Newtons 3rd law and momentum

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Newtons 3rd law and momentum
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S I R I S A A C N E WTON (1647 - 1727)
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• Newton worked in the 1600s. He talked about
  momentum before he talked about force but, maybe
  because momentum is hard to conceptualise, we
  learn Newtons Laws as statements about force.
• Momentum is always conserved. Because momentum is
  always conserved, pairs of forces must be equal
  and opposite.
• Lets look at momentum

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Momentum and collisions

• We use momentum to solve collision problems in
  isolated systems.
• An isolated system has no external forces eg
  games of pool, frictionless surface problems
• momentum mass x velocity
• p mv
• (kgms-1) (kg) x (ms-1)
• Momentum is a vector. Use vector diagrams if
  story not one dimensional!!

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• Calculate momentum two ways
• Actual momentum at one time
• Change in momentum between start and end times

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1 Momentum now

• On icy winter roads a 1500kg car is travelling at
  21ms-1. Calculate the momentum.
• P mv
• 1500 x 21
• 31 500 kgms-1

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2 Changing momentum
A
B

• Initial momentum of A plus initial momentum of B
• mAviA mBviB

• Equals final momentum of A final momentum of B
• mAvfA mBvfB

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QuestionOn icy winter roads a 1500kg car
travelling at 21m/s collides with a 1800kg van
going 15m/s in the opposite direction. The two
vehicles lock together (1D collision) and move
off with a new speed v. Find v.
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1500kg

• Answer
• Draw a diagram
• Find the initial momentum of each and add
  (considering direction)
• Find the combined mass and multiply by new v
• Equate and solve

• Ptruck 1800 x 15
• Pcar -1500 x 21
• Ptotal 4 500 kgms-1
• After
• Mass 3300kg
• 4 500 3300 x v
• V 4500/3300
• 1.4 ms-1 (in direction of car)

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2D change in momentum

• Change in momentum in 2 or 3D needs vectors.
• change final vector initial vector

Use diagrams!
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• Question
• An ice hockey puck of mass 0.8kg moving at
  3.5ms-1 hits the side of a second puck initially
  at rest. The mass of the second puck is 0.70kg.

• After the collision the 0.8kg puck moves off at
  2.8ms-1 at right angles to its original
  direction.
• Find the velocity of the second puck immediately
  after the collision

3.5m/s
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AnswerChange in momentum of 0.8 puck equal and
opposite tochange in momentum of 0.7 puck.

• 0.8 puck
• Final initial Final opposite

2.8kgm/s
3.6kgm/s on an angle of 37o
2.24kgm/s
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So 0.7 puck has equal and opposite change

• Initially at rest so momentum 0
• final initial 3.6 0
• v p /m
• 3.6 / 0.7
• 5.1 m/s on 37o

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NEWTONS THIRD LAW
ACTION AND REACTION ARE ALWAYS EQUAL AND
OPPOSITE
IF A BODY A EXERTS A FORCE ON BODY B, THEN B
EXERTS AN EQUAL AND OPPOSITELY DIRECTED FORCE ON
A
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• Momentum are calculated at one time or over a
  change in time.
• Forces re calculated over a change in time.
  Mathematically, this is in the acceleration
  number.

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ILL PULL HIM
devishly clever
WELL, ACTION FORCE AND REACTION FORCE ARE ALWAYS
EQUAL AND OPPOSITE!!
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Action force and reaction force are always equal
and opposite,
WELL ACTION AND REACTION ARE ALWAYS EQUAL AND
OPPOSITE!!
SO WHY DOES THE GIRL MOVE FASTER?
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NEWTONS THIRD LAW PAIRS

• THEY ARE EQUAL IN MAGNITUDE
• THEY ARE OPPOSITE IN DIRECTION
• THEY ACT ON DIFFERENT BODIES

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NEWTONS THIRD LAW PAIRS
SIMILARITIES
DIFFERENCES
The 2 forces act for the same length of time
The 2 forces act on different bodies
The 2 forces are in opposite directions
The 2 forces are the same size
The 2 forces act along the same line
Both forces are of the same type
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THE CLUB EXERTS A FORCE F ON THE BALL
THE BALL EXERTS A N EQUAL AND OPPOSITE FORCE F ON
THE CLUB