# Impulse of bat on a baseball

1. Oct 16, 2006

### map7s

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. Oct 16, 2006

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.

3. Oct 16, 2006

### map7s

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.

4. Oct 16, 2006

$$F_{x}\vec{i}t + F_{y}\vec{j}t = 0.13\cdot13\vec{j}-0.14\cdot(-37)\vec{i} \Rightarrow F_{x}t = -0.14\cdot(-0.37) , F_{y}t=0.14\cdot0.13$$. 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 $$\left|\vec{I}\right|=\sqrt{(F_{x}t)^2+(F_{y}t)^2}$$.