11. Conservation Laws
Transcript of 11. Conservation Laws
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UNIT 1: CONSERVATION LAWS
Lesson 11:
Conservation Laws (E and p)
CENTRE HIGH: PHYSICS 30
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Recommended Reading
Heath pp. 315 - 328
Careful: This reading is quite challenging.
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G. CONSERVATION LAWS
- we will now consider situations where both conservation lawswill be used
1. Conservation of mechanical energy
2. Conservation of momentum
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G1. Elastic Collisions
- these are special types of collisions
Elastic Collision
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G1. Elastic Collisions
- these are special types of collisions
Elastic Collision
Momentum is conserved
pT = pT'
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G1. Elastic Collisions
- these are special types of collisions
Elastic Collision
Mechanical energy Momentum is conserved
is conserved
METi = METf pT = pT'
- no loss of energy
due to heat and sound
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Most collisions between objects are NOT elastic
To understand why, consider when two cars collide:
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When two cars collide:
Energy lost as heat / sound
Since energy is lost due to heat and sound, mechanical energy
is not conserved. So, it is not an elastic collision.
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When two objects stick together, the collision is called inelastic
Consider two trains colliding, each having equal massesand equal speeds:
M v v M
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When two trains collide (equal speeds, equal masses):
M v v M
All energy lost as
heat / sound
These trains will stick together and immediately come to rest.
All of the mechanical energy is lost.
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Conclusion:
When two objects stick together, it can never be
an elastic collision.
Elastic objects must bounce off of each other.
If they stick together, energy must be lost as heat / sound.
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Examples of Elastic Collisions
If most collisions between objects are not elastic,are there any examples of elastic collisions?
What would elastic collisions look like?
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Elastic Collision 1: Perfect "superball"
Consider a ball that is dropped from a height, and it lands
with a speed v :
Groundv
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If the ball (and ground) was perfectly elastic,
- it would rebound with the same speed
- it would return to the same height
v
Ground
v
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Elastic Collision 2: Newton's Cradle
Animation:http://www.walter-fendt.de/ph11e/ncradle.htm
When one ball comes down, why must only one ball come
back up?
Why can't two balls come up, each with half the momentum?
To satisfy both the laws of conservation of momentum and
mechanical energy, only one ball must come up.
http://www.walter-fendt.de/ph11e/ncradle.htmhttp://www.walter-fendt.de/ph11e/ncradle.htmhttp://www.walter-fendt.de/ph11e/ncradle.htmhttp://www.walter-fendt.de/ph11e/ncradle.htm -
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Elastic Collision 3: All atomic collisions (Classical theory)
- at the atomic level, heat and sound do not exist
- heat and sound are the vibration of atoms, which is
kinetic energy
Thus, in an atomic collision, energy cannot be lost as heat /
sound.
Mechanical energy is conserved
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Ex. 1
50 J 147 J 162 J KE?20 kgm/s 42 kgm/s 36 kgm/s p?
Boom!
If this collision is elastic, then find KE and p.
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50 J 147 J 162 J KE?
20 kgm/s 42 kgm/s 36 kgm/s p?
Boom!
Cons of ME: METi = METf
KE1i + KE2i = KE1f + KE2f
50 J + 147 J = 162 J + KE
KE = 35 J
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Ex. 2
v 20 m/s 33 m/s 14 m/s
Boom!
0.57 kg 0.41 kg
a) The speed v is ________ m/s.
Your 2-digit answer is
b) Is this collision elastic? Show by calculation.
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v 20 m/s 13 m/s 14 m/s
Boom!
0.41 kg 0.57 kga) pT = pT' Ref: Right is positive
p1 + p2 = p1' + p2'
m1v1 + m2v2 = m1v1' + m2v2'
(0.41) v + (0.57) (-20) = (0.41) (-13) + (0.57) (+14)
v = 34 m/s right
NR: 3 4
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b) METi = KE1i + KE2i
= 0.5 mv12 + 0.5 mv2
2
= 240.76 J + 114 J = 354.76 JMETf = KE1f + KE2f
= 0.5 mv12 + 0.5 mv2
2
= 34.65 J + 55.86 J = 90.51 J
Since METi = METf , this is NOT an elastic collision
2.6 x 102
J was lost as heat / sound.
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Practice Problems
TryLadner p. 60 #27, 28
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G2. Choosing Which Law to Use
When do you use:- conservation of mechanical energy?
- conservation of momentum?
- both conservation laws?
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1. Conservation of Mechanical Energy
- use when there is no loss of energy due to heat / sound
- no friction or air resistance- only mechanical forms of energy can change
2. Conservation of Momentum
- use for collisions and explosions- Fnet = 0 on the system
(the objects have a constant velocity before and after
the collision / explosion)
3. Both conservation laws
- use for elastic collisions
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Ex. 3 For each situation below, decide which law(s) to use
I. Cons of ME II. Cons of p III. Both
a) A rocket launch
b) An electron hitting a mercury atom
c) The maximum speed of a kid on a playground swing
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a) A rocket launch
Since this is an explosion,momentum is conserved. Rocket
i.e. FuelThe rocket exerts a downward
impulse on the fuel.
The fuel exerts an equal but upward
impulse on the rocket
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a) A rocket launch
Heat lost
However, since there is a lot
of mechanical energy lost as heat
and sound, ME is not conserved
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b) Collision between an electron and a mercury atom
e-v
e- v
Since all atomic collisions are elastic,
- momentum is conserved
- mechanical energy is conserved
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c) Max speed of a child on a playground swing
Since it is not a collision or explosion,
momentum is not conserved
Why?
Fnet 0 on the system
Thus, the system accelerates.
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c) Max speed of a child on a playground swing
PEg
There is no friction / air resistance.
All of the PEg at the top of the swing
converts to KE as it reaches the bottom
of the swing KEmax
So, mechanical energy is conserved
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Ex. 4 Ballistic Pendulum
Bullet
350 m/s 16.0 kg
70 gBlock of wood
If the bullet embeds into the block (and stays in the block),
find the maximum height the pendulum (and bullet)
will swing to.
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Stage 1: Collision between bullet and block (Cons of p)
350 m/s v ?
STICK!
0.70 kg 16 kg 16.70 kg
Before collision After collision
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Ref: Right is positive
Rest
350 m/s v ?
STICK!
0.70 kg 16 kg 16.70 kg
pT = pT'
p1 + p2 = p'
m1v1 + 0 = mT v
(0.70) (350) = 16.70 v v = 1.525 m/s right
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Stage 2: Swing of pendulum (Cons of ME)
Rest
h
Ref h
1.52 m/s
At the start of the swing, it has only KE
When it reaches its maximum height, it comes to rest
So, all of the KE has converted to PEg
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Rest
h
METi = METf Ref h
PEgi + KEi = PEgf+ KEf 1.52 m/s
0.5 mvi2 = mghf
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METi = METf
PEgi + KEi = PEgf+ KEf
0.5 mvi2 = mghf
0.5 vi2 = g hf
hf = 0.5 vi2
g
hf = 0.5 (1.525 m/s)2 = 0.12 m
9.81 m/s2
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Practice Problems
Try
Ladner p. 61 #32
For more questions that combine momentum and energy:
Ladner pp. 54 - 61 #5, 12, 13, 19, 20