Ch4 - ActivPhysics can help with these problems m...

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Unformatted text preview: ActivPhysics can help with these problems: m Activities in Section 3, Projectile Motion Section +1: Velocity and Acceleration . A skater is gliding along the ice at 2.4 m/s, when 4. she undergoes an acceleration of magnitude "1.1 m/s2 for 3.0 8. At the end of that time she is moving at 5.7 m/s. What must be the angle between the acceleration vector and the initial ‘ velocity vector? tion ‘ . constant acceleration, Equation 4-3 shows that the ' rs v0, aAt and v form a triangle as shown. The of cosines gives '02 = 03 + (aAt)2 -— ZuoaAtx ms(180° — 90). When the given magnitudes are flistituted, one can solve for 00: (5.7 rn/s)2 = 4 m/s)2 + (1.1 m/32)2(3.0 s)2 + 2(2.4 m/s)x m/sz)(3.0 5) cos 60, or cos 00 = 1.00 (exactly), and 0°. Since v0 and a are colinear, the change in 17 51M: T7 90 Problem 1 Solution. Problem 2. In the preceding problem, what would have been ' ' , the magnitude of the skater’s final velocity if the ‘ acceleration had been perpendicular to her initial velocity? Solution When v0 and aAt are perpendicular, Equation 4—3 and the Phythagorean theorem imply v = «no: + (aAt)2 = {(2.4 m/s)2 + (1.1 m/s2)(3.0 s)2 = 4.08 m/s. VHAPTER 4 MOTION IN MORE THAN ONE DIMENSION Problem An Object is moving in the x direction at 1.3 m/s when it is subjected to an acceleration given by a=0.52j m/s . What is its velocity vector after 4.4 s of acceleration? Solution From Equation 4-3, v =vo + aAt : (1.30 m/s)i+ (0.523 m/52)(4.4s) = (1.30i+2.293) m/s. Problem 4. An airliner is flying 'at a velocity of 2601‘ m/s, when a wind gust gives it an acceleration of 0.38i+ 0.72j m/s2 for a period of 24 s. (a) What is its velocity at the end of that time? (b) By what angle has it been deflected from its original course? Solution (a) Equation 4—3 gives v = 2601‘ m/s + (0.38i + 0.723)(m/52)(24 s) = (269i+ 17.33) m/s. (b) Since Va is along the x-axis, the angular deflection is just tan“l(vy/v,) = tan—1(17.3/269) = 3.67°. Section 4-2: Constant Acceleration Problem @The position of an object as a function of time is given by r = (3.2t + 1.8t2)i+ (1.7t -— 2.4t2)j m, where tis the time in seconds. What are the magnitude and direction of the acceleration? Solution One can always find the acceleration by taking the second derivative of the position, a(t) = d2r(t)/dt2. However, collecting terms with the same power of t, one can write the position in meters as r(t) = (3.2i + 1.7j)t + (1.8‘i — 2.4j)t2. Comparison with Equation 4-4 shows that this represents motion with constant acceleration equal to twice the coefficient of the t2 term, or a = (3.6‘i—4.8j) m/sz. The magnitude and direction of a are «(3.6)2 + (~48)? m/s2 = 4.49 m/s2, and tan—1(-4.8/3.6) = 307° (cow from the x-axis, in the fourth quadrant) or —53.1" (CW from the :c-axis). 50 CHAPTER 4 Problem 6. An airplane heads northeastward down a runway, accelerating from rest at the rate of 2.1 m/sz. Express the plane’s velocity and position at t = 30 s in unit vector notation, using a coordinate system with z—axis eastward and y—axis northward, and with origin at the start of the plane’s takeoff roll. Solution Since the acceleration is constant and the airplane starts from rest (v0 = 0) at the origin (to = 0), Equations 4-3 and 4-4 give v(t)=at and r(t) = %at2. Both vectors are in the NE direction, parallel to a = (2.1 rn/sz)(i‘cos45° +jsin 45°) = (1.48 m/sz)x n+3). Thus, v(30 s) = (1.48 m/s2)(30 s)(i‘+j) = (44.5 m/s)(i+j), and r(30 s) = -%(1.48 m/s)(30 s)2x (i+j) = (668 m) (3+3). (Note: (Hp/fl is a unit vector in the NE direction.) Problem An asteroid is heading toward Earth at a steady 21 kin/s. To save their planet, astronauts strap a giant rocket to the asteroid, giving it an acceleration of 0.035 km/s2 at right angles to its original motion. If the rocket firing lasts 250 s, (a) by what angle does the direction of the asteroid’s motion change? (b) How far does it move during the firing? Solution Av - an: ((1) FIGURE 4—2a Problem 7 Solution. (a) The change in velocity, Av =aAt, is at right angles to the initial velocity, v0, so they form the legs of a right triangle with hypotenuse v (see Fig. 41-24;). The angle between v and Va is 0 = tan—lflAvl —:— Ivol) = tan‘1(0.035 km/s)(250 s)/(21 kin/s) = 22.6°. (b) For constant acceleration, the displacement is (see Equation 4-4) Ar = r — r0 = vot + éatz. Again, the two vectors vet and fiatz form the legs of a right triangle, whose hypotenuse, Ar, has magnitude ,/(21 km/s)2(250 S)? + ($110035 km/sz)2(250 s)4 = 5.36x 103 km. (Note: this is the asteroid’s displacement during the rocket firing, not the pathlength of its trajectory.) Problem 8. An object is moving initially in the a: direction at 4.5 m/s, when an acceleration is applied in the y direction for a period of 18 s. If it moves equal distances in the .1: and y directions during this time, what is the magnitude of its acceleration? Solution The :1: component of the displacement is due only to the initial velocity, Arr: = vOIAt. The y component is just due to the acceleration, Ag 2 éaAt2 = Arr. Thus, a = 2110, /At = 2(4.5 m/s)/18 s = 0.5 m/sz. Problem .A hockey puck is moving at 14.5 m/s when a stick imparts a constant acceleration of 78.2 m/s2 at a 90.0" angle to the original direction of motion. If the acceleration lasts 0.120 3, what is the magnitude of the puck’s displacement during this time? Solution Take the z~aads in the direction of the initial velocity, v0 = 14.5 m/s, and the y—axis in the direction of the acceleration, a = 78.23 m/sz. The displacement during the 0.120 s interval of constant acceleration is (Equation 4-4) Ar: 1- — r0 = vot + %at2 = (14.5 m/s)(0.120 s)i+ 315(78-2 m/s’)(0.120 S)?“ = (1.74i+0.563j In. This has magnitude «(1.74)2 + (0.563)2 m = 1.83 m (and makes an angle of tan“1(0.563/ 1.74) = 17.9° with the direction of the initial velocity). Problem 10. Repeat the preceding problem, except that now the acceleration makes a 65.0° angle with the original direction of motion. Solution Now the acceleration is (78.2 m/s2)icos 650° + jsin 65.0°) and Ar = (14.5 m/s)(0.120 s)i+ §(33.03+ 70.9j)(m/s2) (0.120 s)2 = (1.98‘i +0.5103) m. This has magnitude 2.04 m (and makes an angle of 14.5° with the initial velocity). Problem 11. A particle leaves the origin with initial velocity v0 = 111‘ + 143 m/s. It undergoes a constant acceleration given by a: — 1.2i+0.26j‘ m/s2. (a) When does the particle cross the y-axis? (b) What is its y—coordinate at the time? (c) How fast is it moving, and in what direction, at that time? Solution I (a) Since the particle leaves from the origin (to = 0), its position is r(t) = vat + §at2. It crosses the y—axis when z(t) = vozt + %a.,t2 = O, or t = «21103 /a,, = —2(11 m/s)/(—1.2 m/sz) = 18.3 s. (b) w) = toy: + gay? = [14 m/s + §(o.26 m/sz)(18.3 mass 3) = 300 m. (c) v(t) =vo +at = (11i+ 143) m/s + (-1.2’i+0.263 18.3 m/s = (~11i+18.83) m/s. Then lv(t)l = (~11)2 + (18.8)2 m/s = 21.8 m/s and 9, = tan—1(18.8/(—11))= 120°. Problem @ particle starts from the origin with initial , velocity v0 = mi and constant acceleration a: aj. Show that the particle’s distance from the origin and its direction relative to the z-axis are given by d = “/03 + §¢22t2 and 9 = tan-1(at/2'uo). Solution Rom Equation 4-4, the particle’s displacement from its original position is Ar: r — r0 = Vat + éat2 = (vo‘i + %at j)t. This has magnitude t‘lvfi + («é-at)2 and direction (CCW from the :r-axis) 0 = tan“1(at/2vo). Problem 13. Figure 4-27 shoWs a cathode-ray tube, used to display electrical signals in oscilloscopes and other scientific instruments. Electrons are accelerated by the electron gun, then move down the center of the tube at 2.0x 109 cm/s. In the 4.2-cm—long deflecting region they undergo an acceleration directed perpendicular to the long axis of the tube. The acceleration “steers” them to a particular spot on the screen, where they produce a visible glow. (a) What acceleration is nwded to deflect the electrons through 15°, as shown in the figure? (b) What is the shape of an electron’s path in the deflecting region? Solution (a) With z~y axes as drawn on Fig. 4-27, the electrons emerge from the deflecting region with velocity v=voi+atj, after a time t = :r/vo, Where a: = 4.2 cm and '00 = 2x 109 cm/s. The angle of deflection (direction of v) is 15°, so tan 15° = vy/v, = at/vo = arr/113. Thus, a = vgtan 15°/a: = 2.55x1017 cm/s2 (when values are substituted). (b) Since the CHAPTER 4 51 acceleration is assumed constant, the electron trajectory is parabolic in the deflecting region. FIGURE 4-27 Problem 13 Solution. Section 4-3: Projectile Motion Problem 14. You toss an apple horizontally at 8.7 m/s from a height of 2.6 m. Simultaneously, you drop a peach from the same height. How long does each take to reach the ground? Solution The time of flight for either projectile can be determined from the vertical component of the motion, which is the same for both, since coy = 0. Thus, Equation 4-8 gives t = (/2(y - yo)/ = (/2(2.6 m)/(9.8 m/s2) = 0.728 s. Problem 15. A carpenter tosses a shingle off a 8.8-m—high roof, giving it an initially horizontal velocity of 11 m/s. (a) How long does it take to reach the ground? (b) How far does it move horizontally in this time? ‘Uo = (llm/s)i I Problem 15 Solution. 52 CHAPTER 4 Solution (a) The shingle reaches the ground when y(t) = 0 = yo — égtjl 01' _ En: lm= 13‘ g (9.8 m/sz) 1'345' (b) The horizontal displacement is x = not = (11 m/s)(1.34 s) = 14.7 m. Problem 16. An arrow fired horizontally at 41 m/s travels 23 m horizontally before it hits the ground. E'om what height was it fired? Solution From a: - :co = vat, one finds the time of flight t: 23 m/(41 m/s) = 0.561 s, and from yo — y = é-gtz (recall that 120,, = 0) one finds the height yo — y = %(9.8 m/s2)(0.561 s)2 = 1.54 m. Problem @ A kid fires water horizontally from a squirt gun held 1.6 m above the ground. It hits another kid 2.1 m away square in the back, at a point 0.93 m above the ground (see Fig. 4-28). What was the initial speed of the water? FIGURE 4—28 Problem 17. Solution Since the water was fired horizontally (v03, = 0), the time it takes to fall from yo = 1.6m to y z: 0.93m is given by Equation 4-8, t = (/2010 — ill/9 == (/2(1.6 — 0.93) m/(9.8 m/s2) = 0.370 3. Its initial speed, '00 = v03, can be found from Equation 4-7, 110 = (1: — zo)/t = 2.1 m/0.370 s = 5.68 m/s. Problem 18. In a chase scene, a movie Stuntman is supposed to run right off the flat roof of one city building and land on another roof 1.9 m lower. If the gap between the buildings is wide, how fast must he run? Solution The horizontal and vertical distances covered by the stuntman are a: — so = vot and ya — y = £th (since 120, = 1:0, and my = 0). Eliminating t, one finds 00 = (.7: — mo)\/g/2(yo — y) = (4.5 m)“ (9.8 m/sz)/2(1.9 m) = 7.23 m/s. (Note that Equation 4-9 with 00 = 0 and yo = 0 provides an equivalent solution.) Problem 19. Ink droplets in an ink-jet printer are ejected horizontally at 12 m/s, and travel a horizontal distance of 1.0 mm to the paper. How far do they fall in this interval? Solution From a: — $0 = vast, the time of flight can be found. Substitution into yo — y = §gt2 (recall that 001, = 0) yields 90 — y = $9“? - mop/”ii = gas m/SZXIO—a m)2/(12 m/s)2 = 3.40x10-8 m = 34 nm for the distance fallen, practically negligible. Note that this analysis is equivalent to using Equation 4-9 with 90 = 0. Problem 20. Protons in a. particle accelerator drop 1.2 pm over the 1.7-km length of the accelerator. What is their approximate average speed? Solution The horizontal and vertical distances in projectile motion (range and drop) are related by the trajectory equation (Equation 4-9). With 00 = 0, y = -ga:2/2v§, or 120 = :c‘/(—g)/2y = (1.7x103 m) (—9.8 m/s2)/2(—1.2x10‘6 m) =' 3.44x106 m/s. Since 11,, = —gt = —g(z — 2:0)/vo = —4.85x10‘3 m/s is negligible compared to this, 110 is hardly different from the average speed. Problem 21. You’re standing on the ground 3.0 m from the wall of a building, and you want to throw a package from your 1.5-4.2m shoulder level to someone in a second-floor window above the ground. At what speed and angle should you throw it so it just barely reaches the window? 54 CHAPTER 4 the vertical component of the velocity (negative t = hog/g = 2(6.64 kin/s) sin 45” / (9.8 m/sz) = downward) for the given drop, 11;, = _ “3y __ 29y = 958 s = 16.0 min. (c) At a 20° launch angle, . = 2 . O = .. r———_——_———_Q8.1 m /s)2 $112150 _ 2 (9.8 m/s2)(—5.9 m) = 00 l/ (4500 km)(0.0098 km/s )/ s1n40 8.28 km/s. —14.6 m/s . Problem Problem A rescue airplane is flying horizontally at speed on _ . . at an altitude h above the ocean, attempting to 26. Compare the travel times for the prejectiles drop a package of medical supplies to a shipwreck lauPched at 30° and 60° II} F 1a 4-15, Poth 0_f victim in a lifeboat. At what line-of-sight angle a meh have the same starting and ending pomts. (Fig. 4—29) should the pilot release the package? Solution The package, when dropped, is a horizontally launched projectile, like that in Problem 15 (the expressions for t and a: are the same). The line-of-sight angle is a = tan-1(h/x), where :r = vo‘/2h/g is the horizontal distance at drop time. Thus, a = tan‘1 gh/2vo. V0: V0? N FIGURE 4—15 Problem 26. Solution The time of flight (t > 0) for a projectile trajectory beginning and ending at the same height (y(t) = ya) can be found from Equation 48, y(t) — yo = 0 = voyt —- égt2 or t = 2vOy/g = 200 sinQO/g. In Fig. 4-15, on = 50 m/s, so two = 2(50 m/s) sin 30°/(9.8 m/sz) = 5.10 S, and tsoo = fl t3oo = 8.84 8. FIGURE 4-29 Problem 28 Solution. Problem A submarinelaunched missile has a range of 4500 km. (a) What launch speed is needed for this range when the launch angle is 45°? (b) What is Problem the total flight time? (c) What would be the 29. At a circus, a human cannonball is shot from a minimum launch speed at a 20° launch angle, used cannon at 35 km/h at an angle of 40°. If he leaves the cannon 1.0 m 013" the ground, and lands in a to “depress” the trajectory so as to foil 3 net 2.0 m 011’ the ground, how long is he in the air? space-based antimissile defense? Solution The time of flight can be calculated from Equation 4-8, where the trajectory begins at Solution (a) Assuming Equation 4-10 applies (i.e., the trajectory begins and ends at the same height, or y t = y , one finds 11 : fig 531129 ___ yo = 1.0 m and t = 0, ends at y(t) = 2.0 m, and H—‘—oT—./‘: you = vosinflo = (35 m/3.6 s) sin 40° = 6.25 m/s. The (4500 km)(0.0098 km/s )/ 3m 90 = 6.64 km/s. equation is a quadratic, §gt2 _ "01,1 + (y __ yo) ___ 0’ (b) The time of flight is the positive solution of - - _ 2 _ _ _ Equation 4-8 when y(t) - yo = 0 = ”out — égtz. Thus With solutions t _ [”0” i— 009 29 (y yo)]/g ’— [6.25 m/s i ,/ (6.25 m/s)2 — 2(9.8 m/52)(2 m — 1 m)]~:— (9.8 m/sg) = 0.188 s or 1.09 s. The trajectory crosses the height 2.0 in twice, once going up, at the smaller time of flight, and once going down into the net, at the larger time of flight. The latter is the answer to the question asked here. Problem 30. Preparing to give an injection, a physician ejects a drop of medicine to ensure there’s no air in the syringe. The syringe is pointing upward, at 20° to the vertical, with the tip of the needle 45.0 cm above a tabletop. The drop leaves the needle at 1.30 m/s. (a) What maximum height does it reach? (b) How long is it before the drop hits the tabletop? (c) How far does the drop travel horizontally? Solution (a) Take the tip of the syringe to be at yo = 45.0 cm above the tabletop at y = 0. At its maximum height, just the vertical component of the drop’s velocity is instantaneously zero, so '03 = 0 = v33, — Zg(ymmt — yo), or gym“ — yo = nay/2g. The vertical component of the initial velocity is boy 2 (1.30 m/s) cos 20°, which gives ymax — yo = 7.61 cm. This is the maximum height above the tip, corresponding to 52.6 cm above the tabletop. (b) The time of flight is the positive solution of the equation y(t) = 0 = yo + voyt -— égtz, or t = (coy + Mug” + Zgyo)/g. Substituting the numbers from part (a), we obtain t.= 0.452 s. (c) The horizontal distance traveled is a: — :co = volt = (1.30 m/s)(sin 20°)(0.452 s) = 20.1 cm. Problem fyou can hit a golf ball 180 m on Earth, how far can you hit it on the moon? (Your answer is an underestimate, because the distance on Earth is restricted by air resistance as well as by a larger 9.) Solution For given W), the horizontal range is inversely proportional to 9. With surface gravities from Appendix E, we find zmoon = (gEarth/gmoon)$EsI-th = (9.81 /1.62)(180 m) = 1090 m. Problem 32. Prove that a projectile launched on level ground reaches its maximum height midway along its trajectory. CHAPTER 4 55 Solution From the trajectory (Equation 4-9), the maximum height occurs when dy/dx = tanB — gz/vg cos2 00 = 0, or :1: = 0% sin 00 cos 9/9, which is midway, or half of the horizontal range (Equation 4.10). This result can be derived in other ways; for example, the maximum height is reached at time vow/g (when 11,, = 0), whereas the total time of flight is twice this (the solution of O = voyt - égtz). Problem 33. A projectile launched at an angle 00 to the horizontal reaches a maximum height h. Show that its horizontal range is 4h / tan 00. Solution The intermediate expression for the horizontal range (when the initial and final heights are equal) is x = 2123 sin 00 cos 90/9 = vavoy/g (see the equation before Equation 4—10). The components of the initial velocity are related by coy / 1103 = tan 4%. The maximum height, h = gmax - yo, can be found from Equation 2-11 (when 2),, = 0) or 113,, = 29h. Then 2 = ZonUoz/g = 20%(voy/ tan 90)/g == 2(2gh)/g tan 00 : 4h/ tan 00. (This result reflects a classical geometrical property of the parabola, namely, that the latus ‘ rectum is four times the distance from vertex to focus.) Problem 34. You’re 5.0 m from the left-hand wall of the house shown in Fig. 4-30, and you want to throw a ball to a friend 5.0 m from the right-hand wall. (a) What is the minimum speed that will allow the ball to clear the roof? (b) At what angle should it be thrown? Assume the throw and catch both occur 1.0 m above the ground. FIGURE 430 Problem 34. Solution Since the trajectory is symmetrical (begins and ends at the same height), one can use the result of the 56 CHAPTER 4 previous problem with h = ymax — yo = 6 m -— 1 m=5 in and horizontal rangez =5 m+6 m+ 5 m = 16 m. Then (b) 00 = tan“1(4h/a:) = 51.3°, and (a) v0 = coy/sinflo = W/sinflo = 12.7 m/s. Problem 35. A circular fountain has jets of water directed from the circumference inward at an angle of 45°. Each jet reaches a maximum height of 2.2 m. (a) If all the jets converge in the center of the circle and at their initial height, what is the radius of the fountain? (b) If one of the jets is aimed at 10" too low, how far short of the center does it fall? Solution (a) The radius is the horizontal range, 1‘ = rig/g (Equation 410 with 9 = 45°). The maximum height is h = vgy/2g = v3/4g (Equation 2-11 with 0,, = 0 and 210,, = 120 cos 45° = ooh/2). Therefore, 1' = (4gh)/g == 4h = 4(2.2 m) = 8.8 m. (b) If one jet is directed at 35° with the same initial speed (113 = rg), it would fall short by 1' -— 2:, where :r: is given by Equation 4—10. Therefore, r —- :c = r — (cg/g) sin(2 x 35°) = (8.8 m)(l — sin 70°) = 0.531 m. Problem 35 Solution. Problem 36. When the Olympics were held in Mexico City in 1968, many sports fans feared that the high altitude would result in poor performances due to reduced oxygen. To their surprise, new records were set in track and field events, probably as a result of lowered air resistance and a decrease in g to 9.786 m/s2, both ultimately associated with the high altitude. In particular, Robert Beamon set a new world record of 8.90 m in the long jump. Photographs suggest that Beamon started his jump at a 25° angle to the horizontal. If he had jumped at sea level, Where g = 9.81 m/sz, at the same angle and initial speed as in Mexico City, how far would Beamon have gone? Neglect air resistance in both cases (although its effe...
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