Physics 1 st Semester Final Exam Review ANSWER KEY 1 f 11 Frmula Bank Hrizntal mtin frmulas Vertical mtin frmulas Distance frmula (1.) x = ±v x t ± ½at Distance frmula (3.) y = y ±v y t ½gt Velcity frmula (.) v x = ±v x ± at Velcity frmula (4.) v y = ±v y gt 4. 9x v (cs ) Catapult Frmula: (5.) y( x ) y xtan Energy frmulas (6.) TKE = ½mv (7.) GPE = mgh (8.) EPE = ½ kx (9.) W = F x d Mechanical Advantage: (9.) M. A. = F ut /F in DIRECTIONS: Put yur answer in the blank prvided. Shw yur wrk fr partial credit, where applicable. Unless therwise specified, each blank is wrth ne pint. Any time the final ptin n multiple chice prblems are in italics, they are gag answers. Please d nt select them. UNIT 1: SKILLS OF SCIENCE d. 1. TESTING: A thery exists that small bugs spntaneusly generate frm rtting meat. What test culd yu d t shw that the bugs are nt spntaneusly generated but get there frm anther cause? a) get tw pieces f meat that are the same in all respects. Set ne under an airtight jar and the ther in the pen air and let them rt t see if small bugs develp n bth. b) get tw pieces f meat that are the same in all respects. Set ne under a wire mesh and the ther in the pen air and let them rt t see if small bugs develp n bth. c) Sanitize tw pieces f meat that are the same in all respects. Set them bth ut n a table and let them rt t see if small bugs develp. d) Sanitize tw pieces f meat that are the same in all respects. Set ne under an airtight jar and the ther in the pen air and let them rt t see if small bugs develp n bth. e) Eat ne f the pieces f meat and sit there with yur muth pen t see if flies cme ut. Dn t cnfuse them with the flies that want t get int yur muth. If yu dn t sanitize the meat, then even if yu put it in an airtight jar, it might already have germs n it and s they will eventually appear n the meat. a.. OBSERVING. Finish the fllwing pattern: FI, SE, TH, FO, FI, SI,.* a) SE Seventh is next. The first tw letters f the rdinal numbers. b) HO c) MA d) PL e) E Impsible! * If yu detect a pattern but are nt sure it is right, explain yur reasning here and I ll give yu full credit if yur argument is sund.
Physics 1 st Semester Final Exam Review ANSWER KEY f 11 _d. 3. COMMUNICATING: If the units fr are s, the units fr are cheps per s, the units f are s, and the units f are s/cheps, what is a frmula that relates,, and? (NOTE: yu can add, subtract, multiply, divide, take the square rt r square anything and can use the quantities mre than nce) a) b) c) d) chep s s s chep chep chep chep chep chep chep s chep 4 chep chep chep chep chep chep chep 3 chep chep chep 4. (5 pints) TESTING, INTERPRETING DATA, & COMMUNICATING: If yu recall the demnstratin we did in which a vlunteer came up and tried t grab a 50,000 Gs. nte between his/her fingers befre it fell thrugh, yu will remember that he/she (generally) failed miserably t d s. Belw are the currency ntes f several cuntries and their dimensins. In the space fllwing, explain hw yu wuld determine which bills the vlunteer culd catch if his/her reactin time was 0.18 secnds. 16 cm. 15.5 cm. One Sapukai dllar One dllar frm the Republic f El Guaira 18 cm. 15.85 cm. One dllar frm the Cmmnwealth Natin f Capiata One dllar frm the Demcratic State f LamberCity y = 1/gt is the frmula fr free-falling bjects where y is the distance fallen, t is the time fallen and g is gravity which is 9.8 m/s/s If the vlunteer s reactin time is 0.18 secnds then the amunt f bill that culd fall wuld be given by: y = 1/gt y = ½(9.8)(0.18) y = 4.9(0.034) y = 0.15876 m y = 15.876 cm. This means that this much bill will fall thrugh befre the vlunteer can catch it. The bills that are the same r lnger will be caught which means the Sapukai dllar (16 cm) and the dllar frm the Cmmnwealth Natin f Capiata (18 cm) UNIT : HORIZONTAL MOTION
Physics 1 st Semester Final Exam Review ANSWER KEY 3 f 11 MOTION FORMULAS: Fabri s car is mving in a hrizntal directin at a velcity f 3 m/s n his way t a multi-player n-line gamer s cmpetitin. It travels fr fur secnds... 1m 5. hw much distance des the car travel? The mst general hrizntal mtin frmula fr distance is x = v x t + ½at. Since the car is mving at a cnstant velcity f 3 m/s, there is n acceleratin (a = 0). Yu have t plug in yur givens and find yur hedind. Yur initial velcity is 3 m/s and yur time is 4 secnds. x = v x t + ½at x = (3)(4) + ½(0)(4) x = (3)(4) x = 1 m Mate hits the gas pedal accelerating the car at a rate f 4m/s/s fr fur secnds mre. 19 m/s 6. what is the car s final velcity? The mst general hrizntal mtin frmula fr velcity is v = v x + at. This is the cusin f the abve frmula, if yu recall. This time there is acceleratin because Fabri hits the gas pedal. Yu have t plug in yur givens and find yur hedind. Yur initial velcity remains 3 m/s and yur acceleratin is 4 m/s fr a time f 4 secnds. v = v x + at v = (3) + (4)(4) v = 3 + 16 v = 19 m/s 44 m 7. hw much distance did the car travel in thse fur secnds? The mst general hrizntal mtin frmula fr distance is x = v x t + ½at. Since the car has an initial velcity f 3 m/s, has an acceleratin f (a = 4 m/s/s) and has traveled fr 4 secnds, yu have t plug in yur givens and find yur hedind. x = v x t + ½at x = (3)(4) + ½(4)(4) x = 1 + 3 x = 44 m DERIVE YOUR OWN FORMULA: Yu cllect the fllwing data frm Lali s run during the Running (W)Man Experiment. Use the infrmatin fr Prbls. 8 10: time elapsed 0 sec 1 3 4 5 ttal distance traveled 0 m 1.575 4.8579 9.3886 14.984 1.533 d = 1.575t 1.64980964 8. What equatin describes the distance Lali ran vs. time? Yu have t use the very fun, fun d = kt n frmula here. Pick tw pints and plug them int the frmula: EX: (1, 1.575) and (, 4.8579) 1.575 = k(1) n and 4.8579 = k() n Nw slve ne frmula fr k: 1.575 = k(1) n 1.575 = k (1) n Nw substitute this value fr k in the ther frmula: 4.8579 = 1.575() n (1) n Nw simplify in several steps: 4.8579 = 1.575 ( 1 ) n Nw simplify in several steps: Nw simplify in several steps: Nw simplify in several steps: 4.8579 1.575 = ( 1 ) n 4.8579 1.575 = () n 3.08438 = () n
Physics 1 st Semester Final Exam Review ANSWER KEY 4 f 11 Nw take the LOG f bth sides: Nw simplify the left side: Nw simplify the right side. Use the rule f LOGs fr expnents: Simplify the right side mre: Slve fr n: LOG 3.08438 = LOG () n 0.489168015 = LOG () n 0.489168015 = n LOG () 0.489168015 = n (0.301099957) 0.489168015 0.301099957 = n 1.64980964 = n Nw t find k, just substitute the value f n int yur frmula abve fr k : 1.575 = k (1) n 1.575 = k (1) 1.64980964 And slve fr k : 1.575 = k (1) 1.575 = k Nw plug in n and k int yur d = kt n frmula d = 1.575t 1.64980964 v =.559345018t 0.64980964 9. What equatin describes hw fast Lali was ging vs. time? Yu have t use the calculus trick that we learned: if the frmula fr distance is given by the frmula d= kt n, then yu can find the frmula fr velcity by ding the fllwing: v = nkt n-1 S, if d = 1.575t 1.64980964, then v = 1.649809641.575t 1.64980964 1 r, simplifying: v =.559345018t 0.64980964 6.087 m/s 10. Hw fast was Lali ging after 4 secnds? v =.559345018t 0.64980964 v =.559345018(4) 0.64980964
distance (in m) Physics 1 st Semester Final Exam Review ANSWER KEY 5 f 11 INTERPRETING GRAPHS: The graph fllwing represents the distance vs. time f Brun Bgarin s mvements at Ranch after curfew. Use the graph fr prblems 1115: B 75 C 50 A 5 D 0 5 10 15 0 5 time (in sec.) 150 m 11. Hw much ttal distance did Brun travel back and frth (in m)? He went 75 meters away frm the rigin (zer) and then traveled back t zer s: 75 m x 10 s 1. Hw much time (in sec) did Brun spend in mvement (nt staying still)? Segment B represents rest. He is in segment B fr: 15 sec - 5 sec = 10 sec 15 m/s 13. What was Brun s velcity in segment A (in m/s)? Velcity is simply the SLOPE f the line segment. In ther wrds, distance (y-axis) divided by time (x-axis) s v = d/t v = 75 meters/5 secnds 15 m/s 0 m/s/s 14. What was Brun s acceleratin in segment D (in m/s/s)? On a DISTANCE VS. TIME graph, there is nly acceleratin if yu see a curve. This is because the distance every mment is changing, getting bigger and bigger. The straight line segment in D suggests mvement at CONSTANT velcity where the distance traveled every mment stays the same 15. (5 pints) Explain what mvements Brun made that wuld result in the distance vs. time graph that is shwn abve. Yur situatin shuld include fur distinct parts that crrespnd t intervals A, B, C, and D. EX: Brun is sleeping at Ranch and wakes up with Klyns n his face. He gets very angry and starts running ut f his rm t see wh did it. He runs fast fr 5 secnds, at 15 m/s. Then he stps and lks arund the utside f the ldge fr hiding students. He des this fr 10 secnds. He desn t see anybdy but he hears a nise behind him s he runs BACK in the ldge dwn the hall twards his rm at 10 m/s. He des this fr 5 secnds until he realizes that the nise he heard was the giggling f his classmates dwn the hall. He slws his pace and highly irritated, walks the rest f the way t his rm at 1 m/s fr five secnds.
Physics 1 st Semester Final Exam Review ANSWER KEY 6 f 11 UNIT 3: VERTICAL MOTION b. 16. VELOCITY VS. ACCELERATION: A bullet is drpped frm a hand. Which f the fllwing mst adequately describes the bullet's fall? a) the bullet's velcity and acceleratin are increasing every secnd at a nn-cnstant rate. b) the bullet's velcity is increasing every secnd but its acceleratin is staying the same. c) the bullet's velcity and acceleratin are increasing every secnd at a cnstant rate. d) the speed f the bullet is increasing every secnd; its velcity remains the same. Gravity is a cnstant: 9.8 m/s/s - and s it will nt change. Velcity, hwever, increases as a result f gravity s actin. MOTION FORMULAS: The fllwing infrmatin is fr Prbls. 17 19. Magali is thrwn straight upwards int the air by fans at a Ripe Banana Skins cncert with an initial velcity f 4 m/s.. 0.8 s 17. After hw many secnds will she hit the grund? This is a vertical mtin situatin and s requires the use f the vertical mtin frmulas. In this case we want t knw after hw many secnds Magali will hit the grund. It is best t use the vertical mtin frmula fr velcity because we knw that at the tp f Magali s flight, she will be mving at 0 m/s befre starting t fall dwn again. S, v y = v y gt Plug in yur knwn values: 0 = 4 9.8t, where 0 m/s is the velcity at the tp f the flight, 4 m/s is the initial velcity, and gravity is 9.8 m/s/s. Slving fr t: 0 = 4 9.8t 4 4. 4 = 9.8t 9.8 9.8 0.41 sec = t This is the time t g up t the tp. The ttal time in the air then is 0.41 sec x = 0.8 sec 0.41 s 18. When will she be at the tp f his flight? This we already fund frm the previus prblem. 0.8 m 19. Hw high will she g? Nw we use the cusin frmula t vertical velcity, which is y = v y t ½gt. We already knw that it takes Magali 0.41 secnds t get t the tp. Just plug in the infrmatin int the frmula and yu will have yur answer: y = v y t ½gt y = 4(0.41) ½(9.8)(0.41) y = 0.81631 m
Physics 1 st Semester Final Exam Review ANSWER KEY 7 f 11 UNIT 4: CONSERVATION OF ENERGY 0. (5 pints) Describe a prcess in which a certain amunt f energy underges a change f frm fur different times (EX: CPE TKE GPE EPE). Remember: the same energy must g thrugh the fur transfrmatins in a sequence. An acrbat drinks a Dark Dg energy drink and gains CPE (chemical ptential energy). Then he walks up the stairs (ding WORK n his muscles) t an elevated platfrm and gains GPE (gravitatinal ptential energy). Then he jumps ff the platfrm and begins falling (gaining TKE translatinal kinetic energy). Abut half way dwn he des tw spins (gaining RKE) and then falls dwn int a trampline belw (cnverting his TKE int EPE elastic ptential energy). The Class f 013 puts their hand-made catapults in a bx fr when their grandkids cme t ASA and have Mr. K as a teacher. They want a simple way f mving them up int the attic fr strage. There is a limp spring lying n the flr. One end f the spring is cnnected t a wall. Julie B. eats a Granla bar (64,000 Jules), digests it, then places a bx f the catapults against the spring platfrm, cmpresses it, and then releases it. CATAPULTS 011 Time Capsule 1. 6.7 m_ Hw far culd Julie B. cmpress the spring if he used up ALL the energy frm the Granla bar t d s? Assume the spring has a spring cnstant f.5 N/m and that this is a frictinless envirnment. Elastic ptential energy has the frmula EPE = ½kx where k (in N/m) is the spring cnstant and x (in m) is the distance cmpressed r stretched. Plugging in the numbers: EPE = ½kx 64,000 J = ½(.5 N/m)x 64,000 = 1.5x 5100 = x 5100 = x 1.5 1.5 6.7 m = x. _01.56 m/s = v ut _ Suppse the bx has a mass f 0.4 kg. If Julie B. cmpresses the spring a distance f 0.5 m, hw fast will the bx be ging when the spring is fully released? In this case the energy is cmpletely cnverted frm EPE t TKE because the spring is FULLY released. S we set up ur algrithm: E in = E ut EPE in = TKE ut ½kx in = ½mv ut ½(65,000)(0.5) = ½(0.4)v ut 815 =0.v ut 815 =0.v ut 0. 0. 4065 = v ut 4065 = v ut 01.56 m/s = v ut CATAPULTS 011 Time Capsule
Physics 1 st Semester Final Exam Review ANSWER KEY 8 f 11 3. 07.7 m Hw far up the slide will the bx g if frictin is NOT present? In this case, all the bx s energy ges frm being EPE t being GPE. S ur algrithm set up becmes: E in = E ut EPE in = GPE ut ½kx in = mgh ut ½(65,000)(0.5) = (0.4)(9.8)h ut 815 = 3.9h ut 815 = 3.9h ut 3.9 3.9 07.7 m = h ut UNIT 5: SIMPLE MACHINES Answer questins 4 7 with the nt-s-simple machine shwn belw. Since the Class f 013 can t put its fashin shw mney in the schl accunt, it puts all its revenue in a bx fr strage and wants t get the bx up t a platfrm using the Incleaxlley 1.5m 3 m Figure 1 wheel & axle, pulley, inclined plane_ 4. The Incleaxlley is a cmbinatin f which simple machines? _3000 J 5. Hw much des the bx weigh if it requires 4500 jules f wrk t get the bx up t the lading ramp using the Incleaxlley? a) 1000 N b) 500 N c) 3000 N d) 4500 N e) 6000 N Once yu have the WORK, yu are Sctt-free! The wrk applies everywhere since W in = W ut. D nt be duped by the wrd weight. That s a cde wrd fr F ut f the inclined plane part f the machine. W ut = F ut x d ut 4500 J = F ut x 1.5 meters (see Figure 1 fr d ut ) 4500 = 1.5F ut 1.5 1.5 3000 J = F ut
Physics 1 st Semester Final Exam Review ANSWER KEY 9 f 11 6. Hw much frce wuld the wrker need t apply t mve the bx up the inclined plane t the lading dck withut the benefit f the pulley (pretend that nly the inclined plane exists)? a) 1847 N b) 680 N c) 3546 N d) 480 N e) 6060 N Once yu have the WORK, yu are Sctt-free! The wrk applies everywhere since W in = W ut. In this case, the questin is hw much frce is required t lift the bx using the inclined plane, in ther wrds, F in : W in = F in x d in 4500 J = F in x 3 meters (see Figure 1 fr d in ) 4500 = 3F ut 3 3 1500 J = F ut 7. What is the mechanical advantage f the Incleaxlley if the wheel has twice the radius f the axle? a) :1 b) 3.5:1 c) 5:1 d) 6:5:1 e) 8.:1 The mechanical advantage f the inclined plane, then, is :1 because F in :F ut = 3000 J:1500 J. The mechanical advantage f the cmpund pulley is autmatically :1, by definitin. If the wheel is twice the radius f the axle, then the mechanical advantage is, nce again, :1. MA Inclined plane x MA Pulley x MA Wheel & Axle x x = 6:1 UNIT 6: THE CATAPULT CATAPULT FORMULA: Fr prblems 8-30 use the catapult frmula. Make sure yur graphing calculatr is in degree mde. y(x) y 4.9x x tan θ v (cs θ) nd 8. Which f the fllwing water ballns will g further: water balln A launched frm the grund with an initial velcity f 8 m/s at an angle f 3º with the hrizntal r water balln B launched frm the grund with an initial velcity f 3 m/s at an angle f 8º with the hrizntal? Shw yur wrk fr credit. Fr these prblems yu have t insert each equatin in yur graphing calculatr, graph and find the psitive x-intercept f each ne using nd -TRACE-ZERO. The first typed-in equatins lk like this: \Y 1 = 0 + Xtan(3) 4.9(x) /(8 (cs(3)) ) The secnd typed-in equatins lk like this: \Y 1 = 0 + Xtan(8) 4.9(x) /(3 (cs(8)) ) Yielding a graph that lks like this: After nd -TRACE-ZERO, the secnd trajectry shws it reaches further at 86.63 meters.
Physics 1 st Semester Final Exam Review ANSWER KEY 10 f 11 _36.84 m 9. A water balln is launched frm an angle f 45 º with an initial velcity f 38 m/s frm an initial height f 1 m. What was its maximum height? Once again, we use ur graphing calculatrs, plug in ur equatin, graph it and then use nd - TRACE-ZERO t find the maximum. Make sure yu select the y-value. The typed-in equatins lk like this: \Y 1 = 1 + Xtan(45) 4.9(x) /(38 (cs(45)) ) Yielding a graph that lks like this: After nd -TRACE-MAXIMUM, the secnd trajectry shws it reaches a height f Y = 36.84 meters. 30. A water balln is launched frm the grund at an angle f 53º and hits the grund 38 meters later. What was its initial velcity? Plug in all yur values int yur catapult equatin. Nte that y(x), the height f the marble, is 0, when the marble hits the grund. 4.9(38) 0 0 38tan53 (cs 53) ) v Slving fr v we get 0 50.47703... v 7075.6 (0.36318131...) 50.47703 50.47703 7075.6 50.47703... (0.36318131...) v 50.47703... v 50.47703... v 7075.6 (0.36318131...) 19536.07094 50.47703 19536.07094... v 50.47703 50.47703 v 387.4075099... v x = 73.67347 387.4075099... v = 19.68 m/s
Physics 1 st Semester Final Exam Review ANSWER KEY 11 f 11 (15 pints) FORMS OF ENERGY: ESSAY: This essay n yur final exam is wrth 5% f the final exam grade and will be graded accrding t the 6 + 1 Traits f Writing. It will have t d with hw the catapult transfrms energy frm ne frm t anther. Be sure t understand what frms f energy are invlved in a catapult, what the frmulas are fr each kind f energy, and hw t calculate fr any f the energies r quantities in the frmulas. 6 + 1 Grading Rubric /5 (65%) Cntent & Ideas /5 (10%) Organizatin /5 (10%) Wrd Chice /5 (5%) Vice /5 (5%) Sentence Fluency /5 (5%) Cnventins /5 OVERALL GRADE Just write a stry and include all the terms