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# University Physics I PHYS 1321

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This 41 page Class Notes was uploaded by Kayley Corwin on Saturday September 19, 2015. The Class Notes belongs to PHYS 1321 at University of Houston taught by Pei-Herng Hor in Fall. Since its upload, it has received 83 views. For similar materials see /class/208341/phys-1321-university-of-houston in Physics 2 at University of Houston.

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Date Created: 09/19/15

Chapter 4 Concept question 4 You are to launch a rocket from just above the ground with one of the following initial velocity v n Elli l 21 l Elli TUj 3153 ZUi 39quotl39j 4 vectors 1 v D Ul YUJ39 In your coordinate system x runs along level ground andy increases upward a Rank the vectors according to the launch speed of the projectile greatest first b Rank the vectors according to the time of ight of the projectile greatest first Solution aVector1 1 Vector2 1 Vector3 1 Vector4 1 bVector1 Vector2 1 Vector3 3 Vector4 3 Chapter 4 Concept question 6 The only good use of a fruitcake is in catapult practice Curve 1 in Fig 424 gives the height yof a catapulted fruitcake versus the angle L9 between its velocity vector and its acceleration vector during flight a Which of the lettered points on that curve corresponds to the landing of the fruitcake on the ground b Curve 2 is a similar plot for the same launch speed but for a different launch angle Does the fruitcake now land farther away or closer to the launch point Y Solution aA bCloser Hint The midpoint of AB corresponds to S90 a When 645 at the landing point the fruitcake lands at the farthest place Chapter 4 Concept question 13 a Is it possible to be accelerating while traveling at constant speed Is it possible to round a curve with b zero acceleration and c a constant magnitude of acceleration Solution aYes bNo c Yes Chapter 4 Problem 9 The figure gives the path of a squirrel moving about on level ground from point A at time t O to points B at t 5 min C at t 10 min and finally D att 15 min Consider the average velocities of the squirrel from point A to each of the other three points Of them what are the a magnitude and b angle of the one with the least magnitude and c the magnitude and d angle of the one with the greatest magnitude 39 l in Si Solution The xy coordinates in meters of the points are A 15 15 B 30 45 C 20 15 andD 45 45 The respective times are IA 0 t3 300 s to 600 s and ID 900 s Average velocity is de ned by Eq 48 Each displacement A7 is understood to originate at pointA a The average velocity having the least magnitude 50 m600 s is for the displacement ending at point C l 17a l 00083 ms VS b The direction of 17an is 0 measured counterclockwise from the x axis c The average velocity having the greatest magnitude 15 m2 30 m2 300 s is for the displacement ending at point B l 17 l 0 11 ms avg d The direction of 1an is 297 counterclockwise from x or 63 which is equivalent to measuring 63 clockwise from the x axis Chapter 4 Problem 21 A dart is thrown horizontally with an initial speed of 10 ms toward point P the bull39seye on a dart board It hits at point Q on the rim vertically below P 019 s later a What is the distance PQ b How far away from the dart board is the dart released Solution We adopt the positive direction choices used in the textbook so that equations such as Eq 422 are directly applicable The initial velocity is horizontal so that v0y 0 and V 0X v0 10 ms a With the origin at the initial point where the dart leaves the thrower s hand the y coordinate of the dart is given by y gt2 so that with y iPQ we have PQ 98 ms20l9 s2 018m b Fromx vot we obtainx 10 ms0 19 s 19 In Chapter 4 Problem 37 A lowly high diver pushes off horizontally with a speed of 200 ms from the platform edge 100 m above the surface of the water a At what horizontal distance from the edge is the diver 0800 s after pushing off b At what vertical distance above the surface ofthe water is the diverjust then c At what horizontal distance from the edge does the diver strike the water Solution The initial velocity has no vertical component 60 0 7 only an x component Eqs 4 21 and 422 can be simpli ed to x x0 VOxt 2 v t 1 t2 l t y yo 0y 2g 2g where x0 0 v0x 2120 20 ms and yo 100 m taking the water surface to be at y 0 The setup of the problem is shown in the gure below y A v0 water surface x R 0 a At t 080 s the horizontal distance of the diver from the edge is 05110 x x0 v0xt O20ms080 s 2160 m b Similarly using the second equation for the vertical motion we obtain y y0 gt2 100 m 980ms2080 s2 686 m c At the instant the diver strikes the water surface y 0 Solving for t using the equationy y0 gt2 0 leads to t l 200390 m3 143 s g 980 ms During this time the xdisplacement ofthe diver is R x 200 Wsl 43 s 286 m Note Using Eq 425 with 60 0 the trajectory of the diver can also be written as Z 8 2v Part c can also be solved by using this equation 2 22 2 yy00 3xR v0y0 220ms 10m286m 2v0 g 98ms Chapter 4 Problem 45 yzyo In Fig 440 a ball is launched with a velocity of magnitude 100 ms at an angle of 500 to the hotizontal The launch point is at the base of a ramp of hotizontal length 11 Z 5 Uquot m and height d2 360 m A plateau is located at the top of the ramp a Does the ball land on the ramp or the plateau When it lands what are the b magnitude and c angle of its displacement from the launch point Figure 440 Problem 45 Solution d a Let m 72 0600 be the slope of the ramp so y mx there We choose our coordinate 1 origin at the point of launch and use Eq 425 Thus 980 ms2x2 0600 2100 ms2c0s500 2 x y tan500 x which yields x 499 m This is less than d1 so the ball does land on the ramp b Using the value of x found in part a we obtain y mx 299 m Thus the Pythagorean theorem yields a displacement magnitude ofxlx2 y2 582 m c The angle is of course the angle of the ramp tan 1m 3109 Chapter 4 Problem 50 Two seconds after being projected from ground level a projectile is displaced 40 m horizontally and 53 m vertically above its launch point What are the a horizontal and b vertical components of the initial velocity of the projectile c At the instant the projectile achieves its maximum height above ground level how far is it displaced horizontally from the launch point Solution We apply Eq 421 Eq 422 and Eq 423 a From Ax v0xt we nd v0 40 m2s 20 ms b FromAyv0yt gt2 we nd voy 53 m98 ms22 s22 36ms c From vy v0y gt with vy 0 as the condition for maximum height we obtain t39 36 ms 98 msz 37 s During that time the xmotion is constant so 2639 x0 20 ms37 s 74 In Chapter 4 Problem 59 A woman rides a carnival Ferris wheel at radius 15 m completing five turns about its horizontal axis every minute What are a the period of the motion the b magnitude and c direction of her centripetal acceleration at the highest point and the d magnitude and e direction of her centripetal acceleration at the lowest point Solution a Since the wheel completes 5 turns each minute its period is one fth of a minute or 2 s b The magnitude of the centripetal acceleration is given by a vzR where R is the radius of the wheel and v is the speed of the passenger Since the passenger goes a distance 27rR for each revolution his speed is 122115111 v Sgtu1 s 125 and his centripetal acceleration is 785111 5 s a 41m 539 15111 c When the passenger is at the highest point his centripetal acceleration is downward toward the center of the orbit d At the lowest oint the centri etal acceleration is a 41 msz same as art b P P P e The direction is up toward the center of the orbit Chapter 4 Problem 66 A particle moves along a circular path over a horizontal xy coordinate system at constant speed At time t1400s it is at point 500 m 600 m with velocity 300 ms and acceleration in the positive x direction At time t2100s it has velocity 300 ms i and acceleration in the positive y direction What are the a x and b ycoordinates of the center of the circular path if tZtl is less than one period Solution The fact that the velocity is in the y direction and the acceleration is in the x direction at t1 400 s implies that the motion is clockwise The position corresponds to the 900 position On the other hand the position at t2 100 s is in the 600 position since the velocity points in the x direction and the acceleration is in the y direction The time interval At 100 s 400 s 600 s is equal to 34 ofaperiod 600sT 3 T800 s Equation 435 then yields r E 300 ms800 s 27239 27239 382 m a The x coordinate ofthe center ofthe circular path is x 500 m382 m 882 m b The y coordinate of the center of the circular path is y 600 m In other words the center ofthe circle is at xy 882 m 600 m Chapter 4 Problem 79 Two ships A and B leave port at the same time Ship A travels northwest at 24 knots and ship B travels at 28 knots in a direction 40 west of south 1 knot 1 nautical mile per hour see Appendix D What are the a magnitude and b direction of the velocity of ship A relative to B c After what time will the ships be 160 nautical miles apart d What will be the bearing of B the direction of B39s position relative toA at that time Solution Given that EA 45 and SE 40 as de ned in the gure the velocity vectors relative to the shore for ships A andB are given by y north 1 VA c0345 ilr1 si1145 i A VB 1395 sin10 i lvs cos10 with VA 24 knots and VB 28 knots We take 4 east as and north as 1 l The velocity of shipA relative to ship B is simply given by 17 17A 175 a The relative velocity is B 17 i B 1393 sin 40 395 cos 45 li E cos 40 71d sin 15 VJ 103 knotslil 384 knots lj the magnitude of which is if l ll03 knots2 384 knots2 384 knots or 38 knots in 2 signi cant gures b The angle 639 that 17A B makes with north is given by 615 tan 1vAB39X J tan 1 13903kn0tsj 15 VAByy 384 knots which is to say that 17A B points 150 east of north c Since the two ships started at the same time their relative velocity describes at what rate the distance between them is increasing Because the rate is steady we have l Ar l 160 nautical miles 384 knots 42 h d The velocity 17 B does not change with time in this problem and 7A B is in the same direction as 17 B since they started at the same time Reversing the points of view we have 17 17M so that 7 A ie they are 1800 opposite to each other Hence we conclude that B stays at a bearing of 150 west of south relative to A during the journey neglecting the curvature of Earth Note The relative velocity is depicted in the gure below When analyzing relative motion in two dimensions a vector diagram such as the one shown can be very helpful y no h Chapter 4 Problem 83 A woman who can row a boat at 64 kmh in still water faces a long straight river with a width of 64 km and a current of 32 kmh Letipoint directly across the river and 1 point directly downstream If she rows in a straight line to a point directly opposite her starting position a at what angle toimust she point the boat and b how long will she take c How long will she take if instead she rows 32 km down the river and then back to her starting point d How long if she rows 32 km up the river and then back to her starting point e At what angle toi should she point the boat if she wants to cross the river in the shortest possible time f How long is that shortest time Solution We establish coordinates with pointing to the far side of the river perpendicular to the current and pointing in the direction of the current We are told that the magnitude presumed constant of the velocity of the boat relative to the water is l 17W l 64 kmh 9 0 1 The above expression leads to t 54 lnno4 kmh 10 h or 60 min Chapter 4 Problem 101 In the figure a ball is shot directly upward from the ground with an initial speed of v0 700 ms Simultaneously a construction elevator cab begins to move upward from the ground with a constant speed of vC 300 ms What maximum height does the ball reach relative to a the ground and b the cab floor At what rate does the speed of the ball change relative to c the ground and d the cab floor Give the magnitude of the rate of change Solution Using Eq 216 we obtain v2 v 2gh or h v02 v22g a Since v0at the maximum height of an upward motion with v0 700 ms we have h700 ms22980 ms2250 m b The relative speed is v v0 vC 700 ms 300 ms 400 ms with respect to the oor Using the above equation we obtain h 400 ms2 2980 msz 082 m c The acceleration or the rate of change of speed of the ball with respect to the ground is 980 ms2 downward d Since the elevator cab moves at constant velocity the rate of change of speed of the ball with respect to the cab oor is also 980 ms2 downward Chapter 3 Concept Question 2 The two vectors shown in Fig 321 lie in an xy plane What are the signs of the x and y components respectively of a EZ b CIT Z c 32 61 l x positive y positive 2 x positive y negative 3 x negative y positive 4 gtlt negative y negative Solution 3 3 bl4lcl 1 Chapter 3 Concept Question 5 Which of the arrangements of axes in Fig 323 can be labeled righthanded coordinate system As usual each axis label indicates the positive side of the axis x 139 z 3 z 0 b I x z x z y J 2 X J I P f Solution abcdf Chapter 3 Concept Question 9 If F q gtlt and G is perpendicular to E then what is the direction of E in the four situations shown in Fig 324 6 H 2 i ii a situation 1 q positive b situation 1 q negative c situation 2 q positive d situation 2 q negative e situation 3 q positive f situation 3 q negative Solution ax bX cz dz ez fz Chapter 3 Problem 11 a In unitvector notation what is the sum of a 40 111 30 111 and E 130 111 70 111 What are b the magnitude and c the direction of EE relative to i Solution a The x and the y components of 7 are rx ax bx 40 m 713 m 790 m and ry ay by 30 m 70 m 10 m respectively Thus quot 9 10 b The magnitude of 7 is r i fi lr 12 1l 90 m2 10 m2 13 m c The angle between the resultant and the x axis is given by 0tan 1 ri tan 1 48 or 132 r 90m Since the x component of the resultant is negative and the y component is positive characteristic of the second quadrant we nd the angle is 1320 measured counterclockwise from x axis The addition of the two vectors is depicted in the gure below not to scale Indeed we expect 7 to be in the second quadrant Chapter 3 Problem 25 Oasis B is a distance d 25 km east of oasis A along the X axis shown in the figure A confused camel intending to walk directly from A to B instead walks a distance W1 24 km west of due south by angle 91 150 It then walks a distance W2 8 km due north If it is to then walk directly to B a how far in km and b in what direction should it walk relative to the positive direction of the X axis l K r 1 Solution a The strategy is to nd where the camel is C by adding the two consecutive displacements described in the problem and then nding the difference between that a location and the oasis B Using the magnitudeangle notation 6 244 90W404 11225 1 6 254W7142594 which is efficiently implemented using a vectorcapable calculator in polar mode The distance is therefore 3471 km The direction is 25940 north of due east S0 Chapter 3 Problem 32 In the figure a vector at with a magnitude of 170 m is directed at angle 9 56 counterclockwise from the gtlt axis What are the components a ax and b ay of the vector A second coordinate system is inclined by angle 9 18 with respect to the first What are the components c a X and d a y in this primed coordinate system Solution a With a 170 m and 639 560 we nd ax a cos 639 951m b Similarly ay a sin 639 141 m c The angle relative to the new coordinate system is 639 39 560 7 180 380 Thus a acosH39 134 m d Similarly a a sin 6quot 105 In Chapter 3 Problem 43 The three vectors in the figure have magnitudes a 300 m b 400 m and c 100 m and angle 9 300 What are a the X component and b the y component of a c the X component and d the y component of b and e the X component and f the y component of Z If E pZ1qb what are the values of g p and hq Solution From the gure we note that E J 5 which implies that the angle between E and the x axis is 6 90 In unitvector notation the three vectors can be written as Z aj E bibjbcos 6lbsin 6 2 01 Cy c cos690W0 1 The above expressions allow us to evaluate the components of the vectors a The x component of 21 is ac a cos 0 a 300 m b Similarly the ycomponnet of 21 is ay a sin 0 0 c The x component of b is bx b cos 30 400 m cos 30 346 m d and the ycomponent is by b sin 30 400 m sin 30 200 m e The x component of E is C 0 cos 120 100 m cos 120 7500 m f and the ycomponent is cy 0 sin 30 100 m sin 120 866 m g The fact that E pg q implies E c 9 1202 qb 12 pa qbf qbyi or a Pa qb 0y qby Substituting the values found above we have 500mp300mq346 m 866 m q 200 m Solving these equations we ndp 467 h Similarly q 433 note that it s easiest to solve for q rst The numbers p and q have no units Chapter 3 Problem 44 In the cross or vector product I quE we know that q 2 F 40i 20oj12012 G 20406012 E BXEByjBZflt In I the X and y components have the same value What is the value of BZ Solution Applying Eq 323 F qt X E where q is a scalar becomes in E13E qvsz szy i qvB vB j qvBy vyB 12 which 7 plugging in values 7 leads to three equalities 4024OBZ 6053 202603 203 122203y 403 Since we are told that Bx By the third equation leads to By 730 Inserting this value into the rst equation we nd E 30i 303 40h Thus our answer is B 740 Chapter 3 Problem 51 Rock faults are ruptures along which opposite faces of rock have slid past each other In the figure points A and B coincided before the rock in the foreground slid down to the right The net displacement Z3 is along the plane of the fault The horizontal component of is the strike slip AC The component of AB that is directly down the plane of the fault is the dip slip AD a What is the magnitude of the net displacement AB if the strikeslip is 220 m and the dipslip is 170 m b If the plane of the fault is inclined at angle q 52 to the horizontal what is the vertical component of AB Sirikhsli gt t i l39zmli illdlll39 Solution Although we think of this as a threedimensional movement it is rendered effectively twodimensional by referring measurements to its welldefined plane of the fault a The magnitude of the net displacement is lfBl lADl2 lACl2 J170 m2 220 m2 278m a b The magnitude of the vertical component of AB is lADl sin 520O 134 m Chapter 3 Problem 62 A golfer takes three putts to get the ball into the hole The first putt displaces the ball 366 m north the second 183 m southeast and the third 091 m southwest What area the magnitude and b the angle between the direction of the displacement needed to get the ball into the hole on the first putt and the direction due east Solution We choose x east and y north and measure all angles in the standard way positive ones counterclockwise from x negative ones clockwise Thus vector all has magnitude d1 366 with the unit meter and three significant figures assumed and direction 61 90 Also 512 has magnitude d2 183 and direction 62 45 and vector g1 has magnitude d3 091 and direction 63 135 We add the x and ycomponents respectively x d1 cos 01dzcos 02d3cos 03 065 m y all sin 01d2 sin 02d3 sin 0317 m a The magnitude of the direct displacement the vector sum 62 1 3 is 065 m2 17 m2 18 m b The angle understood in the sense described above is tan 1 17065 69 That is the first putt must aim in the direction 69 north of east Chapter 3 Problem 63 Here are three vectors in meters 1 30i30201amp 1 20i 4oj 201 What results from a dll 39 3l b dllldesl 1417 LivJ1 and 13 c d and e forl j and k components respectively Give your answers in standard SI units Solution 3 b C Since 31 173 Olil0i30l we have J M 53 30303201 Joi 1oj301 i 0 3060 30 111 Using Eq 330 we obtain 3 y 73 10l60i 201 Thus E if 73 30i30j201 10i60201 301s40 52 1113 We found 73 111 part a Use of Eq 330 then leads to 31Ma a 30i30j201yroi 1oj301 11 9Uj301Linz So i component is 11ml d component is 90m2 e 1 component is 30m2 Chapter 3 Problem 64 Consider two displacements one of magnitude 30 m and another of magnitude 40 m What angle between the directions of this two displacements give a resultant displacement of magnitude a 70 m b 10 m and c 50 m Solution a The vectors should be parallel to achieve a resultant 7 m long the unprimed case shown below b antiparallel in opposite directions to achieve a resultant l m long primed case shown c and perpendicular to achieve a resultant I32 42 5m long the doubleprimed case shown In each sketch the vectors are shown in a headtotail sketch but the resultant is not shown The resultant would be a straight line drawn from beginning to end the beginning is indicated by A with or without primes as the case may be and the end is indicated by B A gt gt3 BM Chapter 1 Solutions Problem 5 Horses are to race over a certain English meadow for a distance of 40 furlongs What is the race distance in units of a rods and b chains 1 furlong 201168 m 1 rod 50292 m and 1 chain 20117 m a Number I Units I b Number Units 1I 1 Solution Given that 1 furlong 201168 m 1 rod 50292 m and 1 chain 20117 In we nd the relevant conversion factors to be 1 rod 10 furlon 201168 In 201168 40 rods g M 50292111 and lchain 10 furlong 201168 m201168m 10 cha1ns 20117m Note the cancellation of m meters the unwanted unit Using the given conversion factors we nd a the distance d in rods to be d 40 lrlongs 40 fuflongsfoidS 160 rods ong b and that distance in chains to be 10 chains d 40 lrlongs 40 fulongs 1 40 chains ong Problem 9 Antarctica is roughly semicircular with a radius of 2000 km see the gure The average thickness of its ice cover is 3000 m How many cubic centimeters of ice does Antarctica contain Ignore the curvature of Earth VAN Mi 7 1 mm m T Number Uruts the relevance IS VZ shluthm The volume of ree rs grverr by the produet of the semrerreular surface area kness The area of the semrerrele rs A 1922 where I rs the radms Therefore the volume rs E o 3 quot wherezrsthe reethrekhess Smee there are 103m m 1 km and 102 em m 1 mwe have 103m 10 em 5 t 2000hm mm 1 2000x10 em m In these uruts the threkhess beeomes 3000 x102 em IUZEm 23000m 3000m 1 m whreh yrelds V 2000 x105 em2 3000 x102 em 1 9 x1022 em thlzm 13 Three mgm duets45 and Crun at urrrererrt rates and uu hut have sxmultaneuus rmumgs ur zem The gure shuws srmunarreuus rmumgs urr parrs urthe elueks rur fuur ucmsuns At the V kA what dues elueksrmuv dWhen etuekCrmus 15 s what dues etueksrmuv Assume negzuve mumgs rurprezeru trmes m m I to A t 2 u u 1H0 gnu nm 92 H H 17m I a Number I Unit I b Number I Unit Li c Number Unit I d Number I Unit l J Solution The time on any of these clocks is a straightline function of that on another with slopes 1 and yintercepts 0 From the data in the gure we deduce 2 594t t C 75 7 B 40 539 These are used in obtaining the following results a We nd t tB t tA495s when t39A IA 600 s b We obtain 1 tC tg 48 495 141 s c ClockB reads t3 3340400 6625 m 198 s when clockA reads IA 400 s d From to l5 27tB 5947 we get I m 245 s Problem 17 Five clocks are being tested in a laboratory Exactly at noon as determined by the WV time signal on successive days ofa week the clocks read as in the following table Rank the five clocks according to their relative value as good timekeepers best to worst Give the number of the correct clock in each case according to the following Clock A 1 Clock B 2 Clock C 3 Clock 4 D Clock E 5 Which clock is a the best timekeeper b the secondbest timekeeper c the thirdbest timekeeper d the fourthbest timekeeper e the worst timekeeper cm 5m Mm mm mm quotMm PH 5 1 mD xzmu Aznzw mm mun mus mum Dan I h z d a Snllm39nn Nuns arms dunks admce by exacuy 24 h m a 24 pmud but Lhs 15 nut me mug lmpu znl mtmun fur judgng may quamy fur masunng ume mth vm xmpussxble wha m c cunemuns m secunds that mug be appued m me mdmg un 24 pmud The mines was determmed by submung me duck mamg at me and mm mth mm me clunkradmg aims begnnmg Clucks c and D are bum guud umekeepers m me sense that ach 15 Eunsl ml m 115 deny a relauve m WWF ume mus c and D are easily made pa39fe tquot m best The cunemun Lhatmust be appuedm dcqu 15 m me range mm 15 m 175 Fur unk a n 15 me mugs mm 75 s m 1n s fur dunk E n 15 m me mugs mm m s to 2 s After C and D A has the smallest range of correction B has the next smallest range and E has the greatest range From best to worst the ranking of the clocks is C D A B E Problem 22 Gold which has a density of 1932 gcm3 is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber 21 If a sample ofgold with a mass of 2763 g is pressed into a leaf of 1 pm thickness what is the area in m2 of the leaf b1f instead the gold is drawn out into a cylindrical fiber of radius 2500 pm What is the length in m of the fiber a Number I Units J j b Number I 7 Units l Solution The density of gold is 1932 gcm3 V 1cm a We take the volume of the leaf to be its area A multiplied by its thickness 2 With density p 1932 gcm3 and mass m 2763 g the volume ofthe leafis found to be V 5 1430 cm3 p We convert the volume to SI units 3 V1430cm3 1m 1430gtlt10 6m3 100 cm Since V A2 with z 1 X 10396 m metric pre xes can be found in Table 172 we obtain 1430 X 106 m3 A 1gtlt10 6 m 1430 m2 b The volume of a cylinder of length A is V AZ where the crosssection area is that of a circle A 7W2 Therefore with r 2500 X 10 6 m and V 1430 X 10 6 m3 we obtain V 7284 gtlt104 m 7284 km 2 7239 1 Problem 25 Flying Circus of Physics During heavy rain a section of a mountainside measuring 25 km horizontally 08 km up along the slope and 20 m deep slips into a valley in a mud slide Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 040 km x 040 km and that the mass of a cubic meter of mud is 1900 kg What is the mass of the mud sitting above a 40 m2 area of the valley oor Number Units j the tolerance is 2 Solution The volume of the section is 2500 m800 m20 m 40 X 106 m3 Letting 61 stand for the thickness of the mud after it has uniformly distributed in the valley then its volume there would be 400 m400 md Requiring these two volumes to be equal we can solve for 61 Thus at 25 m The volume of a small part of the mud over a patch of area of 40 m2 is 40d 100 m3 Since each cubic meter corresponds to a mass of 1900 kg stated in the problem then the mass of that small part ofthe mud is 19gtlt105 kg Problem 29 On a spending spree in Malaysia you buy an ox with a weight of 289 piculs in the local unit of weights 1 picu1 100 gins 1 gin 16 tahils 1 tahil 10 chees and 1 chee 10 hoons The weight of 1 hoon corresponds to a mass of 03779 g When you arrange to ship the ox home to your astonished family how much mass in kilograms mustyou declare on the shipping manifest Number Units 3 the tolerance is 2 Solution The mass in kilograms is 289 piculs 100 gm 16tah11 10 chee10 hoon 03799g 1 p1cu1 1 gm 1 tah11 1 chee lhoon which yields 1747 X 106 g or roughly 175gtlt 103 kg Problem 30 Water is poured into a container that has a leak The mass m of the water is given as a function of time tby m 500t0 8 7 300t 2000 with t2 0 m in grams and tin secondsa At what time is the water mass greatest and b what is that greatest mass In kilograms per minute what is the rate ofmass change at c t 200 s and d t 500 s I a Number I Unit I b Number I Unit Li c Number Unit I d Number I Unit 1 J Solution To solve the problem we note that the rst derivative of the function with respect to time gives the rate Setting the rate to zero gives the time at which an extreme value of the variable mass occurs here that extreme value is a maximum a Differentiating mt 500108 300t 2000 with respect to 1 gives d m 400z 02 300 dt The water mass is the greatest when abn a 0 or at t 400300102 421s b At t 421 s the water mass is mt 421 S 50042l0 8 30042l 2000 232 g c The rate ofmass change at t 200 s is dm dt 2005 s 1000 g 1 min 289 gtlt10 2 kgmin d Similarly the rate of mass change at t 500 s is 400500 2 300 gs 0101gs 0101 amp 60 2200 s 1000g 1mm dm dt 605 gtlt10 3 kgmin Problem 43 A person on a diet might lose 23 kg per week Express the mass loss rate in milligrams per second as if the dieter could sense the secondbysecond loss Number Units Ll the tolerance is 2 Solution A million milligrams comprise a kilogram so 23 kgweek is 23 X 106 mgweek Figuring 7 days a week 24 hours per day 3600 second per hour we nd 604800 seconds are equivalent to one week Thus 23 x 106 mgweek604800 sweek 38 mgs Chapter 2 Concept Question 3 Figure 216 shows four paths along which objects move from a starting point all in the same time interval The paths pass over a grid of equally spaced straight lines I In the following questions you will need to rank the paths If multiple paths rank equally use the same rank for each then exclude the intermediate ranking ie if objects A B and C must be ranked and Aand B must both be ranked first the ranking would be Al Bl C3 If all paths rank equally rank each as 39139 Rank the paths according to the average velocity of the objects greatest first Solution Velocity is a vector quantity the displacements of object 1234 are equal all in the same time so object 1234 have the same velocity So their ranks are Object 1 1 Object 2 1 Object 3 1 Object 4 1 Rank the paths according to the average speed of the objects greatest first Solution Speed is a scalar quantity Object l234 have the same motion time So the greater the path length is the greater its speed is We can get their ranks as Object l 2 Object 2 2 Object 3 4 Object 4 1 Chapter 2 Concept Question 4 Figure2l7 is a graph of a particle s position along an x axis versus time At time t 0 what is the sign of the particle s position Solution negative Is the particle s velocity positive negative or 0 at t 1 s Solution positive Is the particle s velocity positive negative or 0 at t 2 s Solution zero Is the particle s velocity positive negative or 0 at t 3 s Solution negative How many times does the particle go through the point X 0 Solution 2 at 11 5 and F3 5 Chapter 2 Concept Question 5 Figure 218 gives the velocity of a particle moving along an axis Point 1 is at the highest point on the curve point 4 is at the lowest point and points 2 and 6 are at the same height l What is the direction of travel at time t 0 Solution positive What is the direction of travel at point 4 Solution negative At which of the six numbered points does the particle reverse its direction of travel Multiple answers may be correct Solution Point 3 Point 5 Rank the six points according to the magnitude of the acceleration greatest first If multiple points rank equally use the same rank for each then exclude the intermediate ranking ie if objects A B and C must be ranked and A and B must both be ranked first the ranking would be Al Bl C3 If all points rank equally rank each as 39139 dv Solution a d we can see from the graph the derlvatlves of v have the greatest l magnitude at point 2 and point 6 the second greatest magnitude at point 3 and point 5 the least great magnitude at point 1 and point 4 So we get Point 1 5 Point 2 1 Point 3 3 Point 4 5 Point 5 3 Point 6 1 Chapter 2 Concept Question 1 Hanging over the railing of a bridge you drop an egg no initial velocity as you throw a second egg downward Curves A and B are parallel so are C D and E so are F and G Which curve in Fig 219 give the velocity vt for the dropped egg Solution D v00 and then V increases in the negative direction when t grows Which curve in Fig 219 give the velocity vt for the dropped egg Solution E v0lt0 and then V increases in the negative direction when t grows Chapter 2 Problem 15 A particle39s position is given by X 400 1200t 312 in which X is in meters and tis in seconds a What is its velocity at t 1 s b Is it moving in the positive or negative direction of gtltjust then c What is its speed just then d Is the speed increasing or decreasing just then Try answering the next two questions without further calculation e Is there ever an instant when the velocity is zero If so give the time t if not answer quot0quot f Is there a time after t 3 s when the particle is moving in the negative direction of X If so give the time t if not answer quot0quot Solution We use Eq 24 to solve the problem a The velocity of the particle is v i4 12t3t2 l26t ch ch Thus at t l s the velocity is v 712 6l 4 ms b Since v lt 0 it is moving in the 7x direction at t l s c At t l s the speed is lvl 6 ms d For 0 lt tlt 2 s lvl decreases until it vanishes For 2 lt tlt 3 s lvl increases from zero to the value it had in part c Then lvl is larger than that value for t gt 3 s e Yes since v smoothly changes from negative values consider the t 1 result to positive note that as t gt 00 we have v gt 00 One can check that v 0 when t 2 s t No In fact from v 712 6t we know that v gt 0 for t gt 2 s Chapter 2 Problem 28 On a dry road a car with good tires may be able to brake with a constant deceleration of 492 ms2 a How long does such a car initially traveling at 246 ms take to stopb How far does it travel in this time Solution We take x in the direction of motion so v0 246 ms and a 7 492 msz We also take x0 0 a The time to come to a halt is found using Eq 211 0v0 at 2 t 500s b Although several of the equations in Table 21 will yield the result we choose Eq 2 16 since it does not depend on our answer to part a 246 ms2 2 492 msz61395m39 0v 2ax 3 x Chapter 2 Problem 37 The figure depicts the motion of a particle moving along an X axis with a constant acceleration The figure39s vertical scaling is set by X 600 m What are the a magnitude and b direction of the particle39s acceleration x denote 1 0 otherwise ij 111 Solution a From the gure we see that x0 720 m From Table 21 we can apply xixo v0t at2 with t 10 s and then again with t 20 s This yields two equations for the two unknowns v0 and a 00 20 mv0 10 sal0 s2 60 m 20 mv0 20 sa20 s2 Solving these simultaneous equations yields the results v0 0 and a 40 msz b The fact that the answer is positive tells us that the acceleration vector points in the x direction Chapter 2 Problem 53 A key falls from a bridge that is 45 m above the water It falls directly into a model boat moving with constant velocity that is 12 m from the point of impact when the key is released What is the speed of the boat S olution The speed of the boat is constant given by vb dt Here d is the distance of the boat from the bridge when the key is dropped 12 m and t is the time the key takes in falling To calculate t we put the origin of the coordinate system at the point where the key is dropped and take the y aXis to be positive in the downward direction Taking the time to be zero at the instant the key is dropped we compute the time I when y 45 m Since the initial velocity of the key is zero the coordinate of the key is given by y iglz Thus I 2y lm3o3s g 98ms Therefore the speed of the boat is 12 m 303 s v 40ms b Chapter 2 Problem 54 A stone is dropped into a river from a bridge 439 m above the water Another stone is thrown vertically down 1 s after the first is dropped Both stones strike the water at the same time What is the initial speed of the second stone S olution We neglect air resistance which justi es setting a 7g 798 ms2 taking down as the 7 y direction for the duration of the motion We are allowed to use Eq 215 with Ay replacing Ax because this is constant acceleration motion We use primed variables except I with the rst stone which has zero initial velocity and unprimed variables with the second stone with initial downward velocity ivo so that v0 is being used for the initial speed SI units are used throughout Ay39 0 th Ay lt vOgtltr 1 gltr 1gtz Since the problem indicates Ay Ay 4139 In we solve the rst equation for t nding t 299 s and use this result to solve the second equation for the initial speed of the second stone i 439 m v0l99 s 98ms2l99 s2 which leads to v0 123 ms Chapter 2 Problem 64 A ball is shot vertically upward from the surface of another planet A plot of y versus tfor the ball is shown in the figure where y is the height of the ball above its starting point and t 0 at the instant the ball is shot The figure39s vertical scaling is set by y 250 m What are the magnitudes of a the freefall acceleration on the planet and b the initial velocity of the ball 39 ml Solution The graph shows y 25 m to be the highest point where the speed momentarily vanishes The neglect of air friction or whatever passes for that on the distant planet is certainly reasonable due to the symmetry of the graph a To nd the acceleration due to gravity gP on that planet we use Eq 215 with y up yyo vt gpt2 2 25 m 0025sgp25s2 so that gP 80 msz b That same max point on the graph can be used to nd the initial velocity v0 025 s y y0lv0vt 3 25 m 0 2 Therefore v0 20 ms Chapter 2 Problem 69 How far does the runner whose velocity time graph is shown in the figure travel in 16 s The figure39s vertical scaling is set by S 800 ms 139 inquot 5 u 4 12 16 i 539 Solution Since v dx dt Eq 24 then Ax Iv dt which corresponds to the area under the v vs I graph Dividing the total area A into rectangular base gtlt height and triangular base gtlt height areas we have A A0lttlt2 A2lttlt10 A10lttlt12 A12lttlt16 l l 5lt2gtlt8gtlt8gtlt8gtlt2gtlt4gt3lt2gtlt4gtlt4gtlt4gt with SI units understood In this way we obtain Ax 100 m Chapter 2 Problem 77 A hot rod can accelerate from 0 to 60 kmh in 54 s a What is its average acceleration in ms2 during this time b How far will it travel during the 54 s assuming its acceleration is constant c From rest how much time would it require to go a distance of 025 km if its acceleration could be maintained at the value in a S olution Since the problem involves constant acceleration the motion of the rod can be readily analyzed using the equations in Table 21 We take x in the direction of motion so v60 kmh W l67m s 3600 s h and a gt 0 The location where it starts from rest v0 0 is taken to be x0 0 a Using Eq 27 we nd the average acceleration to be AV VV W309Ws2 z3lmSZ aav 3 At t to 54s 0 b Assuming constant acceleration a czan 309 ms2 the total distance traveled during the 54s time interval is xx0 v0t at2 00309ms254 s2 45 m c Using Eq 215 the time required to travel a distance ofx 250 m is I 2 250 leat2 3 t E 2135 2 a 31ms2 Note that the displacement of the rod as a function of time can be written as xt 609 ms2t2 Also we could have chosen Eq 217 to solve for b xv0 v tl67ms54s45 m Chapter 2 Problem 100 A parachutist bails out and freely falls 50 m Then the parachute opens and thereafter she decelerates at 2 ms She reaches the ground with a speed of 30 ms a How long is the parachutist in the air b At what height does the fall begin S olution During free fall we ignore the air resistance and set a 7g 798 ms2 where we are choosing down to be the 7y direction The initial velocity is zero so that Eq 215 becomes Ay gt2 where Ay represents the negative of the distance d she has fallen Thus we can write the equation as d 3 gt2 for simplicity a The time t1 during which the parachutist is in free fall is using Eq 215 given by 1 2 1 2 2 d1 50m Egt1 E980ms z which yields t1 32 s The speed of the parachutist just before he opens the parachute is given by the positive root v12 2 gal1 or v12ghl 2980ms250 m 3lm s If the nal speed is vz then the time interval t2 between the opening of the parachute and the arrival of the parachutist at the ground level is v 1V2 3lms 30ms 2 14 s a 2 m s This is a result of Eq 211 where speeds are used instead of the negativevalued velocities so that finalvelocity minus initialvelocity turns out to equal initialspeed minus finalspeed we also note that the acceleration vector for this part of the motion is positive since it points upward opposite to the direction of motion 7 which makes it a deceleration The total time of ight is therefore II t l7 s b The distance through which the parachutist falls after the parachute is opened is given by 2 2 V V 3lms 30m s z 240m 61 2a 220ms2 In the computation we have used Eq 216 with both sides multiplied by 71 which changes the negativevalued Ay into the positive d on the lefthand side and switches the order of v1 and v on the righthand side Thus the fall begins at a height of h 50 d m 290 m Chapter 2 Problem 101 A ball is thrown down vertically with an initial speed of 29 fts from a height of 30 ft a What is its speed in fts just before it strikes the ground b How long does the ball take to reach the ground What would be the answers to c part a and d part b if the ball were thrown upward from the same height and with the same initial speed Before solving any equations decide whether the answers to c and d should be greater than less than or the same as in a and b Solution We neglect air resistance which justifies setting a 7g 798 ms2 32 15 fts2 taking down as the 7y direction for the duration of the motion We are allowed to use Table 21 with Ay replacing Ax because this is constant acceleration motion The ground level is taken to correspond to y 0 a With yo h and v0 replaced with ivo Eq 216 leads to V v02 2gy y0 Vvd 2537 so we get v 29fts2 2gtlt3215fts2 X30 5263 ft s The positive root is taken because the problem asks for the speed the magnitude of the velocity b We use the quadratic formula to solve Eq 215 for t with v0 replaced with ivo V0 V V02 2gAy Ay V0t igt2 3 1 2 g where the positive root is chosen to yield tgt 0 With y 0 and yo h this becomes v Zgh v0 g From part awe know quotv 2gh v so we get 5263 s 29 s t 3215ft s2 0735s c If it were thrown upward with that speed from height h then in the absence of air friction it would return to height h with that same downward speed and would therefore yield the same final speed before hitting the ground as in part a An important perspective related to this is treated later in the book in the context of energy conservation d Having to travel up before it starts its descent certainly requires more time than in part b The calculation is quite similar however except for now having v0 in the equation where we had put in ivo in part b The details follow 1 v quotv2 2 A AyV0tEgt2 3 IM g with the positive root again chosen to yield t gt 0 With y 0 and yo h we obtain w v Zgh v0 g So 5263 s29fts 2539 5 3215 ft 52

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