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Impulse of bat on a baseball

  1. Oct 16, 2006 #1
    A 0.14 kg baseball moves toward home plate with a velocity vi = (-37 m/s) x. After striking the bat, the ball moves vertically upward with a velocity vf = (13 m/s) y. Find the direction and magnitude of the impulse delivered to the ball by the bat. Assume that the ball and bat are in contact for 1.5 ms.

    I tried impulse = mv final - mv initial. Numerically, I did ((0.14)(13)) squared - ((0.14)(37)) squared.
     
  2. jcsd
  3. Oct 16, 2006 #2

    radou

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    Again, it would be useful to write down a vector equation first, and then think about the components: [tex]\vec{I} = \vec{F} \cdot t = (F_{x}\cdot\vec{i}+F_{y}\cdot\vec{j})\cdot t=m\vec{v}_{f}- m\vec{v}_{i}= 0.14\cdot13\cdot \vec{j} - 0.14\cdot (-37) \vec{i}[/tex]. Now simply 'read off' the sides of the equation for [tex]\vec{i}[/tex] and for [tex]\vec{j}[/tex] separately.
     
  4. Oct 16, 2006 #3

    I tried that method, but I don't understand the "read off" Am I supposed to calculate i and j separately? I tried just calculating by multiplying the velocities by the masses and subtracting those two products from each other.
     
  5. Oct 16, 2006 #4

    radou

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    By 'reading off' I meant:

    [tex]F_{x}\vec{i}t + F_{y}\vec{j}t = 0.13\cdot13\vec{j}-0.14\cdot(-37)\vec{i} \Rightarrow F_{x}t = -0.14\cdot(-0.37) , F_{y}t=0.14\cdot0.13[/tex]. Since you know the time, you can easily obtain the components Fx and Fy of the force. Now you know everything, since the impulse equals [tex]\vec{I} = F_{x}t\vec{i}+F_{y}t\vec{j}[/tex]. The direction is found from the relation [tex]\tan(\alpha)=\frac{F_{y}}{F_{x}}[/tex], and the magnitude from [tex]\left|\vec{I}\right|=\sqrt{(F_{x}t)^2+(F_{y}t)^2}[/tex].

    I hope you know how to deal with vectors, I'd feel stupid to make such a mess for nothing. :smile:
     
    Last edited: Oct 16, 2006
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