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A ball falls under gravity

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data

    a ball falls under gravity from a height of 10m with an initial downward velocity v.It collides with the ground and looses 50% of its energy in collision and then rises to same height .1) find the initial velocity v 2)the height to which the ball rises after the collision if the initial velocity v is directed upward instead of downward

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 25, 2012 #2

    lewando

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    What relevant equations do you think would apply?
     
  4. Aug 25, 2012 #3
    I think you have a typo there. Should it be "and then rises to some height"?
     
  5. Aug 25, 2012 #4

    lewando

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    I think it is right but could be worded better. I read into the problem that the ball rises to the 10m height with zero velocity at that moment.
     
  6. Aug 25, 2012 #5
    If an object is dropped from 10m it can NEVER bounce 10m's upwards again, it will always be slightly less so I think the wording is "some height"
     
  7. Aug 25, 2012 #6

    lewando

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    If you apply an initial non-zero downward velocity (not just dropping it), it is possible.
     
  8. Aug 25, 2012 #7
    Thanks for the clarification, lewando. I see the error of my ways.

    So, norrh, follow lewando's lead: what equations do you think would apply?
     
  9. Aug 26, 2012 #8
    ooooooooo it has a velocity while falling and it cannot be equivalent to a freely falling bodyand i do not know how to start.And i do not know many formulas for ccollisions i forgot.i just need starting.
     
  10. Aug 26, 2012 #9

    lewando

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    comments in brackets
     
  11. Aug 26, 2012 #10
    Start with conservation of mechanical energy.
    Make at height 10m as reference of potential energy equal zero.
     
  12. Aug 26, 2012 #11
    Well yes if you shot a pingpong ball at the floor from a distance of say 1 meter hit would it the floor and bounce higher but after that it will always get lower. Otherwise we'd all have free energy and the planet would be saved lol
     
  13. Aug 26, 2012 #12
    would it be possible to find out v0 using just the following equation? y=v0t-0.5a(t*t)
     
  14. Aug 26, 2012 #13
    It is about energy.
    50% of total energy before collision use to raise the object, conservative force work.
    Another 50% goes to non-conservative force work.

    From above you can find total energy of the object just before collision.
     
  15. Aug 26, 2012 #14
    So, to do the conservative force work we have to equal the mass to zero, but still, for the non conservative work, how do we find the "v" in the final cinectic energy?
     
  16. Aug 26, 2012 #15
    50% lost, another 50% converted to potential energy.
    The total energy is total kinetic enery before collision.
    Use SUVAT to find intial velocity.
     
    Last edited: Aug 26, 2012
  17. Aug 26, 2012 #16
    Ok,thanks dude
     
  18. Aug 26, 2012 #17

    lewando

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    Why would you not use the ground as the zero PE reference?
     
  19. Aug 26, 2012 #18

    CAF123

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    I attempted this question too and came up with the following;
    Taking the reference level of potential, [itex] U_o = 0 [/itex] to be ground level,
    [itex] \frac{1}{2}mv_o^2 + mgh = \frac{1}{2}mv_a^2, [/itex] where [itex] v_o [/itex] is the initial velocity and [itex] v_a [/itex] is the velocity prior to impact.

    Since there is 50% dissipation, I said that half of the total energy would be converted to potential energy;
    [itex] \frac{1}{2}(\frac{1}{2}mv_a^2) = mgh [/itex]

    Solving for [itex] v_a [/itex] yields [itex] v_a = \sqrt{4gh} [/itex] but this gives an initial velocity of [itex] \sqrt{2gh} [/itex] which is what it would be were it dropped. So I feel there is an error somewhere. Any ideas?
     
  20. Aug 26, 2012 #19

    lewando

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    To summarize you:

    If V0 = 0, Va = [itex] \sqrt{2gh} [/itex]

    If V0 = [itex] \sqrt{2gh} [/itex], Va = [itex] \sqrt{4gh} [/itex]

    I don't see an error.
     
  21. Aug 26, 2012 #20
    Yes i made an error in earlier post.
     
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