Baseball impulse when ball's initial velocity is purely horizontal

[tony873004]

3. [Walker2 9.P.014.] A 0.14 kg baseball moves toward home plate with a velocity vi = (-25 m/s) x. After striking the bat, the ball moves vertically upward with a velocity vf = (13 m/s) y.

(a) 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.

Don't I need to know what direction the baseball is moving in initially if I am to compute the direction of the impulse. Are they assuming that the ball travels completely horizontal? Or do I not understand this?

 

[Andrew Mason]

Don't I need to know what direction the baseball is moving in initially if I am to compute the direction of the impulse. Are they assuming that the ball travels completely horizontal? Or do I not understand this?

The impulse is the change in momentum of the baseball. You are to assume that $\vec{v_i}$ is completely horizontal and $\vec{v_f}$ is completely vertical.

Just draw the momentum vectors for the ball before and after the bat hits it. The impulse or change in momentum is the difference between these two vectors (ie. 'what vector added to $\vec{v_i}$ results in $\vec{v_f}$?'). The impulse is the product of the bat force and the time interval over which it is applied ($m\triangle \vec{v} = \vec{F_{bat}}\triangle t$).

AM

 

[wais]

You are correct in assuming that the initial direction of the baseball is important in calculating the impulse. In this problem, the initial velocity of the baseball is given as (-25 m/s) x, which means it is moving in the negative x-direction (towards home plate). This information is necessary to calculate the magnitude and direction of the impulse delivered by the bat.

To calculate the magnitude of the impulse, we can use the formula I = m(vf - vi), where I is the impulse, m is the mass of the baseball, and vf and vi are the final and initial velocities, respectively. Plugging in the given values, we get:

I = (0.14 kg)(13 m/s - (-25 m/s))
= (0.14 kg)(38 m/s)
= 5.32 kg m/s

To find the direction of the impulse, we can use the direction of the final velocity, which is in the positive y-direction (vertically upward). This means that the impulse is also in the positive y-direction.

In summary, the impulse delivered by the bat to the baseball has a magnitude of 5.32 kg m/s and a direction of positive y (vertically upward).