PHY053 F05 Exam1a Sol

PHY053 F05 Exam1a Sol - PHY53 F-05 Exam 1 A Name: 30 (.077...

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Unformatted text preview: PHY53 F-05 Exam 1 A Name: 30 (.077 0N3 Section: __ I certzfi; that I have abided by the rules and the spirit of the Duke Community Standard MC P1 P2 P3 Total: In questions or problems not requiring numerical answers, express your answers in terms of the symbols given, and standard constants such as g. If numbers are required, use g = 10 m/sz. Record your MC answers here: MC1_E'_ MC22 MC3}; MC4 MCSQ Multiple Choice: 6 Points Each 1. In the absence of wind, raindrops fall down vertically with a speed of 6 m/s. As seen from a forward moving car, they seem to fall at an angle of 30° with respect to the horizontal. What is the relative speed with which they hit the windshield of the car? A) 3.00 m/s I B) 6.93 m/s , V V C) 9.00 m/s V = 55—; = AIR 5): Veal» D) 8.49 m/s ( ° @20 m/s. V = “(S _Vm’ 2. The gravitational acceleration near the Earth’s surface is about 6 times that on the surface of the Moon. Neglecting air resistance, an object thrown from the ground at the same speed and with the same angle of 45° with respect to the vertical will land on the ground A) At the same distance on the Moon and the Earth. '3 B) 6 times farther on Earth. C 36 times farther on Earth. 67* x D)~ . times farther on the Moon. . E) 86 times farther on the Moon. ‘ d g 1 Va: V1 3 . 3. When attempting to stop a car as quickly as possible on a dry pavement, which method will bring the car to a complete stop over the shortest distance: A) Slamming the brakes as hard as possible, “locking” (i.e. keeping them from turning) the wheels and skidding to a stop. raking as hard as possible without locking any of the four wheels. C) Braking as hard as possible without “locking” the front wheels. D) Braking as hard as possible without locking the back wheels. E) It doesn’t matter, it always takes the same distance. 51.4.: 74.24;». > [01.04.22 @450». 4. A flood victim is hauled up to the rescue helicopter by a rope. The upward acceleration does not exceed 6 m/s2. What is the largest mass the victim may have without risking an accident, if the rope can withstand a force of 4,000 N without breaking: A) 100 kg , B) 150 kg F m (3+ 0‘) C 200 kg @250kg ~ mw= F - “W”: = Z5“: (:3 E) 300 kg 7+ 4% l‘ “I5 5. The position x of an object in an inertial reference frame is given by: x = 5 - 2t + t3, where tdenotes time (in seconds). From this relationship you can deduce that the A) initial acceleration (at t = 0) of the object is a = 6 (In/82). @net force applied on the object is always zero. acceleration of the object always increases with time. v: '2‘, 3.5:. D) initial velocity (at t = 0) of the object is zero. E) velocity of the object always decreases with time. 1: a The followin roblem is worth 10 oints: 6. The period of our Moon’s orbit around the Earth is 27.3 days (2.36 x 106 s). The Moon’s distance from the center of the Earth is 386 000 km. One .of Neptune’s moons, Triton, has an orbital period 0 hours and an orbital radius of 355, . 54°35 a 5- Determine the ratio of the mass of the Earth and to the mass of Neptune. ~ =- (5:08,. ’0 5‘5) (Assume that both moons are on circular orbits around their respective planets.) ’ L . TL G T‘G m if» ’3 ‘ H I “L 1:27;; am = 00!?3 = "“1 - l1 Nari-wt: The following problem is worth 40 [=5V-I-10fl-10+10+5|'Qoints: 7. Two children connected by a short rope are sliding down a 10m long slide, which is inclined at an angle 30° with respect to the horizontal. Bob, who is in front, wears clothes with a kinetic friction coefficient of 0.34; Alice,,who is behind, wears clothes with a kinetic friction coefficient of 0.52. Bob is twice as heavy as Alice whose mass is 20 kg. [Assume that both children slide down the same distance of 10m. Neglect the length of the rope] Draw a free body diagram for each of the children. Find the tension in the rope. Calculate the total work done by the friction force on the two children. . Find their combined kinetic energy at the bottom of the slide. Show that the answers to C. and D. satisfy the work-energy theorem. P‘JUOF’?’ (3) «9,4 5. mkgake x-‘T' —Mmfig.m6 g a) “‘4 T: M ' ) 6:248 6;) : 3 A (MMA+M‘M3)’M6L :‘2'976’ (3)) Cambi— dme. 4k Zwojse (f) ‘Frm (754*): Q: 5?:de ' Wgcma (1‘5) Fm h Wit-Mu” Mm: AK ‘ i. = fr+ Us -wwkdm 63 fn‘cfi‘m Md jam? : 11):: (rah-Has) 9M9 1 3,900 T -* AK=KO+ “f “P ‘ “zoom-1P7“: “213’ The following problem is worth 30 |=5+15+10| points: 8. A basketball player standing 20 m away from the opposing goal wants to score last- second points to win a close game. In order to make sure that he “nails” the shot, he wants the ball to enter the hoop at an angle of precisely 30° with respect to the vertical. A. Draw a diagram of the motion of the ball from its release until it passes through the basket. B. If the ball leaves his hands at a height of 2 m and the goal is mounted 3 m above the floor, what is the angle (with respect to the horizontal) at which he has to shoot the ball? m: You need to use the equations for the position and the velocity of the ball as a function of time.] C. How long does it take for the ball to reach the goal? ‘604490= V *— 104494- A: {5+3 = 99: 60V. . d _ (C) Fm W) %3+ 4M9: 2.(+me+g)=gfi - {fie ——, {32— Lg—(Mata‘l) =’;-(J4~.94U {. gnaw) = 2.675 PHY53 F-OS Exam 1 B Name: 50 L U I l W“ Section: _ I certify that I have abided by the rules and the spirit of the Duke Community Standard. MC P1 P2 P3 Total: In questions or problems not requiring numerical answers, express your answers in terms of the symbols given, and standard constants such as g. If numbers are required, use g = 10 m/sz. Record your MC answers here: MC1£MC2£ MC3_C_ MC4_2_E MC5_Z Multi le Choice: 6 Points Each 1. In the absence of wind, raindrops fall down vertically with a speed of 6 m/s. As seen from a forward moving" car, they seem to fall at an angle of 60° with respect to the horizontal. What is the relative speed with which they hit the windshield of the car? A 5.20m/s I v ($56.93 m/s /_ R ' v’ )9.00m/s V" ‘ a: V“ D) 10.4 m/s “730 g a E) 12.0m/s. = 633 w($ ‘er 2. The gravitational acceleration near the Earth’s surface is about 2.56 times that on the surface of Mars. Neglecting air resistance, an object thrown from the ground at the same speed and with the same angle of 30° with respect to the vertical will land on the ground A) At the same distance on Mars and Earth. '3 a , B) 2.56 times farther on Earth. ' C) 1.6 times farther on Earth. (—-—-—+ K D 1.6 times farther on Mars. ‘1 @256 times farther on Mars. 4 3 a VxVy 3. When attempting to stop a car as quickly as possible on a dry pavement, which method will bring the car to a complete stop over the shortest distance: A) Braking as hard as possible without “looking” (i.e. keeping them from turning) the front wheels B Braking as hard as possible without locking the back wheels @Braking as hard as possible without locking any of the four wheels D) Slamming the brakes as hard as possible, locking the wheels and skidding to a stop E) It doesn’t matter, it always takes the same distance :5th fwh’d‘a‘m > kthct‘t‘c fn‘dfibn 4. A flood victim is hauled up to the rescue helicopter by a rope. The upward acceleration does not exceed 5 m/sz. What is the largest mass the victim may have without risking an accident, if the rope can withstand a force of 3,000 N without breaking: ‘. 100kg ‘ ‘33 00kg F “(3+0 C) 3001“; F 3°01»! ED) 400kg “4m — +4 — (g [5‘ s looks ) 600kg 7 m «n 5. The position x of an object in an inertial reference flame is given by: x = 5 - 312 + 2t3, where t denotes time (in seconds). From this relationship you can deduce that the A) initial acceleration (at t = 0) of the object is a = -3 (m/sz). B) net force applied on the object always decreases with time. C acceleration of the object is never zero. 7.. initial velocity (at t = 0) of the object is zero. -‘-' ’ 6* f at E) velocity of the object always increases with time. = _ 6 + {at The followin roblem is worth 10 oints: 6. The period of our Moon’s orbit around the Earth is 27.3 days (2.36 X 106 s). The Moon’s distance from the center of the Earth is 386,000 km. One of Mars’ moons, Phobos, has an orbital period of 0.3 19 days (2.76 X 104 s) and an orbital radius of 9,380 km. Determine the ratio of the mass of the Earth and the mass of Mars. (Assume that both moons are on circular orbits around their respective planets.) 7, mm air: _ 9727:” Ma) r = 6—— —-—, : ——-——- - , ' 1'?- H G (LG I; H ULY'3 I r3 The following problem is worth 40 |=5+10+10+10+5| goints: 7. Two children connected by a short rope are sliding down a 10m long slide, which is inclined at an angle 30° with respect to the horizontal. Bob, who is in front, wears clothes with a kinetic friction coefficient of 0.35; Alice, who is behind, wears clothes with a kinetic friction coefficient of 0.65. Bob is twice as heavy as Alice whose mass is 15 kg. [Assume that both children slide down the same distance of 10m. Neglect the length of the rope.] A. Draw a free body diagram for each of the children. B. Find the tension in the rope. L‘ (O M C. Calculate the total work done by the friction force on the two children. A; mete D. Find their combined kinetic energy at the bottom of the slide. 3 = 3.5 E. Show that the answers to C. and D. satisfy the work-energy theorem. 6 3 o s a .. ‘ .. ca: 0 (3) mid - Mk5 “9 *7 M “A 5 } a) (.9) Cam R 610% «M 26.2435.- firm $3. Q: 3 "‘ wa56 N*“‘; 4 3 ’(Jo «c/Sa’ V3 EJ— = 7.70 (“/5 "" K=€(mkffl5)vzslmfiffls)fll_ K= m g . (5") Fran WL—%UJj Wow-em: AK dig-e— wfr . War-IL done 57 fn‘cfi‘on radian/15. ' 1%: («4+»ng she = 2,853} -+ K: Ko+ as“)? = OfZ.ZSTPJ-l.7$‘g= 776;. (E) 45;: (lg/+51% , «seam/{y frww. (I?) éawl (6) . t.) fwd)? a [UflwU The following problem is worth 30 |=5+15+10| points: 8. A basketball player standing 30 m away from the opposing goal wants to score last- second points to win a close game. In order to make sure that he “nails” the shot, he wants the ball to enter the hoop at an angle of precisely 45° with respect to the horizontal. A. Draw a diagram of the motion of the ball from its release until it passes through the basket. ' B. If the ball leaves his hands at a height of 2 m and the goal is mounted 3 m above the floor, what is the angle (with respect to the horizontal) at which he has to shoot the ball? [1:111]; You need to use the equations for the position and the velocity of the ball as a function of time] C. How long does it take for the ball to reach the goal? d’3°"‘ 19‘!- 645/46: J , (as) a1:th .. #9; h=vajt 'é9£2=%J—éyé€ (n 53.4! wile.- fm€= él-s-fl- +g_ = --_\_’gy_’+ 22’. (ink) 05 .4 \6x L ‘4‘ Add? (X (*) fares: .. v -1. V0 V 1.- .354 a (M6+E;V)d= g fia- {WE-cl {Mat}: {:3 = {Me +23 - 4+3": 34.07. 6;: «4.7". V, (c) FM ("0 ifng = Z{t‘m€ +25):ng = ELIE" d —_’ ‘5‘: 2’?(t“‘e+fil) 3‘ §Z'(d+m0+l‘) é-—¥§(d+z.) = 2.973 . (1515 f“ 7‘7"?) ...
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PHY053 F05 Exam1a Sol - PHY53 F-05 Exam 1 A Name: 30 (.077...

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