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