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Explanation of acceleration of a ball bouncing up and down on the ground

  1. Aug 21, 2010 #1
    I am from HK. Hope you guys can understand my poor English! ^^
    actually, this is not for my homework but my preparation for the public exam.

    1. The problem statement, all variables and given/known data
    The situation is that a ball bouncing up and down on the ground in vertical direction.
    The question is to choose which graphs best describes the variation of its acceleration a.
    It is a MC-question. Althought I know the ans., I want to have a full explanation but I am not sure if I am correct.


    2. Relevant equations
    No


    3. The attempt at a solution
    We all know that there is a sudden change in acceleration to the opposite sign at the moment that the ball hits ground. I want to explain it. Please comment on my explanation.


      u↓ ↑v
       O    ↓+ve
    -------
      ground


    where u ≥ -v
    ps. the ball with mass m.


    Let's consider momentum.
    impact force
    = (mv - mu)/t
    ≥ [m(-u) - mu]/t (since v ≥ -u)
    = -2mu/t which is negative


    net Force = ma
    impact force - mg = ma
    ma ≤ -2mu/t - mg
    a ≤ -2u/t - g

    hence, when the ball hits the ground, it experiences acceleration in the sign opposite to gravitational field.

    Am I correct?
    Can I explain in other way(s)? or simplier way(s)?
    I want explanation as much as possible.
     
    Last edited: Aug 21, 2010
  2. jcsd
  3. Aug 21, 2010 #2
    Well, for a perfect collision between an immovable ground and perfectly elastic ball, the acceleration would be infinite, and all you need are the conservation laws. For a non-ideal situation, you have to consider the ball as some sort of spring, you can get an expression for the time the impulse will be delivered based on the spring constant (or equation).

    The rate at which the acceleration changes direction--which, I think, is what you want to find out--depends sensitively on how you're modeling/thinking about the ball.
     
  4. Aug 22, 2010 #3
    So, you mean that there is different explanation when it is in different cases?
    Is there any wrong or improper concept?

    How about this?
    Since t ∝ kx where k is the spring constant and x is displacement (how should I difine x?)
    => t = μkx where μ is a constant.

    hence, a ≤ -2u/μkx - g
    since x change from -ve to +ve, and then -ve, acceleration changes its direction "gradually".

    is it correct?
     
  5. Aug 26, 2010 #4
    oh, i must make a mistake

    now, let v be the velocity of the ball when it hits ground
    impluse
    = (mv - mu)/t

    net Force = ma
    impluse - mg = ma
    ma = (mv - mu)/t - mg
    a = (v - u)/t - g

    since t ∝ 1/k => t = μ/k where k is spring constant, μ is a constant
    hence, a = (v - u)k/μ - g
    da/dt = (k/μ)(dv/dt) - uk/μ - g

    When the ball hits ground,
    speed of the ball ↓,
    => (dv/dt) is -ve.
    => da/dt is also -ve
    => slope of the a-t graph is -ve

    when the ball leaves ground,
    speed of the ball ↑,
    => (dv/dt) is +ve.
    => da/dt may become +ve
    => slope of the a-t graph may be +ve

    however, it seems that i can't say "may be +ve" in this case, i should say "must be +ve". i don't know whether i made something wrong before. If no, how should I deduce "must"?
     
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