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Two balls (Momentum Question)

  1. Apr 13, 2008 #1
    1. The problem statement, all variables and given/known data

    A 0.25kg tennis ball is placed right on top of a 1kg volleyball and dropped. Both balls hit the ground at a speed of 3 m/s simultaneously. Find the (upward) velocity of the tennis ball right after it bounces up from the volleyball. Assume elastic collisions. (HINT: the tennis ball will move faster than 3m/s)


    2. Relevant equations

    mass of tennis ball=0.25kg
    mass of volleyball = 1kg
    1) Va'=Va(m1-m2)/(m1+m2)
    2) Vb'=Va(2m1/(m1+m2)

    3. The attempt at a solution

    Someone told me that when the two balls hit the ground, the volleyball reverses direction...so -3m/s? Plus it's an elastic collision, so kinetic energy and momentum are both conserved. But initial velocity of volleyball is not zero, so i can't use eq'ns 1 or 2 above? If i set the velocities of the balls to the reference frame of the tennis so that the velocity of the volleyball is zero will that work out?

    Also, i'm not really sure why only the volleyball reverses direction, if both of them are simulataneously released then shouldn't they hit the ground at the same time? and why does the volleyball rebound when the collision occurs? shouldn't all the kinetic energy of the volleyball be transferred to the tennis ball since the volleyball is sandwiched between the earth and tennisball so that it's velocity is zero?
     
  2. jcsd
  3. Apr 13, 2008 #2

    tiny-tim

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    Hi subwaybusker! :smile:
    Not entirely … kinetic energy is conserved, but momentum involving the wall isn't.

    Hint: pretend that the volleyball is well in front of the tennis ball … it bounces off the wall, and hits the tennis ball … then momentum won't be conserved in the first collision (with the wall), but will be conserved in the second one! :smile:
     
  4. Apr 13, 2008 #3
    why won't momentum be conserved in the first collision? if not then why is kinetic energy conserved?
     
  5. Apr 13, 2008 #4

    Doc Al

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    Staff: Mentor

    Momentum is conserved (of course) if you include the ground and attached earth in your calculations. But it's a safe assumption to treat the mass of such as infinitely higher than that of the volleyball. The momentum of the volleyball alone is not conserved.

    KE is conserved because the collision is assumed to be elastic and because the ground/earth can be treated as infinitely massive.
     
    Last edited: Apr 13, 2008
  6. Apr 13, 2008 #5
    okay, i get why KE is conserved. as for momentum, do you mean that when the volleyball hits the ground, some momentum is transferred to the earth?

    what do i do to solve this question?
    if momentum is not conserved for the first equation, then i don't have the momentum of the volleyball after it hits the ground. does that mean i can only use kinetic energy to solve for the second collision?
     
    Last edited: Apr 13, 2008
  7. Apr 13, 2008 #6

    Doc Al

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    Sure.

    You can figure out the momentum using conservation of energy, if you need to.
     
  8. Apr 13, 2008 #7

    tiny-tim

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    The practical answer is that momentum obviously isn't conserved when something bounces off a wall … the wall has the same momentum before as after, while the bouncing thing doesn't!!

    The technical answer is that momentum is conserved, because the wall does move slightly … but it takes the whole of the Earth with it! :smile: If we had a stationary frame of reference in outer space, then we could see that. However, since our frame of reference is attached to the Earth … the examiner, after all, is expecting a speed relative to the Earth … we can't use conservation of momentum!
    erm … because the question tells you it is! :smile:

    (usually, it isn't conserved in collisions.)

    How are you getting on with the equations? :smile:
     
  9. Apr 13, 2008 #8
    volleyball:
    M=1kg
    V=-3m/s [down]
    tennis ball:
    m=0.25kg
    v=3 m/s [down]

    Set the reference frame to that of the tennis ball so that v=0m/s
    V=-6m/s v=0m/s
    Using conservation of momentum for the second collision:

    MV+mv=MV'+mv'
    (1kg)(-6m/s)+0=(1kg)V'+(0.25kg)v'

    I don't know what to do afterwards?
     
  10. Apr 14, 2008 #9

    tiny-tim

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    Hi subwaybusker! :smile:

    Whyever are you changing the frame of reference?

    The question asks you for speed relative to the Earth, doesn't it?

    You'll only have to convert back again … plus there's the chance of making a mistake when you do convert back! :frown:

    Anyway, whichever frame you use, as Doc Al says, you must use both conservation of energy and conservation of momentum!

    Have a go! :smile:
     
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