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Impulse ball dropped

  1. Oct 22, 2014 #1
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    A 1.2kg ball dropped from a height of 3m onto a steel plate rigidly attached to the ground bounces back up to a height of 2.5m.


    (a). What is the impulse delivered to the ball by the plate? (b). What is the coefficient of restitution of the collision?


    I know my instructor wants me to use something along the lines of a=Δv/Δt; ma= m (Δv/Δt)= Δp/Δt; delta p= maΔt= (ΣF)Δt
    Honestly I'm lost on this one because of the whole Δt when I am not given time...
     
    Last edited by a moderator: Oct 23, 2014
  2. jcsd
  3. Oct 22, 2014 #2

    collinsmark

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    Hello KatyM7,

    Welcome to Physics Forums!

    There are two equivalent ways to interpret impulse.

    The first, is impulse is force times time: The force applied to an object multiplied by the time interval that the force was applied. (For a uniform force, this is simply [itex] \vec J = \vec F \Delta t [/itex]. On the other hand if the force varies with time, then a little calculus is involved, but that's more than I want to get into here.)

    That definition is fine and dandy, but it doesn't help you much for this particular problem since you don't know what the force was nor how long it was applied. Fortunately, it's not the only way to view impulse.

    For reasons which you've already worked out in your original post, impulse is can also be viewed as an object's change in momentum. What is the ball's momentum at the instant after it collides with the steel plate relative to its momentum at the instant before the collision?

    (Hint: don't forget that momentum, like force and impulse, is a vector with both magnitude and direction.)
     
  4. Oct 23, 2014 #3
    The ball's acceleration the instant before the collision would be equal to gravity (9.8m/s^2) and after the collision since it is moving in the opposite direction a=-9.8m/s^2. The ball's mass doesn't change, again the unknown is time. If I am not mistaken this in an isolated system? If so then P(final)= P(initial) in the opposite direction, therefore Δp=0. Please correct me if I am mistaken. Thank you for your help! :)
     
  5. Oct 23, 2014 #4

    collinsmark

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    Well, okay. But I was asking more about momentum though.

    But yes, the ball acceleration the instant before the collision is 9.8 m/s2 in the down direction.

    Well, no, that's not true. After the collision, even though the ball is moving up (velocity is up) its acceleration is still 9.8 m/s2 in the down direction.

    But anyway, I'm asking about the ball's change in momentum. Not the ball's change in acceleration.

    Momentum, [itex] \vec p [/itex] is

    [tex] \vec p = m \vec v [/tex]
    The ball is definitely not an isolated system. It encounters the external impulse brought on by the steel plate, and not to mention that it is undergoing an external gravitational force the whole while.

    I believe that answer is mistaken.

    Use your kinematics equations to find the velocity of the ball immediately before impact knowing the initial height of the ball is 3 m. Alternatively, you could use conservation of energy; either way is fine. Then do the same to find the velocity of the ball immediately after the collision, knowing the final height of the ball is 2.5 m.

    Finally, calculate the ball's change in momentum during the collision with the steel plate. [Edit: then move on to part b.]
     
    Last edited: Oct 23, 2014
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