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Each action there is a responsive reaction

  1. Apr 6, 2005 #1
    We all know that to each action there is a responsive reaction. Let's take the case of a falling ball. Ignoring factors such as friction, elasticity and thermal energy, the ball is dropped from a certain distance perpendicularly above the ground with an inexistent initial velocity. At the instant the ball hits the ground with a certain a certain force, the ground applies an equal force on the ball. The ball is then reflected in the opposite direction. But to get from a direction to one that is opposite, the ball has to, at a certain point drop to a velocity of zero. My thoughts were, naturally, that the reaction force reaccelerates the ball until a certain velocity is reached (witch is equivalent to the final velocity of the falling motion). And here is my problem; if really this is the case, then the force needs a certain time to reaccelerate the ball. Since the force is equivalent to the one of gm, so the time needed would be the same as the time the ball was in the air… this is obviously wrong, but by what logic?
     
  2. jcsd
  3. Apr 6, 2005 #2

    Doc Al

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    Staff: Mentor

    What makes you think the force equals the weight? It's much greater than that (at least for ball hitting a hard surface).
     
  4. Apr 6, 2005 #3
    Okay, so the acceleration would be relative to the mass of the floor... thank you for teaching me!
     
  5. Apr 6, 2005 #4
    That is not what Doc Al ment. Hardness [itex]\neq[/itex] Mass. Typical harder the surface the less energy lost from the ball and its return bounce.
     
  6. Apr 6, 2005 #5
    This i know, but I am talking about the force, and ignoring such factors... Wouldnt the time the ball stays on the ground (then again ignoring all other factors) would be:
    T= 2vm/F? Correct me if I am still wrong...
     
  7. Apr 6, 2005 #6
    The time the ball spends on the ground depends on how the ground and ball deform.

    In the ideal case where the ball and the ground do not deform then the ball spends zero time on the ground.

    Your equation only works if the force provided by deformation is a constant this will not typically be true.
    [tex]
    \int_{t_o}^{t_f}\vec{F}dt = 2\vec{v} m
    [/tex]
     
  8. Apr 6, 2005 #7
    Hummm I would be lying if I said I completly understand... thanks for helping, Ill investigate this.
     
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