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Vectors and 2-D Motion

Vectors and 2-D Motion. resultant :. the single vector that has the same effect as several vectors added together. Vector Addition: Graphical Method. 1. Choose a scale. 2. Use ruler and protractor to draw first vector to scale at proper angle. 3. Draw subsequent

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Vectors and 2-D Motion

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  1. Vectors and 2-D Motion

  2. resultant: the single vector that has the same effect as several vectors added together

  3. Vector Addition: Graphical Method 1. Choose a scale. 2. Use ruler and protractor to draw first vector to scale at proper angle. 3. Draw subsequent vectors tip-to-tail. 4. Draw resultant by connecting tail of first vector to tip of last.

  4. measure length in cm measure angle d2 (2.8 cm) d1 (4.2 cm) Add. d1 = 420 m horiz. d2 = 280 m @ 30.o above horiz. scale: 1 cm = 100 m (say, 6.8 cm) dr Q 30o dr = 680 m @ 12o above horiz.

  5. v2 (3.2 cm) v1 (5.3 cm) v3 (4.5 cm) W W N vr Add. v1 = 53 km/h @ 25o N of W v2 = 32 km/h @ 20.o S of W v3 = 45 km/h @ 25o E of S scale: 1 cm = 10 km/h Q vr = 66 km/h S @ 30o S of W

  6. W E E N Add. d1 = 1160 km @ 20.o N of E d2 = 470 km @ 60.o S of E d3 = 280 km @ 70.o N of W scale: 1 cm = 100 km d1 d2 d3 dr Q dr = 1300 km @ 12o N of E

  7. E E N Add. d1 = 65 m @ 15o N of E d2 = 48 m @ 63o N of E scale: 1 cm = 10 m dr d2 Q d1 dr = 110 m @ 35o N of E

  8. If vectors are , use Pythagorean theorem and trigonometry (i.e., tan–1). d1 d1 N d2 d2 Algebraic Method Add. d1 = 75 km S d2 = 45 km W F Q dr dr = 87 km @ 31o W of S (or 59o S of W)

  9. v2 v2 v1 v1 N Add. v1 = 380 km/h N v2 = 145 km/h E vr Q F vr = 410 km/h @ 21o E of N (or 69o N of E)

  10. What if vectors are NOT ? vector resolution: taking a vector and breaking it down (“resolving” it) into two component vectors J “y” Jy Q Jx “x” where… Jx = J cos Q and Jy = J sin Q Use Q along with sin and cos functions, as appropriate.

  11. 68 N 32o Fv Fh Resolve into components. F = 68 N @ 32o above horiz. Fh= F cosQ = 68 cos 32 = 58 N Fv= F sin Q = 68 sin 32 = 36 N

  12. 15o vv 14.7 m/s v = 14.7 m/s @ 15.0o below horiz. vh vh= v cosQ = 14.7 cos 15 = 14.2 m/s vv= v sin Q = 14.7 sin 15 = 3.80 m/s

  13. aw as 3.2 m/s2 N a = 3.2 m/s2 @ 38o S of W 38o aw= a cosQ = 3.2 cos 38 = 2.5 m/s2 W as= a sin Q = 3.2 sin 38 = 2.0 m/s2 S

  14. 4. Vectors resulting from Steps 2 and 3 are . Add them as you would any two vectors, using Pythagoras and tan–1. To add non-perpendicular vectors algebraically… 1. Resolve all vectors into “x” and “y” components. 2. Add “x” components, taking into account direction of each. 3. Add “y” components in the same way.

  15. d1 = 65 m d2 = 48 m d1NS d2NS 15o d1EW 63o d2EW Add algebraically. d1 = 65 m @ 15o N of E d2 = 48 m @ 63o N of E d1EW= d1NS= d2EW= d2NS=

  16. d2 d2NS d2EW d2EW Q d1NS d1NS d1EW drEW= add comp.’s drNS= dr d1 dr = Q = dr = 1.0 x 102 m @ 35o N of E

  17. d1EW d2 d2NS d1NS d2EW d1 d1 dr 28o d2 12o d1EW= d1NS= d2EW= d2NS= d1 = 425 km @ 12o E of N Add algebraically. d2 = 182 km @ 28o E of S

  18. 182 km 425 km dr 255 km 255 km Q 173.8 km 173.8 km add comp.’s… drEW= drNS= dr = = 310 km Q = F dr = 310 km @ 56o N of E (or 34o E of N)

  19. (projectile) Projectile Motion projectile: an airborne object acted upon “only” by gravity -- other forces are NOT significant non-examples: examples: baseball arrow football golf ball bird airplane frisbee hang glider

  20. 0 m/s, but increases as object falls (i.e., object accelerates at g = 9.81 m/s2) Horizontally-Launched Projectiles • Theory: A projectile is launched horizontally • w/velocity vx from height Dy. vx An HL’edproj. stays in the air for the same amt. of time as a dropped proj. from same height. Dy constant (assuming no air resistance) -- vx is… ZERO! -- axis… -- vy starts at…

  21. Change in VERTICAL velocity: From Dv = a Dt, we get… Dvy = g Dt or…vy,f– vy,i= g Dt Change in VERTICAL displacement: If vyi or vyf = zero, then…Dy = ½ g (Dt)2 If not, then…Dy = ½ (vy,f+ vy,i) Dt

  22. A rifle bullet is fired horizontally w/initial speed 850 m/s from height 1.73 m. Find… a. …time bullet is in air Dy = ½ g (Dt)2 Dt = 0.594 s

  23. b. …bullet’s range (max. horizontal displacement) Dx = vxDt = 850 m/s (0.594 s) = 5.0 x 102 m c. …bullet’s height above ground at t = 0.38 s. Bullet is released from “vertical” rest.In 0.38 s, it will fall… Dy = ½ g (Dt)2 = ½ (–9.81)(0.38)2 = –0.71 m Height at t = 0.38 s is… 1.73 m – 0.71 m = 1.02 m

  24. vx = ? . 62 m 238 m Projectile is launched horizontally from height 62 m. Its range is measured to be 238 m. Find launching speed. = 3.56 s Dy = ½ g (Dt)2 = 67 m/s

  25. vx = ? . 62 m 238 m Projectile is launched horizontally from height 62 m. Its range is measured to be 238 m. Find launching speed. Where? g = ? Dt to fall 62 m req’dvx Earth 9.81 m/s2 3.56 s 67 m/s Moon 1.6 m/s2 8.80 s 27 m/s Mars 3.71 m/s2 5.78 s 41 m/s Jupiter 24.8 m/s2 2.24 s 110 m/s

  26. vx Dy Dx = ? Military plane flies horizontally at altitude 8000. m at speed 100. m/s. As measured along ground, how far from target must plane be when it releases target’s provisions? besieged city of G.G. B.G. B.G. Dy = ½ g (Dt)2 Dx = vxDt = 100 m/s (40.4 s) = 4.04 x 103 m

  27. 8000 7000 6000 5000 y (m) 4000 3000 2000 1000 0 0 1000 2000 3000 4000 5000 x (m) Coordinates of Horizontally-Launched Projectiles (and Velocity at Any Time) For the previous problem, what are the coordinates of the provisions (and their speed) at any time t?

  28. 0 0 0 8000 100 100 8000 1000 –491 –98.1 7509 100 140 8000 2000 –1962 –196.2 100 220 8000 6038 3000 3585 100 311 8000 –4415 –294.3 4000 152 –7848 –392.4 8000 100 405 –12263 –490.5 100 501 –4263 5000 8000 ystraight “–” yfall orig.height “pyth” vx and vy vx vy,i “–” g Dt vxDt ½ g (Dt)2 xcoordis “how far horizontally” in each Dt ystraight is “ycoordof projectile without gravity” yfall is “how far object would fall” in each Dt

  29. 8000 7000 6000 5000 y (m) 4000 3000 2000 1000 0 0 1000 2000 3000 4000 5000 x (m)

  30. 8000 7000 6000 5000 y (m) 4000 3000 2000 1000 0 0 1000 2000 3000 4000 5000 x (m) yfall ystraight(orig. height) (½ g (Dt)2) ycoord (ystraight“–” ½ g (Dt)2)

  31. Review Problem from 1-D Motion An object is launched w/initial velocity 37 m/s . a. Find time object is in air. vyf – vyi = g Dt To get to top… 0 – 37 = –9.81Dt Dtair = 7.5 s Dttop = 3.77 s b. Find max. height object attains. Dy = ½ g (Dt)2 = ½ (-9.81 )(3.77)2 = 70. m

  32. 0 b. Find max. height object attains. (alternate methods) Use the “No time” equation for vertical-only motion... vyf2 = vyi2 + 2 g Dy vyi2 = –2 g Dy = 70. m Or use Dy = ½ (vyf + vyi) Dt between the ground and the top… Dy = ½ (37 + 0)(3.77) = 70. m

  33. vyi = vi sin Q, but changes as object accelerates at 9.81 m/s2 vi vyi Q vx Projectiles Launched at an Angle • Theory: A projectile is launched • at initial angle Q with • initial velocity vi. -- vx is… constant (assuming no air resistance) -- vy starts at…

  34. 37.0 m/s 56.4 m/s 41o Projectile is launched w/initial velocity 56.4 m/s at 41o above horizontal. a. Find time object is in air. 42.6 m/s vyi = vi sin Q = 56.4 m/s (sin 41) vx = vi cos Q = 56.4 m/s (cos 41) 7.5 s (See blender prob.) b. Find max. height object attains. (See blender prob.) 70. m

  35. 37.0 m/s 56.4 m/s 41o c. Find projectile’s range. 42.6 m/s = 42.6 m/s (7.5 s) Dx = vxDt = 320 m An angled projectile... stays in the air as long as – and goes as high as – a “straight up” projectile launched upward at the vyi of the angled projectile.

  36. 8.73 m/s 16.4 m/s 18.6 m/s 28.0o Projectile is launched w/initial velocity 18.6 m/s at 28.0o above horizontal. a. Find time object is in air. (Get init. comps of vel.) To get to top… = 0.890 s  1.78 s

  37. 8.73 m/s 18.6 m/s 28.0o 16.4 m/s b. Find max. height object attains. Dy = ½ g (Dt)2 = ½ (-9.81)(0.890)2 = 3.89 m c. Find projectile’s range. Dx = vxDt = (16.4 m/s) (1.78 s) = 29.2 m

  38. 35.58 m/s 80 60 y (m) 55.80o 40 20 0 0 20 40 60 80 100 120 x (m) Coordinates of Projectiles Launched at an Angle Object is launched from ground w/init. vel. 35.58 m/s @ 55.80oabove horizontal. Find its x and y coordinates at each second it is in the air. vyi 29.43 m/s vx 20.00 m/s

  39. 0 0 29.43 0 0 20 35.58 20 –4.91 19.62 24.52 20 28.02 29.43 40 –19.62 9.81 20 22.28 58.86 39.24 60 44.14 20 20.00 88.29 –44.15 0.00 80 39.24 –78.48 –9.81 117.72 20 22.28 –122.63 –19.62 20 28.02 24.52 100 147.15 –29.43 –176.58 20 35.58 0 176.58 120 vyiDt ystraight “–” yfall “pyth” vx and vy vx vy,i “–” g Dt vxDt ½ g (Dt)2 20(Dt) 29.43(Dt) xcoordis “how far horizontally” in each Dt ystraight is “ycoordof projectile without gravity” yfall is “how far object would fall” in each Dt

  40. 80 60 y (m) 40 20 0 0 20 40 60 80 100 120 x (m) ystraight 29.43 m/s 35.58 m/s 55.8o 20 m/s yfall 20 m/s 22.28 m/s 28.02 m/s 35.58 m/s

  41. 84 m/s 105 m 105 m 330 m For how long will the projectile remain airborne? From Dy = ½ g (Dt)2… OR… Find xcoord @ 4.6 s… = 84 m/s (4.6 s) Dx = vxDt = 390 m Dt = 6.5 s

  42. 21.39 m/s 49.5 m/s 25.6o A cannonball is launched off the edge of a cliff at an init. speed of 49.5 m/s at 25.6o above horiz. The ball is in the air for 9.75 s. Find height of cliff, the range, and the impact speed of the cannonball. 49.5 m/s 25.6o 44.64 m/s ycoord KEY: height = –ycoord at impact ycoord = ystraight – yfall ycoord = (21.39)(9.75) – ½ (9.81)(9.75)2 So height = 258 m

  43. 21.39 m/s 49.5 m/s 25.6o (range…) = (44.64 m/s) (9.75 s) Dx = vxDt = 435 m (impact speed…) 44.64 m/s 44.64 m/s vy = 21.39 – 9.81(9.75) Q vy @ 9.75 s = ? (74.26 m/s) vr = ? Pythagorize to get… vr = 86.6 m/s

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