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Posted by on Friday, June 1, 2012 at 11:23pm.

A 0.130kg baseball pitched at 35.5 m/s is hit on a horizontal line drive straight back at the pitcher at 50.5 m/s .
Part a: If the contact time between bat and ball is 4.25×10^−3 , calculate the magnitude of the force (assumed to be constant) between the ball and bat.

part b)Determine the direction of the force exerted on a ball.
a)the force is directed towards the pitcher
b) the force is directed away from the pitcher.

  • physics- please respond - , Saturday, June 2, 2012 at 1:56am

    this exercise is conservation of linear momentum.
    initial momentum: (0.130kg)(35.5)(i)
    "remember linear momentum is vectorial.
    i is vector direction and only the movement is in X"
    = 4.615(i)

    Final momentum is :
    = (0.130)(50.5)(-i)
    =6.565(-i)

    then I=impulse

    I= 6.565(-i) - 4.615(i)
    I= -6.565(i) - 4.615(i)
    I= -11.18(i)
    I= 11.18(-i) Kgm/s

    I=F.t
    F=I/t= 11.18(-i)Kgm/s/4.25×10^−3 s
    F=2630.6 (Kgm/s² o N)

    Enjoy the answer.
    good luck

  • physics- please respond - , Saturday, June 2, 2012 at 1:57am

    the magnitude of the force is 2630.6 N

  • physics- please respond - , Saturday, June 2, 2012 at 2:26am

    thank you !

  • physics- please respond - , Sunday, June 3, 2012 at 3:21am

    What about part B please?

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