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Falling object collision question

  1. Jun 2, 2013 #1
    I've been trying to figure out the physica behind a collision of two objects due to the force of gravity. Here's the situation as I'm picturing it:

    A massive object A is falling due to gravity which impacts a stationary object B being held up by a structure. In this particular case, the structure is unable to support the impact force of object A and fails, causing the now combined object AB to continue falling.

    My difficulty is in picturing how the forces interact during the collision. Object A is falling with a force of F=MaG, which is then applied to Object B. At this instant, object B begins its acceleration... Would a force F be applied, during the collision, to Object A in the opposite direction due to Newtons 3rd law?
  2. jcsd
  3. Jun 2, 2013 #2


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    Hello Stellar1! :smile:
    That isn't how it works.

    Mag is the force on A (from the Earth).

    To find the force on A from B, you find the acceleration of A during the collision, subtract that from g, and multiply by Ma.

    The force from A on B is of course the same (Newton's third law).

    (that's if you analyse it instantaneously …

    the more sensible way would be to use impulse, rather than instantaneous force)
  4. Jun 2, 2013 #3
    If object A is on object B, the it is applying the force MaG onto it, no?
  5. Jun 2, 2013 #4
    The only way I can reconcile this in my head is if I take compressibility into account. That allows for object B to accelerate and object A to decelerate until a common speed is achieved and gravity accelerates the combined object.
  6. Jun 2, 2013 #5


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    no :confused:

    the force from A on B (or vice versa) depends on a lot of things
  7. Jun 2, 2013 #6
    Let me rewrite this now that I'm not doing 3 things at once and able to think about this with minimal distractions:

    I know that A will be falling with a force of Fa=mag onto object B, which can only support a force of, say, Fs. The length of time object A is under this acceleration will allow us to determine the velocity of object B at the instant just prior to collision, which we will call vi

    Now, assuming object B is incompressible and, since we assume Fa>Fs, its perfectly consistent to assume that the force during the collision will break the supports and object B will now accelerate from rest under the collision force of object A, as well as gravity.

    Since we assume that object B is not very compressible, wouldn't this mean that object A has to essentially come to rest as well, so that object B (now joined to A) will both accelerate from rest (since B is starting from an initially stationary position)? This would then imply that the velocity at the instant just after collision is 0, correct? The force exerted by A on B (and by B on A) would then equal F=mavi/t + mag, where t is the duration of the collision.

    Is there any way to determine how long this collision would take?
  8. Jun 4, 2013 #7


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    No, that's completely wrong.

    Before A hits B, it has an acceleration of g downward, and there is a force on A (from the Earth) of Mag.

    A does not exert a force of Mag on anything (except of course the Earth).

    When A hits B, A will slow down a lot, and will also change shape slightly. B will also change shape slightly, and will start accelerating.

    You find the force by finding the acceleration first (on either A or B), and multiplying that by the mass.

    You find the acceleration by using Newton's second law, which is about momentum, and also using your knowledge of the elasticity of A and B.
  9. Jun 4, 2013 #8
    Ahh, yeah. That's what I meant. Falling onto =>Falling towards with a force of F=mag which accelerates it to a certain velocity prior to colliding with B.
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