Softball collision problem

[VinnyCee]

A 0.30 kg softball has a velocity of 15 m / s, 35 degrees below horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of 20 m / s, vertically downward? b) How about 20 m / s, horizontally back towards the pitcher?

I have this so far:
\Delta \vec{p} = \vec{J} The change in an object's momentum is equal to the impulse on the object.

\vec{J} = \int_{t_i}^{t_f} \vec{F}(t) dt Impulse defined.

I have no idea how to go about the rest of this! I've been stuck for 3 hours! I don't know the mass of the bat, I don't have a function for F(x), I don't know the time intervals, and I don't know how to do this problem, help me please!

[jamesrc]

Don't worry about the impulse. The change in momentum is just
\Delta\vec p = m\vec v - m\vec v_o

You have been given all of the quantities on the RHS.

[VinnyCee]

With the formula you listed:
(0.30 kg)(20 m / s) - (0.30 kg)(15 m / s) = 6 kg m /s - 4.5 kg m / s = 1.5 kg m / s

Howeer, the book lists the answer as 5.0 kg m / s. Am I forgetting to do the components? If so, how do I compute them again?

[HallsofIvy]

Draw a picture!
Before hitting the bat, the horizontal component of momentum of the ball is
px= (.3)(.15)(cos(35)) and the vertical component is
py= -(.3)(.15)(sin(35)).

Afterwards, the horizontal component of momentum is 0 (it is going straight down) and the vertical component is -20(0.3)

The change in momentum vector is the difference between those. Then use the Pythagorean theorem to find the magnitude.

In the second part (back toward the pitcher), the momentum after the hit has horizontal component -(.3)(20) and vertical component 0.