Q3.8: A person trying to throw a ball as far as possible will run forward during the throw. Explain why this increases the distance of the throw.


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1 Problem Set 3 Due: 09/3/, Tuesda Chapter 3: Vectors and Moton n Two Dmensons Questons: 7, 8,, 4, 0 Eercses & Problems:, 7, 8, 33, 37, 44, 46, 65, 73 Q3.7: An athlete performn the lon jump tres to acheve the mamum dstance from the pont of takeoff to the frst pont of touchn the round. After the jump, rather than land uprht, she etends her les forward as n the photo. How does ths affect the tme n the ar? How does ths affect ths ve the jumper a loner rane? Q3.7. Reason: B etendn ther les forward, the runners ncrease ther tme n the ar. As ou wll learn n chapter 7, the center of mass of a projectle follows a parabolc path. B rasn ther feet so that ther feet are closer to ther center of mass, the runners ncrease the tme t takes for ther feet to ht the round. B ncreasn ther tme of flht, the ncrease ther rane. Also, havn ther feet ahead of them means that ther feet wll land ahead of where the would have landed otherwse. Assess: B smpl movn ther feet, runners can chane ther tme of flht and chane the spot where ther feet land. Q3.8: A person trn to throw a ball as far as possble wll run forward durn the throw. Eplan wh ths ncreases the dstance of the throw. Q3.8. Reason: Runnn whle thrown a ball ncreases the dstance of the throw because t ncreases the horzontal component of the ball s veloct wthout chann the tme of flht. Relatve to the round, the ball s horzontal veloct component s ncreased b an amount equal to the speed of the runner. Snce the veloct of the runner s purel horzontal, the vertcal veloct component of the ball s unchaned and so the tme of flht of the ball s unchaned. It s nterestn to note that snce, relatve to the round, the ball has a reater horzontal component of veloct, the anle of the ball s veloct s hher relatve to the person thrown the ball than relatve to the round. However, nether relatve to the round, nor relatve to the person thrown the ball wll a launch anle of 45 ve the mamum rane. The anle 45 ves the mamum rane for a ball launched b someone not movn because t ves the best compromse between havn a hh tme of flht and a lare horzontal component of veloct. But when the person s runnn, the horzontal component of veloct ets an advantae so the tme of flht can be reater at the epense of the horzontal component of veloct. Thus the best anle relatve to the person s reater than 45. Assess: Whle the best launch anle s no loner 45, t s stll true that runnn ncreases the rane of the ball. Q3.: In an amusementpark rde, cars rolln alon at hh speed suddenl head up a lon, straht ramp. The roll up the ramp, reverse drecton at the hhest pont, then roll backward back down the ramp. In each of the follown sements of the moton, are the cars acceleratn, or s ther acceleraton zero? If acceleratn, whch wa does ther acceleraton vector pont? a. As the cars roll up the ramp. b. At the hhest pont on the ramp. c. As the cars roll back down the ramp. Q3.. Reason: The acceleraton s due to ravt, so the acceleraton wll alwas act to pull the cars back down the ramp. Snce the roller coaster s constraned to move alon the ramp, the acceleraton must be alon the ramp. So n all three cases the acceleraton s downward alon the ramp. Assess: Gravt alwas acts, even f moton s constraned b an nclned ramp. See Fure 3.4 n the tet. Q3.4: You are drvn our car n a crcular path on level round at a constant speed of 0 mph. At the nstant ou are drvn north, and turnn left, are ou acceleratn? If so,
2 toward what pont of the compass (N, S, E, W) does our acceleraton vector pont? If not, wh not? Q3.4. Reason: Because acceleraton s defned as the chane n the (vector) veloct dvded b the correspondn tme nterval, t, too, s a vector. a = v t A car n unform crcular moton (constant speed) s stll acceleratn because the drecton of the veloct vector s chann. The wa to determne the drecton of the acceleraton vector s to realze that a must alwas pont n the same drecton as v because t s not onl a scalar, t s a postve scalar (as lon as tme doesn t o backward). v v v, f so, to determne the drecton of v, look at the veloct vector arrows just before and just after the car s ponted north, and subtract them usn vector subtracton. v = v v = v ( v ) and, as shown n the fure, v ponts west. f f Assess: It s worth notn that the specfc drectons were not as mportant as the concluson that the v vector and therefore the a vector both pont toward the center of the crcle n unform crcular moton. Q3.0: A car travels at constant speed alon the curved path shown from above n the fure. Fve possble vectors are also shown n the fure; the letter E represents the zero vector. Whch vector best represents a. The car's veloct at poston? b. The car's acceleraton at pont? c. The car's veloct at poston? d. The car's acceleraton at pont? e. The car's veloct at poston 3? f. The car's acceleraton at pont 3? Q3.0. Reason: The car s traveln at constant speed, so the onl possble cause for acceleratons s a chane n drecton. (a) At pont the car s traveln straht to the rht on the daram, so ts veloct s straht to the rht. The correct choce s B. (b) At pont the car s traveln at constant speed and not chann drecton so ts acceleraton s zero. The correct choce s E. (c) The car s veloct at ths pont on the curve s n the drecton of ts moton, whch s n the drecton shown at choce C. (d) The car s movn on a porton of a crcle. The acceleraton of an object movn n a crcle s alwas drectl toward the center of the crcle. The correct choce s D. (e) The car s movn on a porton of a crcle at pont 3. The nstantaneous veloct vector s drectl to the rht, whch s choce B.
3 (f) The car s acceleratn because t s movn on a porton of a crcle. The acceleraton s toward the center of the crcle, whch s n drecton A. Assess: The nstantaneous veloct of a partcle s alwas n the drecton of ts moton at that pont n tme. For moton n a crcle, the drecton of the acceleraton s alwas toward the center of the crcle. P3.: Draw each of the follown vectors, then fnd ts  and components. a. d (.0km, 30 left of as) b. v (5.0 cm/s, as) c. a (0 m/s, 40 left of as) P3.. Prepare: We wll follow rules ven n Tactcs Bo 3.. Solve: (a) d ( km) sn 30 km d ( km) cos 30.7 km (b) v (5 cm/s) sn 90 5 cm/s v (5 cm/s) cos 90 0 cm/s (c) a (0 m/s ) sn m/s a (0 m/s ) cos m/s Assess: The components have the same unts as the vectors. Note the mnus sns we have manuall nserted accordn to the Tactcs Bo 3.3. P3.7: A car traveln at 30 m/s runs out of as whle traveln up a 5.0 slope. How far wll t coast before startn to roll back down? P3.7. Prepare: A vsual overvew of the car s moton that ncludes a pctoral representaton, a moton daram, and a lst of values s shown below. We have labeled the as alon the nclne. Note that the problem ends at a turnn pont, where the car has an nstantaneous speed of 0 m/s before rolln back down. The rolln back moton s not part of ths problem. If we assume the car rolls wthout frcton, then we have moton on a frctonless nclned plane wth acceleraton a = sn = sn 5.0 = m/s. Solve: Constant acceleraton knematcs ves v (30 m/s) vf v a( f ) 0 = v af f 530 m a ( m/s ) Notce how the two neatves canceled to ve a postve value for f. 3
4 Assess: We must nclude the mnus sn because the a vector ponts down the slope, whch s n the neatve drecton. P3.8: A ball wth a horzontal speed of.5 m/s rolls off a bench.00 m above the floor. a. How lon wll t take the ball to ht the floor? b. How far from a pont on the floor drectl below the ede of the bench wll the ball land? P3.8. Prepare: We can use the vertcal part of the moton to calculate the tme t takes the ball to ht the floor and the horzontal part of the moton to fnd out how far from the bench t lands. Solve: Refer to the vsual overvew shown. The ntal vertcal veloct s zero. Take the floor as the orn of coordnates. The ball falls from =.00 m and lands at f = 0 m. (a) We can use the vertcalposton equaton from Equatons 3.5 to fnd the tme t takes the ball to reach the floor. Solvn for t, f ( v) t ( t) 0.00 m.00 m (9.80 m/s )( t) (.00m) t 0.45s 9.80m/s (b) The dstance the ball travels horzontall s overned b the horzontalposton equaton from Equatons 3.5. ( v ) t (.5 m/s)(0.45 s) = m f Assess: Ths seems reasonable. Compare to Eample 3. n the tet, whch s smlar. P3.33: On the Apollo 4 msson to the moon, astronaut Alan Sheppard ht a olf ball wth a olf club mprovsed from a tool. The freefall acceleraton on the moon s /6 of ts value on earth. Suppose he ht the ball wth a speed of 5 m/s at an anle 30 above the horzontal. a. How lon was the ball n flht? b. How far dd t travel? c. Inorn ar resstance, how much farther would t travel on the moon than on earth? P3.33. Prepare: The olf ball s a partcle follown projectle moton. We wll appl the constantacceleraton knematc equatons to the horzontal and vertcal motons as descrbed b Equatons
5 Solve: (a) The dstance traveled s f = (v ) t f = v cos t f. The flht tme s found from the equaton, usn the fact that the ball starts and ends at = 0: Thus the dstance traveled s For = 30, the dstances are The flht tmes are v sn f 0 v sn tf t f ( v sn t f ) tf tf f v cos f earth earth 9.80 m/s v sn v sn cos v sn cos (5 m/s) sn 30cos 30 ( ) 55. m v sn cos v sn cos v sn cos ( ) 6 6( ) 33. m f moon f earth moon 6 earth earth v sn ( tf ) earth.55 s earth v sn v sn ( t ) 6( t ) 5.30 s f moon f earth moon 6 earth The ball spends 5.30 s.55 s =.75 s = 3 s loner n flht on the moon. (b) From part (a), the dstance traveled on the moon s 33 m or 330 m to two snfcant fures. (c) From part (a), the olf ball travels 33. m 55. m = 76 m farther on the moon than on earth. P3.37: A CDROM drve n a computer spns the cmdameter dsks at 0,000 rpm. a. What are the dsk's perod (n s) and frequenc (n rev/s)? b. What would be the speed of a speck of dust on the outsde ede of ths dsk? c. What s the acceleraton n unts of that ths speck of dust eperences? P3.37. Prepare: We are asked to fnd perod, speed and acceleraton. Perod and frequenc are nverses accordn to Equaton 3.6. To fnd speed we need to know the dstance traveled b the speck n one perod. Then the acceleraton s ven b Equaton Solve: (a) The dsk s frequenc can be converted as follows: The perod s the nverse of the frequenc: rev rev mn rev 0,000 0, mn mn 60 sec sec T 6.00 ms f 67 rev/s (b) The speed of the speck equals the crcumference of ts orbt dvded b the perod: whch rounds to 63 m/s. r (6.0 cm) 000 ms m v 6.8 m/s, T 6.00 ms s 00 cm (c) From Equaton 3.30, the acceleraton of the speck s ven b Whch rounds to v / r : v a r 6.0 cm m 66,000 m/s. In unts of, ths s as follows: (6.8 m/s) 00 cm 65,700 m/s, 9.8 m/s 65,700 m/s 65,700 m/s 6,700 5
6 Assess: The speed and acceleraton of the ede of a CD are remarkable. The speed, 63 m/s, s about 40 m/hr. As ou wll learn n chapter 4, ver lare forces are necessar to create lare acceleratons lke 6,700. P3.44: Suppose E A 3B where vector A has components A = 5, A = and vector B has components B = 3, B = 5. a. What are the  and components of vector E? b. Draw a coordnate sstem and on t show vectors A, B, and E? c. What are the mantude and drecton of vector E? P3.44. Prepare: Because A A A, and B B B, so the components of the resultant vector are E = A + 3B and E = A + 3B. The vectors ABand,, E A 3B are shown. Solve: (a) Wth A = 5, A =, B = 3, and B = 5, we have E = and E =. (b) Vectors A, B, and E are shown n the above fure. (c) From the E vector, E and E. Therefore, the mantude and drecton of E are E () ( ) E tan tan 5. E Assess: Note that s the anle made wth the as, and that s wh tan ( E / E ) rather than tan ( E / E) whch would be the case f were the anle made wth the as. P3.46: Let A (3.0 m, 0 south of east), B (.0 m, north), and C (5.0 m, 70 south of west). a. Draw and label A, B, and C wth ther tals at the orn. Use a coordnate sstem wth the as to the east. b. What are the  and components of vectors A, B, and C. c. Fnd the mantude and drecton of vector D A B C. P3.46. Prepare: The vectors ABand,, C are shown. We wll frst calculate the  and components of each vector and then obtan the mantude and the drecton of the vector D. 6
7 Solve: (a) The vectors A, B, and C are drawn above. (b) The components of the vectors A, B, and C are A (3 m) cos 0.8 m and A (3 m) sn 0.03 m; B 0m and B m; C (5 m) cos 70.7m and C (5 m) sn m. (c) We have D A B C D D, whch means D =. and D D (. m) (3.73 m) 3.9 m tan tan The drecton of D s south of east, 73 below the postve as P3.65: In 780, n what s now referred to as "Brad's Leap," Captan Sam Brad of the U.S. Contnental Arm escaped certan death from hs enemes b runnn over the ede of the clff above Oho's Cuahoa Rver, whch s confned at the spot to a ore. He landed safel on the far sde of the rver. It was reported that he leapt ft across whle 0 ft. a. Representn the dstance jumped as L and the vertcal drop as h (as shown n the fure) derve an epresson for the mnmum speed v he would need to make hs leap f he ran straht off the clff? b. Evaluate our epresson for a ft jump wth a 0 ft drop to the other sde? c. Is t reasonable that a person could make ths leap? Use the fact that the world record for the 00 m dash s appromatel 0 s to estmate the mamum speed such a runner would have. P3.65. Prepare: Ths problem s somewhat smlar to Problem 3.7 wth all of the ntal veloct n the horzontal drecton. We wll use the vertcal equaton for constant acceleraton to determne the tme of flht and then see how far Captan Brad can o n that tme. Of nterest s the fact that we wll do ths twostep problem completel wth varables n part (a) and onl plu n numbers n part (b). We could do part (b) n feet (usn = 3 ft/s ), but to compare wth the world record 00 m dash, let s convert to meters. L = ft = 6.7 m and h = 0 ft = 6.0 m. Solve: (a) Gven that (v ) = 0.0 ft/s we can use Equaton.4. Wth up as the postve drecton, Solve for t. = a ( t) s neatve and a = ; those sns cancel leavn h = ( t) t = h 7
8 Now use that epresson for Fnall solve for v = v n terms of L and h. t n the equaton for constant horzontal veloct. L = = v t = v h = L v = L (b) Now plu n the numbers we are ven for L and h. v h h 9.8 m/s = L = (6.7m) = 6.0 m/s h (6.0 m) Compare ths result (v = 6.0 m/s) wth the worldclass sprnter (v = 0 m/s); a ft person could make ths leap. Assess: The results are reasonable, and not obvousl wron. 6.0 m/s 3 mph, and that would be a fast run, but certanl possble. B solvn the problem frst alebracall before plun n an numbers, we are able to substtute other numbers as well, f we desre, wthout resolvn the whole problem. P3.73: A chld sldes down a frctonless 3.0mlon plaround slde tlted upward at an anle of 40. At the end of the slde, there s an addtonal secton that curves so that the chld s launched off the end of the slde horzontall. a. How fast s the chld movn at the bottom of the slde? b. If the end of the slde s 0.40 m above the round, how far from the end does she land? P3.73. Prepare: Frst draw a pcture. In part (a) use tlted aes so the as runs down the slde. The acceleraton wll be a = sn. Part (b) s a famlar twostep projectle moton problem where we use the vertcal drecton to determne the tme of flht and then plu t nto then equaton for constant horzontal veloct. Use aes that are not tlted for part (b). Solve: (a) We use Equaton.3 wth ( v ) = 0.0 m/s. ( v ) = a f o ( v) f = ( sn ) = (9.8 m/s )(sn40 )(3.0 m) = 6.m/s (b) We use Equaton.4 to fnd the tme for an object to fall 0.4 m from rest: = 0.4m. = a ( t) ( 0.4 m) t 0.86 s 9.8 m/s At last we combne ths nformaton nto the equaton for constant horzontal veloct. 8
9 = v t = (6.5 m/s)(0.86 s) =.8 m Assess: We reported the speed at the bottom of the slde to two snfcant fures, but kept track of a thrd to use as a uard dt because ths result s also an ntermedate result for the fnal answer. We also kept a thrd snfcant fure on the t as a uard dt. The result of landn.8 m from the end of the frctonless slde seems just a bt lare because ths slde was frctonless and real sldes aren t, but t doesn t seem to be too far out of epectaton, so our result s probabl correct. 9
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