Topic 9 Motion in Fields 9.1


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Topic 9 Motion in Fields 9.1

  1. 1. Fields and ForcesTopic 9.1 Projectile Motion
  2. 2. Components of Motion When a body is in free motion, (moving through the air without any forces apart from gravity and air resistance), it is called a projectile Normally air resistance is ignored so the only force acting on the object is the force due to gravity This is a uniform force acting downwards
  3. 3. Therefore if the motion of the projectile isresolved into the vertical and horizontalcomponentsThe horizontal component will be unaffectedas there are no forces acting on itThe vertical component will be accelerateddownwards by the force due to gravity
  4. 4. These two components can beconsidered as independent factors inthe motion of a projectile in a uniformfieldIn the absence of air resistance thepath taken by any projectile is parabolic
  5. 5. Solving Problems In solving problems it is necessary to consider the 2 components independently
  6. 6. Therefore the horizontal motion it isnecessary to use the equationspeed = distance timeWhere speed is the horizontalcomponent of the velocity
  7. 7. Therefore the vertical motion it isnecessary to use the kinematicequations for uniform accelerationi.e. Using the s.u.v.a.t equationsWhere u and v are the initial and finalvertical components of the velocity
  8. 8. Example A ball is kicked at an angle of 40.0o with a velocity of 10.0 ms-1. Taking g = 10 ms–2. How far does it travel horizontally? 10ms-1 40o
  9. 9. To be able to calculate the horizontaldistance we need to know thehorizontal speed, and the time.The horizontal distance is easy tocalculate by resolving the velocity 10.0 sin 40.0o 10.0ms-1 40.0o 10.0 cos 40.0o
  10. 10. However, to calculate the time we willneed to use the vertical component andthe s.u.v.a.t. Equations
  11. 11. s=?u = 10.0 sin 40.0o ms-1v=?a = -10.0 ms-2 (Up is positive, thereforeacceleration here is negative)t=?We only have 2 of the values when weneed three to find any other
  12. 12. However, if we ignore air resistance,then the final vertical component of thevelocity will be equal and opposite ofthe initial componenti.e. v = -10.0 sin 40.0o ms-1Looking at the equations for uniformacceleration, we need an equation thatlinks u, v, a and t.
  13. 13. v = u + atRearranging to make t the subjectt=v–u aSubstitute int = -10.0 sin 40.0o – 10.0 sin 40.0o -10t = 1.286 seconds
  14. 14. Now returning to the horizontalcomponentsUsing speed = distance timeRearranging distance = speed x timeDistance = 10.0 cos 40.0o x 1.286Distance = 9.851 = 9.9 metres
  15. 15. Using the Conservation ofEnergy In some situations the use of the conservation of energy can be a much simpler method than using the kinematic equations Solving projectile motion problems makes use of the fact that Ek + Ep = constant at every point in the objects flight (assuming no loss of energy due to friction)
  16. 16. Example A ball is projected at 25.0 ms-1 at an angle of 40.00 to the horizontal. The ball is released 2.00m above the ground. Taking g = 10.0 ms-2. Find the maximum height it reaches.
  17. 17. Solution B v = vhorizontal H 25.0 ms-1A 2.0m
  18. 18. Total energy at A is given byEk + Ep = ½ m (25.0)2 + mg x 2.0 =312.5m + 20m = 332.5m
  19. 19. Next, to find the total energy at B weneed to know the velocity at B, which isgiven by the horizontal component ofthe velocity at ATotal energy at B is given byEk+ Ep = ½ m (25.0 cos 40o)2 + mg x H =183.38m+ 10mHThen using the conservation of energy
  20. 20. Equating the 2 equations332.5m = 183.38m + 10mH332.5 = 183.38 + 10H332.5 – 183.38 = 10H10H = 149.12H = 14.912 = 14.9m