1.1 CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model-Building 1.3 Density and Atomic Mass 1.4 Dimensional Analysis 1.5 Conversion of Units 1.6 Estimates and Order-of- Magnitude Calculations 1.7 Significant Figures Physics and Measurement ANSWERS TO QUESTIONS Q1.1 Atomic clocks are based on electromagnetic waves which atoms emit. Also, pulsars are highly regular astronomical clocks. Q1.2 Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard. Q1.3 People have different size hands. Defining the unit precisely would be cumbersome. Q1.4 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms Q1.5 (b) and (d). You cannot add or subtract quantities of different dimension. Q1.6 A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. If an equation is not dimensionally correct, it cannot be correct. Q1.7 If I were a runner, I might walk or run 101 miles per day. Since I am a college professor, I walk about 100 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on vacation. Q1.8 On February 7, 2001, I am 55 years and 39 days old. 55 365 25 1 39 20 128 86 400 1 1 74 10 109 9 yr d yr d d s d s s . . ~ F HG I KJ+ = F HG I KJ = . Many college students are just approaching 1 Gs. Q1.9 Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10. Q1.10 The mass of the forty-six chapter textbook is on the order of 100 kg . Q1.11 With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr. 1

2. 2 Physics and Measurement SOLUTIONS TO PROBLEMS Section 1.1 Standards of Length, Mass, and Time No problems in this section Section 1.2 Matter and Model-Building P1.1 From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, L L Ldiag = +2 2 . Thus, since the atoms are separated by a distance L = 0 200. nm , the diagonal planes are separated by 1 2 0 1412 2 L L+ = . nm . Section 1.3 Density and Atomic Mass *P1.2 Modeling the Earth as a sphere, we find its volume as 4 3 4 3 6 37 10 1 08 103 6 3 21 3 r = = . .m me j . Its density is then = = = m V 5 98 10 1 08 10 5 52 10 24 21 3 3 3. . . kg m kg m . This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 3 kg m . The average density of the Earth is significantly higher, so higher-density material must be down below the surface. P1.3 With V = base area heighta fb g V r h= 2 e j and = m V , we have = = F HG I KJ = m r h2 2 9 4 3 1 19 5 39 0 10 1 2 15 10 kg mm mm mm m kg m 3 3 . . . . a f a f *P1.4 Let V represent the volume of the model, the same in = m V for both. Then iron kg= 9 35. V and gold gold = m V . Next, gold iron gold kg = m 9 35. and mgold 3 3 3 kg 19.3 10 kg / m kg / m kg= F HG I KJ =9 35 7 86 10 23 03 . . . . P1.5 V V V r ro i= = 4 3 2 3 1 3 e j = m V , so m V r r r r = = F HG I KJ = 4 3 4 3 2 3 1 3 2 3 1 3 e j e j . 3. Chapter 1 3 P1.6 For either sphere the volume is V r= 4 3 3 and the mass is m V r= = 4 3 3 . We divide this equation for the larger sphere by the same equation for the smaller: m m r r r rs s s = = = 4 3 4 3 5 3 3 3 3 . Then r rs= = =5 4 50 1 71 7 693 . . .cm cma f . P1.7 Use 1 u . g= 1 66 10 24 . (a) For He, m0 24 4 00 6 64 10= F HG I KJ = . .u 1.66 10 g 1 u g -24 . (b) For Fe, m0 23 55 9 9 29 10= F HG I KJ = . .u 1.66 10 g 1 u g -24 . (c) For Pb, m0 24 22 207 1 66 10 3 44 10= F HG I KJ = u g 1 u g . . . *P1.8 (a) The mass of any sample is the number of atoms in the sample times the mass m0 of one atom: m Nm= 0. The first assertion is that the mass of one aluminum atom is m0 27 26 27 0 27 0 1 66 10 1 4 48 10= = = . . . .u u kg u kg . Then the mass of 6 02 1023 . atoms is m Nm= = = = 0 23 26 6 02 10 4 48 10 0 027 0 27 0. . . .kg kg g . Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be written m Nm= 0. 0 027 0 6 02 1023 0. .kg = m , so m0 23 260 027 6 02 10 4 48 10= = . . . kg kg , in agreement with the first assertion. (b) The general equation m Nm= 0 applied to one mole of any substance gives M NMg u= , where M is the numerical value of the atomic mass. It divides out exactly for all substances, giving 1 000 000 0 10 1 660 540 2 103 27 . . = kg kgN . With eight-digit data, we can be quite sure of the result to seven digits. For one mole the number of atoms is N = F HG I KJ = +1 1 660 540 2 10 6 022 137 103 27 23 . . . (c) The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one molecule of H O2 is 2 1 008 0 15 999 18 0. . .b g+ =u u. Then the molar mass is 18 0. g . (d) For CO2 we have 12 011 2 15 999 44 0. . .g g g+ =b g as the mass of one mole. 4. 4 Physics and Measurement P1.9 Mass of gold abraded: m = = = F HG I KJ = 3 80 3 35 0 45 0 45 1 4 5 10 4 . . . . .g g g g kg 10 g kg3b g . Each atom has mass m0 27 25 197 197 1 66 10 1 3 27 10= = F HG I KJ = u u kg u kg . . . Now, m N m= 0 , and the number of atoms missing is N m m = = = 0 4 25 214 5 10 3 27 10 1 38 10 . . . kg kg atoms. The rate of loss is N t N t = F HG I KJF HG I KJF HG I KJF HG I KJ = 1 38 10 50 1 1 1 1 8 72 10 21 11 . . . atoms yr yr 365.25 d d 24 h h 60 min min 60 s atoms s P1.10 (a) m L= = = = 3 3 6 3 16 19 7 86 5 00 10 9 83 10 9 83 10. g cm cm g kge je j. . . (b) N m m = = = 0 19 27 79 83 10 55 9 1 66 10 1 06 10 . . . . kg u kg 1 u atoms e j P1.11 (a) The cross-sectional area is A = + = 2 0 150 0 010 0 340 0 010 6 40 10 3 . . . . . . m m m m m2 a fa f a fa f. The volume of the beam is V AL= = = 6 40 10 1 50 9 60 103 3 . . .m m m2 3 e ja f . Thus, its mass is m V= = = 7 56 10 9 60 10 72 63 3 . . .kg / m m kg3 3 e je j . FIG. P1.11 (b) The mass of one typical atom is m0 27 26 55 9 1 66 10 1 9 28 10= F HG I KJ = . . .u kg u kga f . Now m Nm= 0 and the number of atoms is N m m = = = 0 26 2672 6 9 28 10 7 82 10 . . . kg kg atoms . 5. Chapter 1 5 P1.12 (a) The mass of one molecule is m0 27 26 18 0 1 66 10 2 99 10= F HG I KJ = . . .u kg 1 u kg . The number of molecules in the pail is N m m pail kg 2.99 kg molecules= = = 0 26 251 20 10 4 02 10 . . . (b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere. N N m M both pail pail total 25 (4.02 10 molecules) kg kg = F HG I KJ = F HG I KJ 1 20 1 32 1021 . . , or Nboth molecules= 3 65 104 . . Section 1.4 Dimensional Analysis P1.13 The term x has dimensions of L, a has dimensions of LT2 , and t has dimensions of T. Therefore, the equation x ka tm n = has dimensions of L LT T= 2 e j a fm n or L T L T1 0 2 = m n m . The powers of L and T must be the same on each side of the equation. Therefore, L L1 = m and m = 1 . Likewise, equating terms in T, we see that n m 2 must equal 0. Thus, n = 2 . The value of k, a dimensionless constant, cannot be obtained by dimensional analysis . *P1.14 (a) Circumference has dimensions of L. (b) Volume has dimensions of L3 . (c) Area has dimensions of L2 . Expression (i) has dimension L L L2 1 2 2 e j / = , so this must be area (c). Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L L L2 3 e j= , so it is (b). Thus, (a) ii; (b) iii, (c) i= = = . 6. 6 Physics and Measurement P1.15 (a) This is incorrect since the units of ax are m s2 2 , while the units of v are m s. (b) This is correct since the units of y are m, and cos kxa f is dimensionless if k is in m1 . *P1.16 (a) a F m or a k F m = represents the proportionality of acceleration to resultant force and the inverse proportionality of acceleration to mass. If k has no dimensions, we have a k F m = , L T 1 F M2 = , F M L T2 = . (b) In units, M L T kg m s2 2 = , so 1 1newton kg m s2 = . P1.17 Inserting the proper units for everything except G, kg m s kg m 2 L NM O QP= G 2 2 . Multiply both sides by m 2 and divide by kg 2 ; the units of G are m kg s 3 2 . Section 1.5 Conversion of Units *P1.18 Each of the four walls has area 8 00 12 0 96 0. . .ft ft ft2 a fa f= . Together, they have area 4 96 0 1 3 28 35 72 2 . .ft m . ft m2 e jF HG I KJ = . P1.19 Apply the following conversion factors: 1 2 54in cm= . , 1 86 400d s= , 100 1cm m= , and 10 19 nm m= 1 32 2 54 10 10 9 19 2 9 in day cm in m cm nm m 86 400 s day nm s F HG I KJ = . . b ge je j . This means the proteins are assembled at a rate of many layers of atoms each second! *P1.20 8 50 8 50 0 025 4 1 39 10 3 4 . . . .in in m 1 in m3 3 3 = F HG I KJ = 7. Chapter 1 7 P1.21 Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A m m m2 =30 50 1 500a fa f . Categorize: We model the lot as a perfect rectangle to use Area = Length Width. Use the conversion: 1 m 3.281 ft= . Analyze: A LW= = F HG I KJ F HG I KJ = 100 1 3 281 150 1 3 281 1 39 103 ft m ft ft m ft = 1 390 m m2 2 a f a f. . . . Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m2 . Unit conversion is a common technique that is applied to many problems. P1.22 (a) V = = 40.0 m 20.0 m 12.0 m . m3 a fa fa f 9 60 103 V = = 9 60 10 3 39 103 5 3 . m 3.28 ft 1 m ft3 3 b g . (b) The mass of the air is m V= = = air 3 3 kg m 9.60 10 m . kg1 20 1 15 103 4 .e je j . The student must look up weight in the index to find F mgg = = = 1.15 10 kg 9.80 m s 1.13 10 N4 2 5 e je j . Converting to pounds, Fg = = 1 13 10 2 54 105 4 . N 1 lb 4.45 N lbe jb g . . P1.23 (a) Seven minutes is 420 seconds, so the rate is r = = 30 0 420 7 14 10 2. . gal s gal s . (b) Converting gallons first to liters, then to m3 , r r = F HG I KJ F HG I KJ = 7 14 10 3 786 10 2 70 10 2 3 4 . . . . gal s L 1 gal m 1 L m s 3 3 e j (c) At that rate, to fill a 1-m3 tank would take t = F HG I KJ F HG I KJ = 1 2 70 10 1 1 034 m m s h 3 600 h 3 3 . . . 8. 8 Physics and Measurement *P1.24 (a) Length of Mammoth Cave = F HG I KJ = = = 348 1 609 1 560 5 60 10 5 60 105 7 mi km mi km m cm . . . . (b) Height of Ribbon Falls = F HG I KJ = = = 1 612 0 1 491 m 0 491 4 91 104 ft .304 8 m ft km cm. . . (c) Height of Denali = F HG I KJ = = = 20 320 0 1 6 6 19 10 6 19 103 5 ft .304 8 m ft .19 km m cm. . . (d) Depth of Kings Canyon = F HG I KJ = = = 8 200 0 1 2 2 50 10 2 50 103 5 ft .304 8 m ft .50 km m cm. . . P1.25 From Table 1.5, the density of lead is 1 13 104 . kg m3 , so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks. Density is defined as mass per volume, in = m V . We must convert to SI units in the calculation. = F HG I KJF HG I KJ = 23 94 2 10 1 1 000 100 1 1 14 103 3 4. . g cm kg g cm m . kg m3 At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important common-sense check on density values is that objects which sink in water must have a density greater than 1 g cm3 , and objects that float must be less dense than water. P1.26 It is often useful to remember that the 1 600-m race at track and field events is approximately 1 mile in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to 1 1 640 1 609 4 05 10 2 3 acre mi acres m mi m 2 2 a fF HG I KJF HG I KJ = . . *P1.27 The weight flow rate is 1 200 2 000 1 1 667 ton h lb ton h 60 min min 60 s lb s F HG I KJF HG I KJF HG I KJ = . P1.28 1 1 609 1 609mi m km= = . ; thus, to go from mph to km h, multiply by 1.609. (a) 1 1 609mi h km h= . (b) 55 88 5mi h km h= . (c) 65 104 6mi h km h= . . Thus, v = 16 1. km h . 9. Chapter 1 9 P1.29 (a) 6 10 1 1 1 190 12 F HG I KJ F HG I KJF HG I KJF HG I KJ = $ 1 000 $ s h 3 600 s day 24 h yr 365 days years (b) The circumference of the Earth at the equator is 2 6 378 10 4 01 103 7 . . = m me j . The length of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9 30 1011 . m. Thus, the 6 trillion dollars would encircle the Earth 9 30 10 2 32 10 11 4. . = m 4.01 0 m times7 . P1.30 N m m atoms Sun atom kg 1.67 kg atoms= = = 1 99 10 10 1 19 10 30 27 57. . P1.31 V At= so t V A = = = 3 78 10 25 0 1 51 10 151 3 4. . . m m m or m 3 2 b g P1.32 V Bh= = = 1 3 13 0 43 560 3 481 9 08 107 . . , acres ft acre ft ft 2 3 a fe j a f or V = F HG I KJ = 9 08 10 2 83 10 1 2 57 10 7 2 6 . . . ft m ft m 3 3 3 3 e j B h B h FIG. P1.32 P1.33 Fg = = 2 50 2 00 10 2 000 1 00 106 10 . . .tons block blocks lb ton lbsb ge jb g *P1.34 The area covered by water is A A Rw = = = = 0 70 3 6 102 14 . 0.70 4 0.70 4 6.37 10 m . mEarth Earth 6 2 2 a fe j a fa fe j . The average depth of the water is d = = 2.3 miles 1 609 m l mile . ma fb g 3 7 103 . The volume of the water is V A dw= = = 3 6 10 3 7 10 1 3 1014 2 3 18 3 . m . m . me je j and the mass is m V= = = 1 000 1 3 10 1 3 103 18 3 21 kg m . m kge je j . . 10. 10 Physics and Measurement P1.35 (a) d d d d nucleus, scale nucleus, real atom, scale atom, real m ft 1.06 10 m ft= F HG I KJ = F HG I KJ = 2 40 10 300 6 79 1015 10 3 . .e j , or dnucleus, scale ft mm 1 ft mm= = 6 79 10 304 8 2 073 . . .e jb g (b) V V r r d d r r atom nucleus atom nucleus atom nucleus atom nucleus m m times as large = = F HG I KJ = F HG I KJ = F HG I KJ = 4 3 4 3 3 3 10 15 3 13 3 3 1 06 10 2 40 10 8 62 10 . . . *P1.36 scale distance between real distance scale factor km m m km= F HG I KJF HG I KJ = F HG I KJ = 4 0 10 7 0 10 1 4 10 20013 3 9 . . . e j P1.37 The scale factor used in the dinner plate model is S = = 0 25 1 0 10 2 105 6. . m lightyears .5 m lightyears . The distance to Andromeda in the scale model will be D D Sscale actual 6 6 2.0 10 lightyears 2.5 10 m lightyears m= = = e je j 5 0. . P1.38 (a) A A r r r r Earth Moon Earth Moon 2 Earth Moon m cm m cm = = F HG I KJ = F H GG I K JJ = 4 4 6 37 10 100 1 74 10 13 4 2 2 6 8 2 . . . e jb g (b) V V r r r r Earth Moon Earth Moon 3 3 Earth Moon m cm m cm = = F HG I KJ = F H GG I K JJ = 4 3 4 3 6 8 3 3 6 37 10 100 1 74 10 49 1 . . . e jb g P1.39 To balance, m mFe Al= or Fe Fe Al AlV V= Fe Fe Al Al Al Fe Fe Al cm cm 4 3 4 3 2 00 7 86 2 70 2 86 3 3 1 3 1 3 F HG I KJ = F HG I KJ = F HG I KJ = F HG I KJ = r r r r / / . . . . .a f 11. Chapter 1 11 P1.40 The mass of each sphere is m V r Al Al Al Al Al = = 4 3 3 and m V r Fe Fe Fe Fe Fe = = 4 3 3 . Setting these masses equal, 4 3 4 3 3 3 Al Al Fe Fer r = and r rAl Fe Fe Al = 3 . Section 1.6 Estimates and Order-of-Magnitude Calculations P1.41 Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball as a sphere of diameter 0.038 m. The volume of the room is 4 4 3 48 = m3 , while the volume of one ball is 4 3 0 038 2 87 10 3 5 . . m 2 m3F HG I KJ = . Therefore, one can fit about 48 2 87 10 105 6 . ~ ping-pong balls in the room. As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called best packing fraction is 1 6 2 0 74 = . so that at least 26% of the space will be empty. Therefore, the above estimate reduces to 1 67 10 0 740 106 6 . . ~ . P1.42 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus, the tire would make 50 000 5 280 1 3 107 mi ft mi rev 8 ft rev~ 10 rev7 b gb gb g= . P1.43 In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least 1 16 in 43 10 ft2 5 2 = . Since 1 acre 43 560 ft2 = , the number of blades of grass to be expected on a quarter-acre plot of land is about n = = = total area area per blade acre ft acre ft blade 2.5 10 blades blades 2 2 7 0 25 43 560 43 10 105 7 . ~ a fe j . 12. 12 Physics and Measurement P1.44 A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then approximately 4 10 3 in3 . Since 1 acre 43 560 ft2 = , the volume of water required to cover it to a depth of 1 inch is 1 acre 1 inch 1 acre in ft 1 acre in ft 6.3 10 in6 3 a fa f a f= F HG I KJ F HG I KJ 43 560 144 1 2 2 2 . The number of raindrops required is n = = = volume of water required volume of a single drop in in . 6 3 10 4 10 1 6 10 10 6 3 3 3 9 9. ~ . *P1.45 Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then V = =0 5 1 3 0 5 0 3 0 10. . . . .a fa fa fa fm m m m3 . The mass of this volume of water is m Vwater water 3 3 kg m m kg kg= = = 1 000 0 10 100 102 e je j. ~ . Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is m Vcopper copper 3 3 kg m m kg kg= = = 8 920 0 10 892 103 e je j. ~ . P1.46 The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are ~250 million people, and 365 days in a year, so 250 10 365 106 11 cans day days year canse jb g are thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, we estimate this represents 10 0 1 1 1 3 1 1011 5 cans oz can lb 16 oz ton 2 000 lb tons yeare jb gb gb g. . . ~105 tons P1.47 Assume: Total population = 107 ; one out of every 100 people has a piano; one tuner can serve about 1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore, # tuners ~ 1 tuner 1 000 pianos 1 piano 100 people people F HG I KJ F HG I KJ =( )10 1007 . 13. Chapter 1 13 Section 1.7 Significant Figures *P1.48 METHOD ONE We treat the best value with its uncertainty as a binomial 21 3 0 2 9 8 0 1. . . . a f a fcm cm , A = 21 3 9 8 21 3 0 1 0 2 9 8 0 2 0 1. . . . . . . .a f a f a f a fa f cm2 . The first term gives the best value of the area. The cross terms add together to give the uncertainty and the fourth term is negligible. A = 209 42 2 cm cm . METHOD TWO We add the fractional uncertainties in the data. A = + F HG I KJ = = 21 3 9 8 0 2 21 3 0 1 9 8 209 2% 209 42 2 2 . . . . . . cm cm cm cm cma fa f P1.49 (a) r2 2 2 2 2 2 10 5 0 2 10 5 2 10 5 0 2 0 2 346 13 = = + = . m . m m m m m m m a f ( . ) ( . )( . ) ( . ) (b) 2 2 10 5 0 2 66 0 1 3 r = = . m . m m ma f . . P1.50 (a) 3 (b) 4 (c) 3 (d) 2 P1.51 r m m r = = = = 6 50 0 20 6 50 0 20 10 1 85 0 02 2 4 3 3 . . cm . . m . . kg a f a f a f c h also, = + m m r r 3 . In other words, the percentages of uncertainty are cumulative. Therefore, = + = 0 02 1 85 3 0 20 6 50 0 103 . . . . . a f , = = 1 85 6 5 10 1 61 10 4 3 2 3 3 3. . . c h e jm kg m and = = 1 61 0 17 10 1 6 0 2 103 3 . . . .a f a fkg m kg m3 3 . 14. 14 Physics and Measurement P1.52 (a) 756.?? 37.2? 0.83 + 2.5? 796./5/3 = 797 (b) 0 003 2 356 3 1 140 16 1 1. 2 s.f. . 4 s.f. . 2 s.f.a f a f a f = = . (c) 5.620 4 s.f. >4 s.f. 17.656= 4 s.f. 17.66a f a f a f = *P1.53 We work to nine significant digits: 1 1 365 242 199 24 60 60 31 556 926 0yr yr d 1 yr h 1 d min 1 h s 1 min s= F HG I KJF HG I KJF HG I KJF HG I KJ = . . . P1.54 The distance around is 38.44 m 19.5 m 38.44 m 19.5 m 115.88 m+ + + = , but this answer must be rounded to 115.9 m because the distance 19.5 m carries information to only one place past the decimal. 115 9. m P1.55 V V V V V V V V = + = + = + + = = = = + = 2 2 2 17 0 1 0 1 0 1 0 0 09 1 70 10 0 1 0 0 090 0 900 2 1 70 0 900 5 2 1 2 1 2 1 2 3 b g a fa fa f a fa fa f e j . . . . . . . . . . . . . m m m m m m m m m m m m m 3 3 3 3 1 1 1 1 1 1 0 12 0 0063 0 01 0 010 0 1 0 011 0 006 0 010 0 011 0 027 3% = = = = = = U V ||| W ||| = + + = = . . . . . . . . . . m 19.0 m m 1.0 m cm 9.0 cm w w t t V V FIG. P1.55 Additional Problems P1.56 It is desired to find the distance x such that x x100 1 000 m m = (i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that x2 5 100 1 000 1 00 10= = m m m2 a fb g . and therefore x = =1 00 10 3165 . m m2 . 15. Chapter 1 15 *P1.57 Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has mass m0 27 25 197 3 27 10= F HG I KJ = u 1.66 10 kg 1 u kga f . . So, the number of atoms in the cube is N = = 19 300 5 90 1025 28kg 3.27 10 kg . . The imagined cubical volume of each atom is d3 28 291 5 90 10 1 69 10= = m m 3 3 . . . So d = 2 57 10 10 . m . P1.58 A N A V V A V r rtotal drop total drop drop total 4 = = F HG I KJ = F H GG I K JJa fe j e j e j 3 3 2 4 A V r total total 3 2m m m= F HG I KJ = F HG I KJ = 3 3 30 0 10 2 00 10 4 50 6 5 . . . P1.59 One month is 1 30 24 3 600 2 592 106 mo day h day s h s= = b gb gb g . . Applying units to the equation, V t t= +1 50 0 008 00 2 . .Mft mo Mft mo3 3 2 e j e j . Since 1 106 Mft ft3 3 = , V t t= + 1 50 10 0 008 00 106 6 2 . .ft mo ft mo3 3 2 e j e j . Converting months to seconds, V t t= + 1 50 10 0 008 00 106 6 2 2. .ft mo 2.592 10 s mo ft mo 2.592 10 s mo 3 6 3 2 6 e j . Thus, V t tft ft s ft s3 3 3 2 [ ] . .= + 0 579 1 19 10 9 2 e j e j . 16. 16 Physics and Measurement P1.60 (deg) (rad) tan a f sin a f difference 15.0 0.262 0.268 0.259 3.47% 20.0 0.349 0.364 0.342 6.43% 25.0 0.436 0.466 0.423 10.2% 24.0 0.419 0.445 0.407 9.34% 24.4 0.426 0.454 0.413 9.81% 24.5 0.428 0.456 0.415 9.87% 24.6 0.429 0.458 0.416 9.98% 24.6 24.7 0.431 0.460 0.418 10.1% P1.61 2 15 0 2 39 2 39 55 0 3 41 r r h r h = = = = = . m . m tan 55.0 . m ( . ) ma ftan . 5555 h rr h FIG. P1.61 *P1.62 Let d represent the diameter of the coin and h its thickness. The mass of the gold is m V At d dh t= = = + F HG I KJ 2 4 2 where t is the thickness of the plating. m = + L N MM O Q PP = = = = 19 3 2 2 41 4 2 41 0 178 0 18 10 0 003 64 0 003 64 036 4 3 64 2 4 . . . . . . . $10 $0. . a f a fa f e j grams cost grams gram cents This is negligible compared to $4.98. P1.63 The actual number of seconds in a year is 86 400 s day 365.25 day yr 31 557 600 s yrb gb g= . The percent error in the approximation is = 10 31 557 600 31 557 600 100% 0 449% 7 s yr s yr s yr e j b g . . 17. Chapter 1 17 P1.64 (a) V = L3 , A = L2 , h = L V A h= L L L L3 2 3 = = . Thus, the equation is dimensionally correct. (b) V R h R h Ahcylinder = = = 2 2 e j , where A R= 2 V wh w h Ahrectangular object = = =a f , where A w= P1.65 (a) The speed of rise may be found from v D = = = Vol rate of flow (Area: cm s cm s 3 cm a f a f 2 2 4 6 30 4 16 5 0 529 ) . . . . (b) Likewise, at a 1.35 cm diameter, v = = 16 5 11 5 1.35 4 2 . . cm s cm s 3 cma f . P1.66 (a) 1 cubic meter of water has a mass m V= = = 1 00 10 1 00 10 1 0003 3 3 2 3 . kg cm . m cm m kge je je j (b) As a rough calculation, we treat each item as if it were 100% water. cell: kg m m kg kidney: . kg cm cm kg fly: kg cm mm mm cm mm 3 3 3 2 m V R D m V R m D h = = F HG I KJ = F HG I KJ = F HG I KJ = = = F HG I KJ = F HG I KJ = = F HG I KJ = F HG I KJ 4 3 1 6 1 000 1 6 1 0 10 5 2 10 4 3 1 00 10 4 3 4 0 0 27 4 1 10 4 2 0 4 0 10 3 3 6 3 16 3 3 3 2 3 1 e j e j e j e j a f a fe . . ( . ) . . . j3 5 1 3 10= . kg P1.67 V20 mpg 10cars mi yr mi gal 5.0 10 gal yr= = ( )( )10 10 20 8 4 V25 mpg 10cars mi yr mi gal 4.0 10 gal yr= = ( )( )10 10 25 8 4 Fuel saved gal yr25 mpg 20 mpg= = V V 1 0 1010 . 18. 18 Physics and Measurement P1.68 v = F HG I KJ F HG I KJ F HG I KJ F HG I KJF HG I KJF HG I KJ = 5 00 220 1 0 914 4 1 1 14 1 24 1 3 600 8 32 10 4 . . . furlongs fortnight yd furlong m yd fortnight days day hrs hr s m s This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth. P1.69 The volume of the galaxy is r t2 21 2 19 61 10 10 10= m m m3 e j e j~ . If the distance between stars is 4 1016 m, then there is one star in a volume on the order of 4 10 1016 3 50 m m3 e j ~ . The number of stars is about 10 10 10 61 50 11m m star stars 3 3 ~ . P1.70 The density of each material is = = = m V m r h m D h2 2 4 . Al: g cm cm g cm The tabulated value g cm is smaller. Cu: g .23 cm .06 cm g cm The tabulated value g cm is smaller. Brass: .54 cm .69 cm g cm Sn: g .75 cm .74 cm g cm Fe: .89 cm .77 cm g cm 3 3 3 3 3 3 3 = = F HG I KJ = = F HG I KJ = = = = = = 4 51 5 2 52 3 75 2 75 2 70 2% 4 56 3 1 5 9 36 8 92 5% 4 94.4 g 1 5 8 91 4 69 1 1 3 7 68 4 216.1 g 1 9 7 88 2 2 2 2 2 . . . . . . . . . . . . b g a f a f b g a f a f b g a f a f b g a f a f b g a f a f The tabulated value g cm is smaller.3 7 86 0 3%. . F HG I KJ P1.71 (a) 3 600 s hr 24 hr day 365.25 days yr s yrb gb gb g= 3 16 107 . (b) V r V V mm cube mm 18 . m . m m m 1.91 10 micrometeorites = = = = = 4 3 4 3 5 00 10 5 24 10 1 5 24 10 3 7 3 19 3 3 19 3 e j . This would take 1 91 10 3 16 10 6 05 10 18 7 10. . . = micrometeorites micrometeorites yr yr . 19. Chapter 1 19 ANSWERS TO EVEN PROBLEMS P1.2 5 52 103 3 . kg m , between the densities of aluminum and iron, and greater than the densities of surface rocks. P1.34 1 3 1021 . kg P1.36 200 km P1.38 (a) 13.4; (b) 49.1P1.4 23.0 kg P1.40 r rAl Fe Fe Al = F HG I KJ 1 3 P1.6 7.69 cm P1.8 (a) and (b) see the solution, NA = 6 022 137 1023 . ; (c) 18.0 g; P1.42 ~10 rev7 (d) 44.0 g P1.44 ~109 raindrops P1.10 (a) 9 83 10 16 . g ; (b) 1 06 107 . atoms P1.46 ~1011 cans; ~105 tons P1.12 (a) 4 02 1025 . molecules; (b) 3 65 104 . molecules P1.48 209 4 2 a fcm P1.14 (a) ii; (b) iii; (c) i P1.50 (a) 3; (b) 4; (c) 3; (d) 2 P1.16 (a) M L T2 ; (b) 1 1newton kg m s2 = P1.52 (a) 797; (b) 1.1; (c) 17.66 P1.54 115.9 m P1.18 35 7. m2 P1.56 316 m P1.20 1 39 10 4 . m3 P1.58 4 50. m2 P1.22 (a) 3 39 105 3 . ft ; (b) 2 54 104 . lb P1.60 see the solution; 24.6 P1.24 (a) 560 5 60 10 5 60 105 7 km m cm= = . . ; P1.62 3 64. cents ; no(b) 491 m 0 491 4 91 104 = = . .km cm ; (c) 6 6 19 10 6 19 103 5 .19 km m cm= = . . ; P1.64 see the solution (d) 2 2 50 10 2 50 103 5 .50 km m cm= = . . P1.66 (a) 1 000 kg; (b) 5 2 10 16 . kg ; 0 27. kg ; 1 3 10 5 . kg P1.26 4 05 103 . m2 P1.28 (a) 1 1 609mi h km h= . ; (b) 88 5. km h ; P1.68 8 32 10 4 . m s ; a snail(c) 16 1. km h P1.70 see the solution P1.30 1 19 1057 . atoms P1.32 2 57 106 3 . m 20. 2 CHAPTER OUTLINE 2.1 Position, Velocity, and Speed 2.2 Instantaneous Velocity and Speed 2.3 Acceleration 2.4 Motion Diagrams 2.5 One-Dimensional Motion with Constant Acceleration 2.6 Freely Falling Objects 2.7 Kinematic Equations Derived from Calculus Motion in One Dimension ANSWERS TO QUESTIONS Q2.1 If I count 5.0 s between lightning and thunder, the sound has traveled 331 5 0 1 7m s s kmb ga f. .= . The transit time for the light is smaller by 3 00 10 331 9 06 10 8 5. . = m s m s times, so it is negligible in comparison. Q2.2 Yes. Yes, if the particle winds up in the +x region at the end. Q2.3 Zero. Q2.4 Yes. Yes. Q2.5 No. Consider a sprinter running a straight-line race. His average velocity would simply be the length of the race divided by the time it took for him to complete the race. If he stops along the way to tie his shoe, then his instantaneous velocity at that point would be zero. Q2.6 We assume the object moves along a straight line. If its average velocity is zero, then the displacement must be zero over the time interval, according to Equation 2.2. The object might be stationary throughout the interval. If it is moving to the right at first, it must later move to the left to return to its starting point. Its velocity must be zero as it turns around. The graph of the motion shown to the right represents such motion, as the initial and final positions are the same. In an x vs. t graph, the instantaneous velocity at any time t is the slope of the curve at that point. At t0 in the graph, the slope of the curve is zero, and thus the instantaneous velocity at that time is also zero. x tt0 FIG. Q2.6 Q2.7 Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, or is a constant. 21 21. 22 Motion in One Dimension Q2.8 Yes. If you drop a doughnut from rest v = 0a f, then its acceleration is not zero. A common misconception is that immediately after the doughnut is released, both the velocity and acceleration are zero. If the acceleration were zero, then the velocity would not change, leaving the doughnut floating at rest in mid-air. Q2.9 No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past. Q2.10 Yes. Consider throwing a ball straight up. As the ball goes up, its velocity is upward v > 0a f, and its acceleration is directed down a < 0a f. A graph of v vs. t for this situation would look like the figure to the right. The acceleration is the slope of a v vs. t graph, and is always negative in this case, even when the velocity is positive. v t v0 FIG. Q2.10 Q2.11 (a) Accelerating East (b) Braking East (c) Cruising East (d) Braking West (e) Accelerating West (f) Cruising West (g) Stopped but starting to move East (h) Stopped but starting to move West Q2.12 No. Constant acceleration only. Yes. Zero is a constant. Q2.13 The position does depend on the origin of the coordinate system. Assume that the cliff is 20 m tall, and that the stone reaches a maximum height of 10 m above the top of the cliff. If the origin is taken as the top of the cliff, then the maximum height reached by the stone would be 10 m. If the origin is taken as the bottom of the cliff, then the maximum height would be 30 m. The velocity is independent of the origin. Since the change in position is used to calculate the instantaneous velocity in Equation 2.5, the choice of origin is arbitrary. Q2.14 Once the objects leave the hand, both are in free fall, and both experience the same downward acceleration equal to the free-fall acceleration, g. Q2.15 They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity equal to vi. This velocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds will also be the same. Q2.16 With h gt= 1 2 2 , (a) 0 5 1 2 0 707 2 . .h g t= a f . The time is later than 0.5t. (b) The distance fallen is 0 25 1 2 0 5 2 . .h g t= a f . The elevation is 0.75h, greater than 0.5h. 22. Chapter 2 23 Q2.17 Above. Your ball has zero initial speed and smaller average speed during the time of flight to the passing point. SOLUTIONS TO PROBLEMS Section 2.1 Position, Velocity, and Speed P2.1 (a) v = 2 30. m s (b) v x t = = m m s = 16.1 m s 57 5 9 20 3 00 . . . (c) v x t = = = 57 5 0 11 5 . . m m 5.00 s m s *P2.2 (a) v x t = = F HG I KJ F HG I KJ = 20 1 1 3 156 10 2 107 7ft 1 yr m 3.281 ft yr s m s . or in particularly windy times v x t = = F HG I KJ F HG I KJ = 100 1 1 3 156 10 1 107 6ft 1 yr m 3.281 ft yr s m s . . (b) The time required must have been t x v = = F HG I KJF HG I KJ = 3 000 1 609 10 5 10 3 8mi 10 mm yr m 1 mi mm 1 m yr . P2.3 (a) v x t = = = 10 5 m 2 s m s (b) v = = 5 1 2 m 4 s m s. (c) v x x t t = = = 2 1 2 1 5 10 2 2 5 m m 4 s s m s. (d) v x x t t = = = 2 1 2 1 5 5 4 3 3 m m 7 s s m s. (e) v x x t t = = =2 1 2 1 0 0 8 0 0 m s P2.4 x t= 10 2 : For t x s m a f a f = = 2 0 2 1 3 0 40 44 1 90 . . . . (a) v x t = = = 50 50 0 m 1.0 s m s. (b) v x t = = = 4 1 41 0 . . m 0.1 s m s 23. 24 Motion in One Dimension P2.5 (a) Let d represent the distance between A and B. Let t1 be the time for which the walker has the higher speed in 5 00 1 . m s = d t . Let t2 represent the longer time for the return trip in = 3 00 2 . m s d t . Then the times are t d 1 5 00 = . m sb g and t d 2 3 00 = . m sb g. The average speed is: v d d d v d d d = = + + = = = Total distance Total time m s m s m s m s m s m s m s 2 2 2 25 00 3 00 8 00 15 0 2 2 15 0 8 00 3 75 . . . . . . . b g b g b g e j e j (b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0 . Section 2.2 Instantaneous Velocity and Speed P2.6 (a) At any time, t, the position is given by x t= 3 00 2 . m s2 e j . Thus, at ti = 3 00. s: xi = =3 00 3 00 27 0 2 . . .m s s m2 e ja f . (b) At t tf = +3 00. s : x tf = +3 00 3 00 2 . .m s s2 e ja f , or x t tf = + +27 0 18 0 3 00 2 . . .m m s m s2 b g e ja f . (c) The instantaneous velocity at t = 3 00. s is: v x x t t t f i t = F HG I KJ = + = lim lim . . . 0 0 18 0 3 00 18 0m s m s m s2 e je j . P2.7 (a) at ti = 1 5. s, xi = 8 0. m (Point A) at tf = 4 0. s , x f = 2 0. m (Point B) v x x t t f i f i = = = = 2 0 8 0 4 1 5 6 0 2 4 . . . . . a f a f m s m 2.5 s m s (b) The slope of the tangent line is found from points C and D. t xC C= =1 0 9 5. .s, mb g and t xD D= =3 5 0. s,b g, v 3 8. m s . FIG. P2.7 (c) The velocity is zero when x is a minimum. This is at t 4 s . 24. Chapter 2 25 P2.8 (a) (b) At t = 5 0. s, the slope is v 58 23 m 2.5 s m s . At t = 4 0. s, the slope is v 54 18 m 3 s m s . At t = 3 0. s, the slope is v 49 m 14 3.4 s m s . At t = 2 0. s , the slope is v 36 m 9 4.0 s .0 m s . (c) a v t = 23 5 0 4 6 m s s m s2 . . (d) Initial velocity of the car was zero . P2.9 (a) v = ( ) ( ) = 5 0 1 0 5 m s m s (b) v = ( ) ( ) = 5 10 4 2 2 5 m s m s. (c) v = ( ) ( ) = 5 5 5 4 0 m m s s (d) v = ( ) ( ) = + 0 5 8 7 5 m s s m s FIG. P2.9 *P2.10 Once it resumes the race, the hare will run for a time of t x x v f i x = = = 1 000 25 m 800 m 8 m s s . In this time, the tortoise can crawl a distance x xf i = ( )=0 2 25 5 00. .m s s ma f . 25. 26 Motion in One Dimension Section 2.3 Acceleration P2.11 Choose the positive direction to be the outward direction, perpendicular to the wall. v v atf i= + : a v t = = = 22 0 25 0 3 50 10 1 34 103 4. . . . m s m s s m s2a f . P2.12 (a) Acceleration is constant over the first ten seconds, so at the end, v v atf i= + = + ( )=0 2 00 10 0 20 0. . .m s s m s2 c h . Then a = 0 so v is constant from t =10 0. s to t =15 0. s. And over the last five seconds the velocity changes to v v atf i= + = + ( )=20 0 3 00 5 00 5 00. . . .m s m s s m s2 c h . (b) In the first ten seconds, x x v t atf i i= + + = + + ( ) = 1 2 0 0 1 2 2 00 10 0 1002 2 . .m s s m2 c h . Over the next five seconds the position changes to x x v t atf i i= + + = + ( )+ = 1 2 100 20 0 5 00 0 2002 m m s s m. .a f . And at t = 20 0. s, x x v t atf i i= + + = + ( )+ ( ) = 1 2 200 20 0 5 00 1 2 3 00 5 00 2622 2 m m s s m s s m2 . . . .a f c h . *P2.13 (a) The average speed during a time interval t is v t = distance traveled . During the first quarter mile segment, Secretariats average speed was v1 0 250 1 320 52 4 35 6= = = . . . mi 25.2 s ft 25.2 s ft s mi hb g. During the second quarter mile segment, v2 1 320 55 0 37 4= = ft 24.0 s ft s mi h. .b g. For the third quarter mile of the race, v3 1 320 55 5 37 7= = ft 23.8 s ft s mi h. .b g, and during the final quarter mile, v4 1 320 57 4 39 0= = ft 23.0 s ft s mi h. .b g. continued on next page 26. Chapter 2 27 (b) Assuming that v vf = 4 and recognizing that vi = 0 , the average acceleration during the race was a v vf i = = + + +( ) = total elapsed time ft s s ft s257 4 0 25 2 24 0 23 8 23 0 0 598 . . . . . . . P2.14 (a) Acceleration is the slope of the graph of v vs t. For 0 5 00< 10 s . (d) Maximum negative acceleration is at t = 8 s, and is approximately 1 5. m s2 . Section 2.4 Motion Diagrams P2.18 (a) (b) (c) (d) (e) continued on next page 28. Chapter 2 29 (f) One way of phrasing the answer: The spacing of the successive positions would change with less regularity. Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude and direction. Section 2.5 One-Dimensional Motion with Constant Acceleration P2.19 From v v axf i 2 2 2= + , we have 10 97 10 0 2 2203 2 . = + ( )m s mc h a , so that a = 2 74 105 . m s2 which is a g= 2 79 104 . times . P2.20 (a) x x v v tf i i f = + 1 2 c h becomes 40 1 2 2 80 8 50m m s s= + ( )vi . .a f which yields vi = 6 61. m s . (b) a v v t f i = = = 2 80 6 61 8 50 0 448 . . . . m s m s s m s2 P2.21 Given vi =12 0. cm s when x ti = =( )3 00 0. cm , and at t = 2 00. s, x f =5 00. cm, x x v t atf i i = + 1 2 2 : = ( )+ ( )5 00 3 00 12 0 2 00 1 2 2 00 2 . . . . .a = +8 00 24 0 2. . a a = = 32 0 2 16 0 . . cm s2 . *P2.22 (a) Let i be the state of moving at 60 mi h and f be at rest v v a x x a a xf xi x f i x x 2 2 2 2 0 60 2 121 0 1 3 600 242 5 280 1 21 8 21 8 1 609 1 9 75 = + = + F HG I KJ = F HG I KJF HG I KJ = = F HG I KJF HG I KJ = d i b g a fmi h ft mi 5 280 ft mi h ft 1 mi h 3 600 s mi h s mi h s m 1 mi h 3 600 s m s 2 2 . . . . (b) Similarly, 0 80 2 211 0 6 400 5 280 422 3 600 22 2 9 94 2 = + = = = mi h ft mi h s mi h s m s2 b g a f b g b g a a x x . . . (c) Let i be moving at 80 mi h and f be moving at 60 mi h. v v a x x a a xf xi x f i x x 2 2 2 2 2 60 80 2 211 121 2 800 5 280 2 90 3 600 22 8 10 2 = + = + = = = d i b g b g a f b g a fb g mi h mi h ft ft mi h s mi h s m s2 . . . 29. 30 Motion in One Dimension *P2.23 (a) Choose the initial point where the pilot reduces the throttle and the final point where the boat passes the buoy: xi = 0 , x f =100 m, vxi = 30 m s, vxf =?, ax =3 5. m s2 , t =? x x v t a tf i xi x= + + 1 2 2 : 100 0 30 1 2 3 5 2 m m s m s2 = + + a f c ht t. 1 75 30 100 02 . m s m s m2 c h a ft t + = . We use the quadratic formula: t b b ac a = 2 4 2 t = ( ) = = 30 900 4 1 75 100 2 1 75 30 14 1 3 5 12 6 m s m s m s m m s m s m s m s s 2 2 2 2 2 . . . . . c h c h or 4 53. s . The smaller value is the physical answer. If the boat kept moving with the same acceleration, it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s. (b) v v a txf xi x= + = =30 3 5 4 53 14 1m s m s s m s2 . . .e j P2.24 (a) Total displacement = area under the v t,a f curve from t = 0 to 50 s. x x = + + = 1 2 50 15 50 40 15 1 2 50 10 1 875 m s s m s s m s s m b ga f b ga f b ga f (b) From t =10 s to t = 40 s , displacement is x = + + = 1 2 50 33 5 50 25 1 457m s m s s m s s mb ga f b ga f . FIG. P2.24 (c) 0 15 t s: a v t 1 50 0 15 0 3 3= = ( ) = m s s m s2 . 15 40s s< the ball clears the goal on its way down . P4.20 The horizontal component of displacement is x v t v tf xi i i= = cosb g . Therefore, the time required to reach the building a distance d away is t d vi i = cos . At this time, the altitude of the water is y v t a t v d v g d v f yi y i i i i i i = + = F HG I KJ F HG I KJ1 2 2 2 2 sin cos cos . Therefore the water strikes the building at a height h above ground level of h y d gd v f i i i = = tan cos 2 2 2 2 . 90. Chapter 4 91 *P4.21 (a) For the horizontal motion, we have x x v t a t v v f i xi x i i = + + = + + = 1 2 24 0 53 2 2 0 18 1 2 m s m s cos . . . a fa f (b) As it passes over the wall, the ball is above the street by y y v t a tf i yi y= + + 1 2 2 y f = + + =0 18 1 53 2 2 1 2 9 8 2 2 8 13 2 . sin . . . .m s s m s s m2 b ga fa f e ja f . So it clears the parapet by 8 13 7 1 13. .m m m = . (c) Note that the highest point of the balls trajectory is not directly above the wall. For the whole flight, we have from the trajectory equation y x g v xf i f i i f= F HG I KJtan cos b g 2 2 2 2 or 6 53 9 8 2 18 1 53 2 2 2 m m s m s 2 = F H GG I K JJtan . . cos a f b g x xf f . Solving, 0 041 2 1 33 6 01 2 . .m m + =e jx xf f and x f = 1 33 1 33 4 0 0412 6 2 0 0412 2 1 . . . . a fa f e jm . This yields two results: x f = 26 8. m or 5.44 m The ball passes twice through the level of the roof. It hits the roof at distance from the wall 26 8 24 2 79. .m m m = . 91. 92 Motion in Two Dimensions *P4.22 When the bomb has fallen a vertical distance 2.15 km, it has traveled a horizontal distance x f given by x y x gx v f f f f i i i i i i i i = = = = = + = = F H I K = 3 25 2 15 2 437 2 2 150 2 437 9 8 2 437 2 280 2 150 2 437 371 19 1 6 565 4 792 0 1 2 6 565 6 565 4 1 4 792 3 283 3 945 2 2 2 2 2 2 2 2 2 2 2 . . . tan cos tan . cos tan . tan tan . tan . tan . . . . . . km km km m m m s m m s m m m 2 a f a f b g e jb g b g b g a fe j a f a fa f Select the negative solution, since i is below the horizontal. = tan .i 0 662 , i = 33 5. P4.23 The horizontal kick gives zero vertical velocity to the rock. Then its time of flight follows from y y v t a t t t f i yi y= + + = + + = 1 2 40 0 0 0 1 2 9 80 2 86 2 2 . . . . m m s s 2 e j The extra time 3 00 2 86 0 143. . .s s s = is the time required for the sound she hears to travel straight back to the player. It covers distance 343 0 143 49 0 40 02 2 m s s m mb g a f. . .= = +x where x represents the horizontal distance the rock travels. x v t t v xi xi = = + = = 28 3 0 28 3 2 86 9 91 2 . . . . m m s m s 92. Chapter 4 93 P4.24 From the instant he leaves the floor until just before he lands, the basketball star is a projectile. His vertical velocity and vertical displacement are related by the equation v v a y yyf yi y f i 2 2 2= + d i. Applying this to the upward part of his flight gives 0 2 9 80 1 85 1 022 = + vyi . . .m s m2 e ja f . From this, vyi = 4 03. m s. [Note that this is the answer to part (c) of this problem.] For the downward part of the flight, the equation gives vyf 2 0 2 9 80 0 900 1 85= + . . .m s m2 e ja f . Thus the vertical velocity just before he lands is vyf = 4 32. m s. (a) His hang time may then be found from v v a tyf yi y= + : = + 4 32 4 03 9 80. . .m s m s m s2 e jt or t = 0 852. s . (b) Looking at the total horizontal displacement during the leap, x v txi= becomes 2 80 0 852. .m s= vxi a f which yields vxi = 3 29. m s . (c) vyi = 4.03 m s . See above for proof. (d) The takeoff angle is: = F HG I KJ = F HG I KJ = tan tan . .1 1 4 03 50 8 v v yi xi m s 3.29 m s . (e) Similarly for the deer, the upward part of the flight gives v v a y yyf yi y f i 2 2 2= + d i: 0 2 9 80 2 50 1 202 = + vyi . . .m s m2 e ja f so vyi = 5 04. m s. For the downward part, v v a y yyf yi y f i 2 2 2= + d i yields vyf 2 0 2 9 80 0 700 2 50= + . . .m s m2 e ja f and vyf = 5 94. m s. The hang time is then found as v v a tyf yi y= + : = + 5 94 5 04 9 80. . .m s m s m s2 e jt and t = 1 12. s . 93. 94 Motion in Two Dimensions *P4.25 The arrows flight time to the collision point is t x x v f i xi = = = 150 45 50 5 19 m m s s b gcos . . The arrows altitude at the collision is y y v t a tf i yi y= + + = + + = 1 2 0 45 50 5 19 1 2 9 8 5 19 47 0 2 2 m s s m s s m2 b ga f e ja fsin . . . . . (a) The required launch speed for the apple is given by v v a y y v v yf yi y f i yi yi 2 2 2 2 0 2 9 8 47 0 30 3 = + = + = d i e ja f. . . m s m m s 2 (b) The time of flight of the apple is given by v v a t t t yf yi y= + = = 0 30 3 9 8 3 10 . . . . m s m s s 2 So the apple should be launched after the arrow by 5 19 3 10 2 09. . .s s s = . *P4.26 For the smallest impact angle = F HG I KJ tan 1 v v yf xf , we want to minimize vyf and maximize v vxf xi= . The final y-component of velocity is related to vyi by v v ghyf yi 2 2 2= + , so we want to minimize vyi and maximize vxi . Both are accomplished by making the initial velocity horizontal. Then v vxi = , vyi = 0, and v ghyf = 2 . At last, the impact angle is = F HG I KJ = F HG I KJ tan tan1 1 2v v gh v yf xf . FIG. P4.26 94. Chapter 4 95 Section 4.4 Uniform Circular Motion P4.27 a v r c = = = 2 2 20 0 1 06 377 . . m s m m s2b g The mass is unnecessary information. P4.28 a v R = 2 , T = =24 3 600 86 400h s h sb g v R T = = = 2 2 6 37 10 463 6 ( . m) 86 400 s m s a = = 463 6 37 10 0 033 76 2 m s m m s directed toward the center of Earth 2 b g . . . P4.29 r = 0 500. m; v r T a v R t = = = = = = = 2 2 0 500 10 47 10 5 10 47 0 5 219 60 0 2 2 . . . . . . m m s m s m s inward s 200 rev 2 a f a f P4.30 a v r c = 2 v a rc= = =3 9 8 9 45 16 7. . .m s m m s2 e ja f Each revolution carries the astronaut over a distance of 2 2 9 45 59 4 r = =. .m ma f . Then the rotation rate is 16 7 1 0 281. .m s rev 59.4 m rev s F HG I KJ = . P4.31 (a) v r= At 8.00 rev s, v = = =0 600 8 00 2 30 2 9 60. m . rev s rad rev . m s . m sa fb gb g . At 6.00 rev s, v = = =0 900 6 00 2 33 9 10 8. m . rev s rad rev m s m sa fb gb g . . . 6 00. rev s gives the larger linear speed. (b) Acceleration = = = v r 2 2 39 60 0 600 1 52 10 . . . m s m m s2b g . (c) At 6.00 rev s, acceleration = = 10 8 0 900 1 28 10 2 3. . . m s m m s2b g . 95. 96 Motion in Two Dimensions P4.32 The satellite is in free fall. Its acceleration is due to gravity and is by effect a centripetal acceleration. a gc = so v r g 2 = . Solving for the velocity, v rg= = + = 6 400 600 10 8 21 7 58 103 3 , . .a fe je jm m s m s2 v r T = 2 and T r v T = = = = F HG I KJ = 2 2 7 000 10 7 58 10 5 80 10 5 80 10 1 96 7 3 3 3 3 , . . . . . m m s s s min 60 s min e j Section 4.5 Tangential and Radial Acceleration P4.33 We assume the train is still slowing down at the instant in question. a v r a v t a a a c t c t = = = = = = + = + 2 3 1 2 2 2 2 1 29 40 0 10 15 0 0 741 1 29 0 741 . . . . . . m s km h m km s m s m s m s 2 h 3 600 s 2 2 2 b ge je j e j e j at an angle of tan tan F HG I KJ = F HG I KJ1 1a a t c 0.741 1.29 a = 1 48. m s inward and 29.9 backward2 o FIG. P4.33 P4.34 (a) at = 0 600. m s2 (b) a v r r = = = 2 2 4 00 20 0 0 800 . . . m s m m s2b g (c) a a at r= + =2 2 1 00. m s2 = = tan .1 53 1 a a r t inward from path 96. Chapter 4 97 P4.35 r = 2 50. m, a = 15 0. m s2 (a) a ac = = =cos . . cos .30 0 15 0 30 13 0o 2 2 m s m se ja f (b) a v r c = 2 so v rac 2 2 50 13 0 32 5= = =. . .m m s m s2 2 2 e j v = =32 5 5 70. .m s m s FIG. P4.35 (c) a a at r 2 2 2 = + so a a at r= = =2 2 2 15 0 13 0 7 50. . .m s m s m s2 2 2 e j e j P4.36 (a) See figure to the right. (b) The components of the 20.2 and the 22 5. m s2 along the rope together constitute the centripetal acceleration: ac = + =22 5 90 0 36 9 20 2 36 9 29 7. cos . . . cos . .m s m s m s2 2 2 e j a f e j (c) a v r c = 2 so v a rc= = =29 7 1 50 6 67. . .m s m m s tangent to circle2 a f v = 6 67. m s at 36.9 above the horizontal FIG. P4.36 *P4.37 Let i be the starting point and f be one revolution later. The curvilinear motion with constant tangential acceleration is described by x v t a t r a t a r t xi x t t = + = + = 1 2 2 0 1 2 4 2 2 2 a at ar FIG. P4.37 and v v a txf xi x= + , v a t r t f t= + =0 4 . The magnitude of the radial acceleration is a v r r t r r f = = 2 2 2 2 16 . Then tan = = = a a r t t r t r 4 16 1 4 2 2 2 = 4 55. . 97. 98 Motion in Two Dimensions Section 4.6 Relative Velocity and Relative Acceleration P4.38 (a) v a i j v i j v a i j v i j v i j i j v i j v H H 2 H J 2 J HJ H J HJ HJ + . . m s . s . . m s + . + . m s . s . + . m s . . . . m s . . m s m s = = = = = = = = = = + = 0 3 00 2 00 5 00 15 0 10 0 0 1 00 3 00 5 00 5 00 15 0 15 0 10 0 5 00 15 0 10 0 25 0 10 0 25 0 262 2 t t v v j ( . ) ( . ) e j a f e j e j a f e j e j e j .9 m s (b) r a i j r i j r i j i j r r r i j i j r i j r H H 2 2 H J 2 HJ H J HJ HJ m s 5.00 s . . m m s 5.00s m . . . . m . . m m m = + + = = = + = + = = = = + = 0 0 1 2 1 2 3 00 2 00 37 5 25 0 1 2 1 00 3 00 12 5 37 5 37 5 25 0 12 5 37 5 25 0 62 5 25 0 62 5 67 3 2 2 2 2 t . . . . . . . . . e j a f e j e j a f e j e j e j a f a f (c) a a a i j i j a i j HJ H J 2 HJ 2 . . . . m s m s = = = 3 00 2 00 1 00 3 00 2 00 5 00. . e j e j *P4.39 vce = the velocity of the car relative to the earth. vwc = the velocity of the water relative to the car. vwe =the velocity of the water relative to the earth. These velocities are related as shown in the diagram at the right. (a) Since vwe is vertical, v vwc cesin . .60 0 50 0= = km h or vwc = 57 7 60 0. .km h at west of vertical . (b) Since vce has zero vertical component, vce 60 vwe vwc v v vwe ce wc= + FIG. P4.39 v vwe wc= = =cos . . cos . .60 0 57 7 60 0 28 9km h km h downwardb g . 98. Chapter 4 99 P4.40 The bumpers are initially 100 0 100m km= . apart. After time t the bumper of the leading car travels 40.0 t, while the bumper of the chasing car travels 60.0t. Since the cars are side by side at time t, we have 0 100 40 0 60 0. . .+ =t t, yielding t = = 5 00 10 18 03 . .h s . P4.41 Total time in still water t d v = = = 2 000 1 20 1 67 103 . . s . Total time = time upstream plus time downstream: t t up down s s = = = + = 1 000 1 20 0 500 1 43 10 1 000 1 20 0 500 588 3 ( . . ) . . . . Therefore, ttotal s= + = 1 43 10 588 2 02 103 3 . . . P4.42 v = + =150 30 0 1532 2 . km h = F HG I KJ = tan . .1 30 0 150 11 3 north of west P4.43 For Alan, his speed downstream is c + v, while his speed upstream is c v . Therefore, the total time for Alan is t L c v L c v L c v c 1 2 1 2 2 = + + = . For Beth, her cross-stream speed (both ways) is c v2 2 . Thus, the total time for Beth is t L c v L c v c 2 2 2 2 2 1 2 2 = = . Since 1 1 2 2 < v c , t t1 2> , or Beth, who swims cross-stream, returns first. 99. 100 Motion in Two Dimensions P4.44 (a) To an observer at rest in the train car, the bolt accelerates downward and toward the rear of the train. a = + = = = = 2 50 9 80 10 1 2 50 9 80 0 255 14 3 2 2 . . . tan . . . m s m s m s m s m s . to the south from the vertical 2 2 2 b g b g (b) a = 9 80. m s vertically downward2 P4.45 Identify the student as the S observer and the professor as the S observer. For the initial motion in S, we have = = v v y x tan .60 0 3 . Let u represent the speed of S relative to S. Then because there is no x-motion in S, we can write v v ux x= + = 0 so that = = v ux 10 0. m s. Hence the ball is thrown backwards in S. Then, v v vy y x= = =3 10 0 3. m s. Using v ghy 2 2= we find h = = 10 0 3 2 9 80 15 3 2 . . . m s m s m2 e j e j . FIG. P4.45 The motion of the ball as seen by the student in S is shown in diagram (b). The view of the professor in S is shown in diagram (c). *P4.46 Choose the x-axis along the 20-km distance. The y- components of the displacements of the ship and the speedboat must agree: 26 40 15 50 11 0 50 12 71 km h km hb g a f b gt tsin sin sin . . . = = = The speedboat should head 15 12 7 27 7+ = . . east of north . 15 N E 40 25 x y FIG. P4.46 100. Chapter 4 101 Additional Problems *P4.47 (a) The speed at the top is v vx i i= = =cos cos 143 45 101m s m sb g . (b) In free fall the plane reaches altitude given by v v a y y y y yf yi y f i f f 2 2 2 3 2 0 143 45 2 9 8 31 000 31 000 522 3 28 1 3 27 10 = + = + = + F HG I KJ = d i b g e jd im s m s ft ft m ft m ft 2 sin . . . . (c) For the whole free fall motion v v a tyf yi y= + = + = 101 101 9 8 20 6 m s m s m s s 2 . . e jt t (d) a v r c = 2 v a rc= = =0 8 9 8 4 130 180. . ,m s m m s2 e j P4.48 At any time t, the two drops have identical y-coordinates. The distance between the two drops is then just twice the magnitude of the horizontal displacement either drop has undergone. Therefore, d x t v t v t v txi i i i i= = = =2 2 2 2a f b g b gcos cos . P4.49 After the string breaks the ball is a projectile, and reaches the ground at time t: y v t a tf yi y= + 1 2 2 = + 1 20 0 1 2 9 80 2 . .m m s2 e jt so t = 0 495. s. Its constant horizontal speed is v x t x = = = 2 00 4 04 . . m 0.495 s m s so before the string breaks a v r c x = = = 2 2 4 04 0 300 54 4 . . . m s m m s2b g . 101. 102 Motion in Two Dimensions P4.50 (a) y x g v xf i f i i f= tan cos b gd i 2 2 2 2 Setting x df = cos , and y df = sin, we have d d g v di i i sin tan cos cos cos = b gb g b g2 2 2 2 . FIG. P4.50 Solving for d yields, d v g i i i i = 2 2 2 cos sin cos sin cos cos or d v g i i i = 2 2 2 cos sin cos b g . (b) Setting dd d i = 0 leads to i = +45 2 and d v g i max sin cos = 2 2 1 b g . P4.51 Refer to the sketch: (b) x v txi= ; substitution yields 130 35 0= v ti cos .b g . y v t atyi= + 1 2 2 ; substitution yields 20 0 35 0 1 2 9 80 2 . sin . .= + v t tib g a f . Solving the above gives t = 3 81. s . (a) vi = 41 7. m s FIG. P4.51 (c) v v gtyf i i= sin , v vx i i= cos At t = 3 81. s, vyf = = 41 7 35 0 9 80 3 81 13 4. sin . . . .a fa f m s v v v v x f x yf = = = + = 41 7 35 0 34 1 36 72 2 . cos . . . . a f m s m s 102. Chapter 4 103 P4.52 (a) The moons gravitational acceleration is the probes centripetal acceleration: (For the moons radius, see end papers of text.) a v r v v = = = = 2 2 6 6 1 6 9 80 1 74 10 2 84 10 1 69 . . . . m s m m s km s 2 2 2 e j (b) v r T = 2 T r v = = = = 2 2 1 74 10 6 47 10 1 80 6 3 ( . . . m) 1.69 10 m s s h3 P4.53 (a) a v r c = = = 2 2 5 00 1 00 25 0 . . . m s m m s2b g a gt = = 9 80. m s2 (b) See figure to the right. (c) a a ac t= + = + =2 2 2 2 25 0 9 80 26 8. . .m s m s m s2 2 2 e j e j = F HG I KJ = = tan tan . . .1 1 9 80 25 0 21 4 a a t c m s m s 2 2 FIG. P4.53 P4.54 x v t v tf ix i= = cos .40 0 Thus, when x f = 10 0. m, t vi = 10 0 40 0 . cos . m . At this time, y f should be 3.05 m m m =2 00 1 05. . . Thus, 1 05 40 0 10 0 40 0 1 2 9 80 10 0 40 0 2 . sin . . cos . . . cos . m m m s m2 = + L NM O QPv v v i i i b g e j . From this, vi = 10 7. m s . 103. 104 Motion in Two Dimensions P4.55 The special conditions allowing use of the horizontal range equation applies. For the ball thrown at 45, D R v g i = =45 2 90sin . For the bouncing ball, D R R v g g i vi = + = +1 2 2 2 2 2 2sin sin e j where is the angle it makes with the ground when thrown and when bouncing. (a) We require: v g v g v g i i i 2 2 2 2 2 4 2 4 5 26 6 = + = = sin sin sin . FIG. P4.55 (b) The time for any symmetric parabolic flight is given by y v t gt v t gt f yi i i = = 1 2 0 1 2 2 2 sin . If t = 0 is the time the ball is thrown, then t v g i i = 2 sin is the time at landing. So for the ball thrown at 45.0 t v g i 45 2 45 0 = sin . . For the bouncing ball, t t t v g g v g i v i i = + = + = 1 2 22 26 6 2 26 6 3 26 6sin . sin . sin .e j . The ratio of this time to that for no bounce is 3 26 6 2 45 0 1 34 1 41 0 949 v g v g i i sin . sin . . . . = = . 104. Chapter 4 105 P4.56 Using the range equation (Equation 4.14) R v g i i = 2 2sin( ) the maximum range occurs when i = 45 , and has a value R v g i = 2 . Given R, this yields v gRi = . If the boy uses the same speed to throw the ball vertically upward, then v gR gty = and y gR t gt = 2 2 at any time, t. At the maximum height, vy = 0, giving t R g = , and so the maximum height reached is y gR R g g R g R R R max = F HG I KJ = = 2 2 2 2 . P4.57 Choose upward as the positive y-direction and leftward as the positive x-direction. The vertical height of the stone when released from A or B is yi = + =1 50 1 20 30 0 2 10. . . m . msina f (a) The equations of motion after release at A are v v gt t v v y t t x t y i x i A = = = = = = sin cos 60 0 1 30 9 80 60 0 0 750 2 10 1 30 4 90 0 750 2 . . . m s . . m s . + . . m . m a f e j a f vi B A vi 3030 30301.20 m1.20 m FIG. P4.57 When y = 0, t = + = 1 30 1 30 41 2 9 80 0 800 2 . . . . a f . s. Then, xA = =0 750 0 800 0 600. . m ma fa f . . (b) The equations of motion after release at point B are v v gt t v v y t t y i x i i = = = = = sin . . cos . 60 0 1 30 9 80 60 0 0 750 2 10 1 30 4 90 2 . m s . . m s . . . m a f a f e j When y = 0, t = + + = 1 30 1 30 41 2 9 80 0 536 2 . . . . a f . s. Then, xB = =0 750 0 536 0 402. . m ma fa f . . (c) a v r r = = = 2 2 1 50 1 20 1 87 . . . m s m m s toward the center2b g (d) After release, a j= =g .9 80 m s downward2 105. 106 Motion in Two Dimensions P4.58 The football travels a horizontal distance R v g i i = = = 2 2 2 20 0 60 0 9 80 35 3 sin . sin . . . b g a f a f m. Time of flight of ball is t v g i i = = = 2 2 20 0 30 0 9 80 2 04 sin ( . )sin . . . s . FIG. P4.58 The receiver is x away from where the ball lands and x = =35 3 20 0 15 3. . . m. To cover this distance in 2.04 s, he travels with a velocity v = = 15 3 2 04 7 50 . . . m s in the direction the ball was thrown . P4.59 (a) y g t= 1 2 2 ; x v ti= Combine the equations eliminating t: y g x vi = F HG I KJ1 2 2 . From this, x y g vib g2 22 = F HG I KJ FIG. P4.59 thus x v y g i= = = = 2 275 2 300 9 80 6 80 10 6 803( ) . . . km . (b) The plane has the same velocity as the bomb in the x direction. Therefore, the plane will be 3 000 m directly above the bomb when it hits the ground. (c) When is measured from the vertical, tan = x y therefore, = F HG I KJ = F HG I KJ = tan tan .1 1 6 800 3 000 66 2 x y . 106. Chapter 4 107 *P4.60 (a) We use the approximation mentioned in the problem. The time to travel 200 m horizontally is t x vx = = = 200 1 000 0 200 m m s s , . . The bullet falls by y v t a tyi y= + = + = 1 2 0 1 2 9 8 0 2 0 1962 2 . . .m s s m2 e ja f . (b) The telescope axis must point below the barrel axis by = = tan . .1 0 196 200 0 0561 m m . (c) t = = 50 0 1 000 0 050 0 . . m m s s . The bullet falls by only y = = 1 2 9 8 0 05 0 0122 2 . . .m s s m2 e ja f . 50 150 200 250 barrel axis bullet path scope axis FIG. P4.60(b) At range 50 1 4 200m m= a f, the scope axis points to a location 1 4 19 6 4 90. .cm cma f= above the barrel axis, so the sharpshooter must aim low by 4 90 1 22 3 68. . .cm cm cm = . (d) t y = = = = 150 1 000 0 150 1 2 9 8 0 15 0 110 2 m m s s m s s m2 . . . . e ja f Aim low by 150 200 19 6 11 0 3 68. . .cm cm cma f = . (e) t y = = = = 250 1 000 0 250 1 2 9 8 0 25 0 306 2 m m s s m s s m2 . . . . e ja f Aim high by 30 6 250 200 19 6 6. .cm cm .12 cm =a f . (f), (g) Many marksmen have a hard time believing it, but they should aim low in both cases. As in case (a) above, the time of flight is very nearly 0.200 s and the bullet falls below the barrel axis by 19.6 cm on its way. The 0.0561 angle would cut off a 19.6-cm distance on a vertical wall at a horizontal distance of 200 m, but on a vertical wall up at 30 it cuts off distance h as shown, where cos .30 19 6= cm h, h = 22 6. cm. The marksman must aim low by 22 6 19 6 3 03. . .cm cm cm = . The answer can be obtained by considering limiting cases. Suppose the target is nearly straight above or below you. Then gravity will not cause deviation of the path of the bullet, and one must aim low as in part (c) to cancel out the sighting-in of the telescope. barrel axis 30 scope 19.6 cm h 19.6 cm scope axis bullet hits here 30 FIG. P4.60(fg) 107. 108 Motion in Two Dimensions P4.61 (a) From Part (c), the raptor dives for 6 34 2 00 4 34. . . s = undergoing displacement 197 m downward and 10 0 4 34 43 4. . . ma fa f= forward. v d t = + = 197 43 4 4 34 46 5 2 2 a f a f. . . m s (b) = F HG I KJ = tan . .1 197 43 4 77 6 (c) 197 1 2 2 = gt , t = 6 34. s FIG. P4.61 P4.62 Measure heights above the level ground. The elevation yb of the ball follows y R gtb = + 0 1 2 2 with x v ti= so y R gx v b i = 2 2 2 . (a) The elevation yr of points on the rock is described by y x Rr 2 2 2 + = . We will have y yb r= at x = 0, but for all other x we require the ball to be above the rock surface as in y yb r> . Then y x Rb 2 2 2 + > R gx v x R R gx R v g x v x R g x v x gx R v i i i i i F HG I KJ + > + + > + > 2 2 2 2 2 2 2 2 2 4 4 2 2 2 4 4 2 2 2 2 4 4 . If this inequality is satisfied for x approaching zero, it will be true for all x. If the balls parabolic trajectory has large enough radius of curvature at the start, the ball will clear the whole rock: 1 2 > gR vi v gRi > . (b) With v gRi = and yb = 0, we have 0 2 2 = R gx gR or x R= 2 . The distance from the rocks base is x R R = 2 1e j . 108. Chapter 4 109 P4.63 (a) While on the incline v v a x v v at v t v t f i f i f f 2 2 2 2 0 2 4 00 50 0 20 0 0 4 00 20 0 5 00 = = = = = = . . . . . . a fa f m s s FIG. P4.63 (b) Initial free-flight conditions give us vxi = =20 0 37 0 16 0. cos . . m s and vyi = = 20 0 37 0 12 0. sin . . m s v vxf xi= since ax = 0 v a y vyf y yi= + = + = 2 2 9 80 30 0 12 0 27 12 2 . . . .a fa f a f m s v v vf xf yf= + = + = 2 2 2 2 16 0 27 1 31 5. . .a f a f m s at 59.4 below the horizontal (c) t1 5= s ; t v v a yf yi y 2 27 1 12 0 9 80 1 53= = + = . . . . s t t t= + =1 2 6 53. s (d) x v txi= = =1 16 0 1 53 24 5. . .a f m P4.64 Equation of bank: Equations of motion: y x x v t y g t i 2 2 16 1 2 1 2 3 = = = a f a f a f Substitute for t from (2) into (3) y g x vi = F HG I KJ1 2 2 2 . Equate y from the bank equation to y from the equations of motion: FIG. P4.64 16 1 2 4 16 4 16 0 2 2 2 2 4 4 2 3 4 x g x v g x v x x g x vi i i = F HG I KJ L N MM O Q PP = F HG I KJ = . From this, x = 0 or x v g i3 4 2 64 = and x = F HG I KJ =4 10 9 80 18 8 4 2 1 3 . . / m . Also, y g x vi = F HG I KJ = = 1 2 1 2 9 80 18 8 10 0 17 3 2 2 2 2 . . . . a fa f a f m . 109. 110 Motion in Two Dimensions P4.65 (a) Coyote: Roadrunner: x at t x v t v ti i = = = = 1 2 70 0 1 2 15 0 70 0 2 2 ; . . ; . a f Solving the above, we get vi = 22 9. m s and t = 3 06. s. (b) At the edge of the cliff, v atxi = = =15 0 3 06 45 8. . .a fa f m s. Substituting into y a ty= 1 2 2 , we find = = = + = + 100 1 2 9 80 4 52 1 2 45 8 4 52 1 2 15 0 4 52 2 2 2 . . . . . . . a f a fa f a fa f t t x v t a txi x s s s Solving, x = 360 m . (c) For the Coyotes motion through the air v v a t v v a t xf xi x yf yi y = + = + = = + = = 45 8 15 4 52 114 0 9 80 4 52 44 3 . . . . . . a f a f m s m s P4.66 Think of shaking down the mercury in an old fever thermometer. Swing your hand through a circular arc, quickly reversing direction at the bottom end. Suppose your hand moves through one- quarter of a circle of radius 60 cm in 0.1 s. Its speed is 1 4 2 0 6 9 a fa f. m 0.1 s m s and its centripetal acceleration is v r 2 29 0 6 10 ( . ~ m s) m m s 2 2 . The tangential acceleration of stopping and reversing the motion will make the total acceleration somewhat larger, but will not affect its order of magnitude. 110. Chapter 4 111 P4.67 (a) x v txi= , y v t gtyi= + 1 2 2 d tcos . . cos .50 0 10 0 15 0= a f and = + d t tsin . . sin . .50 0 10 0 15 0 1 2 9 80 2 a f a f . Solving, d = 43 2. m and t = 2 88. s . (b) Since ax = 0 , FIG. P4.67 v v v v a t xf xi yf yi y = = = = + = = 10 0 15 0 9 66 10 0 15 0 9 80 2 88 25.6 . . m s m s cos . . sin . . . .a f Air resistance would decrease the values of the range and maximum height. As an airfoil, he can get some lift and increase his distance. *P4.68 For one electron, we have y v tiy= , D v t a t a tix x x= + 1 2 1 2 2 2 , v vyf yi= , and v v a t a txf xi x x= + . The angle its direction makes with the x-axis is given by = = = = tan tan tan tan1 1 1 2 1 2 v v v a t v t a t y D yf xf yi x yi x . FIG. P4.68 Thus the horizontal distance from the aperture to the virtual source is 2D. The source is at coordinate x D= . *P4.69 (a) The ice chest floats downstream 2 km in time t, so that 2 km = v tw . The upstream motion of the boat is described by d v vw= ( )15 min. The downstream motion is described by d v v tw+ = + 2 15km min)( )( . We eliminate t vw = 2 km and d by substitution: v v v v v v v v v v v v v v v w w w w w w w w + = + F HG I KJ + = + = = = b g b g a f a f a f a f a f 15 2 2 15 2 2 2 2 4 00 min km km min 15 min 15 min km km km 15 min 15 min 30 min 2 km km 30 min km h. . (b) Inthereferenceframeofthewater,thechestis motionless.Theboattravelsupstreamfor15min at speed v, and then downstream at the same speed, to return to the same point. Thus it travels for 30 min. During this time, the falls approach the chest at speed vw , traveling 2 km. Thus v x t w = = = 2 4 00 km 30 min km h. . 111. 112 Motion in Two Dimensions *P4.70 Let the river flow in the x direction. (a) To minimize time, swim perpendicular to the banks in the y direction. You are in the water for time t in y v ty= , t = = 80 1 5 53 3 m m s s . . . (b) The water carries you downstream by x v tx= = =2 50 53 3 133. .m s s mb g . (c) vs vw vs vw+ vs vw vs vw+ vs vw vs vw+ To minimize downstream drift, you should swim so that your resultant velocity v vs w+ is perpendicular to your swimming velocity vs relative to the water. This condition is shown in the middle picture. It maximizes the angle between the resultant velocity and the shore. The angle between vs and the shore is given by cos . . = 1 5 2 5 m s m s , = 53 1. . vs vw vs vw+ = 2.5 m/s i (d) Now v vy s= = =sin . sin . . 1 5 53 1 1 20m s m s t y v x v t y x = = = = = = 80 1 2 66 7 2 5 1 5 53 1 66 7 107 m m s s m s m s s m . . . . cos . . .b g 112. Chapter 4 113 *P4.71 Find the highest firing angle H for which the projectile will clear the mountain peak; this will yield the range of the closest point of bombardment. Next find the lowest firing angle; this will yield the maximum range under these conditions if both H and L are > 45 ; x = 2500 m, y = 1800 m, vi = 250 m s. y v t gt v t gt x v t v t f yi i f xi i = = = = 1 2 1 2 2 2 sin cos a f a f Thus t x v f i = cos . Substitute into the expression for y f y v x v g x v x gx v f i f i f i f f i = F HG I KJ = sin cos cos tan cos a f 1 2 2 2 2 2 2 but 1 12 2 cos tan = + so y x gx v f f f i = +tan tan 2 2 2 2 1e j and 0 2 2 2 2 2 2 2 = + + gx v x gx v y f i f f i ftan tan . Substitute values, use the quadratic formula and find tan . = 3 905 or 1.197, which gives H = 75 6. and L = 50 1. . Range at m H i Hv g b g= = 2 32 3 07 10 sin . from enemy ship 3 07 10 2 500 300 2703 . = m from shore. Range at m L i Lv g b g= = 2 32 6 28 10 sin . from enemy ship 6 28 10 2 500 300 3 48 103 3 . . = from shore. Therefore, safe distance is < 270 m or > 3 48 103 . m from the shore. FIG. P4.71 113. 114 Motion in Two Dimensions *P4.72 We follow the steps outlined in Example 4.7, eliminating t d vi = cos cos to find v d v gd v di i i sin cos cos cos cos sin = 2 2 2 2 2 . Clearing of fractions, 2 22 2 2 2 v gd vi icos sin cos cos cos sin = . To maximize d as a function of , we differentiate through with respect to and set dd d = 0: 2 2 2 22 2 2 2 v v g dd d vi i icos cos cos sin sin cos cos cos sin sin + = a f a f . We use the trigonometric identities from Appendix B4 cos cos sin2 2 2 = and sin sin cos2 2 = to find cos cos sin sin 2 2= . Next, sin cos tan = and cot tan 2 1 2 = give cot tan tan2 90 2 = = a f so = 90 2 and = 45 2 . ANSWERS TO EVEN PROBLEMS P4.2 (a) r i j= + 18 0 4 00 4 90 2 . . .t t te j ; P4.8 (a) r i j= +5 00 1 50 2 . .t te jm; v i j= +5 00 3 00. . te j m s;(b) v i j= + 18 0 4 00 9 80. .. ta f ; (c) a j= 9 80. m s2 e j ; (b) r i j= +10 0 6 00. .e jm; 7 81. m s (d) 54 0 32 1. .m ma f a fi j ; (e) 18 0 25 4. .m s m sb g b gi j ; P4.10 7 23 10 1 68 103 3 . . m, me j(f) 9 80. m s2 e jj P4.12 (a) d g h2 horizontally; P4.4 (a) v i j= +5 00 0. e j m s; a i j= +0 5 00 2 . e j m s2 ; (b) tan F HG I KJ1 2h d below the horizontal (b) r j= 4 00. m + 5 00. sin cosm t ti je j; v i j= +5 00. cos sinm t te j; a i j= +5 00 2 . sin cosm t te j; P4.14 0 600. m s2 P4.16 (a) 76.0; (b) the same; (c) 17 8 d P4.18 25 8. m s(c) a circle of radius 5.00 m centered at 0 4 00, . ma f P4.20 d gd v i i i tan cos 2 2 2 2e jP4.6 (a) v j= 12 0. t m s; a j= 12 0. m s2 ; (b) r i j v j= = 3 00 6 00 12 0. . ; .e jm m s 114. Chapter 4 115 P4.22 33.5 below the horizontal P4.48 2v ti icos P4.24 (a) 0.852 s; (b) 3 29. m s; (c) 4.03 m s; P4.50 (a) see the solution; (b) i = +45 2 ; d v g i max sin cos = 2 2 1 b g(d) 50.8; (e) 1.12 s P4.26 tan F HG I KJ1 2gh v P4.52 (a) 1 69. km s; (b) 6 47 103 . s P4.54 10 7. m sP4.28 0 033 7 2 . m s toward the center of the Earth P4.56 R 2P4.30 0 281. rev s P4.58 7 50. m s in the direction the ball was thrown P4.32 7 58 103 . m s; 5 80 103 . s P4.34 (a) 0 600. m s2 forward; P4.60 (a) 19.6 cm; (b) 0 0561. ; (b) 0 800. m s2 inward; (c) aim low 3.68 cm; (d) aim low 3.68 cm; (c) 1 00. m s2 forward and 53.1 inward (e) aim high 6.12 cm; (f) aim low; (g) aim low P4.36 (a) see the solution; (b) 29 7. m s2 ; P4.62 (a) gR ; (b) 2 1e jR(c) 6 67. m s at 36.9 above the horizontal P4.38 (a) 26 9. m s; (b) 67 3. m; P4.64 18 8 17 3. .m; ma f (c) 2 00 5 00. .i je j m s2 P4.66 see the solution; ~102 m s2 P4.40 18.0 s P4.68 x D= P4.42 153 km h at 11.3 north of west P4.70 (a) at 90 to the bank; (b) 133 m; (c) upstream at 53.1 to the bank; (d) 107 m P4.44 (a) 10 1. m s2 at 14.3 south from the vertical; (b) 9 80. m s2 vertically downward P4.72 see the solution P4.46 27.7 east of north 115. 5 CHAPTER OUTLINE 5.1 The Concept of Force 5.2 Newtons First Law and Inertial Frames 5.3 Mass 5.4 Newtons Second Law 5.5 The Gravitational Force and Weight 5.6 Newtons Third Law 5.7 Some Applications of Newtons Laws 5.8 Forces of Friction The Laws of Motion ANSWERS TO QUESTIONS Q5.1 (a) The force due to gravity of the earth pulling down on the ballthe reaction force is the force due to gravity of the ball pulling up on the earth. The force of the hand pushing up on the ballreaction force is ball pushing down on the hand. (b) The only force acting on the ball in free-fall is the gravity due to the earth -the reaction force is the gravity due to the ball pulling on the earth. Q5.2 The resultant force is zero, as the acceleration is zero. Q5.3 Mistake one: The car might be momentarily at rest, in the process of (suddenly) reversing forward into backward motion. In this case, the forces on it add to a (large) backward resultant. Mistake two: There are no cars in interstellar space. If the car is remaining at rest, there are some large forces on it, including its weight and some force or forces of support. Mistake three: The statement reverses cause and effect, like a politician who thinks that his getting elected was the reason for people to vote for him. Q5.4 When the bus starts moving, the mass of Claudette is accelerated by the force of the back of the seat on her body. Clark is standing, however, and the only force on him is the friction between his shoes and the floor of the bus. Thus, when the bus starts moving, his feet start accelerating forward, but the rest of his body experiences almost no accelerating force (only that due to his being attached to his accelerating feet!). As a consequence, his body tends to stay almost at rest, according to Newtons first law, relative to the ground. Relative to Claudette, however, he is moving toward her and falls into her lap. (Both performers won Academy Awards.) Q5.5 First ask, Was the bus moving forward or backing up? If it was moving forward, the passenger is lying. A fast stop would make the suitcase fly toward the front of the bus, not toward the rear. If the bus was backing up at any reasonable speed, a sudden stop could not make a suitcase fly far. Fine her for malicious litigiousness. Q5.6 It would be smart for the explorer to gently push the rock back into the storage compartment. Newtons 3rd law states that the rock will apply the same size force on her that she applies on it. The harder she pushes on the rock, the larger her resulting acceleration. 117 116. 118 The Laws of Motion Q5.7 The molecules of the floor resist the ball on impact and push the ball back, upward. The actual force acting is due to the forces between molecules that allow the floor to keep its integrity and to prevent the ball from passing through. Notice that for a ball passing through a window, the molecular forces werent strong enough. Q5.8 While a football is in flight, the force of gravity and air resistance act on it. When a football is in the process of being kicked, the foot pushes forward on the ball and the ball pushes backward on the foot. At this time and while the ball is in flight, the Earth pulls down on the ball (gravity) and the ball pulls up on the Earth. The moving ball pushes forward on the air and the air backward on the ball. Q5.9 It is impossible to string a horizontal cable without its sagging a bit. Since the cable has a mass, gravity pulls it downward. A vertical component of the tension must balance the weight for the cable to be in equilibrium. If the cable were completely horizontal, then there would be no vertical component of the tension to balance the weight. Some physics teachers demonstrate this by asking a beefy student to pull on the ends of a cord supporting a can of soup at its center. Some get two burly young men to pull on opposite ends of a strong rope, while the smallest person in class gleefully mashes the center of the rope down to the table. Point out the beauty of sagging suspension-bridge cables. With a laser and an optical lever, demonstrate that the mayor makes the courtroom table sag when he sits on it, and the judge bends the bench. Give them I make the floor sag buttons, available to instructors using this manual. Estimate the cost of an infinitely strong cable, and the truth will always win. Q5.10 As the barbell goes through the bottom of a cycle, the lifter exerts an upward force on it, and the scale reads the larger upward force that the floor exerts on them together. Around the top of the weights motion, the scale reads less than average. If the iron is moving upward, the lifter can declare that she has thrown it, just by letting go of it for a moment, so our answer applies also to this case. Q5.11 As the sand leaks out, the acceleration increases. With the same driving force, a decrease in the mass causes an increase in the acceleration. Q5.12 As the rocket takes off, it burns fuel, pushing the gases from the combustion out the back of the rocket. Since the gases have mass, the total remaining mass of the rocket, fuel, and oxidizer decreases. With a constant thrust, a decrease in the mass results in an increasing acceleration. Q5.13 The friction of the road pushing on the tires of a car causes an automobile to move. The push of the air on the propeller moves the airplane. The push of the water on the oars causes the rowboat to move. Q5.14 As a man takes a step, the action is the force his foot exerts on the Earth; the reaction is the force of the Earth on his foot. In the second case, the action is the force exerted on the girls back by the snowball; the reaction is the force exerted on the snowball by the girls back. The third action is the force of the glove on the ball; the reaction is the force of the ball on the glove. The fourth action is the force exerted on the window by the air molecules; the reaction is the force on the air molecules exerted by the window. We could in each case interchange the terms action and reaction. Q5.15 The tension in the rope must be 9 200 N. Since the rope is moving at a constant speed, then the resultant force on it must be zero. The 49ers are pulling with a force of 9 200 N. If the 49ers were winning with the rope steadily moving in their direction or if the contest was even, then the tension would still be 9 200 N. In all of these case, the acceleration is zero, and so must be the resultant force on the rope. To win the tug-of-war, a team must exert a larger force on the ground than their opponents do. 117. Chapter 5 119 Q5.16 The tension in the rope when pulling the car is twice that in the tug-of-war. One could consider the car as behaving like another team of twenty more people. Q5.17 This statement contradicts Newtons 3rd law. The force that the locomotive exerted on the wall is the same as that exerted by the wall on the locomotive. The wall temporarily exerted on the locomotive a force greater than the force that the wall could exert without breaking. Q5.18 The sack of sand moves up with the athlete, regardless of how quickly the athlete climbs. Since the athlete and the sack of sand have the same weight, the acceleration of the system must be zero. Q5.19 The resultant force doesnt always add to zero. If it did, nothing could ever accelerate. If we choose a single object as our system, action and reaction forces can never add to zero, as they act on different objects. Q5.20 An object cannot exert a force on itself. If it could, then objects would be able to accelerate themselves, without interacting with the environment. You cannot lift yourself by tugging on your bootstraps. Q5.21 To get the box to slide, you must push harder than the maximum static frictional force. Once the box is moving, you need to push with a force equal to the kinetic frictional force to maintain the boxs motion. Q5.22 The stopping distance will be the same if the mass of the truck is doubled. The stopping distance will decrease by a factor of four if the initial speed is cut in half. Q5.23 If you slam on the brakes, your tires will skid on the road. The force of kinetic friction between the tires and the road is less than the maximum static friction force. Anti-lock brakes work by pumping the brakes (much more rapidly that you can) to minimize skidding of the tires on the road. Q5.24 With friction, it takes longer to come down than to go up. On the way up, the frictional force and the component of the weight down the plane are in the same direction, giving a large acceleration. On the way down, the forces are in opposite directions, giving a relatively smaller acceleration. If the incline is frictionless, it takes the same amount of time to go up as it does to come down. Q5.25 (a) The force of static friction between the crate and the bed of the truck causes the crate to accelerate. Note that the friction force on the crate is in the direction of its motion relative to the ground (but opposite to the direction of possible sliding motion of the crate relative to the truck bed). (b) It is most likely that the crate would slide forward relative to the bed of the truck. Q5.26 In Question 25, part (a) is an example of such a situation. Any situation in which friction is the force that accelerates an object from rest is an example. As you pull away from a stop light, friction is the force that accelerates forward a box of tissues on the level floor of the car. At the same time, friction of the ground on the tires of the car accelerates the car forward. 118. 120 The Laws of Motion SOLUTIONS TO PROBLEMS The following problems cover Sections 5.15.6. Section 5.1 The Concept of Force Section 5.2 Newtons First Law and Inertial Frames Section 5.3 Mass Section 5.4 Newtons Second Law Section 5.5 The Gravitational Force and Weight Section 5.6 Newtons Third Law P5.1 For the same force F, acting on different masses F m a= 1 1 and F m a= 2 2 (a) m m a a 1 2 2 1 1 3 = = (b) F m m a m a m= + = =1 2 1 14 3 00a f c h. m s2 a = 0 750. m s2 *P5.2 v f = 880 m s, m = 25 8. kg , x f = 6 m v ax x F m f f f 2 2 2= = F HG I KJ F mv x f f = = 2 6 2 1 66 10. N forward P5.3 m m = = + = = + = + = 3 00 2 00 5 00 6 00 15 0 6 00 15 0 16 2 2 2 . . . . . . . . kg m s N N N 2 a i j F a i j F e j e j a f a f 119. Chapter 5 121 P5.4 F mgg = =weight of ball v vrelease = and time to accelerate = t : a i= = = v t v t v t (a) Distance x vt= : x v t vt = F HG I KJ = 2 2 (b) F j ip g g F F v gt = F i jp g g F v gt F= + P5.5 m = 4 00. kg, v ii = 3 00. m s , v i j8 8 00 10 0= +. .e j m s, t = 8 00. s a v i j = = + t 5 00 10 0 8 00 . . . m s2 F a i j= = +m 2 50 5 00. .e jN F = ( ) +( ) =2 50 5 00 5 59 2 2 . . . N P5.6 (a) Let the x-axis be in the original direction of the molecules motion. v v at a a f i= + = + = : . . 670 670 3 00 10 4 47 10 13 15 m s m s s m s2 e j (b) For the molecule, F a= m . Its weight is negligible. F F wall on molecule 2 molecule on wall kg m s N N = = = + 4 68 10 4 47 10 2 09 10 2 09 10 26 15 10 10 . . . . e j 120. 122 The Laws of Motion P5.7 (a) F ma = and v v axf i f 2 2 2= + or a v v x f i f = 2 2 2 . Therefore, F m v v x F f i f = = L NM O QP = 2 2 31 5 2 5 2 18 2 9 11 10 7 00 10 3 00 10 2 0 050 0 3 64 10 e j e j e j b g. . . . . .kg m s m s m N 2 2 (b) The weight of the electron is F mgg = = = 9 11 10 9 80 8 93 1031 30 . . .kg m s N2 c hc h The accelerating force is 4 08 1011 . times the weight of the electron. P5.8 (a) F mgg = = = ( )=120 4 448 120 534lb N lb lb N.a f (b) m F g g = = = 534 54 5 N 9.80 m s kg2 . P5.9 F mgg = = 900 N , m = = 900 91 8 N 9.80 m s kg2 . Fgc h c hon Jupiter 2 kg m s kN= =91 8 25 9 2 38. . . P5.10 Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just canceled out by a glass of tomato juice. By subtraction, F mgg p pc h = and F mgg C Cc h = give F m g gg p C= c h. For a person whose mass is 88.7 kg, the change in weight is Fg = =88 7 9 809 5 9 780 8 2 55. . . .kg Nb g . A precise balance scale, as in a doctors office, reads the same in different locations because it compares you with the standard masses on its beams. A typical bathroom scale is not precise enough to reveal this difference. 121. Chapter 5 123 P5.11 (a) F F F i j = + = +1 2 20 0 15 0. .e jN F a i j a a i j = + = = + m : . . . . . 20 0 15 0 5 00 4 00 3 00e j m s2 or a = = 5 00 36 9. .m s at2 (b) F F m x y 2 2 2 1 2 15 0 60 0 7 50 15 0 60 0 13 0 7 50 13 0 27 5 13 0 5 00 5 50 2 60 6 08 = = = = = + = + = + = = = + = . cos . . . sin . . . . . . . . . . N N N N m s m s at 25.32 2 F i j F F F i j a a a i j e j e j e j FIG. P5.11 P5.12 We find acceleration: r r v a i j a a a i j f i it t = + = = 1 2 4 20 1 20 0 720 5 83 4 58 2 2 . . . . . . m 3.30 m =0+ 1 2 s s m s 2 2 a f e j Now F a = m becomes F F a F i j j F i j g m+ = = + = + 2 2 2 2 80 2 80 9 80 16 3 14 6 . . . . . . kg 5.83 4.58 m s kg m s N 2 2 e j b ge j e j P5.13 (a) You and the earth exert equal forces on each other: m g M ay e e= . If your mass is 70.0 kg, ae = = 70 0 5 98 10 1024 22 . . ~ kg 9.80 m s kg m s 2 2 a fc h . (b) You and the planet move for equal times intervals according to x at= 1 2 2 . If the seat is 50.0 cm high, 2 2x a x a y y e e = x a a x m m xe e y y y e y= = = 70 0 0 500 5 98 10 1024 23. . . ~ kg m kg m a f . 122. 124 The Laws of Motion P5.14 F a = m reads + + =2 00 2 00 5 00 3 00 45 0 3 75. . . . . .i j i j i ae j e jN m s2 m where a represents the direction of a =42 0 1 00 3 75. . .i j ae j e jN m s2 m F = ( ) +( )42 0 1 00 2 2 . . N at tan . . F HG I KJ1 1 00 42 0 below the x-axis F a = =42 0 3 75. .N at 181 m s2 mc h . For the vectors to be equal, their magnitudes and their directions must be equal. (a) a is at 181 counterclockwise from the x-axis (b) m = = 42 0 11 2 . . N 3.75 m s kg2 (d) v v af i t= + = + 0 3 75 10 0. .m s at 181 s2 e j so v f = 37 5. m s at 181 v i jf = + 37 5 181 37 5 181. cos . sinm s m s so v i jf = 37 5 0 893. .e j m s (c) v f = + =37 5 0 893 37 52 2 . . .m s m s P5.15 (a) 15 0. lb up (b) 5 00. lb up (c) 0 Section 5.7 Some Applications of Newtons Laws P5.16 v dx dt tx = =10 , v dy dt ty = = 9 2 a dv dt x x = =10, a dv dt ty y = =18 At t = 2 00. s, a ax y= =10 0 36 0. , .m s m s2 2 F max x = : 3 00 10 0 30 0. . .kg m s N2 e j= F may y = : 3 00 36 0 108. .kg m s N2 e j= F F Fx y = + =2 2 112 N 123. Chapter 5 125 P5.17 m mg = = = = 1 00 9 80 0 200 0 458 . . tan . . kg N m 25.0 m Balance forces, 2 9 80 2 613 T mg T sin . sin = = = N N 50.0 m 0.200 m mg TT FIG. P5.17 P5.18 T Fg3 = (1) T T Fg1 1 2 2sin sin + = (2) T T1 1 2 2cos cos = (3) Eliminate T2 and solve for T1 T F F T F T F T T g g g g 1 2 1 2 1 2 2 1 2 3 1 2 1 1 2 325 25 0 85 0 296 296 60 0 25 0 163 = + = + = = = F HG I KJ = = F HG I KJ = F HG I KJ = cos sin cos cos sin cos sin cos . sin . cos cos cos . cos . b g b g N N N N 1 1 2 F 2 T3 T2 T1 g FIG. P5.18 P5.19 See the solution for T1in Problem 5.18. 124. 126 The Laws of Motion P5.20 (a) An explanation proceeding from fundamental physical principles will be best for the parents and for you. Consider forces on the bit of string touching the weight hanger as shown in the free-body diagram: Horizontal Forces: F max x = : + =T Tx cos 0 Vertical Forces: F may y = : + =F Tg sin 0 FIG. P5.20 You need only the equation for the vertical forces to find that the tension in the string is given by T Fg = sin . The force the child feels gets smaller, changing from T to T cos , while the counterweight hangs on the string. On the other hand, the kite does not notice what you are doing and the tension in the main part of the string stays constant. You do not need a level, since you learned in physics lab to sight to a horizontal line in a building. Share with the parents your estimate of the experimental uncertainty, which you make by thinking critically about the measurement, by repeating trials, practicing in advance and looking for variations and improvements in technique, including using other observers. You will then be glad to have the parents themselves repeat your measurements. (b) T Fg = = = sin . . sin . . 0 132 9 80 46 3 1 79 kg m s N 2 e j P5.21 (a) Isolate either mass T mg ma T mg + = = = 0 . The scale reads the tension T, so T mg= = =5 00 9 80 49 0. . .kg m s N2 e j . (b) Isolate the pulley T T2 1 2 1 2 0 2 2 98 0 + = = = =T T mg . .N (c) F n T g = + + =m 0 Take the component along the incline n T gx x xm+ + = 0 or 0 30 0 0 30 0 2 5 00 9 80 2 24 5 + = = = = = T mg T mg mg sin . sin . . . . . a f N FIG. P5.21(a) FIG. P5.21(b) FIG. P5.21(c) 125. Chapter 5 127 P5.22 The two forces acting on the block are the normal force, n, and the weight, mg. If the block is considered to be a point mass and the x- axis is chosen to be parallel to the plane, then the free body diagram will be as shown in the figure to the right. The angle is the angle of inclination of the plane. Applying Newtons second law for the accelerating system (and taking the direction up the plane as the positive x direction) we have F n mgy = =cos 0: n mg= cos F mg max = =sin : a g= sin FIG. P5.22 (a) When = 15 0. a = 2 54. m s2 (b) Starting from rest v v a x x ax v ax f i f i f f f 2 2 2 2 2 2 2 54 2 00 3 18 = + = = = = d i e ja f. . .m s m m s2 P5.23 Choose a coordinate system with i East and j North. F a = =m 1 00 10 0. .kg m s2 e jat 30 0. 5 00 10 0 30 0 5 00 8 661. . . . .N N N Na f a f a f a fj F j i+ = = + = ( )F1 8 66. N East FIG. P5.23 *P5.24 First, consider the block moving along the horizontal. The only force in the direction of movement is T. Thus, F max = T a= 5 kga f (1) Next consider the block that moves vertically. The forces on it are the tension T and its weight, 88.2 N. We have F may = 88 2 9. N kg =T aa f (2) 5 kg n T 49 N +x 9 kg T +y Fg= 88.2 N FIG. P5.24 Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be added to give 88 2 14. N kg= b ga. Then a T= =6 30 31 5. .m s and N2 . 126. 128 The Laws of Motion P5.25 After it leaves your hand, the blocks speed changes only because of one component of its weight: F ma mg ma v v a x x x x f i f i = = = + sin . . 20 0 22 2 d i Taking v f = 0, vi = 5 00. m s, and a g= sin .20 0a f gives 0 5 00 2 9 80 20 0 0 2 =( ) ( ) . . sin .a fc hx f or x f = ( ) = 25 0 2 9 80 20 0 3 73 . . sin . . a f m . FIG. P5.25 P5.26 m1 2 00= . kg , m2 6 00= . kg, = 55 0. (a) F m g T m ax = =2 2sin and T m g m a a m g m g m m = = + = 1 1 2 1 1 2 3 57 sin . m s2 (b) T m a g= + =1 26 7a f . N FIG. P5.26 (c) Since vi = 0 , v atf = = ( )=3 57 2 00 7 14. . .m s s m s2 c h . *P5.27 We assume the vertical bar is in compression, pushing up on the pin with force A, and the tilted bar is in tension, exerting force B on the pin at 50 . F B B F A A x y = + = = = + = = 0 2 500 30 50 0 3 37 10 0 2 500 30 3 37 10 50 0 3 83 10 3 3 3 : cos cos . : sin . sin . N N N N N Positive answers confirm that B Ais in tension and is in compression. 30 50 A B 2 500 N A 2 500 N cos30 cos50 2 500 N sin30 B sin50B FIG. P5.27 127. Chapter 5 129 P5.28 First, consider the 3.00 kg rising mass. The forces on it are the tension, T, and its weight, 29.4 N. With the upward direction as positive, the second law becomes F may y = : T a =29 4 3 00. .N kga f (1) The forces on the falling 5.00 kg mass are its weight and T, and its acceleration is the same as that of the rising mass. Calling the positive direction down for this mass, we have F may y = : 49 5 00N kg =T a.a f (2) FIG. P5.28 Equations (1) and (2) can be solved simultaneously by adding them: T T a a + = +29 4 49 0 3 00 5 00. . . .N N kg kga f a f (b) This gives the acceleration as a = = 19 6 2 45 . . N 8.00 kg m s2 . (a) Then T = =29 4 3 00 2 45 7 35. . . .N kg m s N2 a fc h . The tension is T = 36 8. N . (c) Consider either mass. We have y v t ati= + = + ( ) = 1 2 0 1 2 2 45 1 00 1 232 2 . . .m s s m2 c h . *P5.29 As the man rises steadily the pulley turns steadily and the tension in the rope is the same on both sides of the pulley. Choose man-pulley- and-platform as the system: F ma T T y y = + = = 950 0 950 N N. The worker must pull on the rope with force 950 N . T 950 N FIG. P5.29 128. 130 The Laws of Motion *P5.30 Both blocks move with acceleration a m m m m g= + F HG I KJ2 1 2 1 : a = + F HG I KJ = 7 2 9 8 5 44 kg kg 7 kg 2 kg m s m s2 2 . . . (a) Take the upward direction as positive for m1. v v a x x x x x xf xi x f i f f f 2 2 2 2 2 0 2 4 2 5 44 0 5 76 2 5 44 0 529 0 529 = + = + = = = d i b g e jd i e j : . . . . . . m s m s m s m s m m below its initial level 2 2 2 (b) v v a t v v xf xi x xf xf = + = + = : . . . . 2 40 5 44 1 80 7 40 m s m s s m s upward 2 e ja f P5.31 Forces acting on 2.00 kg block: T m g m a =1 1 (1) Forces acting on 8.00 kg block: F T m ax = 2 (2) (a) Eliminate T and solve for a: a F m g m m x = + 1 1 2 a F m gx> > =0 19 61for N. . (b) Eliminate a and solve for T: T m m m F m gx= + +1 1 2 2a f T F m gx= =0 78 42for N. . FIG. P5.31 (c) Fx , N 100 78.4 50.0 0 50.0 100 ax , m s2 12.5 9.80 6.96 1.96 3.04 8.04 129. Chapter 5 131 *P5.32 (a) For force components along the incline, with the upward direction taken as positive, F ma mg ma a g x x x x = = = = = : sin sin . sin . . 9 8 35 5 62m s m s2 2 e j For the upward motion, v v a x x x x xf xi x f i f f 2 2 2 2 0 5 2 5 62 0 25 2 5 62 2 22 = + = + = = d i b g e jd i e j m s m s m s m s m 2 2 2 2 . . . . (b) The time t