Exam 1

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Brigham Young University, Idaho *

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105

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Physics

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Oct 30, 2023

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Online PH 105 Exam 1 Please use 3 sig fig, unless otherwise stated Remember units ¢ : opp -1 0pPD lin=2.54cm 1mile=5280ft=1609km g = 9.8™/, sin() = 6 = tan 1;&7 1 X=X+ Uyt +Ea,ct2 Ve =V, gt vt = vyoz +2ay,(y —¥,) Fpr =uFy Fg=mg Conceptual Questions (3pts each) 1. Using the equation v,* = vyoz + 2ay(y o) where a car is moving at an initial speed of 22 m/s slams on the breaks, traveling a distance of 45.6 m before coming to rest. Solving for the acceleration, how many significant figures would it accurately have? a. 1 @ c. 3 d 4 e. 5 2. True/False: In an elevator the acceleration is magically measured to be -3.00 m/s? (where +is up and —is down) | therefore know that the elevator is moving down. (a. True b. False 3. Aparticle is initially traveling east at 5 m/s , then it begins to have an acceleration of 10 m/s? in the north direction. What can you say about its velocity after a long period of time. T a. It will be completely in the northern direction. | b) 1t will be in both the northern and eastern directions. =2y c. Neither of the above can be true, it depends on the amount of time. 4. T?@’ False: In adding two nonzero vectors whose magnitudes are different from each c;iher, | can never obtain the null or zero vector. True 5. If Nemo the fish is able to swim at a speed of 3 m/s and he is in the E.A.C. which is an amazing Current with a constant velocity of 4 m/s. Which of the following best describes the speed Nemo can achieve as relative to the bottom of the ocean? a. 1m/s b. 7m/s c. 5m/s @ AorBbutnotC e. A, BandC
Shorter problems (5pts each) 6. Convert the following into Si units | /\\_,M Sl i JA»J‘_({W:‘_;L i e | I/ 314 gy Sces . = R oD W287 in? x 257 393 mph| beoasic] 1me | 1 | slome (I R [ | ke L (Oprem 7. Which gives a better order of magnitude calculation of final velocity of a rocket starting at rest, accelerating at a rate of 9.80 m/s? for a distance of 197 km? (only one answer here) LatE ok R VooV v sl a. 100 m/s | / agocom= L (44 Y LS+ Gus () V= 0% Q.4 (g00.31) b. 200 m/s pooom= 3 %) /s | 00 m/ ece st o= 19¢y,99 ¢. 1000 m/s o ' //d\ ZOOOm/S It 2 o020y £z 700,51 8. A march:ng band player walks directly north at a speed of 2.50 m/s for a time of 12.0 seconds, turns 35.0°to the east and walks at half that speed for another 15.0 seconds. What is the playerz total displacement and average velocity for the entire trip? \ 26af$ i( /( he. 0s P R 1250 i5of ok 2.5M 124 - ,7 !z,Oj o '%;2,6”"“/5 +1.25 '“/<) /i:’ lg7% ws 204 g, 75, = /L/ T75m 9. The graph at right shows velocity vs. time. The particle at t=0 ol g, T was positioned at Xo = 15. What is the position, velocity and 5 ““\ acceleration at t=8.0 s? (three answers here) Jeloc o PR W -5 E\\\ 10 fp a. Position - 70, -\5 b. Velocity 4,,;,/5 “o c. Acceleration -1.797¢ f«\/cz
10. A sky diver drops his smuggled 1kg weight from physics lab out of the air plane before he jumps, which is flying horizontally going 175 m/s at a height of 2 km. IF WE ASSUME NO AIR RESISTANCE, what is the Velocity (magnitude and direction) when it hits the 3 AT hzL & grpund. I 17§ /s J 5 2 9 V= 264,39 mj /f 2 g T K IHanio0g) sooom: 446 J | anls ) , \ 2198, : - o= 4%.¢" ( 2000m C\\ L e Je-doee SV Vyz 0+ 9.41(20.2) e \ £-7075 B 4% 14 Vo 19% 19 wm fs Af i 2 - Bz te g7 1754 1ag.1a2- = U, = 264 3q 11. If an object initially at rest is given a velocity of: vx =-1.93 m/s and vy =2.51 m/s./\Nhat is the magnitude and direction its displacement after a time of 3.50 seconds? 6/0%\ YEE = e a3 (35) Yz vyt 7 gresi () L= -6.75% 4= 9.75¢ o «.7%8 . WS i e = 4an ( ( 7ee > masmiuc\c L ,\“/s o = Ir (ll"{(4‘{oy")f @ 2.4 e 150-s2.9=1717.5¢ 12. A cannon shell is fired on a flat surface and it is found that the shell is in the air for 2.78 s and reaches a distance of 94.7 m. What was the initial Launch angle and speed? /' \ E,LI\\'\" C\L;Nlé{};&wfi, \/:%, S\, 2 2785 (A8 of 7651500 . 2 s (e —y Han (- >4an < =)= B: 21 _— j——____G=tar <—;§> 7 ¢ Z(%—;} 19 VAR i T—— - - | Louneh Angle L 25“\(?1‘3) ' = 1,27 Coizetk Vo C [ =7 ooy = 3609 R WS 13. In an airport moving walkways help get passengers to their gates faster. In one case a person stands on the walkway and arrives at the gate in 525 seconds. Another time the walkway is broken and he makes the same trip in 357 seconds walking at a constant speed. How fast could he make it, if he walked the same speed on the moving walkway? V= 4/ V, zd/5s7 | “+=—d/ (. d_ 4 T 2 _ 2. = TRy & d 5257 357 00087 21<. 5 ¢
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Longer problems (15pts each) 14. If Ax =4.56 and A, = -2.27 and B has magnitude of 3.65 at 75.0° above the negative x axis. a. What are the x and y components of vector B? By= 365 Cosl78)= ~0.A4S Bor 2066 %a(76% 2-52L 3.53 b. Find the vector components of vector C given as C = 24 - 3E. X Y 3 =N C = Z(L‘/.§Gi~ Z=27y) 25 (- 0SSk + 3.;\:\;) C= AT Hgdy + 2535 ~ 0, 51y L= 1.9 x - '":?i"'s"i‘" ¢. What is the magnitude and direction of vector C? e | /l{»'3 2 . o\2 - B, tan” | —— | = o (= Nt sae ) = 19,29 B o8 \m> oo C\d(jch@m: ~51.7° 0 17" below the y-ans 15. A ball is dropped from rest from the top of a building. Two motion detectors which are positioned outside of two different windows, one above the other, record the velocities of -15.5 m/s and -17.2 m/s as the ball goes past them. (no air resistance) a. How far apart are the motion sensors mounted? Al Top . yoyy 17.2-195F [ g 120w vV - U\' -2 a A A= 2 A( 7 (%) R owet ) )5S l 255 }A:'zlfgm %‘ MS2 V=112 by | - 1203, €355 b. How far from the top motion sensor is the top of the building where the ball was released? VAT A TN orlsg=2(av)p 24025 = 19.¢2 8 Rz12.2m c. If the second motion sensor can also measure the time it takes for the ball to hit the ground after passing it as 3.50 seconds, how tall is the building? C -0 A-B 2.2 . (. - \/&-7}:3‘1 B-c 7“2:%'}“.‘ C-Pz 120.3m c= 1.2 35+ 3 (4%) (3.5) P Total Hc*gm 21253 m
16. You're 6.00 m from one wall of a house. You ‘1 . . N want to toss a ball to your friend who is 6.00 m . A x:}W\ from the opposite wall. The throw and catch > ] {7 30m H Bi %, both occur at 1.00 m above the ground. [1.0m {:3 ! & a. What is Vyg if it just barely clears the roof? \{ N & . 6.0m 6.0 m 6.0m : : LT . . o 30{— 2 a C\: \/] é (c\)(é ) 4on U[S\}-"Z;” ) V7 ) N(:\ ".:1 - 5wz “/kjo (fol\) Tz (\((.%\ '“140§¢ D't + Ji (f.'fal )& 3*,(,\,\(‘155)'— x E’:m: \/j° (2.02‘\‘;55 < e 1 <=3 10 = Vg M‘,() 17 os fotel ch.f)wr of House: 6 Lol dels ¢+ lowz . I\)’\y—; 6{.‘1(2:’1/2] 2€: 2ot .07 %%: v, ¢ b b. How long is it in the air when it just clears the roof? y "(? 2) lq: ° O ] g g d=v €2 (D) I 1095 2 oft) *+ 1 (a9 (E) J €281 02 {j - l.ols c. What minimum launch speed that will allow the ball to clear the roof? Ve las s gay 5 e d. At what angle should you toss the ball? N ‘}'Gx\f\ 6= \/‘/ . 4P Gl o 0= fon ( moqs}

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