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What is the magnitude of the impulse from the spike?

  1. Aug 14, 2007 #1
    1. The problem statement, all variables and given/known data
    A ball is traveling horizontally over a volleyball net when a player spikes it, driving it straight down to the ground. The ball's mass is .22kg, its speed before being hit is 6.4 m/s and its speed immediately after the spike is 21m/s. What is the magnitude of the impulse from the spike? I have two attempts at the solution but I was mainly wondering which attempt(s) is correct.


    2. Relevant equations
    impulse=change in momentum= (mass)(final velocity)-(mass)(initial velocity)


    3. The attempt at a solution
    J=.22(21)-.22(6.4)=3.212kg(m/s)
    I don't know if that is correct because a way I was given to do this was using coordinates:
    .22(0,21)-.22(6.4,0)=(0,4.62)-(1.408,0)=(-1.408,4.62)kg(m/s)
     
  2. jcsd
  3. Aug 14, 2007 #2
    Initially, it was traveling horizontally

    and after the applied force, the direction changed by 90 degrees.

    and impulse is a vector quantity, so you need to take of directions also,.

    so your expression would be:
    J=.22(21[down])-.22(6.4
    )​
     
  4. Aug 14, 2007 #3

    Doc Al

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    Staff: Mentor

    The first way is not correct, since it ignores the fact that momentum (and velocity) is a vector--direction counts. The second method, using components, is correct. (Except for a minus sign, since the final velocity is 21 m/s down.)

    To complete the solution, you need to find the magnitude of the impulse, not just the components.
     
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