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Double ball drop and conservation of momentum

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    The classic double ball drop question: a ball of mass m is placed directly above another ball of mass M, and assume that m is not negligible. Both balls are dropped simultaneously. Find the mass m such that, upon the second collision (m with M), M has a final velocity of zero.

    2. Relevant equations
    Conservation of momentum
    Elastic collision assumed.

    3. The attempt at a solution
    Here, probably my only question is how to justify applying conservation of momentum to the collision. My reasoning is that conservation of momentum shouldn't apply because there is an external force on the two-ball system (gravitation). However, from what I know of the 'impulse approximation', we can treat the background forces (i.e. gravity) as negligible given that the collision occurs across a very short period of time? We can then work with momentum conservation over an infinitesimal time interval? (i.e. the effects of gravitational acceleration on momentum of the system may then be ignored)
    Is that the correct reason for why it may be applied? Thanks :)
     
  2. jcsd
  3. Mar 27, 2016 #2

    andrewkirk

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    You refer to a second collision being between the two balls. What is the first collision?

    Also, if the final velocity of M is zero and it's in a gravitational field, it must be supported by something. What is supporting it?
     
  4. Mar 28, 2016 #3
    Sorry about that. Here's the full question:
    A small ball of mass m is aligned above a larger ball of mass M (with slight separation) and the two are dropped simultaneously from a height h. Assume the radius of each ball is negligible relative to h.
    (a) If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what value of m would result in stopping the larger ball when it collides with the small ball?

    My questions were as above - in this non-closed system (force of gravitation), are we using the impulse approximation to make gravitational forces negligible relative to the collision forces? The method of solving this question (according to a Google) is to equate momentum. I'm personally inclined to doing that too, but I need to be able to justify it.

    Thanks!
     
  5. Mar 28, 2016 #4

    haruspex

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    Yes, the collisions are presumed to take an unknown but very short time, so gravity can be ignored for their duration.
     
  6. Mar 28, 2016 #5

    andrewkirk

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    A much clearer version of this problem would be one in which the balls are masses sliding collinearly on a frictionless surface and M then undergoes an elastic collision at right angles to a wall. That takes gravity out of the equation.
     
  7. Mar 28, 2016 #6
    Yes this makes sense. Thank you!
     
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