Version 001 Review 1: Mechnics tubmn (IBII20142015) 1 This print-out should hve 72 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Displcement Curve 02 001 (prt 1 of 2) 10.0 points Consider moving object whose position x is plotted s function of the time t. The object moved in different wys during the time intervls denoted I, II nd III on the figure. 6 4 2 x I II III 2 4 6 During these three intervls, when ws the object s speed highest? Do not confuse the speed with the velocity. 1. Sme speed during ech of the three intervls. 2. During intervl I 3. Sme speed during intervls II nd III 4. During intervl II 5. During intervl III correct Thevelocityvistheslopeofthex(t)curve; the mgnitude v = v of this slope is the speed. The curve is steepest (in bsolute mgnitude) during the intervl III nd tht is when the object hd the highest speed. 002 (prt 2 of 2) 10.0 points During which intervl(s) did the object s velocity remin constnt? 1. During intervl II only t 2. During ech of the three intervls correct 3. During none of the three intervls 4. During intervl III only 5. During intervl I only orechofthethreeintervlsi,iioriii,the x(t) curve is liner, so its slope (the velocity v) is constnt. Between the intervls, the velocity chnged in n brupt mnner, but it did remin constnt during ech intervl. Distnce Time Grph 01 003 (prt 1 of 6) 10.0 points Consider the following grph of motion. 50 Distnce (m) 40 30 20 10 0 0 1 2 3 4 5 Time (sec) How fr did the object trvel between 2 s nd 4 s? 1. 30 m 2. 10 m 3. 40 m 4. 50 m 5. 20 m correct The prticle moved from 40 m to 20 m, so d = 40 m 20 m = 20 m. 004 (prt 2 of 6) 10.0 points The grph indictes 1. constnt position.
Version 001 Review 1: Mechnics tubmn (IBII20142015) 2 2. decresing velocity. 3. constnt velocity. correct 4. no motion. 5. incresing velocity. The slope of the grph is the sme everywhere, so the grph indictes constnt positive velocity. 005 (prt 3 of 6) 10.0 points Wht is the speed from 2 s to 4 s? 1. 15 m/s 2. 10 m/s correct 3. 5 m/s 4. 20 m/s 5. 0 m/s 3. 400 m 4. 250 m 5. 350 m correct 6. 150 m 7. 450 m 8. 500 m 9. 50 m 10. 100 m The prticle moved from 50 m to 400 m, so the distnce d = 400 m 50 m = 350 m. 007 (prt 5 of 6) 10.0 points The grph indictes 1. no motion. v = d t = 40 m 20 m 2 s = 10 m/s. 2. incresing velocity. correct 3. constnt velocity. 006 (prt 4 of 6) 10.0 points Consider the following grph of motion. Distnce (m) 400 350 300 250 200 150 100 50 0 0 1 2 3 4 5 6 7 8 9 Time (sec) How fr did the object trvel between 3 s nd 9 s? 1. 200 m 2. 300 m 4. constnt position. 5. decresing velocity. The slopes re steeper s time goes on, so the velocities re incresing. 008 (prt 6 of 6) 10.0 points Wht is the verge speed from 3 s to 9 s? 1. 20 m/s 2. 60 m/s 3. 30 m/s 4. 58 m/s correct
Version 001 Review 1: Mechnics tubmn (IBII20142015) 3 5. 47 m/s 6. 40 m/s 4. t 7. 36 m/s 8. 50 m/s 9. 25 m/s 5. t v = d t = 400 m 50 m 6 s = 58 m/s. 6. t Accelertion Time Grph 01 009 (prt 1 of 5) 10.0 points Consider toy cr which cn move to the right (positive direction) or left on horizontl surfce long stright line. cr O + Wht is the ccelertion-time grph if the cr moves towrd the right (wy from the origin), speeding up t stedy rte? 1. t 2. t v 7. t 8. t correct Since the cr speeds up t stedy rte, the ccelertion is constnt. 010 (prt 2 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moves towrd the right, slowing down t stedy rte? 1. t 3. None of these grphs is correct.
Version 001 Review 1: Mechnics tubmn (IBII20142015) 4 2. t 1. t 3. t correct 2. t 4. t 3. t 5. t 4. t 6. None of these grphs is correct. 7. t 5. t 8. t 6. t correct Since the cr slows down, the ccelertion is in the opposite direction. 011 (prt 3 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moves towrds the left (towrd the origin) t constnt velocity? 7. None of these grphs is correct. 8. t Since the cr moves t constnt velocity,
Version 001 Review 1: Mechnics tubmn (IBII20142015) 5 the ccelertion is zero. 012 (prt 4 of 5) 10.0 points Wht is the ccelertion-time grph if the cr movestowrdtheleft,speedinguptstedy rte? 8. t 1. t 2. t 3. None of these grphs is correct. 4. t The sme reson s Prt 1. 013 (prt 5 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moves towrd the right t constnt velocity? 1. t 2. t 5. t 3. t 6. t 4. t 7. t correct 5. t
Version 001 Review 1: Mechnics tubmn (IBII20142015) 6 6. t v x correct y v f θ v y 7. t d 8. None of these grphs is correct. The sme reson s Prt 3. Accelerting Between Wlls 02 014 (prt 1 of 2) 10.0 points A prticle trvels horizontlly between two prllel wlls seprted by 18.4 m. It moves towrd the opposing wll t constnt rte of 8.5 m/s. Also, it hs n ccelertion in the direction prllel to the wlls of 3.2 m/s 2. The horizontl motion will crry the prticle to the opposite wll, so d = v x t f t f = d v x = 18.4 m 8.5 m/s = 2.16471 s is the time for the prticle to rech the opposite wll. Horizontlly, the prticle reches the mximum prllel distnce when it hits the opposite wll t the time of t = d v x, so the finl prllel velocity v y is 3.2 m/s 2 8.5 m/s 18.4 m Wht will be its speed when it hits the opposing wll? Correct nswer: 10.9651 m/s. Let : d = 18.4 m, v x = 8.5 m/s, = 3.2 m/s 2. v y = t = d = (3.2 m/s2 )(18.4 m) v x 8.5 m/s = 6.92706 m/s. The velocities ct t right ngles to ech other, so the resultnt velocity is v f = vx 2 +vy 2 = (8.5 m/s) 2 +(6.92706 m/s) 2 = 10.9651 m/s. 015 (prt 2 of 2) 10.0 points At wht ngle with the wll will the prticle strike? Correct nswer: 50.8218. When the prticle strikes the wll, the verticl component is the side djcent nd the
Version 001 Review 1: Mechnics tubmn (IBII20142015) 7 horizontl component is the side opposite the ngle, so tnθ = v x v y θ = rctn = rctn ( vx v y ) ( ) 8.5 m/s 6.92706 m/s = 50.8218. rustrted With It All 016 (prt 1 of 2) 10.0 points A disgruntled physics student, frustrted with finls, releses his tensions by bombrding the djcent building, 12 m wy, with wter blloons. He fires one t 40 from the horizontl with n initil speed of 24.4 m/s. The ccelertion of grvity is 9.8 m/s 2. or how long is the blloon in the ir? 1. 0.76511 s 2. 1.10657 s 3. 0.642004 s correct 4. 4.97959 s 5. 2.21313 s 6. 0.491803 s 7. 1.56492 s 8. 2.4898 s Bsic Concepts s = s o +v o t+ 1 2 t2 v = v o +t The only thing the horizontl nd verticl motions hve in common is the time the projectile ws in flight. Solution Since the initil velocity is inclined t θ to the horizontl, its sine component is the initil verticl velocity nd its cosine component is the initil horizontl velocity. Thus nd v oy = vsinθ v ox = vcosθ The horizontl motion crries the projectile to its destintion. Since there is no horizontl ccelertion, the horizontl velocity remins constnt nd x = v ox t so tht t = x v ox 017 (prt 2 of 2) 10.0 points How fr bove the initil lunch height does the blloon hit the opposing building? 1. 8.04957 m correct 2. 17.6845 m 3. 14.0196 m 4. 9.98037 m 5. 10.0692 m 6. 2.01963 m 7. 13.6453 m 8. 12.0888 m Verticlly, the initil velocity cts upwrd nd grvity supplies the (downwrd) ccelertion, so y = v oy t 1 2 gt2 Troglodyte Wr 018 (prt 1 of 2) 10.0 points A colony of troglodytes hs been in lengthy feud with its neighbors on the djcent cliff. Colony A finlly develops n importnt militry brekthrough: it rolls bombs off its cliff
Version 001 Review 1: Mechnics tubmn (IBII20142015) 8 t known rtes of speed, thus gining pinpoint ccurcy in its ttcks. The verticl motion crries the bomb to its trget, so COLONY A h COLONY B h = 1 2 gt2 1 2h t 1 = g If the cliffs re seprted by 47.7 m nd bomb is rolled t 6.6 m/s, how fr down the opposite cliff will it lnd? The ccelertion due to grvity is 9.8 m/s 2. Correct nswer: 255.944 m. Let : x = 47.7 m, v o = 6.6 m/s, nd g = 9.8 m/s 2. Horizontlly, there is no ccelertion, so x = v o t;verticlly,there isno initilvelocity, so y = 1 2 gt2. x The horizontl motion crries the bomb to the opposite cliff, so t = x v o nd the distnce it drops is y = 1 2 gt2 = gx2 2v 2 o = 255.944 m. = ( 9.8 m/s 2 ) (47.7 m) 2 2(6.6 m/s) 2 019 (prt 2 of 2) 10.0 points The troglodyte wr continues, nd prticulrly offensive member of colony B is locted 140 m below the top. Atwhtspeedmustbombberolledtoget him? Correct nswer: 8.92385 m/s. Let : h = 140 m. nd the required initil horizontl speed would be v = x g 9.8 m/s = x = (47.7 m) 2 t 1 2h 2( 140 m) = 8.92385 m/s. AP M 1998 MC 11 020 10.0 points AstelliteofmssM movesincirculrorbit of rdius R with constnt speed v. True sttements bout this stellite include which of the following? I) Its ngulr speed is v R. II) Its tngentil ccelertion is zero. III) The mgnitude of its centripetl ccelertion is constnt. 1. I only 2. I, II, nd III correct 3. II only 4. II nd III only 5. I nd III only Here we will only list some fcts of the circulr orbitl movement: 1. The ngulr speed is ω = v R. 2. The ccelertion points to the center of the circulr orbit with mgnitude v2 R = ω2 R. 3. The centripetl force is therefore the mss M times the ccelertion. 4. The period is 2πR v = 2π ω.
Version 001 Review 1: Mechnics tubmn (IBII20142015) 9 Bll of Rope 021 (prt 1 of 3) 10.0 points An thlete swings 3.54 kg bll horizontlly on the end of rope. The bll moves in circle of rdius 0.67 m t n ngulr speed of 0.43 rev/s. Wht is the tngentil speed of the bll? Correct nswer: 1.81019 m/s. v t = rω Let r = 0.67 m nd ω = 0.43 rev/s. = (0.67 m)(0.43 rev/s) 2π 1 rev = 1.81019 m/s. 022 (prt 2 of 3) 10.0 points Wht is its centripetl ccelertion? Correct nswer: 4.8907 m/s 2. c = rω 2 = (0.67 m)(0.43 rev/s) 2 ( 2π 1 rev = 4.8907 m/s 2. ) 2 023 (prt 3 of 3) 10.0 points If the mximum tension the rope cn withstnd before breking is 152.6 N, wht is the mximum tngentil speed the bll cn hve? Correct nswer: 5.37419 m/s. Let : T = 152.6 N. T = mv2 r T r v mx = m (152.6 N)(0.67 m) = 3.54 kg = 5.37419 m/s. Moon Orbit 024 (prt 1 of 2) 10.0 points The orbit of Moon bout its plnet is pproximtely circulr, with men rdius of 4.87 10 8 m. It tkes 26.3 dys for the Moon to complete one revolution bout the plnet. ind the men orbitl speed of the Moon. Correct nswer: 1346.6 m/s. DividingthelengthC = 2πr ofthe trjectory of the Moon by the time T = 26.3 dys = 2.27232 10 6 s of one revolution(in seconds!), we obtin tht the men orbitl speed of the Moon is v = C T = 2πr T = 2π(4.87 108 m) 2.27232 10 6 s = 1346.6 m/s. 025 (prt 2 of 2) 10.0 points ind the Moon s centripetl ccelertion. Correct nswer: 0.00372349 m/s 2. Since the mgnitude of the velocity is constnt, the tngentil ccelertion of the Moon is zero. The centripetl ccelertion is c = v2 r (1346.6 m/s)2 = 4.87 10 8 m = 0.00372349 m/s 2.
Version 001 Review 1: Mechnics tubmn (IBII20142015) 10 Conceptul 04 05 026 (prt 1 of 3) 10.0 points Suzie (of mss 54 kg) is roller-blding down the sidewlk going 24 miles per hour. She notices group of workers down the wlkwy who hve unexpectedly blocked her pth, nd she mkes quick stop in 0.2 seconds. Wht is Suzie s verge ccelertion? Correct nswer: 53.6333 m/s 2. 1. the friction between the ir nd Suzie 2. the grvity 3. All of these 4. the friction between the ground nd the sktes correct 5. the upwrd force exerted by the ground Let : v i = 24 mi/h, t = 0.2 s, nd v f = 0 m/s. irst we need to convert Suzie s initil speed into meters per second: 24 mi/h 1 hour 3600 s 1609 m 1 mile = v = v f v i t t 0 m/s 10.7267 m/s = 0.2 s = 53.6333 m/s 2. = 10.7267 m/s 027 (prt 2 of 3) 10.0 points Wht force ws exerted to stop Suzie? Correct nswer: 2896.2 N. Let : m = 54 kg. = m = (54 kg)( 53.6333 m/s 2 ) = 2896.2 N. The upwrd force exerted by the ground blnces the grvity. The friction between the ir nd Suzie is so smll tht it cn be ignored. It is the friction between the ground nd the sktes tht stopped Suzie. orce nd Motion 04 029 (prt 1 of 8) 10.0 points Consider toy cr which cn move to the right or left long horizontl line (the positive prt of the distnce xis) nd force pplied to tht cr. cr O + Identify the force tht would llow the cr to move towrd the right (wy from the origin) with stedy rte (constnt velocity). Assume friction is so smll tht it cn be ignored. 1. t v 028 (prt 3 of 3) 10.0 points Where did this force come from? 2. None of these
Version 001 Review 1: Mechnics tubmn (IBII20142015) 11 3. t 10. t 4. t 5. t Constnt velocity (in either direction) mens zero ccelertion: = m = 0. 030 (prt 2 of 8) 10.0 points Identify the force tht would llow the cr to remin t rest. cor- 1. t 2. t 3. t 4. t rect correct 6. t 7. t 8. t 9. t 5. t
Version 001 Review 1: Mechnics tubmn (IBII20142015) 12 6. t 3. t 7. t 4. t 8. t 5. None of these 6. t 9. t 10. None of these Zero velocity mens zero ccelertion: = m = 0. 031 (prt 3 of 8) 10.0 points Identify the force tht would llow the cr to move towrd the right nd speed up t stedy rte (constnt ccelertion). 1. t 2. t cor- 7. t rect 8. t 9. t 10. t Positive velocity speeding up t constnt
Version 001 Review 1: Mechnics tubmn (IBII20142015) 13 rte mens constnt positive ccelertion nd constnt positive force: = m > 0. 032 (prt 4 of 8) 10.0 points Identify the force tht would llow the cr to move towrd the left t stedy rte (constnt velocity). 1. t 8. t 9. t 2. None of these 10. t 3. t cor- 4. t 5. t 6. t 7. t rect Constnt velocity (in either direction) mens zero ccelertion: = m = 0. 033 (prt 5 of 8) 10.0 points Identify the force tht would llow the cr to move towrd the right nd slow down t stedy rte (constnt ccelertion). 1. t 2. t 3. t
Version 001 Review 1: Mechnics tubmn (IBII20142015) 14 4. t 1. t 5. t 2. t 6. t 3. t 7. t 4. t 8. t 5. t 9. t 6. t 10. None of these Positive velocity slowing down t stedy rte mens constnt negtive ccelertion nd constnt negtive force: = m < 0. 034 (prt 6 of 8) 10.0 points Identify the force tht would llow the cr to movetowrdtheleftndspeeduptstedy rte (constnt ccelertion). correct correct 7. None of these 8. t
Version 001 Review 1: Mechnics tubmn (IBII20142015) 15 9. t 6. t 10. t 7. t Negtive velocity speeding up t stedy rte mens constnt negtive ccelertion nd constnt negtive force: = m < 0. 035 (prt 7 of 8) 10.0 points Identify the force tht would llow the cr to move towrd the right; first speeding up, then slowing bck down. 1. t 2. t 8. t 9. t 10. t cor- 3. t 4. t rect 5. None of these When the cr speeds up, the force is in the sme direction s the velocity: = m > 0; whenitslowsdown,theforceisintheopposite direction: = m < 0. The cr hs to trnsition from the positive to the negtive force. 036 (prt 8 of 8) 10.0 points The cr is pushed towrd the right nd then relesed. Which grph describes the force fter the cr is relesed?
Version 001 Review 1: Mechnics tubmn (IBII20142015) 16 1. t 9. t 2. t 10. t cor- 3. None of these 4. t rect 5. t 6. t 7. t After the cr isrelesed, no force cts on it. orce on Bullet 02 037 10.0 points A 5.5 g bullet leves the muzzle of rifle with speed of 635.7 m/s. Wht constnt force is exerted on the bullet while it is trveling down the 0.9 m length of the brrel of the rifle? Correct nswer: 1234.79 N. Averge ccelertion cn be found from Since v o = 0, we hve Thus v 2 f = v2 o +2l = v2 2l = m = m v2 2l (5.5 g)(635.7 m/s)2 1 kg = 2(0.9 m) 1000 g = 1234.79 N. 8. t Net orces 02 038 (prt 1 of 3) 10.0 points A 54.7 N object is in free fll. Wht is the mgnitude of the net force which cts on the object? Correct nswer: 54.7 N.
Version 001 Review 1: Mechnics tubmn (IBII20142015) 17 During free fll, the net ccelertion is down, so net = down up. During free fll with no ir resistnce, the only force cting is the weight, so the net force is its weight. 039 (prt 2 of 3) 10.0 points Wht is the mgnitude of the net force when the object encounters 14.3 N of ir resistnce? Correct nswer: 40.4 N. The weight cts down nd the ir resistnce cts up with the net ccelertion cting downwrd, so net = W = 54.7 N 14.3 N = 40.4 N. 040 (prt 3 of 3) 10.0 points Whtisthemgnitudeofthenetforcewhenit flls fst enough to encounter n ir resistnce of 54.7 N? Correct nswer: 0 N. The net force is now zero, since the ir resistnce is the sme s the weight. Prllel nd Series Springs 041 (prt 1 of 5) 10.0 points In the prllel spring system, the springs re positioned so tht the 34 N weight stretches ech spring eqully. The spring constnt for the left-hnd spring is 3 N/cm nd the spring constnt for the right-hnd spring is 5.2 N/cm. Correct nswer: 4.14634 cm. Let : k 1 = 3 N/cm, k 2 = 5.2 N/cm, nd W = 34 N, The springs stretch the sme mount x becuse of the wy they were positioned. Then = kx, so up = down k 1 x+k 2 x = W x = W k 1 +k 2 34 N = 3 N/cm+5.2 N/cm = 4.14634 cm. 042 (prt 2 of 5) 10.0 points In this sme prllel spring system, wht is the effective combined spring constnt k pr of the two springs? Correct nswer: 8.2 N/cm. Considering this s one-spring system, it wouldrectwithforce = kx = W due the lw of ction nd rection, so 3 N/cm 5.2 N/cm 34 N How fr down will the 34 N weight stretch the springs? k pr = W x = W x = 34 N 4.14634 cm = 8.2 N/cm, which is the sum k 1 + k 2 of the individul constnts. 043 (prt 3 of 5) 10.0 points Now consider the sme two springs in series.
Version 001 Review 1: Mechnics tubmn (IBII20142015) 18 3 N/cm 5.2 N/cm 34 N Wht distnce will the spring of constnt 3 N/cm stretch? Correct nswer: 11.3333 cm. In the series system, the springs stretch different mount, but ech crries the full weight W = 34 N. W = k 1 x 1 x 1 = W = 34 N = 11.3333 cm. k 1 3 N/cm 044 (prt 4 of 5) 10.0 points In this sme series spring system, wht distnce will the spring of constnt 5.2 N/cm stretch? k series = W x 1 +x 2 = = 1.90244 N/cm. 34 N 11.3333 cm+6.53846 cm AP M 1998 MC 14 15 046 (prt 1 of 2) 10.0 points A spring hs force constnt of 977 N/m nd n unstretched length of 6 cm. One end is ttched to post tht is free to rotte in the center of smooth tble, s shown in the top view below. The other end is ttched to 3 kg disk moving in uniform circulr motion on the tble, which stretches the spring by 3 cm. Note: riction is negligible. 9 cm 977 N/m 3 kg Whtisthecentripetlforce c onthedisk? Correct nswer: 29.31 N. Correct nswer: 6.53846 cm. W = k 2 x 2 x 2 = W k 2 = 34 N 5.2 N/cm = 6.53846 cm. 045 (prt 5 of 5) 10.0 points In this sme series spring system, wht is the effective combined spring constnt k series of the two springs? Correct nswer: 1.90244 N/cm. x totl = x 1 +x 2, so W = k series x totl Let : r = 6 cm = 0.06 m, r = 3 cm = 0.03 m, m = 3 kg, nd k = 977 N/m. The centripetl force is supplied only by the spring. Given the force constnt nd the extension of the spring, we cn clculte the force s c = k r = (977 N/m)(0.03 m) = 29.31 N. 047 (prt 2 of 2) 10.0 points
Version 001 Review 1: Mechnics tubmn (IBII20142015) 19 Wht is the work done on the disk by the spring during one full circle? 1. W = 11.0496 J r m 2. W = 0 J correct 3. W = 5.5248 J 4. W = 0.00633096 J 5. W = 1.7586 J Since the force is lwys perpendiculr to the movement of the disk, the work done by the spring is zero. Serwy CP 07 25 048 (prt 1 of 3) 10.0 points An ir puck of mss 0.27 kg is tied to string nd llowed to revolve in circle of rdius 0.91 m on horizontl, frictionless tble. The other end of the string psses through hole inthecenterofthetblendmssof0.88kg is tied to it. The suspended mss remins in equilibrium while the puck revolves. v 0.91 m 0.27 kg 0.88 kg Wht is the tension in the string? The ccelertion due to grvity is 9.8 m/s 2. Correct nswer: 8.624 N. Let : M = 0.88 kg nd g = 9.8 m/s 2. v Mg Since the suspended mss is in equilibrium, the tension is T = M g = (0.88 kg) ( 9.8 m/s 2) = 8.624 N. 049 (prt 2 of 3) 10.0 points Wht is the horizontl force cting on the puck? Correct nswer: 8.624 N. The horizontl force cting on the puck is the tension in the string, so c = T = 8.624 N. 050 (prt 3 of 3) 10.0 points Wht is the speed of the puck? Correct nswer: 5.3913 m/s. c = mv2 r v = c r m = = 5.3913 m/s. (8.624 N)(0.91 m) 0.27 kg Work Done on Block 051 10.0 points Lee pushes horizontlly with force of 51 N on 26 kg mss for 27 m cross floor. Clculte the mount of work Lee did. Correct nswer: 1377 J.
Version 001 Review 1: Mechnics tubmn (IBII20142015) 20 Let : = 51 N nd d = 27 m. Themssoftheobjectisignoredsincethere is no friction present, so W = d = (51 N)(27 m) = 1377 J. Sliding Into Loop 052 10.0 points A 5.26 kg block is relesed from A t height 5.3 m on frictionless trck shown. The rdius of the trck is 2.97 m. The ccelertion of grvity is 9.8 m/s 2. A h Determine the mgnitude of the ccelertion for the block t P. Correct nswer: 18.2339 m/s 2. rom conservtion of energy we obtin Therefore R (K +U) A = (K +U) P mgh = mv2 2 +mgr. v = 2g(h R) = 6.75781 m/s. Then the rdil ccelertion t P is r = v2 R = 15.3764 m/s2 nd the tngentil ccelertion is t = g = 9.8 m/s 2. P So, the totl ccelertion t P is = 2 t +2 r = (9.8 m/s 2 ) 2 +(15.3764 m/s 2 ) 2 = 18.2339 m/s 2. Dropping Bsebll 053 10.0 points Youdrop150gbsebllfromwindow15m bove the ground. Wht is the kinetic energy of the bsebll when it hits the ground? The ccelertion due to grvity is 10 m/s 2. 1. 22.5 J correct 2. 0 J 3. 2.25 10 4 J 4. 1.125 10 5 J 5. 1.125 J Let : m = 150 g, h = 15 m, nd g = 10 m/s 2 Energy is conserved: K i +U i = K f +U f The initil kinetic energy nd finl potentil energy re zero, K i = U f = 0, nd the initil potentil energy is the grvittionl potentil energy, U i = mgh, so the kinetic energy of the bsebll when it hits the ground is K f = U i = mgh ( = 150 g kg ) 10 3 (10 m/s 2 )(15 m) g = 22.5 J.
Version 001 Review 1: Mechnics tubmn (IBII20142015) 21 Girl on Swing 02 = 2(9.8 m/s 2 )(2.8 m 0.5 m) 054 (prt 1 of 2) 10.0 points = 6.71416 m/s. Agirlswingsonplygroundswinginsuch wy tht t her highest point she is 2.8 m fromtheground, whilether lowestpointshe is 0.5 m from the ground. 055 (prt 2 of 2) 10.0 points At wht height bove the ground will the girlbemovingtspeedhlfofhermximum speed? 2.8 m 7.4 m 0.5 m Correct nswer: 2.225 m. Let h 1/2 = the height, where v = v 1/2 = 1 2 v mx. Wht is her mximum speed? The ccelertion of grvity is 9.8 m/s 2. r Correct nswer: 6.71416 m/s. v1/2 h 1/2 v h bottom h top Let : r = 7.4 m, h top = 2.8 m, nd h bot = 0.5 m. We cn solve this by using the principle of conservtion of energy. We need to know the kinetic nd potentil energies t two points in time. The girl will be moving the fstest when her kinetic energy is lrgest which occurs when her potentil energy is smllest. This mens tht she will be moving fstest t the bottom of the swing. By conservtion of energy, E top = E bot K top +U top = K bot +U bot 1 2 mv2 top +mgh top = 1 2 mv2 bot +mgh bot. Since v top = 0, 1 2 mv2 bot = mg(h top h bot ) so v mx = v bot = 2g(h top h bot ) rom conservtion of energy, K 1/2 +U 1/2 = K top +U top 1 2 mv2 1/2 +mgh 1/2 = 0+mgh top rom Prt 1, mgh 1/2 = mgh top 1 2 1 2 mv2 mx = mg(h top h bot ). Substitution yields ( ) 1 4 v2 mx. mgh 1/2 = mgh top 1 4 mg(h top h bot ) h 1/2 = 1 4 (3h top +h bot ) = 1 [3(2.8 m)+0.5 m] 4 = 2.225 m. Decying Urnium Nucleus 056 10.0 points
Version 001 Review 1: Mechnics tubmn (IBII20142015) 22 Aurniumnucleus 238 Umystyinonepiece for billions of yers, but sooner or lter it decysintonαprticleofmss6.64 10 27 kg nd 234 Th nucleus of mss 3.88 10 25 kg, nd the decy process itself is extremely fst (ittkesbout10 20 s). Supposetheurnium nucleus ws t rest just before the decy. If the α prticle is emitted t speed of 7.51 10 6 m/s,whtwouldbetherecoilspeed of the thorium nucleus? Correct nswer: 1.28522 10 5 m/s. Let : v α = 7.51 10 6 m/s, M α = 6.64 10 27 kg, nd M Th = 3.88 10 25 kg. Since the speeds involved re less thn 1 10 of the speed of light, we cn ignore the effects of specil reltivity. Use momentum conservtion: Before the decy, the Urnium nucleus hd zero momentum (it ws t rest), nd hence the net momentum vector of the decy products should totl to zero: P tot = M α v α +M Th v Th = 0. This mens tht the Thorium nucleus recoils inthedirectionexctlyoppositetothtofthe α prticle with speed v Th = v α M α M Th = (7.51 106 m/s)(6.64 10 27 kg) 3.88 10 25 kg = 1.28522 10 5 m/s. Cnnon Recoil 057 (prt 1 of 2) 10.0 points A revolutionry wr cnnon, with mss of 2150 kg, fires 19.3 kg bll horizontlly. The cnnonbll hs speed of 123 m/s fter it hs left the brrel. The cnnon crrige is on flt pltform nd is free to roll horizontlly. Wht is the speed of the cnnon immeditely fter it ws fired? Correct nswer: 1.10414 m/s. Let : m = 19.3 kg, M = 2150 kg, nd v = 123 m/s. The cnnon s velocity immeditely fter it ws fired is found by using conservtion of momentum long the horizontl direction: M V +mv = 0 V = m M v where M is the mss of the cnnon, V is the velocity of the cnnon, m is the mss of the cnnonbllndv isthevelocityofthecnnon bll. Thus, the cnnon s speed is V = m M v = 19.3 kg (123 m/s) 2150 kg = 1.10414 m/s. 058 (prt 2 of 2) 10.0 points Thesmeexplosivechrgeisused,sothetotl energy of the cnnon plus cnnonbll system remins the sme. Disregrding friction, how much fster would the bll trvel if the cnnon were mounted rigidly nd ll other prmeters remined the sme? Correct nswer: 0.550836 m/s. By knowing the speeds of the cnnon nd the cnnon bll, we cn find out the totl kinetic energy vilble to the system K net = 1 2 mv2 + 1 2 M V 2. This is the sme mount of energy vilble s when the cnnon is fixed. Let v be the
Version 001 Review 1: Mechnics tubmn (IBII20142015) 23 speed of the cnnon bll when the cnnon is held fixed. Then, 1 2 mv 2 = 1 2 (mv2 +M V 2 ). v = = v v 2 + M m V 2 1+ m M = (123 m/s) = 123.551 m/s. 1+ Thus, the velocity difference is 19.3 kg 2150 kg v v = 123.551 m/s 123 m/s = 0.550836 m/s. Boxing Mtch 059 10.0 points The liner impulse delivered by the hit of boxeris320n sduringthe0.422sofcontct. Wht is the mgnitude of the verge force exerted on the glove by the other boxer? Correct nswer: 758.294 N. Let : t = 0.422 s nd p = 320 N s. The verge force exerted is nd its mgnitude is = p t, = p t 320 N s = 0.422 s = 758.294 N. Your Body in Collision 060 (prt 1 of 2) 10.0 points An impulse of 149 Ns is required to stop person s hed in cr collision. If the fce is in contct with the steering wheel for 0.0279 s, wht is the verge force on the cheekbone? Correct nswer: 5340.5 N. p = 149 Ns nd t = 0.0279 s. orce is relted to impulse by = p t I p = t, so = 149 Ns 0.0279 s = 5340.5 N. 061 (prt 2 of 2) 10.0 points If n verge force of 902 N frctures the cheekbone, how long must it be in contct with the steering wheel in order to frcture? Correct nswer: 0.165188 s. t = p 1 = Let : 1 = 902 N. 149 Ns 902 N = 0.165188 s. Bullet Moves Block 062 10.0 points A(n) 23.3 g bullet is shot into (n) 4976 g wooden block stnding on frictionless surfce. The block, withthebullet init, cquires speed of 1.55 m/s. Clculte the speed of the bullet before striking the block. Correct nswer: 332.571 m/s.
Version 001 Review 1: Mechnics tubmn (IBII20142015) 24 Bsic concepts: Momentum of ny object is p = mv. Solution: The collision is inelstic, nd by conservtion of momentum p before = p fter m b v b +0 = (m b +m w )v f v b = (m b +m w )v f m b. AP B 1993 MC 6 063 10.0 points If Spcecrft X hs twice the mss of Spcecrft Y, then wht is true bout X nd Y? I) On Erth, X experiences twice the grvittionl force tht Y experiences; II) On the Moon, X hs twice the weight of Y; III) When both re in the sme circulr orbit, X hs twice the centripetl ccelertion of Y. 1. I only 2. I, II, nd III 3. III only 4. I nd II only correct 5. II nd III only I) grvittionl force mss. II) weight mss. III) The centripetl ccelertion is determined by c = v2 r, so X nd Y should hve the sme centripetl ccelertion when they re in the sme circulr orbit. AP M 1998 MC 20 064 10.0 points Two identicl strs, fixed distnce D prt, revolve in circle bout their mutul center of mss, s shown below. Ech str hs mss M nd speed v. M v D Which of the following is correct reltionship mong these quntities? G is the universl grvittionl constnt. 1. v 2 = M GD 2. v 2 = 4GM2 D 3. v 2 = 4GM D 4. v 2 = 2GM2 D 5. v 2 = 2GM D 6. v 2 = GM 2D correct 7. v 2 = GM D 2 8. v 2 = GM D rom circulr orbitl movement, the centripetl ccelertion is = v2 D 2 v M = 2v2 D. Using Newton s second lw of motion, the ccelertion is = M = 1 M GM2 D 2 = GM D 2 2v 2 D = = M = GM D 2 v 2 = GM 2D.
Version 001 Review 1: Mechnics tubmn (IBII20142015) 25 Circulr Moon Orbit 065 (prt 1 of 3) 10.0 points A stellite is in circulr orbit just bove the surfce of the Moon. Wht is the stellite s ccelertion? The vlue of grvittionl constnt is 6.67259 10 11 Ncdotm 2 /kg 2 nd the mss of the moon is 7.36 10 22 kg nd its rdius is 2155.2 km. Correct nswer: 1.0573 m/s 2. Let : G = 6.67259 10 11 Ncdotm 2 /kg 2, M moon = 7.36 10 22 kg, nd r moon = 2155.2 km = 2.1552 10 6 m. c = m c = G, so m c = GmM moon r 2 c = GM moon r 2 = (6.67259 10 11 Ncdotm 2 /kg 2 ) 7.36 1022 kg (2.1552 10 6 m) 2 = 1.0573 m/s 2. 066 (prt 2 of 3) 10.0 points Wht is the stellite s speed? Correct nswer: 1509.53 m/s. c = v2 r v = c r = (1.0573 m/s 2 )(2.1552 10 6 m) = 1509.53 m/s. 067 (prt 3 of 3) 10.0 points Wht is the period of the stellite s orbit? Correct nswer: 2.49185 h. T = 2πR v = 2π(2.1552 106 m) 1509.53 m/s = 2.49185 h. 1 h 3600 s Tipler PSE5 11 58 068 10.0 points The grvittionl field t some point is given by g = (2.8 10 6 N/kg)ĵ. Wht is the grvittionl force on mss of 7.12 g t tht point? Correct nswer: 1.9936 10 8 N. Let : g = (2.8 10 6 N/kg)ĵ nd m = 7.12 g = 0.00712 kg. The grvittionl force is g = m = m g = (0.00712 kg)(2.8 10 6 N/kg)ĵ = (1.9936 10 8 N)ĵ. Lunch Energy 069 10.0 points Wht is the minimum kinetic energy needed to lunch pylod of mss m to n ltitude tht is one Erth rdius, R E, bove the surfce of the Erth (the pylod will then fll bck to Erth)? (Notetht M E isthe mss of the Erth.) 1. 0.1 GmM E R E 2. 0.25 GmM E R E
Version 001 Review 1: Mechnics tubmn (IBII20142015) 26 3. 2 GmM E R E 4. 0.5 GmM E R E correct 5. GmM E R E Energy is conserved: K i +U i = K f +U f The kinetic energy fter the pylod hs been lunched to n ltitude R E bove the Erth s surfce is K f = 0. The potentil energy t Erth s surfce before the lunch is U i = GM Em R E, ndthepotentilenergytltituder E bove the Erth s surfce fter the lunch is U i = GM Em 2R E. So, the minimum energy needed to lunch the pylod to R E bove Erth s surfce is K i = GM Em R E GM Em 2R E = GmM E 2R E New Stellite Orbit 02 070 (prt 1 of 3) 10.0 points Wht is the kinetic energy of stellite of mss m which is in circulr orbit of rdius 3R e bout the erth? 5. K = GM em R 2 e 6. K = GM em 3R e 7. K = 3mgR e 8. K = GM em 3R e 9. K = 3GM e m 10. K = GM em 6R 2 e The ccelertion of the stellite in circulr orbit of rdius 3R e is c = v2 3R e, so the force on the stellite is = m c = mv2 = GM em 3R e (3R e ) 2 nd the kinetic energy is mv 2 = GM em er e K = 1 2 mv2 = GM em 6R e. 071 (prt 2 of 3) 10.0 points Wht is the totl energy of the stellite? 1. E = GM em 6R 2 e 2. E = GM em 3R e 3. E = Gm R e 4. E = 3GM e m 1. K = mv2 3R e 2. K = mv2 6R e 3. K = mgr e 4. K = GM em 6R e correct 5. E = mgr e + 1 2 mv2 6. E = GM em R e 7. E = 3mgR e + GM em 3R e 8. E = GM em 3R e
Version 001 Review 1: Mechnics tubmn (IBII20142015) 27 9. E = GM em correct 6R e The potentil energy of the stellite is U = GM em 3R e so the totl energy is E = K +U = GM em GM em 6R e 3R e = GM em. 6R e 072 (prt 3 of 3) 10.0 points How much work must n externl force do on the stellite to move it from circulr orbit of rdius 2R e to 3R e, if its mss is 2000 kg? The universl grvittionl constnt 6.67 10 11 N m 2 /kg 2, the mss of the Erth 5.98 10 24 kg nd its rdius 6.37 10 6 m. Correct nswer: 1.04361 10 10 J. Let : G = 6.67 10 11 N m 2 /kg 2, M e = 5.98 10 24 kg, nd R e = 6.37 10 6 m. Theworkdonebynexternlforcetomove the stellite from the closer orbit to the further orbit will be the work ginst grvity ( positive number which yields the chnge in potentil energy) plus the chnge in kinetic energy ( negtive number since the kinetic energyissmllerintheorbitwiththegretest rdius): W = E f E i = GM em 6R e = GM em 12R e = 1 12 (6.67 10 11 N m 2 /kg 2 ) [ (5.98 10 24 ] kg)(2000 kg) 6.37 10 6 m = 1.04361 10 10 J. ( GM ) em 4R e