Momentum problem!!! PLEASE HELP, any advice will be a big help!?

A baseball player pitches a fastball toward home plate at a speed of 49 m/s. The batter swings, connects with the ball of mass 141 g, and hits it so that the ball leaves the bat with a speed of 37 m/s. Assume that the ball is moving horizontally just before and just after the collision with the bat.








(a) What is the magnitude of the change in momentum of the ball?


?kg路m/s





(b) What is the impulse delivered to the ball by the bat?


?kg路m/s





(c) If the bat and ball are in contact for 3.0 ms, what is the magnitude of the average force exerted on the ball by the bat?


?kNMomentum problem!!! PLEASE HELP, any advice will be a big help!?
v1 = 49m/s


v2 = -37m/s


m = 141g = 0.141kg





* The reason one of the speeds is positive and the other is negative is because they are speeds in opposite directions to each other. One of the speeds has to be negative. You could choose either*





a. p = m * (delta v)


p = 0.141kg*(49m/s-(-37m/s))


p = 0.141kg*86m/s


p = 12kgm/s





b. Impulse equals change in momentum, so it is 12kgm/s also.





c. t = 3*10^-3 s





Impulse (J) = Ft = m*(delta v)


F = P/t


F = (12kgm/s)/(3*10^-3s)


F = 4*10^3 N


F = 4 KNMomentum problem!!! PLEASE HELP, any advice will be a big help!?
I guess this is the help me with my math homework category.
a.)





p before = m * v = .141 kg * 49 m/s = 6.909 kg*m/s





p after = m*v after = .141 kg * 37 m/s = 5.217 kg*m/s





difference in p before and p after is :





6.909-5.217 = 1.692 kg*m/s





b.)


Impulse (J) = change in momentum





so J= 1.692 kg*m/s





c.) Impulse also equals Force*duration of Force = F*t





J=F*t=1.692 and t= .003 s





so F= 1.692/.003 = 564 N





:)