A 0.145 kg baseball pitched at 35.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 58.0 m/s. If the contact time between bat and ball is 2.00 10-3 s, calculate the force (assumed to be constant) between the ball and bat.

Right - finished my poker game and can get around to replying to this then. It is remarkably similar to the problem from a few days ago involving the triangle (delta - remember). Here the momentum of the ball changes due to a force acting on it for a length of time.
It is the same problem and if I am not mistaken Drude has given you the equation to find the force.
F=m(v2-v1)/dt

Your major considerations should include the fact that the ball is being accelerated wrt the pitcher first to zero velocity and then up to 58m/s. So (v2-v1) becomes the sum of the two speeds (93).
F=m*93/2 e-3 - calc out of batteries but you should be able to work it from there.

m = .145 kg , v1=+35.0 m/s, v2=-58.0 m/s, t = 2E-3 second, F=?

These kind of quetions all done using the formula:

F*Δt = m*Δv

most students have problem applying this partly because the book does not delve into this topic, nevertheless it is done using this formula so you have:

F = m (v2-v1)/Δt = (.145)(-58-35)/2E-3 = -27E(-3) N

of course the negative means it is in the direction oppositive to the initial velocity of the ball thrown.


haha, Mosley what a coicidence, we answered it simultaneously. Yes, well you said it well..haha I WOULD NEVER POST over a post you did, so I hope you realized this was just a coicidence. Of course, your answers are always right.

It is easy when I can use your equations from the other thread. I always have an element of doubt about my answers so I appreciate your responses.
Integration/differentiation get transposed and plusses become minusses so always nice to have corroborating testimony.