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Momentum in 2 Dimensions
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Momentum in 2 Dimensions

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  • 1. Momentum in 2D Section 1: Topic 4
  • 2. Newton’s Second Law
    • From previous experiments, we are aware that:
    • a  F and a  1/ m
    • This gives rise to the equation; F = ma
  • 3. Newton’s Second Law
    • In vector form: F = m a
    • I ndicates a relationship between force and acceleration.
      • The acceleration is in the same direction as the net force.
    • I mplies the force on an object determines the change in velocity ( a  F ) and
  • 4. Try Example 1
  • 5. Solution
  • 6. Solution – Part (a)
    •  v = v t - v o
    •  v = 10 ms -1 S - 10 ms -1 W
    • By Pythagoras’ Theorem;
    • (  v ) 2 = ( v t ) 2 + ( - v o ) 2
    • (  v ) 2 = 10 2 + 10 2
    •  v =  200
    •  v = 14.1 m s -1 at 135 o T
  • 7. Solution – Part (b)
    • a = 1.41 x 10 3 m s -2 at 135 o T
  • 8. Solution – Part (c)
    • F = m a
    • F = 50 x 1.41 x 10 3
    • F = 7.07 x 10 4 N at 135 o T
  • 9. Solution – Part (d)
    • By Newton’s Third Law :
    • E qual an opposite force is applied
      • 7.07 x 10 4 N at 315 o T
  • 10. Momentum
    • Is a property of a body that is moving.
    • V ector quantity.
    • If no net force is acting on the body/bodies, momentum is defined as the product of mass and velocity.
  • 11. Momentum
    • p = m v
    • Units are given as kg m s -1 or s N .
    • Direction is the same as the velocity of the object.
  • 12. Momentum
    • Two Astronauts and a Ball
  • 13. Application of Newton II
    • During collisions, objects are deformed.
    • F = m a
    • For constant acceleration
  • 14. Application of Newton II
    • F t = m v f - m v i  
    • F t =  p
    • F t = impulse of the force.
    • I mpulse causes the momentum to change.
    • An impulse is a short duration force .
      • U sually of non constant magnitude.
  • 15. Application of Newton II
    • Units are the same as those for momentum
      • Kg m s -1 or s N.
    • Defined as the product of the force and the time over which the force acts.
    • During collisions,  t is often very small.
      • F av is often very large.
  • 16. Conservation of Momentum
    • The total momentum of all particles in an isolated system remains constant despite internal interactions between the particles.
  • 17. Try Example 2
  • 18. Solution Before After v neutron v neutron v nitrogen m neutron m neutron m nitrogen
  • 19. Solution
    • m neutron = 1.67 x 10 -27 kg
    • m nitrogen = 2.31 x 10 -26 kg
    • v i neutron = 1.50 x 10 7 m s -1
    • v f neutron = -1.30 x 10 7 m s -1
    • v i nitrogen = 0 m s -1
    • v f nitrogen = ? m s -1
  • 20. Solution
    • By the law of conservation of momentum
    • p i =   p f
    • p i = m neutron v i neutron + m nitrogen v i nitrogen
    • (1.67 x 10 -27 ) x (1.50 x 10 7 ) + 0
    • = 2.505 x 10 -20 kg m s -1
  • 21. Solution
    • p f = m neutron v f neutron + m nitrogen v f nitrogen
    • = (1.67 x 10 -27 ) x (-1.30 x 10 7 ) + (2.31 x 10 -26 ) x v nitrogen
    • = (2.31 x 10 -27 x v nitrogen ) - 2.171 x 10 -20
    • As p i = p f
    • 2.505 x 10 -20 =(2.31 x 10 -27 x v nitrogen )-2.171 x 10 -20
    • v nitrogen = 2.02 x 10 6 m s -1
      • in the initial direction of the neutron
  • 22. Animation Example
    • Is momentum conserved?
    • What about kinetic energy?
  • 23. Collision
    • Car Crash
  • 24. Try Example 3
  • 25. Solution 53.1 o truck car
  • 26. Solution
    • m c = 750 kg
    • v i c = ? north
    • m T = 2000 kg
    • v iT = 36 km h -1 east = 10m s -1 east
    • v c+T = ? N53.1 o E
  • 27. Solution
    • By conservation of momentum,
    • m c v i c + m T v i T = m c+T v f c+T
    • 750 v i c N + 2000 x 10 E = 2750 v f c+T N53.1 o E
    • Diagrammatically:
  • 28. Solution
    • = 20.02 m s -1
  • 29. Solution – Part (a)
    • = 72 km h -1
    • The car was exceeding the speed limit.
    • Speed of car = 20.0 m s -1
  • 30. Solution – Part (b)
    • = 9.09 m s -1
    • = 32.7 km h -1
  • 31. Energy
    • The total energy in an isolated system is conserved .
      • E nergy can be transferred from one object to another .
      • Energy can be converted from one form to another.
    • The units are Joules (J) .
      • I s a scalar quantity .
      • D oes not have a direction.  
    • In collisions, total energy is always conserved.
  • 32. Energy
    • The kinetic energy will not always remain constant .
    • M ay be converted to other forms. Could be:
      • R otational kinetic energy
      • S ound
      • H eat.
  • 33. Types of Collisions
    • Elastic collisions
    • Momentum is conserved .
    • N o kinetic energy is lost.
      • O ccurs on the microscopic scale.
        • B etween nuclei.
  • 34. Types of Collisions
    • Inelastic collisions .
    • Momentum is conserved .
    • K inetic energy is lost.
      • All macroscopic collisions are inelastic.
      • Some collisions are almost elastic .
        • Billiard balls .
        • A ir track/table gliders.
  • 35. Types of Collisions
    • Perfectly I nelastic collisions .
    • Momentum is conserved .
    • K inetic energy is lost .
    • B odies stick together after the collision.
  • 36. Types of Collisions
    • Collisions
    • Collisions 2D
    • Rocket
  • 37. Try Example 4
  • 38. Solution
    • m neutron = 1.67 x 10 -27 kg
    • m nitrogen = 2.31 x 10 -26 kg
    • v i neutron = 1.50 x 10 7 m s -1
    • v i nitrogen = 0 ms -1
    • v f neutron = -1.30 x 10 7 m s -1
    • v f nitrogen = ?
  • 39. Solution
    • K i = ½ m neutron v i 2 neutron + ½ m nitrogen v i 2 nitrogen
    • =½(1.67 x 10 -27 ) x (1.50 x 10 7 ) 2 +½(2.31 x 10 -26 ) x 0 2
    •   =1.88 x 10 -13 J
    • K f = ½ m neutron v f 2 neutron + ½ m nitrogen v f 2 nitrogen
    • = ½(1.67 x 10 -27 ) x ( 1.30 x 10 7 ) 2 +½(2.31 x 10 -26 ) x (2.02 x 10 6 ) 2
    •   =1.88 x 10 -13 J
    • Since K i = K f , the collision is elastic.
  • 40. Try Example 5
  • 41. Solution
    • The masses of the incident atom, m i , and the struck atom, m s , are equal since they are both helium atoms.
    • m i = m s = m
  • 42. Solution – Part (a)
    • The collision is isolated .
    • p i = p f
    • m i v ii = m i v fi + m s v fs
    • m v ii = m v fi + m v fs
    • v ii = v fi + v fs
  • 43. Solution – Part (a)
    • The collision is elastic
    • K i = K f
    • ½ m i v i i 2 = ½ m i v fi 2 + ½ m s v fs 2
    •   i.e. ½ m v i i 2 = ½ mv fi 2 + ½ mv fs 2
    •   Thus v i i 2 = v fi 2 + v fs 2
  • 44. Solution – Part (a)
    • Therefore v ii is a hypotenuse .
    •  = 90 o in the vector-addition triangle .
    • Can use Pythagoras’s theorem.
    •   By trigonometry :
  • 45. Solution – Part (a)
    • v fi = 2.6 x 10 2 m s -1
    • Notice that we will always get  = 90 o
    • F rom masses being equal and the collision elastic.
  • 46. Solution – Part (b)
    • By trigonometry in the vector diagram,
    • v s = v i i sin30 o = 300 x ½
    • =150 m s -1
    • = 1.5 x 10 2 m s -1 at 90 o to v i i
  • 47. Flash Photography
    • I s used to analyse interactions between particles :
      • I n the lab .
      • I n either one or two dimensions.
    • R ules for analysis are as follows:
  • 48. Flash Photography
    • 1.   Distance between successive images is a measure of speed.
    • 2. Direction determined from multiple-image photograph.
    • 3. Line joining two successive images represent magnitude and direction of velocity vector.
  • 49. Flash Photography
    • To calculate distance - measure distance between successive images and adjust by the scale.
    • To calculate time - time between flashes =
  • 50. Flash Photography
    • Momentum :
      • - Use velocity vector and let m 1 = 1 unit and m 2 is scaled accordingly.
      • - This doesn’t change the validity of the process, only the scale for the momentum vector.
    • 7. Use vector diagrams for addition.
  • 51. Spacecraft Propulsion
  • 52. Spacecraft Propulsion
    • Who invented jet propulsion?
    • Nature
    • Squids take in water and expelled backward
      • Can control direction to some degree.
  • 53. Spacecraft Propulsion
    • A boat propels itself forward by pushing back on the water.
    • A car pushes back on the road.
    • An aircraft pushes back on the air.
  • 54. Spacecraft Propulsion
    • All vehicles move forward by pushing back on its surroundings.
    • They obey Newton’s Third Law :
      • For every action, there is a reaction.
  • 55. Spacecraft Propulsion
    • What then does a rocket push back against?
    • It can push back against the rocket pad on launch.
    • It can push back against the air when moving through the atmosphere.
  • 56. Spacecraft Propulsion
    • Neither of these options will work when it is in space .
      • N o atmosphere or anything else to push back against.
    •   The law of conservation of momentum can be used to explain what is happening.
  • 57. Spacecraft Propulsion
    • Before a rocket is launched, it is stationary .
      • N o momentum.
    • T otal momentum after the rocket is fired :
      • must also be zero.
  • 58. Spacecraft Propulsion
    • After the rocket is fired
      • G ases are ejected at high speed and,
    • A s the gas has mass,
      • T here is momentum acting in a direction directly opposite that in which the rocket is intended to move.
    • To conserve momentum, there must be an equal momentum acting in the direction in which the rocket moves.
  • 59. Spacecraft Propulsion
    • Mass of the rocket is large compared to the gas ejected, the velocity must be …..
      • much lower.
    • As gas is ejected, mass of the rocket ….
      • becomes less.
    • and the velocity ….
      • becomes greater.
  • 60. Spacecraft Propulsion
  • 61. Spacecraft Propulsion
    • In space, the ‘burn’ time is short.
    • According to Newton’s First Law :
    • R ocket will keep moving at the velocity given to it by ….
      • E xpulsion of the gas in the opposite direction.
  • 62. Spacecraft Propulsion
    • Spacecraft must carry fuel to burn
      • Increases launch mass dramatically
      • 95% of space shuttle launch is fuel, oxidiser and tanks
    • Alternatives exist
  • 63. Spacecraft Propulsion
    • Ion Thrusters
      • Geostationary Satellites
        • Used for station keeping since 1980s
      • LEO
        • Such as Iridium mobile communications cluster
      • Deep space position control
    • Can fire ions in opposite direction to motion
  • 64. Spacecraft Propulsion
    • Solar Sails
    • Converts light energy from the sun into
      • Source of propulsion for spacecraft
    • Giant mirror that reflects sunlight to
      • Transfer momentum from photons to spacecraft
  • 65. Spacecraft Propulsion
    • Ion propulsion is a technique which involves
      • Ionising gas rather than using chemical propulsion
    • Gas such as Xenon
      • Heavy to provide more momentum
      • Is ionised and accelerated
  • 66. Spacecraft Propulsion
    • Ions accelerated to speeds of
    • 3 x 10 4 m s -1
    • Nearly 10 x chemical counterpart
    • Chemical thrusters rely on
      • Energy released from chemical reaction
    • Ion thrusters depend on
      • amount of electrical power available
    • Electric rocket
  • 67. Spacecraft Propulsion
    • Where does an Ion Thruster get the power to ionise the gas?
    • Solar panels.
      • Called Solar Electric Propulsion
        • SEP
  • 68. Spacecraft Propulsion
  • 69. Spacecraft Propulsion
  • 70. Spacecraft Propulsion
  • 71. Spacecraft Propulsion
  • 72. Spacecraft Propulsion
  • 73. Spacecraft Propulsion
  • 74. Spacecraft Propulsion
    • Smart 1 Solar - Electric Propulsion Engine
    • Chemical Rocket vs Ion Thruster
  • 75. Spacecraft Propulsion Sails Unfurled The Cosmos 1 program called for the experimental craft to remain in Earth orbit. It would receive sunlight only on the daylight portions of its journey. The initial acceleration of the craft would be slight, less than 160 kmph by the end of the first day.
  • 76. Spacecraft Propulsion
    • Solar Sails have light
      • As propellant
    • Sun
      • As engine
    • Solar Sail
    • Force of sunlight at the Earth
      • Is approx 4.70  N m -2
  • 77. Spacecraft Propulsion
    • Space Shuttle engine provides
      • 1.67 x 10 6 N of thrust
    • Shuttle mass 2 x 10 6 kg
      • Payload 1 x 10 5 kg
      • 95% gas or spent boosters
    • Solar sail can propel a spacecraft to
      • 5 x faster than chemical rockets
      • Over time
  • 78. Spacecraft Propulsion
    • Photons bounce off (or absorbed) by sail
    • During collision
      • momentum conserved
    • Small mass provides small velocity change
  • 79. Spacecraft Propulsion
    • However, over time
      • Large number of photons
      • Continuous force
    • Large net force
      • eventually
    • Cannot be used to launch spacecraft
      • Still need chemical rocket
  • 80. Spacecraft Propulsion
    • Solar sails do not use the solar wind
    • The density of the solar wind particles
      • So small that wind pressure is less than
      • 0.1% that due to light pressure
  • 81. Spacecraft Propulsion
    • Reflected photons cause greater acceleration
      • than absorbed photons
    • Consider
      • for a given mass
      •  p = m  v
    • As velocity can change direction by 180 o
      •  v can double
      •  p can double
  • 82. Spacecraft Propulsion
    • If photon is absorbed
    • Momentum of spacecraft
    • p is =
    • p ip =
    • Final momentum of system
  • 83. Spacecraft Propulsion
    • If photon reflected
    • Initial momentum of spacecraft
    • p s =
    • p fp =
    • As momentum must be conserved,
      • p of system
    • p sys =
    • p fs =
  • 84. Spacecraft Propulsion
    • As a =  v/t
    • For reflected photon
      •  v can double
      • t is constant
      • Therefore a can double