What is the impulse delivered by a bat to a baseball after contact?

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    Baseball Impulse
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The discussion centers on calculating the impulse delivered to a baseball by a bat after contact. The baseball, with a mass of 0.14 kg, initially moves towards home plate at -37 m/s and, after striking the bat, moves vertically upward at 13 m/s. Participants suggest using vector equations to separate the impulse into its components, emphasizing the importance of calculating the forces in both the x and y directions. The impulse can be determined using the formula I = mv_final - mv_initial, and the direction and magnitude can be derived from the components of the force and the time of contact. Understanding vector calculations is crucial for accurately determining the impulse's characteristics.
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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.
 
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map7s said:
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.

Again, it would be useful to write down a vector equation first, and then think about the components: \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}. Now simply 'read off' the sides of the equation for \vec{i} and for \vec{j} separately.
 
radou said:
Again, it would be useful to write down a vector equation first, and then think about the components: \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}. Now simply 'read off' the sides of the equation for \vec{i} and for \vec{j} separately.


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.
 
By 'reading off' I meant:

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. 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 \vec{I} = F_{x}t\vec{i}+F_{y}t\vec{j}. The direction is found from the relation \tan(\alpha)=\frac{F_{y}}{F_{x}}, and the magnitude from \left|\vec{I}\right|=\sqrt{(F_{x}t)^2+(F_{y}t)^2}.

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