =−0.5 Nx+6.0 Ny

|F |=√

F2x +F2

y = 6.02 N

m =Fa=

6.02 N3.0 m/s2 = 2.0 kg

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11. (a)

T3 =W = mg = 6 kN

∑Fy = T1 sinθ1 +T2 sinθ2−mg = ma = 0 . . . . . .(1)

∑Fx =−T1 cosθ1 +T2 cosθ2 = ma = 0 . . . . . .(2)

T2 = T1cosθ1

cosθ2

Substitute into equation 1:

T1 sinθ1 +

(T1

cosθ1

cosθ2

)sinθ2−mg = 0

T1 =mg

sinθ1 + tanθ2 cosθ1=

6 kNsin30+ tan55 cos30

= 3.45 kN

⇒ T2 = (3.45 kN)cos30

cos55= 5.21 kN

(b)

∑Fy = T1 sinθ1 +T2 sinθ2−mg = ma = 0 . . . . . .(1)

∑Fx =−T1 cosθ1 +T2 cosθ2 = ma = 0 . . . . . .(2)

T2 = T1cosθ1

cosθ2

Substitute into equation 1:

T1 sinθ1 +

(T1

cosθ1

cosθ2

)sinθ2−mg = 0

T1 =mg

sinθ1 + tanθ2 cosθ1

12.

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13. (a)(b)

∑Fx = ma = mgsinθ

→ a = gsinθ

(c)

v2 = v20x+2a4x

v =√

2gsinθ4x =√

2(9.8 m/s2)sin22(3.30 m) = 4.9 m/s

14. The magnitude of the force the teacher is asking for the students to produce is: F = ma = 5.0 kg(15.0 m/s2) =75.0 NIn component form:Fx = 75.0cos20 = 70.5 NFy = 75.0sin20 = 25.7 NThe first student applies a force that is only 50.0 N east. Therefore, the second student needs to apply a force

that is of magnitude: F =√(70.5−50.0)2 +(25.7)2 = 32.9 N at θ = tan−1(25.7

20.5) = 51.4 north of east.

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15. (a)(b)

∑F1,x = T = m1a

∑F2,y =−T +m2g = m2a

Adding the two equations together:

m2g = (m1 +m2)a

a =m2g

m1 +m2

16.

∑F1,x = T = m1a

∑F2,y =−T +m2g = m2a

Adding the two equations together:

m2g = (m1 +m2)a

a =m2g

m1 +m2

Therefore, T = m1

(m2g

m1+m2

).

17. (a)

∑F1,x = T −m1gsinθ = m1a

∑F2,y =−T +m2g = m2a

Adding the two equation together:

m2g−m1gsinθ = (m1 +m2)a

a =(m2−m1 sinθ)g

m1 +m2=

(4.0 kg−2.0 kgsin60)(9.8 m/s2)

6 kg= 3.7 m/s2

(b) T = m2g−m2a = (4.0 kg)(9.8−3.7) = 24.4 N18.

a = 5.9 m/s2

v2 = v2o +2a4y

4y =− v2o

2a=−(−2.4 m/s)2

2(5.9 m/s2)= 0.5 m

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The acceleration the block will experience on the incline is a = gsinθ = 4.9 m/s2. Therefore:

v2 = v2o +2a4x

4x =− v2o

2a=− (5.0 m/s)2

2(−4.9 m/s2)= 2.55 m

19. (a) a = gsinθ

(b) v =√

2a4x =√

2(gsinθ)( y

sinθ

)v0y = vsinθ =

√2gysinθ

(c) The time taken to travel down the inclined plane:

v = at

t =va=

√2(gsinθ)

( ysinθ

)gsinθ

The time it takes to hit the ground is:

0 = y+ vot +12

gt2

t =−voy±

√(voy)2−2gyg

Therefore the total time is:

ttotal =−voy±

√(voy)2−2gyg

+

√2(gsinθ)

( ysinθ

)gsinθ

20.

∑F1,x = T −m1gsinθ = m1a

∑F2,x = m2gsinθ−T = m2a

Adding both equations together:

(m2−m1)gsinθ = (m1 +m2)a

a =(m2−m1)gsinθ

m1 +m2

a =(5.8 kg)(9.8 m/s2)sinθ

14.2 kga = 4 m/s2 sinθ

21.

∑Fx = mgsinθ− f = ma = 0

mgsinθ−µN = 0

∑Fy = N−mgcosθ = ma = 0

N = mgcosθ

→ µ(mgcosθ) = mgsinθ

µ = tanθ

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22. (a)(b)

∑F1 = T − f = m1a

∑F2 = m2g−T = m2a


m2g− f = (m1 +m2)a

a =m2g−µNm1 +m2

=m2g−µ(m1g)

m1 +m2=

(5.0 kg−0.22(2.0 kg))(9.8 m/s2)

(7.0 kg)= 6.4 m/s2

23.

∑Fx = mgsinθ−µN = ma = 0


N = mgcosθ

→ µ(mgcosθ) = mgsinθ

µ = tanθ = 0.65

24.

∑Fx = 60.0 N cosθ−35.0 N = ma = 0

60 N cosθ = 35.0 N

θ = cos−1(

35.060.0

)= 54

25.

∑F1 = T −µN1 = m1a

→ N1 = m1g

∑F2 = m2g−T = m2a


m2g−µm1g = (m1 +m2)a

a =(m2−µm1)g

m1 +m2

⇒ T = m1

((m2−µm1)g

m1 +m2

)+µm1g = 42.1 N

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26. Kid #2: F−T −µm2g = m2aKid #1: T −µm1g = m1aAdding both equations together: F−µm1g−µm2g = (m1 +m2)a

(a) a = F−(m1+m2)µgm1+m2

= 95 N−(45 kg)(0.2)(9.8 m/s2)45.0 kg = 0.15 m/s2

(b)

T = m1a+µm1g

T = (25.0 kg)(0.15 m/s2 +0.2(9.8 m/s2)) = 52.75 N

27. (a)

m1 :

∑F1 = T12−m1g = m1a

m2 :

∑F2 = T23−T21−µm2g = m2a

m3 :

∑F3 = m3g−T23 = m3a

Adding all the equations together:

m3g−m1g−µm2g = (m1 +m2 +m3)a

a =

(m3−m1−µm2

m1 +m2 +m3

)= 0.37 m/s2 (up for m1, to the right for m2, down for m3)

(b)

T12 = m1a+m1g = 25.4 N

T23 = m3g−m3a = 61.3 N

28.

∑Fx = F sinθ−N = ma = 0

N = F sinθ

∑Fy = F cosθ+ f −mg = ma = 0

0 = F(cosθ+µs sinθ)−mg

F =mg

(cosθ+µs sinθ)

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29. (a) The forces must be in the same direction (angle = 0°).(b) For the sum to be 0, the forces must be in opposite directions (angle = 180°).

30.

Fx = F cosθ = (150 N)cos35 = 123 N

Fy = F sinθ = (150 N)sin35 = 86 N

31.

Fx = F cosθ

F =Fx

cosθ

→ Fy =

(Fx

cosθ

)sinθ = Fx tanθ = (75 N) tan22 = 30.3 N

32.

F1 = 225 Ni

F2 = 320cos70 i+320sin70 j

F =√(F1x +F2x)2 +(F1y +F2y)2 = 450 N

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33.

34.35.

∑Fy = N−F sinθ−mg = ma = 0

N = (35.0 N)sin40+(10.2 kg)(9.8 m/s2) = 122.5 N

36.

∑Fx = T −mgsinθ = ma = 0

T = (105.5 kg)(9.8 m/s2)sin33 = 563 N

37. F = ma = (0.02 kg)(−1×105 m/s2) =−2 kN

Ax = 3 N, Ay = 5 N, Bx = 2 N

R =√(Ax +Bx)2 +(Ay +By)2 = 7.1 N

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38. To get the particle to travel along the x-axis, you need to cancel out the y-component of the force due to F1:

∑Fy = F1 sinθ−F2y = ma = 0

F2y = (8.0 N)sin65 = 7.25 N

39. (a)

R =√(−10 N +−9cos20)2 +(−9.0sin20)2 = 18.7 N

θ = tan−1(

−9.0sin20

−10 N +−9cos20

)= 9.5 south of west

(b)

θ = tan−1(

−9.0sin20

−10 N +−9cos20+20.0

)= 63 south of east

40. F = ma = (1.0 kg)(1.0 m/s2) = 1.0 N41.

F1x +F2x +F3x = 0

(2 N)cos45+(2 N)cos135 =−F3x

F3x = 0

F1y +F2y +F3y = 0

F3y =−(2sin45+2sin135) = 2.83 N

Therefore, F3 = 2.83 N, θ3 = 270.42. Since you are not accelerating, the tension is simply your friend’s weight: T = m2g.43.

∑Fx =−mgcosθ+F cosθ = ma = 0

F = mg = (75 kg)(9.8 m/s2) = 735 N

44.

(a)

∑Fx = T sinθ = ma

∑Fy = T cosθ−mg = ma = 0

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Dividing the two equations:

tanθ =ag

a = g tanθ = (9.8 m/s2) tan20 = 3.57 m/s2

(b) From part a:

a = 3.57 m/s2

T =mg

cosθ=

(0.02 kg)(9.8 m/s2)

cos20= 0.21 N

45.

ax =2.0 N2.0 kg

= 1.0 m/s2

az =−0.5 N2.0 kg

=−0.25 m/s2

46. ~F =√

(4.0 N)2 +(2.0 N)2 +(−0.5 N)2 = 4.5 N47. The normal force does NOT equal mg when the dealing with inclined system or where the coordinate system

changes.48. Neither. The rate of an object sliding down a frictionless incline is gsinθ. The acceleration is mass indepen-

dent.49.

∑Fx = mgsinθ− f = ma = 0

f = mgsinθ

50.


N = mgcosθ

∑Fx = mgsinθ−µ(mgcosθ) = ma = 0

µ = tanθ

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5.2 References

1. Haven Giguere. CK-12 Foundation .2. Haven Giguere. CK-12 Foundation .3. Haven Giguere. CK-12 Foundation .4. Haven Giguere and Laura Guerin. CK-12 Foundation .5. Haven Giguere. CK-12 Foundation .6. Haven Giguere. CK-12 Foundation .7. Haven Giguere. CK-12 Foundation .8. Haven Giguere. CK-12 Foundation .9. Haven Giguere and Laura Guerin. CK-12 Foundation .

10. Haven Giguere and Laura Guerin. CK-12 Foundation .11. Haven Giguere. CK-12 Foundation .12. Haven Giguere. CK-12 Foundation .13. Haven Giguere and Laura Guerin. CK-12 Foundation .14. Haven Giguere. CK-12 Foundation .15. Haven Giguere. CK-12 Foundation .16. Haven Giguere. CK-12 Foundation .17. Haven Giguere. CK-12 Foundation .18. Haven Giguere. CK-12 Foundation .19. Haven Giguere. CK-12 Foundation .20. Haven Giguere. CK-12 Foundation .21. Haven Giguere. CK-12 Foundation .22. Haven Giguere. CK-12 Foundation .23. Haven Giguere. CK-12 Foundation .

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www.ck12.org Chapter 6. Work and Energy Problem Sets

CHAPTER 6 Work and Energy ProblemSets

Chapter Outline6.1 WORK AND ENERGY PROBLEM SETS

6.2 REFERENCES

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6.1. Work and Energy Problem Sets www.ck12.org

6.1 Work and Energy Problem Sets

Problems

1. Calculate the amount of work done when you carry a 200 lb. bar across your back for 20 m.2. Two objects each weighing 2 kg are lifted through the same distance in 4 s and 1 s respectively. Which scenario

requires more work?3. Two objects weighing 2 kg and 4 kg are lifted in 4 s and 1 s respectively. Which scenario requires more power?4. A 1.0 kg rock dropped from a tall skyscraper falls 23 m in 1.3 s. Assuming the rock is falling at terminal

speed, what is the expended power?5. A crate weighing 55 pounds is pushed across a factory floor 1.5 m with a horizontal force of 165 N, as seen in

the figure below. Assuming the floor is frictionless, calculate the work done by each of the following forces:

a. The force being applied.b. The gravitational force on the crate.c. The normal force being exerted on the crate.

6. Using the figure below, answer the same questions as in the previous problem, but let θ, the angle below thehorizontal, be 0°.

7. Your car runs out of gas 100.0 m from the gas station. Calculate the work done to push your car if you pushyour 400 kg car with a force of 500 N at an angle of 20° below the horizontal.

8. Prove ~A ·~B = AxBx +AyBy +AzBz.

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9. Determine ~A · (~B−~C) if the vectors are defined as follows: ~A = 3x+2y, ~B = 2x+2y−1z, ~C = 2x−3y+1z.10. Determine the angles between vectors ~A and ~B as well as vectors ~A and ~C in the previous problem. (Hint: Use

the definition of the scalar product.)11. A 500 kg car driving down the street at t=0 s has a speed of 20 m/s. Twenty minutes later, the car has a kinetic

energy of 156 kJ.

a. What was the car’s initial kinetic energy?b. What is the car’s speed at t=20 min?c. Calculate the amount of work done from t=0 min to t=20 min.

12. A 0.2 kg softball has an initial velocity of 35 mph.

a. Calculate the kinetic energy of the softball if the speed is doubled.b. How does this compare to the initial kinetic energy?

13. The initial velocity of a 2.0 kg object flying through the air is given by (2x−4y) m/s. Calculate the total workdone if the velocity is changed to (6x+2y) m/s.

14. A 1000 kg pile driver travels 10 m before coming into contact with a wooden post. If the post is driven intothe ground 1.2 m, calculate the average force exerted on the pile driver from the wooden post.

15. A small 35.0 kg child is pulled from rest with a constant force of 145 N across a horizontal distance of 10.0 mon a sled. If the coefficient of kinetic friction between the child and the sled is 0.300, calculate:

a. The work being done by the person pulling the child.b. The work done by the normal force.c. The child’s final velocity.

16. A 5.0 kg block sitting on a frictionless surface is attached to a spring, as seen in the figure below.

a. If the block is compressed 3.0 cm from equilibrium, calculate the speed of the block when it is releasedif the spring has a spring constant of 350 N/m.

b. Calculate the final speed of the block as it passes through the equilibrium point if the coefficient offriction between the block and the surface it is sitting on is 0.10.

17. A small child is pulled with a 300 N force up a 10.0 m incline that makes an angle of 28° with respect to thehorizontal, as shown in the figure below.

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a. Calculate the energy lost due to friction.b. How much work is being done by the 145 N force that is pulling the child up?c. What is the final speed of the sled?

18. Calculate the kinetic energy of a 1.0 kg kickball at the apex of its trajectory, given that it left the ground withan initial velocity of 52.0 m/s and at an angle of 40° with respect to the horizontal.

19. While moving into your new house, you drop a 25 kg Olympic plate onto a disassembled table, leaving a dent7.0 mm deep. If the plate dropped 3 feet, calculate the average force on the plate.

20. Two blocks of different mass, m1 and m2, are placed on each side of a spring which has a spring constant k, asseen in the figure below. If the total distance the spring is compressed is ∆x, calculate the initial accelerationof each of the blocks when the system is released from rest if the system is frictionless.

21. Assume the blocks in the previous problem (shown in the figure above) rest on a rough surface. If thecoefficient of friction between the blocks and the surface is µk, calculate the initial acceleration of each ofthe blocks when the system is released from rest. Assume the coefficient of static friction is small enough thatthe blocks will move after being released.

22. While working out, you apply a constant horizontal 75 N force on a mass of 100 lb. What is the instantaneouspower required if it takes you 1.5 s to perform the rep from a dead stop?

23. A 2.0 kg object compresses a spring with k=1200 N/m a distance of 0.30 m at the bottom of an incline thatmakes an angle of 45° with respect to the positive x-axis. Calculate:

a. The distance the object travels when it is released if the system is frictionless.b. The distance the object travels when it is released if µk=0.22.

24. A block of mass m compresses a spring a distance ∆x and is released from rest. What is the change in initialvelocity if the block compressed the spring to 2∆x instead?

25. When you kick a soccer ball, is work done while your foot is in contact with the ball?26. An example when a force is exerted but no work is done occurs if you walk horizontally while holding a book

in your arms. What is another?27. A spring with a spring constant k is cut into thirds. What is the spring constant of the new springs? Explain.28. After you solve a quiz question about an exploding star, you find that the kinetic energy is -1.8*1029 J. Why

is this wrong? Explain.

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29. Two balls m1 and m2 are thrown at the same speed but m2 = 2m1. Which has a greater kinetic energy?30. The speed of an object in motion is doubled. What happens to its kinetic energy?31. A force of magnitude 13 N is applied to a 7 kg box through a distance of 3.5 m. Calculate the work done by

the force.32. A 7 kg box resting on a frictionless surface is pushed by a 13 N force through 4.0 m.

a. Calculate the kinetic energy of the system when t=5.0 s.b. Calculate the velocity of the box at t=5.0 s.

33. A sled is pulled a horizontal distance of 3 m along a frictionless 30° incline by a 75 N force that is parallel tothe incline surface. What is the work done on the sled?

34. A small child slides down a 2 m long ice rink with an initial velocity of 2 m/s before coming to a stop.Determine µk.

35. Using the information given in the figure below, derive an equation that gives the final velocity of an objectbeing released from rest.

36. A mass m is attached to a massless string of length l. The other end is attached to the ceiling and the massis pulled back to an angle θ with respect to the vertical. When the mass is released, what is its maximumvelocity?

37. A block of mass m=2 kg is attached to a wall by a spring with a spring constant k=120 N/m. The string ispulled 1.5 m from its equilibrium, as shown in the figure below. Determine:

a. The potential energy before the block is released.b. The velocity of the block as it passes through the equilibrium position.

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38. Two masses are attached to each other by a massless rope that runs over a frictionless pulley, as shown in thefigure below. When the system is released from rest, what is the final velocity of the hanging masses before ithits the ground? (Assume m1 >m.)

39. A 2 kg block is attached to a wall by a spring with a force constant of 5 N/m. If the block is given an initialvelocity of 7 m/s in the direction of the wall, calculate the coefficient of friction if the spring is compressed 3m.

40. A ball of mass m=1 kg is dropped from a cliff 100 m high. Using the conservation of energy, calculate thefinal velocity of the ball before it hits the ground.

41. When looking at the gravitational potential energy of a system, is it:

a. Always positiveb. Always negativec. Positive or negative

42. An object of mass m is dropped from a cliff of height h.

a. Derive an equation for the speed of the object at a height y’ above the ground.b. Instead of being dropped, the object is thrown vertically downward. Derive an equation for the speed of

the object at any height y’ above the ground.

43. While moving into your new house, you slide a 25.0 kg box of physics books from the top of the movingtruck’s ramp. The ramp is 1.5 m long and has an angle of inclination of 34°. Assuming the crate experiences6.5 N of frictional force, use the conservation of energy to determine the speed of the box when it reaches thebottom of the ramp.

44. If two slides on a school playground have the same height, but one is twice as long as the other, what can besaid about their final speeds if friction can be considered negligible?

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45. If the two playground slides in the previous problem have the same coefficients of friction, what can be saidabout their final speeds?

46. Why is it true that kinetic energy always positive, but total energy can be negative? Explain and give anexample.

47. While walking through the park, you notice a child on a swing set, as seen in the figure below.

a. What is the child’s gravitational potential energy when the swing’s chains make a 25° angle with respectto the vertical?

b. What is the child’s gravitational potential energy when the swing makes a 0° angle with respect to thevertical?

48. A cart slides along a track as seen in the figure below. The cart is released from rest at a height that is equal to3.0 R. Calculate the speed of the cart at the highest point on the loop.

49. Using the information presented in the figure below:

a. Find the initial velocity in the y-direction.b. Find the horizontal and vertical components of the velocity before the projectile hits the ground.

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50. Two crates are connected to one another, as seen in the figure below. Using the conservation of energy,determine the height that the lighter mass m2 rises to. (Assume the pulley is frictionless and massless.)

51. A child is sitting at the top of a slide with negligible friction. When the child is released from rest, calculate:

a. The speed of the child at point 1, one-third of the way down the slide.b. The speed of the child at point 2, two-thirds of the way down the slide.c. The work done on the child (m=35.0 kg) from points 1 to 2.

52. A mass m is attached to a light rod of length L, as seen in the figure below. What minimum initial velocity inthe x-direction must be imparted onto the mass to cause it to make a full revolution?

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53. A block of mass m slides down a frictionless incline of length l that makes an angle of θ with respect withthe horizontal. When the block reaches the bottom, it slides a short horizontal distance before coming intocontact with a vertical plate that is attached to the wall by a spring. Assuming the spring has a spring constantk, calculate the distance that the spring is compressed when the block comes to a stop.

54. Wanting to try out a new festival attraction, you jump onto a platform that is attached to a spring. If the springhas a force constant of 3500 N/m and the platform is compressed a total distance of 0.50 m, how far in the airwill you be launched if you weigh 180 pounds?

55. A block of mass m is sitting at rest on top of a table and is connected to a mass m2 by a light string. The stringis placed over a light frictionless pulley and m2 hangs over the edge of a table. Calculate the speed of m2 afterit has fallen a distance y. (Assume the coefficient of kinetic friction between m1 and the table is µk.)

56. During recess a 40.0 kg child run towards a snow covered hill and tries sliding up the hill as far as possible.The child starts approaches the 15° hill with an initial velocity of 5.0 m/s and slides a total distance of 1.5 mup the hill. Determine:

a. The child’s change in potential energy.b. The change in the child’s kinetic energy.c. The coefficient of kinetic friction.

57. While on a mountain expedition your friend (m1=95 kg) accidentally slips over the edge of a cliff. Luckilyfor him, you (m2=75 kg) are attached to him by a strong lightweight rope. Unfortunately for you, you start toslide up the cliff towards the edge he fell over, as shown in the diagram below. Assuming the coefficient ofkinetic friction between you and the snow is 0.33 and the angle that cliff makes with the horizontal is 32.0°,calculate the change in kinetic energy as you slide 25.0 m towards the edge of the cliff.

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58. An 80 kg stunt man jumps from a height of 3.0 meters onto a platform with a large spring under it. What isthe total displacement of the platform if the spring constant is 2500 N/m?

59. A crate of mass m is released from rest on a frictionless ramp that has a height h over the ground. At thebottom of the ramp, the crate slide over a rough surface that has a total distance of x. After passing over thelong rough surface it continues onto a flat frictionless surface until come into contact with a spring that has aforce constant of k and is compressed a distance ∆x. Derive an equation for the coefficient of kinetic frictionbetween the crate and the rough surface.

60. Correct the following statement: “Positive work is done when the force is perpendicular to the displacement.”61. You offer to pay your friend $10 per hour. while he is doing work carrying your textbooks for you from class

to class. When the day is over, you pay him $0. How did you come about this calculation?62. What is the difference between kinetic energy and gravitational potential energy?63. While discussing conservation of energy with your friends, your friend Billy states that a falling object’s

gravitational potential energy disappears. Is he wrong? Explain.64. While cleaning your office, you move a stack of books from your desk to a spare desk in the corner. If the

desks are the same height, how much net work is done? Explain.65. If an object triples its speed, by how much is the kinetic energy changed?66. When you go hiking, your friend wants to take the longer, less steep route. You want to take the more direct

but much steeper route. Which path requires more work? Explain.67. A 500 kg car is accelerating down the freeway at 2 m/s2 over a distance of 5.0 m. If the initial speed of the car

is 2.4 m/s, calculate:

a. The car’s speed after it has traveled 5.0 m.b. The increase in the kinetic energy of the car.

68. Two cars, car A and car B, are racing down the street. Car A is twice as heavy as car B, but has half the kineticenergy of car B. If car A has the same energy after speeding up by 2.3 m/s, what were the original speeds ofboth cars?

69. A crate is pushed 10 m in the x-direction and 7 m in the y-direction by a single force of 30 and 50 in the x-and y-directions respectively. Calculate the work done on the block.

70. A 0.5 kg object is initially moving along the y-axis with velocity of 3.3 m/s. After being acted upon by a 4.0 Nforce, the object has a velocity of -1.0 m/s along the x-axis. Calculate the amount of work done on the object.

71. A cabinet is moved 2.5 m across the floor in a straight line. During its displacement another force is acting onthe cabinet that has a magnitude of 1.5 N and it at an angle of 135° with respect to the positive x-axis. Whatis the work done by the 1.5 N force?

72. Four teams of 2 are trying to get a crate of equipment back to their camp on a popular game show. Using theforces and the angles listed in Table 6.1, calculate the net work done when the crate is displaced 2.0 m in theθ=0° direction.

TABLE 6.1:

F (N) θ (deg)200 235150 180300 050 45

73. A block weighing 4.5 kg slides over a surface of negligible friction a distance of 4.5 m (θ=0°) while threeforces, each at different angles, are acting upon it. Using the information provided in Table 6.2, determine thenet work on the block.

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TABLE 6.2:

F (N) θ (deg)2.0 2704.0 1809.5 45

74. A 20 kg child is trapped at the bottom of a 20 m well. When rescue teams arrive, the child is hoisted up by acable system that has accelerates the child at a rate of 1.0 m/s2.

a. What is the child’s kinetic energy?b. How much work was done on the child by the gravitational force?c. What is the child’s final speed?

75. A mass slides down a 1.0 m incline of θ=35° while experiencing a resistive force due to friction of 35.0 N thatis directed up the ramp. If the mass has a kinetic energy of 75 J at the end of the ramp, what would the kineticenergy be if the ramp was frictionless?

76. On a newly discovered planet, a dropped object falls a distance d with a gravitational acceleration of g/5.

a. What is the work done by the gravitational force?b. What is the kinetic energy of the block?

77. While lifting mining equipment out of a shaft deep in the earth, an elevator system operates in three separatephases. During the first 15.0 m the system accelerates to a speed of 5.5 m/s. Next, the system operates at aconstant speed for 10.0 m. Finally, the system is brought to a halt over the next 15.0 m. If the lift alwaysoperates at a max capacity of 1000 kg, how much work is done during each phase?

78. A block is used to compress a spring 3 cm. If a force of 25 N is needed to keep the block stationary, how muchwork is needed to move the block an additional 2.0 cm?

79. If a spring with spring constant 55 N/m is stretched 3.75 m, how much work is done when the spring reachesits equilibrium position?

80. A 25 kg mass is acted upon by a 6.0 N force over 1.0 m. What is the work done over this distance?81. After getting stuck halfway down a near frictionless incline angled at 48°, your friends toss you a rope and

begin pulling you up at a rate of 1.2 m/s. What is the power if the work done on you while being pulled 10.0m up the incline is 750 J?

82. Calculate the rate at which work is done on a 75 kg mass that is pulled at a constant speed of 7.0 m/s up a 30°incline by a 100 N force.

83. A 1000 kg equipment lift travels up a 200 m shaft in 45 seconds. Assuming the speed is constant, what is theaverage power?

84. While walking to the front counter at the airport, you pull your suitcase with a force of 50 pounds usinga handle that makes an angle of 35° with respect to the horizontal. If you’re walking at a rate of 1.3 m/s,determine the amount of work done in 5 min.

85. You are pushing a 13.0 kg crate up a 15° incline of negligible friction. If you apply 175 N of force parallel tothe incline, how much work is done during the first 2.0 m by your applied force?

86. A car weighing 1000 pounds is driving down the freeway at a constant velocity of 55 mph. What is the kineticenergy?

87. A 5.0 kg block is pulled by an 8.0 N force across the floor at a constant speed. After traveling 5.5 m, what isthe work done by the person pulling the block?

88. A block of mass m is pulled by a force F across the floor at a constant speed. After traveling a distance of ∆x,what is the coefficient of kinetic friction?

89. If an object has a kinetic energy of 320 J, and it has a mass of 22 kg, what is the speed it is traveling at?90. What is the work done by a 5.0 N force applied to a crate sitting on a frictionless surface over 5.0 m?

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Solutions

1. The answer is zero. Since the force on the bar is perpendicular to the displacement, no work is done.2. The both require the same amount of work even though they are lifted in a different amount of time.3. The object that is lifted in the shorter amount of time requires more power. Therefore the object in the second

scenario requires more power.4. P = W

t = mght = (1.0 kg)(9.8 m/s2)(2.3 m)

1.3 s = 7.54 W5. (a) W = F cosθr = (165 N)cos20(1.5 m) = 233 Ne

(b) 0; the gravitational force is perpendicular to the displacement.(c) 0; the normal force is perpendicular to the displacement.

6. (a) W = F cosθr = (165 N)cos0(1.5 m) = 248 N(b) 0; the gravitational force is perpendicular to the displacement.(c) 0; the normal force is perpendicular to the displacement.

7. W = F cosθr = (500 N)cos20(100 m) = 46.98 kJ8. ~A ·~B = (Axx+Ayy+Azz) ·(Bxx+Byy+Bzz)~A ·~B = AxBx(x · x)+AxBy(x · y)+AxBz(x · z)+AyBx(y · x)+AzBx(z ·

x)+AyBx(y · x)+AyBy(y · y)+AyBz(y · z)+AzBy(z · y)+AzBz(z · z)~A ·~B = AxBx +AyBy +AzBz

9.

~B−~C = 0x+5y−2z~A · (~B−~C) = (3x+2y) · (0x+5y−2z)~A · (~B−~C) = (3)(0)+2(5)+0(−2) = 10

10.

~A = 3x+2y~B = 2x+2y−1z~C = 2x−3y+1z

θAB = cos−1

(~A ·~BAB

)= cos−1

((6+4+0)√

(13)(9)

)= 22.4

θAB = cos−1

(~A ·~CAC

)= cos−1

((6−6+0)√

(13)(14)

)= 90

11. (a) K = 12 mv2 = 1

2(500 kg)(20 m/s)2 = 100 kJ(b)

K =12

mv2

→ v =

√2Km

=

√2(156 ·103 J)

500 kg= 25.0 m/s

(c) W = ∆K = K f −Ki = 156 kJ−100 kJ = 56 kJ12. 35 mph = 15.6 m/s

(a) Kv doubled = 12 mv2 = 1

2(0.2 kg)(31.2 m/s)2 = 97.3 J(b) Ki =

12 mv2 = 1

2(0.2 kg)(15.6 m/s)2 = 24.3 JWhen the speed is doubled, the kinetic energy is increased by a factor of 4.

13. The magnitude of the initial velocity is just the dot product of the initial velocity with itself:v = ~vi ·~vi =

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(4+16) = 20 m/sSimilarly for the final velocity,W = ∆K = K f −Ki =

12 m(v2

f − v2i ) =

12(2.0)(402−202) = 1200 J

Let y = 10 m, the distance the pile driver travels, and y′ = 1.2 m, the distance the post travels.W = ∆K⇒Wgravity +Wpost = 0

(mpiledriverg)(y+ y′)+Fy′ cos180 = 0⇒ F =−mhg(y+y′)y′ cos180 =−

(1000 kg)(9.8 m/s2)(10 m+1.2 m)1.2 mcos180 = 91.47 kNup

(a) W = F · x = Fxcosθ = (145 N)(10 m) = 1450 J(b) The work done by the normal force is zero since the force and this displacement are perpendicular to one

another.(c) To calculate the final velocity, you need to know how much work is going towards the frictional effects.

Einternal = fkx = (µN)x = (µ(mg))x = (·3)(35 kg)(9.8 m/s2)(10 m) = 1029 J

K f inal =W −Einternal = 1450 J−1029 J = 421 J

v =

√2Km

=

√2(421 J)

35 kg= 4.9 m/s

14. (a)

W = ∆U =12

kx2f −

12

kx2i

W =12(350 N/m)(0.03 m)2−0 = 0.158 J

→W = ∆K =12

mv2f −

12

mv2i =

12

mv2f −0

⇒ v f =

√2Wm

=

√2(0.158)

5.0= 0.08 m/s

(b)

W = ∆U =12

kx2f −

12

kx2i

W =12(350 N/m)(0.03 m)2−0 = 0.16 J

→W = ∆K + fkd =12

mv2f −

12

mv2i + fkd =

12

mv2 f −0+ fkd

⇒ v f =

√2(W − fkd)

m=

√2(0.16− (0.1)(5.0)(9.8)(0.03))

5.0= 0.07 m/s

15. (a)

E f ric = fkd = (µkmg)d cosθ = (0.3)(35.0 kg)(9.8 m/s2)(10.0 m)cos28

E f ric = 908.6 J

(b) W = Fd = (300 N)(10.0 m) = 3000 J(c) ∆K = ∑W −∆E f ric

the total work being done in the system a sum of the work being done by the person pulling the child andthe work done by gravity.⇒Wgravity = mgd cos(θ+90) = (35 kg)(9.8 m/s2)(10.0 m)cos(28+90)The 90 that is added to the angle takes into account that the force and the displacement are not in the

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same direction.

Wgravity =−1610 J

⇒ ∆K =WF +Wg−∆E f ric = 3000 J−1610 J−908.6 J = 1380.0 J12

mv2f = 1380.0 J

→ v =

√2(1380)

35.0= 8.9 m/s

16. (a)

W = ∆U =12

kx2f −

12

kx2i

W =12(350 N/m)(0.03 m)2−0 = 0.158 J

→W = ∆K =12

mv2f −

12

mv2i =

12

mv2f −0

⇒ v f =

√2Wm

=

√2(0.158)

5.0= 0.08 m/s

(b)

W = ∆U =12

kx2f −

12

kx2i

W =12(350 N/m)(0.03 m)2−0 = 0.16 J

→W = ∆K + fkd =12

mv2f −

12

mv2i + fkd =

12

mv2 f −0+ fkd

⇒ v f =

√2(W − fkd)

m=

√2(0.16− (0.1)(5.0)(9.8)(0.03))

5.0= 0.07 m/s

17. (a) W = Fd cosθ = (85 N)(2.5 m)cos15 = 205 J(b) The work done by the normal force is zero since the force and the displacement are perpendicular to one

another.(c) W = Fd cos(θ+90) = 35.0 kg(9.8 m/s2)cos105 =−88 J

18. At the apex, the y-component of the velocity will be equal to zero.⇒ vx = vcosθ = (52.0 m/s)cos40 =39.8 m/sK = 1

2 mv2 = 12(1.0 kg)(39.8 m/s)2 = 792 J

19.

3 f t.= 0.91 m

Let d′ = (0.91+0.007) m and d = 0.007 m

∑W = ∆K = K f −Ki

Wgravity +Wtable = K f −Ki

(mg)d′ cos0+Ftabled cos180 = 0

Ftable =mgd′

d=

(25 kg)(9.8 m/s2)(0.917)0.007

= (3.2×105 N)y

20. For each mass, the outward force that is experienced is:|F |= k∆xSo the acceleration on each block is simply:∑F1 = m1a1x

⇒ a1x =k∆xm1

to the leftSimilarly for m2:a2x =

k∆xm2

to the right

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21. For each mass, the outward force that is experienced is:

|F |= k∆x

∑F1 = m1a1x

−kx+µkmg = m1a1x

⇒ a1x =k∆x−µkm1g

m1to the left

Similarly for m2:

a2x =k∆x−µkm2g

m2to the right

22. P =(

F2

m

)t = (75 N)2

45.36 kg 1.5 s = 186 watts23. (a)

∑W = ∆K

Wspring +Wgravity =12

k∆x2 =−mgd cos(θ+90)

d =12

k∆x2

mgsin45=

12

(1200 N/m)(.3 m)2

(2.0 kg)(9.8 m/s2)sin45= 3.9 m

(b)

∑W = ∆K +∆E f ric

Wspring +Wgravity = ∆E f ric

12

k∆x2 =−mgd cos(θ+90)+µkmgd cos45

12(1200)(0.30)2 =+(2.0)(9.8)d sin45+(0.22)(2.0)(9.8)d cos45

54 = 13.86 d +3.05d

⇒ d = 3.68 m

24. Since the kinetic energy is proportional to the square of the speed and the work is proportional to the squareof the displacement, doubling the compression doubles the velocity.

25. Yes, work is being done while your foot is in contact with the ball (W = Fd).26. Another example is applying a force to an object that does not budge.27. Each of the small springs would require three times the force to stretch them a distance x compared to the

original spring. So the relationship between the new springs and the original spring is: knew = 3kold .28. It is impossible for the kinetic energy to be negative. The only way this would be possible is if an object had

negative mass, which it cannot.29. Kinetic energy is proportional to the mass and proportional to the square of the speed. Therefore, the more

massive object will have twice as much kinetic energy.30. If the speed is doubled, the kinetic energy would be increased by a factor of 4.31. W = F ·d = (13 N)(3.5 m) = 45.5 J32. (a)

W = ∆K

F ·d = ∆K

∆K = (13 N)(4.0 m) = 52.0 J

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(b)

∆K = 52.0 J12

mv2 = 52.0

→ v =

√2(52)

7= 3.85 m/s

33. (a) W = F ·d = (13 N)(3.5 m) = 45 J(b) Since the system is moving at constant velocity, the forces in both the x- and y-directions are balanced.

∑F = 0

⇒ F− f = 0

F−µkN = 0

µk =FN

=Fapplied

mg=

13 N(7 kg)(9.8 m/s2)

= 0.19

(c) W = f d cosθ = (0.19)(7 kg)(9.8 m/s2)(3.5 m)cos180 =−45 J34. W = Fd cosθ = (75 N)(3 m)cos30 = 194.9 J35.

∑W = ∆K

∆E f ric =12

mv2o

f d =12

mv2o

→ f = µkmg

µk =12

mv2o

mg=

12

(2 m/s)2

(9.8 m/s2)(2 m)= 0.10

∑W = ∆K

Wgrav +Wf ric = K f inal

mgh sinθ+µNh =12

mv2f

h =y

sinθ

mg( y

sinθ

)sinθ+µ(mgcosθ)

( ysinθ

)=

12

mv2f

→ mgy+µ(

mg1

tanθ

)y =

12

mv2f

mgy(

1− µtanθ

)=

12

mv2f

⇒ v f =

√2gy(

1− µtanθ

)For this it is assumed that tanθ > µ for the object to slip.

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36. The maximum velocity will occur when the pendulum makes an angle of 90 with respect to the horizontal or0 with respect to the vertical. Setting the reference point of the potential at the lowest portion that the massreaches:

∆E = 0

E f = Ei

⇒ K f =Ui

12

mv2 = mgl(1− cosθ)

v =√

2gl(1− cosθ)

37. (a) U = 12 kx2 = 1

2(120 N/m)(1.5 m)2 = 135 J(b)

K f =Ui

12

mv2 =12

kx2

→ v =

√km

x2 =

√1202

(1.5)2 = 11.6 m/s

38. Using the conservation of energy:

∆E = 0

E f = Ei

K f +U f = Ki +Ui

The initial kinetic energy is zero since the system is starting from rest.

Ui = K f +U f

m′gh =12

mv2 +12

m′v2 +mghsinθ

Since both blocks are connected, they move at the same velocity:

m′gh =12(m+m′)v2 +mghsinθ

⇒ v =

√2gh(m′−msinθ)

(m+m′)

39.

∆W = ∆E

Wf riction = E f −Ei

− f ∆x =12

k∆x2− 12

mv2

−(µmg)∆x =12

k∆x2− 12

mv2

⇒ µ =mv2− k∆x2

2mg∆x=

(2 kg)(7 m/s)2− (5 N/m)(3 m)2

2(2 kg)(9.8 m/s2)(3 m)= 0.45

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40. The system is closed and there is no losses due to friction. At the top, the ball only has potential energy. Atthe bottom, it only has kinetic energy.

∆E = E f −Ei = 0

E f = Ei

12

mv2 = mgh

v =√

2gh =√

2(9.8 m/s2)(100 m) = 44.3 m/s

41. (c) It can be either positive or negative depending on where the potential is set to zero in the problem.42. (a) The system is closed and there is no losses due to friction. At the top, the ball only has potential energy.

At the point of interest it has kinetic and potential energy.

∆E = E f −Ei = 0

E f = Ei

12

mv2 +mgy′ = mgh

v =√

2g(h− y′)

(b)

∆E = E f −Ei = 0

E f = Ei

12

mv2 +mgy′ = mgh+12

mv2o

v =√

2g(h− y′)+ v2o

43.

E f ric = ∆K +∆U

− f d = K f −Ki +U f −Ui

− f ∆x =12

mv2−mgh

v =

√2(mg∆xsinθ− f ∆x)

m=

√2∆x(mgsinθ− f )

m

v =

√2(1.5 m)[(25 kg)(9.8 m/s2)sin34−6.5 N]

(25 kg)= 3.96 m/s

44. Since there is no friction, the final speeds would be the same.45. The object heading down the longer slide would have a smaller velocity because the distance allows friction

to act on the object for a longer period of time.46. Kinetic energy is always positive due to the velocity squared term, while the potential can be negative

depending on where the zero potential reference point is placed. Since the total energy is the sum of thetwo, it is possible to have a negative total energy.

47. Let the potential energy of the child equal zero when the rope is hanging straight down, θ = 0

(a)

@θ = 25

U = mgl(1− cosθ) = (25 kg)(9.8 m/s2)(2.0 m)(1− cos25)

U = 46 J

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(b)

@0

U = mgl(1− cosθ) = (25 kg)(9.8 m/s2)(2.0 m)(1− cos0)

U = 0 J

48.

E f = Ei

mg(2R)+12

mv2 = mg(3R)

V =√

2gR

49. (a)

∆E = 0

E f = Ei

12

mv2i,x +

12

mv2i,y =

12

mv2f ,x +

12

mv2f ,y +mgh

The velocity in the x-direction is unchanged and the velocity in the y-direction is zero at the highestpoint.⇒ vi,y =

√2gh =

√2(9.8)(10) = 14 m/s

(b) The velocity in the x-direction is unchanged, while the y-component is being acted upon by gravity. So,using the work-kinetic energy theorem:

W = ∆K

mghtotal =12

mv2f −

12

mv2i

→ v f =√

2gytotal + v2i,y =

√2(9.8 m/s2)(14.0 m)+(14 m/s)2 = 21.7 m/s

50. Before we can determine the highest point reached by the lighter mass, we need to figure out the velocity form2 right before m1 hits the ground.

∆E = 0

K f +U f = Ki +Ui

12

m1v2f +

12

m2v2f +m2gh = m1gh

12(m1 +m2)v2 +m2gh = m1gh

v =

√2(m1−m2)gh(m1 +m2)

Now that the velocity right before m1 hits the ground has been solved for, this velocity (which is also the samevelocity for m2, can be used to see how high m2 reaches.

∆Em2 = 012

m2v2 = m2gh′

h′ =v2

2g=

2(m1−m2)gh(m1 +m2)2g

=(m1−m2)h(m1 +m2)

htotal = h+h′

⇒ htotal = h+(m1−m2)h(m1 +m2)

= h(

1+(m1−m2)

(m1 +m2)

)

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51. (a)

@Point 1 :

W = ∆K = K f −Ki

mg∆y =12

mv21

v1 =√

2g∆y =√

2(9.8 m/s2)(15.0 m−5.0 m) = 14 m/s

(b)

@Point 2 :

W = ∆K = K f −Ki

mg∆y =12

mv21

v1 =√

2g∆y =√

2(9.8 m/s2)(15.0−10.0) m = 9.90 m/s

(c) W = mg∆y = (35.0 kg)(9.8 m/s2)(5.0 m) = 1715 J52. The mass needs to be hit with enough energy so that it just reaches the top of the circular path that it will trace

out.Ki +Ui = K f +U f12 mv2 +0 = 0+mg(2l)v =

√4gl

53. ∆E = 0Ui =U f

mg(l sinθ) =12

k∆x2

⇒ ∆x =

√2mgl sinθ

k

54.

180 lb = 81.6 kg

∆E = 0

Ui =U f

12

k∆x2 = mgh

h =12

k∆x2

mg=

12(3500 N/m)(0.5 m)2

(81.6 kg)(9.8 m/s2)= 0.547 m

55.

∆E = 0

U f +K f =Ui +Ki +∆E f ric

12

m1v2 +12

m2v2 = m2g∆y+0− f ∆x

Since the system is connected, ∆x = ∆y and f = µkm1g.

v =

√2g∆y(m2−µkm1)

(m1 +m2)

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56. (a) ∆PE = mgh = (40.0 kg)(9.8 m/s2)(1.5 msin15) = 152 J(b) ∆K = 1/2m2v2

2−1/2m1v21 = 0−1/2(40 kg)(5 m/s2) =−500 J.

(c) E f ric = ∆K +∆U = 152 J−500 J =−348 JE f ric = F ·d = µmgd cosθ

µ =−E f ric

mgd cos(180−θ)=

−348 J(40.0 kg)(9.8 m/s2)(1.5 m)(cos15)

= 0.613

57.

f = µkN = µk(mgcosθ)

∆E = 0

E f ric = ∆Um1 +∆Km1 +∆Um2 +∆Km2

E f ric = m1g(25.0sin32.0)+12

m2(v2f − v2

i )+m2g(−25.0)+12

m2(v2f − v2

i )

⇒ 12

m2(v2f − v2

i ) =

(m1

m1

)12

m2(v2f − v2

i ) =m2

m1∆Km1

⇒ ∆Km1 = E f ric−∆Um1−∆Um2−−9575

∆Km1

∆Km1 +9575

∆Km1 = (−.33)(75)(9.8)cos32(25)− (75)(9.8)(13.25)− (95)(9.8)(−25)

∆Km1 = 3.70 kJ

58.

∆E = 0

Ui =U f

mgh =12

k∆x2

∆x =

√2mgh

k=

√2(9.8 m/s2)(3.0 m)

2500 N/m= 1.37 m

59. Looking at the total energy of the system:

∆E = 0

Ui =U f +E f ric

mgh =12

k∆x2 +µkmgx

⇒ µk =mgh− 1

2 k∆x2

mgx

60. The correct statement would be: “No work is done when the force is perpendicular to the displacement.”61. The force due to the books being carried is in the negative y-direction (downward due to gravity), the dis-

placement (the direction your friend is walking is the x-direction). Since the force and the displacement areperpendicular to one another, the work done is zero.

62. Kinetic energy is energy due to motion while the gravitational potential energy is the energy due to position.63. Yes, Billy is wrong. The potential energy is converted to kinetic energy.64. Zero; the overall work done cancels out.65. The kinetic energy becomes 9 times larger.66. Neither; both require the same amount of work since work is path independent, i.e. all that matters is the

starting and ending points.

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67. (a) v =√

v2o +2a∆x =

√(2.4 m/s)2 +2(2 m/s2)(5.0 m) = 5.08 m/s

(b) ∆K = K f −Ki =12 m(v2

f − v2i ) =

12 m(5.082−2.42) = 5.01×103 J

68. There are two constraints:1) Ki,A = 1

2 KB

2) K f ,A = KB

Therefore, Ki,A = 12 KB.

12

mv2 =12

[12

m(v+2.3 m/s)2]

→ v2− 12

v2−2.3 v−2.65 = 0

⇒ vA = 5.6 m/s

⇒ vB = 11.2 m/s

69.

Wx = (30 N)10 m = 300 J

Wy = (50 N)7 m = 350 J

W =√

3002 +3502 = 461 J

70. W = ∆K = 12 m(v2

f − v2i ) = (0.5)(0.5)(12−3.32)−2.47 J

71. W = Fx∆xcosθ = (1.5 N)cos135(2.5) =−2.65 J72.

W = Fd cosθ

W1 = (200 N)(2.0 m)cos235 =−230 J

W2 = (150 N)(2.0 m)cos180 =−300 J

W3 = (300 N)(2.0 m)cos0 = 600 J

W4 = (50 N)(2.0 m)cos45 = 71 J

73.

W = Fd cosθ

Wnet =W1 +W2 +W3

W1 = (2.0 N)(4.5 m)cos270 = 0 J

W2 = (4.0 N)(4.5 m)cos180 =−18.0 J

W3 = (9.5 N)(4.5 m)cos45 = 30.2 J

Wtotal = 2 =−18.0 J+30.2 J = 12.2 J

74. (a) ∆K =W = Fd = (20 kg)(9.8+1.0 m/s2)(20 m) = 4320 J(b) W =−Fd =−(20 kg)(9.8 m/s2)(20 m) = 3920 J

(c) 12 mv2 = 4320 J−3920 J⇒ v =

√800 J20 kg

= 20.8 m/s

75.

W =Wgravity−Wf riction

→Wg =W +Wf ric = 75 J+35 J = 110 J

76. (a) W = Fd = 15 mgd

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(b) The kinetic energy is the same, 1/5mgd.77.

W1 = ∆PE1 +∆K1 = mgh+12

m(v22− v2

1) = (1000 kg)(9.8 m/s2)(15.0 m)+12(1000 kg)((0 m/s)2− (5.5 m/s)2) = 162kJ

W2 = ∆PE2 +∆K2 = mgh+0 = (1000 kg)(9.8 m/s2)(10.0 m) = 98kJ

W3 = ∆PE3 +∆K3 = mgh+12

m(v22− v2

1) = (1000 kg)(9.8 m/s2)(15.0 m)+12(1000 kg)((5.5 m/s)2− (0 m/s)2) = 132kJ

78. W = 12

F∆x x2 +F ·d = 1

225 N

(0.03 m)(0.02 m)2 +(250 N)(0.02 m) = 0.16 J

79. W = 12 kx2 = 1

2(55 N/m)(3.75)2 = 70.9 kJ80. W = Fd = (6.0 N)(1.0 m) = 6.0 J81. P =W/t = (750 J)

(10.0 m)/(1.2 m/s) = 90 W82. P = Fvcosθ = (100 N)(7.0 m/s)(1) = 700 W83. P = Fv = (1000 kg)(9.8 m/s)

(200 m45 s

)= 4.36×104 W

84. W = Fd cosθ = (222 N)(1.3 m/s(300 s))cos35 = 319.5 J85. W = Fd cosθ = (175 N)(2.0 m) = 350 J86. K = 1

2 mv2 = 12(4448 N)(24.59 m/s)2 = 1.34×106 J

87. W = Fd = (8.0 N)(5.5 m) = 44 J88.

W = Fd

→ F = f = µkN = µk(mg)

W = µk(mg)∆x

µk =F∆x

mg∆x=

Fmg

89. v =

√2Km

=

√2(320 J)

22 kg= 5.40 m/s

90. W = Fd = 25.0 J

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6.2 References

1. Haven Giguere and Laura Guerin. CK-12 Foundation .2. Haven Giguere and Laura Guerin. CK-12 Foundation .3. Haven Giguere and Laura Guerin. CK-12 Foundation .4. Haven Giguere and Laura Guerin. CK-12 Foundation .5. Haven Giguere and Laura Guerin. CK-12 Foundation .6. Haven Giguere. CK-12 Foundation .7. Haven Giguere and Laura Guerin. CK-12 Foundation .8. Haven Giguere. CK-12 Foundation .9. Haven Giguere and Laura Guerin. CK-12 Foundation .

10. Haven Giguere and Laura Guerin. CK-12 Foundation .11. Haven Giguere and Laura Guerin. CK-12 Foundation .12. Haven Giguere. CK-12 Foundation .13. Haven Giguere and Laura Guerin. CK-12 Foundation .14. Haven Giguere and Laura Guerin. CK-12 Foundation .

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www.ck12.org Chapter 7. Momentum Problem Sets

CHAPTER 7 Momentum Problem SetsChapter Outline

7.1 MOMENTUM PROBLEM SETS

7.2 REFERENCES

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7.1. Momentum Problem Sets www.ck12.org

7.1 Momentum Problem Sets

Problems

1. Which has a greater momentum: a car traveling at constant velocity down the road or an M1 Abrams tank atrest?

2. Explain the difference between impulse and force.3. Give two ways to increase the impulse of an object.4. Your friend tries to tell you that impulse and time are inversely related. Explain why your friend is incorrect.5. While trying to help with your physics homework, your dad says, “Impulse is not related to Newton’s Second

Law.” How would you correct him?6. How would you maximize an object’s momentum? Explain in terms of impulse and momentum.7. During the board-breaking competitions, competitors line up to see who can break the most boards in one

minute with their heads. What advantage might a contestant have if he can decrease the amount of time he isapplying to the boards?

8. Combat martial artists are taught to “roll with the punches.” Why is this?9. “There is no change in momentum when the impulse is zero.” Is this statement correct? If not, explain why.

10. Physics students commonly hear the phrase, “Assume momentum is conserved.” Explain what this means?11. What are the differences between elastic and inelastic collisions?12. Two balls of equal mass are sitting on a table. Ball 1 rolls towards ball 2 with an initial velocity v0. After the

collision, ball 1 is stationary, but ball 2 has a velocity v. Ignoring the effects of friction, what can be said aboutthe velocity of ball 2?

13. Two balls of equal mass are sitting on a table. Ball 1 rolls towards ball 2 with an initial velocity v0. Uponcolliding, ball 1 sticks to ball 2, and they roll with a velocity v in the direction of ball 1’s original velocity.Ignoring the effects of friction, what can be said about the velocity of the two-ball system?

14. Calculate the momentum of a 0.5 kg cup sliding across a table at 4 m/s.15. A box of stuffed animals slides across the floor at a speed of 3 m/s. If the mass of the box is 2.5 kg, calculate

the momentum of the sliding box.16. A 2 kg ball that rolls across the floor has a momentum of 6 kg·m/s. What is the ball’s velocity?17. You push against a refrigerator with a force of 150 N for 5 s. Calculate the impulse.18. A 0.5 kg cup with initial velocity 4.0 m/s slides across a table until it comes to a stop. Calculate its impulse.19. If a 0.5 kg cup with an initial velocity of 4 m/s came a stop after being in contact with your hand for 0.5 ms,

calculate the force experienced by the cup.20. Calculate the impulse required to stop a 25 kg object that has an initial velocity of 3 m/s.21. Calculate the time it would take a force of 150 N applied in the opposing direction to bring a 25 kg object with

an initial velocity of 3 m/s to a stop.22. A 1.5 kg block with a needle on one side slides towards a motionless block of the same mass. If the initial

velocity of the moving block is 4 m/s, calculate the final speed of the two blocks after they collide and sticktogether.

23. A 1.5 kg block with a needle on one side slides towards a motionless block. The initial velocity of the movingblock is 4 m/s.

(a) If the mass on the block at rest is 5 kg, calculate the final speed of the blocks.(b) If the block with the needle has a mass of 5 kg and the motionless block has a mass of 1.5 kg, calculate

the final speed of the blocks.

24. Two blocks of mass m1 and m2 are sitting on a frictionless surface. If block m1 is given an initial velocity ofv0 towards m2, which has an initial velocity of 0 m/s. What is the final velocity of the two-block system if the

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collision is completely inelastic?25. Every year, there are approximately 6 million car accidents. How do safety devices such as air bags help

reduce potential injuries during crashes?26. Rock climbers attach themselves to a rope that stretches a considerable amount. Explain how this might be

advantageous as opposed to using an inelastic rope.27. While preparing to go bungee jumping with a friend, you notice that he has packed a steel cable for the

jump. How would you explain to him that using a steel cable for bungee jumping is extremely dangerous andpotentially fatal?

28. In a game of tag with your friends, you prepare to jump off a ledge to avoid getting tagged. Looking below,you spy a sandbox and concrete lot. Which material would you rather land on? Explain your reasoning.

29. When the modern car crashes, the hood of the car, containing the crush zone, will intentionally crumple anddeform. How is this beneficial to the passengers inside?

30. A 2 kg brick and a 2 kg marshmallow are simultaneously thrown at you. If you only have enough time to onlydodge one of them, which would you dodge and why?

31. While drifting in deep space, an astronaut notices he’s out of jetpack fuel. What can the astronaut do to returnto the shuttle?

32. While you’re discussing physics with your friend Allen during lunch, he tells you that there is no relationshipbetween Newton’s Third Law and the conservation of momentum. Is his statement correct? If not, how wouldyou correct Allen?

33. In the previous chapters you dealt with objects sliding down inclines at various angles. As the object rollsdown the incline, does it gain momentum? If so, which force is responsible?

34. Two cars slide towards each other, which results in an elastic collision. If one of the cars has twice the mass,which object has a greater change in momentum?

35. Two cars slide towards each other, which results in an inelastic collision. Would an inelastic collision, ratherthan an elastic collision, be safer for those involved? Why?

36. A 25 kg crate comes to a stop in 2 s after sliding across a rough surface. If the initial velocity of the crate was3.0 m/s, what is the frictional force that acted upon the crate to bring it to a stop?

37. A 25 kg crate sliding across a frictionless surface with an initial velocity of 3 m/s comes to a stop by cominginto contact with a wall. If the crate stops in 0.05 ms, calculate the force that was exerted on the crate by thewall.

38. While racing on a go-kart track, you slam into the track’s walls at 15 m/s to avoid hitting the wandering cat.If you are brought to a stop in 0.075 s, what is the average force that a 35 kg body would feel?

39. If a 14 kN force was applied in 0.075 s to bring you to a stop from 15 m/s, what is your mass?40. Calculate the impulse necessary to stop a 0.2 kg object that travels at 30 m/s.41. Determine the amount of force needed to stop a 0.35 kg object traveling at 35 m/s in 0.02 s.42. While standing on a frictionless patch of ice, Jess catches a 7 kg bag of groceries that was tossed at her at 5.0

m/s. If Jess has a weight of 35 kg, what is her speed after she catches the groceries?43. A 15 kg ball of wax with a velocity of 4.0 m/s experiences a completely inelastic collision with a stationary 3

kg ball of wax. What is their velocity after the collision?44. As a dinosaur slides along a frozen river, it eats berries from the bushes by the riverside. The dinosaur has a

mass of 10 kg, and the fruit that it consumes weighs 1.5 kg. What is the velocity of our dinosaur after eatingthe fruit, if its initial speed is 1.5 m/s?

45. While watching the latest Hollywood superhero movie, you see the star of the movie jumps vertically in theair while holding a car and throw it at his enemies. If the car is 10 times heavier than the hero and thrown witha velocity of 45 m/s, calculate the hero’s velocity if the director wanted the movie to be physically accurate.

46. If two objects have the same kinetic energy, what can be said about their momentum?47. Two objects resting on a frictionless surface are each pushed by a force of equal magnitude over a given

distance. If the only difference between the two is that one is 3 times the mass of the other, what can be saidabout their kinetic energy and their momentum?

48. To achieve the largest impulse possible, the largest force possible needs to be applied. Is this statement correct?If not, explain why it is wrong.

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49. Explain how doubling the velocity of an object affects its momentum and kinetic energy?50. While sliding across a frictionless surface, two crates have equal kinetic energy. Can the same thing be said

about their momentum?51. While watching the latest Hollywood superhero movie, you see the star of the movie jump vertically and throw

a car at his enemies. Upon releasing the car, the hero lands in the same spot he jumped from. How was physicsignored in this scenario?

52. When a bouncy ball bounces off a wall, the wall does not move back. How is this possible if momentum isconserved?

53. When two objects experience an inelastic collision and travel in the same direction, how can you determinewhich object experiences a larger change in kinetic energy?

54. Give an example how is it possible for an object can have the center of mass located in a position where thereis no mass present?

55. A 2 kg object has a velocity of 75 m/s in the x-direction and 45 m/s in the y-direction. Calculate the magnitudeand direction of its momentum.

56. Kinetic energy can also be expressed in terms of momentum. Derive the expression that defines kinetic energyas a function of momentum.

57. You press two blocks that are separated by a spring together with your hands. One block has a mass ofm1 = 0.5 kg and the other has a mass of m2 = 2.0 kg. When you let go of the two blocks, m2 moves with avelocity of 3.5 m/s. Calculate the speed of m1.

58. Many people believe that holding onto a child is safe practice in the event of a plane crash. If you’re holdingonto an 8 kg child during a 100 mph crash that brings you to a stop in 0.5 s, calculate the force that will be onyour arms.

59. While playing racket ball, you hit the 0.04 kg ball flying at you at 65 m/s back to the opponent with a speedof 50 m/s. Calculate the impulse you delivered to the ball.

60. An 15 kg object moving at a speed of 5.5 m/s collides and sticks to a 60 kg object that is traveling in the samedirection at a velocity of 3.0 m/s. Determine the speed of the two objects after the collision.

61. As shown in the following diagram, you press two blocks (m1 = 0.5 kg and m2 = 2.0 kg) against each otherwhile a spring is in the middle. When you let go of the two blocks, m2 moves with a velocity of 3.5 m/s. Ifthe surface the blocks are resting on is frictionless, calculate the height m1 reaches if than angle of the inclineis 55°.

62. Alex claims that his 450 kg car is sturdier than his identical twin brother Brian’s 1350 kg truck. The twodecide to drive at each other at 10 m/s to see if this was true. The completely inelastic collision lasts 0.1 s. Ifeach twin weighs 90.7 kg, calculate the force on each twin.

63. A bullet of mass mb and initially velocity v is fired at a block of wood of mass m that is suspended by string ofnegligible mass whose length is l. If the bullet passes through the block and exits with one third of its initialvelocity, calculate the initial velocity of bullet that causes the block to rotate through one revolution?

64. A piece of gum (mg = 0.5 g) is thrown at a hollow wood block of mass 8 g. The gum sticks to the block andthey both move to the right a total distance of 15 cm. If you’re told the coefficient of kinetic friction betweenthe block and the table is 0.55, what horizontal velocity was the gum thrown with?

65. A 500 kg sedan is heading due west at 24 m/s as it approaches an intersection. Another 600 kg car, and

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velocity 26 m/s heading due north fails to see its stop sign and enters the intersection the same time as the firstcar. If the two vehicles experience a perfectly inelastic collision, calculate the speed and direction of the twovehicles after the collision.

66. A 500 kg sedan is heading due west at 24 m/s as it approaches an intersection. Another 600 kg car, andvelocity 26 m/s heading due north fails to see its stop sign and enters the intersection the same time as the firstcar. If the two vehicles experience a perfectly inelastic collision, determine the amount of energy that was lostduring the crash. Where could have this energy went?

67. A 500 kg sedan is heading due west at 24.0 m/s as it approaches an intersection. Another car of mass 600 kg,and velocity v2 m/s heading due north fails to see its stop sign and enters the intersection the same time as thefirst car. If the two vehicles experience a perfectly inelastic collision, and slide off in a direction that is 45°north of west, what was the initial velocity of the second car?

68. Two blocks of the same mass experience an elastic collision after which the first block has a final velocity ofv1 f = 3.3 m/s and a direction of 42° with respect to the x-axis. Ignoring frictional effects, calculate v2 f if theinitial velocities are v1i = 4.0 m/s x and v2i = 0 m/s.

69. What is the center of mass of a system composed of three objects with the following locations and masses (Table 7.1).

TABLE 7.1:

Mass (kg) x (m)2.0 4.01.0 -5.02.5 2.0

70. Calculate the center of mass of the system of masses in the diagram below, assuming each mass is 2 kg.

71. A two-particle system with masses m1 = 4.0 kg and m2 = 2.0 kg has velocities ~v1 = (1.0x− 2.0y) m/sand~v2 = (2.0x−1.0y) m/s. Determine the velocity of the center of mass.

72. In some action movies the hero is seen jumping onto a stationary cart at a large velocity to escape enemies. Ifour 80 kg hero jumps onto a 150 kg cart with an initial horizontal velocity of 5 m/s, calculate the hero-platformsystem’s final velocity?

73. Vikram and his fiend Jim decide to play catch on a frozen lake, where there is negligible friction between theirfeet and the ice. Vikram tosses Jim the 1 kg ball at 10 m/s. If Jim has a mass of 70 kg, what speed does Jimand the ball have after he catches the ball?

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74. Vikram and his fiend Jim decide to play catch on a frozen lake, where there is negligible friction between theirfeet and the ice. Vikram tosses Jim the 1 kg ball at 10 m/s, but it hits his chest and bounces off. After thecollision, the ball moves horizontally at a speed of 8.5 m/s in the opposite direction. Calculate Jim’s speedafter impact.

75. If an object is thrown vertically downward off a cliff of height h with an initial velocity of v0 = 15 m/s,calculate the object’s velocity 5 s later? (Use the relationship between impulse and momentum)

76. You classmate throws a 0.20 kg rubber ball at you with a horizontal velocity of 15 m/s. Using your physicsbook, you send it back at him with a velocity of 25 m/s. Assuming the ball was in contact with your book for10 ms, calculate the average force exerted on the ball.

77. Upset because you didn’t complete your physics homework in time, you kick a trash so hard that slides 5m at5.5 m/s for 2.0 s. Calculate the coefficient of kinetic friction using the equation for impulse.

78. While checking your text messages while driving, you slam your 1200 lb car into a large dumpster that weighs800 kg. If your initial velocity was 35 mph and the collision was completely inelastic, what is your velocityafter the collision?

79. Despite driving at 35 mph, you decide to check your text messages and fail to pay attention of the 800 kgdumpster in front of you. After the elastic collision between your 1200 lb car and dumpster, your car has afinal velocity of 2 m/s. Determine the final velocity of each object you slam your 1200 lb car into a largedumpster that weighs 800 kg. If your initial velocity was 35 mph and final velocity was 2 m/s, determine thedumpster’s velocity after the crash?

80. Block mA moves to the right and block mB moves to the left. They experience a head on collision as shownin the diagram below. If the collision is elastic and mA = 4 kg, vA = 2.5 m/s, mB = 4 kg, and vB = 3 m/s,calculate the final velocity of the two-mass system.

81. Block mA moves to the right and block mB moves to the left. They experience a head on collision as shownin the diagram below. If the collision is completely inelastic and mA = 4 kg, vA = 2.5 m/s, mB = 4 kg, andvB = 3.0 m/s, calculate the final velocity of the two-mass system.

82. Block mA moves to the right and block mB moves to the left. They experience a head on collision as shownin the diagram below. If the collision is elastic and mA = 50 kg, vA = 25 m/s, mB = 25 kg, and vB = 25 m/s,calculate the final velocity of the two-mass system.

83. Block mA moves to the right and block mB moves to the left. They experience a head on collision as shown inthe diagram below. If the collision is elastic and mA = 100 kg, vA = 15 m/s, mB = 425 kg, and vB = 0 m/s,calculate the final velocity of the two-mass system.

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84. A 150 kg object slides along a frictionless surface into a 400 kg object at rest. Calculate the initial velocity ofthe 150 kg object if the final velocities after the elastic collision is v150 f =−0.5 m/s, v400 f = 0.6 m/s

85. While chasing your brother around the house at a velocity of 5 m/s, you accidentally run head first into acabinet that weighs the same as you. If you and the cabinet both slide together for 1 m, calculate the coefficientof kinetic friction between you and the ground.

86. A bullet of mass m is fired into pendulum of mass M and length l. Derive an expression for the distancetraveled by the pendulum after the collision.

87. Your employer assigned you to calculate the center of mass of some composite objects before they go into thebuilding process. Instead of having to calculate the information over from scratch each time, derive a generalexpression for the locations of the center of mass for both the x and y coordinates.

88. A 10 g bullet is fired into a 5 kg pendulum causing it to rise to a height of 0.2 m. Calculate the initial velocityof the bullet.

89. Calculate the center of mass using the following coordinates and masses ( Table 7.2).

TABLE 7.2:

Mass (kg) x (m) y (m)2.0 2.0 1.55.0 0.0 -3.04.0 -3.0 2.0

90. Given the following coordinates and masses ( Table 7.3), calculate the missing coordinates if the center ofmass is located at (1.0, -1.0).

TABLE 7.3:

Mass (kg) x (m) y (m)5.0 2.0 1.55.0 0.0 -3.05.0 ? ?

91. In the following diagram, two uniform rods that are 10 cm long are in the shape of an upside down "L."Calculate the x and y coordinates of the system’s center of mass.

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92. Jumping down from a ledge that is 0.5 m high, your 100 kg body collides with the floor for 50 ms before beingbrought to a complete stop. What is the magnitude of the impulse that acts on you?

93. A 1 kg ball is thrown at a wall with a velocity of magnitude 4.5 m/s. Before coming into contact with the wall,the ball has an angle of 25° with respect to the negative x-axis. Upon a 5 ms impact, the ball leaves the wallat the same angle and velocity but in the opposite direction. Calculate the components of the average force onthe wall from the rubber ball.

94. While standing on a surface of negligible friction, an 80 kg individual throws a 50 kg pack of equipmenthorizontally outward at a speed of 5 m/s. What speed does the individual have after throwing the equipment?

95. Determine the mass of an object that has a velocity of 10 m/s and a momentum of 32 N-s.96. A bullet is fired with an initial velocity of 25 m/s into a ballistic pendulum of mass 1.5 kg and is embedded

inside. If the pendulum then rises to a height of 1.2 m, calculate the mass of the bullet.97. A 4.0 kg bullet is fired with an initial velocity of 500 m/s into a 400 kg semisoft plastic block. The bullet exits

the block traveling in the same direction with an exit velocity of 365 m/s. Calculate the speed of the block.98. Bored in class, you throw your freshly sharpened pencil into the ceiling tiles above you. However, you threw

the pencil too hard and it causes the tile to raise up and fall back down into place. Even though your teachersees this, he offers to not punish you if you can calculate the height at which the tile raised up. If the pencilhas mass m and the tile has mass M, calculate the height the tile would raise up if the pencil was given aninitial velocity of v0.

99. A 1 kg pendulum is attached to a stiff rod that is 1 m long and held horizontally before being released. Directlyunder the pendulum’s attachment point is a 3 kg block resting on a frictionless surface. If the pendulum andthe block experience an elastic collision, what is the speed of the block and pendulum after they collide?

100. Explain why it is easier to stop a 2 kg crate than a 10 kg crate if both are sliding across a frictionless surfaceand have the same speed.

101. If a 10 kg crate and a 5 kg crate have the same momentum, which one has greater kinetic energy? Explainhow you came to your answer.

102. When the velocity of an object in motion is increased, how does the momentum of the object change? Howdoes the kinetic energy of the object change?

103. Your friend bets you $20 that it is impossible for an object to have kinetic energy without having momentum.Should you take the bet?

104. If the momentum of an object is zero, is the energy of that object zero?

Solutions

1. Since momentum is mass times velocity, the tank at rest would not have any momentum. Therefore, the carhas more momentum than the tank.

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2. Impulse is the change in momentum. Force is the rate of change of momentum.3. The impulse of an object can be increased by either increasing the force that is changing the momentum or

increasing the time for which the force is applied.4. If impulse and time were inversely related, than simply touching the front of a speeding car would bring it to

a dead stop as opposed to applying a force to the front of the car for an extended period of time.5. If you apply a force to a mass, it will accelerate an object. How long the force is applied also changes the

momentum of that object. So impulse and momentum are related to Newton’s second law.6. You would apply the largest force possible for the longest amount of time.7. By decreasing the amount of time their fists are in contact with the boards, the practitioners are able to generate

more force on the boards.8. By being in contact with the incoming attack, you are decreasing the amount of force that is delivered.9. Yes, this statement is correct.

10. Conservation of momentum implies that the total momentum in a system before an event (i.e. a collision) isthe same as the total momentum after a collision.

11. In elastic collisions, both kinetic energy and momentum are conserved. For completely inelastic collisions,only momentum is conserved.

12. Ball 2 has the same velocity that ball 1 had before the collision.13. The two-ball system has half the initial velocity that ball 1 had before the collision.14. p = mv = (0.5 kg)(4 m/s) = 2 N · s15. p = mv = (2.5 kg)(3 m/s) = 7.5 N · s16. m = p

v = 6 kg·m/s2 kg = 3 m/s

17. I = Ft = (150 N)(5.0 s) = 750 N · s18. I = ∆p = p f − pi = 0− (0.5 kg)(4.0 m/s) = 2.0 N · s19.

I = ∆p = p f − pi = 0− (0.5 kg)(4.0 m/s) = 2.0 N · s∆p = 2.0 N · s∆p = Ft

F =∆pt

=(2.0 N · s)

(0.5×10−3s)= 4000 N

20. I = ∆p = 0− p0 =−(25 kg)(3.0 m/s) =−75 N · s21.

I = ∆p = 0− p0 =−(25 kg)(3.0 m/s) =−75 N · s∆p = Ft

t =∆pF

=(−75 N · s)(−150 N)

= 0.5 s

22.

Pi = p f

m(4 m/s) = 2mv f

v f = 2 m/s

23. (a)

pi = p f

(1.5 kg)(4 m/s) = (1.5 kg+5.0 kg)v f

v f = 0.9 m/s

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(b)

(5.0 kg)(4 m/s) = (1.5 kg+5.0 kg)v f

v f = 3.1 m/s

24.

~pi = ~p f

m1~v0 = (m1 +m2)~v f

~v f =m1~v0

(m1 +m2)

25. Air bags increase the time of impact, which reduces the force of impact.26. Rock climbers require ropes that can stretch considerably, to prevent them from getting injured when the ropes

save them from falling.27. The steel cable will not stretch causing his body to deal with a large amount of force being applied to it over

a very short period of time, which could severely injure or even kill him.28. It would be safer to jump to the unpacked sand because it will extend the amount of impact will be applied to

your body.29. Crumple zones allow for the force being transferred to the car to be reduced by increasing the amount of time

the momentum is transferred.30. It would be a better idea to dodge the 2 kg brick because if struck by it, it would impart a larger amount of

force on you in a shorter time than the marshmallow would.31. The astronaut could throw his jetpack in the opposite direction of the spaceship. This would cause him to drift

towards the ship with whatever momentum the imparted on the jetpack.32. When objects interact, the forces that are exerted on one another are equal and opposite. Therefore, the same

rule applies to impulses.33. The momentum is increased due to the force of gravity on the object.34. Both objects will experience the same change in momentum.35. There is a greater change in momentum if the cars are in an elastic collision as opposed to an inelastic collision.

Therefore, less damage would result in an inelastic collision.36. F = ∆p

t =− (25 kg)(3.0 m/s)(2 s) =−37.5 N

37. F = ∆pt = (25 kg)(3.0 m/s)

(0.05×10−3s) = 1.5×106 N38.

Ft = ∆p

F =∆pt

=(35 kg)(15 m/s)

0.075 s= 7 kN

39.

Ft = ∆p

m =Ftv

=(14×103N)(0.075 s)

(15 m/s)= 70.0 kg

40.

I = ∆p

I =−(0.20 kg)(30 m/s) = 6.0 N · s

41. F = ∆pt = −(35 m/s)(0.35 kg)

0.02 s = 613 N

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42.

pi = p f

(7.0 kg)(5.0 m/s) = (35.0 kg+7.0 kg)v f

v f = 0.83 m/s

43.

pi = p f

(15.0 kg)(4.0 m/s) = (15.0 kg+3.0 kg)v f

v f = 3.33 m/s

44.

pi = p f

(10.0 kg)(1.5 m/s) = (10.0 kg+1.5 kg)v f

v f = 1.3 m/s

45.

pi = p f

0 = mhvh +mc(45 m/s)

mc = 10 mh

−mhvh = 10 mh(45 m/s)

vh =−450 m/s

46. No relationship can be determined between the object’s momentum.47. The object with the smaller mass will have a greater acceleration; therefore the time it takes to travel a given

distance will be smaller. Since the time is smaller, the change in momentum is smaller for the smaller object.The kinetic energy on the other hand is the same for both objects (using the work-kinetic energy theorem).

48. This statement is only partially correct. Impulse is defined as force x time. Therefore, the length of time thatthe force is applied matters as well.

49. Doubling the velocity doubles the momentum and quadruples the kinetic energy.50. It is possible for two objects to have the same kinetic energy but different momentums. Therefore, nothing

can be said about the two objects’ momentums without more information.51. The conservation of momentum was ignored because the hero should have been pushed in the opposite

direction of the car that was thrown. (That is, unless he had a significantly larger mass than the car).52. To see that momentum is conserved, you have to look at the whole system; namely the ball-Earth system. It

just appears that momentum isn’t conserved due to the fact that the Earth is so much larger than the ball thatit is colliding with.

53. If two objects are traveling in the same direction along the same line and one overtakes the other, the fastermoving car will lose more kinetic energy and the slower moving car will gain kinetic energy.

54. A donut or a ring would be good examples as to how the center of mass is located at a point where no mass ispresent.

55.

v =√

V 2x + v2

y =√

(75 m/s2)+(45 m/s2) = 87.5 m/s

p = mv = (2.0 kg)(87.5 m/s) = 175 N

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56.

K =12

mv2

p = mv

→ p2 = m2v2

⇒ K =p2

2 m

57.

pi = 0

0 = p f

0 = m1v1 +(2.0 kg)(3.5 m/s)

v1 =−(2.0 kg)(3.5 m/s)

0.5 kg=−14 m/s

58.

100 mph = 44.7 m/s

∆p = Ft

F =∆pt

=(44.7 m/s)(8.0 kg)

0.5 s= 715 N

59. I = ∆p = p f − pi = (0.04 kg)(−50 m/s−65 m/s) = 4.6 N · s60.

∆p = 0

pi = p f

(15 kg)(5.5 m/s)+(60 kg)(3.0 m/s) = (15 kg+60 kg)v f

v f = 3.5 m/s

61.

∆p = 0

pi = p f

0 = p f

p f =−(2.0 kg)(3.5 m/s)+(0.5 kg)v1 f = 0

v1 f = 14 m/s

The distance the block travels up the incline can be described by:

v2 = v20−2a∆x

∆x =v2

02a

=(14 m/s)2

2(9.8)sin(55)= 12.0 m

Find the vertical height the block travels:

sinθ =y

∆xy = 9.8 m

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62. Determine the final velocity of the the two-car system :~v f =m1~v1+m2~v2)(m1+m2)

= (540.7 kg)(10 m/s)+(1440.7 kg)(−10 m/s)(1800 kg) =

−4.54 m/sThe force on Alex is:~F = ∆~p

t = (90.7 kg)(−4.54 m/s−10.0 m/s)0.1 s =−13.2 kN

The force on Brian is:~F = ∆~p

t = (90.7 kg)(−4.54 m/s−(−10.0 m/s))0.1 s = 4.9 kN

63. The block needs just enough energy to rotate the block 180.

Find the block’s velocity through the conservation of energy:

Ki =U f

12

mblockv2block = mblockgl(1− cosθ)

vblock =√

2gl(1− cosθ)

Use the conversation of momentum to determine the initial velocity of the bullet:

∆p = 0

pi = p f

mbv0 = mvblock +mb

(1

3 v0

)v0 =

(3 m2 mb

)vblock =

(3 m2 mb

) √2gl(1− cosθ)

64. After the collision, the kinetic energy is equal to the increase in internal energy:

12(mg +mb)v2

f = f d = µ(mg +mb)gd

v f =√

2µgd

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Use the conservation of momentum:

pi = p f

mgv0 = (mg +mb)v f

v0 =(mg +mb)

mg

√2µgd =

(mg +mb)

mg

√2µgd =

8.5 kg0.5 kg

(1.27 m/s) = 21.6 m/s

65.

pi,y = p f ,y

(600 kg)(26.0 m/s) = (500 kg+600 kg)v f ,y

v f ,y = 14.2 m/s

pi,x = p f ,x

(500 kg)(−24.0 m/s) = (500 kg+600 kg)v f ,x

v f ,x =−10.9 m/s

θ = tan−1(

14.1−10.9

)= 128

v =√

v2x + v2

y = 17.9 m/s

66.

v f = 17.9 m/s

Ki =12[(500 kg)(24 m/s)2 +(600 kg)(26.0 m/s)2]= 3.47×105J

K f =12(1100 kg)(17.9 m/s)2 = 1.76×105J

Ki−K f = 1.71×105J

The energy could have been lost due to heat, sound, etc.67. Determinev f ,y:

tanθ =vy

vx→ vy = vx tanθ

vy = (−24 m/s) tan(135) = 24 m/s

Use the conservation of momentum:

m2v20 = (m1 +m2)vy

v20 =(m1 +m2)

m2vy = 44 m/s

68.

m(4.0 m/s)+0 = m(3.3 m/s)cos(42)+mv2x

v2,x = 1.63 m/s

0 = m(3.3 m/s)sin(42)+mv2y

v2,y =−2.2 m/s

v2 f =√

(−2.2 m/s2)+(1.63 m/s)2 = 2.74 m/s

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69. R = 1M ∑

imiri =

15.5 kg [(2.0 kg)(4.0 m)+(1.0 kg)(−5.0 m)+(2.5 kg)(2.0 m)] = 1.45 m

70.

Rx =1M ∑

imiri =

1(8.0 kg)

[(2.0 kg)(0)+(2.0 kg)(3.0 m)+(2.0 kg)(3.0 m)]

Rx = 1.5 m

Ry =1M ∑

imiri =

1(8.0 kg)

[(2.0 kg)(0)+(2.0 kg)(3.0 m)+(2.0 kg)(3.0 m)]

Ry = 1.5 m

71.

~vcm,x =1M ∑

imi~vi =

16.0 kg

[(4.0 kg)(1.0 m/s)+(2.0 kg)(2.0 m/s)] = 1.3 m/s

~vcm,y =1M ∑

imi~vi =

16.0 kg

[(4.0 kg)(−2.0 m/s)+(2.0 kg)(−1.0 m/s)] =−1.7 m/s

72.

(mball)(vi) = (mball +mperson)v f

(80.0 kg)(5.0 m/s) = (80.0 kg+150.0 kg)v f

v f = 1.74 m/s

73.

(mball)(vi) = (mball +mJim)v f

(1 kg)(10 m/s) = (71 kg)(v f )

v f = 0.141 m/s

74.

(mball)(vi) = (mJim)(v f Jim)+(mball)(v f ball)

(1 kg)(10 m/s) = 70 kg(v f Jim +(1 kg)(8.5 m/s)

v f Jim = 0.021 m/s

75.

p f − pi = Ft

p f = Ft + pi

v f =Ft + pi

m=

mgt +mvi

m= gt + vi = (9.8 m/s2)(5 s)+15 m/s = 64 m/s

76.

∆p = Ft

F =∆pt

=(0.20 kg)(15 m/s)− (0.20 kg)(−25 m/s)

10×10−3s= 800 N

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77.

∆p = f t

F =∆pt

µkmg =−p0

t

µk =−m(5.5 m/s)

m(9.8 m/s2)(2.0 s)= 0.28

78.

1200 lb = 544 kg;

35 mph = 15.6 m/s

pi = p f

(544 kg)(15.6 m/s) = (2000 kg)v f

v f = 4.24 m/s

79.

1200 lb = 544 kg;

35 mph = 15.6 m/s

pi = p f

(544 kg)(15.6 m/s) = (544 kg)(2.0 m/s)+(800 kg)(v f )

v f = 9.2 m/s

80.

v1 f =m1−m2

m1 +m2v10 +

2m2

m1 +m2v20 =

0 kg8.0 kg

(2.5 m/s)+2(4.0 kg)

8.0 kg(−3.0 m/s) =−3.0 m/s

v2 f =2m1

m1 +m2v10 +

m2−m1

m1 +m2v20 = 2.5 m/s

81.

∆p = 0

(4.0 kg)(2.5 m/s)+(4.0 kg)(−3.0 m/s) = (8.0 kg)v f

v f =−0.25 m/s

82.

v1 f =m1−m2

m1 +m2v10 +

2m2

m1 +m2v20 =

25 kg75 kg

(25 m/s)+2(25.0 kg)

75.0 kg(−25.0 m/s) =−8.37 m/s

v2 f =2m1

m1 +m2v10 +

m2−m1

m1 +m2v20 = 25.0 m/s

83.

∆p = 0

(100.0 kg)(15.0 m/s) = (125.0 kg)v f

v f = 12.0 m/s

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84.

v1 f =m1−m2

m1 +m2v10 =−0.5 m/s

150 kg−400 kg550 kg

v0 =−0.5 m/s

v0 = 1.1 m/s

85.

mv0 = 2mv f

v f =12

v0

let v f = v′0Solve for the acceleration:

v2 = v′20 +2ad

a =−v′20

2d=−

v′20

8d∑F =− f = (2m)a

−µk(2m)g = (2m)a

µk =v2

08gd

=(25 m/s)

8(9.8 m/s2)(1.0 m)= 0.32

86.

mvo,b = (m+M)v′

v′ =mvo,b

(m+M)

12

mv′2 = Mgl(1− cosθ)

cosθ = 1− 12

mv′2

Mgl

Define the arc length as s:

s = lθ

→ s = l cos−1

(1− 1

2mv

′2

Mgl

)87.

Rx =1M ∑

imiri,x

Ry =1M ∑

imiri,y

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88.

v′ =(0.01 kg)

(0.01 kg+5 kg)v0

12

Mv′2 = Mgh

12

M[

(0.01 kg)(0.01 kg+5 kg)

v0

]2

= Mgh

v0 =√

1964 m2/s2 = 44.3 m/s

89.

Rx =1M ∑

imiri,x =

111.0 kg

[(2.0 kg)(2.0 m)+(5.0 kg)(0.0 m)+(4.0 kg)(−3.0 m)] =−0.73 m

Ry =1M ∑

imiri,y =

111.0 kg

[(2.0 kg)(1.5 m)+(5.0 kg)(−3.0 m)+(4.0 kg)(2.0 m)] =−0.36 m

90.

Rx =1M ∑

imiri,x =

115.0 kg

[(5.0 kg)(2.0 m)+(5.0 kg)(0.0 m)+(5.0 kg)(x)] = 1.0 m

x = 1.0 m

Ry =1M ∑

imiri,y =

115.0 kg

[(5.0 kg)(1.5 m)+(5.0 kg)(−3.0 m)+(5.0 kg)(y)] =−1.0 m

y =−1.5

91. xcm = L2 ;ycm = L

292. Solve for your velocity upon impact using the given height:

v =√

2gh = 3.13 m/s

F =∆pt

=−(100 kg)(−3.13 m/s)50×10−3 s

= 6.26 kN

93.

∆p = Ft

F =∆pt

=4.5 m/s− (−4.5 m/s)

5×10−3s= 1.8 kN

94.

pi = p f

0 = (50 kg)(5.0 m/s)+(80 kg)v

v =−3.1 m/s

95.

p = mv

m =pv=

32 N · s10 m/s

= 3.2 kg

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96.

12

mv′2 = mgh

v′ =√

2gh

mbvb = (mb +mp)v′

mbvb = (mb +mp)√

2gh

mb =mp√

2ghv0−

√2gh

=(1.5 kg)

√2(9.8 m/s2)(1.2 m)

25 m/s−√

2(9.8 m/s2)(1.2)=

7.2720.2

= 0.36 kg

97. K f −Ki =12 m(v2

f − v2i ) = 0.5(4.0 kg)(3652−5002) =−2.34×10−5J

Kb =12

mv2 = 2.34×10−5J =12(400 kg)v2

v = 34.2 m/s

98.

mv0 = (m+M)v′

v′ =m

(m+M)v0

Use conservation of energy:

12(m+M)v2 = (m+M)gh

h =12

v2

g=

12g

(m

(m+M)v0

)2

99. The velocity of the pendulum before the collision:v =

√2gh = 4.4 m/s

v1 f =m1−m2

m1 +m2v10 =

−2.0 kg4.0 kg

(4.4 m/s) =−2.0 m/s

v2 f =2m1

m1 +m2v10 = 2.2 m/s

100. It is easier to stop the lighter crate because it has less kinetic energy. Since it has less kinetic energy, it willrequire less work to stop.

101. Since both objects have the same momentum, the lighter object has to be traveling faster; therefore the lighterobject has the greater kinetic energy.

102. When the velocity increases, the momentum increases linearly and the kinetic energy increase by the speedsquared.

103. No, you should not take the bet. If the momentum is zero, which implies no velocity, then the kinetic energyis 0 as well.

104. No the energy of an object doesn’t have to be zero since an object could have potential energy, which isindependent of the velocity.

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7.2 References

1. Haven Giguere and Laura Guerin. CK-12 Foundation .2. Laura Guerin. CK-12 Foundation .3. Laura Guerin. CK-12 Foundation .4. Laura Guerin. CK-12 Foundation .5. Laura Guerin. Ck-12 Foundation .6. Laura Guerin. CK-12 Foundation .7. Laura Guerin. CK-12 Foundation .8. Haven Giguere and Laura Guerin. CK-12 Foundation .9. Laura Guerin. CK-12 Foundation .

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www.ck12.org Chapter 8. Angular Motion and Statics Problem Sets

CHAPTER 8 Angular Motion and StaticsProblem Sets

Chapter Outline8.1 ANGULAR MOTION AND STATICS PROBLEM SETS

8.2 REFERENCES

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8.1. Angular Motion and Statics Problem Sets www.ck12.org

8.1 Angular Motion and Statics Problem Sets

Solutions

1. Neither, both will fall at the same rate.2. Torque is a force that rotates objects about an axis.3. The sum of the torques is 0.4. It is able to lean because the center of gravity is still located within the area outlined by the building’s base.5. (a) τ = F× r = (20 N)(0.5 m) = 10 N ·m

(b) τ = F× r = (10 N)(1.0 m) = 10 N ·m(c) If you were working in a limited area where you couldn’t have a lever arm that is 1 m, you could just

shorten the lever arm and apply more force.6. F = τ

r =100 N·m0.75 m = 133 N

7. τ2.0 m = F× r = (25 N)(2.0 m) = 50 N ·mτ1.5 m = F× r = (25 N)(1.5 m) = 37.5 N ·mYou can apply 12.5 more newtons of force.

8. Yes, an object can have balanced forces but unbalanced torques, causing the object to not be in equilibrium.9. In reality, cars are not balanced. The car would start to tip towards the side with the engine.

10. A torque is produced about the car’s center of mass due to friction by the road on the tires.11. Yes, by definition, a lever arm must be present.12. Yes, this is a problem. There is no lever arm about the center of mass. Therefore, no torque could be produced.13. The friction between the ball and the surface provides a torque on the ball causing it to spin.14. With the same weight in each hand, your body is balanced. If you were holding weights that were uneven,

then there would be a torque on your body.15. Looking at the torque on each side: F1D = F2d

Using the constraint: F1 +F2 = 25 N

F1D = (25−F1)d

F1 =25d

D+dD = .25d

→ F1 =25d

1.25d= 20 N

⇒ F2 = 5 N

16. τ = r×F = rF = (0.25 m)(60 N) = 15 N ·m17. r = τ

F = 1×103 N·m200 = 5 m

18.

ω2 = ω

2o +2α∆θ

→ α =∆ω2

2∆θ=

200 rev60 s

2(5 rev)= 0.33 rev/s

τ = [(2.0 kg)(1.2 m)2](2.07 rad/s) = 5.96 N ·m

19. W = (−8.0 N ·m)(31.4 rad) = 251 J20. F = τ

r =10.0 N·m0.0275 m = 364 N

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21.

W = ∆K = K f −Ki;Ki = 0

K f =12

Iω2 =

12(0.5 kg)

(0.52 m2

2

)(104.5 rad/s)2 = 341 J

W = 341 J

22. If room permits, a wrench with a longer handle would allow you to generate more torque using less force.23. No, there is only a torque if its angular momentum changes.24. The helicopter will become unbalanced due to the torque on the system.25. Imagine pushing down on one end of a seesaw and pressing up on the other end with an equal amount of force.26. The total downward force is:

F = 26 N +30 N = 56 N

∑τ = 0

→ Starting from the left:

(−26 N)(0)− (56 N)x+(30 N)(4.0 m) = 0

x = 2.14 m

27.

∑τ = 0

∑F = 0

∑τ =−FpL+LT sinθ = 0

→ T =Fp

sinθ

28. F129. The forces on a block in the vertical direction:

T −mg = maTo determine T , use the sum of the torques:

∑τ = Iα

−RT =

(12

MR2)

α

→ T =−(

12

MR)

α

Since a = αR:

T =−12

Ma

Therefore:

−12

Ma−mg = ma

⇒ a =−2 mg

(2m+M)

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30. Looking at the sums of the torques:

∑τ = 0

(L)(−Fwall)sinθ+

(L2

)(mhandymang)cosθ+

(L3

)(mladderg)cosθ+0 = 0

θ = arcsin(

1023

)= 25.8(

L2

)(mhandymang)cosθ = 80(9.8)(11.5cos25.8) = 8120(

L3

)(mladderg)cosθ = 50(9.8)(7.67cos25.8) = 3380

(L)(Fwall)sinθ = 11500

Fwall =11500

23(sin25.8)= 1150 N

The ladder makes a right triangle with the wall:

→ L2 = h2 + x2

⇒ x =√

L2−h2

Thus:

Fwall =

√L2−h2

(Fh2 + Fl

3

)h

=

√13.02−102

(802 + 50

3

)10

Fwall = 391 N

31. ∆t = ∆Lτ= 200 kg·m2/s

35 N·m = 5.71 s32. τ = ∆L

∆t = 1.2−2.00.33 =−2.42 M ·m

33. ∑τ = 0.9 N ·m34. For a mass attached to a pulley with mass and radius:

a =−2 mg

(2m+M)=

2(2.0 kg)(9.8 m/s2)

2(2.0 kg)+3.3 kg= 5.37 m/s2

α = ar = (5.37 m/s2)(0.25 m) = 1.34 rad/s2

35. F = mat , where at is the tangential acceleration.The torque is defined as τ = Fr = matr.Rewriting the tangential acceleration in radial components:→ τ = m(αr)r = mαr2 = mr2α = Iα

36. F1 should be less than F2.37. (a) For a mass attached a pulley of mass and radius:

a = −2 mg(2m+M) =

−2(2.0 kg)(9.8 m/s2)2(2.0 kg)+3.0 kg =−5.6 m/s2

(b)

a =−2 mg

(2m+M)

Assume M = 0:

→ a =−2 mg

2m=−g

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38. τ = (0.2 kg)(9.8 m/s2)(1.0 m) = 1.96 N ·m39. ∆t = ∆L

τ= 4.8−2.2

3.5 N·m = 0.74 s40. (a) Due to conservation of angular momentum:

Li = L f

→ ω2 =I1

I2 + I1ω1 =

I3(35.0 rad/s) = 11.7 rad/s

(b) The initial kinetic energy of the system is:

Ki =12

I1ω21

K f =12(I1 + I2)ω

2f

Rewrite ω f =I1

I1 + I2ω1

∆KKi

=K f −Ki

Ki= 1− 2

6=

23= 0.67

41.Write out the torques, assuming the rotation point is at the wall:

yTcable− xTmass−12

xmg = 0

Tcable = x(1

2)mg+Mgy

=xg[(1

2)m+M]

y=

(3.0 m)(9.8 m/s2)

1.7 m[(

12)(70.0 kg)+2.0 kg] = 640 N

42. ∆L = L f −Li = 043. L f = τ∆t +Li = (5 N ·m)(2 s)+22 kg ·m2/s = 32 kg ·m2/s44. The rotational speed is the same everywhere, so your rotational speed is 10 rpm.45. Rotational speed is how many radians (or degrees) a point on a circle goes through in a given time, while the

tangential speed is how fast a point is moving a given distance from the axis of rotation.46. Tangential speed is the speed of an object that is tangent to the circular path of an object.47. The tangential speed is proportional to the distance from the axis of rotation.48. Yes. An object that is rotating tends to continue rotating.49. Rotational inertia is defined as the mass times the distance from the axis of rotation squared. Therefore, the

units of rotational inertia are kg ·m2.50. You win, assuming equal mass, as the solid object has less rotational inertia per mass.51. The center of mass is located in the center of the sphere.52. The car has a centripetal force applied to it, causing the car to slide out from under you.

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53. If a space station were able to rotate with the proper angular velocity, there would be a lack of centripetalforce. This would cause everything inside the station to appear to be pulled towards the outside, simulatinggravity.

54. In the simplest sense, angular momentum is the momentum of an object that has rotational motion, whilelinear momentum is the momentum of an object that is moving without rotation.

55. The concept of conservation of angular momentum is the same as that of linear momentum.56. (a) L = mv× r = mvr = (35 kg)(4 m/s)(1.5 m) = 210 N ·m · s

(b) L = mv× r = m(1

2

)vr = (35 kg)

(12

)(4 m/s)(1.5 m) = 105 N ·m · s

57. Both gears have the same tangential speed, as they are attached. The smaller gear must therefore have twicethe rotational speed of the larger gear.

58. Smaller tires have you travel a shorter distance with each revolution of the tire, which means you will bemoving slower than the speedometer.

59. Since v = ωr, the car with the tires with the smaller radius has a greater rotational velocity.60. The basketball and volleyball have the same acceleration because they have the same shape. In this case, shape

means mass distribution. Since they are both hollow, they reach the bottom of the incline at the same time.61. The eight-ball will have a greater acceleration than the volleyball. The moment of inertia of the eight-ball is

likely to be less than the moment of inertia of the volleyball. For the eight-ball, the radius is smaller, and themass is equally distributed throughout the shape. On the volleyball, the radius is larger, and all the mass islocated at the edge of the shape. As a result, the eight-ball will have a greater acceleration down the incline.

62. The object that reaches the bottom first is the one with the least rotational inertia compared to its mass. Thesolid sphere will therefore reach the bottom first.

63. Since both objects have the same mass distribution, both would reach the bottom at the same time.64. The rate of rotation is independent of the radius.65. v = ωr = (2π rad/s)(1.5 m) = 9.42 m/s66. v = ωr =

(2π rad20 s

)(20 m) = 6.28 m/s

67. (a) xcm = ∑mirimtotal

= 1m1+m2

(m1r1 +m2r2) =1

0.5 m+m(0.5 m(0)+m(2)) = 1.33 mxcm = 1.33 m from the leftmost mass

(b) ∑τ = rF =+(1.33)(0.5)(1.0 kg)(9.8 m/s2)− (0.67 m)(1.0 kg)(9.8 m/s2)≈ 068. (a) Due to conservation of angular momentum, the disk will make 0.25 revolutions per second if the disk

expanded.(b) Due to conservation of angular momentum, the disk will make 2 revolutions per second if the disk

shrunk.69.

s = rθ

→ θ =sr=

0.5 m1 m

= .5 rad = 28.6

70. s = rθ = (1.5 m)(2.4 rad) = 3.6 m71.

(23 rotations60 s

)( 2π rad1 rotation

)= 2.41 rad/s

72. v = ωr =(75 rev

hr

)(1 hr60 s

)(2π rad1 rev

)(0.025 m) = 0.196 m/s

73. (a)

27 rev/min = 2.83 rad/s = ω

ω = αt

→ t =ω

α=

2.83 rad/s2.5 rad/s2 = 1.13 s

(b) θ = 12 αt2 = (0.5)(2.5 rad/s2)(1.13 s)2 = 1.60 rad = 91.5

74. t = ω

α= 47 rev/min

15.2 rev/min2 = 3.1 min = 186 s

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75. (a)

ω = ωo +αt→ α =ω

tω = 55 rpm = 5.76 rad/s

α =5.76 rad/s

4s= 1.44 rad/s2

a = αr = (1.44 rad/s2)(0.20 m) = 0.288 m/s2

(b)

ω = ωo +αt→ α =ω

tω = 55 rpm = 5.76 rad/s

α =5.76 rad/s

4s= 1.44 rad/s2

a = αr = (1.44 rad/s2)(0.10 m) = 0.144 m/s2

76.

ω = ωo +αt

α =ω−ωo

t=

230 rev60 s −

100 rev60 s

13 s= 0.167 rev/s2

77.

α =ω−ωo

t=

(2.61−1.05) rad/s1.3 s

= 1.2 rad/s2

θ = ωot +12

αt2 = (1.05 rad/s)(1.3)+12(1.2)(1.3)2 = 2.38 rad = 136

78. (a) I = ∑i

mir2i = m2r2

2 +m4r24 = (2 kg)(1.0 m)2 +(2 kg)(1.0 m)2 = 4 kg ·m2

(b) I = ∑i

mir2i = m2r2

2 +m3r23 = 2[(2 kg)(r)2]

Let r = 12(1.0 m)cos45 , since the configuration is a square.

I = 2 kg ·m2

79. K f −Ki =12 Idiskω2 = 1

2

[(2 kg)

(0.122m2

2

)](52.25 rad/s)2−0 = 19.7 J

80. (a) K f −Ki =12 Ihoopω2−0 = 1

2 [(2 kg)(0.122m2)](52.25 rad/s)2−0 = 39.3 J(b) ∆Khoop

∆Kdisk= 39.3 J

19.7 J ≈ 281.

∆K =12

Iω2 =

12

[2.0 kg

(1.22m2

2

)](20.9 rad/s)2 = 314 J

→ 314 J5 rev

= 62.9 J/rev

82.

τ = Iα

ω2 = ω

2o +2α∆θ

→ α =∆ω2

2∆θ=

(200 rev60 s )2

2(5 rev)= 1.11 rev/s2 = 6.98 rad/s2

τ =

[(4.0 kg)

1.22m2

2

](6.98 rad/s2) = 20.1 N ·m

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83.

I = Icm +Mh2

h = R

→ I =12

MR2 +MR2 =32

MR2

84.

I = Icm +Mh2

h =12

L

→ I =112

ML2 +14

MR2 =13

MR2

85.

~L f =~Li

I f ω f = Iiωi

I f = I +McR

→ (I +McR)ω = Iωi

ω f =Iωi

(I +McR2)=

(1000 kg ·m2)(1.045 rad/s)(1000 kg ·m2 +(30 kg)(2.0 m)2)

= 0.934 rad/s = 8.92 rev/min

86. (a) The angular velocity of each piece of clay must be the same.(b) Since one piece of clay is twice as far away from the center than the other piece, the piece that is further

away will have a tangential velocity that is twice as large.87. The hollow cylinder would have a greater kinetic energy.88.

ω = αt

α =ω

t=

18.29 rad/s2.5 s

= 7.32 rad/s2

89.

0 = ωo +αt

t =−ω

α=

13.06 rad/s2.5 rad/s2 = 5.22 s

90.

25 rev = 157.1 rad

ωo = 1200 rpm = 125.66 rad/s

0 = ω2o +2α∆θ

α =−ω2

o

2∆θ=−(125.66 rad/s)2

2(157.1 rad)=−50.3 rad/s2

91.

25 rotations = 157.1 rad

∆θ = ωot +12

αt2

α =2(157.1 rad)

(5.0 s)2 = 12.6 rad/s2

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92. (a) 800 rev/min = 83.8 rad/s(b) v = ωr = (83.8 rad/s)(0.05 m) = 4.19 m/s

93. (a) I = m1r21 +m2r2

2 = (5.0 kg)(0.5 m)2 +(4.0 kg)(−1.0 m)2 = 5.25 N ·m2

(b)

I = 6.5 N ·m2

K =12

Iω2 =

12(5.25 N ·m2)(2.0 rad/s)2 = 10.5 J

94. 2.27 kg = 5.0 lbKtrans =12(2.27 kg)(10 m/s)2 = 11.35 J

95. K = 12 mv2 + 1

2 Iω2 = 12

(m+ I

R2

)v2

Using conservation of energy:

Ki =U f

→ 12

(m+

IR2

)v2 = mgh

v2 =2gh(

1+( I

mR2

))For a disk, I = 1

2 mR2.

→ v =

√4gh3

A cylinder would have the same answer, as a cylinder can be considered to be composed of many disks rightnext to each other.

96. Assuming 1 complete revolution is every 27.3 days:(1 rev

27.3 days

)(1 day

86400 s

)(2π rad1 rev

)= 2.66×106 rad/s

97.

400 rpm = 41.8 rad/s

v = ωr = (41.8 rad/s)(5.0 m) = 209 m/s

98. v = ωr, ω = vr =

40 m/s175 m = 0.23 rad/s

99.

60 mph = 26.8 m/s

a = αr

v = at

a =vt=

26.8 m/s3.0 s

= 8.93 m/s2

→ α =ar=

8.93 m/s2

(0.275 m)= 32.5 rad/s2

→ ∆θ =12

αt2 =12(32.5 rad/s2)(3.0 s)2 = 146.1 rads = 23.3 revolutions

100. I = ∑i

mir2i = (4.0 kg)(4.0 m)2 +(3.0 kg)(2.5 m)2 = 82.75 kg ·m2

101. Angular displacement is considered negative in the clockwise direction.102. The angular displacement is positive.

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103.

θ =−0.8t +0.5t2

θ(t = 5.0 s) =−0.8(5)+0.5(5.0)2 = 8.5

θ(t = 2.0 s) =−0.8(2)+0.5(2.0)2 = 0.4

∆θ = 8.1

104. R = aα= 2.1 m/s2

0.5 rad/s2 = 4.2 m

105. acom = αR = (1.2 rad/s2)(0.1 m) = 0.12 m/s2

106. The time taken to travel 35 m:t = x

v =35 m

(20 m/s) = 1.75 sThe number of revolutions:1200 rev

60 s

(1.75 s1

)= 35.0 rev

107. acom = g1+ I

MR2= g

1+ (800 g/cm2)(150 g)(0.22 cm)2

= 8.8 cm/s2

108.

K =12

Iω2

I =2Kω2 =

2(1.2×105) J)(123 rad/s)2 = 15.9 kg ·m2

109. t =−ωoα= 15.7 rad/s

2.2 rad/s2 = 7.14 s110.

θ = 0.2t2 +0.07t3

θ(t = 1.2 s) = 0.2(1.2)2 +0.07(1.2)3 = 0.4 rad

111. K = 12 Iω2 = 1

2(1.2)(2.0 rad/s)2 = 2.4 J112.

ω2 = ω

2o +2α∆θ

ω = 0

→ α =−ω2o

∆θ=−(95 rad/s)2

40π rad= 71.8 rad/s2

113.

∆θ = ωot +12

αt2

3.0 rev = 6π rad

6π = (−3.0 rad/s)t +12(0.5 rad/s2)t2

t = 16.6 s

114. Equations b and c have constant acceleration.

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115.

acom =gsinθ

1+ IMR2

→ θ = sin−1

(acom

(1+ I

MR2

)g

)= sin−1

acom

(1+(

23MR2

)MR2

)g

θ = sin−1

(acom

(1+ 2

3

)g

)= sin−1

(5acom

3g

)= 14.8

116.

W = ∆K = K f −Ki = 0−Ki

Ki =12

Iω2 +

12

mv2 =12

[mR2

(v2

R2

)+mv2

]= mv2 = 10.8 J

W =−10.8 J

117. v = ωr = (12 rad/s)(0.05 m) = 0.6 m/s118.

95 = 1.66 rad

ω =1.66 rad

3 s= 0.55

rads

= 0.09 rev/s

119. The time it takes for the stuntman to hit the ground:

t =

√2(30 m)

9.8 m/s2 = 2.47 s

3 rev2.48 s

=3(2π rad)

2.47 s= 7.62 rad/s

120.

ω = 0.4 t +0.21 t2

ω(t = 1.0 s) = 0.4(1.0 s)+0.21(1.0 s)2 = 0.61 rad/s

121. t = ω−ωoα

= 15 rad/s−3 rad/s−2.2 rad/s2 = 5.5 s

122.

1 rev = 2π rad14(2π rad) =

12

π rad

123. For an object that is rolling without slipping, a = αr.

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8.2 References

1. Haven Giguere and Laura Guerin. CK-12 Foundation .2. Haven Giguere and Laura Guerin. CK-12 Foundation .3. Haven Giguere. CK-12 Foundation .4. Haven Giguere. CK-12 Foundation .5. Haven Giguere and Laura Guerin. CK-12 Foundation .6. Haven Giguere. CK-12 Foundation .7. Haven Giguere. CK-12 Foundation .8. Laura Guerin. CK-12 Foundation .9. Haven Giguere. CK-12 Foundation .

10. Haven Giguere. CK-12 Foundation .11. Haven Giguere. CK-12 Foundation .12. Haven Giguere. CK-12 Foundation .13. Haven Giguere. CK-12 Foundation .14. Laura Guerin. CK-12 Foundation .

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www.ck12.org Chapter 9. Newton’s Universal Law of Gravity Problem Sets

CHAPTER 9 Newton’s Universal Law ofGravity Problem Sets

Chapter Outline9.1 NEWTON’S UNIVERSAL LAW OF GRAVITY PROBLEM SETS

9.2 REFERENCES

127

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9.1. Newton’s Universal Law of Gravity Problem Sets www.ck12.org

9.1 Newton’s Universal Law of Gravity ProblemSets

Solutions

1. If the smaller object orbiting the larger object had zero tangential velocity, it would simply be pulled into thelarger object and collide.

2. If one of the masses is tripled, then the force between the two objects is tripled. If both masses increase by afactor of three, then the total force felt between the two is increased by a factor of 9.

3. While the force between the two objects is the same, looking at Newton’s second law we can see that a lightermass would have a greater acceleration

(a = ∑F

m

)since the mass is in the denominator.

4. While there IS a force of attraction between ourselves and all the objects around us, it is negligible due to thefact that the mass difference between these objects and the Earth is so great that we will only notice the forcedue to the Earth.

5. Looking at the equation for gravitational force: F = Gm1m2r2 , if r is doubled, then the force is reduced by a

factor of 4.6. The force is increased by a factor of 4:

Fo =Gm1m2

r2

F =Gm1m2(1

2 r)2 ⇒ F = 4

Gm1m2

r2 = 4Fo = 80N

7. As you fall towards the center of the Earth you would increase your kinetic energy and decrease your potentialenergy. As you pass the center of the Earth, the opposite would happen. Since we are ignoring frictionaleffects, you would continue to move back and forth between the ends of the Earth.

8. The force felt between masses is proportional to the masses but inversely related to the square of the distance.

9. F = Gm1m2r2 =

(6.673×10−11 m3

kg·s2 )(2.0 kg)(2.0 kg)

(2.0 m2)= 6.673×10−11 N

10. The force that we experience as we fall is the force due to gravity. Note there is a difference between referringto gravity as a force (incorrect) and saying that there is a force due to gravity (correct).

11. Your weight is equal to mass times gravity when the only force acting on you in the direction of gravity isgravity itself and there is only the normal force acting in the upward direction.

12. Your weight would be greater than just mg when there is another force acting on you, such as when you are inan elevator that is accelerating upwards.

13.

(a)

F =Gm1m2

r2 =(6.67×10−11 m3 · kg−1 · s−2)(6×1024 kg)(2.0 kg)

(6.4×106 m)2

F = 19.5 N

(b)

F =Gm1m2

r2 =(6.67×10−11 m3 · kg−1 · s−2)(6×1024 kg)(2.0 kg)

(2(6.4×106 m))2

F = 4.88 N

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(c)

Fb

Fa=

4.88 N19.5 N

=14

The force is decreased by a factor of 4.

14.

r =

√Gm1m2

F=

√(6.67×10−11 m3 · kg−1 · s−2)(6×1024 kg)(1.0 kg)

(5.0 N)

r = 8.95×106 m

15.

F =GmmmE

r2 =(6.67×10−11 m3 · kg−1 · s−2)(6×1024 kg)(7.36×1022 kg)

(3.84×108 m)2

F = 2.0×1020 N

16.

F =GmsmE

r2 =(6.67×10−11 m3 · kg−1 · s−2)(6×1024 kg)(1.99×1030 kg)

(1.49×1011 m)2

F = 3.58×1022 N

17.

F =GmpmM

r2 =(6.67×10−11 m3 · kg−1 · s−2)(3.285×1023 kg)(90 kg)

(77×109 m)2

F = 3.33×10−7 N

18. The ratio of the gravitational force felt to the object’s mass is the same for every object. This is why all objectsaccelerate at the same rate.

19. The Earth would follow a straight line tangent to its orbital path if there was no gravitational force on it.20. The force of gravity on each brick is the same.21. The force of gravity is inversely proportional to the square of the distance.22. While the astronauts are in outer space, they are still in Earth’s gravitational pull. This can be seen by the fact

that they orbit the planet as opposed to drifting off. They experience weightlessness due to the fact that thereis no normal or support force present.

23. Yes, there is a point where the gravitational force would be zero, and it would be closer to the Earth.24. The spot where the gravitational pull is zero would be closer to the Moon.25. The field can only be detected by using an object to see if it feels the gravitational force.26. The net gravitational force would be zero.27. Your weight would be increased by four times, since you are half the distance from the center.28. Due to Newton’s Third Law, the Earth would weigh 750 N.29. F = GMm

r2 = G(2M)m(2r)2 = 1

2GMm

r2 , you would weigh half as much.

30. F = GMmr2 = G(3M)m

( 12 r)2 = 12 GMm

r2 , you would weigh 12 times your original weight.

31. F = GMmr2 = GMm

(2r)2 = 14

GMmr2 , you would weigh 1

4 your original weight.32.

F =GM′m

r2 =GM′m(2r)2 =

14

GM′mr2

⇒M′ = 4 M

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33. When a person is in free fall there is no support force; no force is pressing up on them to oppose thegravitational force. When a person has reached terminal velocity, the force due to air resistance is theirsupport force and it is in balance with the gravitational force.

34. If you drop an object while in free fall, gravity would act on both you and the object equally. Therefore, theobject would float right in front of you.

35. BASE jumpers experience weightlessness due to the fact that there is no support force acting on them.36. You would weigh more when the planet has a greater mass, so during the day.37. Astronauts in orbit feel weightlessness because there is no support force acting on them.38. F = GmM

r2 = GmM( 1

5 )2 = 25 GmM

r2 = 25Fo

39. F = GmMr2 = GmM

(2ro)2 =14

GmMr2

o= 1

4 Fo

40. F = GmMr2 = G(2m)M

r2 = 2 GmMr2 = 2Fo

41.

F =GmM

r2

mg =GmM

r2

g =GMr2 =

GM(2ro)2 =

14

go

42. g = GMr2 = (6.67×10−11)(1.98×1030)

(6.96×108)2 = 272.6 m/s2

43. g = GMr2 = (6.67×10−11)(1.35×1022)

1.153×106 = 0.68 m/s2

44. Satellites are never above Earth’s gravity field, as they orbit the Earth due to gravity.45. The velocity is greatest at point A.46. The greatest kinetic energy is at point A. The greatest potential energy is at point D.47. No. In a circular orbit the force on the satellite is always the same, due to the fact that the distance between

the satellite and the body it is orbiting never changes.48. As the distance between two bodies gets closer, the force becomes larger. This results in the acceleration being

larger, thereby making the velocity larger.49. An orbiting satellite in an elliptical orbit has the slowest speed when the distance between the two objects is

the greatest.50. The potential energy is the greatest at apogee.51.

F =GMm

r2

ma =GMm

r2

mv2

r=

GMmr2

v =

√GM

r

52. As a satellite moves in a tangential direction with respect to the Earth, it is pulled inward due to the attractiveforce of gravity. As it is pulled it continues in a tangential path that follows the curvature of the Earth.Therefore, as a satellite falls towards the Earth, it moves in a direction that follows the curvature preventingthe two objects from colliding.

53. Yes, there is a radial acceleration, which is due to gravity.54. The planets closer to the Sun have a smaller orbital period.55. The mass of the body being orbited, and the distance between the two bodies.

56. No, the speed of a satellite does not depend on its mass: v =

√GM

r.

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57. This would not change Earth’s orbital period, as the time it takes for an object to make one complete orbit isnot dependent upon its mass.

58. Firing in an eastern direction allows one to take advantage of the Earth’s spin to give the rocket a little extraboost.

59. When an object is leaving Earth’s atmosphere, it is moving much slower than when it is reentering.60. No work is done because the force on the satellite and satellite’s displacement are perpendicular to one another.61.

v =xt

v =(2πr)

(365 days)

v =(2π(1.5×108 km))

(3.15×107 s)= 29.7 km/s

62. v =

√GM

r=

√6.67×10−11(5.97×1024)

3.8×108 = 1023 m/s

63. v =

√GM

r=

√6.67×10−11(1.99×1030)

5.8×1010 = 48 km/s

64. v =

√GM

r=

√6.67×10−11(1.99×1030)

108×109 = 35 km/s

65. While the force is proportional to the mass, the acceleration felt by all objects is the same due to the fact thatlarger forces are needed to accelerate larger masses.

66. Due to the gravitational force on the Moon being weaker than that on the Earth, less energy is needed to breakfree from the Moon.

67. Since the ratio of the escape velocity from Earth to the escape velocity of the Moon is approximately 5, theamount of fuel needed to escape from the Moon would be 25 times less than that of the Earth since the velocityterm is squared.

68. The net work done on the planet is zero.69. The mass of the object at the center of the Earth would be the same, since mass does not change based on

location.70. 1. Planets move in an elliptical orbit with the Sun at the focus.

2. A line joining any planet to the Sun sweeps out equal areas in equal time.3. The square of the period is equal to the cube of the semi-major axis.

71. Using Kepler’s Third Law:T 2 = 4π2

GMsr3 = 4π2

(6.67×10−11)(1.99×1030)(7.8×1011)3 = 3.75×108 s = 11.8 yr

72.

r = RE + r = 6378 km+500 km = 6878 km

T 2 =4π2

GMEr3 =

4π2

(6.67×10−11)(5.97×1024)(6.88×106)3 = 5682 s = 95 min

73.

T 2 =4π2

GME

r =3

√GMET 2

4π2 =3

√(6.67×10−11)(5.97×1024)(86400 s)2

4π2 = 4.2×107 m

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74.

∆E = 0

K +U = 012

mv2− GMEmRE

= 0

vescape =

√2GME

RE=√

2gRE

75. vescape =

√2GM

R=

√2(6.67×10−11 N ·m2/kg)

6.05×106 m(4.87×1024 kg) = 10.4 km/s

76. No this is incorrect. The force varies inversely with the square of the distance.77. T 2 =Cr3

Using Kepler’s Third Law to relate the unknown object with the Earth:

C =T 2

E

r3E

T 2 =T 2

E

r3E

r3

r3 =

(TTE

)2

r3E =

(5 yr1 yr

)(1 AU)3 = 25 AU

→ r = 2.92 AU = 4.38×1011 m

78. g/479.

F =GMEm

r

2=

mv2

r12

mv2 =GMEm

2r

E = K +U =GMEm

2r− GMEm

r

E =−GMEm2r

80.

(a) U =−GME mRE

=− GME m(6378+8500) km =−8.03×109 J

(b) The total energy is simply 12 U , or 4.02×109 J.

81. T 2 =Cr3


C =T 2

E

r3E

T 2 =T 2

E

r3E

r3

r3 =

(TTE

)2

r3E =

(76 yr1 yr

)2

(1 AU)3 = 5776 AU

→ r = 17.9 AU = 2.70×1012 m

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82. T 2 =Cr3


C =T 2

E

r3E

T 2 =T 2

E

r3E

r3

r3 =

(TTE

)2

r3E =

(2400 yr

1 yr

)2

(1 AU)3 = 5.76×106 AU

→ r = 179 AU = 2.69×1013 m

83.

W = mg

m =Wg

=15 N

9.8 m/s2 = 1.53 kg

84. F = mg = (100 kg)(9.8 m/s) = 980 N

85. WMoon = (WEarth)mMoonmEarth

(RERM

)2= (15 N)(0.012)(13.4) = 2.41 N

86. WSun = (WEarth)mSun

mEarth

(RERM

)2= (15 N)(3.33×105)(8.41×10−5) = 420 N

87. U =−GMmR =− (6.67E−11 N·m2/kg2)(5.98E24 kg)(7.36E22 kg)

3.82E8 m =−7.68E28 J

88. vescape =

√2GME

RE=√

2gRE =√

2(9.8 m/s2)(6.37E6 kg) = 11.2 km/s

89.

vescapeEarth =√

2gRE =√

2(9.8 m/s2)(6.37×106 kg) = 11.2 km/s

vescapeMoon =√

2(6.67×10−11 N ·m2/kg2)(7.36×1022 kg)/(1.74×106 kg) = 2.38 km/s

⇒vescapeEarth

vescapeMoon

= 4.7

90. vm =√

2(6.67×10−11 N ·m2/kg2)(2.0×1030 kg)/(1.0×104 kg) = 1.63×105 km/s91. No, the escape velocity does not depend on the direction, but different methods are available to make attaining

the desired speed easier.

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9.2 References

1. Laura Guerin. CK-12 Foundation .2. Haven Giguere. CK-12 Foundation .3. Haven Giguere. CK-12 Foundation .

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www.ck12.org Chapter 10. Periodic Motion Problem Sets

CHAPTER 10 Periodic Motion ProblemSets

Chapter Outline10.1 PERIODIC MOTION PROBLEM SETS

10.2 REFERENCES

135

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10.1. Periodic Motion Problem Sets www.ck12.org

10.1 Periodic Motion Problem Sets

Solutions

1. The period of a pendulum is defined as the point when the object repeats its motions, one complete cycle.

2. The period of a simple pendulum is defined as T = 2π

√Lg

. The period is proportional to the square root of

the length. Therefore, if the length of the rope is longer, then the period is longer.3. Assuming the system is a simple pendulum, they both would have the same period since the period of a simple

pendulum is independent of mass.4. The period of a pendulum depends on g, the acceleration due to gravity. Therefore, if you move to a place of

higher or lower altitude, then the reading of the clock will be slightly off.5. When shortening a pendulum the period decreases and the frequency increases.6. The frequency would not change since it is not dependent upon the mass of the bucket.7. The period is less for the taller child due to the center of mass being shifted.8. The motion of the pendulum is dependent upon gravity which is the restoring force in this case and that is

independent of mass.9. T = 1

f =13 s

10. (a) A = 0.25 m

(b) f = 12π

[ km

] 12 = 1

2π

[150 N/m

2.0 kg

] 12= 1.38 Hz

11. (a) A = 0.25 m

(b) f = 12π

[ km

] 12 = 1

2π

[300 N/m

4.0 kg

] 12= 1.38 Hz

(c) T = 1f =

11.38 Hz = 0.72 s

12. (a) Looking at the energy of the system at t = 0, Ekinetic =12 mv2 = 1

2(1.5 kg)(4.0 m/s) = 12 J(b) The amplitude can be found from the conservation of energy:

Ekinetic =12

kx2max→ x2

max =2Ekinetic

k

⇒ x =

√2Ekinetic

k=

√2(12 J)

200 N/m= 0.35 m

(c) The velocity can also be found from the conservation of energy:

Ekinetic = Etotal

Etotal = K +U =12

mv2 +12

k∆x2

let ∆x = A3

12

mv2 = Etotal−12

k(

A3

)2

⇒ v =

√2E− k

(A3

)2

m=

√2(12 J)−200 N/m

(0.35 m3

)2

1.5 kg= 3.77 m/s

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13. The initial description gives you a method to figure out the spring constant:

k∆x = mg

⇒ k =mg∆x

=0.15 kg(9.8 m/s2)

0.2 m= 7.35 N/m

Therefore, f = 12π

[k

mdouble

] 12= 1

2π

[7.35 N/m2(0.15 kg)

] 12= 7.8 Hz

14. For small amplitudes:

T = 2π

√Lg(

T2π

)2

=Lg

⇒ L = g(

T2π

)2

= (9.8 m/s2)

(2.0 s2π

)2

= 0.99 m

15. The initial energy in the system is given by:E = mgh = mgL(1− cosθ)At the bottom of the swing, all the energy is kinetic:K = 1

2 mv2

Therefore,

12

mv2 = mgL(1− cosθ)

v =√

2gL(1− cosθ)

θ = sin−1(

0.08 m1.5 m

)= 3.1

v =√

2(9.8 m/s2)(1.5 m)(1− cos(3.1)) = 0.21 m/s

16. Perform the proper calculations:

T = 2π

√Lg(

T2π

)2

=Lg

⇒ L = g(

T2π

)2

= (9.8 m/s2)

(2.0 s2π

)2

= 0.992 m

Since the student was off by a factor of 100, it is possible that she used g = 980 cm/s2 and forgot that heranswer was in centimeters instead of meters.

17. One cycle results in traveling four times the length of the amplitude (A to EQ, EQ to -A, and back.) Thus, theblock travels through a total distance of 20 cm.

18. If the length increases, the period increases.19. The period is unchanged, since it is independent of mass.20. The period of the physical bob decreases.

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21. T = 2π

√mk= 2π

√0.080 kg2.5 N/m

= 1.12 s

22. E = 12 kA2 = 1

2(30.0 N/m)(0.040 m)2 = 0.024 J23. U = 1

2 kx2 = 12(20.0 N/m)(0.025 m)2 = 6.25×10−3 J

24. One reason is that the acceleration of the ball is not proportional to the position. Another reason is that thereis not a linear restoring force acting to pull the ball back towards an equilibrium position.

25. Yes, it is possible for both the velocity and acceleration vectors to face the same direction. This would indicatethat the object is speeding up. It is also possible for both vectors to be facing opposite directions. This wouldindicate that the object is slowing down.

26. Since Etotal =12 kA2 , any change in mass has no effect on the total energy of the system.

27. This is easily solved using ratios. Let T ′ be the period of the pendulum with its length doubled:

T ′

T=

2π

√2Lg

2π

√Lg

→ T ′

T=√

2

⇒ T ′ =√

2T

28. In one cycle the object will travel 4 times its amplitude.4A = 4(7.2 cm) = 28.8 cm

29. T = 1f =

14 cycles

8.0 s= 8.0 s

4 = 2.0 s

30. ω = 2π f = 2π(3 cycles15 s ) = 1.26 rad/s

31. Solve using energy conservation.

E0 = E f

12

mv2 =12

kx2

v2 =kx2

m

v =

√km

x =

√2.5×105 N/m

50 kg(0.020 m) = 1.41 m/s

32. Using conservation of energy:

Etotal = K +U12

kA2 =12

mv2 +12

kx2

→ mv2 = kA2− kx2

⇒ m =kv2 (A

2− x2) =(15.0 N/m)

(0.40 m/s)2

((0.075 m)2−

(12(0.075) m

)2)

= 0.40 kg

33. E = 12 kA2 = 1

2(65.0 N/m)(0.050 m)2 = 0.081 J34. Use energy conservation to solve for kinetic energy K.

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Etotal = K +U

→ K = Etotal−U → K =12

kA2− 12

kx2

⇒ K =12

k(A2− x2) =12(85.0 N/m)((0.070 m)2− (0.025 m)2)

K = 0.18 J

35.

F = kx

⇒ k =Fx=

25.0 N0.20 m

= 125 N/m

36. E = 12 kA2

Since A is squared, doubling it would result in the total energy increasing by a factor of 4.

37. The period of the system is given by T = 2π

√mk

. Since it is independent of amplitude, the period is

unchanged.38.

E0 = E f

12

mv2 +12

kx2 =12

kA2

→ v2 +km

x2 =km

A2→ v2 +ω2x2 = ω

2A2

let v2 =ω2A2

22

→ ω2A2

22 +ω2x2 = ω

2A2→ A2

4+ x2 = A2

x2 =34

A2

⇒ x =

√3

2A

39. Solve using conservation of energy:

K =U12

mv2 = mgl(1− cosθ)

→ v2 = 2gl(1− cosθ)

⇒ v =√

2gl(1− cosθ) =√

2(9.8 m/s2)(0.75 m)(1− cos10)

v = 0.47 m/s

40. F = mg sinθ = (0.5 kg)(9.8 m/s2)sin(15) = 1.27 N

41. f = ω

2π= 1

2π

√km

= 1.59 Hz

42. Using energy conservation:

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12

kA2 =12

mtotalv2max

→ A2 =mtotalv2

max

k

⇒ A =

√mtotalv2

max

k=

√(3.0 kg+4.0 kg)(2.14 m/s)2

12.0 N/m= 1.63 m

∆A = 2.50 m−1.63 m = 0.87 m

43. (a) T = 2π

ω= 2π

√Lg= 2π

√3.3 m

9.8 m/s2 = 3.65 s

(b) At equilibrium, all energy in the system is kinetic:E = K = 1

2 mv2 = 12(5.0 kg)(2.0 m/s)2 = 10.0 J

44. U = mgl(1− cosθ) = (4.4 kg)(3.0 m)(9.8 m/s2)(1− cos13) = 3.32 J45. ∑~F =−2T sinθi

∑~F =−2T(

x√x2 +L2

)i

46.

U = mgl(1− cosθ)

U1 = (4.0 kg)(9.8 m/s2)(2.5 m)(1− cos5.0) = 0.37 J

U2 = (4.0 kg)(9.8 m/s2)(2.5 m)(1− cos10.0) = 1.49 J

∆U = 1.49 J−0.37 J = 1.12 J

47.

f =ω

2π=

12π

√km

ky = mg

→ k =mgy

Therefore, the frequency can be found by:

f = 12π

√mgym

= 12π

√gy= 1

2π

√(9.8 m/s)2

(0.10 m)= 1.58 Hz

48. U = 12 kx2 = 1

2(125 N/m)(0.078 m)2 = 0.38 J49. At equilibrium, the magnitudes of the gravitational force and the force of the spring are equal:

F = kx = mg

→ kmgx

=(2.0 kg)(9.8 m/s2)

0.078 m= 251 N/m

⇒U =12

kx2 =12(251 N/m)(0.078 m)2 = 0.76 J

50. When the displacement angle, θ, is small (less than 10 or 15), sinθ≈ θ.

51. T = 2π

√Lg= 2π

√1.0 m

9.8 m/s2 = 2.0 s

52. Since the velocity is at the maximum value, that means that, at that instant, the object is not "speeding up" or"slowing down." Therefore, a = 0.

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53. No, SHM will only occur for small displacements from equilibrium in a simple pendulum.54.

T = 2π

√Lg

→ T 2 = (2π)2(

Lg

)⇒ L = g

( T2π

)2= 9.8 m/s2

(3.3 s2π

)2= 2.70 m

55.

E =12

kA2

LetA′ = 4A

→ E ′ =12

k(A′)2 =12

k(4 A)2 =12

k16A2

E ′ = 16(

12

kA2)= 16(E)

⇒ E ′ = 16E

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10.2 References

1. Laura Guerin. CK-12 Foundation .2. Laura Guerin. CK-12 Foundation .3. Laura Guerin. CK-12 Foundation .4. Laura Guerin. CK-12 Foundation .

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www.ck12.org Chapter 11. Vibrations and Sound Problem Sets

CHAPTER 11 Vibrations and SoundProblem Sets

Chapter Outline11.1 CHAPTER 11 PROBLEMS

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11.1. Chapter 11 Problems www.ck12.org

11.1 Chapter 11 Problems

Problems

1. If the given period of a wave is 0.210 seconds, what is the frequency?2. The world’s tallest building is reported to sway at a frequency of 0.05 Hz. Determine the period.3. While waiting in line at the Matsudo station in Japan, you notice that one of the trains is going so fast that 4

of the train car’s compartments pass you every second. If each train car is approximately 10 m long, calculatethe speed of the train.

4. For a pop quiz in class your professor creates a wave that has a horizontal peak to peak distance of 3.0 m andoscillates up and down 4 times a second.

(a) Calculate the frequency.(b) Calculate the wavelength.(c) Calculate the speed of the wave.

5. Meta and Aliya are each holding the end of a rope. With proper timing the two girls create traveling waveswith equal amplitude, which combine and double the original amplitude. Is it possible to do the exact oppositeand have the two girls create traveling waves that cancel one another out? Explain your answer.

6. While sitting on the corner, you see a cop car coming towards you with its sirens on. Does the speed of thesound wave increase or decrease as it comes towards you?

7. What is the relationship between frequency and period?8. Why are sine (or cosine) waves used when talking about waves?9. Define the following parts of a wave: period, amplitude, wavelength, frequency. Additionally, use a sketch to

show the amplitude and the wavelength.10. Calculate the period of a wave with frequency 107.8 Hz.11. Describe where mechanical waves originate from.12. Define the relationship between wavelength, wave speed and frequency.13. Is the direction of vibration parallel or perpendicular to the direction of motion for:

(a) A transverse wave?(b) A longitudinal wave?

14. Explain the superposition principle and how it is used.15. Both you and your friend each hold onto one end of a slinky and stretch it a little past its equilibrium point. If

you wanted to create a transverse wave in the slinky, what should you do?16. What kind of wave is created by snapping one end of a whip, sending a wave down it?17. Imagine an object is vibrating at a given frequency f0. If the frequency is doubled, what happens to the period?18. Are you more likely to first hear or smell a gas leak? Why? (Assume the gas leak is large enough that you can

hear it)19. Describe the difference between wave speed and wave frequency? What are their units?20. If frequency is decreased in a wave,

(a) What happens to the period?(b) What happens to the wavelength?

21. The speed of a wave is doubled.

(a) What happens to its wavelength if the frequency is held constant?(b) What happens to its frequency if the wavelength is held constant?

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22. While shooting water from a water gun, how could you have the stream of water exiting the toy gun mimic asine wave?

23. An object that is vibrating has its frequency doubled, what happens to the period?24. While preparing for a quiz, your lab partner uses the terms wave frequency and wave speed interchangeably.

Is he correct in doing so? If not, why?25. (a) Blue light has a greater frequency than red light. Which one has a greater wavelength?

(b) Orange light has a longer wavelength than blue light. Which has a greater frequency?(c) Indigo light has a shorter wavelength than yellow light. Which has a greater frequency?(d) If violet light has a greater frequency than red light, which one has a greater wavelength?

26. A thin rope is attached to a thick rope at the ends. If a wave is sent down the thick rope, what characteristicsof the wave change when the wave travels from the thick rope to the thin rope?

27. Determine the frequency of the second hand of a clock.28. Determine the frequency of the hour hand of a clock.29. While talking on an old fashioned corded phone you get bored with the conversation and begin moving the

cord up and down creating a wave. If you start moving your hand up and down even faster, what happens tothe wavelength?

30. In 1 period, how far does a wave travel. (Answer in terms of wave lengths)31. Demonstrate your knowledge of waves by answering the following question: How many nodes exist in a

standing wave that is 4 wavelengths long? (Do not count the end points)32. When a wave is created in water, what happens to the energy of the waves as the waves dissipate?33. A police car is driving towards a large wall while its siren is on. How is the reflected sound wave heard by the

cop different from the original wave?34. Should a listener move towards or away from a sound source to hear a higher frequency?35. Suppose a cop has turned his siren on to pull you over while speeding on the freeway. If the cop is going the

same speed as you, do you hear a change in the frequency of the siren? Explain.36. What is the basic principle behind radar guns?37. Determine the frequency for the following periods:

a. 0.25 s,b. 25 s.

38. Determine the period for the following frequencies:

a. 10 Hz,b. 60 Hz,c. 0.10 Hz

39. While sitting at the beach and staring towards the ocean, you notice the waves passing by a boat every 8seconds. Each time one of the waves just passes the end of the boat, you notice another wave is at the front ofthe boat. If the boat is 15 m long, what is the speed of the waves passing the boat?

40. A boat adrift in the ocean is seen moving up and down in 1.5 m high waves, 5 times every minute.

(a) What is the boat’s frequency in the vertical direction?(b) What is the amplitude of the boat?

41. A 0.5 kg mass is attached to a spring as seen in the figure below. If the mass has an amplitude of 0.1 m with aperiod of 0.5 seconds while oscillating up and down on the spring, what is the object’s period?

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42. If your favorite radio stations broadcasts at 107.9 FM, what is the wavelength of the waves received? (Assumethe speed of the radio waves is approximately 3*108 m/s and that the station’s frequency is measured in MHz)

43. What is the difference in wavelength from the waves received from the radio signal at 107.9 FM and the radiosignal at 83.5 FM?

44. The annoying sound you hear when a mosquito is near you is due to how fast its wings are moving persecond. If the wavelength of the sound produced by the mosquito is approximately 1.13 m, what is thesound’s frequency? Use 340 m/s as the speed of sound.

45. While sitting in a long line of traffic, the first car moves forward causing the other cars behind it to brieflymove forwards. Is this an example of a transverse or a longitudinal wave?

46. The frequency of a wave is doubled.

(a) What happens to the speed of the wave?(b) If the wave speed is unchanged, what happens to the wavelength of the wave?

47. While standing on the corner of a street you hear the siren of a police car approaching. As the car approaches,what happens to the frequency of the siren?

48. A sinusoidal wave traveling in the positive x direction has an amplitude of 20 cm, a frequency of 10.0 Hz anda wavelength of 55 cm. What is the wave speed and the angular frequency of the wave described?

49. Two emergency vehicles are traveling towards one another with their sirens on. Vehicle A has a speed of 10.0m/s while vehicle B has a speed of 15.0 m/s.

(a) If the frequency of the sirens is 700 Hz, what is the frequency that is detected by a passenger in vehicleA? (Assume the speed of sound is 331 m/s).

(b) After the vehicles pass by one another, what is the frequency that is heard by the passenger as the vehiclesdrive further and further apart?

50. Suppose it was your task in physics lab to create waves on a spring.

(a) Describe how you would create a longitudinal wave.(b) Describe how you would create a transverse wave

51. If a transverse wave in a taut string approaches a wall, will the wave invert upon reflection from the wall?52. What happens to the wavelength of a wave on a string if the frequency of the wave is doubled, assuming

everything else is unchanged?53. If the frequency of a wave on a string is doubled, what happens to the wave speed? (Assume that the tension

is unchanged)54. Will the wave speed change for a wave that is traveling from a heavier rope to a lighter rope? Explain.

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55. How does the wavelength and frequency change of a wave that is traveling from a heavier rope to a lighterrope?

56. If your cellphone rings in a perfect vacuum, will you hear it? Why?57. What is the relationship between the wavelength and the frequency of sound?58. If the wavelength of sound is reduced by a factor of 4, what happens to the wave speed?59. As an object emitting a given frequency approaches you, does the wavelength increase or decrease?60. Describe how an object emitting a given frequency can move without causing a Doppler shift in the frequency

for a nearby observer.61. Is it possible for strong winds to cause a Doppler shift?62. An oscillator generates waves on a string with a frequency of 1.3 Hz. If the speed of the wave is 50.0 cm/s,

what is the wavelength?63. If 5 crests pass a given point ever 15.0 s, calculate the wave speed of a wave that has a wavelength of 1.0 m.64. If a bat emits a sound towards a tree that is 175 m away, how long will it be before the bat hears an echo? Use

343 m/s as the speed of sound.65. Determine the wavelength in air for a frequency at 1.0 kHz.66. What is the result of the combination of two waves traveling towards one another with opposite displacements?67. When a standing wave is created with a string that is attached at both ends, is the number of nodes equal to

the number of antinodes?68. Consider a standing wave on a string attached at both ends. Is the following true or false: The length of the

string is equal to the wavelength multiplied by an integer number?69. What can be said about the center of a string on a standing wave whose ends are both fixed?70. When comparing two pipes (one with both ends closed, the other with one end open), what is the relationship

between the fundamental frequencies?71. What is the definition of a beat frequency?72. Determine the frequency of the 3rd harmonic for a note with a frequency of 260 Hz.73. (a) Determine the frequency of the first two harmonics of an open pipe of length 1.5 m.

(b) If the the pipe is closed at one end, what are the frequencies of the first two harmonics?

Solutions

1. f = 1T = 1

0.21 s = 4.8 Hz2. T = 1

f =1

0.05 s = 20 Hz3. The frequency of train cars is 4 cars/second. That means the period, or the time for one car to pass is .25

seconds. Speed is distance/time, s = 10/.25 = 40 meters.4. From the description of the problem:

(a) λ = 3.0 m(b) f = 4.0 Hz(c) v = λ f = (3.0 m)(4.0 Hz) = 12.0 m/s

5. Yes is it possible. This is the idea behind destructive interference. The parts of the rope the combinedestructively will have no amplitude.

6. Actually, the answer is neither. When the sound is coming towards you the speed of the wave doesn’t change,but the frequency does.

7. Frequency and period are inversely related, f = 1T

8. The cosine and sine functions are mathematical models that can be used to describe various type of wavephenomenon.

9. The frequency is the number of complete cycles while the period is inversely related to this value.

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10. T = 1f =

1107.8 Hz = 9.3 ms

11. Mechanical waves originate from vibrating objects.12. The relationship between wavelength, frequency and wave speed is f λ = v13. (a) For a transverse wave, the direction of vibration is perpendicular to the direction of travel.

(b) For a longitudinal wave, the direction of vibration is parallel to the direction of travel.14. The superposition principle states that linear systems must be additive, and can be used to add wave ampli-

tudes.15. To create a transverse wave, you would move your hand in a direction that is perpendicular to the length of

the slinky.16. When the pulse is sent down the whip, a transverse wave is created.17. Since the relationship between the period and the frequency is T = 1

f , when the frequency is doubled theperiod is halved.

18. You will more likely hear the gas leak before you smell it. The sound wave is transmitted by collisions, whichallows it to travel much faster than the individual gas molecules.

19. Wave speed is a measure of how fast a wave is traveling, while the wave frequency describes how long it takesfor the vibration to repeat. The units for wave speed are distance over time while the units for wave frequencyare inverse seconds.

20. (a) The period becomes larger because you are dividing by a smaller number.(b) The wavelength becomes larger.

21. (a) The wavelength would double as well.(b) The frequency is doubled.

22. If you move the water gun in an up and down motion that is perpendicular to the stream, the resulting streamwould mimic a sine wave. (This is assuming you pay careful attention to the distance you move your handboth above and below equilibrium.)

23. T = 1f , therefore if the frequency is doubled, the period is halved.

24. Wave frequency involves the rate of vibration while wave speed involves the rate of travel.25. (a) Red light has a greater wavelength than blue light.

(b) Blue light has a greater frequency than orange light.(c) Indigo has a greater frequency.(d) Red light has a greater wavelength.

26. When the wave changes from one rope to another both the wavelength and the wave speed will change.However, frequency remains unchanged.

27. f = 160 s = 0.016Hz

28. f = 1(12)(3600 s) = 2.31×10−5Hz

29. The frequency would increase, while the wavelength would decrease.30. 1 wavelength31. There are 7 nodes.32. The energy is spread out over the increasing area as the wave spreads out.33. The cop hears a higher frequency due to the Doppler Effect.34. The listener should move towards the source since the wave crests would be closer together.

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35. No, since you are going the same speed there is no Doppler Effect.36. Radar guns use the shift in the reflected frequencies to determine how fast an object is moving.37. (a) f = 1

T = 10.25 = 4.0 Hz,

(b) f = 1T = 1

25 = 0.04 Hz38. (a) T = 1

f =1

10 = 0.1 s,(b) T = 1

f =1

60 = 0.016 s,(c) T = 1

f =1

0.1 = 10 s39. v = λ f = (15 m)( 1

8 s) = 1.90 m/s40. (a) f = 5

60 s = 0.08 Hz(b) The amplitude is 0.75 m

41. T = 1f =

10.5 = 2 s

42. λ = vf =

3×108

107.9×106 = 2.78 m)

43. ∆λ = ∆vf =

3×108

83.5×106 − 3×108

107.9×106 = 0.812 m

44. f = vλ= 340 m/s

1.13 = 300 Hz45. This would be an example of longitudinal wave.46. (a) The wave speed is unaffected since it is determined by the medium you are in.

(b) The wavelength is cut in half.47. The frequency increases due to the Doppler shift.48.

v = λ f = (55 cm)(10 Hz) = 550 cm/s

ω = 2π f = 2π(10 Hz) = 63 rad/s

49. (a)

f ′ =(

v+ v0

v− vs

)f

f ′ =(

331 m/s+10 m./s331 m/s− (15 m/s)

)700 Hz = 755 Hz

(b)

f ′ =(

v+ v0

v− vs

)f

f ′ =(

331 m/s+(−10 m./s)331 m/s− (−15 m/s)

)700 Hz = 649 Hz

50. (a) You would create a longitudinal wave by stretching or compressing some of the coils of the spring andletting go.

(b) A transverse wave could be created in a spring by simply moving one end side to side or up and downrepeatedly.

51. Yes, it will invert.52. The wavelength will be reduced by 1

2 .53. The speed of the wave remains constant.54. Yes, the speed of the wave will increase due to the decrease in linear mass density.55. The wavelength will increase while the frequency will be unchanged.56. No, there is no medium to transmit the sound, therefore nothing will be heard.57. The two quantities are inversely related.58. The wave speed will be unchanged.59. The frequency increases, therefore the wavelength decreases.60. If both observers were moving with the same velocity, there would be no Doppler shift in the frequency.

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61. No, the winds cannot cause a Doppler shift, but it can affect an existing shift.62. λ = v

f =(50.0 cm/s)

1.3 Hz = 38.5 cm/s63.

f =5

15 s= 0.33 s

v = λ f = (1.0 m)(0.33 s−1) = 0.33 m/s

64. t = dv = (175+175)

343 = 1.02 s65. λ = v

f =343 m/s1.0 kHs = 0.34 m

66. The waves completely cancel each other out when they completely overlap, before they continue on in theiroriginal direction.

67. No, the number of nodes is one greater than the number of antinodes.68. No, this isn’t always the case. Consider an odd number of modes.69. The center is either a node or an antinode.70. fclosed = 1

2 fopen

71. fbeat = f1− f272. f3 = 3 f1 = 3(260 Hz) = 780 Hz73. (a)

f1 =v

2 L=

343 m/s2(1.5 m)

= 114 Hz

f2 = 2 f1 = 228 Hz

(b)

f1 =v

4 L=

343 m/s4(1.5 m)

= 57.2 Hz

f3 = 3 f1 = 171.5 Hz

To be deleted

58

1. Name two parameters that could be manipulated if you wanted to increase the rate of energy that is transferredby a wave down a string.

1. Increasing the tension, frequency, or amplitude would increase the rate of energy transfer.

60

1. A sinusoidal wave traveling in the positive x direction has an amplitude of 20 cm, a frequency of 10.0 Hz anda wavelength of 55 cm. Determine the period and the wave number of the wave.

T =1f=

110 Hz

= 0.1 s

k =2π

λ=

2π rad55 cm

= 0.12 rad/cm

62-64

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1. A 3.0 kg object is suspended by a rope of mass 0.4 kg and length 5.0 m over a pulley with negligible mass. Ifthe other end of the rope is attached securely to an object of mass m=200.0 kg, calculate the speed of a wavein this rope.

2. If the hanging mass in the previous problem is doubled and the length of the rope is cut in half, determine thenew speed of a pulse in the rope.

3. A string has waves traveling along it with an angular frequency of 470.0 s−1, a wave speed of 50.0 m/s andan amplitude of 5.0 cm. If the linear mass density of the string is 5.0*10−1 kg/m, how much power is beingsupplied to generate these waves?

62. v =

√Tµ=

√mweightg

mropel

=

√(3.0 kg)(9.8 m/s2)(5.0 m)

0.4 kg= 19.2 m/s

63. v =

√Tµ=

√mweightg

mropel

=

√(2)(1

2

)(3.0 kg)(9.8 m/s2)(5.0 m)

0.4 kg= 19.2 m/s

64. The result is the same because the changes cancel each other out.65. P = 1

2 µω2A2v = 12(5.0×10−2 kg/m)(470 s−1)2(0.05 m)2(50 m/s) = 690 W

69

How much would the tension in a taut string need to be increased by to double the wave speed?

Due to the square root, the tension would need to be 4 times larger than original.

83-87

1. y = (3.0 cm)sin[(1.5 rad/m)x− (3.0 rad/s)t]2. What is the wavelength and wave speed of the wave in the previous problem?3. A 3.0 m long cord has a mass of 0.30 kg. What is the linear mass density of this cord?4. If the tension on a rope is 5.0 N and the resulting wave speed of a wave traveling along the rope is 25.0 m/s,

what tension would be required to have a wave speed of 35.0 m/s?5. A taut rope with linear mass density of 0.05 kg/m has a series of waves traveling along it with an amplitude of

0.1 m, a wavelength of 0.5 m and a wave speed of 20.0 m/s. How much power must be supplied to generatethese waves?

A = 3.0 cm = 0.03 m

f =ω

2π=

3.02π

= 0.48 Hz

λ =2π

k=

2π

1.5= 4.2 m

v = λ f = (4.2 m)(0.48 s−1) = 2.0 m/s

85. µ = mL = 0.3

3.0 = 0.1 kg/m

86. Since the linear mass density is unchanged

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µ =Tl

v21=

T2

v22

⇒ v2 =T2

µ

T2 = v22µ =

(v2

v1

)2

T1 = (1.96)(5.0 N) = 9.8 N

87. P = 12 µω2A2v = 1

2(0.5 kg/m)(2π(20.0

0.5 s−1)2(0.1 m)2(20 m/s) = 4.7 kJ

(97)

1. If one string has a fundamental frequency of 260 Hz and another string’s fundamental frequency is 480 Hz,what is the ratio of the tensions assuming the linear mass densities are the same?

87. T1T2

=(

f1f2

)2= (480

260)2 = 3.4

(100)

Define antinodes.

100. Antinodes are points of maximum amplitude.

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www.ck12.org Chapter 12. Fluid Mechanics Problem Sets

CHAPTER 12 Fluid Mechanics ProblemSets


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Problems

Note: Densities for various substances can be found at http://en.wikipedia.org/wiki/Density#Various_materials .

1. Is the following statement correct: An object experiences a buoyant force of 5 N if the weight of the fluid thatis displaced weighs 5 N?

2. An object of volume 1000 cm3 has a mass of 25.0 kg. Calculate the buoyant force that acts on this containerwhen it is put into water.

3. When an object sinks, how does the buoyant force change?4. Two objects of identical size and shape but different masses are submerged. Which one experiences a greater

buoyant force?5. Why is it harder to float in a body of fresh water than in a body of salt water?6. Are hydraulic devices used to multiply energy or force?7. How are force and pressure related?8. How are pressure in a liquid and the depth of the liquid related?9. If you compare two points at exactly the same depth in saltwater and freshwater, which would have the greater

pressure?10. What is the relationship between the buoyant force on an object that is submerged and the weight of the water

that is displaced?11. Will an object that is less dense than water will float or sink? What about an object that is denser?12. (a) What is the pressure 50 m down in a tower filled with water? (Ignore atmospheric pressure.)

(b) What is the pressure if you account for atmospheric pressure?13. You’re drinking out of your special yard long drinking glass while working on your physics homework. If the

glass is filled with regular water, what is the pressure at the bottom?14. Calculate the water pressure in the bottom of a 100-story-tall water tower. (Assume that one story is 10 ft.)15. While swimming 75 meters down from the surface from the ocean, how much does the pressure change from

50 m down to 75 m (ρ=1020 kg/m3)?16. What is the relationship between the pressure 3 ft. beneath the surface of a small swimming pool and the

pressure 3 ft. beneath the surface of an Olympic pool twice the size of the small swimming pool?17. While driving through the city, you notice several towers that hold water elevated high in the air. What is their

purpose?18. You have two blocks: one is made out of aluminum, and the other is made out of silver.

(a) If each block has the same volume, what can be said about the amount of liquid they displace when putinto a container of water?

(b) If each block has the same mass, what can be said about the amount of liquid they displace when putinto a container of water?

(c) If each block has a weight of 1 N, what can be said about the amount of liquid they displace when putinto a container of water?

19. In some cartoons you see a character plug a hole in a dam by using his finger. Explain how this could bepossible in real life.

20. A tin ingot floats in mercury but not in water. Explain why.21. Why does an underwater ping pong ball have a buoyant force that is larger than a floating ping pong ball?

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22. While floating down a long river on a raft, your newly discovered gold has the water level dangerously nearthe top of the raft. Your raft mate suggests tying the treasure to a rope and letting it hang under the raft whileyou drift to your home base. Will his idea work? Why?

23. If an object starts sinking in water, will it continue to sink?24. A ferry floating across a lake is carrying several high priced sports cars. If some of the cars accidentally rolled

off the ferry, would the water on the shore rise, fall or remain unchanged?25. Assuming all the cars on the ferry from the previous problem got dumped into the lake, what would happen to

the water level?26. While an object sinks, its buoyant force increases. Is this correct?27. While Billy is testing how far down he can swim in the ocean, he notices that his buoyant force decreases the

further he goes. How is this possible?28. When submerged in water, it is possible to change your density. However, the density of materials like gold

does not change. How is this possible?29. When water is hot, the water molecules have a higher average kinetic energy than cold water. Knowing this

fact, which would you rather have: a leak in a container of cold water, or a leak in a container of hot water?30. The density of water is 1.0 and the approximate density of ice is 0.8. If your friend pours a liquid that forms a

layer between the ice cubes and the water, what can you say about the density of the liquid?31. If one object is floating on top of another, what is the most basic thing that can be said about the relationship

between the two objects?32. What is the pressure at the bottom of a dam which has a water depth of approximately 125 m? (Ignore

atmospheric pressure.)33. A 0.2 m tall glass is half full of water. What is the pressure at the bottom of the glass if you ignore the pressure

due to the atmosphere?34. A 10 kg object displaces 1.0 L of water when it is submerged. Calculate the density of the object.35. If the density of an object is 5×103 kg/m3, what volume of water would be displaced if the mass of the object

was determined to be 25 kg?36. A raft with dimensions 6 m x 3 m floats down a river with no weight on it. Calculate how much of the raft

sinks if 200 kg worth of cargo are loaded onto the raft.37. If the raft in the previous problem only has 5 cm above the water with no cargo on it, how much cargo can be

added until the top of the raft is flush with the water?38. A 2 cm2 hole appears shooting water out at the base of a 5 m tall park dam. Making a split second decision,

you try plugging the hole with your finger. How much force must you apply to the hole to stop the water?39. While digging in your back yard, you find what appears to be a solid 1 kg gold brick. Remembering your

course in physics, you decided to submerge the brick in water to determine its density in the hopes to determineif it really is solid gold. How much water does the brick need to displace to confirm it is gold?

40. Which would be more painful, having your foot stepped on by a 500 lb. elephant, or having your foot steppedon by a 100 lb. woman wearing heels?

41. Express the definition of pressure in equation form and then describe the expression in words.42. Is the pressure at the bottom of a container of water ρ=1030 kg/m3 higher or lower than the pressure at the

bottom of a container that is filled with glycerin ρ=1260 kg/m<sup3?43. If an object is just submerged under the surface of a liquid, how much more force is needed to submerge the

object further?44. Water is flowing through a pipe with a radius of 2 cm before entering another pipe which has a radius of 5 cm.

In which pipe is the speed of the water greater?45. A sealed treasure chest is sitting 10 m below the surface of the water. If a hole suddenly forms in the side of

the treasure chest, at what speed does the outside water enter at?46. Two objects of equal volume are put into a large bucket of water. If the first object is iron and the other is

copper, which one experiences a greater buoyant force?47. At which point, a or b, in the figure below is the pressure greater?

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48. How is it possible to levitate small spherical objects by simply blowing air over them?49. Would you expect the water pressure from a faucet on the first floor of a house to be greater or less than the

pressure on the second floor of the same house?50. If you wanted to increase the buoyant force on your body while you are swimming, would you inhale as much

air as possible or would you want to blow out as much air from your lungs as possible?51. While fishing in a small lake, you throw a cheap makeshift anchor, a large rock, overboard. What happens to

the water level of the small lake you are in? Why?52. Determine the mass of a solid sphere of tin that has a radius of 1.25 cm.53. Determine the volume of a solid iron sphere of mass 100 g.54. You are asked to determine the density of an unknown material in a lab. A cube made of the material has a

mass of 100 g with side length 50 cm.55. The tires of a heavy duty pickup truck have a gauge pressure of 250 kPa. If the area of each of the truck tires

in contact with the ground is 0.03 m2, determine the weight of the pickup truck.56. Assume the vehicle in the previous problem is an 18-wheeler, semi-truck. What would the weight of the semi

be? (Assume same area and tire pressure.)57. Determine the absolute pressure 100 m below the surface of the ocean, assuming that the air pressure above

the surface is 101.3 kPa.58. Calculate the pressure exerted on the floor by a 45.0 kg woman standing on the heels of a pair of stiletto heels

with a radius of 0.25 cm.59. A common demonstration seen on vacuum cleaner commercials is a vacuum picking up heavy objects. If the

diameter of the hose seen on the commercial is 2.5 cm, calculate the heaviest object that the can be pickupwith no attachments. Assume that the vacuum cleaner creates a perfect vacuum.

60. A large pit filled with water has base with dimensions 25 m by 35 m. If the pit is filled with 15 m of water,what is the force caused by the water at the bottom of the pit?

61. For one of your physics labs, the instructor asks you to determine the height h of liquid mercury column seenin the figure below (Assume normal atmospheric pressure.)

62. Calculate the density of an unknown liquid that creates a column of water h=1.0 m high when placed in the

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barometer at normal atmospheric pressure, as shown in the figure below.

63. While outside playing you notice that the height of the mercury barometer on the side of your house decreasesby 25.0 mm. What is the atmospheric pressure?

64. A slab of a given density ρ and thickness h has a mass m on top of it. When placed in fresh water, the top ofthe water is level with the top of the slab. Determine the area of the slab.

65. What force is required to hold a ball of density 0.075 g/cm3 and radius 1.6 cm completely submerged underwater?

66. 1.5 N of force is needed to hold a ball completely submerged underwater. If the density of the ball is 0.05g/cm3, determine the radius of the ball.

67. A small wooden block that is initially floating in a container of water has a force applied on it so it is completelysubmerged under that water. Draw a free body diagram of the forces acting on the wooden block.

68. A 15.0 kg block, suspended from a string is immersed in a container of water as seen in the figure below.Using the following dimensions, calculate the forces acting on the top and bottom of the block: a=11.0 cm,b=10.0 cm, c=15.0 cm, and d=6.0 cm.

69. A wood cube of dimensions with an edge length of 15.0 cm and density of 800 kg/m3 floats in a small containerof water. Calculate the vertical distance of the wooden block above the surface of the water.

70. A small cube of edge length 13.0 cm has a density of 700 kg/m3 floats in a large body of water. Calculate theadditional mass that should be placed on top of the cube to make the top of the cube level with the water.

71. A hose that is used to fill a 2.0 L bottle has a radius of 1.0 cm. If it takes 2.0 minutes to fill the 2.0 L bucket,calculate the speed of the water in the hose.

72. If the hose in the previous problem has a nozzle attached to the end where the radius is one half the originalsize, determine the speed of the water in the nozzle.

73. A 5.0 cm hose has a flow rate of 0.009 m3/s. The end of the hose has a nozzle attached which has a diameterof 1.75 cm. Determine the speed of the water that is exiting the nozzle.

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74. What is the relationship between the flow rate of a fluid inside of a pipe and the velocity of the fluid?75. A large container of height h is filled with water. At a height h’ below the surface, the container is punctured

and water shoots out, shown in the figure below. Derive an expression for the velocity of the water exiting thispoint.

76. A large container of height h is filled with water and placed on the ground. At a height h′ below the surface, thecontainer is punctured and water shoots out, shown in the figure below. Derive an expression for the distancethe water stream travels before hitting the ground.

77. A large container of water open at the top develops a hole below the water level. If the flow rate from the leakis 3.3×10−3 m3/min, determine the diameter of the hole if the speed of the water exiting is 19.0 m/s.

78. The large piston on a hydraulic lift has a radius of 30 cm. What force must be applied to the small piston ofradius 3.0 cm in order to raise a mass of 2000 kg located on the larger piston?

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79. Calculate the force needed in problem 87 if the 2000 kg mass is located on the small piston - how much forceshould be applied to the larger piston to raise the mass on the smaller piston?

80. Convert 100 mmHg to Pascals.81. A block only weighs 6.0 N while submerged in water, but weighs 10 N in air. Calculate the specific gravity of

the material.82. Lead has a specific gravity of 11.3. If the weight of a lead cube is 50 N, what does it weigh when submerged

in water?83. A champagne cork has a density of approximately 200 kg/m3. Calculate the fraction of the cork that is

submerged when the cork floats in water.84. Find the mass of a solid copper cylinder of length 4.0 cm and diameter 3.0 cm.85. Determine the mass of an iron sphere of radius 2.5 cm.86. In a recent horror movie explorers discovered a long lost ruin that had a door made out of solid bone. If the

door is 1.0 m high, 0.80 m wide and 0.5 m thick, determine the mass of the door if the density of bone isapproximately 1.7×103 kg/m3.

87. A block of unknown origin weighs 5.00 N in air and 4.55 N when completely submerged in water. Whatmaterial is the block most likely made of?

88. If ,a solid metal weighs 100 N in air and 62.5 N when submerged in water, calculate the specific gravity of themetal.

89. A small chunk of bread floating in a puddle of water has two-thirds of its volume submerged in water. Whatis its density?

90. A sphere of volume 20 cm3 weighs 100 g according the balance it is hanging from while submerged in water.Calculate the density of the sphere.

91. In some towns, the town’s water pressure is provided by large water towers approximately 100 ft. tall.Calculate the gauge pressure at ground level near one of these towers. (Assume the tower is open at thetop.)

92. A 1.0 kg object is placed in a bucket of water. While in the water, the object weights 0.5 kg. What is thedensity of this object?

93. A metal of unknown density with dimensions 100 x 100 x 100 cm floats in a pool of mercury with only 10 cmvisible above the liquid. If you cut off the top 10 cm that is visible above the surface of the liquid, how muchof the remaining metal would be visible above the surface of the liquid?

94. Swimming into an enemy stronghold a spy needs to take enough gear to complete his mission. If the spy iscarrying so much equipment that his total mass is 90 kg and his density is the same as the water, what is hisvolume?

95. If the spy in the previous problem has to dive 3 meters down to make it into a secret tunnel, what is the buoyantforce acting on the spy?

96. A confused house cat is seen floating in the ocean with 20% of his body above the surface. If the density ofocean water is approximately equal to 1.02*103 kg/m3, what is the cat’s average density?

97. In a system made of two hydraulic pistons with diameters 3 and 9 cm respectively, how much more force isthe larger piston able to exert over the smaller piston?

Solutions

1. Yes. This is similar to force pairs and how things in equilibrium are balanced.2. The weight of the 1000 cm3 (1 liter) of water displaced is what determines the buoyant force. Since the mass

of 1000 cm3 = 1 L is 1.0 kg, the total weight of the displaced fluid is 9.8 N. Therefore, FB = 9.8N upward.3. When an object sinks the buoyant force does not change since it is only dependent upon the fluid that it

displaces not on the position of the object that is submerged.4. Neither. The buoyant force is related to the volume of the water that is displaced.5. It is harder to float in fresh water because it is slightly less dense than salt water.

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6. Hydraulic devices are used to multiply force. This is done through the principal of Pascal’s Law.7. Force can be seen as a push or a pull; while the pressure is defined as the force over a given area.8. The depth of a liquid is proportional to the pressure:P = ρgh, where ρ is the density of the liquid, g is gravity

and h is the distance between the surface and the point of interest in the liquid.9. Freshwater is less dense than saltwater; therefore you would experience a larger pressure in the saltwater.

10. FB =Wwaterdisplaced11. If an object is less dense than the liquid it is being put it, it will float; if it is denser, than it will sink.12. (a) P = ρgh = (1000 kg/m3)(9.8 m/s2)(50 m) = 4.9×105 Pa

(b) P = Po +ρgh = (1.01×105 +4.9×105)Pa = 5.91×105 Pa13.

1 yard = 0.91 m

P = Po +ρgh = (1.01×105 +(1×103)(9.8)(0.91))Pa = 1.09×105 Pa

14.

1000 f t = 304.8 m

P = ρgh = (1.01×105 +(1×103)(9.8)(304.8))Pa = 3.09×106 Pa

15.

ρsaltwater ≈ 1.02×103 N/m3

∆P = ρg∆h = ((1.02×103)(9.8)(25))Pa = 2.50×105 Pa

16. The pressures would be equal, as the pressure is not dependent on the area of the pool.17. The added elevation helps supply the required pressure.18. (a) Since both blocks have the same volume, they will displace equal amounts of water.

(b) Since the silver is denser, a larger amount of aluminum is needed to have the same mass as the silver.Therefore, the aluminum will displace more water.

(c) Using the same logic as (b), the aluminum will displace more liquid.19. Depending on the depth of where the leak is with respect to the water on the other side this could be a

reasonable way to plug a hole in a dam.20. The density of tin is less than that of mercury so it will float, but the density of tin is greater than that of water,

so it will sink.21. It has a larger buoyant force because it has displaced a greater amount of water.22. Your raft mate has a great idea. By placing the treasure in the water, the buoyant force on the container makes

it less heavy, thereby reducing the overall weight of the raft-treasure system.

23. If an object sinks, it will continue to do so because it is less dense than the matter it is in.24. The water level will fall. The submerged cars displace only their volume.25. The water level would still fall.

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26. No, the buoyant force on an object remains unchanged as it sinks because the buoyant force is proportional tothe objects volume.

27. His buoyant force decreases due to the fact that his body is being compressed. As his body is compressed, histotal volume decreases. It is important to remember that there are many compressible air pockets in the body,such as the lungs.

28. It is possible because our bodies being compressible, which changes our total volume. Our bodies containmultiple compressible air pockets, the largest of which is the lungs.

29. It would be better to have a leak in a container of cold water because the rate of liquid leaving would beslightly slower.

30. The density of the liquid is between 0.8 and 1.0. It is denser than ice, but less dense than water31. The object on top is less dense than the object that it is floating on.32. P = ρgh = (1000 kg/m3)(9.8 m/s2)(125 m) = 1225 kPa33. P = ρgh = (1000 kg/m3)(9.8 m/s2)(.2 m) = 1.96 kPa34.

1000 L = 1 m3

ρ =mV

=10 kg

0.0005m3 = 2×104 kg/m3

35. V = mρ= 25 kg

5×103 kg/m3 = 5×10−3 m3

36.

FB = mg

ρV g = mg

ρ(Ah) = m

h =mρA

=200 kg

(1000 kg/m3)(18 m2)= 0.01 m = 1 cm

37.

FB = mg

ρV g = mg

ρ(Ah) = m

m = (1000kg/m3)(18m2)(0.05m) = 900 kg

38.

P = ρgh = (1000 kg/m3)(9.8 m/s2)(5 m) = 49 kPa

F = PA = 49 kPa(0.0002 m2) = 9.8 N

39.

ρgold = 19.3 g/cm3

V =mρ=

1000 g19.3 g/cm3

V = 51.8 cm3

40. It would be more painful to have the 100 lb. woman step on your foot because she would be applying the forceover a smaller area.

41. P = FA ; Pressure is defined as the force divided by the area that the force is applied over.

42. The pressure would be greater at the bottom of the container of glycerin due to it being denser.

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43. When an object is submerged the buoyant force does not depend on depth.44. The water flows more rapidly in the pipe with the smaller radius.45. From Bernoulli’s eqn:

Po = po +12 ρv2 +ρgh

let h =−y under the surface of the waterv =

√2 gy =

√2(9.8 m/s)(10.0 m) = 14 m/s

46. Neither; the buoyant force is the density of water times the volume displaced. Since both objects are of equalvolume, the buoyant force is the same.

47. Neither; pressure in a liquid is due to the vertical distance from the surface of the water. Therefore, both areat the same pressure.

48. By blowing air very rapidly over small, light spherical objects, you cause a region of lower pressure above theball. This causes the higher pressure underneath the ball to lift it up.

49. The water pressure would be greater on the first floor because of its lower height compared to the water’ssource.

50. By inhaling as deeply as possible, you would increase your volume, thereby increases the buoyant force onyour body.

51. The water level in the lake falls, since the anchor is denser than water.52.

ρ =MV

M = ρV =(

7.26g

cm3

)(43

π(1.25 cm)3)= 59.4 g

53. V = mρ= 100 g

7.78 g/cm3 = 12.7 cm3

54. ρ = mV = 1000 g

(5 cm)3 = 8.0 g/cm3

55. F = 4AP = 4(0.03 m2)(250×103 Pa) = 3.00×104 N56. F = 18AP = 18(0.03 m2)(250×103 Pa) = 1.35×105 N57. P = Po +ρgh = 101.3×103 Pa+(1024 kg/m3)(9.8 m/s2)(100 m) = 1.10×106 Pa

58. P = FA =

12 45.0 kg

π(0.0025 m)2 = 1.15×106 N/m2

59. F = PA = (1.013×105 Pa)(π(1.25×10−2 m)2) = 49.7 N60. At the bottom

P = ρgh = (1000 kg/m3)(9.8 m/s2)(15 m) = 1.53×104 PaThe force at the bottom isF = PA = (1.53×104 Pa)(875 m2) = 1.34×107 N

61.

Po = ρgh

h =Po

gρ=

1.013×105 Pa(9.8 m/s2)(13.6×103)

= 0.76 m

62.

Po = ρgh

ρ =Po

gh=

1.013×105 Pa(9.8 m/s2)(1.0 m)

= 1.03×104 kg/m3

63.

∆P = ρg4h = (13.59 g/cm3)(980 cm/s2)(−2.5 cm) =−0.033×105 Pa

→ P = Po−∆P = 0.98 Pa

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64.

(m+ρslabV )g = ρwaterV g

V = Ah

→ (m+ρslabAh)g = ρwaterAhg

⇒ A =m

(ρwater−ρslab)h

65.

F +mg = FB

F = ρwV g−mg

F = ρwV g−ρballV g

F =V g(ρw−ρball)

F =43

π(1.6×10−2)3(9.8 m/s2)(1×103 kg/m3−75.0 kg/m3) = 0.16 N

66.

F =V g(ρw−ρball)

V =43

πr3

→ r = 3

√34

Fg

1(ρw−ρball)

r = 3

√3

4π

1.5 N9.8 m/s2

1(1×103 kg/m3−50.0 kg/m3)

= 0.034 m

67.

Ptop = Po +ρgh = 1.01×105 +1.00×103(9.8 m/s2)(0.06 m) = 1.02×105 Pa

Pbot = Po +ρgh = 1.01×105 +1.00×103(9.8 m/s2)(0.21 m) = 1.03×105 Pa

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68.

ρwaterVwaterg = ρwoodVwoodg

ρwaterA(15.0 cm−h)g = ρwoodVwoodg

h = 15.0 cm− ρwoodVwood

ρwaterA= 15.0 cm− (800 kg/m3)(0.15 m)3

(1×103 kg/m3)(0.15 m)2 = 3.0 cm

69.

FB = Fg +mg

ρwVwg = ρcVcg+mg

(1×103 kg/m3)(0.13 m)3g = (700 kg/m3)(0.13 m)3g+mg

m = 0.66 kg

70.

A1v1 = A2v2

2.0 L120.0 s

(1000 cm3

1 L

)= π(1.0 cm)2v

v = 5.31 cm/s

71.

A1v1 = A2v2

2.0 L120.0 s

(1000 cm3

1 L

)= π(0.5 cm)2v

v = 21.2 cm/s

72.

rate = v1A1 = v2A2

v2 =rate

A=

0.009 m3/sπ(0.00875)2 = 37.4 m/s

73. rate = velocity×Area74. From Bernoulli’s eqn.

Po +ρgh = Po +12

ρv2

h = h′

→ v =√

2gh′

75. From Bernoulli’s eqn., the water exiting is→ v =

√2gh′

Therefore, x = vxt =√

2gh′

√2(h−h′)

g= 2√

h′(h−h′)

76.

rate = Γ = Av =(

πd2

419.0 m/s

)d =

4Γ

π(19.0 m/s)= 2.21×1014 m

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77. The pressure due to the mass

F2 = PA2

→ P =mgA2

Relate this to the small piston

F = PA1 = mgA1

A2= mg

r21

r22= (2000 kg)(9.8 m/s2)

(0.030.3

)2

= 196 N

78. From the previous problem:F = PA1 = mg A2

A1= mg r2

2r2

1= (2000 kg)(9.8 m/s2)

( 0.30.03

)2= 1.96×106 N

79. P = 100 mmHg(

101.325 kPa760 mmHg

)= 13.3 kPa

80. specific gravity =wair

wloss in water= 10 N

4 N = 2.581.

specific gravity = SG =wair

wloss in water

wloss in water =wair

SG

→ wwater = wair−wair

SG= wair

(1− 1

SG

)= 45.6 N

82.

FB = Fg

ρcVcg = ρwV ′g

V ′

V=

ρc

ρw=

2001000

=15

83. m = ρV = 8.93×103 kg/m3(π(0.015 cm)2(0.04 m)) = 0.03 kg84. m = ρV = 7.96×103 kg/m3(4

3 π(0.025 cm)3) = 0.52 kg85. m = ρV = 1.7×103 kg/m3(.4 m3) = 680 kg86. Specific gravity = SG =

wairwloss in water

= 5.45 = 12.5

The Material is Lead87. Specific gravity = SG =

wairwloss in water

= 10037.5 = 2.67

88.

FB = Fg

ρwV ′g = ρbVbg

ρbread = ρwV ′

Vb=

23

ρw = 0.66 g/cm3

89.

FB +Fs = mg

ρ =mV

+ρw =100 g

20 cm3 +1 g/cm3 = 6 g/cm3

90. P−Po = ρgh = (1000 kg/m3)(9.8 m/s2)(30.48 m) = 2.99×105 Pa

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91. Since the object’s mass is reduced by 0.5 kg

0.5 kg(

1 liter1 kg

)(10−3 m3

1 liter

)= 5×10−4 m3

ρ =MV

=(1 kg)

(5×10−4 m3)= 2000 kg /m3

92. Before the system is cut, 10% of the total height of the metal is visible. Therefore, after being cut, only .9 cmwould be seen.

93. V = mρ= (120 kg)

(1000 kg/m3)= 0.12 m3

94. FB = ρV g = (1000 kg/m3)(0.12 m3)(9.8 m/s2) = 1.18 kN95. ρ = (.80)(1.02 kg/m3) = 816 kg/m3

96. The answer can be found by looking at the ratio of the areas.The area of a circle is A = πr2

Therefore, A2A1

=πr2

2πr2

1= (4.5)2

(1.5)2 = 9

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www.ck12.org Chapter 13. Heat Problem Sets

CHAPTER 13 Heat Problem SetsChapter Outline

13.1 HEAT PROBLEM SETS

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13.1. Heat Problem Sets www.ck12.org

13.1 Heat Problem Sets

Solutions

1. Your classmate is incorrect. Temperature is a measure of the average kinetic energy, not the total.2. By definition, the internal energy is the total of ALL energies, kinetic and potential, of a system.3. The temperature would only increase by 2°C since the container is twice the size of the first.4. Ignoring the effects that may be caused by driving a nail into a material, the nail heats up due to the compres-

sion of the atoms inside the nail.5. T = 32F , T = 0C6. Energy due to motion from one direction to another.7. None, all of them affect the kinetic energy.8. Energy is transferred from the hotter object to the colder object.9. Temperature is a measure of the average kinetic energy, while heat is just thermal energy.

10. Heat is the flow of energy, while internal energy is the sum of the total kinetic and potential energies in asystem.

11. The direction is determined by which system has the lower average kinetic energy. Heat flows from a regionof higher average kinetic energy to a region of lower average kinetic energy.

12. 1 Calorie = 1000 calories.13. 1 calorie = 4.184 joules.14. Zinc has a lower specific heat, so it gets warmer more quicker.15. Iron has a higher specific heat, so it takes longer to heat up.16. (a) If a material heats up quickly, it means that it has a low specific heat.

(b) If a substance cools quickly, it must have a low specific heat.17. The heat is transferred to the surrounding area: air, land etc. This explains why the temperature near large

bodies of water is usually constant.18. Ice is less dense than water because the volume of water increases when it freezes, decreasing the overall

density.19. An increase of one degree Celsius.20. No, the molecules only have the same average speed at a given temperature.21. If you’re running a fever, your whole body temperature is increased. Therefore, there will not be a noticeable

temperature difference between your hand and your forehead.22. A calorie with a capital C is worth 1000 calories with a small c, so the Calorie is much larger.23. The temperature change in the pool would be too small to measure due to the mass of the water involved.24. The gas pressure changes.25. As the temperature increases, the kinetic energy of the gas particles increases. This causes a greater pressure

on the walls of the containers.26. Yes, if the masses are not the same, or the objects have different specific heats.27. Silver has a greater temperature change; it has a smaller specific heat when compared to water.28. Silver has a greater temperature change; it has a smaller specific heat when compared to iron.29. An object with smaller specific heat would heat up more rapidly.30. The liquid is still cold because it has a much higher specific heat than the container, so it would not have

warmed as much.31. The sand has a very low specific heat.32. Since both containers have the same amount of liquid, the final temperature would be 30°C.

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33. Q = mc∆T = (20 kg)(4.18 kJ/kgK)(10K) = 835 kJ34. Q = mc∆T = (10 kg)(4.18 kJ/kgK)(20K) = 837 kJ35. Q = mc∆T = (5 kg)(0.386 kJ/kgK)(60K) = 116 kJ36. Let the 50 g of water be identified as 1, while the 30 g of water is represented by 2

Qin =−Qout

m1c(Tf −T1i) =−m2c(Tf −T2i)

Tf =(m2cT2i+m1cT1i)

m1c+m2c = c(m2T2i+m1T1i)c(m1+m2)

= 317 K = 44C37. Let the water be identified as 1, while the copper is represented by 2

Qin =−Qout

m1c1(Tf −T1i) =−m2c2(Tf −T2i)

Tf =(m2c2T2i+m1c1T1i)

m1c1+m2c1= (200)(4.18)(318)+(300)(0.386)(363)

(200)(4.18)+(300)(0.386) = 323 K = 50C38. The difference is in the shift of absolute zero on the scale.39. The internal energy of the gas in container B is half that of container A.40. The kinetic energy is related to the temperature. The kinetic energy of each gas is the same.41. −4C42. 283 K43. The temperature of the metal would have to decrease.44. They are in thermal equilibrium.45. The temperature is 673 K.46. ∆L = αLi∆T = (9×10−6)(0.2 m)(75C) = 0.135 mm47. ∆V = 3αVi∆T = (9×10−6)(4π(0.5)2(0.3))(80C) = 0.679 mm48. TF = 9

5(−20C)+32 =−4F49. TC = 5

9(−20F)−32 =−43C50. TF = 9

5(20 K−273)+32 =−423F

51. Vrms,H2 =

√3RTM

=

√3(8.314 J/mol ·K)(500 K)

2×10−3 kg/mol= 2.50 km/s

52. A change in 1C is larger.53.

TF,low =95(−196 C)+32 =−320 F

TF,high =95(160 C)+32 = 320 F

54.

T =95(T )+32

5T = 9T +160

T =−40

55.

T =95(T +273)+32

5T = 9T +2457+160

4T =−2617

T =−654

There is no number that matches on the two scales.56. No.57. The silver would experience a greater change in temperature due to its lower specific heat.

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58.

Qcold +Qhot = 0

let mw = mass of the water

mwcw(T −Tow) = mbcb(Tob−T )

→ cb =mwcw(Tf −Tow)

mb(Tob−Tf )=

(0.5 kg)(4186 J/kg ·C)(10.2C)

(1.0 kg)(224.8C)= 95.0 J/kg

59. The change in temperature for the mercury is 30x that of water due to the difference in specific heat.60. When dealing with temperatures over 100°C, the water may boil, losing some of its mass.61. The rate of heat transfer is much greater for the hardwood floor than the carpet. Your foot thus loses heat at a

greater rate to the hardwood floor, which makes it feel colder.62. The heavy curtains slow energy transfer into the house.63. The products being stored are kept at nearly constant cool temperatures due to the insulating properties of the

ground.64. Heat must travel a longer distance through the less conductive handle in order to reach your hand. A large

amount of the heat is dissipated by the time it reaches your hand.65. Convection. The bridge has much more surface area than the road, and rapidly loses heat to the air.66.

∆Q = mc∆T

Tf = Ti +∆Qmc

= 54.5C

67. (a)

Qin =−Qout

mc∆Twater =−mc∆Tiron

Let iron→ 2

water→ 1

Tf =m2c2T2i−m1c1T1i

m1c1 +m2c2=

(1.0 kg)(448 J/kg ·C)(500C)+(35.0 kg)(4186 J · kg ·C)(20C)

(1.0 kg)(448 J/kg ·C)+(35.0 kg)(4186 J/kg ·C)

Tf = 21.5C

(b)

Qin = Qout

mc∆Twater =−mc∆Tiron

Let Al→ 2

water→ 1

Tf =m2c2T2i−m1c1T1i

m1c1 +m2c2=

(1.0 kg)(900 J/kg ·C)(500C)+(35.0 kg)(4186 J/kg ·C)(20C)

(1.0 kg)(900 J/kg ·C)+(35.0 kg)(4186 J/kg ·C)

Tf = 23.0C

The final temperature is greater by 1.43C when aluminum is use68.

Q = (mcu +mwcw)∆T

In one minute

Q = [(0.3 kg)(387 J/kg ·C)+(0.5 kg)(4186 J/kg ·C)](−1.2C) =−2650 J

P =Qt=−2650 J

60 s= 44.2 J/s

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69.

Qin =−Qout

(mwcw +mccc)∆T1 =−mac∆T2

ma =(mwcw +mccc)∆T1

c∆T2=−[(0.20 kg)(4186 J/kg ·C)(0.025 kg)(387 J/kg ·C)](25C)

(4186 J/kg ·C)(−50C)= 0.10 kg

70. (a)

2(

12

mv2)= 2mc∆T

Tf =12

v2

c+To =

12

(100 m/s)2

128 J/kg ·C+10C = 49.0C

(b) The energy needed to raise the lead to its melting point (327C):Q = 2mc∆TTherefore, the initial velocity is:

2(

12

mv2)= 2mc∆T

v =√

2c∆T =√

2(128 J/kg ·C)(327−10) = 284.9 m/s

71. Q = mc∆T = (2.0 kg)(128 J/kg ·C)(20C) = 5.12 kJ

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Solutions

1. Classical thermodynamics focuses on macroscopic behavior.2. ∆V = NR∆T

P = NRP

3. 0 K, −459.67F4. Kinetic and potential energy.5. They are exactly the same thing. The energy in a closed system must remain constant.6. When heat is added, the internal energy is increased. This in turn increases the potential for the system to do

external work.7. Assuming no heat is transferred into or out of the system, the temperature will increase.8. An adiabatic process is one where no heat is exchanged.9. The internal energy decreases when work is being done by the system.

10. Heat will flow from a higher temperature body to a lower temperature body.11. The efficiency of a heat engine is defined as

ε = 1− TCQH

Therefore, both methods will increase the efficiency of the heat engine.12. If the temperature of the cold bath is increased, than the efficiency goes down.13. Yes, it is possible to heat up an apartment with a hot oven. It would be extremely inefficient.14. The term used to define the measure of disorder in a system is entropy.15. Heat is added to the system, work is done and heat is removed.16. When the gas is rapidly compressed, work is being done that increases the internal energy.17. Unless the two objects are made of the same material and have the same mass, their respective temperature

changes will be different.18. It does matter, 200K=-73°C but 200°C=473K.19. No, the difference is only 273°. The temperature is approximately 1,000,000 regardless if it is in Celsius or

Kelvin.20. U = Q−W = (140−90) J = 50 J21. Since the volume is held constant, the pressure is increased.22. Assuming constant volume, the pressure decreases.23. The difference is based on time scale. Renewable energy is derived from natural processes that renew

themselves over time.24. The efficiency will increase.25. If a heat engine is attached to a cold bath at 0 K, the efficiency would be 100%.26. The machine performs work to move heat from inside the freezer to outside. However, since the door is open,

no net heat transfer occurs. Instead, the work increases the internal energy of the room, which warms up theroom.

27. Fans increase the rate of evaporation on your skin, thereby cooling you without actually cooling the air that isblown at you.

28. You would want to increase the temperature of the heat bath.29. When the temperature is increased, the volume gets larger. This decreases the density.30. The second law of thermodynamics states that the universe becomes more disordered.31. A liquid is more ordered.32. A liquid is more disordered than a solid.

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33.

U = Q−W

U =W

→W = 2000 J

34.

U = Q−W

|U |= 2500 J

35. (a) The best possible efficiency occurs when engine operates in a Carnot cycle.η = 1− TC

TH= 1− 250

2500 = 0.9 = 90%(b) η = 1− TC

TH= 1− 5

120 = 0.96 = 96%(c) η = 1− QC

QH= 1− 500

3000 = 0.83 = 83%36.

W = 3500 Cal(

1000 cal1 Cal

)(4.186 J1 cal

)= 1.47×107 J

W = N(F ·d)⇒ N =WFd

N =1.47×107 J

(143 kg)(9.8 m/s2)(.71 m)= 14774 times

37. Temperature is a measure of the average kinetic energy of molecules while internal energy is the energy dueto molecular motion and molecular interactions.

38. An object with a high specific heat could be at a lower temperature, but contain more energy compared to ahigh temperature object with a smaller specific heat.

39. The temperature would increase by a small amount.40. No energy enters or leaves an isolated system; it simply changes form.41.

W =−13

A(V 3f −V 3

i ) =−13(A)((2Vi)

3−V 3i ) = A(7V 3

i )

42. W = P∆V = (4×106 Pa)(2 m3) = 8.0 MJ43. W = P∆V = (2×106 Pa)(2 m3)+ 1

2(2 m3)(6×106 Pa) = 10.0 MJ44. W = P∆V = 2P(0) = 0 J ·Work = 0 in an isochoric process.45. W = P∆V = (8.0 Pa)(2 m3)− (4.0 Pa)(2 m3) = 8.0 J46. Wdone = nR(Tf −Ti) = (0.30 mol)(8.314)(230C) = 574 J47. n =− W

R∆T = −700 J8.314(300C) = 0.28 moles

48.

U = Q+W

Q =U−W =−400 J−240 J =−640 J

49. (a) During a cyclic process, U = 0→ Q =−W = 1

2(4.0 m3)(4 Pa) = 8.0 J(b) The answer is the opposite of the previous problem, -8.0 J.

50. U = Q−P∆V = 14.0 KJ− (3.0 kPa)(3.00 m3) = 5.0 kJ51.

V1

T1=

V2

T2

T2 =V2

V1T1 = (2.0)(400 K) = 800 K

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52.

V1

T1=

V2

T2

T1 =V1

V2T2 =

12(500 K) = 250 K

53. (a)

W =WBC +WDA =−3P(4V −V )+−P(V −4V ) =−9PV +3PV =−6PV

(b) From the first law of thermodynamics: U = Q+WFor a cyclic processU = 0⇒ Q =−WTherefore, using the answer from above:Q =−W = 6PV

54. W =−6PV =−6nRT =−6(2.0 mol)(8.314)(283 k) = 28 kJ55. W =−nRT ln

(VfVi

)=−(1.0 mol)(8.314)(350 K)ln

(13

)= 3.2 kJ

56.

U = Q+W

U = 0

→ Q =−W =−3.2 kJ

57. An adiabatic process is a process where no heat is transferred in the system(Q=0).58. Q = nCp∆T = (1.0 mol)(20.8 J/mol ·K)(50 K) = 1.04 kJ59. Q = nCv∆T = (1.0 mol)(12.3 J/mol ·K)(100 K) = 1.23 kJ60. ∆E = 3

2 nR∆T = 32(2.0 mol)(8.314)(5.0 K) = 124.7 J

61. Since the process is at constant volume, W=0.62. ∆E = Q+W = 200 J+0 = 200 J

63. Pf = Pi

(ViVf

)γ

= 4.0 atm(10

20

)1.4= 1.52 atm

64. TfTi=(

ViVf

)γ−1= 300 K

(12

)0.40= 227 K

65. By driving the car, the air inside the tires heats up causing the tire to expand. If the tires are over-inflated, thetires could explode if they heat up too much.

66. The gas company would benefit because gasoline is sold by volume.67. The overflow chambers in a car allow for gas to expand on hot summer days. If this mechanism isn’t allowed,

flammable gasoline could be spilled.68. The number of moles of gas increases after the air conditioner is turned on, due to the decreased temperature.

Gas flows in from the outside.69.

PiVi

Ti=

PfVf

Tf

Vi =Vf

→ Pf = Pi

(Tf

Ti

)= 1000 kPa

70. Tf =Pf VfPiVi

Ti =(8×10−3)(200 kPa)(10×10−3)(100 kPa)250 K = 400 K

71. Tf =Pf VfPiVi

Ti =(10×10−3)(50 kPa)(5×10−3)(300 kPa)200 K = 66 K

72. 15.999 g

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73. 3.0 g74. The volume of the gas would be equal to 0.75. The truck tire contracts.76. Flask A must have half the pressure of flask B.77. V = nRT

P = (1 mol)(0.82 L·atm/mol·K)(273 K)1 atm = 22.4 L

78. (a) m−mn =M−Mn

NA= (15.99 g/mol)−(1.008 g/mol)

6.022×1023atoms/mol = 2.48×10−23 g

(b) mAu =MAuNA

= (196.96 g/mol)6.022×1023atoms/mol = 3.27×10−22 g

(c) mAumH

= 3.27×10−22 g/atom1.67×10−24 g/atom = 196X larger

79. T = PVnR where nis the number of moles.

n =mM

=200 g

(32 g/mol)= 6.25 mol

T =(1 atm)(300 L)

(6.25 mol)(0.082 L ·atm/mol ·K)= 585 K

80.

V =nRT

P

Vf =nR(2T )

P= 2Vi

81.

P =nRT

V

Pf =nRT3V

=13

Pi

82. The pressure remains unchanged

P =nRT

V

Pf =nR(2T )(2V )

= Pi

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www.ck12.org Chapter 15. Electrostatics Problem Sets

CHAPTER 15Electrostatics Problem SetsChapter Outline

15.1 CHAPTER 15 PROBLEMS

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Solutions

1. Like charges repel and unlike charges attract.2. The charges are equal but opposite. The proton has a positive charge while the electron has a negative charge.3. You are actually building up electrons as you shuffle across the floor.4. Newton’s 3rd Law.5. The force felt is decreased by a factor of 4.6. Due to the mass of your head, the attractive forces would not be large enough to keep your head stuck to a

wall.7. The negative charges in the paper get pushed away, leaving the section of the paper that is closer to the rod

more positively charged. Therefore, the paper is attracted to the rod. If the rod’s charge was reversed, it wouldyield the same result.

8. The planets have very low or no net charge, so the gravitational force wins.9. The proton has a positive charge, the electron has a negative charge, and the neutron has no charge.

10. 0, since all electrons have the same charge.11. Most atoms in nature are uncharged.12. A positive ion has lost one or more electrons, while a negative ion has gained one of more electrons.13. When charge is conserved, it means that there was no net gain or loss of charge in the system.14. When something is quantized, it means that it can’t be divided up. You can have a +1 charge but not a +1

2charge.

15. 1 C is 11.6×10−19 = 6.25×1018 electrons.

16. It is an inverse squared law.17. As the positive rod is brought near the electroscope, the charges are repelled from the top and migrate to the

bottom in the leaves, where they repel one another.18. As the positive rod is brought near the negative electroscope, the negative charges are drawn to the top, causing

density of negative charges in the leaves is reduced, causing them to be repelled less.19. A negative ion is an atom that has more electrons than protons, making it negatively charged overall.20. If the distance is reduced by half, the force is 4 times larger.21. If the distance is reduced by one-fourth, the force is 16 times larger.22. If the distance is doubled, the force is reduced by 1

4 .23. F ∼ 1

r2

24. If the charge is doubled, so is the force felt between the particles.25. When an object is electrically charged, it has a surplus or deficit of electrons.26. Protons are more tightly bound in an atom than electrons. The amount of energy to break apart the nucleus is

many orders of magnitude large than the energy needed to remove an electron27. The plastic wrap becomes electrically charged when it is removed from the container. Therefore it is attracted

to the metal container.28. The clothes become electrically charged from rubbing into each one another while tumbling in the dryer.29. If an object is negatively charged, it will weigh more due to the extra electrons.30. This is due to the fact that as the electrons get further away from the nucleus, the attractive force becomes

weaker.31. Since F =

kq1q2r2 r, it would be unchanged.

32. An inverse square law is defined as : y∼ 1x2 , which means that as x doubles, y reduced by a factor of 4.

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33. The force would be reduced by a factor of 9.34. The force felt between them is unchanged since their separation and charge is unchanged.35. The force is now : F = F

0.64 .36. The force is : F = 0.75 F.37. The force would be 9 times its original value.38. Regions with greater electric field strength have denser field lines.39. Electric field lines show the direction of force on a positive test charge, and go from positive to negative.40. The answer should be 1

9 of the value at 2 m.41. Superconductive materials have zero electrical resistance.42. The electric field is zero due to balanced electrical forces.43. F = 10 N

(2)2 = 10 N4 = 2.5 N

44. F = 10 N( 1

2 )2 = 4(10) = 40 N

45. q =

√Fr2

k=

√(10)(.2)2

1.38×10−23 = 6.6 µC

46. F =kq1q2

r2 = 8.99×109(2×10−6)2

.22 = 0.9 N

47. Fg =GMm

r2 = 6.67×10−11(1.67×10−27)(9.1×10−31)(1.0×10−10)2 = 1.0×10−47N

48. E = |F |q = 2.5×10−3

(0.5×10−6 C)= 5000 N/C

49. q = FE = 5×10−3 N

3×103 N/C = 1.67 µC

50. U = qV = (3C)(5×103 V ) = 15 kJ

51.

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52.53. Sphere 1 and sphere 3 must have the same sign.54. Sphere 1 and sphere 3 must have opposite signs.55. F12 = F2156. True.57. F = keq2

r2 = (8.99×109)(1.60×10−19)2

(5×10−11)2 = 9.20×10−8N

58. kq l

(y2+l2)32

⇒ Ex =2kq

y3(

1+ l2

y2

) 32

Since y >> 1, the second term in parenthesis approx. 0.Ex =

2kqy3

59. The force on the particle is equal to F = qE, which gives an acceleration of a = qEm . The kinetic energy of the

particle after it has moved a distance ∆x is

K =12

mv2

v2 = v2o +2a∆d;v2

o = 0

K =12

m(

2qEm

d)= qEd

60. Using Gauss’s Law: ∮ −→E ·d−→A =

qεo

E(4πr2) =qεo

→ E =kqεo

61. The uniformly charged sphere will be seen as a point charge, therefore E = kqεo

62. ∮ −→E .d−→A =

qεo

E(4πr2) =qεo

→ E =q

(4πr2)εo

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Since we are looking at a point inside the sphere, the change within the Gaussian sphere we are considering isless than the total charge.

→ q = q ( 43 )πr3

( 43 )πR3

Therefore,E = 1

(4πr2)εoq ( 4

3 )πr3

( 43 )πR3 =

qr4πεoR3

63. Assuming the charge has constant charge per unit length, the line charge can be surrounded by a cylindricalGaussian surface.∮ −→

E .d−→A = q

εoEA = q

εoFor a cylinder, thearea = 2πrh, and the charge can be written in terms of the linear charge densityλ

→ E(2πrh) =λhεo

E =λ

2πrεo

64. From Gauss’s Law, the Gaussian surface will have two sides where the electric field is going through a surface,see figure below.∮ −→

E .d−→A = q

εo(2EA) = q

εoE = q

2AεoRewriting the charge in terms of the surface charge densityσ

→ E = σA2Aεo

= σ

2εo65. A negatively charged atom is one that has more electrons than it does protons.66. Both laws are inversely proportional to the square of the distance.67. The wall isn’t positively charged, the negatively charged balloon is temporarily polarizing the wall.68. Electric fields are a result of positive or negative charges. The field extends in all directions, so it could exist

in the middle of space.69. There would only be a force if there was a charge present in the field to ’feel’ it.70. You can approximate the charge distribution as a point charge.71. Nothing would change. Everything would be exactly the same.72. You would put the test charge in the center of the two other charges.73. You would put the test charge in the center of the two other charges.74. More lines entering signals that the net charge must be negative.75. Equal number of lines entering and leaving implies that the net charge is 0.76. More lines leaving the Gaussian surface implies that the net charge inside must be positive.77. The net charge would be 0.78. The net flux would be 0.79. Excess charges will be repelled from every other charge in the conductor. The furthest the can be repelled is

on the surface of the conductor.80.

→E = 0

81.→E = k r

R3 r, for a sphere of radius Rat a point r < R

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82.

F =kq1q2

r2 =(8.99×109 N ·m/c2)(1.6×10−19 C)2

(2.5×10−15 m)2

F = 36.8 N

83.

F =kq1q2

r2 =(8.99×109 N ·m/c2)(1.6×10−19 C)2

22(2.5×10−15 m)2

F = 9.2 N

84. No, if both of the charges are positive or if both of the charges are negative, the resulting force would be thesame.

85. Charges can only be found in full integers, not 12 , 3

4 etc.86. Electric field lines point outward for positive charges, but point inward for negative charges.

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CHAPTER 16 Electric Potential ProblemSets


16.2 REFERENCES

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Solutions

1. One of the terminals in a 9-volt battery is 9 V higher in potential than the other terminal.2. The electric potential energy is doubled.3. Each charge has its energy increased by 9 joules.4. 1 Joule

1 Coulomb = 1 V5. .1 Joule

.01 Coulomb = 10 V6. A Volt is the same as a Joule per Coulomb. Therefore, there are 9 Joules per every Coulomb of charge that

flows through a 9-V battery.7. 5 J/C is 5 V.8. 0.5 J

0.1 C = 5 V9. The charges are equal but opposite because the charges go from one plate to another.

10. There are several things you could do to store more charge in a parallel plate capacitor. You could increasethe area of the plates, decrease the distance between the plates, and/or place a dielectric material between theplates.

11. MeV is only a million while Gev is a billion, so MeV is smaller.12. U = qV = (2 C)(200 V ) = 400 J13. V = W

q = 20 J0.50 C = 40 V

14. W = q∆V = (1.0 C)(20 V ) = 20 J15. E = |VB−VA|

d = 12 V2.0×10−3 m = 6.0×103 V/m

16. E = |VB−VA|2d = 12 V

5.0×10−3 m)= 2.4×103 V/m

17. V =−Ed =−(3.0×104 V/m)(0.70 m) =−2.1×104 V18.

∆U = q∆V = e(−Ed)

∆U =−(1.6×10−19 C)(20000 V/m)(0.50 m) =−1.6×10−15 J

19. ∆V =−Ed =−(20000 V/m)(0.50 m) =−1.0×104 V

∆U = q∆V

∆U = (−1.6×10−19 C)(−1.0×104 V ) = 1.6×10−15 J

20. V = kqr = (8.99×109 N·m2/C2)(2.0×10−6 C)

2.5 m = 7.2×103 V21. The electric potential is inversely proportional to the separation.22. Given C = Q

V ⇒Q=CV , the charge doubles when the voltage is doubled. (Capacitance is fixed and is thereforeunchanged.)

23. The smallest equivalent capacitance is achieved by connecting the capacitors in series.Series:

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1Ceq

=1C+

1C

=2C

Ceq =C2

Parallel:Ceq =C+C = 2C

24. To obtain the largest possible amount of stored energy, the capacitors should be connected in parallel to eachother.Series:

1Ceq

= 1C1

+ 1C2

+ 1C3

+ 1C4

Parallel:Ceq =C1 +C2 +C3 +C4

25. The net charge is 0. This is because the net charges on each plate of the capacitor is equal but opposite incharge.

26. Ceq = 4.0 µF +8.0 µF = 12.0 µF27. The potential energy increases.28. If the potential is doubled, the energy is increased by a factor of four.29. Solve using energy conservation:

Ui = K f

qV =12

mv2

v =

√2qVm

=

√2(1.6×10−19 C)(240 V )

(9.11×10−31 kg)= 9.18×106 m/s

30. The velocity is proportional to the square root of the voltage: v ∝ V12

31. W = Q∆V = (−1.6×10−19 C)(−10 J/C) = 1.6×10−18 J(Remember that Volts are Joules per Coulomb.)32. ∆U =−qE∆x =−(10.0×10−6 C)(200 V/m)(0.20 m) =−4.0×10−4 J33. ∆V = ∆U

q = −5.0×10−4 J10×10−6 C =−50 V

34. E = |∆V |d = |20×103 V |

0.010 m = 2.0×106 N/C35. Because E = |∆V |

d , the magnitude of the electric field is halved.36.

∆U =−∆K

q∆V =12

m(v2f − v2

i )

∆V =−12

m(v2f − v2

i )

q=−1

2

((9.11×10−31 kg)((2.0×104 m/s)2− (3.5×103 m/s)2)

−1.6×10−19

)= 1.1×10−3 V

37.

∆U =−∆K

q∆V = ∆K

∆K = (−1.6×10−19 C)(−40.0 V ) = 6.4×10−18 J

38. V = kqr = (8.99×109 N·m2/C2)(−1.6×10−19 C)

0.020 m = 7.2×10−8 V

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39.

V =kqr2− kq

r1= kq

(1r2− 1

r1

)→(8.99×109 N ·m2/C)(−1.60−19 C)

(1

0.050 m− 1

0.020 m

)=−4.32×10−8 V

40. The charges are symmetric about the test charge and of equal charge therefore, the net force is 0.

F =Eq

;F = 0

→ E = 0

41. V = 2 kqr = 2(8.99×109 N ·m2/C2)

(4.00×10−6 C

0.5 m

)= 1.44×105 V

42.

F = kq(

q1

r21+

q2

r22

)= (8.99×109 N ·m2/C2)q

(−3×10−6 C(0.5 m)2 +

4×10−6 C(0.5 m)2

)= 1.6×104 q N/C

E =Fq= 1.6×104 N/C

V = k(

q1

r1+

q2

r2

)= (8.99×109 N ·m2/C2)

(−3.00×10−6 C

0.5 m+

4.00×10−6 C0.5 m

)= 18 kV

43.

E = k(

q1

r21+

q2

r22

)= k(−qx2 +

2q(x−4.0)2

)E = 0

→ qx2 +16 q−8 xq = 2 qx2

qx2 +8 xq−16 q = 0

⇒ x =−9.66 m

44.

U = q1V2 +q1V3 +q1V4 = q1K(

q2

r2+

q3

r3+

q4

r4

)U = (10.0×10−6)2(8.99×109 N ·m2/C2)

(3

0.3 m

)= 8.99 J

45.

U = q1V2 = q1k(

q2

r2

)U = (8.0×10−6 C)(8.99×109 N ·m2/C2)

(4.0×10−6 C

0.2 m

)= 1.44 J

46. Since we are looking at the point at the base of the triangle, the equal charges to either side of the base exertno net force. Thus, we only need to determine the electric potential caused by the charge at the top of thetriangle.First, find the distance from the base to the charge using the Pythagorean theorem:

r =√

c2−a2 =√

3.0 cm2−1.5 cm2

r = 2.6 cm = 2.6×10−2 m

V = kqr = (8.99×109 N·m2/C2)(5.0×10−6 C)

2.6×10−2 m = 1.7×106 V

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47. V = 2 kqr = 2 (8.99×109 N·m2/C2)(2.0×10−6 C)√

1.0 m2 +1.0 m2= 25 kV

48.

V = 2kqr= 2(8.99×109)(2.0×10−6)

(1√

1.02 +1.02

)= 25 kV

U = qV = (−3.0×10−6 C)(25×103 V ) =−0.075 J

49.

U =U12 +U23 +U13 +U14 +U24 +U34

U =kq2

L+

kq2

L

(1√2+1)+

kq2

L

(1√2+2)=

kq2

L

(2√2+4)

50. No, there is no point within the plane of the triangle at which the electric potential is zero.51. V = a+bx2 = 2.0 V +(−1.2 V/m2)(2.5 m)2 =−5.5 V52.

V =kqr

q =rVk

=

((0.10 m)(3.0×103 V )

(8.99×109 N ·m2/C2)

)= 3.34×10−8 C

→ N =3.34×10−8 C

qo= 2.1×1011 electrons

53. E = 054.

−→E =

kqr2 =

(8.99×109)(25×10−6)

0.162 r = 8.8×106r N/C

V =kqr

= 1.4×106 V

55. Q =C∆V = (3.0×10−6 F)(24 V ) = 7.2×10−5 C56. Q =C∆V = (5.5×10−6 F)(12 V ) = 6.6×10−5 C57. C = Q

∆V = (10.0×10−6 C)5.0 V = 2.0×10−6 F

58. ∆V = QC = 150×10−6 C

2.0×10−6 F = 75 V59.

∆V = Ed

→ E =∆Vd

=30 V

0.90×10−3 m= 33 kV/m

60. Ceq =C1 +C2 = 9.0 µF61. Yes, the potential difference across each capacitor is equal to the potential difference across the battery.

Therefore, each capacitor has a potential difference of 6 V.62.

Q2 =C∆V = (2.0 µF)(9.0 V ) = 18.0 µC

Q9 = (9.0 µF)(9.0 V ) = 81.0 µC

63.

1Ceq

=1

C1+

1C2

Ceq =C1C2

C1 +C2=

(2.0)(3.0)5.0

µF = 1.2 µF

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64.

Qeq =Ceq∆V = 14.4 µF

Qeq = Q1 = Q2 = 14.4 µC

→ ∆V1 =Q1

C1=

14.4 µC2.0 µF

= 7.2 V

→ ∆V2 =Q2

C2=

14.4 µC3.0 µF

= 4.8 V

65.

Ceq =C1C2C3

C23 +C13 +C12=

(6.0 µF)(8.0 µF)(2.0 µF)

(8.0 µF)(2.0 µF)+(6.0 µF)(2.0 µF)+(6.0 µF)(8.0 µF)= 1.3 µF

Qeq =Ceq∆V = (1.3 µF)(24 V ) = 31 µC

Qeq = Q1 = Q2 = Q3 = 31 µC

66. ∆V = QC = 35.0 µC

7.0 µF = 5.0 V

67. U = kq1q2r = (8.99×109)(24)(48)(1.6×10−19)2

(4.2×10−15+6.0×10−15)= 2.6×10−11 J = 1.62×108 eV

68. U = −(8.99×109)(1.60×10−19)2

0.0529×10−9 =−4.35×10−18 J69.

U =kq1q2

rr→ ∞

⇒U = 0

70. 0. If the charge is increased, the potential will be increased by the same amount.71. U = 1

2CV 2 = 12(5.0×10−6 F)(80 V )2 = 0.16 J

72. U = 12CV 2 = 1

2

(QV

)V 2 = 1

2 QV = 12(1.0×10−9 C)(24 V ) = 12 nJ

73. Ceq =C1C2

C1+C2= (2.0 µF)(1.0 µF)

3.0 µF = 0.67 µF74. C = 4πεoR = 4πεo(0.1 m) = 1.1×10−11 F75. ∆C = 076. ∆V = 12 V for both capacitors77. n = Q

CV = 8×10−3 C(2.0×10−6 F)(20 V )

= 20078. C1 = 2.0 µF,C2 = 1.0 µF,C3 = 3.0 µF .

In parallel:Ceq = 2.0 µF +1.0 µF +3.0 µF = 6.0 µFIn series:Ceq =

C1C2C3C2C3+C1C3+C1C2

= 6.0 µF3.0 µF+6.0 µF+2.0 µF µF = 0.55 µF

79. In parallel:Ceq,p =C1 +C2 = 4.0 µF +3.0 µF = 7.0 µFIn series:Ceq,s =

C1C2C1+C2

= (4.0 µF)(3.0 µF)7.0 µF = 1.7 µF

∆C =Ceq,p−Ceq,s = 7.0 µF−1.7 µF = 5.3 µF

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16.2 References

1. Haven Giguere. CK-12 Foundation .2. Haven Giguere and Laura Guerin. CK-12 Foundation .3. Laura Guerin. CK-12 Foundation .4. Laura Guerin. CK-12 Foundation .5. Laura Guerin. CK-12 Foundation .6. Laura Guerin. CK-12 Foundation .7. Laura Guerin. CK-12 Foundation .8. Laura Guerin. CK-12 Foundation .

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www.ck12.org

CHAPTER 17 Circuits Problem SetsChapter Outline

17.1 CHAPTER 17 PROBLEMS

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www.ck12.org Chapter 17. Circuits Problem Sets


Problems

1. (a) If two lamps are connected in series, what happens if one of the lamps burns out?(b) If two lamps are connected in parallel, what happens if one of the lamps burns out?

2. (a) What would happen to the intensity of a light bulb if several other light bulbs are added in series with it?(b) What would happen to the intensity of a light bulb if several other light bulbs are added in parallel with

it?3. Some wires become hot to the touch when they carry current. Why is this?4. Each month home owners pay an electric bill. What is being paid for?5. Describe an electric circuit.6. A popular demonstration when talking about electric circuits is to hook up small light bulbs in series and to

see what happens when more light bulbs get added to the system. If two light bulbs are hooked up in series,what is the current through the second light bulb if there is 1 A flowing through the first?

7. Suppose two light bulbs are hooked up in series to a battery source that produces 9 V. If the voltage dropacross the first light bulb is 5 V, what is the voltage across the second light bulb?

8. When designing circuits, what is a possible disadvantage of series circuits compared to parallel circuits?9. If 2 resistors are in parallel and the voltage across one of the resistors is 3 V, what is the voltage in the other

resistor?10. How do currents I1 and I2 compare to the current going through the battery I3 in the figure below?

11. As more registers are opened at the supermarket, the resistance to people being helped is reduced. How is thissimilar to what happens when more paths are added to a parallel circuit?

12. Calculate the power of a toy robot that has a current of 1.5 A when powered by 220 V.13. What is the power in a device that is operates on 120 V with a current of 25 A?14. If device has a rating of 500 Watts while operating on 120 V, how much current will it draw?15. How can you tell if automobile headlights are wired in series or parallel?16. (a) If you wanted to connect two resistors together so that their combined resistance is greater than either

resistor individually, would you connect them in series or parallel?(b) If you wanted to connect two resistors together so that their combined resistance is smaller than either

resistor individually, would you connect them in series or parallel?

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17. (a) If two resistors are in series, which is the same across each resistor: the voltage or the current?(b) If two resistors are in parallel, which is the same across each resistor: the voltage or the current?

18. Experiments have shown that the speed of an electrical signal is significantly faster than the speed of sound.Explain why this is.

19. You have two light bulbs and a battery. What configuration will give you the brightest bulbs?20. Using the figure below, what happens to brightness of light bulb #1 when the switch is closed?

21. When you attach several light bulbs in series, they become hot to the touch but no light is emitted. Explainhow this is possible.

22. Consider the following set of light bulbs. Assume that all the light bulbs are identical.

(a) Which group of light bulbs emits more light?(b) If the light bulb in group 2 suddenly burns out, what happens to the light bulbs in group 1?(c) What happens to the circuit seen if the circuit is broken between the two light bulbs in group 1?

23. When two components are in parallel and one of the components shorts out, why is the other unaffected?24. If you had a handful of 10 ohm resistors but need a 5 ohm resistor, what configuration would you need to

make to get the component you need?25. If you had a handful of 20 ohm resistors but need a 40 ohm resistor, what configuration would you need to

make to get the component you need?26. Which quantity (voltage/current) is the same if:

(a) Resistors are connected in series.(b) Resistors are connected in parallel.

27. Calculate the resistance of a 50 W light bulb designed to be powered by 120 V.28. Determine the power that a light bulb with R=260Ω would draw when operated on a 240 V line.29. A 9 V battery has an internal resistance of 0.2Ω. Calculate the power dissipated when the 9 V battery

discharges through a 12Ω resistor.30. Find the equivalent resistance of the circuit in the figure below.

31. Calculate the equivalent resistance seen in the figure below.

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32. Determine the equivalent resistance between A and B in the figure below.

33. If a potential difference of 16.0 V is maintained between points A and B, calculate the current in each of theresistors seen in the figure below.

34. Calculate the power dissipated in both resistors in the previous problem.35. Find the equivalent resistance between points a and b in the figure below.

36. Calculate the equivalent resistance between A and B in the figure below if the R1 resistor overheats andprevents current from flowing through it.

37. Three 95Ω resistors are connected as shown in the figure below. Determine the equivalent resistance betweenterminals A and B.

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38. If each resistor in the figure below can safely handle 30.0 W, what is the maximum voltage that can be appliedbetween points A and B?

39. The ammeter in the figure below reads 1.0 A. Calculate the magnitude of the currents I1 and I2.

40. Assume that the ammeter in the figure below is removed and the current that flows through the 4.0Ω path, I3,is unknown. Determine all the currents in the circuit.

41. The following equations describe an electric circuit:

−(4 Ω)I2 +5.0 V − (3.0 Ω)I3 = 0

−(5 Ω)I1 +(3.0 Ω)I3 +3.0 V = 0

I1 + I2 = I3

Determine all the currents in the circuit described.42. Calculate the current if 2 C of charge flows through a given area in 4.0 s.43. Calculate the length of wire needed to obtain a resistance of 5.0Ω in a wire of radius 0.75 mm and a resistivity

of ρ=10−6Ω*m.44. A 12-gauge wire has an area of 3.30 mm2 and a length of 2.0 m. If the resistivity of the wire is 10*10−8Ω*m,

calculate the resistance per length.45. Calculate the magnitude of the electric field along the length of an 8-gauge wire with a resistance per unit

length of 5*10−3Ω*m while carrying a current of 2.4 A.

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46. Determine the power dissipated in a 16Ω resistor that carries a current of 3.5 A.47. What is the power put into a wire with R=4.0Ω that carries a current of 5.0 A for 5 seconds.48. How is the internal resistance of a battery related to the current in the figure below?

49. Consider a 10Ω resistor that is connected to a battery with an emf of 9 V and an internal resistance of 0.75Ω.

(a) Calculate the current that passes through the system(b) Determine the terminal voltage of the battery in the previous problem.

50. Calculate the usable power delivered by a battery with an emf of 9 V that generates a current of 1.3 A.51. Determine the power dissipated by a battery’s internal resistor if r=1Ω and the battery produces a current of

0.7 A.52. When the equivalent resistance of a system is the sum of all the resistors, are the resistors in series or parallel?53. Consider the figure below

a. Determine the equivalent resistance of the circuitb. Determine the total current of the circuit seen in the figure below.c. Calculate the power dissipated in each resistor.

54. Consider the circuit drawn below

(a) What is the total current flowing through the circuit seen in the figure below?(b) Using Kirchhoff’s loop rule, write out the equations for the voltage drops in loop A and loop B, assuming

the current through each of the loops is clockwise.(c) Find the values of each of the currents.

55. Determine the power dissipated in each resistor seen in the figure below.

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56. Derive a general expression for N equivalent resistors in parallel.57. Consider the circuit shown below.

(a) Calculate the equivalent resistance of the system(b) Determine the potential drop across each resistor.(c) Determine the current across each of the resistors.

58. What is the significance of a negative current value when you are calculating the currents in a circuit?59. What is Kirchhoff’s loop rule for voltages?60. How is Kirchhoff’s junction rule different than Kirchhoff’s loop rule?61. The power dissipated in a resistor that carries a current I is P. If the current in the resistor is increased to 4 I,

what is the power dissipated?62. Determine the power dissipated in a resistor with R=20Ω that has a potential drop of 120 V.63. Calculate the resistance of a 1.0 kW air compressor that operates at 220 V.64. Calculate the internal resistance of a 12-V battery with a terminal voltage of 11.2 V when it delivers a current

of 15 A.65. The internal resistance of a 9-V battery is 0.02Ω while it is delivering a current of 20 A. Calculate the voltage

across the terminals.66. True or False: Kirchhoff’s loop rule is a result of the conservation of energy.67. Determine the length of 14-gauge copper wire that would result in a resistance of 2.2Ω.68. A wire 50 m long carries a current of 8.3 A. Calculate the potential difference across if the wire is standard

14-gauge copper wire.69. A 14Ω resistor carries a current of 5 A. Determine the power dissipated in the resistor.70. How is the voltage across a wire related to the resistivity of the wire?71. What is the relationship between current and the charge in a circuit?

Solutions

1. (a) If one of the lamps burns out, there would not be a complete circuit. Therefore both lamps would go out.(b) If one of the lamps burns out, the other will stay light because there will still be a complete circuit.

2. (a) By adding more light bulbs in series, the potential drop across each lamp becomes smaller, resulting in areduction of the light bulb intensity.

(b) As other light bulbs are added in parallel, the brightness would be unaffected.3. As electrons transfer their kinetic energy through the wire, heat is released in the collision. This reflects the

resistance of the wire.4. You are paying mostly for the energy. Additionally, power companies maintain electrical networks, which are

often included in the price of the electrical energy.5. An electric circuit is composed of electronic components that are connected by a conductive medium through

which electric current can flow.6. There would be 1 A flowing through the second light bulb as well.7. If the voltage drop across the first bulb is 5 V, then the voltage drop across the second bulb MUST be 4 V,

since the sum of the voltages being zero in a closed loop.

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8. If one of the components in the circuit fails, then the whole system will fail.9. Since the resistors are in parallel, they have the same voltage drop; therefore the answer is 3 V.

10. I3 = I1 + I211. As more paths are added, the overall resistance is reduced.12. P = IV = (1.5 A)(220 V ) = 330 W13. P = IV = (25 A)(120 V ) = 3 kW14. I = P

V = 500 W120 V = 4.16 A

15. Automobile headlights are wired in parallel because it is possible for one of the lights to go out and the otherto stay on.

16. (a) You would connect the resistors in series.(b) You would connect the resistors in parallel.

17. (a) When components are in series, the current is the same across each.(b) When components are parallel to one other, the voltage drop across each is the same.

18. Electrical signals are due to the propagation of electric fields while the speed of sound is a result of molecularcollisions. The electrical signals travel near the speed of light.

19. You want the configuration that will provide each bulb the greatest voltage. Therefore you would attach thelight bulbs in parallel.

20. Nothing, the brightness stays the same.21. When several bulbs are hooked up in series, the drop in voltage across each bulb which may not be enough for

the filament to heat up enough to emit light. Each bulb still dissipates energy, which makes them to the touch.22. (a) The light bulbs in group 2 are dimmer than the light bulb in group 1.

(b) Nothing, a complete circuit is still formed even though the light bulb in group 2 burned out.(c) If a break forms between the two light bulbs in group 1, there is no complete circuit for that path so only

the light bulb in group 2 is lit.23. Each part in a circuit must be part of a complete path that includes a voltage source. Each part in a parallel

circuit represents an independent path through the circuit.24. You would hook two 10 ohm resistors in parallel.25. You would hook two 20 ohm resistors in series.26. (a) When components are connected in series, the current across each is the same.

(b) When components are connected in parallel, the voltage across each is the same.27.

P =V 2

R

→ R =V 2

P=

(120 V )2

50 W= 288 Ω

28. P = V 2

R = 2402

260 Ω= 222 W

29. P = I2R =( E

r+R

)2R =

( 9 V0.2+12

)2 12 Ω = 6.5 W30. For resistors R1 and R2

Req =R1R2

R1+R2= 12

7 Ω = 1.71 Ω

For Req and R3

R f inal =ReqR3

Req+R3= 0.92 Ω

Note that all 3 resistors are in parallel, so the resistance could also be found with:R f inal =

11

R1+ 1

R2+ 1

R3

= 0.92 Ω

31. Req = R1 +R2 +R3 = (5+3+8)Ω = 16 Ω

32.

Req1,2 = R1 +R2 = (3+4)Ω = 7 Ω

Req3,4 =R3R4

R3 +R4= 0.8 Ω

Req,total = 7 Ω+0.8 Ω = 7.8 Ω

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33.

I1 =∆VR1

=16 V5.0

= 3.2 A

I2 =∆VR2

=16 V8.0

= 2.0 A

34.

P1 = I21 R1 = (3.2 A)2(5.0 Ω) = 51.2 W

P2 = I22 R2 = (2.0 A)2(8.0 Ω) = 32.0 W

35.

Req1,2 =R1R2

R1 +R2=

(4.0)(1.0)(4.0+1.0)

= 0.8 Ω

R′eq = Req1,2 +R3 = 0.8+5.0 = 5.8 Ω

36. Req = R2 +R3 = 1.0+5.0 = 6.0 Ω

37. Req =R2R3

R2+R3+R1 = (95+ 952

190)Ω = 142.5 Ω

38. Since all the current must pass through R1

P = I2maxR

→ I =

√PR=

√30.0 W95 Ω

= 0.52 A

→ ∆Vmax = ReqImax = (From previous problem Req = 142.5)

∆Vmax = (142.5 Ω)(0.52 A) = 74.1 V

39. For the leftmost branch (counter clockwise):

−I2(2.0 Ω)+(3.0 A)(4.0 Ω) = 0

→ I2 = 3.5 A

Because I2 + I3 = I1, I1 = 3.5 A+1 A = 4.5 A40. Looking at the relationship between all the currents:

I1 = I2 + I3Looking at the rightmost branch (and going counter clockwise):

12 V −5.0 I1−4.0 I3 = 0

12 V −5.0(I2 + I3)−4.0 I3 = 0

−5.0 I2 = 9 I3−12

For the leftmost branch (counter clockwise):

−I2(2.0 Ω)+(4.0 Ω)I3 = 0

I2(2.0 Ω) = (4.0 Ω)I3

(9I3−12)−5.0

(2.0 Ω) = (4.0 Ω)I3

18.0 I3−24 =−20 I3

I3 = 0.63 A

→ I2 = 1.26

→ I1 = 1.89

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41. From the third equationI2 = I3− I1Sub into the first eqn.

−4(I3− I1)+5.0−3I3 = 0

−7I3 +4I1 +5 = 0

→ I1 =(7I3−5)

4Sub into 2nd eqn.

−5(

7I3−54

)+3I3 +3 = 0

−35I3 +25+12I3 +12 = 0

I3 = 1.61 A

→ I1 = 1.57

I2 = 0.04

42. I = Qt = 2 C

4.0 s = 0.5 A

43. L = RAρ= (5.0 Ω)π(0.75×10−3 m)2

10−6 Ω·m = 8.84 m

44. RL = ρ

A = 10×10−8 Ω·m3.3×10−3 m2 = 3.03×10 Ω/m

45.

E =VL

;V = IR

→ E =VL= I

RL= (2.4 A)(5.0×10−3

Ω ·m) = 0.012 V/m

46. P = I2R = (3.5 A)2(16 Ω) = 196 W47. P = (5.0 A)2(4.0 Ω) = 100 W48. I = ε

R+r ; the internal resistance is inversely related to the current.49. (a) I = ε

R+r =9 V

(10+0.75)Ω = 0.84 A(b) ∆V = ε− Ir = 9 V − (0.84 A)(0.75 Ω) = 8.37 V

50. P = IV = (9V )(1.3A) = 11.7W51. P = I2r = (0.7 A)2(1 Ω) = 0.49 W52. The resistors are in series.53. (a) Req = R1 +R2 = (5+8) Ω = 13 Ω

(b)

V = IReq

→ I =V

Req=

(12 V )

(13 Ω)= 0.92 A

(c) The current is the same in each resistor, therefore

P5Ω = I2R = (0.92 A)2(5 Ω) = 4.23 W

P8Ω = I2R = (0.92 A)2(8 Ω) = 6.77 W

54. (a)

Req = R6Ω +R8Ω = 14 Ω

1R′eq

=1

Req+

1R4Ω

→ R′eq =(14)(4)14+4

Ω = 3.11 Ω

⇒ I =V

R′eq=

(12 V )

3.11 Ω= 3.86 A

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(b)

loop a→ 12 V − I2(4 Ω) = 0

loop b→−I3(6 Ω)− I3(8 Ω)+ I2(4 Ω) = 0

loop a→ 12 V − I2(4 Ω) = 0

loop b→−I3(6 Ω)− I3(8 Ω)+ I2(4 Ω) = 0

(c) Using the junction rule:I1 = I2 + I3From loop aI2 =

12 V4 Ω

= 3 AFrom loop b

−I3(6 Ω)− I3(8 Ω)+(3 A)(4 Ω) = 0

(14 Ω)I3 = 12 V

I3 = 0.86 A

→ I1 = 3.86

55.

P7Ω = I2R = (1.45 A)2(7 Ω) = 14.72 W

P4Ω = I2R = (1.45 A)2(4 Ω) = 8.41 W

56.

1Req

=1

R1+

1R2

+ . . .

⇒ Req =RN!N∑i

Ri

; where N is the number of resistors in parallel.

57. (a) For the resistors in parallel;

Req =(6)(8)(8+6)

Ω = 3.43 Ω

R′eq = (3.43 Ω)+3 Ω = 6.43 Ω

(b) The total current isI = V

R′eq= 24 V

6.43 Ω= 3.73 A

∆V3Ω = IR3Ω = 11.19 VThe potential drop across the resistors in parallel isV = IReq = (3.73 A)(3.43 Ω) = 12.8 V

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(c)for loop a24 V −3I1−6I2 = 0 . . .(1)for loop b−6I2 +8I3 = 0 . . .(2)from the junction ruleI1 = I2 + I3 . . .(3)from loop b

−6I2 +8I3 = 0

→ I3 = 0.75I2

using eqn(1) and eqn(3)

24 V −3(I2 + I3)−6I2 = 0

24 V −3(I2 +0.75I2)−6I2 = 0

24 V = 11.25I2

→ I2 = 2.13 A

I3 = 1.6 A

I1 = 2.73 A

58. The negative value signifies choosing the wrong initial direction of the current when setting up the initialconstraints.

59. The sum of the potentials around any closed loop circuit must be zero.60. The junction rule deals with currents entering or leaving a junction while the loop rule deals with the sum of

the voltages in a closed loop.61. The power dissipated is 16 P.62. P = V 2

R = (120 V )2

20 Ω= 720 W

63. R = V 2

P = (220 V )2

(1×103 W )= 48.4 Ω

64.

∆V = ε− Ir

r =ε−∆V

I=

12 V −11.2 V15 A

= 0.05 Ω

65. ∆V = ε− Ir = 9 V − (20 A)(0.02 Ω) = 8.6 V

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66. True67.

R = ρLA

L =RAρ

=(2.2 Ω)(2.081×10−6 m2)

1.7×10−8 Ω ·m= 270 m

68.

VI= ρ

LA

V = ρILA

= (1.7×10−8Ω ·m)

(8.3 A)(50 m)

2.081×10−6 m2 = 3.40 V

69. P = I2R = (25 A2)(14 Ω) = 350 W70. V

I = ρLA ; they are proportional.

71. I = ∆Q∆t ; the current is the time rate of change of the charge.

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www.ck12.org Chapter 18. Magnetism Problem Sets

CHAPTER 18 Magnetism Problem SetsChapter Outline

18.1 CHAPTER 18 PROBLEMS MAGNETISM

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18.1. Chapter 18 Problems Magnetism www.ck12.org

18.1 Chapter 18 Problems Magnetism

Solutions

1. No, there is no such thing as a magnetic monopole.2. If the magnet is brought close to a material whose domains can be realigned, the object can become magnetized

by the magnet.3. Newton’s 3rd Law.4. The force felt is the smallest when a charged particle is moving parallel to a magnetic field.5. Magnetism is created by moving charges.6. Like poles repel, opposite poles attract.7. There are no magnetic monopoles, so there are no particles that have a magnetic property similar to those seen

when dealing with electric fields.8. The magnetic field changes as the inverse square to the distance from the source.9. Magnetic fields are produced by moving electric charges.

10. The magnetic domain is defined as the range in which the magnetic field has uniform magnetization.11. The magnetic field forms concentric circles surrounding the current-carrying wire.12. The magnetic field still forms concentric circles surrounding the current-carrying wire, but changes direction.13. Inside the loop, due to the reinforcement of the magnetic field.14. (a) You would send the charged particle perpendicular to the magnetic field.

(b) You would send the charged particle parallel to the magnetic field.15. The force is due to the fact that the current-carrying wire is made up of moving charged particles, which

creates a magnetic field.16. When they are perpendicular.17. All magnetic are based on moving charges. Every electron (which is negatively charged) is moving, so it has

a magnetic field associated with it.18. Whether a material is magnetic or not depends on the motion of the charges inside of it. In non-magnetic

materials, the magnetic field lines combine in such a way to cancel each other out.19. The magnet induces an opposite polarity in the iron, which is then attracted to the magnet.20. There is a difference. When the charge is stationary, an electric field is present. When the charge is moving,

there is an electric field as well as a magnetic field.21. A magnet can attract an object if the object in question has magnetic domains that can be induced into

alignment.22. Magnets can be weakened through physical shock, by having the domains knocked out of alignment with one

another.23. You would need to be at the Earth’s north magnetic pole24. Yes, since the needle will still attracted to the north magnetic pole.25. The rotation happens when the loop is perpendicular to the field. In order to prevent rotation, align the loop

with the magnetic field.26. The weaker magnet exerts 20 N, due to Newton’s third law.27. Yes, this is a result of Newton’s third law.28. If the electron is initially at rest, it would not be possible to exert a force on it due to a magnetic field.29. Yes, ~F = q~E30. The equation that governs this relationship is: r = mv

qB . Therefore, as the magnetic field is increased, the radiusdecreases.

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31. You would decrease the magnetic field.32. Yes, as charged particles that are moving have an associated magnetic field.33. The total kinetic energy of the particle is unchanged.34. Work is defined as: W = F · x = Fxcosθ. Which means work is force (in the direction of the displacement)

times the displacement. The magnetic field applies a force that is perpendicular, so no work is being done.35. The two particles must have opposite charge.36. (a) If the currents are running in the same direction, the wires attract one another.

(b) If the currents are running in opposite directions, the wires repel one another.37. (a) The magnetic force points into the page.

(b) The magnetic force points out of the page.38. (a) The radius of particle B is smaller.

(b) The radius of particle B is larger.39. Using the right hand rule, the direction of the magnetic field must be into the page.40. B = µoI

2πr , therefore B > A =C41. No, the magnetic field of a solenoid is independent of the length.42. (a) F = qvBsinθ = (1.6×10−19 C)(4.0×105 m/s)(0.05 T )sin(50) = 2.45×10−15 N

(b) 0, the values would be exactly the same.43.

mv2

r= qvB

v =qBrm

=(1.60×10−19 C)(0.25 T )(0.1 m)

1.67×10−27 kg= 2.40×106 m/s

44. T = 2πmqB = 2π(1.67×10−27 kg)

(1.60×10−19 C)(0.25 T ) = 2.62×10−7 s

45. µ = IA = (8.0×10−3 A)(8.75×10−4 m2) = 7.0×10−6

46. B = µoNI2πr = (4π×10−7 T ·m/A)(15)(5.0 A)

2π(0.1 m) = 1.5×10−4 T47. For the two particles to be deflected in opposite directions, they must have opposite charge.48. The magnetic field is perpendicular to the magnetic force.49. F =−q~v×~B

For the electrons to be deflected downwards (in the negative y−direction), the magnetic field must be directedinto the page.

50. If the current in the conductor is parallel to the magnetic field, no force will be felt.51. No force will be on the loop when the plane of the loop is perpendicular to the magnetic field.52. The magnetic field from wire 1 produces a magnetic field into the page at wire 2. The resulting magnetic force

on wire 2, using the right hand rule, pushes wire 2 in the direction of wire 1.53. The perpendicular current-carrying wires neither attract nor repel one another.54. Since I = 0, inside the tube, there is no magnetic field inside.55. B = µoNI

l56. If the length is doubled, the magnetic field is reduced by a factor of 2.57. The geographic north pole is the magnetic south pole.

(a) Towards the top of the page,(b) Out of the page.

58. The electron is deflected away from the Earth.59. (a) F = qvBsinθ = (1.6×10−19 C)(2.5×106 m/s)(0.2 T )sin(35) = 4.59×10−14 N

(b)

F = qvBsinθ = (1.6×10−19 C)(2.5×106 m/s)(0.2 T )sin(35) = 4.59×10−14 N

a =Fm

=4.59×10−14 N1.67×10−27 kg

= 2.75×1013 m/s2

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18.1. Chapter 18 Problems Magnetism www.ck12.org

60. The minimum value is 0, if the electron enters the magnetic field so that it is parallel to the field.61.

E =12

mv2 = e∆V

evBsin90 =mv2

R

B =1R

√2m∆V

e=

14.0×105 m

√2(1.67×10−27 kg)(5.0×106 V )

1.60×10−19 C= 8.07×10−7 T

62. F = Il×B = (1.8 A)(0.80 m)(1.50 T )(− j) =−2.16 j63. F = Il×B = IlBsinθ = IlBsin(0) = 0

There is no magnetic force on the wire.64. (a) F = Il×B = IlBsinθ = (3.9 A)(2.0 m)(0.5 T )sin(30) = 1.95 N

(b)

F30 = Il×B = IlBsinθ = (3.9 A)(2.0 m)(0.5 T )sin(30) = 1.95 N

F75 = Il×B = IlBsinθ = (3.9 A)(2.0 m)(0.5 T )sin(75) = 3.77 NF75

F30=

3.771.95

= 1.93 times larger

65. µ = IA = (12.0×10−3 A)(π(0.3 m)2) = 3.4×10 mA ·m2

66. τ = BAI sinφ = (.06 T )(0.4 m)2(2.0 A)sin45 = 0.14 N ·m67. I = 2πRB

µo= 2π(100 m)(1 T )

4π×10−7T ·m./A = 5×108 A

68. B = µoI2πR = (4π×10−7 T ·m/A)(5.0 A)

2π(0.5 m) = 2×10−6 T

69. ~B = µoI2πr =

(4π×10−7 T ·m/A)(6.0 A)2π(0.15 m) = 8.0×10−6 T

70.

F = I2~l×~B

~B =µoI2πr

=(4π×10−7 T ·m/A)(8.0 A)

2π(0.05 m)= 3.2×10−6 T

F = (4.0 A)(1.0 m)(3.2×10−6 T )− z = 1.28×10−5 N towards wire 1

71. (a) As seen in the diagram below, wire 1 must carry a downward current to attract wire 3.

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(b) Due to Newton’s Laws

F2,1 = F3,1

µo(2.0 A)(I1)

2πx=

µo(5.0 A)(I1)

2π(x+25 cm)

(2.0 A)(x+25 cm) = (5.0 A)x

x = 8.33 cm

72. r = µoI2πB = (4π×10−7 T ·m/A)(3.0 A)

2π(0.20×10−6 T ) = 3 m = 300 cm

73. (a) B = µoNI2πrin

= (4π×10−7 T ·m/A)(500)(9.0×103 A)2π(0.80 m) = 1.13 T

(b) Doubling the number of turns would simple double the answer from the previous problem.→ B2N =2.26 TTherefore the magnetic field is increased by 1.13 T.

74. B = µoNIl ⇒ I = B

µon = 2.0×10−4 T(4π×10−7 T ·m/A)(833.3) = 0.2 A

75. Assuming the wires are lying along the z-axis.

BI1=100 =µoI1

2π(1

2 r)(−x),BI1=100 =

µo(3I1)

2π(1

2 r)(x)

Btotal =µo2I1

2π(1

2 r)(x) = (4π×10−7 T ·m/A)(100 A)

π(.5)(.05 m)= 1.6×10−3 T

76. T = 2πmqB = 2π(1.67×10−27 kg)

(1.6×10−19 C)(0.45 T ) = 1.46×10−7 s

77. v = rqBm = (0.75 m)(1.6×10−19 C)(1 T )

(1.67×10−27 kg) = 7.19×107 m/s

78. r =

√2m∆VqB2 =

√2(1.3×10−25)

(1.6×10−19 C)(0.15)2 = 8.50×10−3 m

79. f = qB2πm = (1.6×10−19 C)(0.8)

2π((1.67×10−27 kg)) = 1.22×10 Hz

80. K = 12

(q2B2

m

)r2 = 1

2

((1.6×10−19 C)2(1 T )2

(1.67×10−27 kg)

)(0.4)2 = 1.22×10−12 J

81. τ = µBsinθ = (4.3×10−2 A ·m2)(0.7)sin40 = 0.02 N ·m82. No, it is not uniform.

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CHAPTER 19Electromagnetism ProblemSets


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www.ck12.org Chapter 19. Electromagnetism Problem Sets


Solutions

1. The conservation of energy.2. Most consumer glass appears opaque to ultraviolet light because its wavelength is short enough to interact

with and lose energy to electrons in the glass.3. Radio waves and visible light each have different wavelengths and frequency.4. Violet5. λ = c

f

6. λ = vf =

3.0×107 m1.2 Hz = 2.5×107 m

7. The source is vibrating electric charges.8. Visible light has a larger wavelength.9. Gamma rays have a greater frequency.

10. v = c11. Neither, both travel the same speed.12. λ f = c→ f = c

λ= (3.0×108 m/s)

(600×10−9 m)= 5×1014 Hz

13. v = xt ⇒ t = x

v =(20×103 m)

(3.0×108 m/s) = 6.7×10−5 s

14. If the distance to the sun is approx. 150×106 kmt = x

v =(150×109 m)(3.0×108 m/s) = 500 s = 8.33 min

15. If the distance to the moon is approx. 3.84×108 mt = x

v =(3.84×108 m)(3.0×108 m/s) = 1.3 s

16. t = xv =

(4×1016 m)(3.0×108 m/s) = 1.3×108 s = 4.2 yrs

17. f = cλ= (3.0×108 m/s)

(800×10−9 m)= 3.75×1014 Hz

18. f = cλ= (3.0×108 m/s)

(515×10−9 m)= 5.8×1014 Hz

19. λ = cf =

(3.0×108 m/s)(6.0×1014 Hz) = 5×10−7 m

20. f = cλ= (3.0×108 m/s)

(0.015 m) = 20 GHz21. The electric field is in the +y direction, to be in agreement with the Pointing vector.22. The electromagnetic field is propagating along the negative z-axis.23. The intensity is an average, therefore it does not change.24. λ = c

f =(3.0×108 m/s)(1×1010 Hz) = 3 cm

25. Kilometers26. T = 1

f =1

35 Hz = 0.03 s27.

f =1T

→ λ =cf= cT = (3.0×108 m/s)(0.08 s) = 2.4×107 m

28. Bmax =Emax

c = 350 N/C3.0×108 m/s = 1.17×10−6 T z

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29.

E = Emax cos(kx−wt)

B = Bmax cos(kx−wt)

w = 2π f = 2π(25×106 s−1) = 1.57×108 rad/s

k =2π

λ=

(2π)[(3.0×108 m/s)(25×106 s−1)

] = 0.52

→ E = Emax cos(kx−wt) = (920 N/c)cos(0.52 x−1.57×108 t)

→ B = Bmax cos(kx−wt) = (2.8×10−6 T )cos(0.52 x−1.57×108 t)

30. f ′ = f

√c+ vc− v

= 82.0 MHz

√c+ .75 cc− .75 c

= 217 MHz

31. f ′ = f

√c− vc+ v

= 82.0 MHz

√c− .75 cc+ .75 c

= 31.0 MHz

32. Emax = cBmax = 4.2×10−6 T (3×108 m/s) = 1260 N/C33.

I =Pavg

4πr2 =E2

max

2µoc

→ E =

√µocPavg

2πr2 =

√(4π×10−7 T ·m/A)(3.0×108 m/s)(900 W )

2π(2.0 m)2 = 116 V/m

34.

Pavg

4πr2 =E2

max

2µoc

r =

√2µocPavg

4πE2max

=

√2(4π×10−7 T ·m/A)(3.0×108 m/s)(500 W )

4π(800 N/C)2 = 0.22 m

35. I = Pavg4πr2 =

1000 W4π(.3)2 = 884 W/m2

36. Bmax =Emax

c = 1c

√µocPavg

2πr2 = 1(3.0×108 m/s)

√(4π×10−7 T ·m/A)(3.0×108 m/s)(300 W )

2π(0.5 m)2 = 9.0×10−7 T

37. P = IA = (1200 W/m2)(100 m2) = 1.2×105 W38. F = PA =

(Ic

)A =

(1200 W/m2

3.00×108 m/s

)(100 m2) = 4.0×10−4 N

39. P = IC = 800 W/m2

3.0×108 m/s = 2.67×10−6 N/m2

40. Pavg =2Savg

c = 2c

PA = 2

(3.0×108 m/s)4.2×10−3 W1.2×10−3 m = 2.3×10−8 N/m2

41. The electric and magnetic fields move.42.

50 miles = 80467 m

t =xv=

80467(3.0×108 m/s)

= 2.68×10−4 s

43. The Pointing vector is the energy flow associated with the electromagnetic wave.44. Conservation of momentum.

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45. ∮ −→E ·d−→A =

qenc

εo

E(2πrl) =λlεo

→ E =λ

2πεor=

(20×10−9 C/m)

2π(8.85×10−12 C2)(0.12)≈ 3.0×104 N/C outward

46.−→B = µoI

2 πr =(4π×10−7 T ·m/A)[(30×10−9 C/m)(2.0×106 m/s)]

2π(0.15 m) 8.0×10−8 T z

47.−→F = qvBx = (1.6×10−19)(2.5×103)(3.0)x = 1.2×1015 Nx

48. The direction of the force on the charged particle is in the x direction.49. t = x

t =1.3×104 m3×108 m/s = 4.33×10−5 s

50. B = Ec = (120 V/m)

(3.0×108 m/s) = 4×10−7 T

51. E = Bc = (800×10−9 T )(3.0×108 m/s) = 240 V/m52.

Emax = 2.9 kV/m

λ =2π

k=

2π

(1.2×107)= 5.2×10−7 m

53. f = 1

2π

√LC

= 1

2π

√(0.3 H)(4.3×10−6 C)

= 140 Hz

54.

f =1

2π√

LC

f ′ =1

2π√

L′C=

12π√

(4L)C

f ′ =12

f

55.

12

Cε2 =

12

LI2max

Imax = ε

√CL= (9 V )

√1.2×10−6 F

0.2 H= 0.2 A

56. S = Ic =

1000 W/m2

3.0×108 m/s = 3.3×10−6 J/m3

57. S = P4πr2 =

420×103 W4π(305 m)2 = 0.36 W/m2

58. r =

√P

4πS=

√900×103 W

4π(0.2 W/m2)= 598 m

59.

Pin =Pout

ε=

2×106 W0.85

= 2.35×106 W

A =Pin

I= 2353 m2

60.−→E = 3i−2 j+1k, B = 2i+6 j+6k

−→E ·−→B = (3)(2)+(−2)(6)+(1)(6) = 0

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61.−→S = 11.2i−30.0 j

|−→S |= 32.0,θ =−69.5

62. r =

√Pµoc

2πE2max

=

√(80 W )(377)2π(25 V/m)2 = 2.77 m

63.

rEmax=25.0 = 2.77 m

r′ = 5.54 m

→ Emax =

√Pµoc2πr2 = 12.5 V/m

64.

Pπr2 =

E2max

2µoc

Emax =

√2µocP

πr2 = 141 N/C

65. p = Uc = 4×10−11

3.0×108 = 1.3×10−19 kg ·m/s

66. 20.2×10−3 W3.0×108 m/s (0.5 m) = 3.4×10−11 J

67. (a)

t100 km =1.0×105 m3×108 m/s

= 3.3×10−4 s

t1.0 m =1.0 m

343 m/s= 2.9×10−3 s

∆t = t100 km− t1.0 m = 2.6×10−3 s

(b) t100km = 3.3×10−4s = vt100km = (343 ms )(3.3×10−4 s) = 0.1 m

68. They are inversely related.69. 2 kHz = 150 km. Radio wave.70. 2 f m = 1.5×1023 Hz. Gamma rays.71. ∆λ = c

(1f2− 1

f1

)= 0.015 m

72.

λ =cf=

(3.0×108 m/s)(900× 103)

= 333 m

→ 165333

= 0.5

73. x = vt = (3.0×108)(2.5×10−4

2 ) = 3.75×104 m74. I = P

4πr2 =60 W

(4 π)(0.05)2 = 1.9 kW/m2

75. (a) Pr =Ic =

0.53.0×108 m/s = 1.7×10−9 Pa

(b)

P =B2

o

2µo

→ Bo =√

2Pµo =√

2(1.7×10−9)(4π×10−7) = 6.54×10−8 T

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76. Pr =Ic =

EoBo2 µoc =

ErmsBrmsµoc = B2

o2µo

77.−→E ·−→E = E2

x +E2y +E2

z78.

−→A ·−→B = ABcosθ

i f θ = 90

→ cosθ = 0

Therefore,−→A ·−→B = 0

79.

−→A ×−→B = ABsinθ

i f θ = 0

→ sinθ = 0

Therefore,−→A ×−→B = 0

80. Visible rays have smaller frequencies.81. Bo =

Eoc = 1.0×10−6 T

82. The Lorentz force is the force felt on a charged particle when it enters a region with either an electric ormagnetic field or both.

83. EB equals the speed of light.

84. Electromagnetic waves are transverse waves.85. Spherical waves.

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CHAPTER 20 Geometric Optics ProblemSets


20.2 REFERENCES

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www.ck12.org Chapter 20. Geometric Optics Problem Sets


Chapter 21 Solutions

1. No, the frequency of reflected light does not change.2. The camera should be set at 4 m.3. n =

speed of light in vacuumspeed of light in the material

4. Refraction can only occur when there is a change in index of refraction between two objects, which signifiesa difference in the speed of light between the two mediums.

5. Fermat’s principle states that the path taken by light between two points is the path that can be traversed in theleast time.

6. The image in the mirror is the same distance the object is from the mirror.7. It is due to refraction.8. The critical angle is the angle at which all light is reflected.9. A converging lens will bring the light rays to a point, while a diverging lens will spread the light rays further

apart.10. The focal length is defined as the distance from the center to the principal foci.11. Real images can be projected onto a screen while virtual images cannot.12. 60 cm13. 2 m/s.14. Rays: 2, 515. 25

16. θ = 35

17. The angle of reflection is equal to the angle of incidence. Therefore, the angle of reflection is 90.18.

n1 sinθ1 = n2 sinθ2

θ2 = sin−1(

n1

n2sinθ1

)19. The light bends towards the normal.20. n2 = n1

sinθ1sinθ2

= (1.0) sin35sin22 = 1.53

21. v = cn = (3×108 m/s)

1.2 = 2.5×108 m/s22. λ = λo

n = 600 nm1.33 = 451 nm

23. θ = 55

24.



θ3 = sin−1(

n2

n3sinθ2

)= sin−1

(n1

n3sinθ1

)→ θ3 = θ1

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25.

y =t

cosθ2

d = ysinφ =t

cosθ2sin(θ1−θ2)

26. (Phy-Int-Pro-21-26S.png)27. No, the angle of reflection is independent of the wavelength.28. (Phy-Int-Pro-21-28S.png)29. θc = sin−1

(n2n1

)= sin−1 ( 1

1.33

)= 48.8

30. This is a result of the diamond having a higher index of refraction, leading to total internal reflection.31. This is due to diamonds acting like prisms, and splitting the colors since the index of refraction varies with the

frequency of light.32. No, the frequency does not change.33. The beam that is bent less will be the one with the greater speed.34. There is no scenario that would allow this type of process to work, as it is not possible.35. θ2 = sin−1

(n1n2

sinθ1

)= sin−1 ( 1

1.33 sin45)= 32

36.

λ =λo

n=

320 nm1.33

= 240.6 nm

4λ = 240.6−320 =−79.4 nm

37. λglass =λ

n = 630 nm1.5 = 420 nm

38. n = cv =

(3.0×108 m/s)(2.25×108 m/s) = 1.33

39. If you’re in medium 1 : n1 = 1.0θ1 = sin−1((1.33)(sin45)) = 70.1

⇒ 19 with respect to the horizontal.40. θ1 = sin−1(n2 sinθ2) = sin−1(1.66sin21.2) = 37

41. θ1 = sin−1(n2 sinθ2) = sin−1 (1.66sin10.6) = 17.9

42. n2 = n1sinθ1sinθ2

= 1.93

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43.

n2 = 1.93

→ λ =λo

n=

630 nm1.93

= 326 nm

f =c

λo= 4.76×105 Hz

v =cn= 1.55×108 m/s

44.

v f irst =cn=

3×108

1.66= 1.80×108 m/s

vcrown =cn=

3×108

1.52= 1.97×108 m/s

45. v = cn = c

n1 sinθ1sinθ2

= csinθ2n1 sinθ1

= 1.16×108 m/s

46.

θ2 = sin−1((

n1

n2

)sinθ1

)= 23.5

θ3 = sin−1((

n2

n3

)sinθ2

)= 32

47.

tanθ2 =xh

→ h =x

tanθ2

θ2 = sin−1(

n1

n2sinθ2

)= 32.1

→ h =4.0 m

tan(32.1)= 6.4 m

48. n2 =n1 sinθ1

sinθ2= 1.18

49.

n2 =n1 sinθ1

sinθ2= 1.18

φ = sin−1(

n2

n3sinθ2

)= 22

50. θc = sin−1 n2n1

= sin−1 2.421 = 24.4

51.

θc = sin−1 n2

n1= sin−1 1

1.33= 48.6

d = 2(2.0 m) tanθc = (4.0 m) tan48.6 = 4.54 m

r =d2= 2.27 m

52. No, your image is 2.0 m away while the spider is only 1.0 m away.53. 1

s +1s′ =

1f

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54. When the image is positive, the image is in front.55. The mirror is a convex mirror.56. When the magnification is negative, the image is inverted.57. A concave mirror will focus the lights rays.58.

1s+

1s′=

1f

s′ =s f

s− f=

(20 cm)(10 cm)

10 cm= 20 cm

59.

1s+

1s′=

1f

s′ =s f

s− f=

(20.0 cm)(10.0 cm)

10.0 cm= 20.0 cm

m =−s′

s=−20.0 cm

20.0 cm=−1.00 inverted

60.

1s+

1s′=

1f

1s′=

15.0 cm

− 15.0 cm

→ s′ = ∞

This means that an image will not be formed.61.

1s+

1s′=

1f

s′ =s f

s− f=

(8.0 cm)(10.0 cm)

−2.0 cm=−40 cm

m =−s′

s=−−40 cm

8.0 cm= 5.0

62. s′ = s fs− f =

(3.0 cm)(9.0 cm)−6.0 cm =−4.5 cm

63.

m =−s′

s=−−2.5 cm

8.0 cm= 0.3

h′ = mh = 0.3(2.5) = 0.75

64.

m′ =−s′

s

s′ =−ms =−(

12

)(4.2) =−2.1 cm

65. s′ =−n2n1

s =− 1.01.33 d =−0.75 d

66. It is a virtual image.67. It is a converging lens.

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68. The image is upright.69. It is infinite due to the infinite radius of curvature.

70.71.

1f= (n−1)

(1

R1− 1

R2

)= 0.016

f = 60 cm

72. s′ = s fs− f =

(22.0 cm)(12.0 cm)(22.0 cm−12.0 cm) = 26.4 cm

73.

s′ =s f

s− f=

(30.0 cm)(10.0 cm)

(30.0 cm−10.0 cm)= 15.0 cm

m =−s′

s=−

(15.0 cm30.0 cm

)=−1

2

74. The image location is at infinity. (IMG)

75.s′ = s f

s− f =(8.0 cm)(11.0 cm)(8.0 cm−11.0 cm) =−29 cm

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76.

s′1 =s1 f1

s1− f1=

(35.0 cm)(8.0 cm)

(35.0 cm−8.0 cm)= 10.4 cm

m1 =−s′1s1

=−0.3

The image formed by the first lens acts as the object for the second lens.let s′1 = s2

s′2 =s2 f2

s2− f2=

(10.4 cm)(20.0 cm)

10.4 cm−20.0 cm=−21.7 cm

m2 =−(−21.7 cm)

10.4 cm= 2.1

Mtotal = m1m2 = (−0.3)(2.1) =−0.6

77. (a) When the object is farther than the focal length(b) When the object is closer than the focal length(c) When the object is farther than twice the focal length

78. (a) Never(b) Always(c) Always

79. Since R = 0

1s+

1s′=

1f= 0

s′ =−s

Using the magnification equation:

m = 1 =y′

yy′ = y = 72 inches

Using the geometry:

y′(

ss− s′

)=

y′

2

→ The mirror needs to be 36 inches tall.80. The first closest image is 5.0 m in the mirror on the left, or 10.0 m from your location. The second closest

image is located 10.0 m in the mirror on the right, or 20.0 m from your position. Lastly, the third image islocated 20.0 m in the mirror on the right or 40.0 m from your location.

81. The first image in the left mirror is 5.0 m in or a total distance of 10.0 from you. The second closest imagein the left mirror is due to the first image in the right mirror. That image is 25.0 m from the left mirror so itcreates an image that is 25.0 m in the left mirror. Therefore the total distance between you and the secondimage is 50.0 m

82.

f =R2=

25.0 cm2

= 12.5 cm

s′ =s f

s− f=

(40.0 cm)(12.5 cm)

40.0 cm−12.5 cm= 18.2 cm, real

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83.

f =R2=

40.0 cm2

= 20.0 cm

s′ =s f

s− f=

(25.0 cm)(20.0 cm)

25.0 cm−20.0 cm= 100 cm, real

m =−s′

s=−100 cm25.0 cm

=−4.0, inverted

84. s′ = s fs− f =

(15.0 cm)(0.65 cm)15.0 cm−0.65 cm =−0.68 m

85.

f =−R2

=−40.0 cm

2=−20.0 cm

s′ =s f

s− f=

(30.0 cm)(−20.0 cm)

30.0 cm+20.0 cm=−12.0 cm

m =−s′

s=

12.0 cm30.0 cm

= 0.400

86.

f =R2= 25.0 cm

s′ =s( f )

s− ( f )=

(80.0 cm)(25.0 cm)

80.0 cm−25.0 cm= 36.4 cm

m =−s′

s=−0.45

87.

m =−5 =−s′

s→ s′ = 5 s

s′− s = 0.6 m = 5 s− s

⇒ s = 0.15 m; s′ = 0.75

f =ss′

s+ s′= 0.125

88.

m =12=−s′

s→ s =−2 s′

|s′|−s = 10.0 cm =−s′−2 s′

⇒ s′ =−3.33 cm; s = 6.67 cm

f =ss′

s+ s′=−6.67 cm

89.

m =y′

y=

35= 0.6 =

−s′

ss′ =−0.6 s

s+ |s′|= 28.0 cm = s− s′

s = 28.0+ s′ = 28.0−0.6 s

s = 40.0 cm

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90.

1f= (n−1)

[1

R1− 1

R2

]= (0.5)

[1

2.0 cm− 1

3.0 cm

]= 0.083 cm−1

f = 12.0 cm

91.

s′ =s f

s− f=

(18.0 cm)(30.0 cm)

(18.0 cm−30.0 cm)=−45.0 cm

m =−s′

s=

45.0 cm18.0 cm

= 2.50 Virtual and upright

92.

s′ =s f

s− f=

(35.0 cm)(30.0 cm)

(35.0 cm−30.0 cm)= 210 cm

m =−s′

s=

210 cm35.0 cm

= 6.0 Real and inverted

93.

s′ =s f

s− f=

(10.0 cm)(20.0 cm)

(10.0 cm−20.0 cm)=−20.0 cm

m =−s′

s=

20.0 cm10.0 cm

= 2.0 inverted

94. Diverging lenses have negative focal lengths.95. Two rays is the minimum, the third is just a check.96. It is assumed that the difference between the distance between the surface of the lens and the focal point and

the distance between the center of the lens and the focal point is negligible.97. A virtual image is an image in which light rays do not pass through an image point.98. R→ ∞

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20.2 References

1. Haven Giguere. CK-12 Foundation .2. Haven Giguere. CK-12 Foundation .3. Haven Giguere. CK-12 Foundation .4. Haven Giguere. CK-12 Foundation .5. Haven Giguere. CK-12 Foundation .6. Haven Giguere. CK-12 Foundation .7. Haven Giguere. CK-12 Foundation .8. Haven Giguere. CK-12 Foundation .9. Laura Guerin. CK-12 Foundation .

10. Laura Guerin. CK-12 Foundation .

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www.ck12.org

CHAPTER 21 Physical Optics ProblemSets


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www.ck12.org Chapter 21. Physical Optics Problem Sets



1. Diffraction causes light to bend around obstacles and small openings.2. The spacing would be larger if red light was used instead of blue light.3. Diffraction would be more pronounced through a small opening. Additionally, optimal diffraction occurs

when the light’s wavelength is the same as the width of the opening.4. Young’s double slit experiment demonstrates the wave nature of light.5. The phase difference between the waves is due to the difference in path length between the two waves.6. (a) θ = sin−1

(mλ

d

).m = 0,1,2, ...

(b) θ = sin−1((m+ 1

2 λ

d

).m = 0,1,2, ...

7. Using the relationshiptanθ = y

Lusing the small angle approx.θ = y

LThe distance to the first maxima is theny = mλL

dThe distance to the next maxima is

ym+1 =(m+1)λL

d

∆y =λLd

=(800×10−9 m)(0.5 m)

0.1×10−3 m= 4.0×10−3 m

8. y = mλLd = (4)(600×10−9 m)(0.6 m)

0.2×10−3 m = 7.2 mm9. Yes, the reflected wave has a different phase than the incident phase.

10. For constructive interference in a thin film

2L =

(m+

12

)λ

n

⇒ λ =2 nL

m+(1

2

) = 2(1.33)(300 nm)

1+(1

2

) = 532 nm

11. Since the reflections have the same phase shift, the thickness needs to have a path difference of 2L

2L =

(m+

12

)λ

n

→ L =

(0+

12

)600 nm2(1.4)

= 107 nm

12.

2L = mλ

n

→ L = (1)600 nm2(1.4)

= 214 nm

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13. The spacing between the fringes decreases.14. The interference pattern would show a minimum(dark fringe).15. Dark fringes are produced when the path length difference is nλ.16. You should decrease the distance between the slits.17.

ymax = L tanθ

→ θ = tan−1( y

L

)λ =

d sinθ

m=

27×10−6 sin(tan−1

( yL

))2

= 599 nm

18.

λ

2= 2 nt

t =λ

4 n=

550 nm4(1.42)

= 97 nm

19. d = 12500 = 0.4 mm

20. d = 0.1 mm⇒ 10.1 mm = 1000 lines/cm

21.

d =1

5000= 2×10−6 m

θ = sin−1(

λ

d

)= sin−1

(720×10−9 m

2×10−6 m

)= 21

22.

d = 1.32×10−6 m

→ 1d= 7576 lines/cm

23. (a) The path difference must be multiple of half a wavelength.(b) The path difference between the two waves must be a multiple of the wavelength.

24. The fringe spacing would decrease.25. Monochromatic light is used since using light of multiple wavelengths would super impose all the patterns on

the screen.26.

d sinθ = mλ

λ = d sinθ = (0.4×10−3 m)sin(

tan−1(

3.3×10−3

3.0

))= 440 nm

27.

d sinθ = mλ

λ = d sinθ = (0.25×10−3 m)sin(

tan−1(

2.9×10−3

3.0

))= 242 nm

28.

d sinθ = mλ

d =mλ

sinθ=

2(500 nm)

sin(tan−1

(2.2×10−3 m

2 m

)) = 0.9 mm

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29.

λ =vf= 0.1 m

d sinθ = mλ

θ = sin−1(

mλ

d

)= sin−1

(0.1.25

)= 23.6

30.

d sinθ = mλ

λ =d sinθ

m=

0.285 m(sin22)3

= 0.036 m

f =vλ=

343 m/s0.036 m

= 9528 Hz

31.

3000 Hz⇒ λ = 0.11 m

d sinθ =

(m+

12

)λ

θ = sin−1((

m+12

)λ

d

)= sin−1

((0+

12

)0.110.28

)= 11.3

32.

d sinθ =

(m+

12

)λ

d =

(m+

12

)λ

sinθ=

(6+

12

)650 nm

sin(

tan−1(

8.0×10−3

2.2

)) = 1.16 mm

33.

tanθ =yL

y = L tanθ = (1.3 m) tan(3) = 0.07 m

34. 4y = λLd

(0+ 1

2

)= (550×10−9 m)(2.5 m)

2(2×10−3 m)= 3.4×10−4 m

35.

d sinθ = (m+12)λ

m =d sinθ

λ− 1

2=

(0.33×10−3)sin(20)500×10−9 m

− 12= 225

Therefore, the total on both sides of the central maximum is 250 minimum.36. For the number of dark fringes

d sinθ =

(m+

12

)λ

m =d sinθ

λ− 1

2=

(1.8×10−3)sin(20)550×10−9 m

− 12= 1118

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The number of dark fringes on both sides is 2236The number of bright fringes

d sinθ = mλ

m =(1.8×10−3)sin(20)

550×10−9 m= 1119

The total number of bright fringes plus the center is max is then 2239mtotal = 4475

37. No38. The phase difference is defined as

φ = 2π

λd sinθ = 2π

500×10−9 m(0.1×10−3 m)sin(0.6) = 13.2 rad39. φ = 2π

λd sinθ = 2π

(500×10−9 m)(0.2×10−3 m)sin(0.5) = 22

40. φ = 2π

λd( y

L

)= 2π

(550×10−9 m)(0.08×10−3 m)

(4.75×10−3 m

1.2 m

)= 3.6 rad

41.

φ =2π

λd sinθ

→ θ = sin−1(

λφ

2πd

)= sin−1

((550×10−9 m)( π

0.25)

2π(1.3×10−3 m)

)= 0.05 rad

42.

φ = 2cos−1√

IImax

= 2cos−1√

0.55 = 1.47

δ =λφ

2π=

(550 nm)(1.47 rad)2π

= 129 nm

43.

I = Imax cos(

φ

2

)I

Imax= cos

(φ

2

)= cos

(1.70 rad

2

)= 0.66→ 66%

44.

I = Imax cos(

πydλL

)I

Imax= cos

(πydλL

)= cos

(π(0.005 m)(0.2×10−3 m)

(650×10−9 m)(0.8 m)

)= 0.97

45. For constructive interference

λn

2+2t = λn

⇒ λn =λ

n

2t =λ

2n→ λ = 4nt = 4(1.33)(120 nm) = 638 nm

46. t =(1

2 +m)

λ

2n =(1

2

) 550 nm2(1.3) = 106 nm

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47.

d sinθ = mλ

λ =d sinθ

m=

( 14300

)sin(tan−1 0.5

1.9

)1

= 592 nm

48. d = 12300 line/cm = 4.35 µm

49. 14.0×10−6 m = 2500 lines/cm

50.

d =λ

sinθ

→ 1d=

sinθ

λ=

sin15800 nm

= 0.8 µm

51. False52. Yes.53.

d =1

3300= 3.0 µm

θ = sin−1(

λ

d

)= sin−1

(700×10−9

3×10−6

)= 0.24

54.

2d sinθ = mλ

θ = sin−1 mλ

2d= sin−1

(0.13

2(0.3)

)= 0.22 rad = 12.5

55.

2d sinθ = mλ

λ =2d sinθ

m=

2(0.18 nm)sin(9.2)1

= 0.06 nm

56.

δ = 180 =4rλ

360

4 r =λδ

360=

12

λ =12

500 nm = 250 nm

57. The energy goes into the components that interfere constructively.58. True.59. Differences in path length in the traveling wave fronts.

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CHAPTER 22 The Special Theory ofRelativity Problem Sets


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www.ck12.org Chapter 22. The Special Theory of Relativity Problem Sets


Solutions

1. (a) The laws of physics are independent of the reference frame.(b) The speed of light in a vacuum is constant.

2. The first postulate states that absolute uniform motion cannot be detected.3. γ = 1√

1− v2

c2

4. Yes, from your point of view, your copy’s pulse would be slower.5. It would be the same if they are moving at the same velocity.6. With respect to your reference frame, the other person’s measurement will seem slower.7. 6 flashes8. No9.

L = Lo

√1− v2

c2

v2 = c

√1−(

LLo

)2

= (3.0×108 m/s)

√1−(

12

)2

= 2.60×108 m/s

10. Since v «c, relativistic effects can be ignored. 3 + 5 = 8 m/s11. 0.5 m/s12. The experiment disproved the existence of ether (or medium that transmits wave motion).13. Time.14. Longer.15. γ = 1√

1− v2

c2

= 1√1− (0.50 c)2

c2

= 1.15

16.

γ75 =1√

1− v2

c2

→ v = c

√1−(

1γ

)2

= c

√1−(

11.33

2)= 0.67 c

17. The frequency changes increases.18. The maximum possible value is 1. This could only be possible if the numerator equals c2.

19. (a) L = Lo

√1− v2

c2 = (1.0 m)

√1− (0.999 c)2

c2 = 0.0447 m

(b) 1 m since it is in the same inertial reference frame.20. p = ∞

21. You wouldn’t be able to tell the difference unless there was an acceleration.22. The bola is moving faster.23. The ball has a faster velocity.24. (a) The twin that stayed behind would be older.

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(b) No, time will only travel forward.25. Any reference frame that is moving relative to your own will detect a change.26. The probe has a speed of uPinO =

us−up

1− usupc2

= 0.85 c−0.425 c

1− (0.85)(0.425) c2

c2

= 0.67c

27. t = toγ = (10 min) 1√1− v2

c2

= (10 min) 1√1− (0.5 c)2

c2

= 11.5 min

28. (a) L = Loγ= 500 m

(√1− v2

c2

)= 500 m

(√1− (0.7 c)2

c2

)= 357 m

(b) L = Loγ= 500 m

(√1− v2

c2

)= 500 m

(√1− (0.35 c)2

c2

)= 468 m

29. L50%L99%

= γ99%γ50%

=

(√1− v2

c2

)(√

1− v2

c2

) =

√1− (.50)2√1− (.99)2

= 6.1

30. Classical velocity addition is used in non-relativistic cases where similar components are added or subtractedfrom one another to determine changes in velocity.

31.

f ′ = fo

√1− V

c

1+ Vc(

f ′

fo

)2

=1− V

c

1+ Vc

=

(λo

λ

)2

=

(6001500

)2

= 0.16

Vc= 0.16

32. ux =u′x+v

1+ vu′xc2

= c+v1+ vc

c2=

c(1+ vc)

1+ vc

= c

33. E = mc2√1− v2

c2

= mc2√1− (0.8 c)2

c2

= 0.7 MeV

34. The rest mass is mc2 = 0.5 MeV . p = mv√1− v2

c2

= m(0.85 c)√1− (0.85 c)2

c2

= m(0.85 c)0.53

cc =

(0.85)(0.5 MeV )0.53 c = 0.8 MeV/c

35. (a) The proton has a rest mass: mc2 = 940 MeV .

E =moc2√1− v2

c2√1− v2

c2 =mo c2

E

v = c

√1− (

moc2

E)2 = .34c

(b)

E2 = (pc)2 +(mc2)2

p =1c

√E2− (mc2)2 =

1c

√(1000 MeV )2− (940 MeV )2 = 341 MeV/c

36.

E2 = (pc)2 +(mc2)2

pc =√

E2− (mc2)2 =√((1 MeV +2 MeV )+3 MeV )2− (1 MeV +3 MeV )2

p = 3.46 MeV/c

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37.

E2 = (pc)2 +(mc2)2

m =1c2

√[E2− (pc)2] = 6.7 MeV/c2

38. (a) Your heart beat would still measure at the same rate since they are in the same reference frame.(b)

t ′ = γt =1√

1− v2

c2

(1.02 s) = 2.3 s

→ 60s/min2.3s/beat

= 26 beats/min

39. uAB = uB−uA1− uBuA

c2= −0.8 c−0.6 c

1− (−0.8)(0.6) c2

c2

=−0.95c

40. Relativistic momentum: pr =mv√1− v2

c2

= (9.11×10−31 kg)(0.7)(3×108 m/s)√1− (0.7 c)2

c2

= 2.68×10−22 kg ·m/s

Classical momentum:

pc = mv = (9.11×10−31 kg)(0.7)(3×108 m/s) = 1.91×10−22

pr

pc= 1.40 times larger

41. Dependent on the mass but independent of the momentum.42. E = mc2 = (9.11×10−31 kg)(3.0×108 m/s)2

(1.0 eV

1.6×10−19 J

)= 0.5 MeV

43. (a)

E = 2mc2 =mc2√1− v2

c2

2mc2 =mc2√1− v2

c2

⇒ v = c

√1−(

12

)2

= 0.87 c

(b) K = E−mc2 = 2mc2−mc2 = mc2 = 0.5 MeV44. They will both agree on the speed of light and their relative speeds45. The momentum is increased by p = m(4uo)√

1− 16u2o

c2

46. As the speed of light is approached, the amount of work needed to bring the particle to v = c is infinite.47. Photons are massless.48.

t = 2to

v = c

√1−(

to2to

)2

= 0.87 c

49. L = Lo

√1− v2

c2 = (1.0 m)

√1− (0.9 c)2

c2 = 0.44 m

50. (a) t = xv =

15 c yr0.5 c = 30 years

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(b)

t = γto

to =ty= (30 yrs)

√1− v2

c2 = (30 yrs)

√1− (0.5 c)2

c2 = 26.0 yrs

51. t = xv =

5000 m0.9 c = 1.85×10−5 s

52. Non-relativistic speeds: p = mv = (9.11×10−31 kg)(0.01)(3.0×108 m/s) = 2.73×10−24 kg ·m/s53.

4E = (γ2− γ1) mc2

(mc2)electron = 0.511 MeV

4E =

(1√

1−0.72−1)(0.511 MeV ) = 0.45 MeV

54. The first postulate of Special Relativity is that “All laws of nature are the same in all uniformly moving framesof reference”. An example of this would be a plane that is flying in a straight line with a constant speed. Ifyou try to perform an experiment, such as throwing a ball, in the plane, you’ll find that the ball follows thesame rules of trajectories as if it were thrown on the surface of the Earth.

55. You and your friend will have the same frame of reference if you are not moving relative to one another. Forexample, if you are travelling in a car together, you are both moving (relative to the surface of the Earth), butnot moving relative to each other since you both have the same speed. You will not be in the same referenceframe if you have different velocities.

56. There would be no way to tell the difference between being at rest and being in motion if you were movingat a constant speed. However, if you accelerate, objects will appear to move in the direction opposite of theacceleration. As an example, consider something hanging from the rear-view mirror swinging as you turn,speed up, or slow down.

57. Relative to an outside observer, the paintball has a faster velocity when the car is moving. If someone is insidethe car with her, the paintball appears to move at the same speed as it would if it were fired by Jeanette whenshe is standing outside on the ground.

58. No, there is no way to have observers in a different frame of reference see the fireworks going off in differentplaces or at different times. Relativity allows for the adjustment of distances and/or times, so that if there is adifference between the distances (or times) of two events in one frame, it can be translated to what is observedin a different frame of reference. If, however, the difference between the distance and times of two events in aparticular frame of reference is zero, there is no translation to a difference frame of reference. The differencein distance and time between the two events will still be zero in all frames of reference.

59. Time passes for each twin throughout the trip at different speeds. However, to have a twin return before theother twin was born would require travelling backwards in time, which is not possible without travelling fasterthan the speed of light.

60. Since the situation is relative, there isn’t a preferred frame of reference. So while objects and people appear tobe shrunk along your direction of motion, you appear to be shrunk along your direction of motion to them.

61. Since E =mc2, the reaction would produce: (2 kg)(3×108 m/s)2 = 1.8×1017 Joules (2 kg)(3×108 m/s)2 =1.8×1017 Joules

62. V = v1+v21+ v1v2

c2= 1.7 c

1.723 = 0.99 c

63.

γ =1√

1− v2

c2

=1√

1− (0.95 c)2

c2

=1√

1− .9025= 3.203

γt = (3.203)(1) = 3.203 yrs

64. (a) t = γto =(

1√1−0.995

2

)(2.2×10−6 s) = 2.2×10−5 s

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(b) No. Relativity shows that either length or time can be different according to different observers.65. In this case, “l0” is the length measured at rest relative to the sled, so that is the length someone riding on the

sled would measure. “l” is the length measured by someone watching the sled go by. Solve the equation forrelativistic length for “l0” and plug in the appropriate values:

l = l0

√1− u2

c2

lo =l√

1− u2

c2

=2√

1− (0.8)2= 3.33 meters

66. (a) Pam measures the proper time interval, since this is the interval between events taken from the perspec-tive of someone at rest in the frame of the car.

(b) Pam measures the proper length of the car, since this is the distance measured at rest in the frame of themoving vehicle.

(c) Ted.67. No, this does not contradict Einstein’s postulate. Remember that the equation for the speed of a light wave is

λ f = c. Since “c” is a constant, when λ goes up, f must go down to compensate, and the speed of the wave“c” is unchanged.

68. 4t = γt0 = 1√1− (0.01)2

(2) = 2.00

69. The pilot measures the proper time since it is the interval between events taken from the perspective ofsomeone at rest in the sign’s reference frame.

70. l0 = l√1− u2

c2

= 110√1−(0.75 c

c

)2= 166.3 m

71. u = c

√1−(

ll0

)2

= c

√1−(

0.401

)2

= 0.917c

72. (a)

l0 = 7000

l = l0

√1− u2

c2 = (7000)

√1− (2000∗343)2

(3×108)2 ≈ 7000 f t

(b) 4t0 = l0u = 7000

2000×343 = 198 = 0.010 s

(c) 4t = lu = 7000 m

2000(343) m/s ≈ 0.010 s

73. v = v′+u1+ uv′

c2= 0.8c+0.75 c

1+ (0.8)(0.75) c2

c2

= 0.969 c

74. 4t = γt0 =

1√1−(0.55 c

c

)2

(7 s) = 8.38 s

75. v′ = v−u1− uv

c2= −0.96 c−0.96 c

1− (0.96 c)(−0.96 c)c2

=−0.999 c

76. Only the side along the direction of motion is shortened. V = 1000

√1− (0.8 c)2

c2 = 600The box is now 400 m3smaller.

77. 4t =( c

0.925 c

)(50 yr) = 54.1 yr To measure the time dilation,

4t0 =4t

√1− u2

c2 = (54.1 yr)√

1− (0.925)2 = 20.56 yr

ttotal = 38 years old

78. pr− pclassical =γmv−mv

mv = mv(γ−1)mv = γ−1 = 1√

1− v2

c2

−1 = 4.4×10−14

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79.

L =Lo

γ

L = Lo

√1− v2

c2 = 2

√1− (0.7 c)2

c2 = 1.43 m

80. 4t = xv =

15 c·yr0.9 c = 16.7 yrs

Mirella ages by→ to =4tγ= 16.7 yr

√1−0.92 = 7.3 yrs

Similarly for Sam: to =15 c·y0.75 c

√1−0.752 = 13.2 yrs

When Mirella lands, she has to wait, so she ages by 15 c·y0.75 c −

15 c·y0.90 c = 3.3 yrs

Therefore, Sam is older by 13.2− (7.3+3.3) = 2.6 yrs

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CHAPTER 23 Quantum Physics ProblemSets


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TABLE 23.1:

Vs (V) λ(nm)1.9 2800.72 450

25. When 500 nm light strikes a surface, the stopping potential for the ejected electrons is 0.70 V. What would thewavelength need to be for the stopping potential to be twice as large?

26. Calculate the magnitude of the momentum of a photon that has energy equal to the rest energy of an electron.27. What is the wavelength of a photon with a momentum of 2.7 * 10−22 kg m/s?28. Determine the energy and frequency associated with a photon whose wavelength is 42.0 pm.29. Determine the de Broglie wavelength of a 2.0 keV photon.30. Determine the de Broglie wavelength of a 1.0 keV neutron.31. Determine the de Broglie wavelength of electrons that are accelerated through a 15.0 kV potential difference.32. While performing a photoelectric experiment, if the frequency of the light used is increased but the intensity

is held constant, does the stopping potential increase or decrease?33. Determine the cutoff wavelength for a sodium surface (ϕ =2.5 eV).34. A 2.0 kg brick is thrown with a speed of 22.0 m/s. What is its de Broglie wavelength?35. Show that a charged particle of mass m that is accelerated from rest through a potential difference of ∆V has

a de Broglie wavelength of: h√2mp4V

36. Determine the energy of a photon of frequency 500 GHz in electron volts.37. A transmitter with a power output of 100 W sends out a signal at 98.5 MHz. Determine the number of photons

emitted per second.38. Tungsten has a work function of 4.6 eV in vacuum. Determine the cutoff wavelength for the photoelectric

effect.39. Does the discovery of the cutoff frequency in the photoelectric effect give a stronger case for a particle or

wave-like theory of light?40. Electrons are released from a surface with a maximum velocity of 4.0 * 105 when 600 nm light is used.

Calculate the work function of the surface the electrons are ejected from.41. What is the momentum and energy of a 650 nm photon?42. What is the de Broglie wavelength of a proton that has a speed of 1.6 * 106 m/s?43. Determine λm for a body at a temperature of 2400 K44. An incandescent light bulb has a temperature of approximately 2000 K. Explain why an incandescent light

bulb is not a very efficient source of light.45. What is the frequency of a photon with a 400 nanometer wavelength?46. How much energy (in joules and electron volts) does a photon with a wavelength of 300 nanometers have?47. A photon of heat energy (infrared light) has a wavelength of 730 nanometers. Find the frequency and

momentum of this photon.48. A gamma-ray photon striking Earth’s upper atmosphere has an energy of 2.77 MeV. What is the wavelength

and frequency of this photon?49. A given material has a photoelectric threshold wavelength of 305 nm. Calculate the maximum kinetic energy

(in electron volts) of the electrons ejected from the surface of the material if the material is struck by radiationwith a frequency of 1.77 * 1015 Hz.

50. A metal surface is exposed to 350 nm light. What is the maximum speed of the electrons emitted from thissurface if the surface’s work function is 2.7 eV?

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51. What is the work function for a metal that is just capable of ejecting electrons when 350 nm light shines on it?52. If a monochromatic light bulb emits 500 nm light, calculate the frequency of the light.53. If a monochromatic 100 Watt light bulb generates 500 nm light, how many photons per second is it emitting?54. How is the energy of a photon changed if its momentum is doubled?55. If the photoelectric work function of a particular metal is 3.1 eV, and 325 nm light falls on it, calculate the

stopping potential in volts.56. Can a particle be diffracted?57. If the photoelectric work function of a particular metal is 2.7 eV, and 400 nm light falls on it, calculate the

stopping potential in electron volts.58. Calculate the velocity of the ejected electrons created in the previous problem.59. If the stopping potential of a metal for electrons ejected by 320 nm photons is 0.220 V, what is the photoelectric

threshold wavelength for the metal?60. If the stopping potential of a metal for electrons ejected by 425 nm is 0.20 V, what is the work function for the

metal?61. If a photon has a momentum of 7.5 * 10−27 kg m/s, what is the energy of the photon expressed in electron

volts?62. If a photon has a momentum of 8.1 * 10−27 kg m/s, what is the wavelength of the photon?63. If a radio station broadcasts on a frequency of 94.7 MHz, what is the momentum of each photon emitted by

the radio transmitter?64. If a radio station broadcasts on a frequency of 97.9 MHz and a power of 40 kW, how many photons does the

radio transmitter emit per second?65. If a particular metal ejects electrons with a maximum kinetic energy of 7.1 eV when light of 220 nm falls on

it, what is the work function of the metal in eV?66. What would change if the value of Planck’s constant h were increased by a significant amount?67. Waves often travel in a medium. Water waves travel in water. Sound waves travel in air. What do quantum

mechanical wavefunctions travel in?68. What does the wave nature of matter have to say about why objects appear solid?69. What potential difference would you have to accelerate electrons through to have a wavelength of .22 nanome-

ters?70. How fast would a neutron have to be moving to effectively measure the distance between atoms in a crystal

lattice with atoms approximately .25 nm apart?71. Are protons or electrons better for probing very small scale structures?72. A beam of electrons is scattered by atoms with a lattice spacing of .115 nm. The first intensity maximum

occurs at 35 degrees. What is the kinetic energy of each of the electrons in the beam in Joules?

Solutions

1. The gamma ray has greater photon energy.2. R = P

E = Pλ

hc = (250 W )(580×10−9 m)(6.6×10−34)(3.0×108)

= 7.32×1020 photons/s

3. λ = R(hc)P = (3.2×1020 photons/s)(6.6×10−34)(3.0×108)

150 W = 422 nm4. Proportional.5. Φ = h fo = (6.63×10−34J · s)(5.0×1014Hz) = 3.32×10−19J

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6.

f raction =h f −h f ′

h f=4λ

4λ+λ

4λ =h

mc(1− cosφ) =

(6.63×10−34 J · s)(9.11×10−31 kg)(3.0×108 m/s)

(1− cos85) = 2.21 pm

f rac =2.21

2.21+50= 0.042 = 4.2%

7. No the wavelength isn’t the same. The electron has a greater wavelength.8. No, the wavelength isn’t the same. The electron has a greater de Broglie wavelength.9. λ = h

p = h√2mK

= 6.63×10−34 J·s√2(9.11×10−31 kg)(100 eV )(1.6×10−19 J/eV )

= 123 pm

10.

4x =h4p

→4p = (0.001)p = (0.001)(9.11×10−31 kg)(1.78×105 m/s) = 1.62×10−28

4x =(6.63×10−34 J·s)

2π

1.62×10−28 = 651 nm

11. RA = 4λP

hc(πd2)= 4(652×10−9 m)(5.0×10−3 W )

π(6.63×10−34 J·s)(3×108 m/s)(3,0×10−3 m)2 = 2.31×1021 phtons/m2·12. E = hc

λ= (1240 eV ·nm)

590 nm = 2.10 eV

13. R = λPhc = (590×10−9)(120 W )

(6.63×10−34 J·s)(3×108 m/s) = 3.56×1020 photons/s14.

P =Rhc

λ

→ Rate =λPhc

The greater the wavelength, the higher the rate. Therefore, 800 nm lamp emits more photons/s.15. λ = hc

eVs+Φ= 1240 eV ·nm

5.0 eV+2.0 eV = 177 nm16. E = hc

λ= 1240 eV ·nm

250 nm = 4.96 eV17. K = h f −Φ = (4.14×10−15 eV · s)(3.2×1015 Hz)−2.0 eV = 11.2 eV

18. v = c

√2(E−Φ)

mc2 = (3×108 m/s)

√2(6.0−4.5)511×103 eV

= 7.30×105 m/s

19. Vs =h f−Φ

e =( 1240 eV ·nm

450 nm )−1.5 eVe = 1.26 V

20.

Vs =h f −Φ

e=

(1240 eV ·nm225 nm

)−1.5 eV

e= 4.01 V

v =

√2eVs

m= c

√2eVs

mc2 = (3×108 m/s)

√2e(4.01 V )

511×103 eV= 1.19×106 m/s

21. Kmax =hcλ− hc

λmax= 1240 eV ·nm

255 nm − 1240 nm270 nm = 0.27 eV

22. The slowest electron just breaks free of the surface and has zero kinetic energy.23. Vs =

Kme = 1

e

(hcλ−λ)= 1

e

(1240 eV ·nm300 −4.0

)= 0.13 V

24. Φ = e(λ1V1−λ2V2)λ1−λ2

= e((1.75)(280 V )−(450)(0.80))(450−280) = 0.76 eV

25.hcλ2

=hcλ1−K1 +K2

→ λ2 =hcλ1

hc+λ1(K2−K1)=

(1240 eV ·nm)(500 nm)

1240 eV ·nm+(500 nm)(1.4−0.7)= 390 nm

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26. p = mec = 2.7×10−22 kg ·m/s27. λ = h

p = 2.4 pm28.

f =cλ= 7.14×1018 Hz

E = h f = 2.96×104 eV

29. λ = hcE = 1240 eV ·nm

2.0 keV = 0.62 nm30. λ = h√

2mK= 6.63×10−34 J·s√

2(1.675×10−27 kg)(1.6×10−16 J)= 0.9 pm

31. λ = h√2meeV

= (6.626×10−34)√2(9.11×10−31 kg)(1.602×10−19 C)(15×103 V )

= 10.0 pm

32. The magnitude of the stopping potential increases.33. λc =

hcΦ= 1240 eV ·nm

2.5 eV = 496 nm34. λ = h

mv =6.63×10−34 J

(2.0 kg)(22.0 m/s) = 1.5×10−35 m35.

E f = Ei

K f +U f = 012

mv2 +(−4V )q = 0

p2

2 m=4V q⇒ p =

√2 mq4V

⇒ λ =h√

2 mp4V

36. E = h f = (6.626×10−34 J · s)(500×109 Hz)(

1.0 eV1.6×10−19 J

)= 2.1×10−3 eV

37.

E = h f = (6.626×10−34J · s)(98.5×106Hz) = 6.53×10−26J

→ PE

=100 J/s

6.53×10−26J/photon= 1.53×1027 photons

38. λ = hcφ= 1240 eV ·nm

(4.6 eV ) = 270 nm39. The photoelectric effect gives a stronger case for the particle nature of light.40. Φ = E−Km = 1240 eV · nm

600 nm − 12(9.11×10−31)(4.0×105 m/s)2 = 2.07 eV

41.

p =hλ= 1.02×10−27 kg ·m/s

E =hcλ

= 1.91 eV

42. λ = hp = h

mv = 2.48×10−13m

E = K =12

mv2 = h f

vph = f λ =mv2h2 hmv

=u2

43. λm = 2.898×10−3 m·K2400 K = 1.2075×10−6 m

44. Because using the equation to calculate the peak emission wavelength for 2000 K shows that the light bulbemits primarily in the infrared portion of the EM spectrum. Therefore most of the energy is emitted as heatinstead of light.

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45. f = cλ= 7.5×1014 Hz

46. E = h f = 6.63×10−19 J = 4.14 eV47.

f =cλ= 4.29×1014 Hz

p =hλ= 9.47×10−28 kg ·m/s

48.

f =Eh= 6.70×1020 Hz

λ =cλ= 4.48×10−13 m

49. K = h f −φ = h f −h cλth

= 4.136×10−15eV · s(

1.77×1015Hz− 3×108 m/s305×10−9m

)= 3.25 eV

50.

12

mv2 =hcλ−φ

v =

√2

hcλm−2

φ

m=

√2

1240 eV ·nmme350 nm

−22.7 eV

me=√

1.68 eV/me = 5.43×105 m/s

51. φ = hcλ= 3.55 eV

52. f = cλ= 600 pHz

53.

E = h f = h( c

λ

)= 2.48 eV

⇒ 1.26×1019 photons/s

54. If the energy is doubled, the momentum is doubled.55. eVo = h f −φ = 0.72 eV56. Yes, a particle can be diffracted since it has wave properties through the rules of quantum mechanics.57. eVo =

12 mv2 = h f −φ = 0.402 eV

58.

12

mv2 = 0.402 eV

→ v = 3.75×105 m/s

59. λ = hcφ= hc

hcλ−eVo

= 1240 eV ·nm3.875 eV−0.220 eV = 339 nm

60.

eVo = h f −φ

φ =hcλ− eVo =

1240 eV ·nm425 nm

−0.2 eV = 2.72 eV

61. E = pc = (7.5×10−27 kg ·m/s)(3×108 m/s)(1.602×1019 J) = 36.0 eV62. λ = h/p = 80.7 nm63. p = h f

c = 2.09×10−34

64.

p =h fc

= 2.16×10−26

# =40000 J/s

2.16×10−26 J= 18.52×1029 photons/s

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65. φ = hcλ−Km = 3.51 eV

66. If the value of Planck’s constant were larger, atoms would be bigger. This can be seen from the equation forthe deBroglie wavelength of matter waves. As the “h” in the numerator increases, so does the length of thedeBroglie waves. Therefore, the wavelengths of electrons making up orbitals in atoms would be larger as well,thereby making atoms bigger.

67. Quantum mechanical wavefunctions do not travel in a medium. This was first seen with light. It was originallythought that light travelled through a medium called the “ether”. It was discovered that light did not need amaterial to travel through. This was the first case of a quantum mechanical wavefunction (in this case, forphotons) that did not require a medium for propagation. Today we know that all quantum mechanical objectsare capable of travelling without a medium.

68. The wave nature of matter keeps the electrons and protons in atoms apart. Otherwise, atoms would collapse inon themselves, with the negatively charged electrons falling onto the positively charged protons in the nucleus.However, because the electrons can only exist at certain wavelengths around the nucleus, they are forced tostay away from the nucleus by a fixed amount, giving matter its structure.

69. 4V = h2

2 emλ2 =(6.626×10−34 J·s)2

2(1.60×10−19 C)(9.11×10−31 kg)(0.22×10−9 m)2 = 31.12 V70. The momentum of the neutron can be set equal to its q.m momentum

p =hλ

→ v =h

mλ=

(6.626×10−34)

(1.674×10−27 kg)(0.25×10−9)= 1.58×103 m/s

71. A larger mass of probe would then need to be accelerated to a higher velocity to get a desired wavelength.Therefore, the least massive particle (the electron) is the best probe.

72.

λ = d sinθ =h√

2 mE

→ E =h2

2 m2d2 sin2θ= 5.06×10−17 J

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CHAPTER 24 Atomic Physics ProblemSets

Chapter Outline24.1 CHAPTER 24 PROBLEMS ATOMIC PHYSICS

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www.ck12.org Chapter 24. Atomic Physics Problem Sets

24.1 Chapter 24 Problems Atomic Physics

Solutions

1. It can either fall to the n = 2 or n = 1 state.2. The electrical force.3. The energy absorbed is the same as the energy emitted.4. Blue light has more energy.5. Rutherford’s experiment demonstrated that an atom is mostly made up of empty space.6. When the particles were back-scattered, it shows that atoms had a nucleus that was much heavier that an alpha

particle, and were mostly made up of empty space.7. Atomic number.8. There are 2 protons.9. 1.6×10−19C

10. The atomic radius is on the order of 0.1 nm.11. Assuming visible light is in the range: 400 < λ < 700 nm

→ Elow =hcλ

=1240 eV ·nm

400 nm= 3.1 eV

→ Ehigh =hcλ

=1240 eV ·nm

700 nm= 1.77 eV

12. E = hcλ= 1240 eV ·nm

0.01 = 1.24×105 eV13.

1λ= R

(112 −

122

)= 8.25×10−3nm−1

→ λ = 121 nm

14.

Ephoton = E3p−E2s

→ E3p = Ephoton +E2s = 2.5+20.6 eV = 23.1 eV

15. λ = hcE = 1240 eV ·nm

2.2 eV = 564 nm16. −13.6 eV .17.

1λ= R

(19− 1

16

)= 5.35×10−4

→ λ = 1.87 µm

18. The force felt between the two charges is due to the Coulomb force

F =kq1q2

r2

q1 =−q2

→ F =−kq2

r2

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24.1. Chapter 24 Problems Atomic Physics www.ck12.org

Rewriting the LHS in terms of centripetal motion

mv2

r=−kq2

r2

→ mv2 =−kq2

r

→ K =−kq2

2r

19. E =−Z2(−13.6 eV )(1− 1

22

)= 91.8 eV

20.

E =−(82)2(−13.6 eV )

(1− 1

22

)= 68.6 keV

λ =hcE

=1240 eV ·nm68.6×103eV

= 0.02 nm, these are X− rays

21. E = hcλ= 1240 eV ·nm

550 nm = 2.25 eV22.

E =hcλ

λ =hcE

=1240eV ·nm

3.2 eV= 387.5 nm

f =cλ= 7.7×105Hz

23.

λ =hp→ K =

12

p2

m

⇒ p =√

2 mK

λ =h√

2 mK=

hc√2 mcK

=1240 eV ·nm√

2(0.51×106eV )(100 eV )= 0.12 nm

24. λ = h√2 mK

= hc√2 mcK

= 1240 eV ·nm√2(0.51×106eV )(400zeV )

= 0.06 nm

25. For blue light λ≈ 450 nmK = (λh)2

2 m = [1240 eV ·nm]2

2(0.511×106)450 nm = 3.3 meV

26. The muon has a mass that is approx. 105.7 Mev/c2

λ = h√2 mK

= hc√2 mcK

= 1240 eV ·nm√2(105.7×106eV )(20×103eV )

= 6.0×10−4nm

27. λ = hcE = 590 nm

28. E = (−13.6 eV )(1− 1

4

)= 10.2 eV

29. The energy for the transition is 10.2 eVTherefore,λ = hc

E = 122 nm30. E = (−13.6 eV )

(1− 1

25

)= 13.06 eV

Therefore,λ = hc

E = 95 nm31. E = (13.6 eV )

(14 −

19

)= 1.89 eV

Therefore,λ = hc

E = 656 nm32. Increases

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www.ck12.org Chapter 24. Atomic Physics Problem Sets

33. E = (13.6 eV )(1

9 −1

39

)= 1.13 eV

34. E4

35. 4px ≥ h2π4x =

6.626×10−34 J·s2π(20×10−2 m)

= 5×10−34 kg ·m/sDividing the momentum by mass gives the velocity:4vx =

4pxm = 5×10−34 kg·m/s

103 kg = 5×10−37 m/s36. Since we know that the ball is somewhere on the field, its maximum uncertainty in position is 25 meters.37. We know that the Heisenberg Uncertainty Principle says that (4x)(m4vx)≥ h

2π. Setting4vx = (0.025)vx we

find that:

(4x)(m)(.025)(vx) =h

2π

vx =6.63×10−34

2π(.025)(9.11×10−31)(.45×10−3)= 10.3011 m/s

38.

4x =h4p

→4p = (0.001)p = (0.001)(9.11×10−31 kg)(1.78×105 m/s) = 1.62×10−28

4x =(6.63×10−34 J·s)

2π

1.62×10−28 = 651 nm

39. A fluorescent compound might be excited several energy levels by a high frequency photon(ie n=1 to n=4),then emit two photons of lesser energy(n=4 to n=3, n=3 to n=1), which would emit a photon with lowerfrequency.

40. The wall’s recoil is subject to the same uncertainty as the photon itself, even though the wall is much larger.41. There is no lower limit. It is possible to know nothing about a particle.42. The student’s statement is relatively accurate. The nucleus contains the majority of the atom’s mass, and is

much smaller (105) the size of the radius of the atom.43. Quantum effects apply at all sizes. However, these effects are very small on the macroscopic scale.44. An incandescent bulb can be modeled as a black-body radiator, which emits light across all wavelengths.

However, a gas discharge tube is made of a single substance, which shares similarly quantized electrons.45. The square of a particle’s amplitude is proportional to its probability of existing at a location. Thus, the particle

is four times as likely to be found at y than at x.46. The emission spectra would be the combination of the each gas’s spectra, as each gas absorbs light indepen-

dently.47. The smallest possible transition is a n=2 to n=1 transition for the Lyman series. This generates a 10.6 eV

photon. However, visible light has energies between 1.6 and 3.4 eV.48. These measurements are not taken at the same time. The second measurement will destroy the results of the

first measurement, reducing the precision to an amount consistent with the uncertainty principle.49. All matter can be represented with wave equations. However, electrons have very small masses, which means

that wave equations are the best way to describe them.50. The hydrogen atom will be ionized, as the energy needed to ionize the hydrogen atom’s electron is 13.6 eV.

The remaining 1.4 eV become part of the electron’s kinetic energy

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CHAPTER 25 Nuclear Physics ProblemSets


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www.ck12.org Chapter 25. Nuclear Physics Problem Sets



1. True.2. Alpha rays – positive

Beta rays – negativeGamma rays – neutral

3. There are 6 protons.4. The half-life of an element is the time it takes for half of the original quantity to decay.5. 1

4 of the original amount remains.6. The isotope becomes a different element.7. Assumed the same starting amounts, the 0.1 s substance because it decays faster, releasing more radiation in a

shorter amount of time.8. 56

26Fe→5627 Co+0

−1 e9. 64

29Cu→6027 CO+4

2 He10. Gamma rays.11. Nucleus.12. The atomic number is reduced by 2.13. No14. Gamma radiation gives off electromagnetic energy while alpha and beta radiation give off particles.15. Mendelevium.16. 162

65 T b17. The new atomic number is 23.18. 1

419. 9 years.20. There is 1

32 of a gram remaining.21.

Eb =4mc2

→4m = (M3He +Mn)−M4He = (3.016 u+1.009 u)−4.003 u = 0.022 u

⇒ Eb =4mc2 = (0.022 u)c2(

931.5 MeV/c2

1 u

)= 20.5 MeV

22. One equation is: N =(1

2

)nNo

Another equation using the number e is: N = Noe(−0.693

Tot)

23. First, determine the number of half-lives:13.5 days

2.36 days/half-life = 5.72 half-lives

Now solve using: N =(1

2

)nNo(1

2

)5.72 half-lives×5.23 mCi = 0.0992 mCi = 99.2 µCi

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24.

First, find the number of half-lives:

N = No

(12

)n

NNo

=

(12

)n

ln[(

12

)n]= ln

[NNo

]n ln

[(12

)]= ln

[NNo

]

n =ln[

NNo

]ln[(1

2

)] = ln[5.2 mCi

6.0 mCi

]ln[(1

2

)] = 0.206 half-lives

Then, convert into number of days:

0.206 half-lives(

5.0 days1 half-life

)= 1.03 days

25.

E = NEnuc

N =6.02×1023 nuc/mol

235 g/mol(1.2 g) = 3.07×1021 nuclei

E = (3.07×1021 nuclei)(200×106 eV )

(1.6×10−19 J

1 eV

)(1 h

3600 s

)(1 kW

1000 J/s

)= 2.73×104 kW ·h

26. The two atoms have the same number of neutrons.27. They all have the same number of protons.28. In one year, there are 36.5 half-lives. Therefore, the activity measures

(12

)36.5(1.00 µCi) = 1.03×10−11 µCi.

29. If we find the half-life of the sample, we can determine the isotope.

N = No

(12

)n

NNo

=

(12

)n

ln[(

12

)n]= ln

[NNo

]n ln

[(12

)]= ln

[NNo

]

n =ln[

NNo

]ln[(1

2

)] = ln[

2.64 g2.65 g

]ln[(1

2

)] = 0.00545 half-lives

31 years0.00545 half-lives

≈ 5700 years

Since the half-life of the sample is around 5700 years, it must be carbon-14.30. 14

6 C→147 N + e−+ v

31. 127 N→12

6 C+ e++ v32. Due to electrostatic repulsion, the amount of kinetic energy required to have an alpha particle make contact

with the nucleus of an atom is extremely large.

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33. Yes.34. Fusion is the process in which lighter nuclei form a heavier nucleus, thereby releasing energy.35. Eb

A = [146(1.0087) u+92(1.0078) u−238.05 u]/235 = 7.42 MeV/nucleon36. A

ZX →A−4Z−2 Y +4

2 He37. They have the same number of nucleons.38. Electromagnetic.39. Higgs boson.40. They could produce 2 gamma rays.41. Yes.42. The electroweak force is composed of electromagnetism and the nuclear weak force.43. The strong force dominates.44. Nucleons are made of quarks. Yes, quarks have a fractional charge.45. The energy and photons generated by fusion resist the compressional force of gravity.46. The energy released by decay(E=∆mc2) heats the plutonium, which makes it warm to the touch.47. The most stable element is iron-56. Since iron-56 is the most stable element, no energy can be released by

iron undergoing fusion. Therefore, the star will implode because there is not enough energy to resist gravity.48. Ingesting the alpha radiation emitting substance is more dangerous. Because alpha radiation is easily absorbed

by matter, the vast majority of the radiation will be absorbed by the body.49. There is a 50% chance that the atom has not decayed, and a 50% chance that the atom has decayed.50. Neutrons are uncharged, which allows them to more easily approach the nucleus. This lowers the energies

needed to accomplish fission.

Not Covered/Beyond Scope:

Questions:

1. True or False: All baryons and anti-baryons consist of 3 quarks?2. Is the following reaction possible? p+n→ p+ p+n+ p3. What is the sum of the baryon number for the following particles? p+ p+ p4. Can the following reaction occur? µ−→ e−+ ve + vµ

5. Show that the following decay process is not possible: π+→ µ++ vµ + ve

6. Show that the following reaction is possible: π+n→ K++∑−

7. Determine whether the following reaction is possible: π−+ p→ π−+∑+

8. If a star has a velocity of 0.25 c in a direction away from Earth, how far away is it?9. What type of neutrinos are produced when electrons and muons collide?

10. When a neutral pion decays into two photons, calculate the energy of each photon. (Assume the pion is atrest)

11. Determine the frequency of the photon that was created in the previous problem.12. Using the conservation laws, determine the missing information: p+?→ n+µ+

13. Give a possible decay path for an anti-omega particle.14. Show one possible decay path for an anti-neutron.15. Determine whether the following reaction is possible. If it is not possible, which conservation law is vio-

lated? p+ p→ p+π+

16. Is the following reaction possible? If not, which conservation law is broken? p+π−→ p+π+

17. Is strangeness conserved in the following reaction? p+π+→ π++∑+

18. Using the conservation laws, what type of neutrino is missing from the reaction below? n+?→ e−+ p.19. What type of neutrinos is missing from the following decay process?µ−→ e−+?+?.20. One of the following reactions can occur, identify which one it is and explain why the other reaction is not

possible.

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(a) p→ π++π

(b) π+→ µ++ vµ

21. Determine if the strangeness is conserved in the following reaction given: Λ→ π−+ p.22. Is strangeness conserved in the decay process given? Ξ→ π−+Λ

23. What is conserved in the following process e−+ e+→ µ−+µ+?24. Determine the radius of an alpha particle.25. What is the radius of a Uranium-235 nucleus?26. What is the Larmor frequency for neutrons in a B-field of magnitude 0.5 T?27. Determine the Larmor frequency for a proton in the magnetic field given in the previous problem.28. Determine the binding energy per nucleon for 2H.29. What is the name of the particle that interacts via the strong force? What are the different types?30. Mesons are known to decay into what four particles?31. Which of the Hadron particles have an odd half-integer spin?32. What is the law of conservation for Baryon numbers?33. What is the law of conservation of electron-lepton number?34. True or False: The law of conservation of meson number is similar to the law of conservation of Baryon

number?35. Does a meson or a baryon consist of three quarks?36. Does a meson or an antibaryon consist of one quark and one anti-quark?37. List the 6 quarks that are have been currently classified.38. What three quarks have a charge of 2/3 e?39. Describe the difference between hadrons and leptons.40. What is the baryon number of kaons?41. Derive an expression for the mass of a nucleus of mass number A.42. Calculate the distance of closest approach when an alpha particle with K=7.0 MeV is fired at a gold nucleus.43. If the distance of closet approach between an alpha particle and a gold nucleus is 98.5 fm, determine the

kinetic energy of the alpha particle.44. Calculate the Q value when 226Ra nucleus emits an alpha particle.45. What is the field particle that is associated with the nuclear force?46. A radioactive sample that has a half life of 30 seconds has a counting rate of 1000 counts/s. Determine the

new rate at t = 2 minutes.47. A radioactive sample with a half-life of 30.0 seconds has an initial counting rate of 1000 counts/s. If the

detector used has an efficiency of 10%, calculate the number of radioactive nuclei at t0.48. Calculate the number of nuclei that are remaining from the previous problem after 1 minute.49. How many different types of quarks are there and what are they named?50. If the half-life of radium is 1.6 * 103 yr, calculate the decay constant.51. A radioactive sample containing 2.5 * 1010 nuclei at t=0, determine the activity of the sample if its decay

constant is 1.2 * 10−11 s−1.52. A radioactive sample with a half-life of 5.0 days has an initial activity of 6.0 mCi. Several days later, the

sample has an activity of 5.2 mCi. Determine how much time has elapsed.53. Define an isobar.54. What is the decay constant for a radioactive isotope that has an initial activity of 12.0 mCi, but after 2.0 hours

it has an activity of 9.0 mCi.55. A radioactive sample with an initial activity is measured to be 15 mCi. Determine the activity 24 hours later

if the activity is 12.5 mCi 4 hours after the initial measurement.56. Determine the decay constant of a sample that has 0.5 x 1015 atoms and an activity of 4.4 x 1011 Bq.57. Determine the half life of the sample in the previous problem.58. The initial activity for a sample is measured to be 2.2 x 1013 Bq. If the number of atoms in the sample when

the activity is measured is 0.2 x 1016, determine the half-life.

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Solutions

1. True.2. Using the conservation law for baryon number, yes, the reaction can occur.3. 1+1+(−1) = 14. Using the conservation of electron-lepton numbers, the decay reaction listed is possible.5.

LHS = RHS

0 =−1+1+1

0 6= 1

6. Using the conservation of strangeness:

LHS = RHS

0+0 = 1+−1

0 = 0

7. Using the conservation of strangeness:

LHS = RHS

0+0 = 0−1

0 6=−1

8. Using Hubble’s Law,9. Since after the collision, the muon-lepton number is -1 and the electron-lepton number is 1, the resulting

neutrinos must be ve and vµ.10. E = mc2 = (135 MeV/c2)c2 Since there are two photons E

2 = 67.5 MeV11. f = Eγ

h = 1.6×1022 Hz12. By looking at the charge on the RHS, the unknown charge must be neutral. Using the conservation laws, the

particle must be a vµ.13. Ω+→ Λo +K+

14. n→ p+ e++ ve

15. No, the reaction is not possible. The reaction violates the conservation of baryon number.16. No, it isn’t possible. Charge conservation is violated.17. No, strangeness is not conserved in the following reaction.18. Using the lepton numbers, the missing particle is electron-neutrino.19. Using the conservation laws, both electron and muon neutrinos are needed to complete the given decay

process.20. Reaction (a) is not possible due to it violating the baryon conservation law. Reaction (b) is possible.21. −1→ 0+0, strangeness is not conserved22. Ξ−→ π−+Λo−2→ 0+(−1), strangeness is not conserved.23. Baryon number, Strangeness, Lepton number and charge.24. r = roA

13 = (1.20×10−15 m)(4)

13 = 1.90×10−15 m

25. r = roA13 = (1.20×10−15 m)(235)

13 = 7.4×10−15 m

26. f = 2µBh = 2(1.91)(5.05×10−27 J/T )(0.5 T )

6.6×10−34 J·s = 1.47×107 Hz

27. f = 2µBh = 2(2.79)(5.05×10−27 J/T )(0.5 T )

6.6×10−34 J·s = 2.13×107 Hz

28. EbA = [−2.014+1(1.0087)+1(1.0078)

2 u](931.5 MeV

u

)= 1.16 MeV/nucleon

29. The particles are called hadrons and they are either mesons or baryons.30. Mesons are known to decay into electrons, positrons, neutrinos and photons.

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31. Baryons.32. The sum of the baryon numbers of the system must be equal before and after a process.33. The sum of electron-leptons numbers before an event must equal the sum of the electron-lepton numbers after

the event.34. False, there is no law of conservation of meson number.35. Baryon.36. Meson.37. Up, down, charmed, bottom, top, strange.38. Up, charmed and top.39. Hadrons are particles with structure and size while leptons are light particles with no structure or size. The

leptons interact via the weak force while the hadrons interact via the strong force.40. The baryon number is 0.41. Since the protons and neutrons are approximately the same mass, the answer is Am.42. dmin =

2kZe2

Eα= 2(8.99×109)(79)(1.60×10−19 C)2

7.0(1.60×10−13)= 32.5 f m

43. Eα = 2kZe2

dmin= 2(8.99×109)(79)(1.60×10−19 C)2

98.5×10−15 m(1.6×10−19)= 2.3 MeV

44. Q = (226.025 u−222.018 u−4.00260 u)(931.5 MeV/u) = 4.10 MeV45. Gluon.46. The decay rate can be written as

R = (12)

nRo

Where n is the number of half lives→ In 2 minutes, there are 4 half livesR =

(12

)4(1000 counts/s) = 62.5 counts/s

47. The decay rate can be determined from the counting rate:

Ro = 10(1000 counts/s) = 104 s−1

No =Ro

λ= Ro

( t 12

0.693

)= 4.33×106

48. N =(1

2

)2(4.33×106) = 1.08×106

49. There are 6 types of quarks: up, down, top, bottom, strange, and charm.50.

t 12= (1.6×103 yr)

(3.16×107 s

1 yr

)= 5.1×1010 s

λ =0.693

5.1×1010 s= 1.36×10−11 s−1

51. Ro = λNo = (1.2×10−11 s−1)(2.5×1010) = 0.3(

1 Ci3.7×1010

)= 8.1×10−12 Ci

52.

RRo

= e−λt

→ t =−1λ

ln(

RRo

)λ =

0.693t 1

2

=0.693

5.0 days= 0.139 d−1

⇒ t =− 1(0.139 d−1)

ln(

5.26.0

)= 1.03 days

53. Isobars are atoms that have the same number of nucleons, which differ in atomic number but not in the massnumber.

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54. λ = 1t ln

(RoR

)= 1

2.0 hr ln(12.0 mCi

9.0 mCi

)= 0.14 hr−1

55.

λ =1t

ln(

Ro

R

)=

14 hr

ln(

15.012.5

)= 0.0456 hr−1

→ R = Roe−λt = (15.0 mCi) e−(0.0456)(24.0) = 5.02 mCi

56. λ = 1N

(−dN

dt

)= 8.8×10−4

57. T12= ln2

λ= ln (2)(8.8×104) = 1017 min

58.

λ =1N

(−dN

dt

)= (5.0×10−16)(2.2×1013) = 0.011

T12=

ln2λ

= 1.05 min

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CK-12 Physics Intermediated3r1z7wkgqhj9d.cloudfront.net/Physics Rhetoric Resources... · 2018-06-01 · 1. What Is Science? Problem Sets Professor A in the previous problem. 9.True