8. cons of p (1-d)
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UNIT 1: CONSERVATION LAWS
Lesson 8:
Conservation of Momentum (1-D)
CENTRE HIGH: PHYSICS 30
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Recommended Reading:
Heath pp. 244 - 248
Ladner pp. 43 - 44
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Momentum is conserved for collisions and explosions
Collisions:
STICK!
Explosions:
BOOM!
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F. Law of Conservation of Momentum
1 2
When two objects collide:
3rd Law: They exert equal and opposite forces
for the same period of time
So,
Fon 1 = - Fon 2
Fon 1 t = - Fon 2 t
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Fon 1 = - Fon 2
Fon 1 t = - Fon 2 t
Based on the equation F t = m v, it follows
m v1 = - m v2
p1
= - p2
or
p1 + p2 = 0
No overall change in momentum
(conserved)
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F1. Conservation of Momentum (1-D)
For collisions and explosions, there are two equations:
1. Total Momentum is conserved (constant)
pT = pT '
- the total momentum before the collision
is the same as the total momentum after the collision
- this is also true for explosions
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F1. Conservation of Momentum (1-D)
For collisions and explosions, there are two equations:
1. Total Momentum is conserved (constant)
pT = pT '
These are vector sums
Need a reference direction
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2. No overall change in momentum
p1 + p2 = 0
or
p1 = - p2
Equal but opposite impulses
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Conditions for momentum to be conserved :
1. A closed system
- no objects lost / gained
2. Fnet = 0 on the whole system
- the objects must move at a constant velocity
before and after the collision
e.g.
No friction
Level surface
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Consider a collision:
10 kgm/s 6 kgm/s 3 kgm/s p?
Boom!
What is the total momentum before the collision?
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Ref: Right is positive
10 kgm/s -6 kgm/s -3 kgm/s p?
Boom!
Remember, for 1-D vectors,
you must determine a positive direction (i.e. a ref system)
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10 kgm/s -6 kgm/s -3 kgm/s p?
Boom!
pT = 10 - 6 = 4 kgm/s
What is the total momentum after the collision?
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10 kgm/s -6 kgm/s -3 kgm/s p?
Boom!
pT = 10 - 6 = 4 kgm/s pT ' = 4 kgm/s
pT = pT ' = 4 kg m/s
Total momentum remains constant throughout the collision!
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10 kgm/s -6 kgm/s -3 kgm/s
Boom!
pT = 10 - 6 = 4 kgm/s pT ' = 4 kgm/s
So, what is the momentum of the blue object after the collision?
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10 kgm/s -6 kgm/s -3 kgm/s 7 kgm/s
Boom!
pT = 10 - 6 = 4 kgm/s pT ' = 4 kgm/s
So, p = +7 kgm/s
The momentum of the two objects must add up to 4 kgm/s
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10 kgm/s -6 kgm/s -3 kgm/s 7 kgm/s
Boom!
Calculate the change in momentum for object 1 and 2
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10 kgm/s -6 kgm/s -3 kgm/s 7 kgm/s
Boom!
Also,
p1 = p1 ' - p1 = (-3 kgm/s) - (10 kgm/s)
= -13 kgm/s
p2 = p2 ' - p2 = (7 kgm/s) - (-6 kgm/s)
= +13 kgm/s
Equal but opposite impulses (no overall change)
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Animation:
Air Track collisionshttp://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassM
echanics/AirTrack/AirTrack.html
Attempt to predict the answers before you run the animation- use the equations to calculate
Be aware that an inelastic collision is when they stick
together
http://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.htmlhttp://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.htmlhttp://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.htmlhttp://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.html -
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Ex. 1 A 12.5 kg object, moving East at 4.0 m/s, collides with
a 13.0 kg object moving West at 7.0 m/s. If they stick
together (an inelastic collision), then the magnitude oftheir velocity is _________ m/s.
Your 2-digit answer is
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Solution:
4.0 m/s 7.0 m/s v ?
Stick!12.5 kg 13.0 kg 25.5 kg
Draw a picture first.
Show clearly before and after the collision.
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Solution:
+4.0 m/s -7.0 m/s v ?
Stick!12.5 kg 13.0 kg 25.5 kg
Ref: East is +
pT = pT'
p1 + p2 = pT'
m1 v1 + m2 v2 = mT v
The two masses combine
Treat as one (total) mass
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m1 v1 + m2 v2 = mT v
(12.5 (4.0) + (13.0) (-7.0) = 25.5 v
-41 = 25.5 v
v = -1.6 m/s
So, v = 1.6 m/s West
Numerical Response:
1 . 6
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Ex. 2 A 750 kg car is moving East and collides with a 580 kg car
moving West at 20.00 m/s. After the collision, the 580 kg car is
moving East at 5.00 m/s.
The magnitude of the impulse on Car 1 is b x 10w N s.
Your 3-digit value for b is
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Solution:
Recall, for momentum to be conserved,the impulses have to be equal and opposite
i.e.
Both cars experience the same magnitude of impulse,but in opposite directions
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Ref: East is positive
750 kg 580 kg 1 2
-20.00 m/s Boom! +5.00 m/s
Find impulse on Car 2:
m2 v2 = m2 (v2' - v2)
= 580 kg [ (+5.00 m/s) - (-20.00 m/s) ]
= +1.45 x 104 kg m/s (East)
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The cars will experience equal but opposite impulses
(Cons of momentum)
Thus, Impulse on Car 1 = -1.45 x 104 kg m/s
= 1.45 x 104 Ns West
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Numerical Response format (b 10w)
1.45 x 10
4
w
b
So, the 3-digit value for b is
1 . 4 5
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Practice Problems:
TryLadner p. 46 #1 - 12
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SUMMARY
1. Conservation of momentum is based on which law?
2. What are the two equations for Cons of p?
3. What conditions are required for Cons of p?
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SUMMARY
1. Conservation of momentum is based on which law?
Newton's 3rd Law- equal and opposite forces
- so, equal and opposite impulses
2. What are the two equations for Cons of p?Total p is constant
pT = pT'
No overall change in momentum
p1 + p2 = 0
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3. What conditions are required for Cons of p?
Closed system - no objects lost or gained
Fnet = 0 on system- objects have constant velocity before and after
the collision / explosion
- no friction / level surface
4. Review Animations:
Air Track collisions
2 Block collisions
http://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.htmlhttp://www.geocities.com/thesciencefiles/collisions/collisions.htmlhttp://www.geocities.com/thesciencefiles/collisions/collisions.htmlhttp://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.html
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