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Double Ball Drop Ratio

  1. May 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball of mass m1 is aligned above a ball of mass M2 (with slight separation), and the two are dropped simultaneously from height h. (Assume the radius of each ball is negligible compared to h.) (a) If M2 rebounds elastically from the floor and then m1 rebounds from M2, what ratio m1/M2 results in M2 stopping upon its collision with m1? (b) What height h does m1 reach?

    2. Relevant equations
    KE = .5mv^2
    P = mv

    3. The attempt at a solution
    Okay, so I drew a diagram and used cons. of momentum and energy to find VM for both and set them equal to each other, but there's something wrong in my math because I know my ratio should be 3:1 and I'm not getting that. Here's what I did as best as I can type out on a computer:

    Cons. Momentum)

    (m1-M2) sqrt(ugh)=m1v+M2V, with (m1-M2) because M2 has a negative momentum after it strikes the ground

    ((m1-M2) sqrt(2gh) - m1v)/M2 = V

    Cons. Energy)
    1/2(m1+M2)(sqrt(2gh))^2 = 1/2(m1v^2) + 1/2(M2V^2)

    sqrt( (2gh(m1+M2)-mv^2)/M2) = V

    I think it's a math issue since when I'm setting them equal to each other my work is only getting more complex…

    I don't know what I'm doing wrong because I thought my equations were okay, so if someone could explain how you get through this I would be very thankful.
  2. jcsd
  3. May 21, 2015 #2


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    Hi Pud,

    I have difficulty understanding why you set the equations equal to each other. Where do you use the given that M2 stops after colliding with m1 ?
  4. May 21, 2015 #3


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    As BvU writes, you have not used that M2 stops. But I wonder about the signs in your first equation. Which direction are you taking as positive?
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