θ Vman V ship α φ V β

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1 Answer, Key { Homework 3 { Rubin H Landau 1 This print-out should hae 9 uestions. Check that it is complete before leaing the printer. Also, multiple-choice uestions may continue on the next column or page: nd all choices before making your selection. Note that only a few (usually 4) of the problems will hae their scores kept for a grade. You may make multiple tries to get a problem right, although it's worth less each time. Worked solutions to a number of these problems (een some of the scored ones) may be found on the Ph211 home page. Walking on a Ship 03:01, calculus, numeric, > 1 min. 001 A ship cruises forward at 7 ms relatie to the water. On deck, a man walks diagonally toward the bow such that his path forms an angle 16 with a line perpendicular to the boat's direction of motion. He walks at 4 ms relatie to the boat. θ Vman V ship At what speed does he walk relatie tothe water? Correct answer: 8:96859 ms. θ α Vman φ V ship When you complete the parallelogram, the resultant elocity with respect to the water is the side of the triangle opposite the obtuse angle, which has a measure of Let ~ s be the elocity of the ship, ~ m be the elocity of the man, and ~ be the resultant elocity of the man relatie to the water (Earth). By the law of cosines V β 2 2 m + 2 s, 2 m s cos 2 m + 2 s, 2 m s cos 12 (4 ms) 2 +(7ms) 2,2(4 ms)(7 ms) cos(106 )] 12 (4 ms) 2 +(7ms) 2 +(15:4357 m 2 s 2 ) 12 8:96859 ms : Alternatiely: We can analyze the ector addition using the components of the ectors. Note: or ~ ~ s + ~ m ; x s + m sin and y m cos : Hence the speed of the man with respect to the water is 2 x + 2 y (8:10255 ms) 2 +(3:84505 ms) 2 8:96859 ms : 002 At what angle to his intended path does the man walk with respect to the water? Correct answer: 48:6134. The law of sines can be used to compute the reuested angle, which is the angle opposite the ship's path and elocity. sin sin s sin s sin h s i arcsin sin 7ms arcsin 8:96859 ms sin(106 ) 48:6134 : Alternatie Solution: Using ector components from Part 1, we hae y arctan x 3:84505 ms arctan 8:10255 ms 25:3866 :

2 Answer, Key { Homework 3 { Rubin H Landau 2 Therefore the angle between ~ m and ~ is 90,, m 2:0 + s 2:0 + V h4i 2:0 + h7i 2:0 + h15:4357i (13) 90, (16 ), (25:3866 ) 8:96859 ms 48:6134 : hmsi hmsi 2:0 + hmsi 2:0 + hm 2 s 2 i u arctan2 ( y ; x ) (14) hdegi 0: raddeg (1) rad arctan2 (h3:84505i; h8:10255i) h deg radi 57:2958 degrad (2) s 7ms 0: m 4ms hi arctanhi (hmsi; hmsi) u h deg radi (15) 10 (5) h0:443079i h57:2958i 70 90:0+ 25: :0+h16i 106 h i hi + h i u h rad deg (7) h106i h0: i 1:85005 hi h ihraddegi u hdegi rad (8) h16i h0: i 0: hi h ihraddegi x m sin ( u )+ s (9) h4i sin (h0:279253i)+ h7i 8:10255 ms hmsi hmsi sin (hi)+ hmsi y m cos ( u ) (10) h4i cos (h0:279253i) 3:84505 ms hmsi hmsi cos (hi) V x 2:0 + y 2:0 (11) h8:10255i 2:0 + h3:84505i 2:0 hmsi 8:96859 ms hmsi 2:0 + hmsi 2:0 V (,2:0 m s cos ( u )) (12) (,2:0 h4i h7i cos (h1:85005i)) 15:4357 m 2 s 2 hm 2 s 2 i (,hi hmsi hmsi cos (hi)) h i hi hdegradi s sin ( u ) u arcsin (16) h7i sin (h1:85005i) arcsin h8:96859i 0: hmsi sin (hi) hi arcsin hmsi u h deg radi (17) h0:848465i h57:2958i 48:6134 h i hi hdegradi Alpha 90,, (18) 90,h16i,h25:3866i 48:6134 h i hi, h i,h i Vector Addition 05 03:03, trigonometry,multiple choice, > 1 min. 003 A P C R B The ector R ~ shown in the sketch may be expressed in terms of A, ~ B, ~ C, ~ and D, ~ which are the edges of a parallelogram, as 1. ~ R ~ B + ~ D D

3 Answer, Key { Homework 3 { Rubin H Landau 3 2. ~ R ~ B, ~ A 3. ~ R ~ A, ~ B 4. ~ R ~ C + ~ B 5. ~ R ~ C + ~ D 6. ~ R ~ A + ~ D 7. ~ R ~ A, ~ D 8. ~ R ~ D, ~ A 9. ~ R ~ A, ~ C 10. ~ R ~ B + ~ A correct The resultant, or sum of seeral ectors, is the single ector leading from the tail of the rst ector to the head of the last ector when the ectors are drawn tail to head in any order. 004 The ector P ~ shown in the sketch may be expressed in terms of A, ~ B, ~ C, ~ and D ~ as 1. ~ P ~ A, ~ B correct 2. ~ P ~ B, ~ A 3. ~ P ~ A + ~ D 4. ~ P ~ C + ~ B 5. ~ P ~ C + ~ D 6. ~ P ~ B + ~ A 7. ~ P ~ A, ~ D 8. ~ P ~ D, ~ A 9. ~ P ~ C, ~ A 10. ~ P ~ B + ~ D Wind and a Jet 03:04, calculus, numeric, > 1 min. 006 What is the direction of the aircraft (use counterclockwise from due East as the positie angular direction, between the limits of,180 and +180 )? Correct answer: 11:0616. The direction is h rad arctan y x degi 0: raddeg (1) radi 57:2958 degrad (2) h deg mph mph (5) 80 r hdegi rad h17i h0: i 0: rad hradi h ihraddegi x cos ( r ) (7) h488i + h905i cos (h0:296706i) 1353:46 mph hmphi hmphi + hmphi cos (hradi) y 2 sin ( r ) (8) h905i sin (h0:296706i) 264:596 mph hmphi hmphi sin (hradi) x 2:0 + y 2:0 (9) h1353:46i 2:0 + h264:596i 2:0 1379:08 mph hmphi hmphi 2:0 + hmphi 2:0 : r arctan2 ( y ; x ) (10) arctan2 (h264:596i; h1353:46i) 0: rad hradi arctanhi (hmphi; hmphi) h deg radi r (11) h57:2958i h0:193062i

4 11:0616 h i hdegradi hradi Answer, Key { Homework 3 { Rubin H Landau During a baseball game, a batter hits a popup to a elder 90 m away. If the ball remains in the air for 7 s, how high does it rise? Correct answer: 60:025 m. The distance to the elder is extraneous information and can be ignored. Consider the motion as the ball rises to its maximum height h: t t. The height is 2 dened by h 1 2 g(t)2 1 2 g t2 4 gt2 8 t 7s 5 7 (1) d 90m (2) g 9:8 ms 2 h gt2:0 8:0 h9:8i h7i2:0 8:0 60:025 m hmi hms2 ihsi 2:0 hi Dropping Medical Supplies 009 A plane drops a hamper of medical supplies from a height of 5940 m during a practice run oer the ocean. The plane's horizontal elocity was 143 ms at the instant the hamper was dropped. What is the magnitude of the oerall elocity of the hamper at the instant it strikes the surface of the ocean? Correct answer: 369:964 ms. Basic Concepts Motion in graity eld 2 2gh Solution: This is a projectile motion problem. The motion of the dropping hamper consists of two parts: horizontally, it moes with the initial elocity of the plane, i.e. h 143 ms; ertically, due to Graity, itmoes as a freely falling body. Applying the euation aboe gies the ertical elocity as p 2 gh 341:21 ms Thus the oerall elocity at the instant the hamper strikes the surface of the ocean is f h (341:21 ms) 2 + (143 ms) 2 369:964 ms g 9:8 ms 2 (1) h 5940 m (2) 143 ms h 143 ms p 2:0 gh (5) hmsi p 2:0 h9:8i h5940i 341:21 ms hi hms 2 ihmi 2:0 + h 2:0 h341:21i 2:0 + h143i 2:0 f hmsi 369:964 ms hmsi 2:0 + hmsi 2:0 Moon Orbit 010 The orbit of a Moon about its planet is approximately circular, with a mean radius of 6: m. It takes 39:3 days for the Moon to complete one reolution about the planet.

5 Answer, Key { Homework 3 { Rubin H Landau 5 Find the mean orbital speed of the Moon. Correct answer: 1123:21 ms. Diiding the length C 2r of the trajectory of the Moon by the time T 39:3 days 3: s of one reolution (in seconds!), we obtain that the mean orbital speed of the Moon is C T 2 r T 2 (6: m) 3: s 1123:21 ms : Moon Orbit 011 Find the Moon's centripetal acceleration. Correct answer: 0: ms 2. Since the magnitude of the elocity is constant, the tangential acceleration of the Moon is zero. For the centripetal acceleration we use the formula a c 2 r (1123:21 ms)2 6: m 0: ms 2 : i sday (1) h s day r 6: m 1: : (2) T 39:3 days 10:92 43:68 T u h s i T day h86400i h39:3i 3: s hsi hsdayi hdaysi 2:0 r (5) T u 2:0 h3: i h6: i h3: i 1123:21 ms hmsi a c 2:0 hi hi hmi r hsi h1123:21i2:0 h6: i 0: ms 2 hms 2 i hmsi2:0 hmi